ch1 - algorithms with numbers basic arithmetic basic arithmetic addition addition multiplication...

Ch1 - Algorithms with numbers

Basic arithmetic Addition Multiplication Division

Modular arithmetic RSA –factoring is hard Primality testing

Addition

53+35=88

Cost? (n – number of bits) O(n)

Multiplication 13x11=143

Cost? O(n2)

al-Khwārizmī

Operations determining parity (even or odd) addition duplation (doubling a number, left shift) mediation (halving a number, rounding down,

right shift)

al-Khwārizmī

Cost? O(n2) Can we do better?

Division

Cost?

Modular arithmetic A system for dealing with restricted

ranges of integers

Addition x+y mod N, assuming x, y <N O(n), n - number of bits N has (size of

input)(x+y mod N = x+y or x+y-N)

Multiplication x*y mod N ?

Modular arithmetic

RSA Ron Rivest, Adi Shamir, Leonard

Adleman (1977) Algorithm for public-key cryptography,

based on the presumed difficulty of the factoring problem.

2002 A.M. Turing Award RSA is one of the most used

cryptographic protocols on the net. Your browser uses it to establish a secure session with a site.

Needed for implementing RSA: FLT (Fermat’s Little Theorem) Fast Exponentiation Extended Euclidean Algorithm Modular inverses CRT (Chinese Remainder Theorem)

Turing Lecture on Early RSA Days, Ronald

L. Rivest

Turing Lecture on Early RSA Days, Ronald L. Rivest

Turing Lecture on Early RSA Days, Ronald L. Rivest

In April 2012, the factorization of 143 is achieved.

RSA public-key cryptosystem

In a public-key cryptosystem, everyone has a public key and a secret key. Suppose Alice and Bob are two participants.

Alice PA , SA

Bob PB , SB

The keys specify 1-1 functions from message M to itself:

M= SA (PA (M))M= PA (SA (M))

Communicationchannel

PA(M)

encrypt decrypt

Encryption:

M PA SA

M

Bob Alice

RSADigital signatures:

Communicationchannel

SA(M)

Accept=?

SA PA

M MBobAlice

RSA algorithm Select at random 2 large prime numbers p &

q;(p & q might be, say, 100 decimal digits each.)

Compute n: n = pq; Select an odd integer e that is relatively prime

to (n) = (p-1)(q-1);

Compute d as the multiplicative inverse of e, modulo (n);

(de 1 mod (n)) Publish P = (e, n) as the RSA public key; Keep secret S = (d, n) as the RSA secret key.

If M Zn ={0,1,…,n-1},

P(M) = Me mod nS(C) = Cd mod n, C=P(M).

RSA examplePick p = 47, q=71.n=pq=3337.(n) = (p-1)(q-1)=46*70=3220, choose e=79 (at random).d =79-1 mod 3220 = 1019.PA=(79, 3337).

SA=(1019, 3337).Message: M = 6882326879666683

= 688 232 687 966 668 3M1 = 688 68879 mod 3337 = 1570 =C1M2 = 232 23279 mod 3337 = 2756 =C2…C = 1570 2756 2091 2276 2423 158C1 = 1570 15701019 mod 3337 = 688 =M1…C2 = 158 1581019 mod 3337 = 3 =M2

Another example

n = 4559, e = 13. Smiley Transmits: “Last name

Smiley” L A S T N A M E S M I L E

Y 1201 1920 0014 0113 0500 1913 0912

0525

120113 mod 4559, 192013 mod 4559, …

1074 0116 1478 2150 3906 4256 1445 2462

m e mod n

RSA

Bob receives the encrypted blocks c = m e mod n. He have a private decryption exponent d which when applied to c recovers the original blocks m : (m e mod n )d mod n = m

For n = 4559, e = 13 the decryptor d = 3397.

RSA

n = 4559, d = 3397 1074 0116 1478 2150 3906 4256 1445

2462

1074 3397 mod 4559, 01163397 mod 4559, …

1201 1920 0014 0113 0500 1913 0912 0525

L A S T N A M E S M I L E Y

RSA

Technical difficulties:

How do we know the algorithm works correctly?

How to pick large prime numbers? Compute pq How to choose e Compute d How to compute Me, Cd

Can any one break the code?

RSA

If I want to encrypt credit card numbers, how big my p and q should be?

If I want to encrypt words of four random characters from ASCII set, how big my p and q should be?

How to pick large prime numbers ?

Primality testing Hard, but much easier than factoring. Fermat’s Little Theorem(~1640):

If p is prime, then a, s.t. 1≤a<p, ap-11 (mod p).

The numbers make us fail are called Fermat pseudoprime -extremely rare (ex. 2340=1mod341; Carmichael number 561, 2560=1mod561)

?

Lagrange’s Prime Number Theorem

Theorem: The number of prime numbers between 1 and x is “about” x/lnx .

Not only are primes easy to detect, but they are also relatively abundant.

Carmichael number

A number c is a Carmichael number if it is not a prime, and still for all prime divisors d of c it so happens that d-1divides c-1. The smallest Carmichael number is 561 = 31117 .

If c is a Carmichael number and a is relatively prime to c, then ac-1 1 mod c.

Primality testing

Primality testing

Fermat's Last Theorem

Fermat's Last Theorem states that

xn + yn = zn has no non-zero

integer solutions for x, y and z when n > 2.

RSA

Technical difficulties:

How do we know the algorithm works correctly?

How to pick large prime numbers? Compute pq How to choose e Compute d How to compute Me, Cd? Can any one break the code?

How to compute Me, Cd ?

Modular exponentiation

In order to implement RSA, exponentiation relative some modulo needs to be done a lot. So this operation better be doable, and fast.

Q: How is it even possible to compute 28533397 mod 4559 ? After all, 28533397 has approximately 3397·4 digits!

Modular exponentiation

A: By taking the mod after each multiplication.

For example:

233 mod 30 -73 (mod 30) (-7)2 ·(-7) (mod 30) 49 · (-7) (mod

30) 19·(-7) (mod 30) -133 (mod 30) 17 (mod 30)

Modular exponentiation

Therefore, 233 mod 30 = 17.

Q: What if had to figure out 2316 mod 30. Same way tedious: need to multiply 15 times.

Is there a better way?

Modular exponentiation

A: Better way. Notice that 16 = 2·2·2·2 so that 2316 = 232·2·2·2 = (((232)2)2)2

Therefore:2316 mod 30 (((-72)2)2)2 (mod 30) (((49)2)2)2 (mod 30) (((-11)2)2)2 (mod 30) ((121)2)2 (mod 30) ((1)2 )2 (mod 30) (1)2 (mod 30) 1(mod 30)

Which implies that 2316 mod 30 = 1.Q: How about 2325 mod 30 ?

Modular exponentiation

A: The previous method of repeated squaring works for any exponent that’s a power of 2. 25 isn’t. However, we can break 25 down as a sum of such powers: 25 = 16 + 8 + 1. Apply repeated squaring to each part, and multiply the results together. Previous calculation:

238 mod 30 = 2316 mod 30 = 1 Thus: 2325 mod 30 2316+8+1 (mod 30)

Modular exponentiation

x25 mod N

Cost? – polynomial time (n=logN)

Modular exponentiation

How do we compute xy mod m , m>0?

repeated squaring algorithm:

mod-exp(x, y, m)if y = 0 then return(1)else

z = mod-exp(x, y div 2, m)if y mod 2 = 0 then return(z * z mod m)else return(x * z * z mod m)

Compute d ?

Modular Inverse

GCD

Greatest common divisor

Example:

Euclid AlgorithmIf a,bZ+, apply division (mod) repeatedly

as follows:a = q1b + r1, where 0 < r1 < b

b = q2r1 + r2, where 0 < r2 < r1

r1 = q3r2 + r3, where 0 < r3 < r2

……rk-2 = qkrk-1+ rk, where 0 < rk-1 <

rk

rk-1 = qk+1rk

Then, rk = GCD(a,b).  

Proof: (1) rk|a, rk|b

(2) if d|a, d|b, then d| rk.

Recursion Theorem

a,b N, b0, gcd(a,b) = gcd(b, a mod b).

Proof :

Let d = gcd(a,b). d|a, d|b.d|a-qb = a mod b d|b, d|a mod b d|gcd(b, a mod b).

Let d = gcd(b, a mod b). d|b, d| a mod b.d|a-qb, d|b d|a d|gcd(a,b).

gcd(a,b) = gcd(b, a mod b).

Computing GCD

Euclid gcd(x,y) {if y = 0 then return(x)else return(gcd(y,x mod y))

}

Euclid AlgorithmExample: Computing gcd(125, 87)

125 = 1*87 + 38 87 = 2*38 + 11 38 = 3*11 + 5 11 = 2*5 + 1 5 = 5*1

gcd(125,87)=1

gcd(125,87) = 111 - 2*5 = 111 - 2*(38-3*11) = 1 - 2*38 + 7*11 = 1- 2*38 + 7*(87 - 2*38) = 17*87 - 16*38 = 17*87 - 16*(125-1*87) = 1- 16*125 + 23*87 = 1

1 = 125*(-16) + 87*231 = as + bt

Extended Euclidean Algorithm

obtain gcd(a,b) and x,y, s.t. gcd(a,b) = ax+by.

Extended-Euclid (a,b)if (b= =0)

return (a,1,0);(d’,x’,y’)=Extended-Euclid(b, a mod b);(d,x,y)=(d’, y’, x’-a/by’);return (d,x,y);

Ex:

4

1

0

5

4

20

4

0

1

0

4

4

5

-2

2

44

108

4

-7

5 1

4

-2

-7

4

12

12

4

-19

2

20

44

1

152

260

1

260

412

1

108

152

x

q

d

y

b

a

demo

Cost?

Theorem: The algorithm above correctly computes the gcd of x and y in time O(n), where n is the total number of bits in the input (x; y)

Multiplicative Inverse

Multiplicative inverse x of a, modulo n:

ax = 1 mod n.ax = kn+1If gcd(a,n)=1, ax-kn = gcd(a,n).

ax+ny = gcd(a,n).Therefore, x can be found using

extended Euclidean algorithm. Is the multiplicative inverse unique?

Multiplicative Inverse

Theorem: n>1, if gcd(a,n)=1, then ax=1 (mod n) has a unique positive solution, modulo n.

Example:a = 79; n = 3220.x = 1019.ax = 80501 = 25*3220+1.

x = -2201. ax = -173879 = -54*3220+1.

RSA

Technical difficulties:

How do we know the algorithm works correctly?

How to pick large prime numbers? Compute pq How to choose e Compute d How to compute Me, Cd? Can any one break the code?

How do we know RSA works correctly?

Chinese Remainder Theorem (~1700 old)

http://en.wikipedia.org/wiki/RSA_Factoring_Challenge#The_prizes_and_records