nxn application to cryptography

8/10/2019 nxn application to cryptography

1/34

ENGG2013 Unit 10

n n determinant and

an application to cryptographyFeb, 2011.


2/34

YesterdayA formula

for matrix inverse using cofactors

kshum ENGG2013 2

Suppose that det Ais nonzero.Three steps in computing above formula

1. for i,j = 1,2,3, replace each aijby cofactor Cij2. Take the transpose of the resulting matrix.

3. divide by the determinant of A.

Usually called the adjoint of A

cofactors


3/34

Outline

nxn determinant

Caesar Cipher

Modulo arithmetic Hill Cipher

kshum ENGG2013 3


4/34

DETERMINANT IN GENERAL

kshum ENGG2013 4


5/34

A pattern

Arrange the products so that the first

subscripts are in ascending order.

All possible orderings of the second subscripts

appear once and only once.kshum ENGG2013 5


6/34

Transposition

A transpositionis an exchange of two objects

in a list of objects.

kshum ENGG2013 6

A B C D

A C B D

Examples:

2 1 4 5 3

1 2 4 5 3

Transposition is another

mathematical term, and is

not the same as matrix tranpose.


7/34

Another pattern

The sign of each term is closely related to the

number of transpositions required to obtain

the second subscripts, starting from (1,2) for

the 2x2 case or (1,2,3) for the 3x3 case.kshum ENGG2013 7


8/34

The sign

Let p(1), p(2), , p(n) be an order of 1,2,,n.

For example p(1)=3, p(2) = 2, p(3)=1 is an ordering

of 1, 2, 3.

Starting from (1,2,,n), if we need an oddno.

of transpositions to get ( p(1), p(2), , p(n) ),

we define the sign of (p(1), p(2),,p(n)) be1.

Otherwise, if we need an evenno. of

transpositions to get ( p(1), p(2), , p(n) ), we

define the sign of (p(1), p(2),,p(n)) be +1.

kshum ENGG2013 8


9/34

Definition of nn determinant

The summation is over all n! possible

orderings p= ( p(1), p(2), , p(n) ) of 1,2,,n.

There are n! terms.

sgn(p) is either +1 or1, usually called the

signature or signumof p.

kshum ENGG2013 9

http://en.wikipedia.org/wiki/Determinant

1
http://en.wikipedia.org/wiki/Determinanthttp://en.wikipedia.org/wiki/Determinant


10/34

Properties of determinant

Determinant of nn identity matrix equals 1.

Exchange two rows (or columns)multiply

determinant by1.

Multiply a row (or a column) by a constant k

multiply the determinant by k.

Add a constant multiple of a row (column) to

another row (column)no change

Additive property as in the 33 and 22 case.

kshum ENGG2013 10


11/34

Cofactor and the adjoint formula

for matrix inverse Cofactors are defined in a similar way as in the 3x3 case.

The cofactor of the (i,j)-entry of a matrix A, denoted by Cij, isdefined as (1)i+jAij, where A is the determinant of the sub-matrix obtained by removing the i-th row and the j-th column.

We have similar expansion along a row or a column (alsocalled the Laplace expansion) as in the 3x3 case.

The adjoint formula:

kshum ENGG2013 11

nxn identityA adjoint of A

The formula in this form holds when det A= 0 also

transpose


12/34

CAESAR CIPHER

kshum ENGG2013 12


13/34

Caesar and his army

kshum ENGG2013 13

ATTACK

Soldier carrying the

message ATTACK

Message may be intercepted

by enemy


14/34

Caesar cipher

kshum ENGG2013 14

http://en.wikipedia.org/wiki/Caesar_cipher

ATTACK

Soldier carrying the

encrypted message

DWWDFN

The encrypted message

looks random and meaningless
http://en.wikipedia.org/wiki/Caesar_cipherhttp://en.wikipedia.org/wiki/Caesar_cipher


15/34

Private key encryption

kshum ENGG2013 15

Plain text Encryption

function Ciphertext

Plain text Decryptionfunction Ciphertext

Key

key

The value of key is keptsecret


16/34

Mathematical description

kshum ENGG2013 16

ATTACK Shift to the right

by 3 DWWDFN

ATTACK Shift to the leftby 3 DWWDFN

Key =3

Key = 3

Caesar cipher is not secure

enough, because the numberof keys is too small.


17/34

MODULO ARITHMETIC

kshum ENGG2013 17


18/34

Mod 12

Clock arithmetic

kshum ENGG2013 18

121

2

9 3

6

4

57

8

10

11

6+8= 2 mod 12

5+12 = 5 mod 12


19/34

Mod 7

Week arithmetic

kshum ENGG2013 19

6

1+9 = 3 mod 7

2+3 = 5 mod 7

Sun Mon Tue Wed Thr Fri Sat

1

2 3 4 5 6 7 8

9 10 11 12 13 14 15

16 17 18 19 20 21 2223 24 25 26 27 28 29

30 31

0 1 2 3 4 5 6


20/34

Mod 60

arithmetic

kshum ENGG2013 20

http://www.h

ko.gov.h

k/gts/time/stemsandbranchesc.h

tm

1 2 3 4 5 6 7 8 9 10 11 12

13 14 15 16 17 18 19 20 21 22 23 24

25 26 27 28 29 30 31 32 33 34 35 36

37 38 39 40 41 42 43 44 45 46 47 48

49 50 51 52 53 54 55 56 57 58 59 60

Year of rabbit
http://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htmhttp://www.hko.gov.hk/gts/time/stemsandbranchesc.htm


21/34

Mod nformal definition

nis a fixed positive integer

Definition: amod nis the remainder of aafter

division by n.

Example: 25 = 1 mod 12.

Addition and multiplication: If the sum or

product of two integers is larger than or equal

to n, divide by n and take the remainder.

Example: 2+10 = 0 mod 12.

Example: 25 = 3 mod 12.kshum ENGG2013 21


22/34

More examples

10 mod 7 = 3

4+5 mod 7 = 2

6+7 mod 7 = 6 27 mod 7 = 0

kshum ENGG2013 22


23/34

Mod 26

A B C D E F G H I J K L M

0 1 2 3 4 5 6 7 8 9 10 11 12

kshum ENGG2013 23

N O P Q R S T U V W X Y Z13 14 15 16 17 18 19 20 21 22 23 24 25

Fix a one-to-one correspondence between the English alphabets

and the integers mod 26.

Caesars cipher: shifting a letter to the right by 3

is the same as adding 3 in mod 26 arithmetic.


24/34

Examples of mod 26 calculations

3+19 = ? mod 26

13+20 = ? mod 26

34 = ? Mod 26

134 = ? Mod 26

kshum ENGG2013 24


0 1 2 3 4 5 6 7 8 9 10 11 12

N O P Q R S T U V W X Y Z

13 14 15 16 17 18 19 20 21 22 23 24 25


25/34

Peculiar phenomena

in modulo arithmetic

Non-zero times non-zero may be zero

49 = 0 mod 12

22 = 0 mod 4

Multiplicative inverse may not exist

Cannot find an integer x such that 4x = 1 mod 12.

4-1does not exist mod 12.

kshum ENGG2013 25


26/34

No fraction in modulo arithmetic

In mod 12, dont write 1/3 or 3-1because it

does not exist.

But 5-1is well-defined mod 12, because we

can solve 5x=1 mod 12.

Indeed, we have 55 = 1 mod 12.

Therefore 5-1 = 5 mod 12.

kshum ENGG2013 26

FractionFact from number theory:multiplicative inverse of x mod n exists

if and only the gcd of x and n is 1.


27/34

HILL CIPHER

kshum ENGG2013 27


28/34

Hill cipher

Invented by L. S. Hill in 1929. Inputs : String of English letters, A,B,,Z.

An nnmatrix K, with entries drawn from 0,1,,25.(The matrix Kserves as the secret key. )

Divide the input string into blocks of size n. Identify A=0, B=1, C=2, , Z=25.

Encryption: Multiply each block by Kand thenreduce mod 26.

Decryption: multiply each block by the inverse ofK, and reduce mod 26.

kshum ENGG2013 28

http://en.wikipedia.org/wiki/Hill_cipher
http://en.wikipedia.org/wiki/Hill_cipherhttp://en.wikipedia.org/wiki/Hill_cipher


29/34

Note

The decryption must be the inverse function of

the encryption function.

It is required that K-1K= Inmod 26.

Provided that det(K) has a multiplicative inversemod 26, i.e., if det(K) and n has no common

factor, the inverse of Kcan be computed by the

adjoint formula for matrix inverse. Inverse of an integer mod 26 can be obtained by

trial and error.

kshum ENGG2013 29


30/34

Example

Plain text: LOVE, Secret Key: LO

VE

2, 3, 16, 5 are transformed to cipher text

CDQF

kshum ENGG2013 30


0 1 2 3 4 5 6 7 8 9 10 11 12

N O P Q R S T U V W X Y Z

13 14 15 16 17 18 19 20 21 22 23 24 25


31/34

How to decode?

Given CDQF, and the encryption matrix

How do we decrypt?

We need to compute the inverse of

Remind that all arithmetic are mod 26. There

is no fraction and care should be taken in

computing multiplicative inverse mod 26.

kshum ENGG2013 31


32/34

Determinant

The determinant of equals 20(7)-3(15),

which is 17 mod 26.

Find the multiplicative inverse of 17 mod 26,

i.e., find integer x such that 17x = 1 mod 26.

Just try all 26 possibilities for x:

kshum ENGG2013 32

171 = 17 mod 26

172= 8 mod 26173 = 25 mod 26

174 = 16 mod 26

175 = 7 mod 26

176 = 24 mod 26

177 = 15 mod 26

178 = 6 mod 26

179= 23 mod 261710 = 14 mod 26

1711 = 5 mod 26

1712 = 22 mod 26

1713 = 13 mod 26

1714 = 4 mod 26

1715 = 21 mod 26

1716= 12 mod 261717 = 3 mod 26

1718 = 20 mod 26

1719 = 11 mod 26

1720 = 2 mod 26

1721 = 19 mod 26

1722 = 10 mod 261723= 1 mod 26

1724 = 18 mod 26

1725 = 9 mod 26

170 = 0 mod 26


33/34

Computing the inverse mod 26

From 1723= 1 mod 26, we know that the

multiplicative inverse of 17 mod 26 is 23.

Using the formula for 2 2 matrix inverse

we get

kshum ENGG2013 33

Replace (17)-1mod 26 by 23


34/34

Decryption

Given the ciphertext CDQF, we decrypt by

multiplying by

From the table in p.23, 11, 14, 21, 4 is LOVE.

kshum ENGG2013 34

nxn application to cryptography

Documents