number-theoretic algorithms (rsa and related...

p1.

Number-Theoretic Algorithms(RSA and related algorithms)

Chapter 31, CLRS book

p2.

Outline

• Modular arithmetic

• RSA encryption scheme

• Miller-Rabin algorithm (a probabilistic algorithm)

p3.

Modular Arithmetic

p4.

| : divides , is a divisor of .

gcd( , ): greatest common divisor of and .

Coprime or relatively prime: gcd( , ) 1.

Euclid's algorithm: compute gcd( , ).

Extented Eucli

Integers

a b a b a b

a b a b

a b

a b

d's algorithm: compute integers

and such that gcd( , ).x a b ay x y b

p5.

Let 2 be an integer.

Definition: is congruent to modulo , written

, if | ( ), i.e., and have the

same remainder when divided by .

mo

Note:

d

mo

Integers modulo

n

a b n

n a b a ba

n

b

a b

n

n

and are different.

Definition: [ ] : mod .

[ ] is called a residue class modulo , and is a

representative of that

d mo

class.

d

n

n

a x

n a

Z x a n

n

n

b

a a

p6.

There are exactly residue classes modulo :

[0], [1], [2], , [ 1].

If [ ], [ ], then [ ] and [ ].

Define addition and multiplication for residue classes:

[ ] [n

n n

n

x a y b x y a b x y a b

a

] [ ]

[ ] [ ] [ ].n

b a b

a b a b

p7.

A group, denoted by ( , ), is a set with a

binary operation such that

1. , , (closure)

1. ( ) ( ) (associativity)

2. s.t. , ( )

identi

3.

y

t

Group

G G

x y G x y

e

G

x y z x y z

e G x G x ex x

, s.t. ( )

A group ( , ) is if , , .

Examples:

invers

( , ), ( , ), ( \{0}, ), ( , ),

( \{0},

abel a

e

).

i n

x G y G x y y x

G x y G x y y x

Z Q Q R

R

e

p8.

Define [0], [1], ..., [ 1] .

Or, more conveniently, 0, 1, ..., 1 .

, forms an abelian group.

For ,

,

additiv

( ) mod . (Or, [ ] [ ] [ ] [ mod ].)

0 is

e

th

n

n

n

n

Z n

Z n

Z

a b Z

a b a b n a b a b a b n

10

e identity element.

The inverse of , denoted by , is .

When doing addition/substraction in , just do the regular

addition/substraction and reduce the result modulo .

In , 5

n

a a n a

Z

n

Z

5 9 4 6 2 8 3 ?

p9.

1

1

1

, is not a group, because 0 does not exist.

Even if we exclude 0 and consider only \ {0},

, is not necessarily a group; some may not exist.

For , exists if and on

n

n n

n

n

Z

Z Z

Z a

a Z a

ly if gcd( , ) 1.a n

p10.

*

1

Let : gcd( , ) 1 .

, is an abelian multiplicative group.

mod .

mod .

1 is the identity elemen

t.

The inverse of , written , can be computed by

n n

n

Z a Z a n

Z

a b ab n

a b ab n

a a

*

12

*

the

Extended Euclidean Algorithm.

For example, 1,5,7,

Q: How many e

11 . 5 7 35

lements ar

mod12 1

e there in ?

1.

n

Z

Z

p11.

*

1

Euler's totient function:

Fac

( )

= : and gcd( , ) 1

1. ( ) ( 1) for prime

2. ( ) ( ) ( ) if gc

ts:

d( , ) 1

n

n

e e

n Z

a a Z a n

p p p p

ab a b a b

p12.

*

| |

mod

Let be a (multiplicative) group.

Lagrange's theorem: For any element ,

Co

.

rollary: For any element ,

Euler's theorem:

If (for any

f

inite

1), the

.

n

G

m Gm

n

G

a G

a G

a Z

e

a

a

a

n

a

( ) *

* ( ) 1 *

1 in .

Fermat's little theorem:

If ( a prime), then 1 in .

n

n

p p

p p

Z

a Z p a a Z

p13.

*

15

*

15

*

15

( ) 8

816243240481

= 1, 2, 4, 7, 8, 11, 13, 14

(15) (3) (5) 2 4 8

: 1 2 4 7 8 11 13 14

ord( ) : 1 4 2 4 4 2 4 2

ord( ) : smallest integer such that 1.

1

13

?

Example: 15

k

n

Z

Z

a Z

a

a k a

a a

n

p14.

Algorithms

1

3

gcd ,

mod

mod

Running time: log

Here we assume , .

k

n

a b

a n

a n

O n

a b Z

p15.

Given 0, compute gcd( , ).

Theorem: If 0, gcd( , ) .

If 0, gcd( , ) gcd( , mod )

Euclid( , )

0

if

then

( ,

retur

)

Euclid's Algorithm

nn a b a b

b a b a

b a b b

a b Z

a b

a b

b

2

n( )

return Euclid( , mod )

The number of recursive calls to Euclid is (log ).

Computing mo

d takes (l

els

og

e

) .

a

b a b

O n

a b O n

15

p16.

Example: gcd(299,221)

Given 0, compute , such that gcd( , ) .

1 78

2 65

1 13

65 5 13 0

gcd(229,221) 13

299 221

221 78

78 65

78 65

78 221 78 78 2( 2 ) 3

21

3

?

Extended Euclidean Algorithm

a b x y d a b ax by

299 221) 221

299 221

( 1

3 4

p17.

Given 0, compute , , such that gcd( , ) .

Extende

if

d - Euclid( , )

0 t

return( ,1,0)

( , , ) Extended - Eucl

hen

els

e

i

Extended Euclidean Algorithm

a b d x y a b a b

a b

b

a

d x

d x y

y

d( , mod )

( , , ) , ,

return( , , )

b a b

d x y d y x a b y

d x y

p18.

If 0, gcd( , ) .

The returned answer is correct.

If ( , , ) is correct,

1 0

( ,1,0)

gcd( , mod ) ( mod )

gcd( , mod )

Correctness Proof

b a b a b

d x y

b a b d b x a b y

b a b d b x a a

a

b b y

a

gcd( , )

( , , ) , , is correct

a b a b

d x y d y x a

d y y

y

x a b

b

p19.

1

1 *

1

Compute in .

exists if and only if gcd( , ) 1.

Use extended Euclidean algorithm to find ,

such that gcd( , ) 1 (in )

mod

[ ]

?How to compute

na Z

a a n

x y

ax ny a n

a n

Z

a

1

[ ] [ ][ ] [1]

[ ][ ] [1] (since [ ] [0])

[ ] [ ].

Note: may omit [ ], but reduce everything modulo .

x n y

a x n

a x

n

p20.

1

1

1 *

47

Compute 15 mod 47.

Using extended Euclidean algorithm, we obtain

gcd(15,47) 1 15 47 7

15 mod 47 22

2

That is, 15 22 in

2

Example

Z

p21.

1 0

2

Comment: compute mod , where in binary.

1

for downto 0 do

mod

if 1 then mod

Algorithm: Square-and-Multiply( , , )

c

k k

i

x n c c c c

z

i k

z z n

c z z x n

x c n

22

22

if c i

retu

s even Note:

if is

rn (

o

)

dd

c

c

c

xx

x x

z

c

p22.

2

2

2

2

3

2

23 10111

1

11 mod 187 11 (square and multiply)

mod 187 121 (square)

11 mod 187 44 (square and multiply)

11 mod 187 165 (square and

11 mod187

mu

Example:

b

z

z z

z z

z z

z z

2

ltiply)

11 mod 187 88 (square and multiply)z z

p23.

RSA Encryption

p24.

mcE D

Bob Alice

m

Alice’s Alice’s

public key secret key

Public-key Encryption

plaintext encryption ciphertext decryption

algorithm algorithm

p25.

By ivest, hamir & dleman of MIT in 1977.

Best known and most widely used public-key scheme.

Based on the one-way property

of mo

R S

du

lar

powering:

A

assumed

The RSA Cryptosystem

1

: mod (easy)

: mod

In turn based on the hardness

(hard)

of integer factorization.

e

e

f x x n

f x x n

p26.

1

1

RSA

RSA

RSA

* *

Encryption (easy):

Decryption

It works in group . Let be

(hard):

Decryption (easy with "trapdoor"):

a messa

Lookin

ge

g

.

Idea behind RSA

e

n

e

e

n

x x

x x

x x

Z x Z

( ) 1 ( )

for a "trapdoor": ( ) .

If is a number such that 1mod ( ), then

( ) 1 for some , and

( ) 1 .

e

ke d n k

d

ed n

x x

d ed n

ed k n k

x x x x x x x

p27.

1

(a) Choose large primes and , and let : .

(b) Choose (1 ( )) coprime to ( ), and

compute : mod ( ). ( .)

(c) Public ke

Key generation:

1 mod ( )

RSA Cryptosystem

p q n pq

e e n n

d nn ede

*

*

y: . Secret key: .

( ) : mod , w

( , ) ( , )

here .

( ) : mod , where .

E

ncryption:

Decryptio

n:

e

pk n

d

sk n

E x x n x Z

D y y n y

pk n e sk n d

Z

p28.

Select two primes: 17, 11.

Compute the modulus 187.

Compute ( ) ( 1)( 1) 160.

Select between 0 and 160 such that gcd( ,160) 1.

Say 7.

Compute

RSA Example: Key Setup

p q

n pq

n p q

e e

e

d

1 1mod ( ) 7 mod160 23

(using extended Euclid's algorithm).

Public key: .

Secret ke

( ,

y:

) (7, 187)

( , ) (23 ., 7 18 )

pk e n

e

s n

n

k d

p29.

7

23

23

23

Suppose 88.

Encryption: mod 88 mod187 11.

Decryption: mod 11 mod187 88.

When computing 11 mod187, we first

compute 11 and

d

the

o

n

ot

n

RSA Example: Encryption & Decryption

e

d

m

c m n

m c n

reduce it modulo 187.

Rather, use , and reduce intermediate

results modulo 187 whenever they g

square-a

et bigge

nd-mult

r than

iply

187.

p30.

Attacks on RSA

p31.

There are many attacks on RSA:

brute-force key search

mathematical attacks

timing attacks

chosen ciphertext attack

s

The

m

ost important one is intege

Attacks on RSA

1

r factorization:

If the adversary can Then he can

calculate ( ) ( 1)( 1) and the secret key

m

factor into .

d ( ). o

n p q

n pq

d e n

p32.

A difficult problem.

More and more efficient algorithms have been developed.

In 1977, RSA challenged researchers to decode a

ciphertext encrypted with a modulus of 129

Integer Factorization

n

digits (428 bits).

Prize: $100. RSA thought it would take quadrillion years

to break the code using fastest algorithms and computers

of that time. Solved in 1994.

In 1991, RSA put forw ard more challenges (called RSA numbers),

with prizes, to encourage research on factorization.

p33.

Each RSA number is a semiprime. (A number is

semiprime if it is the product of two primes.)

There are two labeling schemes.

by the number of decimal digits:

RSA-100, .

RSA Numbers

.., RSA-500, RSA-617.

by the number of bits:

RSA-576, 640, 704, 768, 896, , 1536, 210 .24 048

p34.

RSA-100 ( bits), 1991, 7 MIPS-year, Quadratic Sieve.

RSA-110 ( bits), 1992, 75 MIPS-year, QS.

RSA-120

332

365

3 ( bits), 1993, 830 MIPS-year, QS.

RSA-129

98

4(

RSA Numbers which have been factored

bits), 1994, 5000 MIPS-year, QS.

RSA-130 ( bits), 1996, 1000 MIPS-year, GNFS.



28

4

31

465

5

RSA-16

1

0 (

2

530

576

6

bits), 2003, Lattice Sieve.

RSA- (174 digits), 2003, Lattice Sieve.

RSA- (193 digits), 2005, Lattice Sieve.

RSA-200 ( bits), 2005, Lattice

40

663 Sieve.

p35.

RSA-200 =

27,997,833,911,221,327,870,829,467,638,

722,601,621,070,446,786,955,428,537,560,

009,929,326,128,400,107,609,345,671,052,

955,360,856,061,822,351,910,951,365,788,

637,105,954,482,006,576,775,098,580,557,

613,579,098,734,950,144,178,863,178,946,

295,187,237,869,221,823,983.

p36.

In light of current factorization technologies,

RSA recommends using an of 1024-2048 bits.

Remark

n

p37.

Generating large primes

To set up an RSA cryptosystem,

we need two large primes p and q.

p38.

Generate a random odd number of desired size.

Test if is prime.

If not, discard it and try a different number.

How to generate a large prime number?

n

n

p39.

12

10.5

Can it be solved in polynomial time?

A long standing open problem until 2002.

AKS(Agrawal, Kayal, Saxena) : log .

Later improved by others to log ,

Primality test : Is a prime?

O n

O n

n

6

3

and then

to log .

In practice, Miller-Rabin's probabilistic algorithm is still

the most popular --- much faster, log .

O n

O n

p40.

Using some characteristic property of prime numbers:

is prime 2.. , does not divide .

Miller-Rabin's idea: look for some property ( ) s

Miller-Rabin primality test : Is a prime?

n a n a n

P a

n

*

*

*

.t.

is prime For , ( )

not prime For a portion of elements

, ( )

Algorithm: Randomly pick

a

elements

ll

at most 1

.

n

n

n

n a Z P a true

n

a Z P a true

k

t a Z

If ( ) is true for all of them then return

else return .

A "prime" answer may be incorrect with

prime

compos

probabilit

ite

y

1t

P a

k

p41.

*

nZ

*If is prime, then for all , ( ) is true.nn a Z P a

( )P a true

p42.

*

nZ

*

*

If is , then there are

which are elements s.t

not prime strong

.

Say, at most 1 of are

witnesse

bl .

,

a

s

( )

ck

n

n

P a faa ls

k Z

e

n

Z

( )P a true

p43.

2 1

2

2 2 2

Write 1 2 , where is odd.

1 mod Let ( )

1 mod for some , 0 1

Consider the sequence

, , , ,

r

o

The property ( )

i

k

k

u

u

u u u u

n u u

a nP a

a n i i k

a a a a

P a

p44.

*

If is prime, then ( ) for all .

If is an odd composite and not a prime power,

then of the elements are

black (i.e., ( ) ).

A composite num

at most one h

be

f

al

r

n

n

n P a true a Z

n

a Z

P a true

n

is a if

for some prime and integer 2; a

if

prime power

perfect power

for some integer and . 2 )

e

e

n p

p e

n k k e

p45.

Input: integer 2 and parameter

Output: a decision as to whether is prime or

if is even, return "composit

composite

1. e"

if is a per2

. fect

Algorithm: Miller-Rabin primality test

n t

n

n

n

power, return "composite"

for : 1 to do

choose a random integer , 2 1

if gcd( , ) 1, return "composite"

if is a strong witness, ret

3

urn "com

. i t

a a n

a n

a

posite"

return ("pri4. me")

p46.

If the algorithm answers "composite", it is always correct.

If the algorithm answers "prime", it may or may not be correct.

The algorithm gives a wrong answ

Analysis: Miller-Rabin primality test

er if is composite but

the algorithm fails to find a strong witness in iterations.

This may happen with probability at most 2 .

Actually, at most 4 , by a more sophisticated analysis.

t

t

n

t

p47.

A is a probabilistic algorithm

which always gives an answer

but sometimes the answer may be inco

Mo

rr

nte

ect.

Carlo a

A

lgorithm

Monte Carlo algorithm for a decisi

Monte Carlo algorithms

on problem is

if its “yes” answer is always correct but a “no” answer may

be incorrect with some error probability.

A -iteration Miller-Rabin is a “composite”-biased Mon

yes-bias

te Carl

ed

o

t

algorithm with error probability at most 1 4 .t

p48.

A is a probabilistic algorithm

which may sometimes fail to give an answer

but never gives an incorrect

Las Ve

one

gas algori

A Las Vegas algorithm can be conver

thm

Las Vegas algorithms

ted into a

Monte Carlo algorithm.

number-theoretic algorithms (rsa and related...

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