a new secure distributed storage scheme for...

Shing-Tung Yau High School Applied Mathematical Sciences Award 2013

1

A New Secure Distributed Storage Scheme for Cloud

- Mathematical Designs and Implementation

Cheong Hou Teng and Tan Chih Wei

Pui Ching Middle School Macau

Supervisor Mr Wong Chan Lam

Abstract

In this report we have mainly developed an abstract framework of k bit (tn) secret sharing framework for cloud storage Such framework is clean and it can be implemented A successful implementation of the framework would provide users with protection when the system is under the attack on its confidentiality integrity and reliability Furthermore such system has its own encryption by using permutations and tailor made error detection location and data rescue

We make use of Lagrange polynomials and take the advantages of the algebraic property ldquot distinct points on the plane can uniquely determine a polynomial function of degree t-1rdquo to design a k bit (tn) -secret sharing distributed storage We employ the set with unique factorization property (UFP) so that we simply need to calculate the y intercept of a Lagrange polynomial and then use a look up table to recover a secret Moreover the set which has minimum UFP would help us to design storage with smallest containers

In addition to the algebraic methods we can utilize the geometric facts that three non collinear points determine a unique circle and four non coplanar points determine a unique sphere to construct k bit (3n) and k bit (4n) secret sharing storage respectively To generalize to arbitrary case it is straight forward if we have defined the Haar measure on the higher dimensional unit sphere

The last method is an application of Chinese Remainder Theorem (CRT) and we have designed k bit (tn) secret sharing distributed storage and one of the designs can produce containers with the half size of the original secret However such k is no longer unrestricted and it has to be chosen from an interval

We have developed a C program for implementing both algebraic geometric k bit (3n) secret sharing distributed storages as well as the CRT method The performances of both algebraic and geometric designs are satisfactory in term of processing time and compressed container size The container size is even half of the size of the original secret in the CRT case and it is also very speedy

1 Introduction

The world keeps evolving so as our life Alvin Toffler an America futurist described in his famous book The Third Wave [1] that human progress could be divided into three lsquowavesrsquo The Agricultural Revolution constitutes the First wave the Industrial Revolution the Second Wave and the Third Wave which is a different world we have just entered comprises the Information Age based on the revolution brought by Computer Technology To review from the past IBM developed the mainframe computer

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Page - 437

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ical Sciences

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we are living loud computic developm

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still hesitateRef Figure 2t the securityInformationWof the top cl

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5

Reference

1 Toffler A ldquoThe third waverdquo Bantam Books 1989

2 httpwwwmicrosoftcom httpwwwedgestrategiescom

3 HKPC ldquoCloud guide book for HK SMErdquo 《香港中小企业云端方案应指南》 ISBN 9789881-200617 HKPC 2012

4 httpreportsinformationweekcomabstract249898Storage

5 Lan T Lee RB Chiang M Reliable and Secure Distributed Storage of Critical Information Princeton University Department of Electrical Engineering Technical Report CE-L2008-017 2008

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6

2 Literature Revision of (tn)-Secret Sharing

Secret sharing [1] is currently a very popular topic which is a method of distributing a secret among a group of users requiring a cooperative effort to determine the secret Secret sharing schemes are designed with specific parameters that determine the number of shares needed to uncover the secret and the overall number of shares in the scheme The ultimate goal of the scheme is to divide the secret being hidden into n shares but any subset of t shares can be used together to solve for the value of the secret Additionally any subset of 1t minus shares will prevent the secret from being reconstructed [2] This is defined as a ( )t n threshold scheme meaning that the secret is dispersed into n overall pieces with any t pieces being able to recreate the original secret We are interested in (3 )n -threshold scheme in this research

As in [3] we would like to consider the following problem

4 scientists are working on a secret project They wish to lock up the documents in a cabinet so that the cabinet can be opened if and only if 3 or more of the scientists are present What is the smallest number of locks needed What is the smallest number of keys to the locks each scientist must carry

It is not hard to show that the minimal solution for this (3 4) problem uses 4 locks and 2 keys per scientist However if we increase the number of scientist to 11 and at least 6 of the scientists have to present in order to open the cabinet We can show that the minimal solution 462 locks and 252 keys per scientist It is clearly not practical Moreover the numbers will increase exponentially as the number of scientist increases Therefore other schemes which are more innovative have been proposed

21 Shamirs Scheme

In the Shamirrsquos scheme [3] [4] any t out of n shares may be used to recover the secret The method is based on the fact that you can fit a unique polynomial of degree ( 1)t minus to any set of t points that lie on the polynomial The method is to create a polynomial of degree 1t minus with the secret as the first coefficient and the remaining coefficients picked at random Next find n points on the curve of the polynomial and give one to each of the shares When at least t out of the n shares reveal their points there is sufficient information to fit a ( 1)t minus th degree polynomial to them the first coefficient being the secret

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7

Example 211 (A (36)-Shamirrsquos scheme)

Suppose that our secret 1234S = We obtain 2 numbers 166 and 94 randomly Define a quadratic function

2 2( ) 166 94 1234 166 94 f x S x x x x= + + = + +

We construct 6 points form f as below

(11494) (21942) (3 2578) (43402) (5 4414) (65614)

In order to reconstruct the secret S any three of the above points will be enough Let us consider 0 0 1 1( ) (21942) ( ) (43402)x y x y= = and 2 2( ) (5 4414)x y = We will compute Lagrange basis polynomials

21 10

0 1 0 1

20 21

1 0 1 2

20 12

2 0 2 1

1 3 106 2 3

1 7 5 2 21 823 3

x x x xL x xx x x xx x x xL x xx x x xx x x xL x xx x x x

minus minus= = minus +

minus minus

minus minus minus= = + minus

minus minusminus minus

= = minus +minus minus

Therefore2

2

0( ) ( ) 1234 166 94j j

jf x y L x x x

=

= = + +sum Hence (0)S f=

22 Blakleys Scheme

Blakleys secret sharing scheme [5] is geometric in nature For a ( )t n secret sharing we use the fact that any t nonparallel ( 1)t minus -dimensional hyperplanes intersect at a specific point So suppose the secret S is a point in the t dimensional space Just create n nonparallel ( 1)t minus -dimensional hyperplanes as keys Then any t of them will uniquely determine a point which is the secret point

Example 221 (A (26) Blakleys secret sharing scheme)

Let S=(00) We can create a (26) Blakleys secret sharing Make 6 keys for each share as below

1 2 3 4 5 6 2 3 4 5 6K x K x K x K x K x K x= = = = = =

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8

For any iK and jK we are able to solve the intercept point out which is the secret point

(00)

23 Using the Chinese Remainder Theorem

We would like to only illustrate the idea by an example It is based on the Asmuth-Blooms Scheme [6] Let 3t = and 4n = Let 0 1 2 33 11 13 17m m m m= = = = and

4 19m = Let the secret 2S = Pick 51α = according to Asmuth-Blooms Scheme Then

2 51 3 155+ times = Assign ( )155(mod )i im m to the i th share for 1 23 4i = To recover

the secret we have one possible 3 of the shares for example (111) (1312) and (17 2) Then apply to solve the system of equations

( )( )( )

1 mod1112 mod132 mod17

xxx

===

[ie Consider the solutions of the following systems of equations

( )( )( )

( )( )( )

( )( )( )

1 mod11 0 mod11 0 mod110 mod13 1 mod13 0 mod130 mod17 0 mod17 1 mod17

x x xx x xx x x

= = == = == = =

which are 221 1496 and 715 respectively] So the solution is 1 221 12 1496 2 715 19603times + times + times = and 19603 155(mod11 13 17)= times times Finally

2 155(mod 3)S = =

It is worthy to note that the above method normally cannot be applied directly to practical problems Further development modification and design are needed to be made so that a real world problem can be solved

Reference

1 Martin R Introduction to secret sharing [Online] Available httpwwwcsritedu~rfm6038Paperpdf

2 Stinson D R ldquoCryptography Theory and Practicerdquo 2nd ed CRC Press

2006

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9

3 Shamir Adi How to share a secret Communications of the ACM 22 (11) 612ndash613 1979

4 Wikipedia Shamirs Secret Sharing

5 Blakley G R Safeguarding cryptographic keys Proceedings of the National Computer Conference 48 313ndash317 1979

6 Wikipedia Secret sharing using the Chinese remainder theorem

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10

3 Framework of k bit t Secret Sharing Distributed Storage

In this Chapter we will introduce the various notions related to proposed framework Besides we will give simplified examples in order to illustrate the ideas The algorithms of generating containers and recovering the original secret are also given A tailor made encryption for container is given by the end of the chapter

31 Secret Shareable Pairs

Let k be the set of all binary strings with length k ie

1 2 0 or 1 12 k k i i k= = =ε ε ε ε

Definition 311 Let 3t ge be an integer and letΛ be a family of subsets of d such that

if

a) 1 2C C isinΛ and 1 2C C tcap ge then 1 2C C=

b) t is the smallest integer which has property a) c) Let 1 2 tp p p Cisin and CisinΛ There is a method Ψ (or an algorithm)

enable us to obtain C from the given t points

Then Λ is t -secret shareable

Notation We would like to denote the set of all the graphs of polynomials of degree 1t minus by tΛ Hence 3Λ is the set of all the graphs of quadratic functions

Example 312 Recall that a classical algebra result two quadratic functions agree with each other at three distinct points if and only if they are equal Then 3Λ has properties a)

and b) in the definition 311 Assume that 1 1( )x y 2 2( )x y and 3 3( )x y Let

2 3 1 3 1 21 2 3

1 2 1 3 2 1 2 3 3 1 3 2

( )( ) ( )( ) ( )( )( )( )( ) ( )( ) ( )( )

x x x x x x x x x x x xL x y y yx x x x x x x x x x x xminus minus minus minus minus minus

= + +minus minus minus minus minus minus

Then 1 1 2 2( ) ( )L x y L x y= = and 3 3( )L x y= Hence 1 1( )x y 2 2( )x y and 3 3( )x y belong

to the graph of ( )L x and 3Λ is 3 -secret shareable

Let kΦ rarrΛ be an one to one onto function from k to Λ Then it is called an encoding function and the ordered pair ( )Λ Φ is called a k bit t-secret shareable pair

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ecret shareab

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ear points dete

e define a fu

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00011011

CCCC

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Page - 447


12

32 k bit (tn) Secret Sharing Distributed Storage framework

Assume that the ordered pair ( )Λ Φ is a k bit t-secret shareable pair and n tge is an integer

Definition 321 Let 1 2 nπ π π be a sequence of functions from Λ to d such that for 1 2 j j n= and CisinΛ

1 ( )j C Cπ isin

2 ( ) ( )j jC Cπ π= if and only if j j=

Then 1 2 nπ π π is said to be a sequence of choice functions of Λ

Example 322 Consider kΛ andCisinΛ such that C is a graph of a polynomial ( )P x We define

( ) ( ( ))j C j P jπ =

for 12 j n=

Let Mk= for some positive integer M Then we call s is a secret if 1 2s = isinε ε ε Therefore s can also be written as

1 2 Ms s s s=

where ( 1) 1 ( 1) 2i i k i k iks minus + minus += ε ε ε and 12 i M=

Note that it is sometime useful if we index is as below

1 2i i i

kis = ε ε ε

and so

1 2 11 1 1 2 2

2 1 12

2 2 Mi i ik k

M Mk ks = ε ε ε ε ε ε ε ε ε ε ε ε

321 Creating Containers from a Secret

Given a secret 1 2 Ms s s s= isin a container array [ ]s is an M by n array of points ind such that

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Page - 448


13

[ ] ( ( ))ij j is sπ= Φ

for 12 i M= and 12 j n= The j th container [ ] js of the secret s is the j

column of the container array [ ]s where 12 j n=

Example 323 if 100111s = then 1 2 310 01 11s s s= = = Then

( )( ) ( )( ) ( )( )( ) ( )( ) ( )( )( ) ( )( ) ( )

1

2

3

10 ( 2)( 1)

01 ( 1)( 3)

11 ( 2)( 3)

j j

j j

j j

s j j j

s j j j

s j j j

π π

π π

π π

Φ = Φ = minus minus



where 123j = So the container array of the secret s is

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

10 20 32 46 512[ ] 10 23 30 43 58

12 20 30 42 56s

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

Finally we can obtain 5 containers

( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

1 2 3 4 5

10 20 32 46 512[ ] 10 [ ] 23 [ ] 30 [ ] 43 [ ] 58

12 20 30 42 56s s s s s

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Remark The size of containers depends on the parameter k However it is not always the

case that the bigger k the smaller the size of containers would be theatrically

322 Recover the Secret from t Containers

To recover the secret 1 2 1 2 Ms s s s= =ε ε ε we have to have at least t distinct containers By the property c) of definition 331 without loss of generality we can assume that they are 1 2[ ] [ ] [ ]ts s s First of all we form the M by t collector array

[ ] [ ]1 212[ ] [ ] [ ][ ] i

jtts s ss p= =

The i th row of the collector array consists of t distinct points 1 2 i i itp p p of ( )isΦ By

the properties a) and c) of Definition 311 we have

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Page - 449


14

1 2( ) ( )i i ii ts p p pΦ = Ψ

and

11 2( )( )i i i

i ts p p pminus= Φ Ψ

for 1 2 i M=

Finally we recover the secret

1 2

1 1 1 1 1 2 2 2 11 2 1 2 1 2( ( )) ( ( )) ( ( ))

M

n n nt t t

s s s s

p p p p p p p p pminus minus minus

=

=Φ Ψ Φ Ψ Φ Ψ

Moreover from property b) of Definition 311 the number of distinct containers needed to recover the secret s should be at least t

323 Algorithms of k bit (tn) Secret Sharing Distributed Storage

Assume a ( )Λ Φ is a k bit t-secret shareable pair n tge is an integer and 1 2 nπ π π is a sequence of choice functions of Λ

Algorithm for Distributed Storage

Step 1 Input a secret 1 2 Ms s s s= isin

Mk=

Step 2 Form the container array [ ]s by

[ ] ( ( ))ij j is sπ= Φ

where 12 i M= and 12 j n=

Step 3 Store the i th container [ ]is into the i th storage

Step 4 End

Algorithm for (tn) Secret Sharing Recovery

Step 1 Get t containers 1 2

[ ] [ ] [ ]tj j js s s from t distinct storages

Step 2 Form the collector array [ ]221 1

[ ] [ ] [ ]tt

j j jj jjs ss s⎡ ⎤= ⎣ ⎦

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Page - 450


15

Step 3 For 1 2 i M= recover is from the i th row of the collector array

as in section 322

Step 4 Output 1 2 Mss s s=

Step 5 END

33 Permutations and encryptions

Definition 331 Let σ be an one to one onto function from 123 n Then we call the function σ a permutation The set of all the permutations on 123 n is denoted by nS Let id be a permutation in nS such that id maps every element of 123 n to

itself For any nSσ isin we define 0 idσ = and 1j jσ σ σminus=

Given a permutation nSσ isin and M ntimes container array [ ]s of a secret s we define σ

encrypted container array [ ]s σ to be M ntimes array of points in d such that

[ ] [ ] ( )ij i js sσ

σ=

for 1 2 i M= and 12 j n= The i th column of [ ]s σ is said to be the i th σ

encrypted container denoted by [ ]is σ

Let 2 m nle le A permutation σ in nS denoted by 1 2( )mn n n is called a cycle if

there exist 1 2 mn n n are distinct numbers in 123 n such that

mod ( 1)( )mr rn nσ +prime =

for 1 2 r m= and σ maps other element to itself The number m is the length of σ

denoted by σ

Example 332 Let 5 (251) Sσ = isin Then (2) 5 (5) 1 (1) 2 (3) 3σ σ σ σ= = = = and (4) 4σ = The length of (251) is now equal to3

The period of a permutation σ is the smallest positive integer T such that Tσ σ= Obviously the period of a circle 1 2 ( )mn n nσ = is its length mσ = Since a permutation σ can be factorized uniquely into a product of cycles we have the period of σ is the L C M of the lengths of the cycles in the product It is also known as Ruffini

S20

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16

Theorem (1799) [1] We would like to use a permutation with biggest period for encryption for highest complexity of containers

Example 333 Consider 10S Since 10 5 3 2= + + the pattern of permutations in 10S with maximum period 30 5 3 2= times times is ()()() There are 120960 of them in total and and we list some of them below

(0 8 6 7 2)(1 4 9)(3 5) (2 4 7 6 9)(0 8 5)(1 3) (0 3 7 9 2)(4 5 8)(1 6)

(1 6 9 4 5)(0 2 8)(3 7) (0 1 8 9 4)(2 5 6)(3 7) (0 2 8 4 9)(1 6 5)(3 7)

(0 8 9 5 4)(1 6 7)(2 3) (0 2 7 8 9)(1 4 3)(5 6) (0 9 5 4 7)(1 8 3)(2 6)

(0 2 9 5 3)(1 8 4)(6 7)

Example 334 Let

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

10 20 3 2 46 512[ ] 10 23 30 43 58

12 20 30 42 56s

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

and (123)(45)σ = Then 2 (132)(54)σ = and 3 (45)σ = Hence

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

32 10 20 512 46[ ] 23 30 10 43 58

12 20 30 56 42s σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

and the σ encrypted containers are

( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

1 2 3 4 5

3 2 10 20 512 46[ ] 23 [ ] 30 [ ] 10 [ ] 43 [ ] 58

12 20 30 56 4 2s s s s sσ σ σ σ σ

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Assume a ( )Λ Φ is a k bit t-secret shareable pair n tge is an integer nSσ isin and

1 2 nπ π π is a sequence of choice functions of Λ

Algorithm for Encrypted Distributed Storage


Mk=

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Page - 452


17


[ ] ( ( ))ij j is sπ= Φ


Step 3 Obtain the σ encrypted container array [ ]s σ from [ ]s

Step 4 Store the i th encrypted container [ ]is into the i th storage

Step 5 End

Note that the property c) of definition 311 is true regardless the order of the points

1 2 tp p p Therefore it is no need to modify the recovery algorithm in subsection 323 and it also works well with encrypted containers

34 Features of k bit (tn) Secret Sharing Distributed Storage

In this section we will mention some theoretical features of the framework of k bit (tn) Secret Sharing Distributed Storage

Recall that in Chapter 1 we mention that distributed storages will be under three major kinds of attacks

-The attack on confidentiality reveals stored server contents to attackers

-The attack on integrity modifies data in victim storage servers without being noticed

-The attack on reliability makes storage server unavailable to legitimate users

The propose framework will provide the following protections

a) Ensuring confidentiality

The original secret is a binary string However a container is a column of points in d which carries only partial information of the secret Even attackers successfully obtain less than t containers They are not able to extract any information of the secret

b) Ensuring integrity

Before the discussion we would like to mention first that practically t will not be a large number and it is very likely less than 5 and n is much larger then t Now assume that the storage system is under attack and the content of container is modified without notification and authorization We are able to detect such modification and correct it

S20

Page - 453


18

when a defected container is used for recovery For example we have t containers of a secret s namely 1[ ]s 2[ ]s [ ]ts such that the first entry of the first container has been modified illegally When we recover s from 1 2[ ] [ ]s s [ ]ts the first row of collector array will determine an element in Λ which is not in the range of Φ It is because the size of Λ is very much larger than the size of the range of Φ Besides in practice we would like to make the elements of the range of Φ ldquoaway from each otherrdquo So we are able to know that the first row of 1 2[ ] [ ]s s [ ]ts has been modified illegally To identify the location and correct the illegal modification we need another t good containers Check the containers one by one with 1t minus good containers Then 1[ ]s is the bad container So we conclude that the first entry of the first container has been modified illegally To correct this firstly we use the first row of the t good containers to recover 1s and then by using the choice function we can recover the bad entry on 1[ ]s (ie the first entry in this case)

c) Ensuring reliability

Assume attackers have damaged a container Administrators have no problem to restore the impaired container First of all we obtain t containers from non-compromised storages Secondly we recover the original secret s from those containers and finally we can utilize the secret s to generate the damaged container again Furthermore we should note that getting t good containers is possible since n is much larger than t For

example we choose 10n = and 3t = in this project Therefore we have 9 8463

= many

of combinations of three good containers ready for recovering the damaged container Besides with big n the framework provides user with more access availability for t containers and hence the secret Even because of the fact that it is ( )t n secret sharing it still maintains the good confidentiality while the number of containers or storages increases

Reference

1 Gallian J ldquoContemporary Abstract Algebrardquo Brooks Cole 6 edition 2004

S20

Page - 454


19

4 Algebraic Methods

Let tΛ be the set of the graphs of all the polynomials of degree 1t minus By the elementary algebra result [1]

Assume that functions f and g are polynomial functions of degree 1t minus and they agree with each other on t distinct points then f g= and their coefficients are also equal

Therefore tΛ satisfies properties a) and b) of definition of 311 To show that tΛ also satisfies property c) of the definition we need Lagrange polynomials

41 Lagrange Polynomials

Given 1 1( )x y 2 2( )x y ( )t tx y such that their x coordinates are distinct let

1

1

( )

( )( )

t

iii j

j t

j iii j

x x

L xx x

=ne

=ne

minus

=minus

prod

prod

where 123 j = So ( )jL x is a polynomial of degree 1t minus such that

1( )

0j i

i jL x

i j=⎧

= ⎨ ne⎩

where 1 23 i j t= and they are called Lagrange basis functions Hence the Lagrange polynomial [2]

1 1 2 2 3 3( ) ( ) ( ) ( )L x y L x y L x y L x= + + +

will pass through all the given points

Note that the y intercept of ( )L x (0)L can be evaluated by

1 1 2 2 3 3(0) (0) (0)y L y L y L+ + +

where

S20

Page - 455


20

1

1 1

1 1

(0 ) ( 1)

(0)( ) ( )

t tt

i ii ii j i j

j t t

j i j ii ii j i j

x x

Lx x x x

minus

= =ne ne

= =ne ne

minus minus

= =minus minus

prod prod

prod prod

for 1 23j =

Therefore tΛ is t secret shareable if Ψ maps 1 1( )x y 2 2( )x y ( )t tx y to the graph

of the Lagrange polynomial 1 1 2 2 3 3( ) ( ) ( )y L x y L x y L x+ + +

42 Unique Factorization Property (UFP)

Let 1 2 kA m m m= be a set of positive integers bigger than 1 and we always assume

that they are in increasing order ie 1 2 km m mlt lt lt The set A (or the finite increasing

sequence 1 2 km m m ) has unique factorization property (UFP) if for any binary strings

1 2 kε ε ε and 1 2 kε ε ε

1 2 1 21 2 1 2

k kk km m m m m m=ε εε ε ε ε

implies i i=ε ε for all 12 i k=

Also the span of an UFP set A denoted by A is defined to be

1 21 2 1 2 for some binary string k

k kA m m m m m= = εε ε ε ε ε

Suppose now 1 21 kA m m m= le le le le le is an infinite set of integers For any

123k = a k th segment kA of A is defined to be the set of the first k elements of

A We call the set A has UFP if for any 123k = kA has UFP The span of A

denoted by A is the union of the span of all the segments of A

Example 421 The sets 23 p and 234 are of UFP where p is a prime which is larger than 3

Lemma 422 Let k be a positive integer For any integer 0 2kmle lt there is a unique binary string 1 2 kε ε ε such that

11 21 2 2k

km minus= + +ε ε ε

Conversely for any binary string 1 2 kε ε ε we have 11 20 1 2 2 2k

kkminus ltle + +ε ε ε

S20

Page - 456


21

Example 423 Let p be a prime number and let k be a positive integer Define

2 3 11 2 2 2 2( ) k

kX p p p p p pminus

=

Assume that

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k

k kp p p p p p p p p pminus minus

=ε ε ε εε ε ε ε ε ε

So

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k

k kp pminus minus+ ++ + + ++ + + +=ε ε ε εε ε ε ε ε ε

By the Lemma 422 we have i i=ε ε for all 1 2 i k= So ( )kX p has UFP

From the lemma again the span of ( )kX p is the set 0 1 2 2 2 2 1 k k

p p p p pminus minus

Therefore 2 3 11 2 2 2 2 2( )

k k

X p p p p p ppminus

= has UFP and its span is 0 1 2 3 p p p p

Example 424 Let ( ) is a primeX X p p=cup We arrange and index the elements of

A in increasing order and

1 2 3 234579111316171923 X m m m= =

from now on

Let 1 2 3 k km m mX m= be the set of first k elements of X where 1 23k = We claim that kX has UFP Assume that kX consist of the powers of primes 1 2 rp p p Hence kX can be written as below

1 21 1 12 2 2 2 2 21 1 1 2 2 2

k k kr

k r r rp p p p p p p p pXminus minus minus

= where 1 2 rk k k k+ + + = Let

1 2

1 1 2 21 1 1 r

r rk k kε ε ε ε ε ε and

1 2

1 1 2 21 1 1 r

r rk k kε ε ε ε ε ε

such that

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr

r r rp p p p p p p p pminus minus minusε ε εε ε ε ε ε ε

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr

r r rp p p p p p p p pminus minus minus

= ε ε εε ε ε ε ε ε So

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr

rp p pminus minus minus

+ + + + + + + + +ε ε εε ε ε ε ε ε

S20

Page - 457


22

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr


+ + + + + + + + += ε ε εε ε ε ε ε ε Since such factorization is unique and ( )

ik iX p has UFP for all 12 i r= we conclude that

i ij j=ε ε

for all i j and hence kX has UFP and hence X has UFP Theorem 425 Let kX be the same as the one in above example and let m be a positive

integer less than km Then m is in the span of kX

Proof Assume that 1 21 2

rrm p p pα α α= Since km xlt we have 1 2

1 2 rr kp p p mα α α lt and

hence 1 2 r kp p p Xisin For any 1 23 i r= there is a positive integer ik such that

12 2i ik k

i k ixp pminus

lelt

Since ( )i ik kpX Xsub and 2ki

ii ip pα lt is in its span

12 331 2 4 221 2 2

i iii i i kki i

i i i i i ip p p p p pαminus

= εεε ε ε for

some binary string 1 2 i

i i ikε ε ε Therefore m is in the span of kX

Let k be a positive integer and let A be an UFP set with A k= Define AΠ to be the

product of all its elements A is minimum if for any UFP set Aprime with A kprime = we have

A AprimeΠ le Π A is completely minimum if every segment of A has minimum UFP

Example 426 The first eight minimum UFP sets are 1 2X = 2 23X =

3 234X = 4 2345X = 5 23457X = 6 234579X =

7 23457911X = and 8 2345791113X = Also 1 2XΠ = 2 6XΠ =

3 24XΠ = and 4 120XΠ =

Note that

4X is useful in this project since it has minimum UFP and 4 120XΠ = is smaller than 127 the absolute limit of a character type variables

We conclude the section by two interesting conjectures For 1 23k =

Conjecture 1) kX is the unique set which has completely minimum UFP

Conjecture 2) kX is the unique set which has minimum UFP

S20

Page - 458


23

43 Proposed Encoding Function ΦΛt

Recall that tΛ is the set of the graphs of all the polynomials of degree 1t minus For k bits

1 2 kε ε ε and 1 2 1 1tk k k minus ge such that 1 2 1tk k k kminus+ + + =

1 2( )t kΛΦ ε ε ε is the graph of

1 21 2 1 11 2 1 1 1 2

1 1 1 2 1 11 2 1 2 1 2( )( ) ( )k kk k k k t t k

t tk k k k k k kx m m m x m m m x m m m+ ++ + minus minus

minus minus+ + + +minus minus minus ε εε ε ε ε εε ε

Since kX is of UFP we have the function tΛ

Φ is one to one Hence ( )tt ΛΛ Φ is k bit t

secret sharing pair

Example 431 Let 4 2345X = and 1 2 2k k= = Therefore 4k = Let 3Λ be the set

of the graphs of all the quadratic polynomials For any 4 bits 1 2 3 4ε ε ε ε Φ maps 1 2 3 4ε ε ε ε to

the graph of the quadratic polynomial 31 2 4( 2 3 )( 4 5 )x xminus minus εε ε ε For convenience we would like to identify the graph of 1 2 3 4( )Φ ε ε ε ε with the polynomial 1 2 3 4( )Φ ε ε ε ε in this case

So the y intercepts of 1 2 3 4( )Φ ε ε ε ε ( )1 2 3 4( ) 0Φ ε ε ε ε can be summarized in the table below

1 2 3 4ε ε ε ε 1 2 3 4( )Φ ε ε ε ε ( )1 2 3 4( ) 0Φ ε ε ε ε 1 2 3 4ε ε ε ε 1 2 3 4( )Φ ε ε ε ε ( )1 2 3 4( ) 0Φ ε ε ε ε0000 (x-2030) (x-4050) 1 0110 (x-2031) (x-4150) 12 1000 (x-2130) (x-4050) 2 0101 (x-2031) (x-4051) 15 0100 (x-2031) (x-4050) 3 0011 (x-2030) (x-4151) 20 0010 (x-2030) (x-4150) 4 1110 (x-2131) (x-4150) 24 0001 (x-2030) (x-4051) 5 1101 (x-2131) (x-4051) 30 1100 (x-2131) (x-4050) 6 1011 (x-2130) (x-4151) 40 1010 (x-2130) (x-4150) 8 0111 (x-2031) (x-4151) 60 1001 (x-2130) (x-4051) 10 1111 (x-2131) (x-4151) 120

Then 3( )Λ Φ is a 4 bit 3-secret sharing pair

Remark We would like to choose a encoding pair in order to making the size of containers to be as small as possible In the example 431 the size of resulted containers is double of the size of origin secret

44 Choice functions of Λ and the Calculation of Φ Λt -1(Ψ)

Let ( )tt ΛΛ Φ be the k bit t secret sharing pair as above and n tge For any CisinΛ such

that C is a graph of a polynomial ( )P x in tΛ We define

S20

Page - 459


24

( ) ( ( ))j C j P jπ =

for 12 j n=

Given 1 1( )x y 2 2( )x y ( )t tx y we are going to find

11 1 2 2( ) ( ) ( )( ( ))

t t tx y x y x yminusΛΦ Ψ

First of all we build up a look up table which show the one to one correspondence between k bits onto y intercept of polynomial in the range of Φ By using Lagrange polynomials we can find a polynomial of degree 1t minus ( )L x passing through all the given points Find the y intercept of ( )L x by evaluating (0)L Then we can find

1 2 kε ε ε such that

1 2 1 1 2 2( ) ( ) ( )( ) ( )t k t tx y x y x yΛΦ Ψ=ε ε ε

by looking up from a table which gives the correspondence between 1 2 kε ε ε and

1 2( )(0)t kΛΦ ε ε ε

Exmaple 441 Look up table for 22( )ΛΛ Ψ is

1 2 3 4ε ε ε ε ( )2 1 2 3 4( ) 0ΛΦ ε ε ε ε 1 2 3 4ε ε ε ε ( )

2 1 2 3 4( ) 0ΛΦ ε ε ε ε 0000 1 0110 12 1000 2 0101 15 0100 3 0011 20 0010 4 1110 24 0001 5 1101 30 1100 6 1011 40 1010 8 0111 60 1001 10 1111 120

45 Robustness Analysis of Lagrange Polynomial when t=3

Let 3t = Given 1 1 2 2( ) ( )x y x y and 3 3( )x y the y intercept of the Lagrange polynomial passing given points is given by

( )( ) ( )( ) ( )( )3 2 1 3 1 2

1 2 31 3 1 2 2 1 2 3 1 33 2

(0) x x x x x xL y y yx x x x x x x c x x x x


We assume that noise presents on 1 2 3 x x x which is bounded by ε and the noise presents

S20

Page - 460


25

on 1 2 3 y y y which is bounded byδ Also 1 2 3 x x x is bounded by M and the distance between any pair of them is bigger than λ We wish to found an error bound of the calculation of noisy (0) (0)L L Observing that

( ) ( )( )( )

( )( )

2 1 2 2 1 1

2 1 2 1 2 1

2 1 2 1 2 1

2 1

2 1 2 1 2 1

1 1x x x x

x x x xx x x x

x x x x

ε ε

ε εε ε

ε εε ε

+ minus

+ minus minus=

+ minus

minus=

+

minusminus minus


minusminus minus

where 1 2ε ε are noises associated with 1 2x x respectively If 1 2x x λminus ge then

2 2 1 1 2x xε ε λ ε+ minus gt minusminus Therefore we have

( )2 1 2 2 1 1

1 1 22xx xx

εε ε λ λ ε

minusminus minus

le+ minus minus

Observe that

( )( )( )( )

( )( )

( )( )( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )

3 23 2

3 1 2 1 3 1 2 1

3 23 2

3 1 2 1 3 1 2 1

3 2 3 2

3 1 2

3 2

3 1 2 1

3 2

3 2 3 2

3 1

3 2 3 2

3 1

1 3 1 2 1

3 2 3 2

3 1 2 1 3 1 2 13 1 2 1

3 1 2

1

x xx xx x x x x x x x


x x x xx x x x x x x x


x x x

ε εε ε ε ε

ε ε

ε ε ε εε ε

ε ε ε εε ε ε ε ε ε

minusminus minus minus minus

le minusminus minus minus minus

+ minusminus minus minus minus

+ minusminus minus minus

+ ++ minus + minus

+ +

+ + + ++ minus

+ + + ++ minus

=minus

+ minus + minus minus

( )( )( )

( ) ( ) ( )( )( )( ) ( ) ( )

3 2 3 2

3 2

3 1

3 2

2 31

3 2

2 1 3 1

3 1

3 1

3 2

3 1 2 1 2 12 1

1

1 1

1

x xx

x xx x x x x x

x xx x x x x x

ε ε ε ε

ε εε ε

ε εε ε ε ε

minus


+ minusminus minus + minus

+ +

+ ++ minus

+ ++ minus minus

S20

Page - 461


26

( )( )( )( )

( )( )

( ) ( ) ( )( )

3 2

3 1

3 23 2

3 1 2 1 3 1 2 12 1

222 2

2

2 2

2

1 (2 )

1 2( )

1 2 4

2 22 2

2 22 2 2


M M M

M M M

ε εε ε ε ε

ε εε ε ε ελ

ε ε εελ

ω ω ε

λ λ ε λ λ ε

ε ελ ε λ ε λ ε

λ

+ ++ minus + minus

minus minus

leminus


le + + + + +

⎡ ⎤+ + +⎛ ⎞ ⎛ ⎞+ +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤le

minus

⎣

minus

+ ⎦

Therefore

where 2

M εε

ωλ

+=

minus So we have

( )( )( )( )

( )( ) ( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( ) ( )

( )( )( )

3 23 21 1 1

3 1 2 1 3 1 2 1

3 23 21 1

3 1 2 1 3 1 2 1

3 2

3 1 2 1

3 2

3 1 2

3 2 3 21 1 1

3 1 2 1 3 1 2 1

33 2

3 1 2 1

1

3 2 3 2

3 1 2 1 3 1 2 1

3

x xx x y yx x x x x x x x


x x x xy y

x x x x x x x x

xx xx x x x

ε εε ε ε ε

ε εε ε ε ε

ε ε ε εε ε ε ε ε ε ε

δε

δ

ε

+ ++ minus + minus

+ ++ minus + minus

+ +

minus +minus minus minus minus


+ minus+ +

+ minus + minus + minus ++

minus minus minus minus

le minusminus

minus

+minus

( )( )( )

( )( )( )( )

( )

21

3 1 2 1

3 21

3 1 2 1

2

3 1 2 1

3 2

3 1 2 1

2 21 2 4

xy

x x x x

x xx x x x

M

εε ε ε ε

ε εε ε ε ε

δ

ω ω ε ω δλ

++ minus + minus

+ ++ minus + minus

minus minus

+minus minus

le + +

where M is an upper bound of absolute values of all possible y

Theorem 451 Given 1 1 2 2( ) ( )x y x y and 3 3( )x y noise presents on 1 2 3 x x x which is bounded by ε and on 1 2 3 y y y with is bounded byδ Also 1 2 3 x x x and 1 2 3 y y y are bounded by M and M respectively Also suppose that the distance between any pair of them is bigger than λ Let (0)L be a noisy evaluation of (0)L Then

( )2 21(0) (0) 6 12 3L L Mω ω ε ω δλ

minus le + +

where 2

M εε

ωλ

+=

minus

S20

Page - 462


27

The following corollary gives a relatively simple inequality when 1 2 nx x x are equally spaced and noise is not present on them

Corollary 452 If 0=ε Mn

λ = where n is the number of containers then nω = and

2(0) (0) 3L L n δminus le

We assume 10n = in our study and employ the choice functions in 44 From corollary 452 we have

(0) (0) 300L L δminus le

Therefore keep the noise bounded by a ldquonot very smallrdquo number 0001δ = Then (0) (0) 1 3L Lminus le Note that from the look up table in Example 441 the smallest

distance between those y intercept is 1 which is bigger than 1 3 So the calculation of (0)L is still stable in this case

46 Implementation of the Algebra Method

All the programs developed in this report is in C and their interface is Qt from Qt Project [3] They are all running under testing environment as described below

OS Windows 8 CPU Inter Core i5-3337U(18Ghz Turbo27Ghz) RAM 4GB DDR3 Harddisk 128GB SSD

The testing group is randomly generated text files with different sizes 2 4 6 8 and 10 MB (megabytes) and the compression is performed by 7-Zip (httpwww7-ziporg)

Let 2( )Λ Φ be the 4 bit 3 secret sharing pair as in Example 431 Therefore

Φ maps 4 bits 1 2 3 4ε ε ε ε to the graph of a quadratic function

31 2 4( 2 3 )( 4 5 )x xminus minus εε ε ε

The entries of containers can be stored in the character type variables for obtaining smaller size of containers It can be done since the range of quadratic function

S20

Page - 463


28

31 2 4( 2 3 )( 4 5 )x xminus minus εε ε ε is from -120 to 120 and it is contained in the range of character which is from -127 to 127 One character occupies 1 byte or 8 bits and the method takes 4 bits at a time Therefore the size of the containers is double of that of the original secret file

461 Speed Tests

a) Producing Containers

The following table shows how much time needed for producing 10 uncompressed containers

Figure 7 Time required for various sizes of secrets to generate 10 containers

b) Recover the Original File

The following table shows how much time needed for recovering original files from their containers in a)

Secret Size ( in MB)

2 4 6 8 10

Rcovert Time (in Sec)

0237 0471 0708 0935 1176

0

1

2

3

4

5

6

7

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

Secret Size ( in MB) 2 4 6 8 10 Generating Time for 10 Containers ( in Sec)

1183 2349 358 473 5917

S20

Page - 464


29

Figure 8 Time required for various sizes of secrets to recover the secret from three containers

462 Size Tests

The following table shows that average sizes of the compressed container files of the testing group

Secret Size (in MB) 2 4 6 8 10

Compressed Secret Size (in Kb)

987 1973 2960 3947 4934

Compressed Container Size (in Kb)

9845 19502 29195 38791 46789

Size Ratio of Compressed Secret and Compressed Container (in )

100 99 99 98 95

0

02

04

06

08

1

12

14

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 465


30

Figure 9 Comparison of Sizes of compressed secret

and compressed container for various sizes of secrets

Reference

1 Fine H B ldquo A College Algebrardquo Ginn amp company 1904

2 Hildebrand F B ldquoIntroduction to Numerical Analysisrdquo New York McGraw-Hill 1956

3 Qt Project httpqt-projectorgsearchtagqtgui

0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

File

Siz

e in

KB

Secret Size in MB

CompressedSecret

CompressedContainer

S20

Page - 466


31

5 Geometric Methods

51 3 Secret Shareable Set Λ3

Let 3Λ be the set of all circles on the plane Recall that 3 non-collinear distinct points determine a circle Since three points on a circle cannot be collinear in 2 we have 3Λ satisfies a) and b) Indeed it also satisfies property c) To see that we consider the following calculation [1]

Let 1 1( )x y 2 2( )x y and 3 3( )x y lie on a circle

2 2C Ax By Cx Dy E+ + + + or 2 2 ( ) ( )C C CC x x y y rminus + minus =

Consider the following determinant equation

2 2

2 21 1 1 12 22 2 2 22 23 3 3 3

11

( ) 011

x y x yx y x y

J x yx y x yx y x y

++

= =++

By evaluating the cofactors 1 jM for the first row of the determinant the determinant can

be written as an equation of these cofactors

2 211 12 13 14( ) 0M x y M x M y M+ minus + minus =

or

2 2 1312 14

11 11 11

( ) 0MM Mx y x yM M M

+ minus + minus =

Since 1 1 2 2 3 3( ) ( ) ( ) 0J x y J x y J x y= = = we have the above equation also represents the

circle C Extending 2 2( ) ( )C C Cx x y y rminus + minus = into

( )2 2 2 2 22 2C C C C Cx y x x y y x y r+ minus minus + + minus

and comparing the coefficients of

2 2 1312 14

11 11 11


+ minus + minus =

we have

S20

Page - 467


32

12

11

13

11

2 2 14

11

2

2

C

C

C C C

MxM

MyM

Mr x yM

=

=

= + +

Hence Ψ maps 1 1( )x y 2 2( )x y and 3 3( )x y to the circle with center

( ) 1312

11 11

2 2C C

MMx yM M

⎛ ⎞= ⎜ ⎟⎝ ⎠

and radius 2 2 14

11C C

Mx yM

+ + Therefore 3Λ is 3 secret

shareable

52 Encoding Function ΦΛ3

Let 1k 2k and 3k be positive integers such that 1 2 3k k k k= + + Since a circle C can be

determined by its center ( )C Cx y and its radius Cr we would like to define 3ΛΦ to be a

function that maps a k bit string 1 2 kε ε ε to a circle C such that

21

1 1 2

1

1 11 2 1 2( ) ( 2 2 )2 2k kC C k k k k

kkx y +

+++= + + + + + +ε ε ε ε ε ε

and

1 2 1 21 21 2 2k

k k k kC kr + + + += + + + +ε ε ε

Then 3ΛΦ is one to one since binary representation of integers is unique and hence

33( )

ΛΛ Ψ is a k bit 3 secret sharing pair

53 Choice functions of Λ3 and the Calculation of Φ Λ3 -1(Ψ)

Let 3n ge and 02nπθlt le For any 3CisinΛ such that C has center ( )C Cx y and its radius

Cr We define

( ) ( sin( ) cos( ))j c c c cC x r j y r jπ θ θ= + +

for 12 j n=

S20

Page - 468

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar

3 if two sphhey are equa

hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4

heres have foal It is becau

Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4

our points in use these fou

Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched

common (Rur distinct po

spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b

Ref Figure 1oints may be

our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s

0) it does ne coplanar (

ommon

s Award 201

3

with center

sharing pair

not implies Ref Figure

13

33

S20

Page - 469


34

Therefore the set of all the spheres in 3 is not 4 secret shareable However by Theorem 541 we can see that it is ldquoalmostrdquo 4 secret shareable

Theorem 541 The probability of picking n distinct points randomly and independently on the 3 dimensional sphere such that there are 4 point among them lying on the same plane is equal to zero

Proof We would like to prove it by induction

Assume that we have chosen 3 points from the sphere The area of the intersection of the sphere and the plane determined by chosen points is equal to zero So the event picking a point randomly such that it lies on the intersection has zero probability

Let A be the event such that n th point is chosen and there is 4 points among them are coplanar and B be the event such that 1n minus point has been chosen and there is 4 points among them are coplanar We wish to prove that ( ) 0P A = by using the formula of conditional probabilities

( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +

where the event cB is the negation of the event B By the induction hypothesis ( ) 0P B = so ( ) ( | ) ( )c cP A P A B P B=

Now suppose that we have picked 1n minus points and any 4 points among them are not coplanar Similarly the area of the intersection the sphere and the union of all planes determined by 4 points in the 1n minus points is zero since the number of the planes is

14

n minus⎛ ⎞⎜ ⎟⎝ ⎠

and the area of the intersection of the sphere and a plane is zero Hence

( | ) 0cP A B = So we conclude that ( ) 0P A = and by the principle of Mathematical induction the proof is complete

According to Theorem 541 a non-painful way is that just let Λ be the set of all the spheres in 3 since the probability of making mistake is zero However how safe would it be probability 0

If we would like to go for a safe approach pick enough points from the unit sphere and form a set Ω Note that 1 1 1 2 2 2( ) ( )x y z x y z and 3 3 3( )x y z are coplanar if and only if

1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35

Check all the combinations of four distinct points in Ω if they are coplanar By the above Theorem you are almost safe but it takes time If not for all combinations then stop and keep the set Ω If yes for a combination then choose n points and repeat the checking again unit we finally have a non coplanar set Ω of points on the sphere

Let 0r gt and let v be a vector in 3 Define if D is in 3 the r dilation and v translation of D is

rD p p Dα= isin and v D p v p D+ = + isin

Since coplanarity is invariant under dilation and then translation we define

4 3 0 and v r r vΛ = + Ω gt isin

Note that v r+ Ω is a subset of the sphere with center v and radius r Therefore we can consider v r+ Ω as a discrete model of the sphere such that any combination of four distinct points in v r+ Ω are not coplanar We would like to denote v r+ Ω by ( )S v r

Also given four non-coplanar distinct points 1 1 1 2 2 2 3 3 3 4 4 4( ) ( ) ( ) ( )x y z x y z x y z x y z and similarly consider the determinant

2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111

x y z x y zx y z x y z

J x y z x y z x y zx y z x y zx y z x y z

+ ++ +

= =+ ++ ++ +

1 1 1 2 2 2 3 3 3 4 4 4( ) ( ) ( ) ( )x y z x y z x y z x y z should belong to 1 2 3(( ) )S v v v r such that

121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +

where 11 12 13 14 M M M M and 15M are cofactors corresponding to the first row of

( )J x y z Hence 4Λ is 4 secret sharing

S20

Page - 471


36


Let k be an integer bigger than or equal to 4 and let 1k 2 3k k and 4k be positive integers

such that 1 2 3 4k k kk k= + + + Then the function 4ΛΦ maps a k bit string 1 2 kε ε ε to

( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε

Again 4ΛΦ is one to one since binary representation of integers is unique Hence

44( )

ΛΛ Φ is a k bit 4 secret sharing pair


Suppose that 1 2 lu u uΩ = such that 4l nge ge For any 3( )S v r isinΛ we define

( )( ( ))j jS v r v r uπ = +

for 12 j n=

Similarly 41 ( )minusΛ

Φ Ψ maps a k bits 1 2 kε ε ε to

2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2

M M MM M M MM M M M M M M

⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠

We can implement the k bit ( )4n secret sharing storage for the pair 44( )

ΛΛ Φ

Note that to generate the geometric method to the case with 5t ge we only need the finite dimensional version of Theorem 341 Of course it is true but we need the notion of ldquoareardquo on the high dimensional sphere first It is a topic of advanced Mathematics called

S20

Page - 472


37

Haar measure [2] If we take it for granted the generalization of the theorem for arbitrary t is straight forward from the 4t = case

56 Implementation of the Geometric Method

Indeed we have implemented the circle and sphere methods and their performance is satisfactory as we expect In this section we only present testing results for 3

3( )Λ

Λ Ψ

with 24k = and 1 2 3 8k k k= = = A point ( )x y is stored in two floating type variables

which occupy 64 bits So the size of the container is 64 2224 3

= times of that of the original

secret file before compression

561 Speed Test




Generating Time for 10 Containers ( in Sec)

1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference

1 Center and Radius of a Circle from Three Points httpwwwabecedaricalcomzenosampleszs_circle3ptshtml

2 Halmos R ldquoMeasure Theoryrdquo Springer 1974

0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB

CompressedContainerCompressed Secret

S20

Page - 475


40

6 Number Theory-Chinese Reminder Theorem (CRT) Methods

In this chapter we will develop a k bit ( )3 n secret sharing storage which is following

the idea of Asmuth and Bloom [1] Note that such n is not any integer larger than 3 It actually depends on the primes prepared for such secret sharing method The Asmuth and Bloom secret sharing is based on Chinese remainder theorem (CRT) Firstly we are going to demonstrate the basic idea of CRT by solving the following problem

Find the smallest whole number that when divided by 35 and 7 gives remainders of 12 and 3respectively

Formally the above problem is asking for the smallest whole number such that

1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv

Instead of solving the above system we would like to solve three much simpler systems

1 mod 3) 0 mod 3) 0 mod 3)0 mod ) 1 mod ) 0 mod )0 mod ) 0 mod )

( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x

equiv equiv equivequiv equiv equivequiv equiv equiv

The solutions of the first second and third system are 70 21 and 15 respectively We call these numbers base solutions Consider

70 1 21 2 15 3 157times + times + times =

Although 157 divided by 35 and 7 gives remainders of 1 2 and 3respectively it is too big Hence 157x = minusLCM(357) 157 105 52= minus = is the answer Moreover if we want to solve a system such as

1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv

then the general solutions are 1 2 370 21 15 105y y y ρtimes + times + times minus times where ρ isin Therefore the key is obtaining the base solution Formally we can state the Chinese Remainder Theorem as below

S20

Page - 476


41

Theorem (Chinese Remainder Theorem) Let 1 2 p p and mp be increasing distinct

primes For any integers 1 2 ma a a there is an integer x with

( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv

and x is unique mod 1 2 mp p p

61 Three Shareable Set Λ(p1p2hellippm)

Let 1 2 mp p p be the set of increasing prime numbers such that

1 1 2 3m mp p p p pminus lt

and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus

minus= For 1 1 2 3m mp p l p p pminus lt lt we define

( ) 2 ( mod ) 12 l i iC p l p i m= = sub

and

21 2 1 1 2 3( ) m l m mp p p C p p l p p pminusΛ = sub lt lt

Suppose that 1l

C and 2l

C has three distinct points in common Hence there are three

prime 1p 2p and 3p such that

( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1

mod modmod modmod mod

p l p p l pp l p p l pp l p p l p

===

Since by CRT 1l and 2l are the unique solutions of the systems

( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42

and the systems in above are equivalent we have 1 2l l= Hence 21l lC C=

Ψ can be defined as follow Given three distinct points ( )11p y ( )22 p y and ( )33p y in

1 2( )mC p p pisinΛ by applying CRT to the system

2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv

we have a unique solution 1 2 30 l p p plt le Since 1 2 3 1 2 3p p p p p pge we have lC =C

Lastly Let ( )11p y and ( )22 p y be two distinct points in lC Suppose that l is the

unique solution of the system

2 2

1 1modmod

xx

y py p

equivequiv

which is between 0 and 1 2p p Since 1 2 1m mpp pp minusle there exist at least numbers

between 1m mp pminus and 1 2 3p p p equivalent to l in mod 1 2p p Therefore lC cannot be determined

Hence 1 2( )mp p pΛ is 3 shareable

62 Encoding Function ΦΛ(p1p2hellippm)

Let k be the biggest positive integer such that 11 1 2 32 1k

m mp p p p p+minus lt minus lt We would like

to define 1 2( )mp p pΛΦ to be a function that maps a k bit string 1 2 kε ε ε to a circle lC

such that

1 2 2 2kkl = + + +ε ε ε

Then 1 2( )mp p pΛΦ is one to one since binary representation of integers is unique and

hence 1 21 2 ( )( ( ) )

mm p p pp p p ΛΛ Φ is a k bit 3 secret sharing pair

63 Choice functions of Λ(p1p2hellippm) and the Calculation of Φ Λ(p1p2hellippm) -1(Ψ)

Let 3 n mlt le For any 12 j n= we define

( ) ( (mo ))dj l j jC p l pπ =

S20

Page - 478


43

for all 1 2 31m mp p l p p pminus ltle

Given three distinct points ( )11p y ( )22 p y and ( )33p y in

1 2( ) 1 2 1 2( ) ( )mp p p k mp p pΛΦ isinΛε ε ε

Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε

where l is the unique solution of the system

2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv

between 1m mp pminus and 1 2 3p p p Then 1 2 1 mod2( )k m ml p pminus= minusε ε ε

Hence a k bit ( )3n secret sharing storage can be launched for the k bit 3 secret

sharing pair 1 21 2 ( )( ( ) )

mm p p pp p p ΛΛ Φ

Example 631 Consider the following table

1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10

8884p p p p pp pminus

= =

Example 632 Also look at

1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44

64 Implementation of the CRT Method

We consider the following sequence of prime numbers

1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10


= = The size of containers is now half of the original

secret file

641 Speed tests



Secrect Size (in MB) 2 4 6 8 10 Generating Time for 10 containers (in sec)

0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45

c) Recover the Original File


Secret Size (in MB) 2 4 6 8 10 Recover Time (in sec) 0037 0078 0104 0150 0188


642 Size Tests



Compressed Secret Size (in KB)

987 1973 2960 3947 4934

Compressed Container Size (in KB)

3941 7885 11834 1577 19719

Size Ratio of Compressed Secret and Compressed Containers (in )

40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference

1 Asmuth CA and Bloom J ldquoA modular approach to key safeguardingrdquo IEEE Transactions on Information Theory IT-29(2)208-210 1983

0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret

CompressedContainers

S20

Page - 482


47

7 Conclusion

In this report we have mainly developed an abstract framework of k bit t secret sharing framework for cloud storage This framework is clean and it can be implemented A successful implementation of the framework would provide users with protection when the system is under the attack on its confidentiality integrity and reliability Furthermore such system has its own encryption by using permutations and tailor made error detection location and data rescue

We make use of Lagrange polynomials and take the advantages of the algebraic property ldquo t distinct points on the plane can uniquely determine a polynomial function of degree

1t minus rdquo to design a k bit ( )t n -secret sharing distributed storage We employ the set with unique factorization property (UFP) so that we simply need to calculate the y intercept of a Lagrange polynomial and then use a look up table to recover a secret Moreover the set which has minimum UFP would help us to design storage with smallest containers

In addition to the algebraic methods we can utilize the geometric facts that

a) three non collinear points determine a unique circle b) four non coplanar points determine a unique sphere

to construct k bit (3 )n and k bit (4 )n secret sharing storage respectively To generalize to arbitrary case it is straight forward if we have defined the Haar measure on the higher dimensional unit sphere

The last method is an application of Chinese Remainder Theorem and we have designed k bit (tn) -secret sharing distributed storage and one of the designs can offer containers with the same size of the original secret However such k is no longer unrestricted and it is chosen within a certain range

We have developed a C program for implementing both algebraic and geometric k bit (3 )n secret sharing distributed storages The performances of both algebraic and geometric designs are satisfactory in term of processing time and compressed container size The container size reaches 50 of size of the original secret and 40 after compression in the CRT case Besides it is very speedy

Finally concerning the framework of distributed storage and its techniques the notion is clean cute and mostly original Also it is proved to be working efficiently

S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery

ical Sciencess Award 201

4

13

49

S20

Page - 485


50

所獲獎項

年份比賽獎勵名稱頒發機構所獲獎勵備注

2013 第 28 屆年全國青少年科技創

新大賽

The 26th China Adolescents Science and Technology Innovation Contest

周培源基金會和中國

科協教育部科技

部國家環保總局

國家體育總局國家

自然科學基金委全

國少工委全國婦

聯國家自然科學

基金委和南京人民政

府共同主辦

周培源青少

年科技創新

獎和二等銀

獎

一等獎缺二等獎兩

名

2013 紅藍之光澳門培正中學得獎者因在數學及科學上有

優異表現為澳為校

爭光而獲獎

2013 科普活動傑出獎學金澳門培正中學得獎者

2013 Intel 國際科學與工程大獎賽 Intel International Science and Engineering Fair

Society for Science amp the Public (SSP)

Finalist Nominated -The IEEE Foundation Presidents Scholarship Award

2013 科技創新成果競賽 2013 澳門教育暨青年局高中組個人

項目優異獎

2012 第 3屆丘成桐中學應用數學

科學獎

The 3th Shing ndashTung Yau Mathematical Science Award

丘成桐教授泰康人壽保險股份有

限公司

優勝獎

全球決賽

2012 澳門科學與工程大獎賽 2012 澳門教育暨青年局優異獎

2012 第 5 屆丘成桐中學數學科學

獎-南部賽區

中山大學一等獎中國南部賽區決賽

2012 紅藍之光澳門培正中學得獎者




Finalist 國際大賽


新大賽

The 26th China Adolescents

中國科協教育部

科技部國家環保總

局國家體育總局

內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51

Science and Technology Innovation Contest

國家自然科學基金

委全國少工委全

國婦聯國家自然

科學基金委和內蒙

古自治區人民政府共

同主辦

2011 紅藍之光澳門培正中學得獎者因在科學數學寫

作及音樂上有優異表

現為澳為校爭光而

獲獎

2010 首屆澳門十大傑出少年選舉

1st Macau 10 outstanding teenager election

澳門基督教青年會

Y M C A Macau 十大傑出少

年

2010 第 8 屆走進美妙的數學花園-

中國青少年數學論壇

中國少年科學院一等一名金

獎

全國獎項

2010 第 5 屆中國中學生作文比賽-

澳區比賽

澳門青年聯合會優異獎獲選全國賽澳區代表

2010 聖庇護十世音樂院獎學金聖庇護十世音樂院高階組得獎

者

各階選一人得獎

2009 第 27 屆澳門青年音樂比賽- 鋼琴四手聯彈初級

澳門文化局第二名

公開賽獎項

2009 區師達神父獎學金聖庇護十世音樂院得獎者

2008 第 25 屆全澳學生朗誦比賽- 普通話高小獨誦

澳門中華教育會二等獎校際賽獎項

2007 ldquo我與世界遺產rdquo中國校際作

文徵集活動中國聯合國教科文組

織全國委員會

一等獎全國獎項

2007 澳門青年交響樂團 ndash 優秀團員獎 (少年團)

澳門青年交響樂團得獎者

2006 第 23 屆全澳學生朗誦比賽- 普通話初小獨誦


2005 第 23 屆澳門青年音樂比賽- 鋼琴獨奏 B 組

澳門文化局第三名公開賽獎項

2005 -2013

鋼琴成績優異獎聖庇護十世音樂院得獎者

S20

Page - 487


52

著作

譚知微曾學為 ldquoldquo眾妙之門rdquomdashmdash數學原理在電子門禁系統上的應用rdquo 《中国科技

教育1》2012 年第 2 期 26-28 頁及第二十六屆全國青少年科技創新大賽獲獎作品集科學普及出版社 180-183 頁

公演及展示

23-29082013 香港第四十六屆聯校科學展覽 (澳門區唯一所獲邀項目)

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httpswwwyoutubecomwatchv=2azl-6OncQA

20012013 扶樂普天頌-區師達神父作品音樂會-鋼琴獨奏

httpswwwyoutubecomwatchv=bOPpoQQmzuk (由 0020 開始)

httpswwwyoutubecomwatchv=Lmpw21-s5BQ

(樂跡區師達神父生平及手稿展覽宣傳片 Opening Music)

11092010 澳門青年交響樂團- 13 週年會慶音樂會

08072010 培正中學管弦樂團新加坡公演

03042009 培正創校 120 周年紀念音樂會-管弦樂團之大提琴鋼琴四手聯彈

20122008 澳門回歸九週年-澳門青年交響樂團

29112008 踏出第一步青少年音樂會-大提琴獨奏

30082008 澳門青年交響樂團十一週年會慶音樂會-大提琴

27082008 澳門青年交響樂團-深圳之行-深圳市保利劇院

02082008 聖庇護十世音樂學院四十五週年慶典音樂會-雙鋼琴八手聯彈

1 《中國科技教育》是中國科學技術協會主管中國青少年科技輔導員協會主辦的一本關於科技教

育的國家級專業科普刊物

S20

Page - 488

in thefollowserveaccorWavebecomfrom cloud

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S20

Page - 438


Sh

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gure 3 Survey


nocent victim

Yau High Sch

Survey by HK

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13

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e

S20

Page - 439

Usuanodes

-The

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-The


Besiddetec

Sh


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hing-Tung Y


onfidentiality

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Yau High Sch


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hool Applied


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13

4

ge

e

f be e

S20

Page - 440


5

Reference






S20

Page - 441


6






21 Shamirs Scheme


S20

Page - 442


7



2 2( ) 166 94 1234 166 94 f x S x x x x= + + = + +


(11494) (21942) (3 2578) (43402) (5 4414) (65614)


21 10

0 1 0 1

20 21

1 0 1 2

20 12

2 0 2 1

1 3 106 2 3

1 7 5 2 21 823 3



minus minus




Therefore2

2

0( ) ( ) 1234 166 94j j

jf x y L x x x

=


22 Blakleys Scheme




1 2 3 4 5 6 2 3 4 5 6K x K x K x K x K x K x= = = = = =

S20

Page - 443


8


(00)






( )( )( )


xxx

===


( )( )( )

( )( )( )

( )( )( )


x x xx x xx x x

= = == = == = =


2 155(mod 3)S = =


Reference



2006

S20

Page - 444


9





S20

Page - 445


10





1 2 0 or 1 12 k k i i k= = =ε ε ε ε


if








2 3 1 3 1 21 2 3

1 2 1 3 2 1 2 3 3 1 3 2

( )( ) ( )( ) ( )( )( )( )( ) ( )( ) ( )( )






S20

Page - 446

Exam

funct

We w

Examtheorfigure

Henc

wher

Then

Sh

mple 313 F

tion 1( 2 )x + ε

would like to


ce 3Λ is 3 s

e xyC is a ci

n 3( )Λ Φ is a

hing-Tung Y


o give anothe


Figure 5 Th

secret sharea

rcle such tha

Figure 6 Th

a 2 bit 3 se

Yau High Sch

2isin we def

hen 33( ΛΛ Φ

er example o


hree non-collin

able Now w

ΦΦΦΦ

at its center i

he function Φ m

ecret shareab

hool Applied


) is a 2 bit

of secret shar

ircles in 2

near determi

ear points dete

e define a fu

( )( )( )( )

00

01

10

11

00011011

CCCC

===

Φ =

is (xy) and i

maps two bits

ble pair as we

d Mathemati

2 ) to be the g

3 secret sha

reable pair


ermine a uniqu

unction Φ fr

its radius is

ε1 ε2 to a circle

ell

ical Sciences

graph of the

areable pair


ue circle

rom 2 to Λ

1 3 (Ref F

e C ε1 ε2

s Award 201

1

quadratic


3Λ by

Figure 6)

13

11

y

S20

Page - 447


12




1 ( )j C Cπ isin




( ) ( ( ))j C j P jπ =

for 12 j n=


1 2 Ms s s s=



1 2i i i

kis = ε ε ε

and so

1 2 11 1 1 2 2

2 1 12

2 2 Mi i ik k




S20

Page - 448


13

[ ] ( ( ))ij j is sπ= Φ




( )( ) ( )( ) ( )( )( ) ( )( ) ( )( )( ) ( )( ) ( )

1

2

3

10 ( 2)( 1)

01 ( 1)( 3)

11 ( 2)( 3)

j j

j j

j j

s j j j

s j j j

s j j j

π π

π π

π π





( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

10 20 32 46 512[ ] 10 23 30 43 58

12 20 30 42 56s

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

1 2 3 4 5

10 20 32 46 512[ ] 10 [ ] 23 [ ] 30 [ ] 43 [ ] 58

12 20 30 42 56s s s s s

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦





[ ] [ ]1 212[ ] [ ] [ ][ ] i

jtts s ss p= =



S20

Page - 449


14

1 2( ) ( )i i ii ts p p pΦ = Ψ

and

11 2( )( )i i i


for 1 2 i M=


1 2

1 1 1 1 1 2 2 2 11 2 1 2 1 2( ( )) ( ( )) ( ( ))

M

n n nt t t

s s s s


=

=Φ Ψ Φ Ψ Φ Ψ






Mk=


[ ] ( ( ))ij j is sπ= Φ



Step 4 End





[ ] [ ] [ ]tt


S20

Page - 450


15


as in section 322


Step 5 END






[ ] [ ] ( )ij i js sσ

σ=







denoted by σ



S20

Page - 451


16



(0 8 6 7 2)(1 4 9)(3 5) (2 4 7 6 9)(0 8 5)(1 3) (0 3 7 9 2)(4 5 8)(1 6)

(1 6 9 4 5)(0 2 8)(3 7) (0 1 8 9 4)(2 5 6)(3 7) (0 2 8 4 9)(1 6 5)(3 7)

(0 8 9 5 4)(1 6 7)(2 3) (0 2 7 8 9)(1 4 3)(5 6) (0 9 5 4 7)(1 8 3)(2 6)

(0 2 9 5 3)(1 8 4)(6 7)

Example 334 Let

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

10 20 3 2 46 512[ ] 10 23 30 43 58

12 20 30 42 56s

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

32 10 20 512 46[ ] 23 30 10 43 58

12 20 30 56 42s σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

1 2 3 4 5

3 2 10 20 512 46[ ] 23 [ ] 30 [ ] 10 [ ] 43 [ ] 58

12 20 30 56 4 2s s s s sσ σ σ σ σ

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦





Mk=

S20

Page - 452


17


[ ] ( ( ))ij j is sπ= Φ




Step 5 End














S20

Page - 453


18





= many


Reference


S20

Page - 454


19

4 Algebraic Methods






1

1

( )

( )( )

t

iii j

j t

j iii j

x x

L xx x

=ne

=ne

minus

=minus

prod

prod


1( )

0j i

i jL x

i j=⎧

= ⎨ ne⎩


1 1 2 2 3 3( ) ( ) ( ) ( )L x y L x y L x y L x= + + +



1 1 2 2 3 3(0) (0) (0)y L y L y L+ + +

where

S20

Page - 455


20

1

1 1

1 1

(0 ) ( 1)

(0)( ) ( )

t tt

i ii ii j i j

j t t

j i j ii ii j i j

x x

Lx x x x

minus

= =ne ne

= =ne ne

minus minus

= =minus minus

prod prod

prod prod

for 1 23j =








1 2 1 21 2 1 2












11 21 2 2k




S20

Page - 456


21


2 3 11 2 2 2 2( ) k

kX p p p p p pminus

=

Assume that

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k



So

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k





Therefore 2 3 11 2 2 2 2 2( )

k k

X p p p p p ppminus




1 2 3 234579111316171923 X m m m= =

from now on


1 21 1 12 2 2 2 2 21 1 1 2 2 2

k k kr



1 2

1 1 2 21 1 1 r


1 2

1 1 2 21 1 1 r


such that

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr


1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr



1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr


+ + + + + + + + +ε ε εε ε ε ε ε ε

S20

Page - 457


22

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr




i ij j=ε ε







12 2i ik k

i k ixp pminus

lelt



12 331 2 4 221 2 2

i iii i i kki i


= εεε ε ε for







3 234X = 4 2345X = 5 23457X = 6 234579X =

7 23457911X = and 8 2345791113X = Also 1 2XΠ = 2 6XΠ =

3 24XΠ = and 4 120XΠ =

Note that





S20

Page - 458


23





1 21 2 1 11 2 1 1 1 2

1 1 1 2 1 11 2 1 2 1 2( )( ) ( )k kk k k k t t k





secret sharing pair











S20

Page - 459


24

( ) ( ( ))j C j P jπ =

for 12 j n=


11 1 2 2( ) ( ) ( )( ( ))






1 2( )(0)t kΛΦ ε ε ε


1 2 3 4ε ε ε ε ( )2 1 2 3 4( ) 0ΛΦ ε ε ε ε 1 2 3 4ε ε ε ε ( )

2 1 2 3 4( ) 0ΛΦ ε ε ε ε 0000 1 0110 12 1000 2 0101 15 0100 3 0011 20 0010 4 1110 24 0001 5 1101 30 1100 6 1011 40 1010 8 0111 60 1001 10 1111 120



( )( ) ( )( ) ( )( )3 2 1 3 1 2

1 2 31 3 1 2 2 1 2 3 1 33 2




S20

Page - 460


25


( ) ( )( )( )

( )( )

2 1 2 2 1 1

2 1 2 1 2 1

2 1 2 1 2 1

2 1

2 1 2 1 2 1

1 1x x x x

x x x xx x x x

x x x x

ε ε

ε εε ε

ε εε ε

+ minus

+ minus minus=

+ minus

minus=

+

minusminus minus


minusminus minus



( )2 1 2 2 1 1

1 1 22xx xx

εε ε λ λ ε

minusminus minus

le+ minus minus

Observe that

( )( )( )( )

( )( )

( )( )( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )

3 23 2

3 1 2 1 3 1 2 1

3 23 2

3 1 2 1 3 1 2 1

3 2 3 2

3 1 2

3 2

3 1 2 1

3 2

3 2 3 2

3 1

3 2 3 2

3 1

1 3 1 2 1

3 2 3 2

3 1 2 1 3 1 2 13 1 2 1

3 1 2

1





x x x

ε εε ε ε ε

ε ε

ε ε ε εε ε






+ ++ minus + minus

+ +

+ + + ++ minus

+ + + ++ minus

=minus


( )( )( )

( ) ( ) ( )( )( )( ) ( ) ( )

3 2 3 2

3 2

3 1

3 2

2 31

3 2

2 1 3 1

3 1

3 1

3 2

3 1 2 1 2 12 1

1

1 1

1

x xx

x xx x x x x x

x xx x x x x x

ε ε ε ε

ε εε ε

ε εε ε ε ε

minus



+ +

+ ++ minus

+ ++ minus minus

S20

Page - 461


26

( )( )( )( )

( )( )

( ) ( ) ( )( )

3 2

3 1

3 23 2

3 1 2 1 3 1 2 12 1

222 2

2

2 2

2

1 (2 )

1 2( )

1 2 4

2 22 2

2 22 2 2


M M M

M M M

ε εε ε ε ε

ε εε ε ε ελ

ε ε εελ

ω ω ε

λ λ ε λ λ ε


λ

+ ++ minus + minus

minus minus

leminus


le + + + + +

⎡ ⎤+ + +⎛ ⎞ ⎛ ⎞+ +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤le

minus

⎣

minus

+ ⎦

Therefore

where 2

M εε

ωλ

+=

minus So we have

( )( )( )( )

( )( ) ( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( ) ( )

( )( )( )

3 23 21 1 1

3 1 2 1 3 1 2 1

3 23 21 1

3 1 2 1 3 1 2 1

3 2

3 1 2 1

3 2

3 1 2

3 2 3 21 1 1

3 1 2 1 3 1 2 1

33 2

3 1 2 1

1

3 2 3 2

3 1 2 1 3 1 2 1

3



x x x xy y

x x x x x x x x

xx xx x x x

ε εε ε ε ε

ε εε ε ε ε


δε

δ

ε

+ ++ minus + minus

+ ++ minus + minus

+ +



+ minus+ +



le minusminus

minus

+minus

( )( )( )

( )( )( )( )

( )

21

3 1 2 1

3 21

3 1 2 1

2

3 1 2 1

3 2

3 1 2 1

2 21 2 4

xy

x x x x

x xx x x x

M

εε ε ε ε

ε εε ε ε ε

δ

ω ω ε ω δλ

++ minus + minus

+ ++ minus + minus

minus minus

+minus minus

le + +



( )2 21(0) (0) 6 12 3L L Mω ω ε ω δλ

minus le + +

where 2

M εε

ωλ

+=

minus

S20

Page - 462


27

















S20

Page - 463


28


461 Speed Tests







2 4 6 8 10


0237 0471 0708 0935 1176

0

1

2

3

4

5

6

7

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB


1183 2349 358 473 5917

S20

Page - 464


29


462 Size Tests




987 1973 2960 3947 4934


9845 19502 29195 38791 46789


100 99 99 98 95

0

02

04

06

08

1

12

14

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 465


30



Reference




0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

File

Siz

e in

KB

Secret Size in MB

CompressedSecret

CompressedContainer

S20

Page - 466


31

5 Geometric Methods






2 2

2 21 1 1 12 22 2 2 22 23 3 3 3

11

( ) 011

x y x yx y x y

J x yx y x yx y x y

++

= =++




or

2 2 1312 14

11 11 11


+ minus + minus =





2 2 1312 14

11 11 11


+ minus + minus =

we have

S20

Page - 467


32

12

11

13

11

2 2 14

11

2

2

C

C

C C C

MxM

MyM

Mr x yM

=

=

= + +


( ) 1312

11 11

2 2C C

MMx yM M

⎛ ⎞= ⎜ ⎟⎝ ⎠

and radius 2 2 14

11C C

Mx yM


shareable





21

1 1 2

1

1 11 2 1 2( ) ( 2 2 )2 2k kC C k k k k

kkx y +

+++= + + + + + +ε ε ε ε ε ε

and

1 2 1 21 21 2 2k

k k k kC kr + + + += + + + +ε ε ε


33( )




Cr We define


for 12 j n=

S20

Page - 468

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar


hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4


Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4


Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched


spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b


our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s


ommon

s Award 201

3

with center

sharing pair


13

33

S20

Page - 469


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


Sh

Fig




hing-Tung Y

Figure 2

gure 3 Survey


nocent victim

Yau High Sch

Survey by HK

by Infromation


ms themselve

hool Applied

KPC on Not Us

nWeek on Clou


d Mathemati

sing Cloud in 2

ud Storage Con


hurts benefits

ical Sciences

2012

ncern in 2013



s Award 201



nies privacy

13

3

le

e

S20

Page - 439

Usuanodes

-The

-The

-The


Besiddetec

Sh


attack on co

attack on in

attack on re




hing-Tung Y


onfidentiality

tegrity modi

liability mak


Fi


Yau High Sch


y reveals sto

ifies data in

kes storage s



igure 4 The id

d to have morction

hool Applied


red server co

victim stora

server unava



dea of the propo

re features su

d Mathemati


ontents to at

age servers w

ailable to leg



osed system

uch as unaut

ical Sciences


ttackers

without being

gitimate user


thorized mo

s Award 201

n one storag5]

g noticed

rs


dification

13

4

ge

e

f be e

S20

Page - 440


5

Reference






S20

Page - 441


6






21 Shamirs Scheme


S20

Page - 442


7



2 2( ) 166 94 1234 166 94 f x S x x x x= + + = + +


(11494) (21942) (3 2578) (43402) (5 4414) (65614)


21 10

0 1 0 1

20 21

1 0 1 2

20 12

2 0 2 1

1 3 106 2 3

1 7 5 2 21 823 3



minus minus




Therefore2

2

0( ) ( ) 1234 166 94j j

jf x y L x x x

=


22 Blakleys Scheme




1 2 3 4 5 6 2 3 4 5 6K x K x K x K x K x K x= = = = = =

S20

Page - 443


8


(00)






( )( )( )


xxx

===


( )( )( )

( )( )( )

( )( )( )


x x xx x xx x x

= = == = == = =


2 155(mod 3)S = =


Reference



2006

S20

Page - 444


9





S20

Page - 445


10





1 2 0 or 1 12 k k i i k= = =ε ε ε ε


if








2 3 1 3 1 21 2 3

1 2 1 3 2 1 2 3 3 1 3 2

( )( ) ( )( ) ( )( )( )( )( ) ( )( ) ( )( )






S20

Page - 446

Exam

funct

We w

Examtheorfigure

Henc

wher

Then

Sh

mple 313 F

tion 1( 2 )x + ε

would like to


ce 3Λ is 3 s

e xyC is a ci

n 3( )Λ Φ is a

hing-Tung Y


o give anothe


Figure 5 Th

secret sharea

rcle such tha

Figure 6 Th

a 2 bit 3 se

Yau High Sch

2isin we def

hen 33( ΛΛ Φ

er example o


hree non-collin

able Now w

ΦΦΦΦ

at its center i

he function Φ m

ecret shareab

hool Applied


) is a 2 bit

of secret shar

ircles in 2

near determi

ear points dete

e define a fu

( )( )( )( )

00

01

10

11

00011011

CCCC

===

Φ =

is (xy) and i

maps two bits

ble pair as we

d Mathemati

2 ) to be the g

3 secret sha

reable pair


ermine a uniqu

unction Φ fr

its radius is

ε1 ε2 to a circle

ell

ical Sciences

graph of the

areable pair


ue circle

rom 2 to Λ

1 3 (Ref F

e C ε1 ε2

s Award 201

1

quadratic


3Λ by

Figure 6)

13

11

y

S20

Page - 447


12




1 ( )j C Cπ isin




( ) ( ( ))j C j P jπ =

for 12 j n=


1 2 Ms s s s=



1 2i i i

kis = ε ε ε

and so

1 2 11 1 1 2 2

2 1 12

2 2 Mi i ik k




S20

Page - 448


13

[ ] ( ( ))ij j is sπ= Φ




( )( ) ( )( ) ( )( )( ) ( )( ) ( )( )( ) ( )( ) ( )

1

2

3

10 ( 2)( 1)

01 ( 1)( 3)

11 ( 2)( 3)

j j

j j

j j

s j j j

s j j j

s j j j

π π

π π

π π





( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

10 20 32 46 512[ ] 10 23 30 43 58

12 20 30 42 56s

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

1 2 3 4 5

10 20 32 46 512[ ] 10 [ ] 23 [ ] 30 [ ] 43 [ ] 58

12 20 30 42 56s s s s s

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦





[ ] [ ]1 212[ ] [ ] [ ][ ] i

jtts s ss p= =



S20

Page - 449


14

1 2( ) ( )i i ii ts p p pΦ = Ψ

and

11 2( )( )i i i


for 1 2 i M=


1 2

1 1 1 1 1 2 2 2 11 2 1 2 1 2( ( )) ( ( )) ( ( ))

M

n n nt t t

s s s s


=

=Φ Ψ Φ Ψ Φ Ψ






Mk=


[ ] ( ( ))ij j is sπ= Φ



Step 4 End





[ ] [ ] [ ]tt


S20

Page - 450


15


as in section 322


Step 5 END






[ ] [ ] ( )ij i js sσ

σ=







denoted by σ



S20

Page - 451


16



(0 8 6 7 2)(1 4 9)(3 5) (2 4 7 6 9)(0 8 5)(1 3) (0 3 7 9 2)(4 5 8)(1 6)

(1 6 9 4 5)(0 2 8)(3 7) (0 1 8 9 4)(2 5 6)(3 7) (0 2 8 4 9)(1 6 5)(3 7)

(0 8 9 5 4)(1 6 7)(2 3) (0 2 7 8 9)(1 4 3)(5 6) (0 9 5 4 7)(1 8 3)(2 6)

(0 2 9 5 3)(1 8 4)(6 7)

Example 334 Let

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

10 20 3 2 46 512[ ] 10 23 30 43 58

12 20 30 42 56s

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

32 10 20 512 46[ ] 23 30 10 43 58

12 20 30 56 42s σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

1 2 3 4 5

3 2 10 20 512 46[ ] 23 [ ] 30 [ ] 10 [ ] 43 [ ] 58

12 20 30 56 4 2s s s s sσ σ σ σ σ

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦





Mk=

S20

Page - 452


17


[ ] ( ( ))ij j is sπ= Φ




Step 5 End














S20

Page - 453


18





= many


Reference


S20

Page - 454


19

4 Algebraic Methods






1

1

( )

( )( )

t

iii j

j t

j iii j

x x

L xx x

=ne

=ne

minus

=minus

prod

prod


1( )

0j i

i jL x

i j=⎧

= ⎨ ne⎩


1 1 2 2 3 3( ) ( ) ( ) ( )L x y L x y L x y L x= + + +



1 1 2 2 3 3(0) (0) (0)y L y L y L+ + +

where

S20

Page - 455


20

1

1 1

1 1

(0 ) ( 1)

(0)( ) ( )

t tt

i ii ii j i j

j t t

j i j ii ii j i j

x x

Lx x x x

minus

= =ne ne

= =ne ne

minus minus

= =minus minus

prod prod

prod prod

for 1 23j =








1 2 1 21 2 1 2












11 21 2 2k




S20

Page - 456


21


2 3 11 2 2 2 2( ) k

kX p p p p p pminus

=

Assume that

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k



So

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k





Therefore 2 3 11 2 2 2 2 2( )

k k

X p p p p p ppminus




1 2 3 234579111316171923 X m m m= =

from now on


1 21 1 12 2 2 2 2 21 1 1 2 2 2

k k kr



1 2

1 1 2 21 1 1 r


1 2

1 1 2 21 1 1 r


such that

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr


1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr



1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr


+ + + + + + + + +ε ε εε ε ε ε ε ε

S20

Page - 457


22

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr




i ij j=ε ε







12 2i ik k

i k ixp pminus

lelt



12 331 2 4 221 2 2

i iii i i kki i


= εεε ε ε for







3 234X = 4 2345X = 5 23457X = 6 234579X =

7 23457911X = and 8 2345791113X = Also 1 2XΠ = 2 6XΠ =

3 24XΠ = and 4 120XΠ =

Note that





S20

Page - 458


23





1 21 2 1 11 2 1 1 1 2

1 1 1 2 1 11 2 1 2 1 2( )( ) ( )k kk k k k t t k





secret sharing pair











S20

Page - 459


24

( ) ( ( ))j C j P jπ =

for 12 j n=


11 1 2 2( ) ( ) ( )( ( ))






1 2( )(0)t kΛΦ ε ε ε


1 2 3 4ε ε ε ε ( )2 1 2 3 4( ) 0ΛΦ ε ε ε ε 1 2 3 4ε ε ε ε ( )

2 1 2 3 4( ) 0ΛΦ ε ε ε ε 0000 1 0110 12 1000 2 0101 15 0100 3 0011 20 0010 4 1110 24 0001 5 1101 30 1100 6 1011 40 1010 8 0111 60 1001 10 1111 120



( )( ) ( )( ) ( )( )3 2 1 3 1 2

1 2 31 3 1 2 2 1 2 3 1 33 2




S20

Page - 460


25


( ) ( )( )( )

( )( )

2 1 2 2 1 1

2 1 2 1 2 1

2 1 2 1 2 1

2 1

2 1 2 1 2 1

1 1x x x x

x x x xx x x x

x x x x

ε ε

ε εε ε

ε εε ε

+ minus

+ minus minus=

+ minus

minus=

+

minusminus minus


minusminus minus



( )2 1 2 2 1 1

1 1 22xx xx

εε ε λ λ ε

minusminus minus

le+ minus minus

Observe that

( )( )( )( )

( )( )

( )( )( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )

3 23 2

3 1 2 1 3 1 2 1

3 23 2

3 1 2 1 3 1 2 1

3 2 3 2

3 1 2

3 2

3 1 2 1

3 2

3 2 3 2

3 1

3 2 3 2

3 1

1 3 1 2 1

3 2 3 2

3 1 2 1 3 1 2 13 1 2 1

3 1 2

1





x x x

ε εε ε ε ε

ε ε

ε ε ε εε ε






+ ++ minus + minus

+ +

+ + + ++ minus

+ + + ++ minus

=minus


( )( )( )

( ) ( ) ( )( )( )( ) ( ) ( )

3 2 3 2

3 2

3 1

3 2

2 31

3 2

2 1 3 1

3 1

3 1

3 2

3 1 2 1 2 12 1

1

1 1

1

x xx

x xx x x x x x

x xx x x x x x

ε ε ε ε

ε εε ε

ε εε ε ε ε

minus



+ +

+ ++ minus

+ ++ minus minus

S20

Page - 461


26

( )( )( )( )

( )( )

( ) ( ) ( )( )

3 2

3 1

3 23 2

3 1 2 1 3 1 2 12 1

222 2

2

2 2

2

1 (2 )

1 2( )

1 2 4

2 22 2

2 22 2 2


M M M

M M M

ε εε ε ε ε

ε εε ε ε ελ

ε ε εελ

ω ω ε

λ λ ε λ λ ε


λ

+ ++ minus + minus

minus minus

leminus


le + + + + +

⎡ ⎤+ + +⎛ ⎞ ⎛ ⎞+ +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤le

minus

⎣

minus

+ ⎦

Therefore

where 2

M εε

ωλ

+=

minus So we have

( )( )( )( )

( )( ) ( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( ) ( )

( )( )( )

3 23 21 1 1

3 1 2 1 3 1 2 1

3 23 21 1

3 1 2 1 3 1 2 1

3 2

3 1 2 1

3 2

3 1 2

3 2 3 21 1 1

3 1 2 1 3 1 2 1

33 2

3 1 2 1

1

3 2 3 2

3 1 2 1 3 1 2 1

3



x x x xy y

x x x x x x x x

xx xx x x x

ε εε ε ε ε

ε εε ε ε ε


δε

δ

ε

+ ++ minus + minus

+ ++ minus + minus

+ +



+ minus+ +



le minusminus

minus

+minus

( )( )( )

( )( )( )( )

( )

21

3 1 2 1

3 21

3 1 2 1

2

3 1 2 1

3 2

3 1 2 1

2 21 2 4

xy

x x x x

x xx x x x

M

εε ε ε ε

ε εε ε ε ε

δ

ω ω ε ω δλ

++ minus + minus

+ ++ minus + minus

minus minus

+minus minus

le + +



( )2 21(0) (0) 6 12 3L L Mω ω ε ω δλ

minus le + +

where 2

M εε

ωλ

+=

minus

S20

Page - 462


27

















S20

Page - 463


28


461 Speed Tests







2 4 6 8 10


0237 0471 0708 0935 1176

0

1

2

3

4

5

6

7

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB


1183 2349 358 473 5917

S20

Page - 464


29


462 Size Tests




987 1973 2960 3947 4934


9845 19502 29195 38791 46789


100 99 99 98 95

0

02

04

06

08

1

12

14

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 465


30



Reference




0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

File

Siz

e in

KB

Secret Size in MB

CompressedSecret

CompressedContainer

S20

Page - 466


31

5 Geometric Methods






2 2

2 21 1 1 12 22 2 2 22 23 3 3 3

11

( ) 011

x y x yx y x y

J x yx y x yx y x y

++

= =++




or

2 2 1312 14

11 11 11


+ minus + minus =





2 2 1312 14

11 11 11


+ minus + minus =

we have

S20

Page - 467


32

12

11

13

11

2 2 14

11

2

2

C

C

C C C

MxM

MyM

Mr x yM

=

=

= + +


( ) 1312

11 11

2 2C C

MMx yM M

⎛ ⎞= ⎜ ⎟⎝ ⎠

and radius 2 2 14

11C C

Mx yM


shareable





21

1 1 2

1

1 11 2 1 2( ) ( 2 2 )2 2k kC C k k k k

kkx y +

+++= + + + + + +ε ε ε ε ε ε

and

1 2 1 21 21 2 2k

k k k kC kr + + + += + + + +ε ε ε


33( )




Cr We define


for 12 j n=

S20

Page - 468

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar


hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4


Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4


Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched


spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b


our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s


ommon

s Award 201

3

with center

sharing pair


13

33

S20

Page - 469


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488

Usuanodes

-The

-The

-The


Besiddetec

Sh


attack on co

attack on in

attack on re




hing-Tung Y


onfidentiality

tegrity modi

liability mak


Fi


Yau High Sch


y reveals sto

ifies data in

kes storage s



igure 4 The id

d to have morction

hool Applied


red server co

victim stora

server unava



dea of the propo

re features su

d Mathemati


ontents to at

age servers w

ailable to leg



osed system

uch as unaut

ical Sciences


ttackers

without being

gitimate user


thorized mo

s Award 201

n one storag5]

g noticed

rs


dification

13

4

ge

e

f be e

S20

Page - 440


5

Reference






S20

Page - 441


6






21 Shamirs Scheme


S20

Page - 442


7



2 2( ) 166 94 1234 166 94 f x S x x x x= + + = + +


(11494) (21942) (3 2578) (43402) (5 4414) (65614)


21 10

0 1 0 1

20 21

1 0 1 2

20 12

2 0 2 1

1 3 106 2 3

1 7 5 2 21 823 3



minus minus




Therefore2

2

0( ) ( ) 1234 166 94j j

jf x y L x x x

=


22 Blakleys Scheme




1 2 3 4 5 6 2 3 4 5 6K x K x K x K x K x K x= = = = = =

S20

Page - 443


8


(00)






( )( )( )


xxx

===


( )( )( )

( )( )( )

( )( )( )


x x xx x xx x x

= = == = == = =


2 155(mod 3)S = =


Reference



2006

S20

Page - 444


9





S20

Page - 445


10





1 2 0 or 1 12 k k i i k= = =ε ε ε ε


if








2 3 1 3 1 21 2 3

1 2 1 3 2 1 2 3 3 1 3 2

( )( ) ( )( ) ( )( )( )( )( ) ( )( ) ( )( )






S20

Page - 446

Exam

funct

We w

Examtheorfigure

Henc

wher

Then

Sh

mple 313 F

tion 1( 2 )x + ε

would like to


ce 3Λ is 3 s

e xyC is a ci

n 3( )Λ Φ is a

hing-Tung Y


o give anothe


Figure 5 Th

secret sharea

rcle such tha

Figure 6 Th

a 2 bit 3 se

Yau High Sch

2isin we def

hen 33( ΛΛ Φ

er example o


hree non-collin

able Now w

ΦΦΦΦ

at its center i

he function Φ m

ecret shareab

hool Applied


) is a 2 bit

of secret shar

ircles in 2

near determi

ear points dete

e define a fu

( )( )( )( )

00

01

10

11

00011011

CCCC

===

Φ =

is (xy) and i

maps two bits

ble pair as we

d Mathemati

2 ) to be the g

3 secret sha

reable pair


ermine a uniqu

unction Φ fr

its radius is

ε1 ε2 to a circle

ell

ical Sciences

graph of the

areable pair


ue circle

rom 2 to Λ

1 3 (Ref F

e C ε1 ε2

s Award 201

1

quadratic


3Λ by

Figure 6)

13

11

y

S20

Page - 447


12




1 ( )j C Cπ isin




( ) ( ( ))j C j P jπ =

for 12 j n=


1 2 Ms s s s=



1 2i i i

kis = ε ε ε

and so

1 2 11 1 1 2 2

2 1 12

2 2 Mi i ik k




S20

Page - 448


13

[ ] ( ( ))ij j is sπ= Φ




( )( ) ( )( ) ( )( )( ) ( )( ) ( )( )( ) ( )( ) ( )

1

2

3

10 ( 2)( 1)

01 ( 1)( 3)

11 ( 2)( 3)

j j

j j

j j

s j j j

s j j j

s j j j

π π

π π

π π





( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

10 20 32 46 512[ ] 10 23 30 43 58

12 20 30 42 56s

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

1 2 3 4 5

10 20 32 46 512[ ] 10 [ ] 23 [ ] 30 [ ] 43 [ ] 58

12 20 30 42 56s s s s s

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦





[ ] [ ]1 212[ ] [ ] [ ][ ] i

jtts s ss p= =



S20

Page - 449


14

1 2( ) ( )i i ii ts p p pΦ = Ψ

and

11 2( )( )i i i


for 1 2 i M=


1 2

1 1 1 1 1 2 2 2 11 2 1 2 1 2( ( )) ( ( )) ( ( ))

M

n n nt t t

s s s s


=

=Φ Ψ Φ Ψ Φ Ψ






Mk=


[ ] ( ( ))ij j is sπ= Φ



Step 4 End





[ ] [ ] [ ]tt


S20

Page - 450


15


as in section 322


Step 5 END






[ ] [ ] ( )ij i js sσ

σ=







denoted by σ



S20

Page - 451


16



(0 8 6 7 2)(1 4 9)(3 5) (2 4 7 6 9)(0 8 5)(1 3) (0 3 7 9 2)(4 5 8)(1 6)

(1 6 9 4 5)(0 2 8)(3 7) (0 1 8 9 4)(2 5 6)(3 7) (0 2 8 4 9)(1 6 5)(3 7)

(0 8 9 5 4)(1 6 7)(2 3) (0 2 7 8 9)(1 4 3)(5 6) (0 9 5 4 7)(1 8 3)(2 6)

(0 2 9 5 3)(1 8 4)(6 7)

Example 334 Let

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

10 20 3 2 46 512[ ] 10 23 30 43 58

12 20 30 42 56s

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

32 10 20 512 46[ ] 23 30 10 43 58

12 20 30 56 42s σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

1 2 3 4 5

3 2 10 20 512 46[ ] 23 [ ] 30 [ ] 10 [ ] 43 [ ] 58

12 20 30 56 4 2s s s s sσ σ σ σ σ

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦





Mk=

S20

Page - 452


17


[ ] ( ( ))ij j is sπ= Φ




Step 5 End














S20

Page - 453


18





= many


Reference


S20

Page - 454


19

4 Algebraic Methods






1

1

( )

( )( )

t

iii j

j t

j iii j

x x

L xx x

=ne

=ne

minus

=minus

prod

prod


1( )

0j i

i jL x

i j=⎧

= ⎨ ne⎩


1 1 2 2 3 3( ) ( ) ( ) ( )L x y L x y L x y L x= + + +



1 1 2 2 3 3(0) (0) (0)y L y L y L+ + +

where

S20

Page - 455


20

1

1 1

1 1

(0 ) ( 1)

(0)( ) ( )

t tt

i ii ii j i j

j t t

j i j ii ii j i j

x x

Lx x x x

minus

= =ne ne

= =ne ne

minus minus

= =minus minus

prod prod

prod prod

for 1 23j =








1 2 1 21 2 1 2












11 21 2 2k




S20

Page - 456


21


2 3 11 2 2 2 2( ) k

kX p p p p p pminus

=

Assume that

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k



So

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k





Therefore 2 3 11 2 2 2 2 2( )

k k

X p p p p p ppminus




1 2 3 234579111316171923 X m m m= =

from now on


1 21 1 12 2 2 2 2 21 1 1 2 2 2

k k kr



1 2

1 1 2 21 1 1 r


1 2

1 1 2 21 1 1 r


such that

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr


1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr



1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr


+ + + + + + + + +ε ε εε ε ε ε ε ε

S20

Page - 457


22

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr




i ij j=ε ε







12 2i ik k

i k ixp pminus

lelt



12 331 2 4 221 2 2

i iii i i kki i


= εεε ε ε for







3 234X = 4 2345X = 5 23457X = 6 234579X =

7 23457911X = and 8 2345791113X = Also 1 2XΠ = 2 6XΠ =

3 24XΠ = and 4 120XΠ =

Note that





S20

Page - 458


23





1 21 2 1 11 2 1 1 1 2

1 1 1 2 1 11 2 1 2 1 2( )( ) ( )k kk k k k t t k





secret sharing pair











S20

Page - 459


24

( ) ( ( ))j C j P jπ =

for 12 j n=


11 1 2 2( ) ( ) ( )( ( ))






1 2( )(0)t kΛΦ ε ε ε


1 2 3 4ε ε ε ε ( )2 1 2 3 4( ) 0ΛΦ ε ε ε ε 1 2 3 4ε ε ε ε ( )

2 1 2 3 4( ) 0ΛΦ ε ε ε ε 0000 1 0110 12 1000 2 0101 15 0100 3 0011 20 0010 4 1110 24 0001 5 1101 30 1100 6 1011 40 1010 8 0111 60 1001 10 1111 120



( )( ) ( )( ) ( )( )3 2 1 3 1 2

1 2 31 3 1 2 2 1 2 3 1 33 2




S20

Page - 460


25


( ) ( )( )( )

( )( )

2 1 2 2 1 1

2 1 2 1 2 1

2 1 2 1 2 1

2 1

2 1 2 1 2 1

1 1x x x x

x x x xx x x x

x x x x

ε ε

ε εε ε

ε εε ε

+ minus

+ minus minus=

+ minus

minus=

+

minusminus minus


minusminus minus



( )2 1 2 2 1 1

1 1 22xx xx

εε ε λ λ ε

minusminus minus

le+ minus minus

Observe that

( )( )( )( )

( )( )

( )( )( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )

3 23 2

3 1 2 1 3 1 2 1

3 23 2

3 1 2 1 3 1 2 1

3 2 3 2

3 1 2

3 2

3 1 2 1

3 2

3 2 3 2

3 1

3 2 3 2

3 1

1 3 1 2 1

3 2 3 2

3 1 2 1 3 1 2 13 1 2 1

3 1 2

1





x x x

ε εε ε ε ε

ε ε

ε ε ε εε ε






+ ++ minus + minus

+ +

+ + + ++ minus

+ + + ++ minus

=minus


( )( )( )

( ) ( ) ( )( )( )( ) ( ) ( )

3 2 3 2

3 2

3 1

3 2

2 31

3 2

2 1 3 1

3 1

3 1

3 2

3 1 2 1 2 12 1

1

1 1

1

x xx

x xx x x x x x

x xx x x x x x

ε ε ε ε

ε εε ε

ε εε ε ε ε

minus



+ +

+ ++ minus

+ ++ minus minus

S20

Page - 461


26

( )( )( )( )

( )( )

( ) ( ) ( )( )

3 2

3 1

3 23 2

3 1 2 1 3 1 2 12 1

222 2

2

2 2

2

1 (2 )

1 2( )

1 2 4

2 22 2

2 22 2 2


M M M

M M M

ε εε ε ε ε

ε εε ε ε ελ

ε ε εελ

ω ω ε

λ λ ε λ λ ε


λ

+ ++ minus + minus

minus minus

leminus


le + + + + +

⎡ ⎤+ + +⎛ ⎞ ⎛ ⎞+ +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤le

minus

⎣

minus

+ ⎦

Therefore

where 2

M εε

ωλ

+=

minus So we have

( )( )( )( )

( )( ) ( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( ) ( )

( )( )( )

3 23 21 1 1

3 1 2 1 3 1 2 1

3 23 21 1

3 1 2 1 3 1 2 1

3 2

3 1 2 1

3 2

3 1 2

3 2 3 21 1 1

3 1 2 1 3 1 2 1

33 2

3 1 2 1

1

3 2 3 2

3 1 2 1 3 1 2 1

3



x x x xy y

x x x x x x x x

xx xx x x x

ε εε ε ε ε

ε εε ε ε ε


δε

δ

ε

+ ++ minus + minus

+ ++ minus + minus

+ +



+ minus+ +



le minusminus

minus

+minus

( )( )( )

( )( )( )( )

( )

21

3 1 2 1

3 21

3 1 2 1

2

3 1 2 1

3 2

3 1 2 1

2 21 2 4

xy

x x x x

x xx x x x

M

εε ε ε ε

ε εε ε ε ε

δ

ω ω ε ω δλ

++ minus + minus

+ ++ minus + minus

minus minus

+minus minus

le + +



( )2 21(0) (0) 6 12 3L L Mω ω ε ω δλ

minus le + +

where 2

M εε

ωλ

+=

minus

S20

Page - 462


27

















S20

Page - 463


28


461 Speed Tests







2 4 6 8 10


0237 0471 0708 0935 1176

0

1

2

3

4

5

6

7

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB


1183 2349 358 473 5917

S20

Page - 464


29


462 Size Tests




987 1973 2960 3947 4934


9845 19502 29195 38791 46789


100 99 99 98 95

0

02

04

06

08

1

12

14

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 465


30



Reference




0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

File

Siz

e in

KB

Secret Size in MB

CompressedSecret

CompressedContainer

S20

Page - 466


31

5 Geometric Methods






2 2

2 21 1 1 12 22 2 2 22 23 3 3 3

11

( ) 011

x y x yx y x y

J x yx y x yx y x y

++

= =++




or

2 2 1312 14

11 11 11


+ minus + minus =





2 2 1312 14

11 11 11


+ minus + minus =

we have

S20

Page - 467


32

12

11

13

11

2 2 14

11

2

2

C

C

C C C

MxM

MyM

Mr x yM

=

=

= + +


( ) 1312

11 11

2 2C C

MMx yM M

⎛ ⎞= ⎜ ⎟⎝ ⎠

and radius 2 2 14

11C C

Mx yM


shareable





21

1 1 2

1

1 11 2 1 2( ) ( 2 2 )2 2k kC C k k k k

kkx y +

+++= + + + + + +ε ε ε ε ε ε

and

1 2 1 21 21 2 2k

k k k kC kr + + + += + + + +ε ε ε


33( )




Cr We define


for 12 j n=

S20

Page - 468

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar


hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4


Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4


Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched


spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b


our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s


ommon

s Award 201

3

with center

sharing pair


13

33

S20

Page - 469


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


5

Reference






S20

Page - 441


6






21 Shamirs Scheme


S20

Page - 442


7



2 2( ) 166 94 1234 166 94 f x S x x x x= + + = + +


(11494) (21942) (3 2578) (43402) (5 4414) (65614)


21 10

0 1 0 1

20 21

1 0 1 2

20 12

2 0 2 1

1 3 106 2 3

1 7 5 2 21 823 3



minus minus




Therefore2

2

0( ) ( ) 1234 166 94j j

jf x y L x x x

=


22 Blakleys Scheme




1 2 3 4 5 6 2 3 4 5 6K x K x K x K x K x K x= = = = = =

S20

Page - 443


8


(00)






( )( )( )


xxx

===


( )( )( )

( )( )( )

( )( )( )


x x xx x xx x x

= = == = == = =


2 155(mod 3)S = =


Reference



2006

S20

Page - 444


9





S20

Page - 445


10





1 2 0 or 1 12 k k i i k= = =ε ε ε ε


if








2 3 1 3 1 21 2 3

1 2 1 3 2 1 2 3 3 1 3 2

( )( ) ( )( ) ( )( )( )( )( ) ( )( ) ( )( )






S20

Page - 446

Exam

funct

We w

Examtheorfigure

Henc

wher

Then

Sh

mple 313 F

tion 1( 2 )x + ε

would like to


ce 3Λ is 3 s

e xyC is a ci

n 3( )Λ Φ is a

hing-Tung Y


o give anothe


Figure 5 Th

secret sharea

rcle such tha

Figure 6 Th

a 2 bit 3 se

Yau High Sch

2isin we def

hen 33( ΛΛ Φ

er example o


hree non-collin

able Now w

ΦΦΦΦ

at its center i

he function Φ m

ecret shareab

hool Applied


) is a 2 bit

of secret shar

ircles in 2

near determi

ear points dete

e define a fu

( )( )( )( )

00

01

10

11

00011011

CCCC

===

Φ =

is (xy) and i

maps two bits

ble pair as we

d Mathemati

2 ) to be the g

3 secret sha

reable pair


ermine a uniqu

unction Φ fr

its radius is

ε1 ε2 to a circle

ell

ical Sciences

graph of the

areable pair


ue circle

rom 2 to Λ

1 3 (Ref F

e C ε1 ε2

s Award 201

1

quadratic


3Λ by

Figure 6)

13

11

y

S20

Page - 447


12




1 ( )j C Cπ isin




( ) ( ( ))j C j P jπ =

for 12 j n=


1 2 Ms s s s=



1 2i i i

kis = ε ε ε

and so

1 2 11 1 1 2 2

2 1 12

2 2 Mi i ik k




S20

Page - 448


13

[ ] ( ( ))ij j is sπ= Φ




( )( ) ( )( ) ( )( )( ) ( )( ) ( )( )( ) ( )( ) ( )

1

2

3

10 ( 2)( 1)

01 ( 1)( 3)

11 ( 2)( 3)

j j

j j

j j

s j j j

s j j j

s j j j

π π

π π

π π





( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

10 20 32 46 512[ ] 10 23 30 43 58

12 20 30 42 56s

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

1 2 3 4 5

10 20 32 46 512[ ] 10 [ ] 23 [ ] 30 [ ] 43 [ ] 58

12 20 30 42 56s s s s s

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦





[ ] [ ]1 212[ ] [ ] [ ][ ] i

jtts s ss p= =



S20

Page - 449


14

1 2( ) ( )i i ii ts p p pΦ = Ψ

and

11 2( )( )i i i


for 1 2 i M=


1 2

1 1 1 1 1 2 2 2 11 2 1 2 1 2( ( )) ( ( )) ( ( ))

M

n n nt t t

s s s s


=

=Φ Ψ Φ Ψ Φ Ψ






Mk=


[ ] ( ( ))ij j is sπ= Φ



Step 4 End





[ ] [ ] [ ]tt


S20

Page - 450


15


as in section 322


Step 5 END






[ ] [ ] ( )ij i js sσ

σ=







denoted by σ



S20

Page - 451


16



(0 8 6 7 2)(1 4 9)(3 5) (2 4 7 6 9)(0 8 5)(1 3) (0 3 7 9 2)(4 5 8)(1 6)

(1 6 9 4 5)(0 2 8)(3 7) (0 1 8 9 4)(2 5 6)(3 7) (0 2 8 4 9)(1 6 5)(3 7)

(0 8 9 5 4)(1 6 7)(2 3) (0 2 7 8 9)(1 4 3)(5 6) (0 9 5 4 7)(1 8 3)(2 6)

(0 2 9 5 3)(1 8 4)(6 7)

Example 334 Let

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

10 20 3 2 46 512[ ] 10 23 30 43 58

12 20 30 42 56s

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

32 10 20 512 46[ ] 23 30 10 43 58

12 20 30 56 42s σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

1 2 3 4 5

3 2 10 20 512 46[ ] 23 [ ] 30 [ ] 10 [ ] 43 [ ] 58

12 20 30 56 4 2s s s s sσ σ σ σ σ

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦





Mk=

S20

Page - 452


17


[ ] ( ( ))ij j is sπ= Φ




Step 5 End














S20

Page - 453


18





= many


Reference


S20

Page - 454


19

4 Algebraic Methods






1

1

( )

( )( )

t

iii j

j t

j iii j

x x

L xx x

=ne

=ne

minus

=minus

prod

prod


1( )

0j i

i jL x

i j=⎧

= ⎨ ne⎩


1 1 2 2 3 3( ) ( ) ( ) ( )L x y L x y L x y L x= + + +



1 1 2 2 3 3(0) (0) (0)y L y L y L+ + +

where

S20

Page - 455


20

1

1 1

1 1

(0 ) ( 1)

(0)( ) ( )

t tt

i ii ii j i j

j t t

j i j ii ii j i j

x x

Lx x x x

minus

= =ne ne

= =ne ne

minus minus

= =minus minus

prod prod

prod prod

for 1 23j =








1 2 1 21 2 1 2












11 21 2 2k




S20

Page - 456


21


2 3 11 2 2 2 2( ) k

kX p p p p p pminus

=

Assume that

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k



So

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k





Therefore 2 3 11 2 2 2 2 2( )

k k

X p p p p p ppminus




1 2 3 234579111316171923 X m m m= =

from now on


1 21 1 12 2 2 2 2 21 1 1 2 2 2

k k kr



1 2

1 1 2 21 1 1 r


1 2

1 1 2 21 1 1 r


such that

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr


1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr



1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr


+ + + + + + + + +ε ε εε ε ε ε ε ε

S20

Page - 457


22

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr




i ij j=ε ε







12 2i ik k

i k ixp pminus

lelt



12 331 2 4 221 2 2

i iii i i kki i


= εεε ε ε for







3 234X = 4 2345X = 5 23457X = 6 234579X =

7 23457911X = and 8 2345791113X = Also 1 2XΠ = 2 6XΠ =

3 24XΠ = and 4 120XΠ =

Note that





S20

Page - 458


23





1 21 2 1 11 2 1 1 1 2

1 1 1 2 1 11 2 1 2 1 2( )( ) ( )k kk k k k t t k





secret sharing pair











S20

Page - 459


24

( ) ( ( ))j C j P jπ =

for 12 j n=


11 1 2 2( ) ( ) ( )( ( ))






1 2( )(0)t kΛΦ ε ε ε


1 2 3 4ε ε ε ε ( )2 1 2 3 4( ) 0ΛΦ ε ε ε ε 1 2 3 4ε ε ε ε ( )

2 1 2 3 4( ) 0ΛΦ ε ε ε ε 0000 1 0110 12 1000 2 0101 15 0100 3 0011 20 0010 4 1110 24 0001 5 1101 30 1100 6 1011 40 1010 8 0111 60 1001 10 1111 120



( )( ) ( )( ) ( )( )3 2 1 3 1 2

1 2 31 3 1 2 2 1 2 3 1 33 2




S20

Page - 460


25


( ) ( )( )( )

( )( )

2 1 2 2 1 1

2 1 2 1 2 1

2 1 2 1 2 1

2 1

2 1 2 1 2 1

1 1x x x x

x x x xx x x x

x x x x

ε ε

ε εε ε

ε εε ε

+ minus

+ minus minus=

+ minus

minus=

+

minusminus minus


minusminus minus



( )2 1 2 2 1 1

1 1 22xx xx

εε ε λ λ ε

minusminus minus

le+ minus minus

Observe that

( )( )( )( )

( )( )

( )( )( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )

3 23 2

3 1 2 1 3 1 2 1

3 23 2

3 1 2 1 3 1 2 1

3 2 3 2

3 1 2

3 2

3 1 2 1

3 2

3 2 3 2

3 1

3 2 3 2

3 1

1 3 1 2 1

3 2 3 2

3 1 2 1 3 1 2 13 1 2 1

3 1 2

1





x x x

ε εε ε ε ε

ε ε

ε ε ε εε ε






+ ++ minus + minus

+ +

+ + + ++ minus

+ + + ++ minus

=minus


( )( )( )

( ) ( ) ( )( )( )( ) ( ) ( )

3 2 3 2

3 2

3 1

3 2

2 31

3 2

2 1 3 1

3 1

3 1

3 2

3 1 2 1 2 12 1

1

1 1

1

x xx

x xx x x x x x

x xx x x x x x

ε ε ε ε

ε εε ε

ε εε ε ε ε

minus



+ +

+ ++ minus

+ ++ minus minus

S20

Page - 461


26

( )( )( )( )

( )( )

( ) ( ) ( )( )

3 2

3 1

3 23 2

3 1 2 1 3 1 2 12 1

222 2

2

2 2

2

1 (2 )

1 2( )

1 2 4

2 22 2

2 22 2 2


M M M

M M M

ε εε ε ε ε

ε εε ε ε ελ

ε ε εελ

ω ω ε

λ λ ε λ λ ε


λ

+ ++ minus + minus

minus minus

leminus


le + + + + +

⎡ ⎤+ + +⎛ ⎞ ⎛ ⎞+ +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤le

minus

⎣

minus

+ ⎦

Therefore

where 2

M εε

ωλ

+=

minus So we have

( )( )( )( )

( )( ) ( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( ) ( )

( )( )( )

3 23 21 1 1

3 1 2 1 3 1 2 1

3 23 21 1

3 1 2 1 3 1 2 1

3 2

3 1 2 1

3 2

3 1 2

3 2 3 21 1 1

3 1 2 1 3 1 2 1

33 2

3 1 2 1

1

3 2 3 2

3 1 2 1 3 1 2 1

3



x x x xy y

x x x x x x x x

xx xx x x x

ε εε ε ε ε

ε εε ε ε ε


δε

δ

ε

+ ++ minus + minus

+ ++ minus + minus

+ +



+ minus+ +



le minusminus

minus

+minus

( )( )( )

( )( )( )( )

( )

21

3 1 2 1

3 21

3 1 2 1

2

3 1 2 1

3 2

3 1 2 1

2 21 2 4

xy

x x x x

x xx x x x

M

εε ε ε ε

ε εε ε ε ε

δ

ω ω ε ω δλ

++ minus + minus

+ ++ minus + minus

minus minus

+minus minus

le + +



( )2 21(0) (0) 6 12 3L L Mω ω ε ω δλ

minus le + +

where 2

M εε

ωλ

+=

minus

S20

Page - 462


27

















S20

Page - 463


28


461 Speed Tests







2 4 6 8 10


0237 0471 0708 0935 1176

0

1

2

3

4

5

6

7

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB


1183 2349 358 473 5917

S20

Page - 464


29


462 Size Tests




987 1973 2960 3947 4934


9845 19502 29195 38791 46789


100 99 99 98 95

0

02

04

06

08

1

12

14

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 465


30



Reference




0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

File

Siz

e in

KB

Secret Size in MB

CompressedSecret

CompressedContainer

S20

Page - 466


31

5 Geometric Methods






2 2

2 21 1 1 12 22 2 2 22 23 3 3 3

11

( ) 011

x y x yx y x y

J x yx y x yx y x y

++

= =++




or

2 2 1312 14

11 11 11


+ minus + minus =





2 2 1312 14

11 11 11


+ minus + minus =

we have

S20

Page - 467


32

12

11

13

11

2 2 14

11

2

2

C

C

C C C

MxM

MyM

Mr x yM

=

=

= + +


( ) 1312

11 11

2 2C C

MMx yM M

⎛ ⎞= ⎜ ⎟⎝ ⎠

and radius 2 2 14

11C C

Mx yM


shareable





21

1 1 2

1

1 11 2 1 2( ) ( 2 2 )2 2k kC C k k k k

kkx y +

+++= + + + + + +ε ε ε ε ε ε

and

1 2 1 21 21 2 2k

k k k kC kr + + + += + + + +ε ε ε


33( )




Cr We define


for 12 j n=

S20

Page - 468

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar


hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4


Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4


Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched


spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b


our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s


ommon

s Award 201

3

with center

sharing pair


13

33

S20

Page - 469


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


6






21 Shamirs Scheme


S20

Page - 442


7



2 2( ) 166 94 1234 166 94 f x S x x x x= + + = + +


(11494) (21942) (3 2578) (43402) (5 4414) (65614)


21 10

0 1 0 1

20 21

1 0 1 2

20 12

2 0 2 1

1 3 106 2 3

1 7 5 2 21 823 3



minus minus




Therefore2

2

0( ) ( ) 1234 166 94j j

jf x y L x x x

=


22 Blakleys Scheme




1 2 3 4 5 6 2 3 4 5 6K x K x K x K x K x K x= = = = = =

S20

Page - 443


8


(00)






( )( )( )


xxx

===


( )( )( )

( )( )( )

( )( )( )


x x xx x xx x x

= = == = == = =


2 155(mod 3)S = =


Reference



2006

S20

Page - 444


9





S20

Page - 445


10





1 2 0 or 1 12 k k i i k= = =ε ε ε ε


if








2 3 1 3 1 21 2 3

1 2 1 3 2 1 2 3 3 1 3 2

( )( ) ( )( ) ( )( )( )( )( ) ( )( ) ( )( )






S20

Page - 446

Exam

funct

We w

Examtheorfigure

Henc

wher

Then

Sh

mple 313 F

tion 1( 2 )x + ε

would like to


ce 3Λ is 3 s

e xyC is a ci

n 3( )Λ Φ is a

hing-Tung Y


o give anothe


Figure 5 Th

secret sharea

rcle such tha

Figure 6 Th

a 2 bit 3 se

Yau High Sch

2isin we def

hen 33( ΛΛ Φ

er example o


hree non-collin

able Now w

ΦΦΦΦ

at its center i

he function Φ m

ecret shareab

hool Applied


) is a 2 bit

of secret shar

ircles in 2

near determi

ear points dete

e define a fu

( )( )( )( )

00

01

10

11

00011011

CCCC

===

Φ =

is (xy) and i

maps two bits

ble pair as we

d Mathemati

2 ) to be the g

3 secret sha

reable pair


ermine a uniqu

unction Φ fr

its radius is

ε1 ε2 to a circle

ell

ical Sciences

graph of the

areable pair


ue circle

rom 2 to Λ

1 3 (Ref F

e C ε1 ε2

s Award 201

1

quadratic


3Λ by

Figure 6)

13

11

y

S20

Page - 447


12




1 ( )j C Cπ isin




( ) ( ( ))j C j P jπ =

for 12 j n=


1 2 Ms s s s=



1 2i i i

kis = ε ε ε

and so

1 2 11 1 1 2 2

2 1 12

2 2 Mi i ik k




S20

Page - 448


13

[ ] ( ( ))ij j is sπ= Φ




( )( ) ( )( ) ( )( )( ) ( )( ) ( )( )( ) ( )( ) ( )

1

2

3

10 ( 2)( 1)

01 ( 1)( 3)

11 ( 2)( 3)

j j

j j

j j

s j j j

s j j j

s j j j

π π

π π

π π





( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

10 20 32 46 512[ ] 10 23 30 43 58

12 20 30 42 56s

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

1 2 3 4 5

10 20 32 46 512[ ] 10 [ ] 23 [ ] 30 [ ] 43 [ ] 58

12 20 30 42 56s s s s s

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦





[ ] [ ]1 212[ ] [ ] [ ][ ] i

jtts s ss p= =



S20

Page - 449


14

1 2( ) ( )i i ii ts p p pΦ = Ψ

and

11 2( )( )i i i


for 1 2 i M=


1 2

1 1 1 1 1 2 2 2 11 2 1 2 1 2( ( )) ( ( )) ( ( ))

M

n n nt t t

s s s s


=

=Φ Ψ Φ Ψ Φ Ψ






Mk=


[ ] ( ( ))ij j is sπ= Φ



Step 4 End





[ ] [ ] [ ]tt


S20

Page - 450


15


as in section 322


Step 5 END






[ ] [ ] ( )ij i js sσ

σ=







denoted by σ



S20

Page - 451


16



(0 8 6 7 2)(1 4 9)(3 5) (2 4 7 6 9)(0 8 5)(1 3) (0 3 7 9 2)(4 5 8)(1 6)

(1 6 9 4 5)(0 2 8)(3 7) (0 1 8 9 4)(2 5 6)(3 7) (0 2 8 4 9)(1 6 5)(3 7)

(0 8 9 5 4)(1 6 7)(2 3) (0 2 7 8 9)(1 4 3)(5 6) (0 9 5 4 7)(1 8 3)(2 6)

(0 2 9 5 3)(1 8 4)(6 7)

Example 334 Let

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

10 20 3 2 46 512[ ] 10 23 30 43 58

12 20 30 42 56s

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

32 10 20 512 46[ ] 23 30 10 43 58

12 20 30 56 42s σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

1 2 3 4 5

3 2 10 20 512 46[ ] 23 [ ] 30 [ ] 10 [ ] 43 [ ] 58

12 20 30 56 4 2s s s s sσ σ σ σ σ

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦





Mk=

S20

Page - 452


17


[ ] ( ( ))ij j is sπ= Φ




Step 5 End














S20

Page - 453


18





= many


Reference


S20

Page - 454


19

4 Algebraic Methods






1

1

( )

( )( )

t

iii j

j t

j iii j

x x

L xx x

=ne

=ne

minus

=minus

prod

prod


1( )

0j i

i jL x

i j=⎧

= ⎨ ne⎩


1 1 2 2 3 3( ) ( ) ( ) ( )L x y L x y L x y L x= + + +



1 1 2 2 3 3(0) (0) (0)y L y L y L+ + +

where

S20

Page - 455


20

1

1 1

1 1

(0 ) ( 1)

(0)( ) ( )

t tt

i ii ii j i j

j t t

j i j ii ii j i j

x x

Lx x x x

minus

= =ne ne

= =ne ne

minus minus

= =minus minus

prod prod

prod prod

for 1 23j =








1 2 1 21 2 1 2












11 21 2 2k




S20

Page - 456


21


2 3 11 2 2 2 2( ) k

kX p p p p p pminus

=

Assume that

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k



So

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k





Therefore 2 3 11 2 2 2 2 2( )

k k

X p p p p p ppminus




1 2 3 234579111316171923 X m m m= =

from now on


1 21 1 12 2 2 2 2 21 1 1 2 2 2

k k kr



1 2

1 1 2 21 1 1 r


1 2

1 1 2 21 1 1 r


such that

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr


1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr



1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr


+ + + + + + + + +ε ε εε ε ε ε ε ε

S20

Page - 457


22

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr




i ij j=ε ε







12 2i ik k

i k ixp pminus

lelt



12 331 2 4 221 2 2

i iii i i kki i


= εεε ε ε for







3 234X = 4 2345X = 5 23457X = 6 234579X =

7 23457911X = and 8 2345791113X = Also 1 2XΠ = 2 6XΠ =

3 24XΠ = and 4 120XΠ =

Note that





S20

Page - 458


23





1 21 2 1 11 2 1 1 1 2

1 1 1 2 1 11 2 1 2 1 2( )( ) ( )k kk k k k t t k





secret sharing pair











S20

Page - 459


24

( ) ( ( ))j C j P jπ =

for 12 j n=


11 1 2 2( ) ( ) ( )( ( ))






1 2( )(0)t kΛΦ ε ε ε


1 2 3 4ε ε ε ε ( )2 1 2 3 4( ) 0ΛΦ ε ε ε ε 1 2 3 4ε ε ε ε ( )

2 1 2 3 4( ) 0ΛΦ ε ε ε ε 0000 1 0110 12 1000 2 0101 15 0100 3 0011 20 0010 4 1110 24 0001 5 1101 30 1100 6 1011 40 1010 8 0111 60 1001 10 1111 120



( )( ) ( )( ) ( )( )3 2 1 3 1 2

1 2 31 3 1 2 2 1 2 3 1 33 2




S20

Page - 460


25


( ) ( )( )( )

( )( )

2 1 2 2 1 1

2 1 2 1 2 1

2 1 2 1 2 1

2 1

2 1 2 1 2 1

1 1x x x x

x x x xx x x x

x x x x

ε ε

ε εε ε

ε εε ε

+ minus

+ minus minus=

+ minus

minus=

+

minusminus minus


minusminus minus



( )2 1 2 2 1 1

1 1 22xx xx

εε ε λ λ ε

minusminus minus

le+ minus minus

Observe that

( )( )( )( )

( )( )

( )( )( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )

3 23 2

3 1 2 1 3 1 2 1

3 23 2

3 1 2 1 3 1 2 1

3 2 3 2

3 1 2

3 2

3 1 2 1

3 2

3 2 3 2

3 1

3 2 3 2

3 1

1 3 1 2 1

3 2 3 2

3 1 2 1 3 1 2 13 1 2 1

3 1 2

1





x x x

ε εε ε ε ε

ε ε

ε ε ε εε ε






+ ++ minus + minus

+ +

+ + + ++ minus

+ + + ++ minus

=minus


( )( )( )

( ) ( ) ( )( )( )( ) ( ) ( )

3 2 3 2

3 2

3 1

3 2

2 31

3 2

2 1 3 1

3 1

3 1

3 2

3 1 2 1 2 12 1

1

1 1

1

x xx

x xx x x x x x

x xx x x x x x

ε ε ε ε

ε εε ε

ε εε ε ε ε

minus



+ +

+ ++ minus

+ ++ minus minus

S20

Page - 461


26

( )( )( )( )

( )( )

( ) ( ) ( )( )

3 2

3 1

3 23 2

3 1 2 1 3 1 2 12 1

222 2

2

2 2

2

1 (2 )

1 2( )

1 2 4

2 22 2

2 22 2 2


M M M

M M M

ε εε ε ε ε

ε εε ε ε ελ

ε ε εελ

ω ω ε

λ λ ε λ λ ε


λ

+ ++ minus + minus

minus minus

leminus


le + + + + +

⎡ ⎤+ + +⎛ ⎞ ⎛ ⎞+ +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤le

minus

⎣

minus

+ ⎦

Therefore

where 2

M εε

ωλ

+=

minus So we have

( )( )( )( )

( )( ) ( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( ) ( )

( )( )( )

3 23 21 1 1

3 1 2 1 3 1 2 1

3 23 21 1

3 1 2 1 3 1 2 1

3 2

3 1 2 1

3 2

3 1 2

3 2 3 21 1 1

3 1 2 1 3 1 2 1

33 2

3 1 2 1

1

3 2 3 2

3 1 2 1 3 1 2 1

3



x x x xy y

x x x x x x x x

xx xx x x x

ε εε ε ε ε

ε εε ε ε ε


δε

δ

ε

+ ++ minus + minus

+ ++ minus + minus

+ +



+ minus+ +



le minusminus

minus

+minus

( )( )( )

( )( )( )( )

( )

21

3 1 2 1

3 21

3 1 2 1

2

3 1 2 1

3 2

3 1 2 1

2 21 2 4

xy

x x x x

x xx x x x

M

εε ε ε ε

ε εε ε ε ε

δ

ω ω ε ω δλ

++ minus + minus

+ ++ minus + minus

minus minus

+minus minus

le + +



( )2 21(0) (0) 6 12 3L L Mω ω ε ω δλ

minus le + +

where 2

M εε

ωλ

+=

minus

S20

Page - 462


27

















S20

Page - 463


28


461 Speed Tests







2 4 6 8 10


0237 0471 0708 0935 1176

0

1

2

3

4

5

6

7

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB


1183 2349 358 473 5917

S20

Page - 464


29


462 Size Tests




987 1973 2960 3947 4934


9845 19502 29195 38791 46789


100 99 99 98 95

0

02

04

06

08

1

12

14

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 465


30



Reference




0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

File

Siz

e in

KB

Secret Size in MB

CompressedSecret

CompressedContainer

S20

Page - 466


31

5 Geometric Methods






2 2

2 21 1 1 12 22 2 2 22 23 3 3 3

11

( ) 011

x y x yx y x y

J x yx y x yx y x y

++

= =++




or

2 2 1312 14

11 11 11


+ minus + minus =





2 2 1312 14

11 11 11


+ minus + minus =

we have

S20

Page - 467


32

12

11

13

11

2 2 14

11

2

2

C

C

C C C

MxM

MyM

Mr x yM

=

=

= + +


( ) 1312

11 11

2 2C C

MMx yM M

⎛ ⎞= ⎜ ⎟⎝ ⎠

and radius 2 2 14

11C C

Mx yM


shareable





21

1 1 2

1

1 11 2 1 2( ) ( 2 2 )2 2k kC C k k k k

kkx y +

+++= + + + + + +ε ε ε ε ε ε

and

1 2 1 21 21 2 2k

k k k kC kr + + + += + + + +ε ε ε


33( )




Cr We define


for 12 j n=

S20

Page - 468

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar


hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4


Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4


Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched


spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b


our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s


ommon

s Award 201

3

with center

sharing pair


13

33

S20

Page - 469


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


7



2 2( ) 166 94 1234 166 94 f x S x x x x= + + = + +


(11494) (21942) (3 2578) (43402) (5 4414) (65614)


21 10

0 1 0 1

20 21

1 0 1 2

20 12

2 0 2 1

1 3 106 2 3

1 7 5 2 21 823 3



minus minus




Therefore2

2

0( ) ( ) 1234 166 94j j

jf x y L x x x

=


22 Blakleys Scheme




1 2 3 4 5 6 2 3 4 5 6K x K x K x K x K x K x= = = = = =

S20

Page - 443


8


(00)






( )( )( )


xxx

===


( )( )( )

( )( )( )

( )( )( )


x x xx x xx x x

= = == = == = =


2 155(mod 3)S = =


Reference



2006

S20

Page - 444


9





S20

Page - 445


10





1 2 0 or 1 12 k k i i k= = =ε ε ε ε


if








2 3 1 3 1 21 2 3

1 2 1 3 2 1 2 3 3 1 3 2

( )( ) ( )( ) ( )( )( )( )( ) ( )( ) ( )( )






S20

Page - 446

Exam

funct

We w

Examtheorfigure

Henc

wher

Then

Sh

mple 313 F

tion 1( 2 )x + ε

would like to


ce 3Λ is 3 s

e xyC is a ci

n 3( )Λ Φ is a

hing-Tung Y


o give anothe


Figure 5 Th

secret sharea

rcle such tha

Figure 6 Th

a 2 bit 3 se

Yau High Sch

2isin we def

hen 33( ΛΛ Φ

er example o


hree non-collin

able Now w

ΦΦΦΦ

at its center i

he function Φ m

ecret shareab

hool Applied


) is a 2 bit

of secret shar

ircles in 2

near determi

ear points dete

e define a fu

( )( )( )( )

00

01

10

11

00011011

CCCC

===

Φ =

is (xy) and i

maps two bits

ble pair as we

d Mathemati

2 ) to be the g

3 secret sha

reable pair


ermine a uniqu

unction Φ fr

its radius is

ε1 ε2 to a circle

ell

ical Sciences

graph of the

areable pair


ue circle

rom 2 to Λ

1 3 (Ref F

e C ε1 ε2

s Award 201

1

quadratic


3Λ by

Figure 6)

13

11

y

S20

Page - 447


12




1 ( )j C Cπ isin




( ) ( ( ))j C j P jπ =

for 12 j n=


1 2 Ms s s s=



1 2i i i

kis = ε ε ε

and so

1 2 11 1 1 2 2

2 1 12

2 2 Mi i ik k




S20

Page - 448


13

[ ] ( ( ))ij j is sπ= Φ




( )( ) ( )( ) ( )( )( ) ( )( ) ( )( )( ) ( )( ) ( )

1

2

3

10 ( 2)( 1)

01 ( 1)( 3)

11 ( 2)( 3)

j j

j j

j j

s j j j

s j j j

s j j j

π π

π π

π π





( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

10 20 32 46 512[ ] 10 23 30 43 58

12 20 30 42 56s

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

1 2 3 4 5

10 20 32 46 512[ ] 10 [ ] 23 [ ] 30 [ ] 43 [ ] 58

12 20 30 42 56s s s s s

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦





[ ] [ ]1 212[ ] [ ] [ ][ ] i

jtts s ss p= =



S20

Page - 449


14

1 2( ) ( )i i ii ts p p pΦ = Ψ

and

11 2( )( )i i i


for 1 2 i M=


1 2

1 1 1 1 1 2 2 2 11 2 1 2 1 2( ( )) ( ( )) ( ( ))

M

n n nt t t

s s s s


=

=Φ Ψ Φ Ψ Φ Ψ






Mk=


[ ] ( ( ))ij j is sπ= Φ



Step 4 End





[ ] [ ] [ ]tt


S20

Page - 450


15


as in section 322


Step 5 END






[ ] [ ] ( )ij i js sσ

σ=







denoted by σ



S20

Page - 451


16



(0 8 6 7 2)(1 4 9)(3 5) (2 4 7 6 9)(0 8 5)(1 3) (0 3 7 9 2)(4 5 8)(1 6)

(1 6 9 4 5)(0 2 8)(3 7) (0 1 8 9 4)(2 5 6)(3 7) (0 2 8 4 9)(1 6 5)(3 7)

(0 8 9 5 4)(1 6 7)(2 3) (0 2 7 8 9)(1 4 3)(5 6) (0 9 5 4 7)(1 8 3)(2 6)

(0 2 9 5 3)(1 8 4)(6 7)

Example 334 Let

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

10 20 3 2 46 512[ ] 10 23 30 43 58

12 20 30 42 56s

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

32 10 20 512 46[ ] 23 30 10 43 58

12 20 30 56 42s σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

1 2 3 4 5

3 2 10 20 512 46[ ] 23 [ ] 30 [ ] 10 [ ] 43 [ ] 58

12 20 30 56 4 2s s s s sσ σ σ σ σ

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦





Mk=

S20

Page - 452


17


[ ] ( ( ))ij j is sπ= Φ




Step 5 End














S20

Page - 453


18





= many


Reference


S20

Page - 454


19

4 Algebraic Methods






1

1

( )

( )( )

t

iii j

j t

j iii j

x x

L xx x

=ne

=ne

minus

=minus

prod

prod


1( )

0j i

i jL x

i j=⎧

= ⎨ ne⎩


1 1 2 2 3 3( ) ( ) ( ) ( )L x y L x y L x y L x= + + +



1 1 2 2 3 3(0) (0) (0)y L y L y L+ + +

where

S20

Page - 455


20

1

1 1

1 1

(0 ) ( 1)

(0)( ) ( )

t tt

i ii ii j i j

j t t

j i j ii ii j i j

x x

Lx x x x

minus

= =ne ne

= =ne ne

minus minus

= =minus minus

prod prod

prod prod

for 1 23j =








1 2 1 21 2 1 2












11 21 2 2k




S20

Page - 456


21


2 3 11 2 2 2 2( ) k

kX p p p p p pminus

=

Assume that

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k



So

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k





Therefore 2 3 11 2 2 2 2 2( )

k k

X p p p p p ppminus




1 2 3 234579111316171923 X m m m= =

from now on


1 21 1 12 2 2 2 2 21 1 1 2 2 2

k k kr



1 2

1 1 2 21 1 1 r


1 2

1 1 2 21 1 1 r


such that

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr


1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr



1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr


+ + + + + + + + +ε ε εε ε ε ε ε ε

S20

Page - 457


22

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr




i ij j=ε ε







12 2i ik k

i k ixp pminus

lelt



12 331 2 4 221 2 2

i iii i i kki i


= εεε ε ε for







3 234X = 4 2345X = 5 23457X = 6 234579X =

7 23457911X = and 8 2345791113X = Also 1 2XΠ = 2 6XΠ =

3 24XΠ = and 4 120XΠ =

Note that





S20

Page - 458


23





1 21 2 1 11 2 1 1 1 2

1 1 1 2 1 11 2 1 2 1 2( )( ) ( )k kk k k k t t k





secret sharing pair











S20

Page - 459


24

( ) ( ( ))j C j P jπ =

for 12 j n=


11 1 2 2( ) ( ) ( )( ( ))






1 2( )(0)t kΛΦ ε ε ε


1 2 3 4ε ε ε ε ( )2 1 2 3 4( ) 0ΛΦ ε ε ε ε 1 2 3 4ε ε ε ε ( )

2 1 2 3 4( ) 0ΛΦ ε ε ε ε 0000 1 0110 12 1000 2 0101 15 0100 3 0011 20 0010 4 1110 24 0001 5 1101 30 1100 6 1011 40 1010 8 0111 60 1001 10 1111 120



( )( ) ( )( ) ( )( )3 2 1 3 1 2

1 2 31 3 1 2 2 1 2 3 1 33 2




S20

Page - 460


25


( ) ( )( )( )

( )( )

2 1 2 2 1 1

2 1 2 1 2 1

2 1 2 1 2 1

2 1

2 1 2 1 2 1

1 1x x x x

x x x xx x x x

x x x x

ε ε

ε εε ε

ε εε ε

+ minus

+ minus minus=

+ minus

minus=

+

minusminus minus


minusminus minus



( )2 1 2 2 1 1

1 1 22xx xx

εε ε λ λ ε

minusminus minus

le+ minus minus

Observe that

( )( )( )( )

( )( )

( )( )( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )

3 23 2

3 1 2 1 3 1 2 1

3 23 2

3 1 2 1 3 1 2 1

3 2 3 2

3 1 2

3 2

3 1 2 1

3 2

3 2 3 2

3 1

3 2 3 2

3 1

1 3 1 2 1

3 2 3 2

3 1 2 1 3 1 2 13 1 2 1

3 1 2

1





x x x

ε εε ε ε ε

ε ε

ε ε ε εε ε






+ ++ minus + minus

+ +

+ + + ++ minus

+ + + ++ minus

=minus


( )( )( )

( ) ( ) ( )( )( )( ) ( ) ( )

3 2 3 2

3 2

3 1

3 2

2 31

3 2

2 1 3 1

3 1

3 1

3 2

3 1 2 1 2 12 1

1

1 1

1

x xx

x xx x x x x x

x xx x x x x x

ε ε ε ε

ε εε ε

ε εε ε ε ε

minus



+ +

+ ++ minus

+ ++ minus minus

S20

Page - 461


26

( )( )( )( )

( )( )

( ) ( ) ( )( )

3 2

3 1

3 23 2

3 1 2 1 3 1 2 12 1

222 2

2

2 2

2

1 (2 )

1 2( )

1 2 4

2 22 2

2 22 2 2


M M M

M M M

ε εε ε ε ε

ε εε ε ε ελ

ε ε εελ

ω ω ε

λ λ ε λ λ ε


λ

+ ++ minus + minus

minus minus

leminus


le + + + + +

⎡ ⎤+ + +⎛ ⎞ ⎛ ⎞+ +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤le

minus

⎣

minus

+ ⎦

Therefore

where 2

M εε

ωλ

+=

minus So we have

( )( )( )( )

( )( ) ( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( ) ( )

( )( )( )

3 23 21 1 1

3 1 2 1 3 1 2 1

3 23 21 1

3 1 2 1 3 1 2 1

3 2

3 1 2 1

3 2

3 1 2

3 2 3 21 1 1

3 1 2 1 3 1 2 1

33 2

3 1 2 1

1

3 2 3 2

3 1 2 1 3 1 2 1

3



x x x xy y

x x x x x x x x

xx xx x x x

ε εε ε ε ε

ε εε ε ε ε


δε

δ

ε

+ ++ minus + minus

+ ++ minus + minus

+ +



+ minus+ +



le minusminus

minus

+minus

( )( )( )

( )( )( )( )

( )

21

3 1 2 1

3 21

3 1 2 1

2

3 1 2 1

3 2

3 1 2 1

2 21 2 4

xy

x x x x

x xx x x x

M

εε ε ε ε

ε εε ε ε ε

δ

ω ω ε ω δλ

++ minus + minus

+ ++ minus + minus

minus minus

+minus minus

le + +



( )2 21(0) (0) 6 12 3L L Mω ω ε ω δλ

minus le + +

where 2

M εε

ωλ

+=

minus

S20

Page - 462


27

















S20

Page - 463


28


461 Speed Tests







2 4 6 8 10


0237 0471 0708 0935 1176

0

1

2

3

4

5

6

7

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB


1183 2349 358 473 5917

S20

Page - 464


29


462 Size Tests




987 1973 2960 3947 4934


9845 19502 29195 38791 46789


100 99 99 98 95

0

02

04

06

08

1

12

14

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 465


30



Reference




0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

File

Siz

e in

KB

Secret Size in MB

CompressedSecret

CompressedContainer

S20

Page - 466


31

5 Geometric Methods






2 2

2 21 1 1 12 22 2 2 22 23 3 3 3

11

( ) 011

x y x yx y x y

J x yx y x yx y x y

++

= =++




or

2 2 1312 14

11 11 11


+ minus + minus =





2 2 1312 14

11 11 11


+ minus + minus =

we have

S20

Page - 467


32

12

11

13

11

2 2 14

11

2

2

C

C

C C C

MxM

MyM

Mr x yM

=

=

= + +


( ) 1312

11 11

2 2C C

MMx yM M

⎛ ⎞= ⎜ ⎟⎝ ⎠

and radius 2 2 14

11C C

Mx yM


shareable





21

1 1 2

1

1 11 2 1 2( ) ( 2 2 )2 2k kC C k k k k

kkx y +

+++= + + + + + +ε ε ε ε ε ε

and

1 2 1 21 21 2 2k

k k k kC kr + + + += + + + +ε ε ε


33( )




Cr We define


for 12 j n=

S20

Page - 468

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar


hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4


Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4


Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched


spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b


our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s


ommon

s Award 201

3

with center

sharing pair


13

33

S20

Page - 469


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

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- Stu

hing-Tung Y

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996 年 09 月

正中學

庇護十世音樂

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ember of Socie

udent Member

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履

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月 27 日

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八級 (Pass w

樂團 A 團

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e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

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c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


8


(00)






( )( )( )


xxx

===


( )( )( )

( )( )( )

( )( )( )


x x xx x xx x x

= = == = == = =


2 155(mod 3)S = =


Reference



2006

S20

Page - 444


9





S20

Page - 445


10





1 2 0 or 1 12 k k i i k= = =ε ε ε ε


if








2 3 1 3 1 21 2 3

1 2 1 3 2 1 2 3 3 1 3 2

( )( ) ( )( ) ( )( )( )( )( ) ( )( ) ( )( )






S20

Page - 446

Exam

funct

We w

Examtheorfigure

Henc

wher

Then

Sh

mple 313 F

tion 1( 2 )x + ε

would like to


ce 3Λ is 3 s

e xyC is a ci

n 3( )Λ Φ is a

hing-Tung Y


o give anothe


Figure 5 Th

secret sharea

rcle such tha

Figure 6 Th

a 2 bit 3 se

Yau High Sch

2isin we def

hen 33( ΛΛ Φ

er example o


hree non-collin

able Now w

ΦΦΦΦ

at its center i

he function Φ m

ecret shareab

hool Applied


) is a 2 bit

of secret shar

ircles in 2

near determi

ear points dete

e define a fu

( )( )( )( )

00

01

10

11

00011011

CCCC

===

Φ =

is (xy) and i

maps two bits

ble pair as we

d Mathemati

2 ) to be the g

3 secret sha

reable pair


ermine a uniqu

unction Φ fr

its radius is

ε1 ε2 to a circle

ell

ical Sciences

graph of the

areable pair


ue circle

rom 2 to Λ

1 3 (Ref F

e C ε1 ε2

s Award 201

1

quadratic


3Λ by

Figure 6)

13

11

y

S20

Page - 447


12




1 ( )j C Cπ isin




( ) ( ( ))j C j P jπ =

for 12 j n=


1 2 Ms s s s=



1 2i i i

kis = ε ε ε

and so

1 2 11 1 1 2 2

2 1 12

2 2 Mi i ik k




S20

Page - 448


13

[ ] ( ( ))ij j is sπ= Φ




( )( ) ( )( ) ( )( )( ) ( )( ) ( )( )( ) ( )( ) ( )

1

2

3

10 ( 2)( 1)

01 ( 1)( 3)

11 ( 2)( 3)

j j

j j

j j

s j j j

s j j j

s j j j

π π

π π

π π





( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

10 20 32 46 512[ ] 10 23 30 43 58

12 20 30 42 56s

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

1 2 3 4 5

10 20 32 46 512[ ] 10 [ ] 23 [ ] 30 [ ] 43 [ ] 58

12 20 30 42 56s s s s s

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦





[ ] [ ]1 212[ ] [ ] [ ][ ] i

jtts s ss p= =



S20

Page - 449


14

1 2( ) ( )i i ii ts p p pΦ = Ψ

and

11 2( )( )i i i


for 1 2 i M=


1 2

1 1 1 1 1 2 2 2 11 2 1 2 1 2( ( )) ( ( )) ( ( ))

M

n n nt t t

s s s s


=

=Φ Ψ Φ Ψ Φ Ψ






Mk=


[ ] ( ( ))ij j is sπ= Φ



Step 4 End





[ ] [ ] [ ]tt


S20

Page - 450


15


as in section 322


Step 5 END






[ ] [ ] ( )ij i js sσ

σ=







denoted by σ



S20

Page - 451


16



(0 8 6 7 2)(1 4 9)(3 5) (2 4 7 6 9)(0 8 5)(1 3) (0 3 7 9 2)(4 5 8)(1 6)

(1 6 9 4 5)(0 2 8)(3 7) (0 1 8 9 4)(2 5 6)(3 7) (0 2 8 4 9)(1 6 5)(3 7)

(0 8 9 5 4)(1 6 7)(2 3) (0 2 7 8 9)(1 4 3)(5 6) (0 9 5 4 7)(1 8 3)(2 6)

(0 2 9 5 3)(1 8 4)(6 7)

Example 334 Let

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

10 20 3 2 46 512[ ] 10 23 30 43 58

12 20 30 42 56s

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

32 10 20 512 46[ ] 23 30 10 43 58

12 20 30 56 42s σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

1 2 3 4 5

3 2 10 20 512 46[ ] 23 [ ] 30 [ ] 10 [ ] 43 [ ] 58

12 20 30 56 4 2s s s s sσ σ σ σ σ

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦





Mk=

S20

Page - 452


17


[ ] ( ( ))ij j is sπ= Φ




Step 5 End














S20

Page - 453


18





= many


Reference


S20

Page - 454


19

4 Algebraic Methods






1

1

( )

( )( )

t

iii j

j t

j iii j

x x

L xx x

=ne

=ne

minus

=minus

prod

prod


1( )

0j i

i jL x

i j=⎧

= ⎨ ne⎩


1 1 2 2 3 3( ) ( ) ( ) ( )L x y L x y L x y L x= + + +



1 1 2 2 3 3(0) (0) (0)y L y L y L+ + +

where

S20

Page - 455


20

1

1 1

1 1

(0 ) ( 1)

(0)( ) ( )

t tt

i ii ii j i j

j t t

j i j ii ii j i j

x x

Lx x x x

minus

= =ne ne

= =ne ne

minus minus

= =minus minus

prod prod

prod prod

for 1 23j =








1 2 1 21 2 1 2












11 21 2 2k




S20

Page - 456


21


2 3 11 2 2 2 2( ) k

kX p p p p p pminus

=

Assume that

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k



So

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k





Therefore 2 3 11 2 2 2 2 2( )

k k

X p p p p p ppminus




1 2 3 234579111316171923 X m m m= =

from now on


1 21 1 12 2 2 2 2 21 1 1 2 2 2

k k kr



1 2

1 1 2 21 1 1 r


1 2

1 1 2 21 1 1 r


such that

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr


1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr



1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr


+ + + + + + + + +ε ε εε ε ε ε ε ε

S20

Page - 457


22

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr




i ij j=ε ε







12 2i ik k

i k ixp pminus

lelt



12 331 2 4 221 2 2

i iii i i kki i


= εεε ε ε for







3 234X = 4 2345X = 5 23457X = 6 234579X =

7 23457911X = and 8 2345791113X = Also 1 2XΠ = 2 6XΠ =

3 24XΠ = and 4 120XΠ =

Note that





S20

Page - 458


23





1 21 2 1 11 2 1 1 1 2

1 1 1 2 1 11 2 1 2 1 2( )( ) ( )k kk k k k t t k





secret sharing pair











S20

Page - 459


24

( ) ( ( ))j C j P jπ =

for 12 j n=


11 1 2 2( ) ( ) ( )( ( ))






1 2( )(0)t kΛΦ ε ε ε


1 2 3 4ε ε ε ε ( )2 1 2 3 4( ) 0ΛΦ ε ε ε ε 1 2 3 4ε ε ε ε ( )

2 1 2 3 4( ) 0ΛΦ ε ε ε ε 0000 1 0110 12 1000 2 0101 15 0100 3 0011 20 0010 4 1110 24 0001 5 1101 30 1100 6 1011 40 1010 8 0111 60 1001 10 1111 120



( )( ) ( )( ) ( )( )3 2 1 3 1 2

1 2 31 3 1 2 2 1 2 3 1 33 2




S20

Page - 460


25


( ) ( )( )( )

( )( )

2 1 2 2 1 1

2 1 2 1 2 1

2 1 2 1 2 1

2 1

2 1 2 1 2 1

1 1x x x x

x x x xx x x x

x x x x

ε ε

ε εε ε

ε εε ε

+ minus

+ minus minus=

+ minus

minus=

+

minusminus minus


minusminus minus



( )2 1 2 2 1 1

1 1 22xx xx

εε ε λ λ ε

minusminus minus

le+ minus minus

Observe that

( )( )( )( )

( )( )

( )( )( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )

3 23 2

3 1 2 1 3 1 2 1

3 23 2

3 1 2 1 3 1 2 1

3 2 3 2

3 1 2

3 2

3 1 2 1

3 2

3 2 3 2

3 1

3 2 3 2

3 1

1 3 1 2 1

3 2 3 2

3 1 2 1 3 1 2 13 1 2 1

3 1 2

1





x x x

ε εε ε ε ε

ε ε

ε ε ε εε ε






+ ++ minus + minus

+ +

+ + + ++ minus

+ + + ++ minus

=minus


( )( )( )

( ) ( ) ( )( )( )( ) ( ) ( )

3 2 3 2

3 2

3 1

3 2

2 31

3 2

2 1 3 1

3 1

3 1

3 2

3 1 2 1 2 12 1

1

1 1

1

x xx

x xx x x x x x

x xx x x x x x

ε ε ε ε

ε εε ε

ε εε ε ε ε

minus



+ +

+ ++ minus

+ ++ minus minus

S20

Page - 461


26

( )( )( )( )

( )( )

( ) ( ) ( )( )

3 2

3 1

3 23 2

3 1 2 1 3 1 2 12 1

222 2

2

2 2

2

1 (2 )

1 2( )

1 2 4

2 22 2

2 22 2 2


M M M

M M M

ε εε ε ε ε

ε εε ε ε ελ

ε ε εελ

ω ω ε

λ λ ε λ λ ε


λ

+ ++ minus + minus

minus minus

leminus


le + + + + +

⎡ ⎤+ + +⎛ ⎞ ⎛ ⎞+ +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤le

minus

⎣

minus

+ ⎦

Therefore

where 2

M εε

ωλ

+=

minus So we have

( )( )( )( )

( )( ) ( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( ) ( )

( )( )( )

3 23 21 1 1

3 1 2 1 3 1 2 1

3 23 21 1

3 1 2 1 3 1 2 1

3 2

3 1 2 1

3 2

3 1 2

3 2 3 21 1 1

3 1 2 1 3 1 2 1

33 2

3 1 2 1

1

3 2 3 2

3 1 2 1 3 1 2 1

3



x x x xy y

x x x x x x x x

xx xx x x x

ε εε ε ε ε

ε εε ε ε ε


δε

δ

ε

+ ++ minus + minus

+ ++ minus + minus

+ +



+ minus+ +



le minusminus

minus

+minus

( )( )( )

( )( )( )( )

( )

21

3 1 2 1

3 21

3 1 2 1

2

3 1 2 1

3 2

3 1 2 1

2 21 2 4

xy

x x x x

x xx x x x

M

εε ε ε ε

ε εε ε ε ε

δ

ω ω ε ω δλ

++ minus + minus

+ ++ minus + minus

minus minus

+minus minus

le + +



( )2 21(0) (0) 6 12 3L L Mω ω ε ω δλ

minus le + +

where 2

M εε

ωλ

+=

minus

S20

Page - 462


27

















S20

Page - 463


28


461 Speed Tests







2 4 6 8 10


0237 0471 0708 0935 1176

0

1

2

3

4

5

6

7

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB


1183 2349 358 473 5917

S20

Page - 464


29


462 Size Tests




987 1973 2960 3947 4934


9845 19502 29195 38791 46789


100 99 99 98 95

0

02

04

06

08

1

12

14

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 465


30



Reference




0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

File

Siz

e in

KB

Secret Size in MB

CompressedSecret

CompressedContainer

S20

Page - 466


31

5 Geometric Methods






2 2

2 21 1 1 12 22 2 2 22 23 3 3 3

11

( ) 011

x y x yx y x y

J x yx y x yx y x y

++

= =++




or

2 2 1312 14

11 11 11


+ minus + minus =





2 2 1312 14

11 11 11


+ minus + minus =

we have

S20

Page - 467


32

12

11

13

11

2 2 14

11

2

2

C

C

C C C

MxM

MyM

Mr x yM

=

=

= + +


( ) 1312

11 11

2 2C C

MMx yM M

⎛ ⎞= ⎜ ⎟⎝ ⎠

and radius 2 2 14

11C C

Mx yM


shareable





21

1 1 2

1

1 11 2 1 2( ) ( 2 2 )2 2k kC C k k k k

kkx y +

+++= + + + + + +ε ε ε ε ε ε

and

1 2 1 21 21 2 2k

k k k kC kr + + + += + + + +ε ε ε


33( )




Cr We define


for 12 j n=

S20

Page - 468

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar


hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4


Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4


Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched


spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b


our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s


ommon

s Award 201

3

with center

sharing pair


13

33

S20

Page - 469


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


9





S20

Page - 445


10





1 2 0 or 1 12 k k i i k= = =ε ε ε ε


if








2 3 1 3 1 21 2 3

1 2 1 3 2 1 2 3 3 1 3 2

( )( ) ( )( ) ( )( )( )( )( ) ( )( ) ( )( )






S20

Page - 446

Exam

funct

We w

Examtheorfigure

Henc

wher

Then

Sh

mple 313 F

tion 1( 2 )x + ε

would like to


ce 3Λ is 3 s

e xyC is a ci

n 3( )Λ Φ is a

hing-Tung Y


o give anothe


Figure 5 Th

secret sharea

rcle such tha

Figure 6 Th

a 2 bit 3 se

Yau High Sch

2isin we def

hen 33( ΛΛ Φ

er example o


hree non-collin

able Now w

ΦΦΦΦ

at its center i

he function Φ m

ecret shareab

hool Applied


) is a 2 bit

of secret shar

ircles in 2

near determi

ear points dete

e define a fu

( )( )( )( )

00

01

10

11

00011011

CCCC

===

Φ =

is (xy) and i

maps two bits

ble pair as we

d Mathemati

2 ) to be the g

3 secret sha

reable pair


ermine a uniqu

unction Φ fr

its radius is

ε1 ε2 to a circle

ell

ical Sciences

graph of the

areable pair


ue circle

rom 2 to Λ

1 3 (Ref F

e C ε1 ε2

s Award 201

1

quadratic


3Λ by

Figure 6)

13

11

y

S20

Page - 447


12




1 ( )j C Cπ isin




( ) ( ( ))j C j P jπ =

for 12 j n=


1 2 Ms s s s=



1 2i i i

kis = ε ε ε

and so

1 2 11 1 1 2 2

2 1 12

2 2 Mi i ik k




S20

Page - 448


13

[ ] ( ( ))ij j is sπ= Φ




( )( ) ( )( ) ( )( )( ) ( )( ) ( )( )( ) ( )( ) ( )

1

2

3

10 ( 2)( 1)

01 ( 1)( 3)

11 ( 2)( 3)

j j

j j

j j

s j j j

s j j j

s j j j

π π

π π

π π





( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

10 20 32 46 512[ ] 10 23 30 43 58

12 20 30 42 56s

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

1 2 3 4 5

10 20 32 46 512[ ] 10 [ ] 23 [ ] 30 [ ] 43 [ ] 58

12 20 30 42 56s s s s s

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦





[ ] [ ]1 212[ ] [ ] [ ][ ] i

jtts s ss p= =



S20

Page - 449


14

1 2( ) ( )i i ii ts p p pΦ = Ψ

and

11 2( )( )i i i


for 1 2 i M=


1 2

1 1 1 1 1 2 2 2 11 2 1 2 1 2( ( )) ( ( )) ( ( ))

M

n n nt t t

s s s s


=

=Φ Ψ Φ Ψ Φ Ψ






Mk=


[ ] ( ( ))ij j is sπ= Φ



Step 4 End





[ ] [ ] [ ]tt


S20

Page - 450


15


as in section 322


Step 5 END






[ ] [ ] ( )ij i js sσ

σ=







denoted by σ



S20

Page - 451


16



(0 8 6 7 2)(1 4 9)(3 5) (2 4 7 6 9)(0 8 5)(1 3) (0 3 7 9 2)(4 5 8)(1 6)

(1 6 9 4 5)(0 2 8)(3 7) (0 1 8 9 4)(2 5 6)(3 7) (0 2 8 4 9)(1 6 5)(3 7)

(0 8 9 5 4)(1 6 7)(2 3) (0 2 7 8 9)(1 4 3)(5 6) (0 9 5 4 7)(1 8 3)(2 6)

(0 2 9 5 3)(1 8 4)(6 7)

Example 334 Let

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

10 20 3 2 46 512[ ] 10 23 30 43 58

12 20 30 42 56s

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

32 10 20 512 46[ ] 23 30 10 43 58

12 20 30 56 42s σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

1 2 3 4 5

3 2 10 20 512 46[ ] 23 [ ] 30 [ ] 10 [ ] 43 [ ] 58

12 20 30 56 4 2s s s s sσ σ σ σ σ

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦





Mk=

S20

Page - 452


17


[ ] ( ( ))ij j is sπ= Φ




Step 5 End














S20

Page - 453


18





= many


Reference


S20

Page - 454


19

4 Algebraic Methods






1

1

( )

( )( )

t

iii j

j t

j iii j

x x

L xx x

=ne

=ne

minus

=minus

prod

prod


1( )

0j i

i jL x

i j=⎧

= ⎨ ne⎩


1 1 2 2 3 3( ) ( ) ( ) ( )L x y L x y L x y L x= + + +



1 1 2 2 3 3(0) (0) (0)y L y L y L+ + +

where

S20

Page - 455


20

1

1 1

1 1

(0 ) ( 1)

(0)( ) ( )

t tt

i ii ii j i j

j t t

j i j ii ii j i j

x x

Lx x x x

minus

= =ne ne

= =ne ne

minus minus

= =minus minus

prod prod

prod prod

for 1 23j =








1 2 1 21 2 1 2












11 21 2 2k




S20

Page - 456


21


2 3 11 2 2 2 2( ) k

kX p p p p p pminus

=

Assume that

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k



So

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k





Therefore 2 3 11 2 2 2 2 2( )

k k

X p p p p p ppminus




1 2 3 234579111316171923 X m m m= =

from now on


1 21 1 12 2 2 2 2 21 1 1 2 2 2

k k kr



1 2

1 1 2 21 1 1 r


1 2

1 1 2 21 1 1 r


such that

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr


1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr



1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr


+ + + + + + + + +ε ε εε ε ε ε ε ε

S20

Page - 457


22

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr




i ij j=ε ε







12 2i ik k

i k ixp pminus

lelt



12 331 2 4 221 2 2

i iii i i kki i


= εεε ε ε for







3 234X = 4 2345X = 5 23457X = 6 234579X =

7 23457911X = and 8 2345791113X = Also 1 2XΠ = 2 6XΠ =

3 24XΠ = and 4 120XΠ =

Note that





S20

Page - 458


23





1 21 2 1 11 2 1 1 1 2

1 1 1 2 1 11 2 1 2 1 2( )( ) ( )k kk k k k t t k





secret sharing pair











S20

Page - 459


24

( ) ( ( ))j C j P jπ =

for 12 j n=


11 1 2 2( ) ( ) ( )( ( ))






1 2( )(0)t kΛΦ ε ε ε


1 2 3 4ε ε ε ε ( )2 1 2 3 4( ) 0ΛΦ ε ε ε ε 1 2 3 4ε ε ε ε ( )

2 1 2 3 4( ) 0ΛΦ ε ε ε ε 0000 1 0110 12 1000 2 0101 15 0100 3 0011 20 0010 4 1110 24 0001 5 1101 30 1100 6 1011 40 1010 8 0111 60 1001 10 1111 120



( )( ) ( )( ) ( )( )3 2 1 3 1 2

1 2 31 3 1 2 2 1 2 3 1 33 2




S20

Page - 460


25


( ) ( )( )( )

( )( )

2 1 2 2 1 1

2 1 2 1 2 1

2 1 2 1 2 1

2 1

2 1 2 1 2 1

1 1x x x x

x x x xx x x x

x x x x

ε ε

ε εε ε

ε εε ε

+ minus

+ minus minus=

+ minus

minus=

+

minusminus minus


minusminus minus



( )2 1 2 2 1 1

1 1 22xx xx

εε ε λ λ ε

minusminus minus

le+ minus minus

Observe that

( )( )( )( )

( )( )

( )( )( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )

3 23 2

3 1 2 1 3 1 2 1

3 23 2

3 1 2 1 3 1 2 1

3 2 3 2

3 1 2

3 2

3 1 2 1

3 2

3 2 3 2

3 1

3 2 3 2

3 1

1 3 1 2 1

3 2 3 2

3 1 2 1 3 1 2 13 1 2 1

3 1 2

1





x x x

ε εε ε ε ε

ε ε

ε ε ε εε ε






+ ++ minus + minus

+ +

+ + + ++ minus

+ + + ++ minus

=minus


( )( )( )

( ) ( ) ( )( )( )( ) ( ) ( )

3 2 3 2

3 2

3 1

3 2

2 31

3 2

2 1 3 1

3 1

3 1

3 2

3 1 2 1 2 12 1

1

1 1

1

x xx

x xx x x x x x

x xx x x x x x

ε ε ε ε

ε εε ε

ε εε ε ε ε

minus



+ +

+ ++ minus

+ ++ minus minus

S20

Page - 461


26

( )( )( )( )

( )( )

( ) ( ) ( )( )

3 2

3 1

3 23 2

3 1 2 1 3 1 2 12 1

222 2

2

2 2

2

1 (2 )

1 2( )

1 2 4

2 22 2

2 22 2 2


M M M

M M M

ε εε ε ε ε

ε εε ε ε ελ

ε ε εελ

ω ω ε

λ λ ε λ λ ε


λ

+ ++ minus + minus

minus minus

leminus


le + + + + +

⎡ ⎤+ + +⎛ ⎞ ⎛ ⎞+ +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤le

minus

⎣

minus

+ ⎦

Therefore

where 2

M εε

ωλ

+=

minus So we have

( )( )( )( )

( )( ) ( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( ) ( )

( )( )( )

3 23 21 1 1

3 1 2 1 3 1 2 1

3 23 21 1

3 1 2 1 3 1 2 1

3 2

3 1 2 1

3 2

3 1 2

3 2 3 21 1 1

3 1 2 1 3 1 2 1

33 2

3 1 2 1

1

3 2 3 2

3 1 2 1 3 1 2 1

3



x x x xy y

x x x x x x x x

xx xx x x x

ε εε ε ε ε

ε εε ε ε ε


δε

δ

ε

+ ++ minus + minus

+ ++ minus + minus

+ +



+ minus+ +



le minusminus

minus

+minus

( )( )( )

( )( )( )( )

( )

21

3 1 2 1

3 21

3 1 2 1

2

3 1 2 1

3 2

3 1 2 1

2 21 2 4

xy

x x x x

x xx x x x

M

εε ε ε ε

ε εε ε ε ε

δ

ω ω ε ω δλ

++ minus + minus

+ ++ minus + minus

minus minus

+minus minus

le + +



( )2 21(0) (0) 6 12 3L L Mω ω ε ω δλ

minus le + +

where 2

M εε

ωλ

+=

minus

S20

Page - 462


27

















S20

Page - 463


28


461 Speed Tests







2 4 6 8 10


0237 0471 0708 0935 1176

0

1

2

3

4

5

6

7

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB


1183 2349 358 473 5917

S20

Page - 464


29


462 Size Tests




987 1973 2960 3947 4934


9845 19502 29195 38791 46789


100 99 99 98 95

0

02

04

06

08

1

12

14

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 465


30



Reference




0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

File

Siz

e in

KB

Secret Size in MB

CompressedSecret

CompressedContainer

S20

Page - 466


31

5 Geometric Methods






2 2

2 21 1 1 12 22 2 2 22 23 3 3 3

11

( ) 011

x y x yx y x y

J x yx y x yx y x y

++

= =++




or

2 2 1312 14

11 11 11


+ minus + minus =





2 2 1312 14

11 11 11


+ minus + minus =

we have

S20

Page - 467


32

12

11

13

11

2 2 14

11

2

2

C

C

C C C

MxM

MyM

Mr x yM

=

=

= + +


( ) 1312

11 11

2 2C C

MMx yM M

⎛ ⎞= ⎜ ⎟⎝ ⎠

and radius 2 2 14

11C C

Mx yM


shareable





21

1 1 2

1

1 11 2 1 2( ) ( 2 2 )2 2k kC C k k k k

kkx y +

+++= + + + + + +ε ε ε ε ε ε

and

1 2 1 21 21 2 2k

k k k kC kr + + + += + + + +ε ε ε


33( )




Cr We define


for 12 j n=

S20

Page - 468

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar


hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4


Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4


Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched


spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b


our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s


ommon

s Award 201

3

with center

sharing pair


13

33

S20

Page - 469


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


10





1 2 0 or 1 12 k k i i k= = =ε ε ε ε


if








2 3 1 3 1 21 2 3

1 2 1 3 2 1 2 3 3 1 3 2

( )( ) ( )( ) ( )( )( )( )( ) ( )( ) ( )( )






S20

Page - 446

Exam

funct

We w

Examtheorfigure

Henc

wher

Then

Sh

mple 313 F

tion 1( 2 )x + ε

would like to


ce 3Λ is 3 s

e xyC is a ci

n 3( )Λ Φ is a

hing-Tung Y


o give anothe


Figure 5 Th

secret sharea

rcle such tha

Figure 6 Th

a 2 bit 3 se

Yau High Sch

2isin we def

hen 33( ΛΛ Φ

er example o


hree non-collin

able Now w

ΦΦΦΦ

at its center i

he function Φ m

ecret shareab

hool Applied


) is a 2 bit

of secret shar

ircles in 2

near determi

ear points dete

e define a fu

( )( )( )( )

00

01

10

11

00011011

CCCC

===

Φ =

is (xy) and i

maps two bits

ble pair as we

d Mathemati

2 ) to be the g

3 secret sha

reable pair


ermine a uniqu

unction Φ fr

its radius is

ε1 ε2 to a circle

ell

ical Sciences

graph of the

areable pair


ue circle

rom 2 to Λ

1 3 (Ref F

e C ε1 ε2

s Award 201

1

quadratic


3Λ by

Figure 6)

13

11

y

S20

Page - 447


12




1 ( )j C Cπ isin




( ) ( ( ))j C j P jπ =

for 12 j n=


1 2 Ms s s s=



1 2i i i

kis = ε ε ε

and so

1 2 11 1 1 2 2

2 1 12

2 2 Mi i ik k




S20

Page - 448


13

[ ] ( ( ))ij j is sπ= Φ




( )( ) ( )( ) ( )( )( ) ( )( ) ( )( )( ) ( )( ) ( )

1

2

3

10 ( 2)( 1)

01 ( 1)( 3)

11 ( 2)( 3)

j j

j j

j j

s j j j

s j j j

s j j j

π π

π π

π π





( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

10 20 32 46 512[ ] 10 23 30 43 58

12 20 30 42 56s

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

1 2 3 4 5

10 20 32 46 512[ ] 10 [ ] 23 [ ] 30 [ ] 43 [ ] 58

12 20 30 42 56s s s s s

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦





[ ] [ ]1 212[ ] [ ] [ ][ ] i

jtts s ss p= =



S20

Page - 449


14

1 2( ) ( )i i ii ts p p pΦ = Ψ

and

11 2( )( )i i i


for 1 2 i M=


1 2

1 1 1 1 1 2 2 2 11 2 1 2 1 2( ( )) ( ( )) ( ( ))

M

n n nt t t

s s s s


=

=Φ Ψ Φ Ψ Φ Ψ






Mk=


[ ] ( ( ))ij j is sπ= Φ



Step 4 End





[ ] [ ] [ ]tt


S20

Page - 450


15


as in section 322


Step 5 END






[ ] [ ] ( )ij i js sσ

σ=







denoted by σ



S20

Page - 451


16



(0 8 6 7 2)(1 4 9)(3 5) (2 4 7 6 9)(0 8 5)(1 3) (0 3 7 9 2)(4 5 8)(1 6)

(1 6 9 4 5)(0 2 8)(3 7) (0 1 8 9 4)(2 5 6)(3 7) (0 2 8 4 9)(1 6 5)(3 7)

(0 8 9 5 4)(1 6 7)(2 3) (0 2 7 8 9)(1 4 3)(5 6) (0 9 5 4 7)(1 8 3)(2 6)

(0 2 9 5 3)(1 8 4)(6 7)

Example 334 Let

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

10 20 3 2 46 512[ ] 10 23 30 43 58

12 20 30 42 56s

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

32 10 20 512 46[ ] 23 30 10 43 58

12 20 30 56 42s σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

1 2 3 4 5

3 2 10 20 512 46[ ] 23 [ ] 30 [ ] 10 [ ] 43 [ ] 58

12 20 30 56 4 2s s s s sσ σ σ σ σ

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦





Mk=

S20

Page - 452


17


[ ] ( ( ))ij j is sπ= Φ




Step 5 End














S20

Page - 453


18





= many


Reference


S20

Page - 454


19

4 Algebraic Methods






1

1

( )

( )( )

t

iii j

j t

j iii j

x x

L xx x

=ne

=ne

minus

=minus

prod

prod


1( )

0j i

i jL x

i j=⎧

= ⎨ ne⎩


1 1 2 2 3 3( ) ( ) ( ) ( )L x y L x y L x y L x= + + +



1 1 2 2 3 3(0) (0) (0)y L y L y L+ + +

where

S20

Page - 455


20

1

1 1

1 1

(0 ) ( 1)

(0)( ) ( )

t tt

i ii ii j i j

j t t

j i j ii ii j i j

x x

Lx x x x

minus

= =ne ne

= =ne ne

minus minus

= =minus minus

prod prod

prod prod

for 1 23j =








1 2 1 21 2 1 2












11 21 2 2k




S20

Page - 456


21


2 3 11 2 2 2 2( ) k

kX p p p p p pminus

=

Assume that

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k



So

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k





Therefore 2 3 11 2 2 2 2 2( )

k k

X p p p p p ppminus




1 2 3 234579111316171923 X m m m= =

from now on


1 21 1 12 2 2 2 2 21 1 1 2 2 2

k k kr



1 2

1 1 2 21 1 1 r


1 2

1 1 2 21 1 1 r


such that

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr


1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr



1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr


+ + + + + + + + +ε ε εε ε ε ε ε ε

S20

Page - 457


22

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr




i ij j=ε ε







12 2i ik k

i k ixp pminus

lelt



12 331 2 4 221 2 2

i iii i i kki i


= εεε ε ε for







3 234X = 4 2345X = 5 23457X = 6 234579X =

7 23457911X = and 8 2345791113X = Also 1 2XΠ = 2 6XΠ =

3 24XΠ = and 4 120XΠ =

Note that





S20

Page - 458


23





1 21 2 1 11 2 1 1 1 2

1 1 1 2 1 11 2 1 2 1 2( )( ) ( )k kk k k k t t k





secret sharing pair











S20

Page - 459


24

( ) ( ( ))j C j P jπ =

for 12 j n=


11 1 2 2( ) ( ) ( )( ( ))






1 2( )(0)t kΛΦ ε ε ε


1 2 3 4ε ε ε ε ( )2 1 2 3 4( ) 0ΛΦ ε ε ε ε 1 2 3 4ε ε ε ε ( )

2 1 2 3 4( ) 0ΛΦ ε ε ε ε 0000 1 0110 12 1000 2 0101 15 0100 3 0011 20 0010 4 1110 24 0001 5 1101 30 1100 6 1011 40 1010 8 0111 60 1001 10 1111 120



( )( ) ( )( ) ( )( )3 2 1 3 1 2

1 2 31 3 1 2 2 1 2 3 1 33 2




S20

Page - 460


25


( ) ( )( )( )

( )( )

2 1 2 2 1 1

2 1 2 1 2 1

2 1 2 1 2 1

2 1

2 1 2 1 2 1

1 1x x x x

x x x xx x x x

x x x x

ε ε

ε εε ε

ε εε ε

+ minus

+ minus minus=

+ minus

minus=

+

minusminus minus


minusminus minus



( )2 1 2 2 1 1

1 1 22xx xx

εε ε λ λ ε

minusminus minus

le+ minus minus

Observe that

( )( )( )( )

( )( )

( )( )( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )

3 23 2

3 1 2 1 3 1 2 1

3 23 2

3 1 2 1 3 1 2 1

3 2 3 2

3 1 2

3 2

3 1 2 1

3 2

3 2 3 2

3 1

3 2 3 2

3 1

1 3 1 2 1

3 2 3 2

3 1 2 1 3 1 2 13 1 2 1

3 1 2

1





x x x

ε εε ε ε ε

ε ε

ε ε ε εε ε






+ ++ minus + minus

+ +

+ + + ++ minus

+ + + ++ minus

=minus


( )( )( )

( ) ( ) ( )( )( )( ) ( ) ( )

3 2 3 2

3 2

3 1

3 2

2 31

3 2

2 1 3 1

3 1

3 1

3 2

3 1 2 1 2 12 1

1

1 1

1

x xx

x xx x x x x x

x xx x x x x x

ε ε ε ε

ε εε ε

ε εε ε ε ε

minus



+ +

+ ++ minus

+ ++ minus minus

S20

Page - 461


26

( )( )( )( )

( )( )

( ) ( ) ( )( )

3 2

3 1

3 23 2

3 1 2 1 3 1 2 12 1

222 2

2

2 2

2

1 (2 )

1 2( )

1 2 4

2 22 2

2 22 2 2


M M M

M M M

ε εε ε ε ε

ε εε ε ε ελ

ε ε εελ

ω ω ε

λ λ ε λ λ ε


λ

+ ++ minus + minus

minus minus

leminus


le + + + + +

⎡ ⎤+ + +⎛ ⎞ ⎛ ⎞+ +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤le

minus

⎣

minus

+ ⎦

Therefore

where 2

M εε

ωλ

+=

minus So we have

( )( )( )( )

( )( ) ( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( ) ( )

( )( )( )

3 23 21 1 1

3 1 2 1 3 1 2 1

3 23 21 1

3 1 2 1 3 1 2 1

3 2

3 1 2 1

3 2

3 1 2

3 2 3 21 1 1

3 1 2 1 3 1 2 1

33 2

3 1 2 1

1

3 2 3 2

3 1 2 1 3 1 2 1

3



x x x xy y

x x x x x x x x

xx xx x x x

ε εε ε ε ε

ε εε ε ε ε


δε

δ

ε

+ ++ minus + minus

+ ++ minus + minus

+ +



+ minus+ +



le minusminus

minus

+minus

( )( )( )

( )( )( )( )

( )

21

3 1 2 1

3 21

3 1 2 1

2

3 1 2 1

3 2

3 1 2 1

2 21 2 4

xy

x x x x

x xx x x x

M

εε ε ε ε

ε εε ε ε ε

δ

ω ω ε ω δλ

++ minus + minus

+ ++ minus + minus

minus minus

+minus minus

le + +



( )2 21(0) (0) 6 12 3L L Mω ω ε ω δλ

minus le + +

where 2

M εε

ωλ

+=

minus

S20

Page - 462


27

















S20

Page - 463


28


461 Speed Tests







2 4 6 8 10


0237 0471 0708 0935 1176

0

1

2

3

4

5

6

7

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB


1183 2349 358 473 5917

S20

Page - 464


29


462 Size Tests




987 1973 2960 3947 4934


9845 19502 29195 38791 46789


100 99 99 98 95

0

02

04

06

08

1

12

14

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 465


30



Reference




0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

File

Siz

e in

KB

Secret Size in MB

CompressedSecret

CompressedContainer

S20

Page - 466


31

5 Geometric Methods






2 2

2 21 1 1 12 22 2 2 22 23 3 3 3

11

( ) 011

x y x yx y x y

J x yx y x yx y x y

++

= =++




or

2 2 1312 14

11 11 11


+ minus + minus =





2 2 1312 14

11 11 11


+ minus + minus =

we have

S20

Page - 467


32

12

11

13

11

2 2 14

11

2

2

C

C

C C C

MxM

MyM

Mr x yM

=

=

= + +


( ) 1312

11 11

2 2C C

MMx yM M

⎛ ⎞= ⎜ ⎟⎝ ⎠

and radius 2 2 14

11C C

Mx yM


shareable





21

1 1 2

1

1 11 2 1 2( ) ( 2 2 )2 2k kC C k k k k

kkx y +

+++= + + + + + +ε ε ε ε ε ε

and

1 2 1 21 21 2 2k

k k k kC kr + + + += + + + +ε ε ε


33( )




Cr We define


for 12 j n=

S20

Page - 468

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar


hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4


Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4


Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched


spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b


our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s


ommon

s Award 201

3

with center

sharing pair


13

33

S20

Page - 469


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488

Exam

funct

We w

Examtheorfigure

Henc

wher

Then

Sh

mple 313 F

tion 1( 2 )x + ε

would like to


ce 3Λ is 3 s

e xyC is a ci

n 3( )Λ Φ is a

hing-Tung Y


o give anothe


Figure 5 Th

secret sharea

rcle such tha

Figure 6 Th

a 2 bit 3 se

Yau High Sch

2isin we def

hen 33( ΛΛ Φ

er example o


hree non-collin

able Now w

ΦΦΦΦ

at its center i

he function Φ m

ecret shareab

hool Applied


) is a 2 bit

of secret shar

ircles in 2

near determi

ear points dete

e define a fu

( )( )( )( )

00

01

10

11

00011011

CCCC

===

Φ =

is (xy) and i

maps two bits

ble pair as we

d Mathemati

2 ) to be the g

3 secret sha

reable pair


ermine a uniqu

unction Φ fr

its radius is

ε1 ε2 to a circle

ell

ical Sciences

graph of the

areable pair


ue circle

rom 2 to Λ

1 3 (Ref F

e C ε1 ε2

s Award 201

1

quadratic


3Λ by

Figure 6)

13

11

y

S20

Page - 447


12




1 ( )j C Cπ isin




( ) ( ( ))j C j P jπ =

for 12 j n=


1 2 Ms s s s=



1 2i i i

kis = ε ε ε

and so

1 2 11 1 1 2 2

2 1 12

2 2 Mi i ik k




S20

Page - 448


13

[ ] ( ( ))ij j is sπ= Φ




( )( ) ( )( ) ( )( )( ) ( )( ) ( )( )( ) ( )( ) ( )

1

2

3

10 ( 2)( 1)

01 ( 1)( 3)

11 ( 2)( 3)

j j

j j

j j

s j j j

s j j j

s j j j

π π

π π

π π





( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

10 20 32 46 512[ ] 10 23 30 43 58

12 20 30 42 56s

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

1 2 3 4 5

10 20 32 46 512[ ] 10 [ ] 23 [ ] 30 [ ] 43 [ ] 58

12 20 30 42 56s s s s s

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦





[ ] [ ]1 212[ ] [ ] [ ][ ] i

jtts s ss p= =



S20

Page - 449


14

1 2( ) ( )i i ii ts p p pΦ = Ψ

and

11 2( )( )i i i


for 1 2 i M=


1 2

1 1 1 1 1 2 2 2 11 2 1 2 1 2( ( )) ( ( )) ( ( ))

M

n n nt t t

s s s s


=

=Φ Ψ Φ Ψ Φ Ψ






Mk=


[ ] ( ( ))ij j is sπ= Φ



Step 4 End





[ ] [ ] [ ]tt


S20

Page - 450


15


as in section 322


Step 5 END






[ ] [ ] ( )ij i js sσ

σ=







denoted by σ



S20

Page - 451


16



(0 8 6 7 2)(1 4 9)(3 5) (2 4 7 6 9)(0 8 5)(1 3) (0 3 7 9 2)(4 5 8)(1 6)

(1 6 9 4 5)(0 2 8)(3 7) (0 1 8 9 4)(2 5 6)(3 7) (0 2 8 4 9)(1 6 5)(3 7)

(0 8 9 5 4)(1 6 7)(2 3) (0 2 7 8 9)(1 4 3)(5 6) (0 9 5 4 7)(1 8 3)(2 6)

(0 2 9 5 3)(1 8 4)(6 7)

Example 334 Let

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

10 20 3 2 46 512[ ] 10 23 30 43 58

12 20 30 42 56s

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

32 10 20 512 46[ ] 23 30 10 43 58

12 20 30 56 42s σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

1 2 3 4 5

3 2 10 20 512 46[ ] 23 [ ] 30 [ ] 10 [ ] 43 [ ] 58

12 20 30 56 4 2s s s s sσ σ σ σ σ

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦





Mk=

S20

Page - 452


17


[ ] ( ( ))ij j is sπ= Φ




Step 5 End














S20

Page - 453


18





= many


Reference


S20

Page - 454


19

4 Algebraic Methods






1

1

( )

( )( )

t

iii j

j t

j iii j

x x

L xx x

=ne

=ne

minus

=minus

prod

prod


1( )

0j i

i jL x

i j=⎧

= ⎨ ne⎩


1 1 2 2 3 3( ) ( ) ( ) ( )L x y L x y L x y L x= + + +



1 1 2 2 3 3(0) (0) (0)y L y L y L+ + +

where

S20

Page - 455


20

1

1 1

1 1

(0 ) ( 1)

(0)( ) ( )

t tt

i ii ii j i j

j t t

j i j ii ii j i j

x x

Lx x x x

minus

= =ne ne

= =ne ne

minus minus

= =minus minus

prod prod

prod prod

for 1 23j =








1 2 1 21 2 1 2












11 21 2 2k




S20

Page - 456


21


2 3 11 2 2 2 2( ) k

kX p p p p p pminus

=

Assume that

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k



So

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k





Therefore 2 3 11 2 2 2 2 2( )

k k

X p p p p p ppminus




1 2 3 234579111316171923 X m m m= =

from now on


1 21 1 12 2 2 2 2 21 1 1 2 2 2

k k kr



1 2

1 1 2 21 1 1 r


1 2

1 1 2 21 1 1 r


such that

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr


1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr



1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr


+ + + + + + + + +ε ε εε ε ε ε ε ε

S20

Page - 457


22

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr




i ij j=ε ε







12 2i ik k

i k ixp pminus

lelt



12 331 2 4 221 2 2

i iii i i kki i


= εεε ε ε for







3 234X = 4 2345X = 5 23457X = 6 234579X =

7 23457911X = and 8 2345791113X = Also 1 2XΠ = 2 6XΠ =

3 24XΠ = and 4 120XΠ =

Note that





S20

Page - 458


23





1 21 2 1 11 2 1 1 1 2

1 1 1 2 1 11 2 1 2 1 2( )( ) ( )k kk k k k t t k





secret sharing pair











S20

Page - 459


24

( ) ( ( ))j C j P jπ =

for 12 j n=


11 1 2 2( ) ( ) ( )( ( ))






1 2( )(0)t kΛΦ ε ε ε


1 2 3 4ε ε ε ε ( )2 1 2 3 4( ) 0ΛΦ ε ε ε ε 1 2 3 4ε ε ε ε ( )

2 1 2 3 4( ) 0ΛΦ ε ε ε ε 0000 1 0110 12 1000 2 0101 15 0100 3 0011 20 0010 4 1110 24 0001 5 1101 30 1100 6 1011 40 1010 8 0111 60 1001 10 1111 120



( )( ) ( )( ) ( )( )3 2 1 3 1 2

1 2 31 3 1 2 2 1 2 3 1 33 2




S20

Page - 460


25


( ) ( )( )( )

( )( )

2 1 2 2 1 1

2 1 2 1 2 1

2 1 2 1 2 1

2 1

2 1 2 1 2 1

1 1x x x x

x x x xx x x x

x x x x

ε ε

ε εε ε

ε εε ε

+ minus

+ minus minus=

+ minus

minus=

+

minusminus minus


minusminus minus



( )2 1 2 2 1 1

1 1 22xx xx

εε ε λ λ ε

minusminus minus

le+ minus minus

Observe that

( )( )( )( )

( )( )

( )( )( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )

3 23 2

3 1 2 1 3 1 2 1

3 23 2

3 1 2 1 3 1 2 1

3 2 3 2

3 1 2

3 2

3 1 2 1

3 2

3 2 3 2

3 1

3 2 3 2

3 1

1 3 1 2 1

3 2 3 2

3 1 2 1 3 1 2 13 1 2 1

3 1 2

1





x x x

ε εε ε ε ε

ε ε

ε ε ε εε ε






+ ++ minus + minus

+ +

+ + + ++ minus

+ + + ++ minus

=minus


( )( )( )

( ) ( ) ( )( )( )( ) ( ) ( )

3 2 3 2

3 2

3 1

3 2

2 31

3 2

2 1 3 1

3 1

3 1

3 2

3 1 2 1 2 12 1

1

1 1

1

x xx

x xx x x x x x

x xx x x x x x

ε ε ε ε

ε εε ε

ε εε ε ε ε

minus



+ +

+ ++ minus

+ ++ minus minus

S20

Page - 461


26

( )( )( )( )

( )( )

( ) ( ) ( )( )

3 2

3 1

3 23 2

3 1 2 1 3 1 2 12 1

222 2

2

2 2

2

1 (2 )

1 2( )

1 2 4

2 22 2

2 22 2 2


M M M

M M M

ε εε ε ε ε

ε εε ε ε ελ

ε ε εελ

ω ω ε

λ λ ε λ λ ε


λ

+ ++ minus + minus

minus minus

leminus


le + + + + +

⎡ ⎤+ + +⎛ ⎞ ⎛ ⎞+ +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤le

minus

⎣

minus

+ ⎦

Therefore

where 2

M εε

ωλ

+=

minus So we have

( )( )( )( )

( )( ) ( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( ) ( )

( )( )( )

3 23 21 1 1

3 1 2 1 3 1 2 1

3 23 21 1

3 1 2 1 3 1 2 1

3 2

3 1 2 1

3 2

3 1 2

3 2 3 21 1 1

3 1 2 1 3 1 2 1

33 2

3 1 2 1

1

3 2 3 2

3 1 2 1 3 1 2 1

3



x x x xy y

x x x x x x x x

xx xx x x x

ε εε ε ε ε

ε εε ε ε ε


δε

δ

ε

+ ++ minus + minus

+ ++ minus + minus

+ +



+ minus+ +



le minusminus

minus

+minus

( )( )( )

( )( )( )( )

( )

21

3 1 2 1

3 21

3 1 2 1

2

3 1 2 1

3 2

3 1 2 1

2 21 2 4

xy

x x x x

x xx x x x

M

εε ε ε ε

ε εε ε ε ε

δ

ω ω ε ω δλ

++ minus + minus

+ ++ minus + minus

minus minus

+minus minus

le + +



( )2 21(0) (0) 6 12 3L L Mω ω ε ω δλ

minus le + +

where 2

M εε

ωλ

+=

minus

S20

Page - 462


27

















S20

Page - 463


28


461 Speed Tests







2 4 6 8 10


0237 0471 0708 0935 1176

0

1

2

3

4

5

6

7

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB


1183 2349 358 473 5917

S20

Page - 464


29


462 Size Tests




987 1973 2960 3947 4934


9845 19502 29195 38791 46789


100 99 99 98 95

0

02

04

06

08

1

12

14

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 465


30



Reference




0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

File

Siz

e in

KB

Secret Size in MB

CompressedSecret

CompressedContainer

S20

Page - 466


31

5 Geometric Methods






2 2

2 21 1 1 12 22 2 2 22 23 3 3 3

11

( ) 011

x y x yx y x y

J x yx y x yx y x y

++

= =++




or

2 2 1312 14

11 11 11


+ minus + minus =





2 2 1312 14

11 11 11


+ minus + minus =

we have

S20

Page - 467


32

12

11

13

11

2 2 14

11

2

2

C

C

C C C

MxM

MyM

Mr x yM

=

=

= + +


( ) 1312

11 11

2 2C C

MMx yM M

⎛ ⎞= ⎜ ⎟⎝ ⎠

and radius 2 2 14

11C C

Mx yM


shareable





21

1 1 2

1

1 11 2 1 2( ) ( 2 2 )2 2k kC C k k k k

kkx y +

+++= + + + + + +ε ε ε ε ε ε

and

1 2 1 21 21 2 2k

k k k kC kr + + + += + + + +ε ε ε


33( )




Cr We define


for 12 j n=

S20

Page - 468

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar


hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4


Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4


Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched


spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b


our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s


ommon

s Award 201

3

with center

sharing pair


13

33

S20

Page - 469


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


12




1 ( )j C Cπ isin




( ) ( ( ))j C j P jπ =

for 12 j n=


1 2 Ms s s s=



1 2i i i

kis = ε ε ε

and so

1 2 11 1 1 2 2

2 1 12

2 2 Mi i ik k




S20

Page - 448


13

[ ] ( ( ))ij j is sπ= Φ




( )( ) ( )( ) ( )( )( ) ( )( ) ( )( )( ) ( )( ) ( )

1

2

3

10 ( 2)( 1)

01 ( 1)( 3)

11 ( 2)( 3)

j j

j j

j j

s j j j

s j j j

s j j j

π π

π π

π π





( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

10 20 32 46 512[ ] 10 23 30 43 58

12 20 30 42 56s

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

1 2 3 4 5

10 20 32 46 512[ ] 10 [ ] 23 [ ] 30 [ ] 43 [ ] 58

12 20 30 42 56s s s s s

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦





[ ] [ ]1 212[ ] [ ] [ ][ ] i

jtts s ss p= =



S20

Page - 449


14

1 2( ) ( )i i ii ts p p pΦ = Ψ

and

11 2( )( )i i i


for 1 2 i M=


1 2

1 1 1 1 1 2 2 2 11 2 1 2 1 2( ( )) ( ( )) ( ( ))

M

n n nt t t

s s s s


=

=Φ Ψ Φ Ψ Φ Ψ






Mk=


[ ] ( ( ))ij j is sπ= Φ



Step 4 End





[ ] [ ] [ ]tt


S20

Page - 450


15


as in section 322


Step 5 END






[ ] [ ] ( )ij i js sσ

σ=







denoted by σ



S20

Page - 451


16



(0 8 6 7 2)(1 4 9)(3 5) (2 4 7 6 9)(0 8 5)(1 3) (0 3 7 9 2)(4 5 8)(1 6)

(1 6 9 4 5)(0 2 8)(3 7) (0 1 8 9 4)(2 5 6)(3 7) (0 2 8 4 9)(1 6 5)(3 7)

(0 8 9 5 4)(1 6 7)(2 3) (0 2 7 8 9)(1 4 3)(5 6) (0 9 5 4 7)(1 8 3)(2 6)

(0 2 9 5 3)(1 8 4)(6 7)

Example 334 Let

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

10 20 3 2 46 512[ ] 10 23 30 43 58

12 20 30 42 56s

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

32 10 20 512 46[ ] 23 30 10 43 58

12 20 30 56 42s σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

1 2 3 4 5

3 2 10 20 512 46[ ] 23 [ ] 30 [ ] 10 [ ] 43 [ ] 58

12 20 30 56 4 2s s s s sσ σ σ σ σ

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦





Mk=

S20

Page - 452


17


[ ] ( ( ))ij j is sπ= Φ




Step 5 End














S20

Page - 453


18





= many


Reference


S20

Page - 454


19

4 Algebraic Methods






1

1

( )

( )( )

t

iii j

j t

j iii j

x x

L xx x

=ne

=ne

minus

=minus

prod

prod


1( )

0j i

i jL x

i j=⎧

= ⎨ ne⎩


1 1 2 2 3 3( ) ( ) ( ) ( )L x y L x y L x y L x= + + +



1 1 2 2 3 3(0) (0) (0)y L y L y L+ + +

where

S20

Page - 455


20

1

1 1

1 1

(0 ) ( 1)

(0)( ) ( )

t tt

i ii ii j i j

j t t

j i j ii ii j i j

x x

Lx x x x

minus

= =ne ne

= =ne ne

minus minus

= =minus minus

prod prod

prod prod

for 1 23j =








1 2 1 21 2 1 2












11 21 2 2k




S20

Page - 456


21


2 3 11 2 2 2 2( ) k

kX p p p p p pminus

=

Assume that

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k



So

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k





Therefore 2 3 11 2 2 2 2 2( )

k k

X p p p p p ppminus




1 2 3 234579111316171923 X m m m= =

from now on


1 21 1 12 2 2 2 2 21 1 1 2 2 2

k k kr



1 2

1 1 2 21 1 1 r


1 2

1 1 2 21 1 1 r


such that

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr


1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr



1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr


+ + + + + + + + +ε ε εε ε ε ε ε ε

S20

Page - 457


22

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr




i ij j=ε ε







12 2i ik k

i k ixp pminus

lelt



12 331 2 4 221 2 2

i iii i i kki i


= εεε ε ε for







3 234X = 4 2345X = 5 23457X = 6 234579X =

7 23457911X = and 8 2345791113X = Also 1 2XΠ = 2 6XΠ =

3 24XΠ = and 4 120XΠ =

Note that





S20

Page - 458


23





1 21 2 1 11 2 1 1 1 2

1 1 1 2 1 11 2 1 2 1 2( )( ) ( )k kk k k k t t k





secret sharing pair











S20

Page - 459


24

( ) ( ( ))j C j P jπ =

for 12 j n=


11 1 2 2( ) ( ) ( )( ( ))






1 2( )(0)t kΛΦ ε ε ε


1 2 3 4ε ε ε ε ( )2 1 2 3 4( ) 0ΛΦ ε ε ε ε 1 2 3 4ε ε ε ε ( )

2 1 2 3 4( ) 0ΛΦ ε ε ε ε 0000 1 0110 12 1000 2 0101 15 0100 3 0011 20 0010 4 1110 24 0001 5 1101 30 1100 6 1011 40 1010 8 0111 60 1001 10 1111 120



( )( ) ( )( ) ( )( )3 2 1 3 1 2

1 2 31 3 1 2 2 1 2 3 1 33 2




S20

Page - 460


25


( ) ( )( )( )

( )( )

2 1 2 2 1 1

2 1 2 1 2 1

2 1 2 1 2 1

2 1

2 1 2 1 2 1

1 1x x x x

x x x xx x x x

x x x x

ε ε

ε εε ε

ε εε ε

+ minus

+ minus minus=

+ minus

minus=

+

minusminus minus


minusminus minus



( )2 1 2 2 1 1

1 1 22xx xx

εε ε λ λ ε

minusminus minus

le+ minus minus

Observe that

( )( )( )( )

( )( )

( )( )( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )

3 23 2

3 1 2 1 3 1 2 1

3 23 2

3 1 2 1 3 1 2 1

3 2 3 2

3 1 2

3 2

3 1 2 1

3 2

3 2 3 2

3 1

3 2 3 2

3 1

1 3 1 2 1

3 2 3 2

3 1 2 1 3 1 2 13 1 2 1

3 1 2

1





x x x

ε εε ε ε ε

ε ε

ε ε ε εε ε






+ ++ minus + minus

+ +

+ + + ++ minus

+ + + ++ minus

=minus


( )( )( )

( ) ( ) ( )( )( )( ) ( ) ( )

3 2 3 2

3 2

3 1

3 2

2 31

3 2

2 1 3 1

3 1

3 1

3 2

3 1 2 1 2 12 1

1

1 1

1

x xx

x xx x x x x x

x xx x x x x x

ε ε ε ε

ε εε ε

ε εε ε ε ε

minus



+ +

+ ++ minus

+ ++ minus minus

S20

Page - 461


26

( )( )( )( )

( )( )

( ) ( ) ( )( )

3 2

3 1

3 23 2

3 1 2 1 3 1 2 12 1

222 2

2

2 2

2

1 (2 )

1 2( )

1 2 4

2 22 2

2 22 2 2


M M M

M M M

ε εε ε ε ε

ε εε ε ε ελ

ε ε εελ

ω ω ε

λ λ ε λ λ ε


λ

+ ++ minus + minus

minus minus

leminus


le + + + + +

⎡ ⎤+ + +⎛ ⎞ ⎛ ⎞+ +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤le

minus

⎣

minus

+ ⎦

Therefore

where 2

M εε

ωλ

+=

minus So we have

( )( )( )( )

( )( ) ( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( ) ( )

( )( )( )

3 23 21 1 1

3 1 2 1 3 1 2 1

3 23 21 1

3 1 2 1 3 1 2 1

3 2

3 1 2 1

3 2

3 1 2

3 2 3 21 1 1

3 1 2 1 3 1 2 1

33 2

3 1 2 1

1

3 2 3 2

3 1 2 1 3 1 2 1

3



x x x xy y

x x x x x x x x

xx xx x x x

ε εε ε ε ε

ε εε ε ε ε


δε

δ

ε

+ ++ minus + minus

+ ++ minus + minus

+ +



+ minus+ +



le minusminus

minus

+minus

( )( )( )

( )( )( )( )

( )

21

3 1 2 1

3 21

3 1 2 1

2

3 1 2 1

3 2

3 1 2 1

2 21 2 4

xy

x x x x

x xx x x x

M

εε ε ε ε

ε εε ε ε ε

δ

ω ω ε ω δλ

++ minus + minus

+ ++ minus + minus

minus minus

+minus minus

le + +



( )2 21(0) (0) 6 12 3L L Mω ω ε ω δλ

minus le + +

where 2

M εε

ωλ

+=

minus

S20

Page - 462


27

















S20

Page - 463


28


461 Speed Tests







2 4 6 8 10


0237 0471 0708 0935 1176

0

1

2

3

4

5

6

7

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB


1183 2349 358 473 5917

S20

Page - 464


29


462 Size Tests




987 1973 2960 3947 4934


9845 19502 29195 38791 46789


100 99 99 98 95

0

02

04

06

08

1

12

14

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 465


30



Reference




0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

File

Siz

e in

KB

Secret Size in MB

CompressedSecret

CompressedContainer

S20

Page - 466


31

5 Geometric Methods






2 2

2 21 1 1 12 22 2 2 22 23 3 3 3

11

( ) 011

x y x yx y x y

J x yx y x yx y x y

++

= =++




or

2 2 1312 14

11 11 11


+ minus + minus =





2 2 1312 14

11 11 11


+ minus + minus =

we have

S20

Page - 467


32

12

11

13

11

2 2 14

11

2

2

C

C

C C C

MxM

MyM

Mr x yM

=

=

= + +


( ) 1312

11 11

2 2C C

MMx yM M

⎛ ⎞= ⎜ ⎟⎝ ⎠

and radius 2 2 14

11C C

Mx yM


shareable





21

1 1 2

1

1 11 2 1 2( ) ( 2 2 )2 2k kC C k k k k

kkx y +

+++= + + + + + +ε ε ε ε ε ε

and

1 2 1 21 21 2 2k

k k k kC kr + + + += + + + +ε ε ε


33( )




Cr We define


for 12 j n=

S20

Page - 468

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar


hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4


Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4


Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched


spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b


our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s


ommon

s Award 201

3

with center

sharing pair


13

33

S20

Page - 469


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


13

[ ] ( ( ))ij j is sπ= Φ




( )( ) ( )( ) ( )( )( ) ( )( ) ( )( )( ) ( )( ) ( )

1

2

3

10 ( 2)( 1)

01 ( 1)( 3)

11 ( 2)( 3)

j j

j j

j j

s j j j

s j j j

s j j j

π π

π π

π π





( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

10 20 32 46 512[ ] 10 23 30 43 58

12 20 30 42 56s

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

1 2 3 4 5

10 20 32 46 512[ ] 10 [ ] 23 [ ] 30 [ ] 43 [ ] 58

12 20 30 42 56s s s s s

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦





[ ] [ ]1 212[ ] [ ] [ ][ ] i

jtts s ss p= =



S20

Page - 449


14

1 2( ) ( )i i ii ts p p pΦ = Ψ

and

11 2( )( )i i i


for 1 2 i M=


1 2

1 1 1 1 1 2 2 2 11 2 1 2 1 2( ( )) ( ( )) ( ( ))

M

n n nt t t

s s s s


=

=Φ Ψ Φ Ψ Φ Ψ






Mk=


[ ] ( ( ))ij j is sπ= Φ



Step 4 End





[ ] [ ] [ ]tt


S20

Page - 450


15


as in section 322


Step 5 END






[ ] [ ] ( )ij i js sσ

σ=







denoted by σ



S20

Page - 451


16



(0 8 6 7 2)(1 4 9)(3 5) (2 4 7 6 9)(0 8 5)(1 3) (0 3 7 9 2)(4 5 8)(1 6)

(1 6 9 4 5)(0 2 8)(3 7) (0 1 8 9 4)(2 5 6)(3 7) (0 2 8 4 9)(1 6 5)(3 7)

(0 8 9 5 4)(1 6 7)(2 3) (0 2 7 8 9)(1 4 3)(5 6) (0 9 5 4 7)(1 8 3)(2 6)

(0 2 9 5 3)(1 8 4)(6 7)

Example 334 Let

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

10 20 3 2 46 512[ ] 10 23 30 43 58

12 20 30 42 56s

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

32 10 20 512 46[ ] 23 30 10 43 58

12 20 30 56 42s σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

1 2 3 4 5

3 2 10 20 512 46[ ] 23 [ ] 30 [ ] 10 [ ] 43 [ ] 58

12 20 30 56 4 2s s s s sσ σ σ σ σ

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦





Mk=

S20

Page - 452


17


[ ] ( ( ))ij j is sπ= Φ




Step 5 End














S20

Page - 453


18





= many


Reference


S20

Page - 454


19

4 Algebraic Methods






1

1

( )

( )( )

t

iii j

j t

j iii j

x x

L xx x

=ne

=ne

minus

=minus

prod

prod


1( )

0j i

i jL x

i j=⎧

= ⎨ ne⎩


1 1 2 2 3 3( ) ( ) ( ) ( )L x y L x y L x y L x= + + +



1 1 2 2 3 3(0) (0) (0)y L y L y L+ + +

where

S20

Page - 455


20

1

1 1

1 1

(0 ) ( 1)

(0)( ) ( )

t tt

i ii ii j i j

j t t

j i j ii ii j i j

x x

Lx x x x

minus

= =ne ne

= =ne ne

minus minus

= =minus minus

prod prod

prod prod

for 1 23j =








1 2 1 21 2 1 2












11 21 2 2k




S20

Page - 456


21


2 3 11 2 2 2 2( ) k

kX p p p p p pminus

=

Assume that

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k



So

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k





Therefore 2 3 11 2 2 2 2 2( )

k k

X p p p p p ppminus




1 2 3 234579111316171923 X m m m= =

from now on


1 21 1 12 2 2 2 2 21 1 1 2 2 2

k k kr



1 2

1 1 2 21 1 1 r


1 2

1 1 2 21 1 1 r


such that

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr


1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr



1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr


+ + + + + + + + +ε ε εε ε ε ε ε ε

S20

Page - 457


22

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr




i ij j=ε ε







12 2i ik k

i k ixp pminus

lelt



12 331 2 4 221 2 2

i iii i i kki i


= εεε ε ε for







3 234X = 4 2345X = 5 23457X = 6 234579X =

7 23457911X = and 8 2345791113X = Also 1 2XΠ = 2 6XΠ =

3 24XΠ = and 4 120XΠ =

Note that





S20

Page - 458


23





1 21 2 1 11 2 1 1 1 2

1 1 1 2 1 11 2 1 2 1 2( )( ) ( )k kk k k k t t k





secret sharing pair











S20

Page - 459


24

( ) ( ( ))j C j P jπ =

for 12 j n=


11 1 2 2( ) ( ) ( )( ( ))






1 2( )(0)t kΛΦ ε ε ε


1 2 3 4ε ε ε ε ( )2 1 2 3 4( ) 0ΛΦ ε ε ε ε 1 2 3 4ε ε ε ε ( )

2 1 2 3 4( ) 0ΛΦ ε ε ε ε 0000 1 0110 12 1000 2 0101 15 0100 3 0011 20 0010 4 1110 24 0001 5 1101 30 1100 6 1011 40 1010 8 0111 60 1001 10 1111 120



( )( ) ( )( ) ( )( )3 2 1 3 1 2

1 2 31 3 1 2 2 1 2 3 1 33 2




S20

Page - 460


25


( ) ( )( )( )

( )( )

2 1 2 2 1 1

2 1 2 1 2 1

2 1 2 1 2 1

2 1

2 1 2 1 2 1

1 1x x x x

x x x xx x x x

x x x x

ε ε

ε εε ε

ε εε ε

+ minus

+ minus minus=

+ minus

minus=

+

minusminus minus


minusminus minus



( )2 1 2 2 1 1

1 1 22xx xx

εε ε λ λ ε

minusminus minus

le+ minus minus

Observe that

( )( )( )( )

( )( )

( )( )( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )

3 23 2

3 1 2 1 3 1 2 1

3 23 2

3 1 2 1 3 1 2 1

3 2 3 2

3 1 2

3 2

3 1 2 1

3 2

3 2 3 2

3 1

3 2 3 2

3 1

1 3 1 2 1

3 2 3 2

3 1 2 1 3 1 2 13 1 2 1

3 1 2

1





x x x

ε εε ε ε ε

ε ε

ε ε ε εε ε






+ ++ minus + minus

+ +

+ + + ++ minus

+ + + ++ minus

=minus


( )( )( )

( ) ( ) ( )( )( )( ) ( ) ( )

3 2 3 2

3 2

3 1

3 2

2 31

3 2

2 1 3 1

3 1

3 1

3 2

3 1 2 1 2 12 1

1

1 1

1

x xx

x xx x x x x x

x xx x x x x x

ε ε ε ε

ε εε ε

ε εε ε ε ε

minus



+ +

+ ++ minus

+ ++ minus minus

S20

Page - 461


26

( )( )( )( )

( )( )

( ) ( ) ( )( )

3 2

3 1

3 23 2

3 1 2 1 3 1 2 12 1

222 2

2

2 2

2

1 (2 )

1 2( )

1 2 4

2 22 2

2 22 2 2


M M M

M M M

ε εε ε ε ε

ε εε ε ε ελ

ε ε εελ

ω ω ε

λ λ ε λ λ ε


λ

+ ++ minus + minus

minus minus

leminus


le + + + + +

⎡ ⎤+ + +⎛ ⎞ ⎛ ⎞+ +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤le

minus

⎣

minus

+ ⎦

Therefore

where 2

M εε

ωλ

+=

minus So we have

( )( )( )( )

( )( ) ( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( ) ( )

( )( )( )

3 23 21 1 1

3 1 2 1 3 1 2 1

3 23 21 1

3 1 2 1 3 1 2 1

3 2

3 1 2 1

3 2

3 1 2

3 2 3 21 1 1

3 1 2 1 3 1 2 1

33 2

3 1 2 1

1

3 2 3 2

3 1 2 1 3 1 2 1

3



x x x xy y

x x x x x x x x

xx xx x x x

ε εε ε ε ε

ε εε ε ε ε


δε

δ

ε

+ ++ minus + minus

+ ++ minus + minus

+ +



+ minus+ +



le minusminus

minus

+minus

( )( )( )

( )( )( )( )

( )

21

3 1 2 1

3 21

3 1 2 1

2

3 1 2 1

3 2

3 1 2 1

2 21 2 4

xy

x x x x

x xx x x x

M

εε ε ε ε

ε εε ε ε ε

δ

ω ω ε ω δλ

++ minus + minus

+ ++ minus + minus

minus minus

+minus minus

le + +



( )2 21(0) (0) 6 12 3L L Mω ω ε ω δλ

minus le + +

where 2

M εε

ωλ

+=

minus

S20

Page - 462


27

















S20

Page - 463


28


461 Speed Tests







2 4 6 8 10


0237 0471 0708 0935 1176

0

1

2

3

4

5

6

7

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB


1183 2349 358 473 5917

S20

Page - 464


29


462 Size Tests




987 1973 2960 3947 4934


9845 19502 29195 38791 46789


100 99 99 98 95

0

02

04

06

08

1

12

14

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 465


30



Reference




0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

File

Siz

e in

KB

Secret Size in MB

CompressedSecret

CompressedContainer

S20

Page - 466


31

5 Geometric Methods






2 2

2 21 1 1 12 22 2 2 22 23 3 3 3

11

( ) 011

x y x yx y x y

J x yx y x yx y x y

++

= =++




or

2 2 1312 14

11 11 11


+ minus + minus =





2 2 1312 14

11 11 11


+ minus + minus =

we have

S20

Page - 467


32

12

11

13

11

2 2 14

11

2

2

C

C

C C C

MxM

MyM

Mr x yM

=

=

= + +


( ) 1312

11 11

2 2C C

MMx yM M

⎛ ⎞= ⎜ ⎟⎝ ⎠

and radius 2 2 14

11C C

Mx yM


shareable





21

1 1 2

1

1 11 2 1 2( ) ( 2 2 )2 2k kC C k k k k

kkx y +

+++= + + + + + +ε ε ε ε ε ε

and

1 2 1 21 21 2 2k

k k k kC kr + + + += + + + +ε ε ε


33( )




Cr We define


for 12 j n=

S20

Page - 468

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar


hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4


Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4


Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched


spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b


our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s


ommon

s Award 201

3

with center

sharing pair


13

33

S20

Page - 469


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


14

1 2( ) ( )i i ii ts p p pΦ = Ψ

and

11 2( )( )i i i


for 1 2 i M=


1 2

1 1 1 1 1 2 2 2 11 2 1 2 1 2( ( )) ( ( )) ( ( ))

M

n n nt t t

s s s s


=

=Φ Ψ Φ Ψ Φ Ψ






Mk=


[ ] ( ( ))ij j is sπ= Φ



Step 4 End





[ ] [ ] [ ]tt


S20

Page - 450


15


as in section 322


Step 5 END






[ ] [ ] ( )ij i js sσ

σ=







denoted by σ



S20

Page - 451


16



(0 8 6 7 2)(1 4 9)(3 5) (2 4 7 6 9)(0 8 5)(1 3) (0 3 7 9 2)(4 5 8)(1 6)

(1 6 9 4 5)(0 2 8)(3 7) (0 1 8 9 4)(2 5 6)(3 7) (0 2 8 4 9)(1 6 5)(3 7)

(0 8 9 5 4)(1 6 7)(2 3) (0 2 7 8 9)(1 4 3)(5 6) (0 9 5 4 7)(1 8 3)(2 6)

(0 2 9 5 3)(1 8 4)(6 7)

Example 334 Let

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

10 20 3 2 46 512[ ] 10 23 30 43 58

12 20 30 42 56s

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

32 10 20 512 46[ ] 23 30 10 43 58

12 20 30 56 42s σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

1 2 3 4 5

3 2 10 20 512 46[ ] 23 [ ] 30 [ ] 10 [ ] 43 [ ] 58

12 20 30 56 4 2s s s s sσ σ σ σ σ

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦





Mk=

S20

Page - 452


17


[ ] ( ( ))ij j is sπ= Φ




Step 5 End














S20

Page - 453


18





= many


Reference


S20

Page - 454


19

4 Algebraic Methods






1

1

( )

( )( )

t

iii j

j t

j iii j

x x

L xx x

=ne

=ne

minus

=minus

prod

prod


1( )

0j i

i jL x

i j=⎧

= ⎨ ne⎩


1 1 2 2 3 3( ) ( ) ( ) ( )L x y L x y L x y L x= + + +



1 1 2 2 3 3(0) (0) (0)y L y L y L+ + +

where

S20

Page - 455


20

1

1 1

1 1

(0 ) ( 1)

(0)( ) ( )

t tt

i ii ii j i j

j t t

j i j ii ii j i j

x x

Lx x x x

minus

= =ne ne

= =ne ne

minus minus

= =minus minus

prod prod

prod prod

for 1 23j =








1 2 1 21 2 1 2












11 21 2 2k




S20

Page - 456


21


2 3 11 2 2 2 2( ) k

kX p p p p p pminus

=

Assume that

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k



So

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k





Therefore 2 3 11 2 2 2 2 2( )

k k

X p p p p p ppminus




1 2 3 234579111316171923 X m m m= =

from now on


1 21 1 12 2 2 2 2 21 1 1 2 2 2

k k kr



1 2

1 1 2 21 1 1 r


1 2

1 1 2 21 1 1 r


such that

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr


1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr



1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr


+ + + + + + + + +ε ε εε ε ε ε ε ε

S20

Page - 457


22

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr




i ij j=ε ε







12 2i ik k

i k ixp pminus

lelt



12 331 2 4 221 2 2

i iii i i kki i


= εεε ε ε for







3 234X = 4 2345X = 5 23457X = 6 234579X =

7 23457911X = and 8 2345791113X = Also 1 2XΠ = 2 6XΠ =

3 24XΠ = and 4 120XΠ =

Note that





S20

Page - 458


23





1 21 2 1 11 2 1 1 1 2

1 1 1 2 1 11 2 1 2 1 2( )( ) ( )k kk k k k t t k





secret sharing pair











S20

Page - 459


24

( ) ( ( ))j C j P jπ =

for 12 j n=


11 1 2 2( ) ( ) ( )( ( ))






1 2( )(0)t kΛΦ ε ε ε


1 2 3 4ε ε ε ε ( )2 1 2 3 4( ) 0ΛΦ ε ε ε ε 1 2 3 4ε ε ε ε ( )

2 1 2 3 4( ) 0ΛΦ ε ε ε ε 0000 1 0110 12 1000 2 0101 15 0100 3 0011 20 0010 4 1110 24 0001 5 1101 30 1100 6 1011 40 1010 8 0111 60 1001 10 1111 120



( )( ) ( )( ) ( )( )3 2 1 3 1 2

1 2 31 3 1 2 2 1 2 3 1 33 2




S20

Page - 460


25


( ) ( )( )( )

( )( )

2 1 2 2 1 1

2 1 2 1 2 1

2 1 2 1 2 1

2 1

2 1 2 1 2 1

1 1x x x x

x x x xx x x x

x x x x

ε ε

ε εε ε

ε εε ε

+ minus

+ minus minus=

+ minus

minus=

+

minusminus minus


minusminus minus



( )2 1 2 2 1 1

1 1 22xx xx

εε ε λ λ ε

minusminus minus

le+ minus minus

Observe that

( )( )( )( )

( )( )

( )( )( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )

3 23 2

3 1 2 1 3 1 2 1

3 23 2

3 1 2 1 3 1 2 1

3 2 3 2

3 1 2

3 2

3 1 2 1

3 2

3 2 3 2

3 1

3 2 3 2

3 1

1 3 1 2 1

3 2 3 2

3 1 2 1 3 1 2 13 1 2 1

3 1 2

1





x x x

ε εε ε ε ε

ε ε

ε ε ε εε ε






+ ++ minus + minus

+ +

+ + + ++ minus

+ + + ++ minus

=minus


( )( )( )

( ) ( ) ( )( )( )( ) ( ) ( )

3 2 3 2

3 2

3 1

3 2

2 31

3 2

2 1 3 1

3 1

3 1

3 2

3 1 2 1 2 12 1

1

1 1

1

x xx

x xx x x x x x

x xx x x x x x

ε ε ε ε

ε εε ε

ε εε ε ε ε

minus



+ +

+ ++ minus

+ ++ minus minus

S20

Page - 461


26

( )( )( )( )

( )( )

( ) ( ) ( )( )

3 2

3 1

3 23 2

3 1 2 1 3 1 2 12 1

222 2

2

2 2

2

1 (2 )

1 2( )

1 2 4

2 22 2

2 22 2 2


M M M

M M M

ε εε ε ε ε

ε εε ε ε ελ

ε ε εελ

ω ω ε

λ λ ε λ λ ε


λ

+ ++ minus + minus

minus minus

leminus


le + + + + +

⎡ ⎤+ + +⎛ ⎞ ⎛ ⎞+ +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤le

minus

⎣

minus

+ ⎦

Therefore

where 2

M εε

ωλ

+=

minus So we have

( )( )( )( )

( )( ) ( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( ) ( )

( )( )( )

3 23 21 1 1

3 1 2 1 3 1 2 1

3 23 21 1

3 1 2 1 3 1 2 1

3 2

3 1 2 1

3 2

3 1 2

3 2 3 21 1 1

3 1 2 1 3 1 2 1

33 2

3 1 2 1

1

3 2 3 2

3 1 2 1 3 1 2 1

3



x x x xy y

x x x x x x x x

xx xx x x x

ε εε ε ε ε

ε εε ε ε ε


δε

δ

ε

+ ++ minus + minus

+ ++ minus + minus

+ +



+ minus+ +



le minusminus

minus

+minus

( )( )( )

( )( )( )( )

( )

21

3 1 2 1

3 21

3 1 2 1

2

3 1 2 1

3 2

3 1 2 1

2 21 2 4

xy

x x x x

x xx x x x

M

εε ε ε ε

ε εε ε ε ε

δ

ω ω ε ω δλ

++ minus + minus

+ ++ minus + minus

minus minus

+minus minus

le + +



( )2 21(0) (0) 6 12 3L L Mω ω ε ω δλ

minus le + +

where 2

M εε

ωλ

+=

minus

S20

Page - 462


27

















S20

Page - 463


28


461 Speed Tests







2 4 6 8 10


0237 0471 0708 0935 1176

0

1

2

3

4

5

6

7

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB


1183 2349 358 473 5917

S20

Page - 464


29


462 Size Tests




987 1973 2960 3947 4934


9845 19502 29195 38791 46789


100 99 99 98 95

0

02

04

06

08

1

12

14

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 465


30



Reference




0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

File

Siz

e in

KB

Secret Size in MB

CompressedSecret

CompressedContainer

S20

Page - 466


31

5 Geometric Methods






2 2

2 21 1 1 12 22 2 2 22 23 3 3 3

11

( ) 011

x y x yx y x y

J x yx y x yx y x y

++

= =++




or

2 2 1312 14

11 11 11


+ minus + minus =





2 2 1312 14

11 11 11


+ minus + minus =

we have

S20

Page - 467


32

12

11

13

11

2 2 14

11

2

2

C

C

C C C

MxM

MyM

Mr x yM

=

=

= + +


( ) 1312

11 11

2 2C C

MMx yM M

⎛ ⎞= ⎜ ⎟⎝ ⎠

and radius 2 2 14

11C C

Mx yM


shareable





21

1 1 2

1

1 11 2 1 2( ) ( 2 2 )2 2k kC C k k k k

kkx y +

+++= + + + + + +ε ε ε ε ε ε

and

1 2 1 21 21 2 2k

k k k kC kr + + + += + + + +ε ε ε


33( )




Cr We define


for 12 j n=

S20

Page - 468

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar


hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4


Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4


Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched


spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b


our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s


ommon

s Award 201

3

with center

sharing pair


13

33

S20

Page - 469


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


15


as in section 322


Step 5 END






[ ] [ ] ( )ij i js sσ

σ=







denoted by σ



S20

Page - 451


16



(0 8 6 7 2)(1 4 9)(3 5) (2 4 7 6 9)(0 8 5)(1 3) (0 3 7 9 2)(4 5 8)(1 6)

(1 6 9 4 5)(0 2 8)(3 7) (0 1 8 9 4)(2 5 6)(3 7) (0 2 8 4 9)(1 6 5)(3 7)

(0 8 9 5 4)(1 6 7)(2 3) (0 2 7 8 9)(1 4 3)(5 6) (0 9 5 4 7)(1 8 3)(2 6)

(0 2 9 5 3)(1 8 4)(6 7)

Example 334 Let

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

10 20 3 2 46 512[ ] 10 23 30 43 58

12 20 30 42 56s

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

32 10 20 512 46[ ] 23 30 10 43 58

12 20 30 56 42s σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

1 2 3 4 5

3 2 10 20 512 46[ ] 23 [ ] 30 [ ] 10 [ ] 43 [ ] 58

12 20 30 56 4 2s s s s sσ σ σ σ σ

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦





Mk=

S20

Page - 452


17


[ ] ( ( ))ij j is sπ= Φ




Step 5 End














S20

Page - 453


18





= many


Reference


S20

Page - 454


19

4 Algebraic Methods






1

1

( )

( )( )

t

iii j

j t

j iii j

x x

L xx x

=ne

=ne

minus

=minus

prod

prod


1( )

0j i

i jL x

i j=⎧

= ⎨ ne⎩


1 1 2 2 3 3( ) ( ) ( ) ( )L x y L x y L x y L x= + + +



1 1 2 2 3 3(0) (0) (0)y L y L y L+ + +

where

S20

Page - 455


20

1

1 1

1 1

(0 ) ( 1)

(0)( ) ( )

t tt

i ii ii j i j

j t t

j i j ii ii j i j

x x

Lx x x x

minus

= =ne ne

= =ne ne

minus minus

= =minus minus

prod prod

prod prod

for 1 23j =








1 2 1 21 2 1 2












11 21 2 2k




S20

Page - 456


21


2 3 11 2 2 2 2( ) k

kX p p p p p pminus

=

Assume that

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k



So

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k





Therefore 2 3 11 2 2 2 2 2( )

k k

X p p p p p ppminus




1 2 3 234579111316171923 X m m m= =

from now on


1 21 1 12 2 2 2 2 21 1 1 2 2 2

k k kr



1 2

1 1 2 21 1 1 r


1 2

1 1 2 21 1 1 r


such that

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr


1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr



1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr


+ + + + + + + + +ε ε εε ε ε ε ε ε

S20

Page - 457


22

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr




i ij j=ε ε







12 2i ik k

i k ixp pminus

lelt



12 331 2 4 221 2 2

i iii i i kki i


= εεε ε ε for







3 234X = 4 2345X = 5 23457X = 6 234579X =

7 23457911X = and 8 2345791113X = Also 1 2XΠ = 2 6XΠ =

3 24XΠ = and 4 120XΠ =

Note that





S20

Page - 458


23





1 21 2 1 11 2 1 1 1 2

1 1 1 2 1 11 2 1 2 1 2( )( ) ( )k kk k k k t t k





secret sharing pair











S20

Page - 459


24

( ) ( ( ))j C j P jπ =

for 12 j n=


11 1 2 2( ) ( ) ( )( ( ))






1 2( )(0)t kΛΦ ε ε ε


1 2 3 4ε ε ε ε ( )2 1 2 3 4( ) 0ΛΦ ε ε ε ε 1 2 3 4ε ε ε ε ( )

2 1 2 3 4( ) 0ΛΦ ε ε ε ε 0000 1 0110 12 1000 2 0101 15 0100 3 0011 20 0010 4 1110 24 0001 5 1101 30 1100 6 1011 40 1010 8 0111 60 1001 10 1111 120



( )( ) ( )( ) ( )( )3 2 1 3 1 2

1 2 31 3 1 2 2 1 2 3 1 33 2




S20

Page - 460


25


( ) ( )( )( )

( )( )

2 1 2 2 1 1

2 1 2 1 2 1

2 1 2 1 2 1

2 1

2 1 2 1 2 1

1 1x x x x

x x x xx x x x

x x x x

ε ε

ε εε ε

ε εε ε

+ minus

+ minus minus=

+ minus

minus=

+

minusminus minus


minusminus minus



( )2 1 2 2 1 1

1 1 22xx xx

εε ε λ λ ε

minusminus minus

le+ minus minus

Observe that

( )( )( )( )

( )( )

( )( )( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )

3 23 2

3 1 2 1 3 1 2 1

3 23 2

3 1 2 1 3 1 2 1

3 2 3 2

3 1 2

3 2

3 1 2 1

3 2

3 2 3 2

3 1

3 2 3 2

3 1

1 3 1 2 1

3 2 3 2

3 1 2 1 3 1 2 13 1 2 1

3 1 2

1





x x x

ε εε ε ε ε

ε ε

ε ε ε εε ε






+ ++ minus + minus

+ +

+ + + ++ minus

+ + + ++ minus

=minus


( )( )( )

( ) ( ) ( )( )( )( ) ( ) ( )

3 2 3 2

3 2

3 1

3 2

2 31

3 2

2 1 3 1

3 1

3 1

3 2

3 1 2 1 2 12 1

1

1 1

1

x xx

x xx x x x x x

x xx x x x x x

ε ε ε ε

ε εε ε

ε εε ε ε ε

minus



+ +

+ ++ minus

+ ++ minus minus

S20

Page - 461


26

( )( )( )( )

( )( )

( ) ( ) ( )( )

3 2

3 1

3 23 2

3 1 2 1 3 1 2 12 1

222 2

2

2 2

2

1 (2 )

1 2( )

1 2 4

2 22 2

2 22 2 2


M M M

M M M

ε εε ε ε ε

ε εε ε ε ελ

ε ε εελ

ω ω ε

λ λ ε λ λ ε


λ

+ ++ minus + minus

minus minus

leminus


le + + + + +

⎡ ⎤+ + +⎛ ⎞ ⎛ ⎞+ +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤le

minus

⎣

minus

+ ⎦

Therefore

where 2

M εε

ωλ

+=

minus So we have

( )( )( )( )

( )( ) ( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( ) ( )

( )( )( )

3 23 21 1 1

3 1 2 1 3 1 2 1

3 23 21 1

3 1 2 1 3 1 2 1

3 2

3 1 2 1

3 2

3 1 2

3 2 3 21 1 1

3 1 2 1 3 1 2 1

33 2

3 1 2 1

1

3 2 3 2

3 1 2 1 3 1 2 1

3



x x x xy y

x x x x x x x x

xx xx x x x

ε εε ε ε ε

ε εε ε ε ε


δε

δ

ε

+ ++ minus + minus

+ ++ minus + minus

+ +



+ minus+ +



le minusminus

minus

+minus

( )( )( )

( )( )( )( )

( )

21

3 1 2 1

3 21

3 1 2 1

2

3 1 2 1

3 2

3 1 2 1

2 21 2 4

xy

x x x x

x xx x x x

M

εε ε ε ε

ε εε ε ε ε

δ

ω ω ε ω δλ

++ minus + minus

+ ++ minus + minus

minus minus

+minus minus

le + +



( )2 21(0) (0) 6 12 3L L Mω ω ε ω δλ

minus le + +

where 2

M εε

ωλ

+=

minus

S20

Page - 462


27

















S20

Page - 463


28


461 Speed Tests







2 4 6 8 10


0237 0471 0708 0935 1176

0

1

2

3

4

5

6

7

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB


1183 2349 358 473 5917

S20

Page - 464


29


462 Size Tests




987 1973 2960 3947 4934


9845 19502 29195 38791 46789


100 99 99 98 95

0

02

04

06

08

1

12

14

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 465


30



Reference




0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

File

Siz

e in

KB

Secret Size in MB

CompressedSecret

CompressedContainer

S20

Page - 466


31

5 Geometric Methods






2 2

2 21 1 1 12 22 2 2 22 23 3 3 3

11

( ) 011

x y x yx y x y

J x yx y x yx y x y

++

= =++




or

2 2 1312 14

11 11 11


+ minus + minus =





2 2 1312 14

11 11 11


+ minus + minus =

we have

S20

Page - 467


32

12

11

13

11

2 2 14

11

2

2

C

C

C C C

MxM

MyM

Mr x yM

=

=

= + +


( ) 1312

11 11

2 2C C

MMx yM M

⎛ ⎞= ⎜ ⎟⎝ ⎠

and radius 2 2 14

11C C

Mx yM


shareable





21

1 1 2

1

1 11 2 1 2( ) ( 2 2 )2 2k kC C k k k k

kkx y +

+++= + + + + + +ε ε ε ε ε ε

and

1 2 1 21 21 2 2k

k k k kC kr + + + += + + + +ε ε ε


33( )




Cr We define


for 12 j n=

S20

Page - 468

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar


hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4


Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4


Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched


spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b


our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s


ommon

s Award 201

3

with center

sharing pair


13

33

S20

Page - 469


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


16



(0 8 6 7 2)(1 4 9)(3 5) (2 4 7 6 9)(0 8 5)(1 3) (0 3 7 9 2)(4 5 8)(1 6)

(1 6 9 4 5)(0 2 8)(3 7) (0 1 8 9 4)(2 5 6)(3 7) (0 2 8 4 9)(1 6 5)(3 7)

(0 8 9 5 4)(1 6 7)(2 3) (0 2 7 8 9)(1 4 3)(5 6) (0 9 5 4 7)(1 8 3)(2 6)

(0 2 9 5 3)(1 8 4)(6 7)

Example 334 Let

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

10 20 3 2 46 512[ ] 10 23 30 43 58

12 20 30 42 56s

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

32 10 20 512 46[ ] 23 30 10 43 58

12 20 30 56 42s σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

1 2 3 4 5

3 2 10 20 512 46[ ] 23 [ ] 30 [ ] 10 [ ] 43 [ ] 58

12 20 30 56 4 2s s s s sσ σ σ σ σ

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦





Mk=

S20

Page - 452


17


[ ] ( ( ))ij j is sπ= Φ




Step 5 End














S20

Page - 453


18





= many


Reference


S20

Page - 454


19

4 Algebraic Methods






1

1

( )

( )( )

t

iii j

j t

j iii j

x x

L xx x

=ne

=ne

minus

=minus

prod

prod


1( )

0j i

i jL x

i j=⎧

= ⎨ ne⎩


1 1 2 2 3 3( ) ( ) ( ) ( )L x y L x y L x y L x= + + +



1 1 2 2 3 3(0) (0) (0)y L y L y L+ + +

where

S20

Page - 455


20

1

1 1

1 1

(0 ) ( 1)

(0)( ) ( )

t tt

i ii ii j i j

j t t

j i j ii ii j i j

x x

Lx x x x

minus

= =ne ne

= =ne ne

minus minus

= =minus minus

prod prod

prod prod

for 1 23j =








1 2 1 21 2 1 2












11 21 2 2k




S20

Page - 456


21


2 3 11 2 2 2 2( ) k

kX p p p p p pminus

=

Assume that

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k



So

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k





Therefore 2 3 11 2 2 2 2 2( )

k k

X p p p p p ppminus




1 2 3 234579111316171923 X m m m= =

from now on


1 21 1 12 2 2 2 2 21 1 1 2 2 2

k k kr



1 2

1 1 2 21 1 1 r


1 2

1 1 2 21 1 1 r


such that

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr


1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr



1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr


+ + + + + + + + +ε ε εε ε ε ε ε ε

S20

Page - 457


22

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr




i ij j=ε ε







12 2i ik k

i k ixp pminus

lelt



12 331 2 4 221 2 2

i iii i i kki i


= εεε ε ε for







3 234X = 4 2345X = 5 23457X = 6 234579X =

7 23457911X = and 8 2345791113X = Also 1 2XΠ = 2 6XΠ =

3 24XΠ = and 4 120XΠ =

Note that





S20

Page - 458


23





1 21 2 1 11 2 1 1 1 2

1 1 1 2 1 11 2 1 2 1 2( )( ) ( )k kk k k k t t k





secret sharing pair











S20

Page - 459


24

( ) ( ( ))j C j P jπ =

for 12 j n=


11 1 2 2( ) ( ) ( )( ( ))






1 2( )(0)t kΛΦ ε ε ε


1 2 3 4ε ε ε ε ( )2 1 2 3 4( ) 0ΛΦ ε ε ε ε 1 2 3 4ε ε ε ε ( )

2 1 2 3 4( ) 0ΛΦ ε ε ε ε 0000 1 0110 12 1000 2 0101 15 0100 3 0011 20 0010 4 1110 24 0001 5 1101 30 1100 6 1011 40 1010 8 0111 60 1001 10 1111 120



( )( ) ( )( ) ( )( )3 2 1 3 1 2

1 2 31 3 1 2 2 1 2 3 1 33 2




S20

Page - 460


25


( ) ( )( )( )

( )( )

2 1 2 2 1 1

2 1 2 1 2 1

2 1 2 1 2 1

2 1

2 1 2 1 2 1

1 1x x x x

x x x xx x x x

x x x x

ε ε

ε εε ε

ε εε ε

+ minus

+ minus minus=

+ minus

minus=

+

minusminus minus


minusminus minus



( )2 1 2 2 1 1

1 1 22xx xx

εε ε λ λ ε

minusminus minus

le+ minus minus

Observe that

( )( )( )( )

( )( )

( )( )( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )

3 23 2

3 1 2 1 3 1 2 1

3 23 2

3 1 2 1 3 1 2 1

3 2 3 2

3 1 2

3 2

3 1 2 1

3 2

3 2 3 2

3 1

3 2 3 2

3 1

1 3 1 2 1

3 2 3 2

3 1 2 1 3 1 2 13 1 2 1

3 1 2

1





x x x

ε εε ε ε ε

ε ε

ε ε ε εε ε






+ ++ minus + minus

+ +

+ + + ++ minus

+ + + ++ minus

=minus


( )( )( )

( ) ( ) ( )( )( )( ) ( ) ( )

3 2 3 2

3 2

3 1

3 2

2 31

3 2

2 1 3 1

3 1

3 1

3 2

3 1 2 1 2 12 1

1

1 1

1

x xx

x xx x x x x x

x xx x x x x x

ε ε ε ε

ε εε ε

ε εε ε ε ε

minus



+ +

+ ++ minus

+ ++ minus minus

S20

Page - 461


26

( )( )( )( )

( )( )

( ) ( ) ( )( )

3 2

3 1

3 23 2

3 1 2 1 3 1 2 12 1

222 2

2

2 2

2

1 (2 )

1 2( )

1 2 4

2 22 2

2 22 2 2


M M M

M M M

ε εε ε ε ε

ε εε ε ε ελ

ε ε εελ

ω ω ε

λ λ ε λ λ ε


λ

+ ++ minus + minus

minus minus

leminus


le + + + + +

⎡ ⎤+ + +⎛ ⎞ ⎛ ⎞+ +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤le

minus

⎣

minus

+ ⎦

Therefore

where 2

M εε

ωλ

+=

minus So we have

( )( )( )( )

( )( ) ( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( ) ( )

( )( )( )

3 23 21 1 1

3 1 2 1 3 1 2 1

3 23 21 1

3 1 2 1 3 1 2 1

3 2

3 1 2 1

3 2

3 1 2

3 2 3 21 1 1

3 1 2 1 3 1 2 1

33 2

3 1 2 1

1

3 2 3 2

3 1 2 1 3 1 2 1

3



x x x xy y

x x x x x x x x

xx xx x x x

ε εε ε ε ε

ε εε ε ε ε


δε

δ

ε

+ ++ minus + minus

+ ++ minus + minus

+ +



+ minus+ +



le minusminus

minus

+minus

( )( )( )

( )( )( )( )

( )

21

3 1 2 1

3 21

3 1 2 1

2

3 1 2 1

3 2

3 1 2 1

2 21 2 4

xy

x x x x

x xx x x x

M

εε ε ε ε

ε εε ε ε ε

δ

ω ω ε ω δλ

++ minus + minus

+ ++ minus + minus

minus minus

+minus minus

le + +



( )2 21(0) (0) 6 12 3L L Mω ω ε ω δλ

minus le + +

where 2

M εε

ωλ

+=

minus

S20

Page - 462


27

















S20

Page - 463


28


461 Speed Tests







2 4 6 8 10


0237 0471 0708 0935 1176

0

1

2

3

4

5

6

7

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB


1183 2349 358 473 5917

S20

Page - 464


29


462 Size Tests




987 1973 2960 3947 4934


9845 19502 29195 38791 46789


100 99 99 98 95

0

02

04

06

08

1

12

14

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 465


30



Reference




0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

File

Siz

e in

KB

Secret Size in MB

CompressedSecret

CompressedContainer

S20

Page - 466


31

5 Geometric Methods






2 2

2 21 1 1 12 22 2 2 22 23 3 3 3

11

( ) 011

x y x yx y x y

J x yx y x yx y x y

++

= =++




or

2 2 1312 14

11 11 11


+ minus + minus =





2 2 1312 14

11 11 11


+ minus + minus =

we have

S20

Page - 467


32

12

11

13

11

2 2 14

11

2

2

C

C

C C C

MxM

MyM

Mr x yM

=

=

= + +


( ) 1312

11 11

2 2C C

MMx yM M

⎛ ⎞= ⎜ ⎟⎝ ⎠

and radius 2 2 14

11C C

Mx yM


shareable





21

1 1 2

1

1 11 2 1 2( ) ( 2 2 )2 2k kC C k k k k

kkx y +

+++= + + + + + +ε ε ε ε ε ε

and

1 2 1 21 21 2 2k

k k k kC kr + + + += + + + +ε ε ε


33( )




Cr We define


for 12 j n=

S20

Page - 468

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar


hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4


Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4


Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched


spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b


our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s


ommon

s Award 201

3

with center

sharing pair


13

33

S20

Page - 469


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


17


[ ] ( ( ))ij j is sπ= Φ




Step 5 End














S20

Page - 453


18





= many


Reference


S20

Page - 454


19

4 Algebraic Methods






1

1

( )

( )( )

t

iii j

j t

j iii j

x x

L xx x

=ne

=ne

minus

=minus

prod

prod


1( )

0j i

i jL x

i j=⎧

= ⎨ ne⎩


1 1 2 2 3 3( ) ( ) ( ) ( )L x y L x y L x y L x= + + +



1 1 2 2 3 3(0) (0) (0)y L y L y L+ + +

where

S20

Page - 455


20

1

1 1

1 1

(0 ) ( 1)

(0)( ) ( )

t tt

i ii ii j i j

j t t

j i j ii ii j i j

x x

Lx x x x

minus

= =ne ne

= =ne ne

minus minus

= =minus minus

prod prod

prod prod

for 1 23j =








1 2 1 21 2 1 2












11 21 2 2k




S20

Page - 456


21


2 3 11 2 2 2 2( ) k

kX p p p p p pminus

=

Assume that

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k



So

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k





Therefore 2 3 11 2 2 2 2 2( )

k k

X p p p p p ppminus




1 2 3 234579111316171923 X m m m= =

from now on


1 21 1 12 2 2 2 2 21 1 1 2 2 2

k k kr



1 2

1 1 2 21 1 1 r


1 2

1 1 2 21 1 1 r


such that

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr


1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr



1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr


+ + + + + + + + +ε ε εε ε ε ε ε ε

S20

Page - 457


22

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr




i ij j=ε ε







12 2i ik k

i k ixp pminus

lelt



12 331 2 4 221 2 2

i iii i i kki i


= εεε ε ε for







3 234X = 4 2345X = 5 23457X = 6 234579X =

7 23457911X = and 8 2345791113X = Also 1 2XΠ = 2 6XΠ =

3 24XΠ = and 4 120XΠ =

Note that





S20

Page - 458


23





1 21 2 1 11 2 1 1 1 2

1 1 1 2 1 11 2 1 2 1 2( )( ) ( )k kk k k k t t k





secret sharing pair











S20

Page - 459


24

( ) ( ( ))j C j P jπ =

for 12 j n=


11 1 2 2( ) ( ) ( )( ( ))






1 2( )(0)t kΛΦ ε ε ε


1 2 3 4ε ε ε ε ( )2 1 2 3 4( ) 0ΛΦ ε ε ε ε 1 2 3 4ε ε ε ε ( )

2 1 2 3 4( ) 0ΛΦ ε ε ε ε 0000 1 0110 12 1000 2 0101 15 0100 3 0011 20 0010 4 1110 24 0001 5 1101 30 1100 6 1011 40 1010 8 0111 60 1001 10 1111 120



( )( ) ( )( ) ( )( )3 2 1 3 1 2

1 2 31 3 1 2 2 1 2 3 1 33 2




S20

Page - 460


25


( ) ( )( )( )

( )( )

2 1 2 2 1 1

2 1 2 1 2 1

2 1 2 1 2 1

2 1

2 1 2 1 2 1

1 1x x x x

x x x xx x x x

x x x x

ε ε

ε εε ε

ε εε ε

+ minus

+ minus minus=

+ minus

minus=

+

minusminus minus


minusminus minus



( )2 1 2 2 1 1

1 1 22xx xx

εε ε λ λ ε

minusminus minus

le+ minus minus

Observe that

( )( )( )( )

( )( )

( )( )( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )

3 23 2

3 1 2 1 3 1 2 1

3 23 2

3 1 2 1 3 1 2 1

3 2 3 2

3 1 2

3 2

3 1 2 1

3 2

3 2 3 2

3 1

3 2 3 2

3 1

1 3 1 2 1

3 2 3 2

3 1 2 1 3 1 2 13 1 2 1

3 1 2

1





x x x

ε εε ε ε ε

ε ε

ε ε ε εε ε






+ ++ minus + minus

+ +

+ + + ++ minus

+ + + ++ minus

=minus


( )( )( )

( ) ( ) ( )( )( )( ) ( ) ( )

3 2 3 2

3 2

3 1

3 2

2 31

3 2

2 1 3 1

3 1

3 1

3 2

3 1 2 1 2 12 1

1

1 1

1

x xx

x xx x x x x x

x xx x x x x x

ε ε ε ε

ε εε ε

ε εε ε ε ε

minus



+ +

+ ++ minus

+ ++ minus minus

S20

Page - 461


26

( )( )( )( )

( )( )

( ) ( ) ( )( )

3 2

3 1

3 23 2

3 1 2 1 3 1 2 12 1

222 2

2

2 2

2

1 (2 )

1 2( )

1 2 4

2 22 2

2 22 2 2


M M M

M M M

ε εε ε ε ε

ε εε ε ε ελ

ε ε εελ

ω ω ε

λ λ ε λ λ ε


λ

+ ++ minus + minus

minus minus

leminus


le + + + + +

⎡ ⎤+ + +⎛ ⎞ ⎛ ⎞+ +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤le

minus

⎣

minus

+ ⎦

Therefore

where 2

M εε

ωλ

+=

minus So we have

( )( )( )( )

( )( ) ( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( ) ( )

( )( )( )

3 23 21 1 1

3 1 2 1 3 1 2 1

3 23 21 1

3 1 2 1 3 1 2 1

3 2

3 1 2 1

3 2

3 1 2

3 2 3 21 1 1

3 1 2 1 3 1 2 1

33 2

3 1 2 1

1

3 2 3 2

3 1 2 1 3 1 2 1

3



x x x xy y

x x x x x x x x

xx xx x x x

ε εε ε ε ε

ε εε ε ε ε


δε

δ

ε

+ ++ minus + minus

+ ++ minus + minus

+ +



+ minus+ +



le minusminus

minus

+minus

( )( )( )

( )( )( )( )

( )

21

3 1 2 1

3 21

3 1 2 1

2

3 1 2 1

3 2

3 1 2 1

2 21 2 4

xy

x x x x

x xx x x x

M

εε ε ε ε

ε εε ε ε ε

δ

ω ω ε ω δλ

++ minus + minus

+ ++ minus + minus

minus minus

+minus minus

le + +



( )2 21(0) (0) 6 12 3L L Mω ω ε ω δλ

minus le + +

where 2

M εε

ωλ

+=

minus

S20

Page - 462


27

















S20

Page - 463


28


461 Speed Tests







2 4 6 8 10


0237 0471 0708 0935 1176

0

1

2

3

4

5

6

7

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB


1183 2349 358 473 5917

S20

Page - 464


29


462 Size Tests




987 1973 2960 3947 4934


9845 19502 29195 38791 46789


100 99 99 98 95

0

02

04

06

08

1

12

14

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 465


30



Reference




0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

File

Siz

e in

KB

Secret Size in MB

CompressedSecret

CompressedContainer

S20

Page - 466


31

5 Geometric Methods






2 2

2 21 1 1 12 22 2 2 22 23 3 3 3

11

( ) 011

x y x yx y x y

J x yx y x yx y x y

++

= =++




or

2 2 1312 14

11 11 11


+ minus + minus =





2 2 1312 14

11 11 11


+ minus + minus =

we have

S20

Page - 467


32

12

11

13

11

2 2 14

11

2

2

C

C

C C C

MxM

MyM

Mr x yM

=

=

= + +


( ) 1312

11 11

2 2C C

MMx yM M

⎛ ⎞= ⎜ ⎟⎝ ⎠

and radius 2 2 14

11C C

Mx yM


shareable





21

1 1 2

1

1 11 2 1 2( ) ( 2 2 )2 2k kC C k k k k

kkx y +

+++= + + + + + +ε ε ε ε ε ε

and

1 2 1 21 21 2 2k

k k k kC kr + + + += + + + +ε ε ε


33( )




Cr We define


for 12 j n=

S20

Page - 468

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar


hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4


Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4


Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched


spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b


our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s


ommon

s Award 201

3

with center

sharing pair


13

33

S20

Page - 469


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


18





= many


Reference


S20

Page - 454


19

4 Algebraic Methods






1

1

( )

( )( )

t

iii j

j t

j iii j

x x

L xx x

=ne

=ne

minus

=minus

prod

prod


1( )

0j i

i jL x

i j=⎧

= ⎨ ne⎩


1 1 2 2 3 3( ) ( ) ( ) ( )L x y L x y L x y L x= + + +



1 1 2 2 3 3(0) (0) (0)y L y L y L+ + +

where

S20

Page - 455


20

1

1 1

1 1

(0 ) ( 1)

(0)( ) ( )

t tt

i ii ii j i j

j t t

j i j ii ii j i j

x x

Lx x x x

minus

= =ne ne

= =ne ne

minus minus

= =minus minus

prod prod

prod prod

for 1 23j =








1 2 1 21 2 1 2












11 21 2 2k




S20

Page - 456


21


2 3 11 2 2 2 2( ) k

kX p p p p p pminus

=

Assume that

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k



So

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k





Therefore 2 3 11 2 2 2 2 2( )

k k

X p p p p p ppminus




1 2 3 234579111316171923 X m m m= =

from now on


1 21 1 12 2 2 2 2 21 1 1 2 2 2

k k kr



1 2

1 1 2 21 1 1 r


1 2

1 1 2 21 1 1 r


such that

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr


1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr



1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr


+ + + + + + + + +ε ε εε ε ε ε ε ε

S20

Page - 457


22

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr




i ij j=ε ε







12 2i ik k

i k ixp pminus

lelt



12 331 2 4 221 2 2

i iii i i kki i


= εεε ε ε for







3 234X = 4 2345X = 5 23457X = 6 234579X =

7 23457911X = and 8 2345791113X = Also 1 2XΠ = 2 6XΠ =

3 24XΠ = and 4 120XΠ =

Note that





S20

Page - 458


23





1 21 2 1 11 2 1 1 1 2

1 1 1 2 1 11 2 1 2 1 2( )( ) ( )k kk k k k t t k





secret sharing pair











S20

Page - 459


24

( ) ( ( ))j C j P jπ =

for 12 j n=


11 1 2 2( ) ( ) ( )( ( ))






1 2( )(0)t kΛΦ ε ε ε


1 2 3 4ε ε ε ε ( )2 1 2 3 4( ) 0ΛΦ ε ε ε ε 1 2 3 4ε ε ε ε ( )

2 1 2 3 4( ) 0ΛΦ ε ε ε ε 0000 1 0110 12 1000 2 0101 15 0100 3 0011 20 0010 4 1110 24 0001 5 1101 30 1100 6 1011 40 1010 8 0111 60 1001 10 1111 120



( )( ) ( )( ) ( )( )3 2 1 3 1 2

1 2 31 3 1 2 2 1 2 3 1 33 2




S20

Page - 460


25


( ) ( )( )( )

( )( )

2 1 2 2 1 1

2 1 2 1 2 1

2 1 2 1 2 1

2 1

2 1 2 1 2 1

1 1x x x x

x x x xx x x x

x x x x

ε ε

ε εε ε

ε εε ε

+ minus

+ minus minus=

+ minus

minus=

+

minusminus minus


minusminus minus



( )2 1 2 2 1 1

1 1 22xx xx

εε ε λ λ ε

minusminus minus

le+ minus minus

Observe that

( )( )( )( )

( )( )

( )( )( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )

3 23 2

3 1 2 1 3 1 2 1

3 23 2

3 1 2 1 3 1 2 1

3 2 3 2

3 1 2

3 2

3 1 2 1

3 2

3 2 3 2

3 1

3 2 3 2

3 1

1 3 1 2 1

3 2 3 2

3 1 2 1 3 1 2 13 1 2 1

3 1 2

1





x x x

ε εε ε ε ε

ε ε

ε ε ε εε ε






+ ++ minus + minus

+ +

+ + + ++ minus

+ + + ++ minus

=minus


( )( )( )

( ) ( ) ( )( )( )( ) ( ) ( )

3 2 3 2

3 2

3 1

3 2

2 31

3 2

2 1 3 1

3 1

3 1

3 2

3 1 2 1 2 12 1

1

1 1

1

x xx

x xx x x x x x

x xx x x x x x

ε ε ε ε

ε εε ε

ε εε ε ε ε

minus



+ +

+ ++ minus

+ ++ minus minus

S20

Page - 461


26

( )( )( )( )

( )( )

( ) ( ) ( )( )

3 2

3 1

3 23 2

3 1 2 1 3 1 2 12 1

222 2

2

2 2

2

1 (2 )

1 2( )

1 2 4

2 22 2

2 22 2 2


M M M

M M M

ε εε ε ε ε

ε εε ε ε ελ

ε ε εελ

ω ω ε

λ λ ε λ λ ε


λ

+ ++ minus + minus

minus minus

leminus


le + + + + +

⎡ ⎤+ + +⎛ ⎞ ⎛ ⎞+ +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤le

minus

⎣

minus

+ ⎦

Therefore

where 2

M εε

ωλ

+=

minus So we have

( )( )( )( )

( )( ) ( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( ) ( )

( )( )( )

3 23 21 1 1

3 1 2 1 3 1 2 1

3 23 21 1

3 1 2 1 3 1 2 1

3 2

3 1 2 1

3 2

3 1 2

3 2 3 21 1 1

3 1 2 1 3 1 2 1

33 2

3 1 2 1

1

3 2 3 2

3 1 2 1 3 1 2 1

3



x x x xy y

x x x x x x x x

xx xx x x x

ε εε ε ε ε

ε εε ε ε ε


δε

δ

ε

+ ++ minus + minus

+ ++ minus + minus

+ +



+ minus+ +



le minusminus

minus

+minus

( )( )( )

( )( )( )( )

( )

21

3 1 2 1

3 21

3 1 2 1

2

3 1 2 1

3 2

3 1 2 1

2 21 2 4

xy

x x x x

x xx x x x

M

εε ε ε ε

ε εε ε ε ε

δ

ω ω ε ω δλ

++ minus + minus

+ ++ minus + minus

minus minus

+minus minus

le + +



( )2 21(0) (0) 6 12 3L L Mω ω ε ω δλ

minus le + +

where 2

M εε

ωλ

+=

minus

S20

Page - 462


27

















S20

Page - 463


28


461 Speed Tests







2 4 6 8 10


0237 0471 0708 0935 1176

0

1

2

3

4

5

6

7

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB


1183 2349 358 473 5917

S20

Page - 464


29


462 Size Tests




987 1973 2960 3947 4934


9845 19502 29195 38791 46789


100 99 99 98 95

0

02

04

06

08

1

12

14

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 465


30



Reference




0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

File

Siz

e in

KB

Secret Size in MB

CompressedSecret

CompressedContainer

S20

Page - 466


31

5 Geometric Methods






2 2

2 21 1 1 12 22 2 2 22 23 3 3 3

11

( ) 011

x y x yx y x y

J x yx y x yx y x y

++

= =++




or

2 2 1312 14

11 11 11


+ minus + minus =





2 2 1312 14

11 11 11


+ minus + minus =

we have

S20

Page - 467


32

12

11

13

11

2 2 14

11

2

2

C

C

C C C

MxM

MyM

Mr x yM

=

=

= + +


( ) 1312

11 11

2 2C C

MMx yM M

⎛ ⎞= ⎜ ⎟⎝ ⎠

and radius 2 2 14

11C C

Mx yM


shareable





21

1 1 2

1

1 11 2 1 2( ) ( 2 2 )2 2k kC C k k k k

kkx y +

+++= + + + + + +ε ε ε ε ε ε

and

1 2 1 21 21 2 2k

k k k kC kr + + + += + + + +ε ε ε


33( )




Cr We define


for 12 j n=

S20

Page - 468

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar


hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4


Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4


Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched


spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b


our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s


ommon

s Award 201

3

with center

sharing pair


13

33

S20

Page - 469


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


19

4 Algebraic Methods






1

1

( )

( )( )

t

iii j

j t

j iii j

x x

L xx x

=ne

=ne

minus

=minus

prod

prod


1( )

0j i

i jL x

i j=⎧

= ⎨ ne⎩


1 1 2 2 3 3( ) ( ) ( ) ( )L x y L x y L x y L x= + + +



1 1 2 2 3 3(0) (0) (0)y L y L y L+ + +

where

S20

Page - 455


20

1

1 1

1 1

(0 ) ( 1)

(0)( ) ( )

t tt

i ii ii j i j

j t t

j i j ii ii j i j

x x

Lx x x x

minus

= =ne ne

= =ne ne

minus minus

= =minus minus

prod prod

prod prod

for 1 23j =








1 2 1 21 2 1 2












11 21 2 2k




S20

Page - 456


21


2 3 11 2 2 2 2( ) k

kX p p p p p pminus

=

Assume that

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k



So

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k





Therefore 2 3 11 2 2 2 2 2( )

k k

X p p p p p ppminus




1 2 3 234579111316171923 X m m m= =

from now on


1 21 1 12 2 2 2 2 21 1 1 2 2 2

k k kr



1 2

1 1 2 21 1 1 r


1 2

1 1 2 21 1 1 r


such that

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr


1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr



1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr


+ + + + + + + + +ε ε εε ε ε ε ε ε

S20

Page - 457


22

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr




i ij j=ε ε







12 2i ik k

i k ixp pminus

lelt



12 331 2 4 221 2 2

i iii i i kki i


= εεε ε ε for







3 234X = 4 2345X = 5 23457X = 6 234579X =

7 23457911X = and 8 2345791113X = Also 1 2XΠ = 2 6XΠ =

3 24XΠ = and 4 120XΠ =

Note that





S20

Page - 458


23





1 21 2 1 11 2 1 1 1 2

1 1 1 2 1 11 2 1 2 1 2( )( ) ( )k kk k k k t t k





secret sharing pair











S20

Page - 459


24

( ) ( ( ))j C j P jπ =

for 12 j n=


11 1 2 2( ) ( ) ( )( ( ))






1 2( )(0)t kΛΦ ε ε ε


1 2 3 4ε ε ε ε ( )2 1 2 3 4( ) 0ΛΦ ε ε ε ε 1 2 3 4ε ε ε ε ( )

2 1 2 3 4( ) 0ΛΦ ε ε ε ε 0000 1 0110 12 1000 2 0101 15 0100 3 0011 20 0010 4 1110 24 0001 5 1101 30 1100 6 1011 40 1010 8 0111 60 1001 10 1111 120



( )( ) ( )( ) ( )( )3 2 1 3 1 2

1 2 31 3 1 2 2 1 2 3 1 33 2




S20

Page - 460


25


( ) ( )( )( )

( )( )

2 1 2 2 1 1

2 1 2 1 2 1

2 1 2 1 2 1

2 1

2 1 2 1 2 1

1 1x x x x

x x x xx x x x

x x x x

ε ε

ε εε ε

ε εε ε

+ minus

+ minus minus=

+ minus

minus=

+

minusminus minus


minusminus minus



( )2 1 2 2 1 1

1 1 22xx xx

εε ε λ λ ε

minusminus minus

le+ minus minus

Observe that

( )( )( )( )

( )( )

( )( )( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )

3 23 2

3 1 2 1 3 1 2 1

3 23 2

3 1 2 1 3 1 2 1

3 2 3 2

3 1 2

3 2

3 1 2 1

3 2

3 2 3 2

3 1

3 2 3 2

3 1

1 3 1 2 1

3 2 3 2

3 1 2 1 3 1 2 13 1 2 1

3 1 2

1





x x x

ε εε ε ε ε

ε ε

ε ε ε εε ε






+ ++ minus + minus

+ +

+ + + ++ minus

+ + + ++ minus

=minus


( )( )( )

( ) ( ) ( )( )( )( ) ( ) ( )

3 2 3 2

3 2

3 1

3 2

2 31

3 2

2 1 3 1

3 1

3 1

3 2

3 1 2 1 2 12 1

1

1 1

1

x xx

x xx x x x x x

x xx x x x x x

ε ε ε ε

ε εε ε

ε εε ε ε ε

minus



+ +

+ ++ minus

+ ++ minus minus

S20

Page - 461


26

( )( )( )( )

( )( )

( ) ( ) ( )( )

3 2

3 1

3 23 2

3 1 2 1 3 1 2 12 1

222 2

2

2 2

2

1 (2 )

1 2( )

1 2 4

2 22 2

2 22 2 2


M M M

M M M

ε εε ε ε ε

ε εε ε ε ελ

ε ε εελ

ω ω ε

λ λ ε λ λ ε


λ

+ ++ minus + minus

minus minus

leminus


le + + + + +

⎡ ⎤+ + +⎛ ⎞ ⎛ ⎞+ +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤le

minus

⎣

minus

+ ⎦

Therefore

where 2

M εε

ωλ

+=

minus So we have

( )( )( )( )

( )( ) ( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( ) ( )

( )( )( )

3 23 21 1 1

3 1 2 1 3 1 2 1

3 23 21 1

3 1 2 1 3 1 2 1

3 2

3 1 2 1

3 2

3 1 2

3 2 3 21 1 1

3 1 2 1 3 1 2 1

33 2

3 1 2 1

1

3 2 3 2

3 1 2 1 3 1 2 1

3



x x x xy y

x x x x x x x x

xx xx x x x

ε εε ε ε ε

ε εε ε ε ε


δε

δ

ε

+ ++ minus + minus

+ ++ minus + minus

+ +



+ minus+ +



le minusminus

minus

+minus

( )( )( )

( )( )( )( )

( )

21

3 1 2 1

3 21

3 1 2 1

2

3 1 2 1

3 2

3 1 2 1

2 21 2 4

xy

x x x x

x xx x x x

M

εε ε ε ε

ε εε ε ε ε

δ

ω ω ε ω δλ

++ minus + minus

+ ++ minus + minus

minus minus

+minus minus

le + +



( )2 21(0) (0) 6 12 3L L Mω ω ε ω δλ

minus le + +

where 2

M εε

ωλ

+=

minus

S20

Page - 462


27

















S20

Page - 463


28


461 Speed Tests







2 4 6 8 10


0237 0471 0708 0935 1176

0

1

2

3

4

5

6

7

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB


1183 2349 358 473 5917

S20

Page - 464


29


462 Size Tests




987 1973 2960 3947 4934


9845 19502 29195 38791 46789


100 99 99 98 95

0

02

04

06

08

1

12

14

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 465


30



Reference




0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

File

Siz

e in

KB

Secret Size in MB

CompressedSecret

CompressedContainer

S20

Page - 466


31

5 Geometric Methods






2 2

2 21 1 1 12 22 2 2 22 23 3 3 3

11

( ) 011

x y x yx y x y

J x yx y x yx y x y

++

= =++




or

2 2 1312 14

11 11 11


+ minus + minus =





2 2 1312 14

11 11 11


+ minus + minus =

we have

S20

Page - 467


32

12

11

13

11

2 2 14

11

2

2

C

C

C C C

MxM

MyM

Mr x yM

=

=

= + +


( ) 1312

11 11

2 2C C

MMx yM M

⎛ ⎞= ⎜ ⎟⎝ ⎠

and radius 2 2 14

11C C

Mx yM


shareable





21

1 1 2

1

1 11 2 1 2( ) ( 2 2 )2 2k kC C k k k k

kkx y +

+++= + + + + + +ε ε ε ε ε ε

and

1 2 1 21 21 2 2k

k k k kC kr + + + += + + + +ε ε ε


33( )




Cr We define


for 12 j n=

S20

Page - 468

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar


hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4


Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4


Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched


spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b


our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s


ommon

s Award 201

3

with center

sharing pair


13

33

S20

Page - 469


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


20

1

1 1

1 1

(0 ) ( 1)

(0)( ) ( )

t tt

i ii ii j i j

j t t

j i j ii ii j i j

x x

Lx x x x

minus

= =ne ne

= =ne ne

minus minus

= =minus minus

prod prod

prod prod

for 1 23j =








1 2 1 21 2 1 2












11 21 2 2k




S20

Page - 456


21


2 3 11 2 2 2 2( ) k

kX p p p p p pminus

=

Assume that

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k



So

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k





Therefore 2 3 11 2 2 2 2 2( )

k k

X p p p p p ppminus




1 2 3 234579111316171923 X m m m= =

from now on


1 21 1 12 2 2 2 2 21 1 1 2 2 2

k k kr



1 2

1 1 2 21 1 1 r


1 2

1 1 2 21 1 1 r


such that

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr


1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr



1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr


+ + + + + + + + +ε ε εε ε ε ε ε ε

S20

Page - 457


22

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr




i ij j=ε ε







12 2i ik k

i k ixp pminus

lelt



12 331 2 4 221 2 2

i iii i i kki i


= εεε ε ε for







3 234X = 4 2345X = 5 23457X = 6 234579X =

7 23457911X = and 8 2345791113X = Also 1 2XΠ = 2 6XΠ =

3 24XΠ = and 4 120XΠ =

Note that





S20

Page - 458


23





1 21 2 1 11 2 1 1 1 2

1 1 1 2 1 11 2 1 2 1 2( )( ) ( )k kk k k k t t k





secret sharing pair











S20

Page - 459


24

( ) ( ( ))j C j P jπ =

for 12 j n=


11 1 2 2( ) ( ) ( )( ( ))






1 2( )(0)t kΛΦ ε ε ε


1 2 3 4ε ε ε ε ( )2 1 2 3 4( ) 0ΛΦ ε ε ε ε 1 2 3 4ε ε ε ε ( )

2 1 2 3 4( ) 0ΛΦ ε ε ε ε 0000 1 0110 12 1000 2 0101 15 0100 3 0011 20 0010 4 1110 24 0001 5 1101 30 1100 6 1011 40 1010 8 0111 60 1001 10 1111 120



( )( ) ( )( ) ( )( )3 2 1 3 1 2

1 2 31 3 1 2 2 1 2 3 1 33 2




S20

Page - 460


25


( ) ( )( )( )

( )( )

2 1 2 2 1 1

2 1 2 1 2 1

2 1 2 1 2 1

2 1

2 1 2 1 2 1

1 1x x x x

x x x xx x x x

x x x x

ε ε

ε εε ε

ε εε ε

+ minus

+ minus minus=

+ minus

minus=

+

minusminus minus


minusminus minus



( )2 1 2 2 1 1

1 1 22xx xx

εε ε λ λ ε

minusminus minus

le+ minus minus

Observe that

( )( )( )( )

( )( )

( )( )( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )

3 23 2

3 1 2 1 3 1 2 1

3 23 2

3 1 2 1 3 1 2 1

3 2 3 2

3 1 2

3 2

3 1 2 1

3 2

3 2 3 2

3 1

3 2 3 2

3 1

1 3 1 2 1

3 2 3 2

3 1 2 1 3 1 2 13 1 2 1

3 1 2

1





x x x

ε εε ε ε ε

ε ε

ε ε ε εε ε






+ ++ minus + minus

+ +

+ + + ++ minus

+ + + ++ minus

=minus


( )( )( )

( ) ( ) ( )( )( )( ) ( ) ( )

3 2 3 2

3 2

3 1

3 2

2 31

3 2

2 1 3 1

3 1

3 1

3 2

3 1 2 1 2 12 1

1

1 1

1

x xx

x xx x x x x x

x xx x x x x x

ε ε ε ε

ε εε ε

ε εε ε ε ε

minus



+ +

+ ++ minus

+ ++ minus minus

S20

Page - 461


26

( )( )( )( )

( )( )

( ) ( ) ( )( )

3 2

3 1

3 23 2

3 1 2 1 3 1 2 12 1

222 2

2

2 2

2

1 (2 )

1 2( )

1 2 4

2 22 2

2 22 2 2


M M M

M M M

ε εε ε ε ε

ε εε ε ε ελ

ε ε εελ

ω ω ε

λ λ ε λ λ ε


λ

+ ++ minus + minus

minus minus

leminus


le + + + + +

⎡ ⎤+ + +⎛ ⎞ ⎛ ⎞+ +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤le

minus

⎣

minus

+ ⎦

Therefore

where 2

M εε

ωλ

+=

minus So we have

( )( )( )( )

( )( ) ( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( ) ( )

( )( )( )

3 23 21 1 1

3 1 2 1 3 1 2 1

3 23 21 1

3 1 2 1 3 1 2 1

3 2

3 1 2 1

3 2

3 1 2

3 2 3 21 1 1

3 1 2 1 3 1 2 1

33 2

3 1 2 1

1

3 2 3 2

3 1 2 1 3 1 2 1

3



x x x xy y

x x x x x x x x

xx xx x x x

ε εε ε ε ε

ε εε ε ε ε


δε

δ

ε

+ ++ minus + minus

+ ++ minus + minus

+ +



+ minus+ +



le minusminus

minus

+minus

( )( )( )

( )( )( )( )

( )

21

3 1 2 1

3 21

3 1 2 1

2

3 1 2 1

3 2

3 1 2 1

2 21 2 4

xy

x x x x

x xx x x x

M

εε ε ε ε

ε εε ε ε ε

δ

ω ω ε ω δλ

++ minus + minus

+ ++ minus + minus

minus minus

+minus minus

le + +



( )2 21(0) (0) 6 12 3L L Mω ω ε ω δλ

minus le + +

where 2

M εε

ωλ

+=

minus

S20

Page - 462


27

















S20

Page - 463


28


461 Speed Tests







2 4 6 8 10


0237 0471 0708 0935 1176

0

1

2

3

4

5

6

7

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB


1183 2349 358 473 5917

S20

Page - 464


29


462 Size Tests




987 1973 2960 3947 4934


9845 19502 29195 38791 46789


100 99 99 98 95

0

02

04

06

08

1

12

14

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 465


30



Reference




0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

File

Siz

e in

KB

Secret Size in MB

CompressedSecret

CompressedContainer

S20

Page - 466


31

5 Geometric Methods






2 2

2 21 1 1 12 22 2 2 22 23 3 3 3

11

( ) 011

x y x yx y x y

J x yx y x yx y x y

++

= =++




or

2 2 1312 14

11 11 11


+ minus + minus =





2 2 1312 14

11 11 11


+ minus + minus =

we have

S20

Page - 467


32

12

11

13

11

2 2 14

11

2

2

C

C

C C C

MxM

MyM

Mr x yM

=

=

= + +


( ) 1312

11 11

2 2C C

MMx yM M

⎛ ⎞= ⎜ ⎟⎝ ⎠

and radius 2 2 14

11C C

Mx yM


shareable





21

1 1 2

1

1 11 2 1 2( ) ( 2 2 )2 2k kC C k k k k

kkx y +

+++= + + + + + +ε ε ε ε ε ε

and

1 2 1 21 21 2 2k

k k k kC kr + + + += + + + +ε ε ε


33( )




Cr We define


for 12 j n=

S20

Page - 468

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar


hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4


Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4


Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched


spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b


our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s


ommon

s Award 201

3

with center

sharing pair


13

33

S20

Page - 469


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


21


2 3 11 2 2 2 2( ) k

kX p p p p p pminus

=

Assume that

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k



So

2 1 2 13 33 31 2 4 1 2 42 2 2 21 2 2 1 2 2k k





Therefore 2 3 11 2 2 2 2 2( )

k k

X p p p p p ppminus




1 2 3 234579111316171923 X m m m= =

from now on


1 21 1 12 2 2 2 2 21 1 1 2 2 2

k k kr



1 2

1 1 2 21 1 1 r


1 2

1 1 2 21 1 1 r


such that

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr


1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 1 1 2 2 2

k k kr rr rk k kr



1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr


+ + + + + + + + +ε ε εε ε ε ε ε ε

S20

Page - 457


22

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr




i ij j=ε ε







12 2i ik k

i k ixp pminus

lelt



12 331 2 4 221 2 2

i iii i i kki i


= εεε ε ε for







3 234X = 4 2345X = 5 23457X = 6 234579X =

7 23457911X = and 8 2345791113X = Also 1 2XΠ = 2 6XΠ =

3 24XΠ = and 4 120XΠ =

Note that





S20

Page - 458


23





1 21 2 1 11 2 1 1 1 2

1 1 1 2 1 11 2 1 2 1 2( )( ) ( )k kk k k k t t k





secret sharing pair











S20

Page - 459


24

( ) ( ( ))j C j P jπ =

for 12 j n=


11 1 2 2( ) ( ) ( )( ( ))






1 2( )(0)t kΛΦ ε ε ε


1 2 3 4ε ε ε ε ( )2 1 2 3 4( ) 0ΛΦ ε ε ε ε 1 2 3 4ε ε ε ε ( )

2 1 2 3 4( ) 0ΛΦ ε ε ε ε 0000 1 0110 12 1000 2 0101 15 0100 3 0011 20 0010 4 1110 24 0001 5 1101 30 1100 6 1011 40 1010 8 0111 60 1001 10 1111 120



( )( ) ( )( ) ( )( )3 2 1 3 1 2

1 2 31 3 1 2 2 1 2 3 1 33 2




S20

Page - 460


25


( ) ( )( )( )

( )( )

2 1 2 2 1 1

2 1 2 1 2 1

2 1 2 1 2 1

2 1

2 1 2 1 2 1

1 1x x x x

x x x xx x x x

x x x x

ε ε

ε εε ε

ε εε ε

+ minus

+ minus minus=

+ minus

minus=

+

minusminus minus


minusminus minus



( )2 1 2 2 1 1

1 1 22xx xx

εε ε λ λ ε

minusminus minus

le+ minus minus

Observe that

( )( )( )( )

( )( )

( )( )( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )

3 23 2

3 1 2 1 3 1 2 1

3 23 2

3 1 2 1 3 1 2 1

3 2 3 2

3 1 2

3 2

3 1 2 1

3 2

3 2 3 2

3 1

3 2 3 2

3 1

1 3 1 2 1

3 2 3 2

3 1 2 1 3 1 2 13 1 2 1

3 1 2

1





x x x

ε εε ε ε ε

ε ε

ε ε ε εε ε






+ ++ minus + minus

+ +

+ + + ++ minus

+ + + ++ minus

=minus


( )( )( )

( ) ( ) ( )( )( )( ) ( ) ( )

3 2 3 2

3 2

3 1

3 2

2 31

3 2

2 1 3 1

3 1

3 1

3 2

3 1 2 1 2 12 1

1

1 1

1

x xx

x xx x x x x x

x xx x x x x x

ε ε ε ε

ε εε ε

ε εε ε ε ε

minus



+ +

+ ++ minus

+ ++ minus minus

S20

Page - 461


26

( )( )( )( )

( )( )

( ) ( ) ( )( )

3 2

3 1

3 23 2

3 1 2 1 3 1 2 12 1

222 2

2

2 2

2

1 (2 )

1 2( )

1 2 4

2 22 2

2 22 2 2


M M M

M M M

ε εε ε ε ε

ε εε ε ε ελ

ε ε εελ

ω ω ε

λ λ ε λ λ ε


λ

+ ++ minus + minus

minus minus

leminus


le + + + + +

⎡ ⎤+ + +⎛ ⎞ ⎛ ⎞+ +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤le

minus

⎣

minus

+ ⎦

Therefore

where 2

M εε

ωλ

+=

minus So we have

( )( )( )( )

( )( ) ( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( ) ( )

( )( )( )

3 23 21 1 1

3 1 2 1 3 1 2 1

3 23 21 1

3 1 2 1 3 1 2 1

3 2

3 1 2 1

3 2

3 1 2

3 2 3 21 1 1

3 1 2 1 3 1 2 1

33 2

3 1 2 1

1

3 2 3 2

3 1 2 1 3 1 2 1

3



x x x xy y

x x x x x x x x

xx xx x x x

ε εε ε ε ε

ε εε ε ε ε


δε

δ

ε

+ ++ minus + minus

+ ++ minus + minus

+ +



+ minus+ +



le minusminus

minus

+minus

( )( )( )

( )( )( )( )

( )

21

3 1 2 1

3 21

3 1 2 1

2

3 1 2 1

3 2

3 1 2 1

2 21 2 4

xy

x x x x

x xx x x x

M

εε ε ε ε

ε εε ε ε ε

δ

ω ω ε ω δλ

++ minus + minus

+ ++ minus + minus

minus minus

+minus minus

le + +



( )2 21(0) (0) 6 12 3L L Mω ω ε ω δλ

minus le + +

where 2

M εε

ωλ

+=

minus

S20

Page - 462


27

















S20

Page - 463


28


461 Speed Tests







2 4 6 8 10


0237 0471 0708 0935 1176

0

1

2

3

4

5

6

7

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB


1183 2349 358 473 5917

S20

Page - 464


29


462 Size Tests




987 1973 2960 3947 4934


9845 19502 29195 38791 46789


100 99 99 98 95

0

02

04

06

08

1

12

14

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 465


30



Reference




0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

File

Siz

e in

KB

Secret Size in MB

CompressedSecret

CompressedContainer

S20

Page - 466


31

5 Geometric Methods






2 2

2 21 1 1 12 22 2 2 22 23 3 3 3

11

( ) 011

x y x yx y x y

J x yx y x yx y x y

++

= =++




or

2 2 1312 14

11 11 11


+ minus + minus =





2 2 1312 14

11 11 11


+ minus + minus =

we have

S20

Page - 467


32

12

11

13

11

2 2 14

11

2

2

C

C

C C C

MxM

MyM

Mr x yM

=

=

= + +


( ) 1312

11 11

2 2C C

MMx yM M

⎛ ⎞= ⎜ ⎟⎝ ⎠

and radius 2 2 14

11C C

Mx yM


shareable





21

1 1 2

1

1 11 2 1 2( ) ( 2 2 )2 2k kC C k k k k

kkx y +

+++= + + + + + +ε ε ε ε ε ε

and

1 2 1 21 21 2 2k

k k k kC kr + + + += + + + +ε ε ε


33( )




Cr We define


for 12 j n=

S20

Page - 468

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar


hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4


Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4


Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched


spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b


our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s


ommon

s Award 201

3

with center

sharing pair


13

33

S20

Page - 469


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


22

1 1 11 21 21 1 2 21 2 1 1 2 2 1 22 2 21 2 1 2 1 2

1 2

k k kr rr rk k kr




i ij j=ε ε







12 2i ik k

i k ixp pminus

lelt



12 331 2 4 221 2 2

i iii i i kki i


= εεε ε ε for







3 234X = 4 2345X = 5 23457X = 6 234579X =

7 23457911X = and 8 2345791113X = Also 1 2XΠ = 2 6XΠ =

3 24XΠ = and 4 120XΠ =

Note that





S20

Page - 458


23





1 21 2 1 11 2 1 1 1 2

1 1 1 2 1 11 2 1 2 1 2( )( ) ( )k kk k k k t t k





secret sharing pair











S20

Page - 459


24

( ) ( ( ))j C j P jπ =

for 12 j n=


11 1 2 2( ) ( ) ( )( ( ))






1 2( )(0)t kΛΦ ε ε ε


1 2 3 4ε ε ε ε ( )2 1 2 3 4( ) 0ΛΦ ε ε ε ε 1 2 3 4ε ε ε ε ( )

2 1 2 3 4( ) 0ΛΦ ε ε ε ε 0000 1 0110 12 1000 2 0101 15 0100 3 0011 20 0010 4 1110 24 0001 5 1101 30 1100 6 1011 40 1010 8 0111 60 1001 10 1111 120



( )( ) ( )( ) ( )( )3 2 1 3 1 2

1 2 31 3 1 2 2 1 2 3 1 33 2




S20

Page - 460


25


( ) ( )( )( )

( )( )

2 1 2 2 1 1

2 1 2 1 2 1

2 1 2 1 2 1

2 1

2 1 2 1 2 1

1 1x x x x

x x x xx x x x

x x x x

ε ε

ε εε ε

ε εε ε

+ minus

+ minus minus=

+ minus

minus=

+

minusminus minus


minusminus minus



( )2 1 2 2 1 1

1 1 22xx xx

εε ε λ λ ε

minusminus minus

le+ minus minus

Observe that

( )( )( )( )

( )( )

( )( )( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )

3 23 2

3 1 2 1 3 1 2 1

3 23 2

3 1 2 1 3 1 2 1

3 2 3 2

3 1 2

3 2

3 1 2 1

3 2

3 2 3 2

3 1

3 2 3 2

3 1

1 3 1 2 1

3 2 3 2

3 1 2 1 3 1 2 13 1 2 1

3 1 2

1





x x x

ε εε ε ε ε

ε ε

ε ε ε εε ε






+ ++ minus + minus

+ +

+ + + ++ minus

+ + + ++ minus

=minus


( )( )( )

( ) ( ) ( )( )( )( ) ( ) ( )

3 2 3 2

3 2

3 1

3 2

2 31

3 2

2 1 3 1

3 1

3 1

3 2

3 1 2 1 2 12 1

1

1 1

1

x xx

x xx x x x x x

x xx x x x x x

ε ε ε ε

ε εε ε

ε εε ε ε ε

minus



+ +

+ ++ minus

+ ++ minus minus

S20

Page - 461


26

( )( )( )( )

( )( )

( ) ( ) ( )( )

3 2

3 1

3 23 2

3 1 2 1 3 1 2 12 1

222 2

2

2 2

2

1 (2 )

1 2( )

1 2 4

2 22 2

2 22 2 2


M M M

M M M

ε εε ε ε ε

ε εε ε ε ελ

ε ε εελ

ω ω ε

λ λ ε λ λ ε


λ

+ ++ minus + minus

minus minus

leminus


le + + + + +

⎡ ⎤+ + +⎛ ⎞ ⎛ ⎞+ +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤le

minus

⎣

minus

+ ⎦

Therefore

where 2

M εε

ωλ

+=

minus So we have

( )( )( )( )

( )( ) ( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( ) ( )

( )( )( )

3 23 21 1 1

3 1 2 1 3 1 2 1

3 23 21 1

3 1 2 1 3 1 2 1

3 2

3 1 2 1

3 2

3 1 2

3 2 3 21 1 1

3 1 2 1 3 1 2 1

33 2

3 1 2 1

1

3 2 3 2

3 1 2 1 3 1 2 1

3



x x x xy y

x x x x x x x x

xx xx x x x

ε εε ε ε ε

ε εε ε ε ε


δε

δ

ε

+ ++ minus + minus

+ ++ minus + minus

+ +



+ minus+ +



le minusminus

minus

+minus

( )( )( )

( )( )( )( )

( )

21

3 1 2 1

3 21

3 1 2 1

2

3 1 2 1

3 2

3 1 2 1

2 21 2 4

xy

x x x x

x xx x x x

M

εε ε ε ε

ε εε ε ε ε

δ

ω ω ε ω δλ

++ minus + minus

+ ++ minus + minus

minus minus

+minus minus

le + +



( )2 21(0) (0) 6 12 3L L Mω ω ε ω δλ

minus le + +

where 2

M εε

ωλ

+=

minus

S20

Page - 462


27

















S20

Page - 463


28


461 Speed Tests







2 4 6 8 10


0237 0471 0708 0935 1176

0

1

2

3

4

5

6

7

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB


1183 2349 358 473 5917

S20

Page - 464


29


462 Size Tests




987 1973 2960 3947 4934


9845 19502 29195 38791 46789


100 99 99 98 95

0

02

04

06

08

1

12

14

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 465


30



Reference




0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

File

Siz

e in

KB

Secret Size in MB

CompressedSecret

CompressedContainer

S20

Page - 466


31

5 Geometric Methods






2 2

2 21 1 1 12 22 2 2 22 23 3 3 3

11

( ) 011

x y x yx y x y

J x yx y x yx y x y

++

= =++




or

2 2 1312 14

11 11 11


+ minus + minus =





2 2 1312 14

11 11 11


+ minus + minus =

we have

S20

Page - 467


32

12

11

13

11

2 2 14

11

2

2

C

C

C C C

MxM

MyM

Mr x yM

=

=

= + +


( ) 1312

11 11

2 2C C

MMx yM M

⎛ ⎞= ⎜ ⎟⎝ ⎠

and radius 2 2 14

11C C

Mx yM


shareable





21

1 1 2

1

1 11 2 1 2( ) ( 2 2 )2 2k kC C k k k k

kkx y +

+++= + + + + + +ε ε ε ε ε ε

and

1 2 1 21 21 2 2k

k k k kC kr + + + += + + + +ε ε ε


33( )




Cr We define


for 12 j n=

S20

Page - 468

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar


hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4


Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4


Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched


spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b


our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s


ommon

s Award 201

3

with center

sharing pair


13

33

S20

Page - 469


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


23





1 21 2 1 11 2 1 1 1 2

1 1 1 2 1 11 2 1 2 1 2( )( ) ( )k kk k k k t t k





secret sharing pair











S20

Page - 459


24

( ) ( ( ))j C j P jπ =

for 12 j n=


11 1 2 2( ) ( ) ( )( ( ))






1 2( )(0)t kΛΦ ε ε ε


1 2 3 4ε ε ε ε ( )2 1 2 3 4( ) 0ΛΦ ε ε ε ε 1 2 3 4ε ε ε ε ( )

2 1 2 3 4( ) 0ΛΦ ε ε ε ε 0000 1 0110 12 1000 2 0101 15 0100 3 0011 20 0010 4 1110 24 0001 5 1101 30 1100 6 1011 40 1010 8 0111 60 1001 10 1111 120



( )( ) ( )( ) ( )( )3 2 1 3 1 2

1 2 31 3 1 2 2 1 2 3 1 33 2




S20

Page - 460


25


( ) ( )( )( )

( )( )

2 1 2 2 1 1

2 1 2 1 2 1

2 1 2 1 2 1

2 1

2 1 2 1 2 1

1 1x x x x

x x x xx x x x

x x x x

ε ε

ε εε ε

ε εε ε

+ minus

+ minus minus=

+ minus

minus=

+

minusminus minus


minusminus minus



( )2 1 2 2 1 1

1 1 22xx xx

εε ε λ λ ε

minusminus minus

le+ minus minus

Observe that

( )( )( )( )

( )( )

( )( )( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )

3 23 2

3 1 2 1 3 1 2 1

3 23 2

3 1 2 1 3 1 2 1

3 2 3 2

3 1 2

3 2

3 1 2 1

3 2

3 2 3 2

3 1

3 2 3 2

3 1

1 3 1 2 1

3 2 3 2

3 1 2 1 3 1 2 13 1 2 1

3 1 2

1





x x x

ε εε ε ε ε

ε ε

ε ε ε εε ε






+ ++ minus + minus

+ +

+ + + ++ minus

+ + + ++ minus

=minus


( )( )( )

( ) ( ) ( )( )( )( ) ( ) ( )

3 2 3 2

3 2

3 1

3 2

2 31

3 2

2 1 3 1

3 1

3 1

3 2

3 1 2 1 2 12 1

1

1 1

1

x xx

x xx x x x x x

x xx x x x x x

ε ε ε ε

ε εε ε

ε εε ε ε ε

minus



+ +

+ ++ minus

+ ++ minus minus

S20

Page - 461


26

( )( )( )( )

( )( )

( ) ( ) ( )( )

3 2

3 1

3 23 2

3 1 2 1 3 1 2 12 1

222 2

2

2 2

2

1 (2 )

1 2( )

1 2 4

2 22 2

2 22 2 2


M M M

M M M

ε εε ε ε ε

ε εε ε ε ελ

ε ε εελ

ω ω ε

λ λ ε λ λ ε


λ

+ ++ minus + minus

minus minus

leminus


le + + + + +

⎡ ⎤+ + +⎛ ⎞ ⎛ ⎞+ +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤le

minus

⎣

minus

+ ⎦

Therefore

where 2

M εε

ωλ

+=

minus So we have

( )( )( )( )

( )( ) ( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( ) ( )

( )( )( )

3 23 21 1 1

3 1 2 1 3 1 2 1

3 23 21 1

3 1 2 1 3 1 2 1

3 2

3 1 2 1

3 2

3 1 2

3 2 3 21 1 1

3 1 2 1 3 1 2 1

33 2

3 1 2 1

1

3 2 3 2

3 1 2 1 3 1 2 1

3



x x x xy y

x x x x x x x x

xx xx x x x

ε εε ε ε ε

ε εε ε ε ε


δε

δ

ε

+ ++ minus + minus

+ ++ minus + minus

+ +



+ minus+ +



le minusminus

minus

+minus

( )( )( )

( )( )( )( )

( )

21

3 1 2 1

3 21

3 1 2 1

2

3 1 2 1

3 2

3 1 2 1

2 21 2 4

xy

x x x x

x xx x x x

M

εε ε ε ε

ε εε ε ε ε

δ

ω ω ε ω δλ

++ minus + minus

+ ++ minus + minus

minus minus

+minus minus

le + +



( )2 21(0) (0) 6 12 3L L Mω ω ε ω δλ

minus le + +

where 2

M εε

ωλ

+=

minus

S20

Page - 462


27

















S20

Page - 463


28


461 Speed Tests







2 4 6 8 10


0237 0471 0708 0935 1176

0

1

2

3

4

5

6

7

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB


1183 2349 358 473 5917

S20

Page - 464


29


462 Size Tests




987 1973 2960 3947 4934


9845 19502 29195 38791 46789


100 99 99 98 95

0

02

04

06

08

1

12

14

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 465


30



Reference




0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

File

Siz

e in

KB

Secret Size in MB

CompressedSecret

CompressedContainer

S20

Page - 466


31

5 Geometric Methods






2 2

2 21 1 1 12 22 2 2 22 23 3 3 3

11

( ) 011

x y x yx y x y

J x yx y x yx y x y

++

= =++




or

2 2 1312 14

11 11 11


+ minus + minus =





2 2 1312 14

11 11 11


+ minus + minus =

we have

S20

Page - 467


32

12

11

13

11

2 2 14

11

2

2

C

C

C C C

MxM

MyM

Mr x yM

=

=

= + +


( ) 1312

11 11

2 2C C

MMx yM M

⎛ ⎞= ⎜ ⎟⎝ ⎠

and radius 2 2 14

11C C

Mx yM


shareable





21

1 1 2

1

1 11 2 1 2( ) ( 2 2 )2 2k kC C k k k k

kkx y +

+++= + + + + + +ε ε ε ε ε ε

and

1 2 1 21 21 2 2k

k k k kC kr + + + += + + + +ε ε ε


33( )




Cr We define


for 12 j n=

S20

Page - 468

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar


hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4


Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4


Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched


spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b


our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s


ommon

s Award 201

3

with center

sharing pair


13

33

S20

Page - 469


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


24

( ) ( ( ))j C j P jπ =

for 12 j n=


11 1 2 2( ) ( ) ( )( ( ))






1 2( )(0)t kΛΦ ε ε ε


1 2 3 4ε ε ε ε ( )2 1 2 3 4( ) 0ΛΦ ε ε ε ε 1 2 3 4ε ε ε ε ( )

2 1 2 3 4( ) 0ΛΦ ε ε ε ε 0000 1 0110 12 1000 2 0101 15 0100 3 0011 20 0010 4 1110 24 0001 5 1101 30 1100 6 1011 40 1010 8 0111 60 1001 10 1111 120



( )( ) ( )( ) ( )( )3 2 1 3 1 2

1 2 31 3 1 2 2 1 2 3 1 33 2




S20

Page - 460


25


( ) ( )( )( )

( )( )

2 1 2 2 1 1

2 1 2 1 2 1

2 1 2 1 2 1

2 1

2 1 2 1 2 1

1 1x x x x

x x x xx x x x

x x x x

ε ε

ε εε ε

ε εε ε

+ minus

+ minus minus=

+ minus

minus=

+

minusminus minus


minusminus minus



( )2 1 2 2 1 1

1 1 22xx xx

εε ε λ λ ε

minusminus minus

le+ minus minus

Observe that

( )( )( )( )

( )( )

( )( )( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )

3 23 2

3 1 2 1 3 1 2 1

3 23 2

3 1 2 1 3 1 2 1

3 2 3 2

3 1 2

3 2

3 1 2 1

3 2

3 2 3 2

3 1

3 2 3 2

3 1

1 3 1 2 1

3 2 3 2

3 1 2 1 3 1 2 13 1 2 1

3 1 2

1





x x x

ε εε ε ε ε

ε ε

ε ε ε εε ε






+ ++ minus + minus

+ +

+ + + ++ minus

+ + + ++ minus

=minus


( )( )( )

( ) ( ) ( )( )( )( ) ( ) ( )

3 2 3 2

3 2

3 1

3 2

2 31

3 2

2 1 3 1

3 1

3 1

3 2

3 1 2 1 2 12 1

1

1 1

1

x xx

x xx x x x x x

x xx x x x x x

ε ε ε ε

ε εε ε

ε εε ε ε ε

minus



+ +

+ ++ minus

+ ++ minus minus

S20

Page - 461


26

( )( )( )( )

( )( )

( ) ( ) ( )( )

3 2

3 1

3 23 2

3 1 2 1 3 1 2 12 1

222 2

2

2 2

2

1 (2 )

1 2( )

1 2 4

2 22 2

2 22 2 2


M M M

M M M

ε εε ε ε ε

ε εε ε ε ελ

ε ε εελ

ω ω ε

λ λ ε λ λ ε


λ

+ ++ minus + minus

minus minus

leminus


le + + + + +

⎡ ⎤+ + +⎛ ⎞ ⎛ ⎞+ +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤le

minus

⎣

minus

+ ⎦

Therefore

where 2

M εε

ωλ

+=

minus So we have

( )( )( )( )

( )( ) ( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( ) ( )

( )( )( )

3 23 21 1 1

3 1 2 1 3 1 2 1

3 23 21 1

3 1 2 1 3 1 2 1

3 2

3 1 2 1

3 2

3 1 2

3 2 3 21 1 1

3 1 2 1 3 1 2 1

33 2

3 1 2 1

1

3 2 3 2

3 1 2 1 3 1 2 1

3



x x x xy y

x x x x x x x x

xx xx x x x

ε εε ε ε ε

ε εε ε ε ε


δε

δ

ε

+ ++ minus + minus

+ ++ minus + minus

+ +



+ minus+ +



le minusminus

minus

+minus

( )( )( )

( )( )( )( )

( )

21

3 1 2 1

3 21

3 1 2 1

2

3 1 2 1

3 2

3 1 2 1

2 21 2 4

xy

x x x x

x xx x x x

M

εε ε ε ε

ε εε ε ε ε

δ

ω ω ε ω δλ

++ minus + minus

+ ++ minus + minus

minus minus

+minus minus

le + +



( )2 21(0) (0) 6 12 3L L Mω ω ε ω δλ

minus le + +

where 2

M εε

ωλ

+=

minus

S20

Page - 462


27

















S20

Page - 463


28


461 Speed Tests







2 4 6 8 10


0237 0471 0708 0935 1176

0

1

2

3

4

5

6

7

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB


1183 2349 358 473 5917

S20

Page - 464


29


462 Size Tests




987 1973 2960 3947 4934


9845 19502 29195 38791 46789


100 99 99 98 95

0

02

04

06

08

1

12

14

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 465


30



Reference




0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

File

Siz

e in

KB

Secret Size in MB

CompressedSecret

CompressedContainer

S20

Page - 466


31

5 Geometric Methods






2 2

2 21 1 1 12 22 2 2 22 23 3 3 3

11

( ) 011

x y x yx y x y

J x yx y x yx y x y

++

= =++




or

2 2 1312 14

11 11 11


+ minus + minus =





2 2 1312 14

11 11 11


+ minus + minus =

we have

S20

Page - 467


32

12

11

13

11

2 2 14

11

2

2

C

C

C C C

MxM

MyM

Mr x yM

=

=

= + +


( ) 1312

11 11

2 2C C

MMx yM M

⎛ ⎞= ⎜ ⎟⎝ ⎠

and radius 2 2 14

11C C

Mx yM


shareable





21

1 1 2

1

1 11 2 1 2( ) ( 2 2 )2 2k kC C k k k k

kkx y +

+++= + + + + + +ε ε ε ε ε ε

and

1 2 1 21 21 2 2k

k k k kC kr + + + += + + + +ε ε ε


33( )




Cr We define


for 12 j n=

S20

Page - 468

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar


hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4


Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4


Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched


spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b


our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s


ommon

s Award 201

3

with center

sharing pair


13

33

S20

Page - 469


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


25


( ) ( )( )( )

( )( )

2 1 2 2 1 1

2 1 2 1 2 1

2 1 2 1 2 1

2 1

2 1 2 1 2 1

1 1x x x x

x x x xx x x x

x x x x

ε ε

ε εε ε

ε εε ε

+ minus

+ minus minus=

+ minus

minus=

+

minusminus minus


minusminus minus



( )2 1 2 2 1 1

1 1 22xx xx

εε ε λ λ ε

minusminus minus

le+ minus minus

Observe that

( )( )( )( )

( )( )

( )( )( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )

3 23 2

3 1 2 1 3 1 2 1

3 23 2

3 1 2 1 3 1 2 1

3 2 3 2

3 1 2

3 2

3 1 2 1

3 2

3 2 3 2

3 1

3 2 3 2

3 1

1 3 1 2 1

3 2 3 2

3 1 2 1 3 1 2 13 1 2 1

3 1 2

1





x x x

ε εε ε ε ε

ε ε

ε ε ε εε ε






+ ++ minus + minus

+ +

+ + + ++ minus

+ + + ++ minus

=minus


( )( )( )

( ) ( ) ( )( )( )( ) ( ) ( )

3 2 3 2

3 2

3 1

3 2

2 31

3 2

2 1 3 1

3 1

3 1

3 2

3 1 2 1 2 12 1

1

1 1

1

x xx

x xx x x x x x

x xx x x x x x

ε ε ε ε

ε εε ε

ε εε ε ε ε

minus



+ +

+ ++ minus

+ ++ minus minus

S20

Page - 461


26

( )( )( )( )

( )( )

( ) ( ) ( )( )

3 2

3 1

3 23 2

3 1 2 1 3 1 2 12 1

222 2

2

2 2

2

1 (2 )

1 2( )

1 2 4

2 22 2

2 22 2 2


M M M

M M M

ε εε ε ε ε

ε εε ε ε ελ

ε ε εελ

ω ω ε

λ λ ε λ λ ε


λ

+ ++ minus + minus

minus minus

leminus


le + + + + +

⎡ ⎤+ + +⎛ ⎞ ⎛ ⎞+ +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤le

minus

⎣

minus

+ ⎦

Therefore

where 2

M εε

ωλ

+=

minus So we have

( )( )( )( )

( )( ) ( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( ) ( )

( )( )( )

3 23 21 1 1

3 1 2 1 3 1 2 1

3 23 21 1

3 1 2 1 3 1 2 1

3 2

3 1 2 1

3 2

3 1 2

3 2 3 21 1 1

3 1 2 1 3 1 2 1

33 2

3 1 2 1

1

3 2 3 2

3 1 2 1 3 1 2 1

3



x x x xy y

x x x x x x x x

xx xx x x x

ε εε ε ε ε

ε εε ε ε ε


δε

δ

ε

+ ++ minus + minus

+ ++ minus + minus

+ +



+ minus+ +



le minusminus

minus

+minus

( )( )( )

( )( )( )( )

( )

21

3 1 2 1

3 21

3 1 2 1

2

3 1 2 1

3 2

3 1 2 1

2 21 2 4

xy

x x x x

x xx x x x

M

εε ε ε ε

ε εε ε ε ε

δ

ω ω ε ω δλ

++ minus + minus

+ ++ minus + minus

minus minus

+minus minus

le + +



( )2 21(0) (0) 6 12 3L L Mω ω ε ω δλ

minus le + +

where 2

M εε

ωλ

+=

minus

S20

Page - 462


27

















S20

Page - 463


28


461 Speed Tests







2 4 6 8 10


0237 0471 0708 0935 1176

0

1

2

3

4

5

6

7

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB


1183 2349 358 473 5917

S20

Page - 464


29


462 Size Tests




987 1973 2960 3947 4934


9845 19502 29195 38791 46789


100 99 99 98 95

0

02

04

06

08

1

12

14

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 465


30



Reference




0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

File

Siz

e in

KB

Secret Size in MB

CompressedSecret

CompressedContainer

S20

Page - 466


31

5 Geometric Methods






2 2

2 21 1 1 12 22 2 2 22 23 3 3 3

11

( ) 011

x y x yx y x y

J x yx y x yx y x y

++

= =++




or

2 2 1312 14

11 11 11


+ minus + minus =





2 2 1312 14

11 11 11


+ minus + minus =

we have

S20

Page - 467


32

12

11

13

11

2 2 14

11

2

2

C

C

C C C

MxM

MyM

Mr x yM

=

=

= + +


( ) 1312

11 11

2 2C C

MMx yM M

⎛ ⎞= ⎜ ⎟⎝ ⎠

and radius 2 2 14

11C C

Mx yM


shareable





21

1 1 2

1

1 11 2 1 2( ) ( 2 2 )2 2k kC C k k k k

kkx y +

+++= + + + + + +ε ε ε ε ε ε

and

1 2 1 21 21 2 2k

k k k kC kr + + + += + + + +ε ε ε


33( )




Cr We define


for 12 j n=

S20

Page - 468

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar


hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4


Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4


Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched


spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b


our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s


ommon

s Award 201

3

with center

sharing pair


13

33

S20

Page - 469


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


26

( )( )( )( )

( )( )

( ) ( ) ( )( )

3 2

3 1

3 23 2

3 1 2 1 3 1 2 12 1

222 2

2

2 2

2

1 (2 )

1 2( )

1 2 4

2 22 2

2 22 2 2


M M M

M M M

ε εε ε ε ε

ε εε ε ε ελ

ε ε εελ

ω ω ε

λ λ ε λ λ ε


λ

+ ++ minus + minus

minus minus

leminus


le + + + + +

⎡ ⎤+ + +⎛ ⎞ ⎛ ⎞+ +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤le

minus

⎣

minus

+ ⎦

Therefore

where 2

M εε

ωλ

+=

minus So we have

( )( )( )( )

( )( ) ( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( ) ( )

( )( )( )

3 23 21 1 1

3 1 2 1 3 1 2 1

3 23 21 1

3 1 2 1 3 1 2 1

3 2

3 1 2 1

3 2

3 1 2

3 2 3 21 1 1

3 1 2 1 3 1 2 1

33 2

3 1 2 1

1

3 2 3 2

3 1 2 1 3 1 2 1

3



x x x xy y

x x x x x x x x

xx xx x x x

ε εε ε ε ε

ε εε ε ε ε


δε

δ

ε

+ ++ minus + minus

+ ++ minus + minus

+ +



+ minus+ +



le minusminus

minus

+minus

( )( )( )

( )( )( )( )

( )

21

3 1 2 1

3 21

3 1 2 1

2

3 1 2 1

3 2

3 1 2 1

2 21 2 4

xy

x x x x

x xx x x x

M

εε ε ε ε

ε εε ε ε ε

δ

ω ω ε ω δλ

++ minus + minus

+ ++ minus + minus

minus minus

+minus minus

le + +



( )2 21(0) (0) 6 12 3L L Mω ω ε ω δλ

minus le + +

where 2

M εε

ωλ

+=

minus

S20

Page - 462


27

















S20

Page - 463


28


461 Speed Tests







2 4 6 8 10


0237 0471 0708 0935 1176

0

1

2

3

4

5

6

7

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB


1183 2349 358 473 5917

S20

Page - 464


29


462 Size Tests




987 1973 2960 3947 4934


9845 19502 29195 38791 46789


100 99 99 98 95

0

02

04

06

08

1

12

14

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 465


30



Reference




0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

File

Siz

e in

KB

Secret Size in MB

CompressedSecret

CompressedContainer

S20

Page - 466


31

5 Geometric Methods






2 2

2 21 1 1 12 22 2 2 22 23 3 3 3

11

( ) 011

x y x yx y x y

J x yx y x yx y x y

++

= =++




or

2 2 1312 14

11 11 11


+ minus + minus =





2 2 1312 14

11 11 11


+ minus + minus =

we have

S20

Page - 467


32

12

11

13

11

2 2 14

11

2

2

C

C

C C C

MxM

MyM

Mr x yM

=

=

= + +


( ) 1312

11 11

2 2C C

MMx yM M

⎛ ⎞= ⎜ ⎟⎝ ⎠

and radius 2 2 14

11C C

Mx yM


shareable





21

1 1 2

1

1 11 2 1 2( ) ( 2 2 )2 2k kC C k k k k

kkx y +

+++= + + + + + +ε ε ε ε ε ε

and

1 2 1 21 21 2 2k

k k k kC kr + + + += + + + +ε ε ε


33( )




Cr We define


for 12 j n=

S20

Page - 468

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar


hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4


Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4


Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched


spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b


our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s


ommon

s Award 201

3

with center

sharing pair


13

33

S20

Page - 469


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


27

















S20

Page - 463


28


461 Speed Tests







2 4 6 8 10


0237 0471 0708 0935 1176

0

1

2

3

4

5

6

7

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB


1183 2349 358 473 5917

S20

Page - 464


29


462 Size Tests




987 1973 2960 3947 4934


9845 19502 29195 38791 46789


100 99 99 98 95

0

02

04

06

08

1

12

14

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 465


30



Reference




0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

File

Siz

e in

KB

Secret Size in MB

CompressedSecret

CompressedContainer

S20

Page - 466


31

5 Geometric Methods






2 2

2 21 1 1 12 22 2 2 22 23 3 3 3

11

( ) 011

x y x yx y x y

J x yx y x yx y x y

++

= =++




or

2 2 1312 14

11 11 11


+ minus + minus =





2 2 1312 14

11 11 11


+ minus + minus =

we have

S20

Page - 467


32

12

11

13

11

2 2 14

11

2

2

C

C

C C C

MxM

MyM

Mr x yM

=

=

= + +


( ) 1312

11 11

2 2C C

MMx yM M

⎛ ⎞= ⎜ ⎟⎝ ⎠

and radius 2 2 14

11C C

Mx yM


shareable





21

1 1 2

1

1 11 2 1 2( ) ( 2 2 )2 2k kC C k k k k

kkx y +

+++= + + + + + +ε ε ε ε ε ε

and

1 2 1 21 21 2 2k

k k k kC kr + + + += + + + +ε ε ε


33( )




Cr We define


for 12 j n=

S20

Page - 468

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar


hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4


Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4


Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched


spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b


our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s


ommon

s Award 201

3

with center

sharing pair


13

33

S20

Page - 469


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


28


461 Speed Tests







2 4 6 8 10


0237 0471 0708 0935 1176

0

1

2

3

4

5

6

7

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB


1183 2349 358 473 5917

S20

Page - 464


29


462 Size Tests




987 1973 2960 3947 4934


9845 19502 29195 38791 46789


100 99 99 98 95

0

02

04

06

08

1

12

14

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 465


30



Reference




0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

File

Siz

e in

KB

Secret Size in MB

CompressedSecret

CompressedContainer

S20

Page - 466


31

5 Geometric Methods






2 2

2 21 1 1 12 22 2 2 22 23 3 3 3

11

( ) 011

x y x yx y x y

J x yx y x yx y x y

++

= =++




or

2 2 1312 14

11 11 11


+ minus + minus =





2 2 1312 14

11 11 11


+ minus + minus =

we have

S20

Page - 467


32

12

11

13

11

2 2 14

11

2

2

C

C

C C C

MxM

MyM

Mr x yM

=

=

= + +


( ) 1312

11 11

2 2C C

MMx yM M

⎛ ⎞= ⎜ ⎟⎝ ⎠

and radius 2 2 14

11C C

Mx yM


shareable





21

1 1 2

1

1 11 2 1 2( ) ( 2 2 )2 2k kC C k k k k

kkx y +

+++= + + + + + +ε ε ε ε ε ε

and

1 2 1 21 21 2 2k

k k k kC kr + + + += + + + +ε ε ε


33( )




Cr We define


for 12 j n=

S20

Page - 468

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar


hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4


Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4


Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched


spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b


our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s


ommon

s Award 201

3

with center

sharing pair


13

33

S20

Page - 469


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


29


462 Size Tests




987 1973 2960 3947 4934


9845 19502 29195 38791 46789


100 99 99 98 95

0

02

04

06

08

1

12

14

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 465


30



Reference




0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

File

Siz

e in

KB

Secret Size in MB

CompressedSecret

CompressedContainer

S20

Page - 466


31

5 Geometric Methods






2 2

2 21 1 1 12 22 2 2 22 23 3 3 3

11

( ) 011

x y x yx y x y

J x yx y x yx y x y

++

= =++




or

2 2 1312 14

11 11 11


+ minus + minus =





2 2 1312 14

11 11 11


+ minus + minus =

we have

S20

Page - 467


32

12

11

13

11

2 2 14

11

2

2

C

C

C C C

MxM

MyM

Mr x yM

=

=

= + +


( ) 1312

11 11

2 2C C

MMx yM M

⎛ ⎞= ⎜ ⎟⎝ ⎠

and radius 2 2 14

11C C

Mx yM


shareable





21

1 1 2

1

1 11 2 1 2( ) ( 2 2 )2 2k kC C k k k k

kkx y +

+++= + + + + + +ε ε ε ε ε ε

and

1 2 1 21 21 2 2k

k k k kC kr + + + += + + + +ε ε ε


33( )




Cr We define


for 12 j n=

S20

Page - 468

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar


hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4


Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4


Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched


spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b


our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s


ommon

s Award 201

3

with center

sharing pair


13

33

S20

Page - 469


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


30



Reference




0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

File

Siz

e in

KB

Secret Size in MB

CompressedSecret

CompressedContainer

S20

Page - 466


31

5 Geometric Methods






2 2

2 21 1 1 12 22 2 2 22 23 3 3 3

11

( ) 011

x y x yx y x y

J x yx y x yx y x y

++

= =++




or

2 2 1312 14

11 11 11


+ minus + minus =





2 2 1312 14

11 11 11


+ minus + minus =

we have

S20

Page - 467


32

12

11

13

11

2 2 14

11

2

2

C

C

C C C

MxM

MyM

Mr x yM

=

=

= + +


( ) 1312

11 11

2 2C C

MMx yM M

⎛ ⎞= ⎜ ⎟⎝ ⎠

and radius 2 2 14

11C C

Mx yM


shareable





21

1 1 2

1

1 11 2 1 2( ) ( 2 2 )2 2k kC C k k k k

kkx y +

+++= + + + + + +ε ε ε ε ε ε

and

1 2 1 21 21 2 2k

k k k kC kr + + + += + + + +ε ε ε


33( )




Cr We define


for 12 j n=

S20

Page - 468

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar


hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4


Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4


Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched


spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b


our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s


ommon

s Award 201

3

with center

sharing pair


13

33

S20

Page - 469


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


31

5 Geometric Methods






2 2

2 21 1 1 12 22 2 2 22 23 3 3 3

11

( ) 011

x y x yx y x y

J x yx y x yx y x y

++

= =++




or

2 2 1312 14

11 11 11


+ minus + minus =





2 2 1312 14

11 11 11


+ minus + minus =

we have

S20

Page - 467


32

12

11

13

11

2 2 14

11

2

2

C

C

C C C

MxM

MyM

Mr x yM

=

=

= + +


( ) 1312

11 11

2 2C C

MMx yM M

⎛ ⎞= ⎜ ⎟⎝ ⎠

and radius 2 2 14

11C C

Mx yM


shareable





21

1 1 2

1

1 11 2 1 2( ) ( 2 2 )2 2k kC C k k k k

kkx y +

+++= + + + + + +ε ε ε ε ε ε

and

1 2 1 21 21 2 2k

k k k kC kr + + + += + + + +ε ε ε


33( )




Cr We define


for 12 j n=

S20

Page - 468

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar


hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4


Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4


Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched


spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b


our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s


ommon

s Award 201

3

with center

sharing pair


13

33

S20

Page - 469


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


32

12

11

13

11

2 2 14

11

2

2

C

C

C C C

MxM

MyM

Mr x yM

=

=

= + +


( ) 1312

11 11

2 2C C

MMx yM M

⎛ ⎞= ⎜ ⎟⎝ ⎠

and radius 2 2 14

11C C

Mx yM


shareable





21

1 1 2

1

1 11 2 1 2( ) ( 2 2 )2 2k kC C k k k k

kkx y +

+++= + + + + + +ε ε ε ε ε ε

and

1 2 1 21 21 2 2k

k k k kC kr + + + += + + + +ε ε ε


33( )




Cr We define


for 12 j n=

S20

Page - 468

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar


hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4


Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4


Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched


spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b


our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s


ommon

s Award 201

3

with center

sharing pair


13

33

S20

Page - 469


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488

G

and r

Then

is

A k 3( Λ Φ

54 4

In 3

that t10)

Sh

Given 1 1( )x y

radius

n the k bits ε

2MM

⎛⎜⎝

bit ( )3n se

3 )Λ

Φ

Secret Shar


hing-Tung Y

) 2 2( )x y an

1 2 kε ε ε such

(3ΛΦ ε

12

11 mod 22MM

M M⎞ ⎛⎟ ⎜⎠ ⎝

ecret sharing

reable Set Λ4


Figure 10 T

Yau High Sch

nd 3 3( )x y

( )C Cx y

Cr x=

h that

)1 2 k = Ψε ε ε

13

11 mod 2

MM

⎛⎞ ⎛⎜⎟ ⎜⎜⎠ ⎝⎝

storage can

4


Two different s

hool Applied

Ψ maps the

12

11

2 2

MMM M

⎛= ⎜⎝

2 2C C

Mx yM

+ +

( 1 1 2( ) (x y xΨ

2

12

112MM

⎛ ⎞ ⎛+⎜ ⎟ ⎜

⎝ ⎠ ⎝

be launched


spheres have fo

d Mathemati

ese points to

13

11

MM

⎞⎟⎠

14

11

MM

2 2 3 3 ) ( y x y

2

13

112M MM M

⎛ ⎞+⎟

⎝ ⎠

d for the k b


our points in co

ical Sciences

o the circle w

))

14

11mod 2

1MM

⎞⎟minus⎟⎠

bit 3 secret s


ommon

s Award 201

3

with center

sharing pair


13

33

S20

Page - 469


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


34






( ) ( | ) ( ) ( | ) ( )c cP A P A B P B P A B P B= +



14

n minus⎛ ⎞⎜ ⎟⎝ ⎠





1 1 1

2 2 2

3 3 3

0x y zx y zx y z

ne

S20

Page - 470


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


35








2 2 2

2 2 21 1 1 1 1 12 2 22 2 2 2 2 22 2 23 3 3 3 3 32 2 24 4 4 4 4 4

11

( ) 0111



+ ++ +

= =+ ++ ++ +


121

11

132

11

143

11

2 2 2 151 2 3

11

2

2

2

MvM

MvM

MvM

Mr v v vM

=

=

=

= + + +



S20

Page - 471


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


36




( )S v r such that

2

2

2 3

2 2 2 3

1

1

1

1 1 1

1

1 1 1

1 2

1 2 3 1 2

1 2

2 2

( ) 2 2

2 2

kk

k kk k

kk

kTk k k

kk k kk k

v v v ++

+ ++ +

+ +

+ + + +

⎡ ⎤+ + +⎢ ⎥

= + + +⎢ ⎥⎢ ⎥+ + +⎢ ⎥⎣ ⎦

ε ε ε

ε ε ε

ε ε ε

and

1 1 23 32 1 21 2 2k

k k k k k k kr + + + ++ += + + + +ε ε ε


44( )




( )( ( ))j jS v r v r uπ = +

for 12 j n=



2 2 2

13 13 1512 14 12 14

11 11 11 11 11 11 11mod 2 mod 2 mod 2mod 2

12 2 2 2 2 2


⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟+ + + minus⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠


ΛΛ Φ


S20

Page - 472


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


37




3( )Λ

Λ Ψ





561 Speed Test





1139 2264 3378 4506 5676


0

1

2

3

4

5

6

2 4 6 8 10

Tim

e in

sec

Secret size in MB

S20

Page - 473


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


38




2 4 6 8 10


0151 0313 0457 0607 0763


562 Size Tests




987 1973 2960 3947 4934


169337 33176 491733 65017 807347


172 168 166 165 164

0010203040506070809

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 474


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


39



Reference



0

2000

4000

6000

8000

10000

12000

14000

2 4 6 8 10

Size

in K

B

Secret size in MB


S20

Page - 475


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


40






1 mod 3)2 mo

(( 5d )

3 mod )( 7

xxx

equivequivequiv



( ( (

1 m( 5 ( 5 ( 5( 7 ( 7d( o )7

x x xx x xx x x





1

2

3

mod3)mod )

(( 5mod )( 7

x yx yx y

equivequivequiv


S20

Page - 476


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


41



( )( )( )

( )

1 1

2 2

3 3

modmodmod

modm m

x a px a px a p

x a p

equivequivequiv

equiv





and let 1 2 3 1

1

m m

m m

p p p p pp p

minus

minus



and


Suppose that 1l

C and 2l



( ) ( )( ) ( )( ) ( )

1 1 1 2 1

2 2 2 2 2

2 2 2 2 2

1

1

1



===


( )( )( )

1 1

1 2

1 3

modmodmod

x l px l px l p

equivequivequiv

and ( )( )( )

2

3

2

2

2

1modmodmod

x l px l px l p

equivequivequiv

respectively

S20

Page - 477


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


42




2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv




2 2

1 1modmod

xx

y py p

equivequiv








such that

1 2 2 2kkl = + + +ε ε ε


hence 1 21 2 ( )( ( ) )





S20

Page - 478


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


43




Then

1 2( ) 1 2( )mp p p k lCΛΦ =ε ε ε


2

3

1 1

2

3

modmodmod

y py p

x yx

p

x equivequivequiv



sharing pair 1 21 2 ( )( ( ) )



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 31 37 41 43 47 53 59 61 67 71

Therefore we have

1 2 3 47027p p p = and 9 10 4757p p =

Hence 1 2 3 9 10

9 10


= =


1p 2p 3p 4p 5p 6p 7p 8p 9p 37 41 43 47 53 59 61 67 71

So

1 2 3 65231p p p = and 8 9 4757p p =

Hence 1 2 3 8 9

8 9


= =

S20

Page - 479


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


44



1p 2p 3p 4p 5p 6p 7p 8p 9p 10p 41 43 47 53 59 61 67 71 73 79

and

1 2 3 82861p p p = and 9 10 5767p p =

Hence 1 2 3 9 10

9 10



secret file

641 Speed tests




0161 0304 0454 0588 0718


000001000200030004000500060007000800

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 480


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


45





642 Size Tests




987 1973 2960 3947 4934


3941 7885 11834 1577 19719


40 40 40 40 40

00000020004000600080010001200140016001800200

2 4 6 8 10

Tim

e in

Sec

Secret Size in MB

S20

Page - 481


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


46



Reference


0

1000

2000

3000

4000

5000

6000

2 4 6 8 10

Size

in K

B

Secrect Size in MB

Compressed Secret


S20

Page - 482


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


47

7 Conclusion










S20

Page - 483

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

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會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

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ety for Science

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c

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4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

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名



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項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

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S20

Page - 486


51







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年




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全國獎項


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者




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2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488

姓名

性別

出生

學歷

年份

2013

2013

2013

2011

2010

2009

2008

2007

2009

2008

2007

Sh

名張顥霆

別男

生日期19

歷澳門培

比賽

港澳數學

公開賽

全國青少

奧林匹克

澳門信息

克選拔賽

全澳校際

全澳校際

港澳數學

公開賽

hing-Tung Y

霆

996 年 12 月

培正中學(

賽活動

學奧林匹克

少年信息學

克競賽

息學奧林匹

賽

際數學比賽

際數學比賽

學奧林匹克

au High Sch

履

月 4 日

高三年級

獲

主

克香港數

克協會

學中國計

匹澳門電

賽澳門教

局

賽澳門教

局

克香港數

克協會

hool Applied

履歷表

)

獲獎經驗

主辦單位

數學奧林匹

會

計算機學會

電腦學會

教育暨青年

教育暨青年

數學奧林匹

會

d Mathemati

所獲

匹金

會決賽

一

年二

年一

匹金

ical Sciences

獲獎項

金獎

賽選手

一等獎

二等獎

一等獎

金獎

s Award 201

4

備註

第二名

13

48

S20

Page - 484

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488

姓名

英文

性別

出生

學歷

2008

2001

2001

才藝

鋼琴

大提琴

參加團

2008

2007

2006

會員

2012

2013

Sh

名譚知微

文姓名拼音

別女

生日期19

- 培正

- 聖庇

-2008 培正

皇家

琴

團體

澳門

澳門

- 2010 澳門

- Me

- Stu

hing-Tung Y

微

音TAN Ch

996 年 09 月

正中學

庇護十世音樂

正中學小學部

家音樂學院八

門青年交響樂

門青年交響樂

門培正中學絃

ember of Socie

udent Member

au High Sch

履

hih Wei

月 27 日

樂學院

部

八級 (Pass w

樂團 A 團

樂團 B 團

絃樂團

ety for Science

r of Association

hool Applied

履歷表

with Distinct

大提

大提

大提

e amp the Public

n for Computin

d Mathemati

tion)

提琴手

提琴手

提琴手

c

ng Machinery


4

13

49

S20

Page - 485


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


50

所獲獎項



新大賽










府共同主辦

周培源青少

年科技創新

獎和二等銀

獎


名



爭光而獲獎






項目優異獎


科學獎



限公司

優勝獎

全球決賽



獎-南部賽區








新大賽





內蒙古自治

區主席獎和

一等金獎

全國獎項

S20

Page - 486


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


51







同主辦




獲獎





年




獎

全國獎項


澳區比賽



者




公開賽獎項






織全國委員會








2005 -2013


S20

Page - 487


52

著作



公演及展示


















S20

Page - 488


52

著作



公演及展示


















S20

Page - 488

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