UCI ICS/Math 6D 3-Integers-1

Integers

Number Theory = Properties of Integers(For this part, assume all values are integers.)“a|b” = “a divides b” = nZ (b=na)

“b is a multiple of a.” “a is a factor of b.”“Multiple” always means “integer multiple”

Thrm: If a|b and a|c, then a|(b+c).

Thrm: If a|b, then m a|mb.

Thrm: If a|b and b|c, then a|c.

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Division Algorithm

Thrm:Thrm:

If a,dIf a,dZ Z d>0, then d>0, then ! q,rZZ (0≤r<d a=qd+r)d is the “divisor” (a is the “dividend”)q is the “quotient,” q = a div d

(quotient = # of multiples of d which fit into a, if a≥0)r is the “remainder,” r = a mod d (“a modulo d”)

a d q = a div d r = a mod d

17 5 3 2

5 17 0 5

51 17 3 0

0 17 0 0

-17 5 -4 3

There is a unique.There is one and only one.

Functions on pairs (a,d)

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Congruent ... Modulo

For a, b, m integers with m>0, we say“a is congruent to b modulo m,” writtena b (mod m)b (mod m), iff m | (a-b)

Thrm: For a, b, m integers with m>0, a b (mod m)b (mod m) iff kZ a=b+km Z a=b+km

Thrm: For a, b, m integers with m>0, a b (mod m)b (mod m) iff (a mod m) = (b mod m)

Thrm: For a, b, c, d, m integers with m>0, if a b (mod m)b (mod m) and c d (mod m), thend (mod m), thena+c b+d (mod m)b+d (mod m) and ac bd (mod m).bd (mod m).

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Applications of Congruences

Hashing Functions: hm(n) = (n mod m)

Range(hm) = {n | 0≤n<m}

Not injective (not one-to-one) Collisions

{0,1,2,3,...,m-1} = “Zm”

Pseudorandom Number Generator:nk+1 =(ank+c) mod m

Example: (a,c,m)=(3,4,7), i.e. nk+1 =(3nk+4) mod 7n1=0 ; n2=4 ; n3=2 ; n4=3 ; n5=6 ; n6=1 ; n7=0 ; …

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Applications of Congruences (cont)

Example: (a,c,m)=(3,4,7), i.e. nk+1 =(3nk+4) mod 7n1=0 ; n2=4 ; n3=2 ; n4=3 ; n5=6 ; n6=1 ; n7=0 ; …

Ceasar’s Cipher (“Shift Cipher”): p = plaintext, encoded as integer in Z26

c = ciphertext, encoded as integer in Z26

Encrypt each letter using a fixed offset k from the alphabet’s start, e.g.:

c = Ek (p) = (p+k) mod 26

Actually, any bijection, f:Z26Z26, provides an encryption algorithm:Examples: E(p) = (3n+13) mod 26

E(p) = (15n+7) mod 26

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Primes

n>1 is “prime” iff the only positive divisors of n are 1 and n itself.n is “composite” = n is not prime.

We say “d is a factor of n” iff d is positive and d is a divisor of n.

We call d a trivial factor of n if d = 1 or n.

=> n is prime if it has no non-trivial factors.

Fundamental Theorem of Arithmetic:Every integer n>1 is either a prime or can be written uniquely as

the product of prime factors.(“Uniquely” means “in exactly one way ignoring differences in ordering”.e.g. 30=2·3·5 and 30=5·3·2 are same factorizations.)

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Factorization into Primes

e.g.420 = 42·10 = 6·7·2·5 = 2·3·7·2·5 = 2·2·3·5·717 is prime

Sieve of EratosthenesThrm: If n is composite, n has a prime factor

whose square is at most n.e.g.

289 is not prime:just test for i=1 to 20, if i2 | 289(can do it only for i = 11,13,17,19…)

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Prime Facts

Thrm: There are infinitely many primes.Equivalently: There is no largest prime.

Prime Number Theorem: If H(n)=|{kN | k<n N | k<n k is prime}|, then

loge(n)·H(n) / n gets arbitrarily close to 1 as n grows large.

Consequently, H(n) ≈ n / log(n) Proportion of numbers in [0,n] which are prime is about 1/log(n)

How to pick a 100-bit prime (e.g. for hash or a cryptosystem)?Answer: Try random 100-bit number, test for primality.

Probability of success ≈ 1/100 => Expected number of attempts before

success ≈ 100

Thrm: If f is a (non-constant) polynomial with integer coefficients, there is an integer n s.t. f(n) is composite.

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Prime Conjectures

Goldbach’s Conjecture:Every even integer greater than 2 can be written

as the sum of two primes. http://en.wikipedia.org/wiki/Goldbach's_conjecture

The Twin Prime Conjecture:There are infinitely many primes p such that

p+2 is also prime.http://en.wikipedia.org/wiki/Twin_prime_conjecture

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Greatest Common Divisor (gcd)

When a and b are integers, not both 0, the “greatest common divisor” of a and b, denoted gcd(a,b),is the largest integer d such that d|a and d|b.Note: If a≠0, gcd(a,0)=|a|

Thrm: When a and b are integers, not both 0,if d|a and d|b, then d|gcd(a,b).

Thrm: If a and b are integers, not both 0, gcd(a,b)=gcd(b,a)

Thrm: If a and b are integers, not both 0, gcd( a , b ) = gcd( a , b mod a ) = gcd( a mod b , b

)

Ref: http://en.wikipedia.org/wiki/Greatest_common_divisor

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Least Common Multiple (lcm)

If a,b>0, the “least common multiple” of a and b, denoted lcm(a,b), is the smallest m>0 such that a|m and b|m.

Thrm: If a,b>0, then a · b = gcd(a,b) · lcm(a.b)

Integers a and b are said to be “relatively prime” iff gcd(a,b)=1.

Set S of integers is said to be “pairwise relatively prime” iff each pair of (different) elements in S is relatively prime.

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Finding gcd’s and lcm’s

Method 1: Factor each number into primesa=p1

j1·p2j2·...·pn

jn, b=p1k1·p2

k2·...·pnkn.

Thengcd(a,b)=p1

min(j1,k1)·p2min(j2,k2)·...·pn

min(jn,kn).

lcm(a,b)=p1max(j1,k1)·p2

max(j2,k2)·...·pnmax(jn,kn).

Method 2: Euclidean Algorithm: Find gcd(a,b) [using gcd(a,b)=gcd(a mod b,b)=gcd(b,a mod b)]

Can then compute lcm(a,b)=a·b/gcd(a,b).

Ref: http://en.wikipedia.org/wiki/Euclidean_algorithm

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Euclidean Algorithm

procedure gcd(a,b: positive integers)x := a; y := b;repeat

r := x mod y;x := y;y := r

until y=0;{gcd(a,b) is x}

procedure gcd(a,b: positive integers)(x,y) := (a,b);repeat

(x,y) := (y, x mod y);until y=0;gcd := x

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Euclidean Algorithm Example

gcd(309,171)= gcd(171,138)= gcd(138,33)= gcd(33,6)= gcd(6,3)= gcd(3,0) = 3

309=1·171+138171=1·138+33138=4·33+633=5·6+36=2·3+0

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Greatest Common DivisorRepresented as Linear Combination of a & b:

Thrm: If a and b are integers, not both 0, then s,tZZ sa + tb = gcd(a,b)

(s,t) can be found by an Extended (version of the) Euclidean Algorithm.

Ref: http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm

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Extended Euclidean Algorithm: Example

gcd(309,171)= gcd(171,138)= gcd(138,33)= gcd(33,6)= gcd(6,3)= gcd(3,0) = 3

309=1·171+138171=1·138+33138=4·33+633=5·6+36=2·3+0

3 = 33-5·6 = 33-5·(138-4·33) = -5·138+21·33= -5·138+21·(171-1·138) = 21·171-26·138 = 21·171-26·(309-1·171) = -26·309+47·171 [= -8034+8037=3]

You can represent the final gcd (= 3) as a linear combination of value (a,b) at each step, going bottom up, i.e.

(a,b) = (33,6), (138,33), (171,138), (309,171), and finally (309,171)

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Representations of Integers

Thrm: If b is an integer greater than 1, then any positive integer n can be written uniquely as

n=akbk+ak-1b

k-1+...+a1b+a0,where ak≠0, 0≤ai<b for all i

(akak-1...a1a0) is a “base b expansion of n”, (or “base b representation of n”)

Notation: (akak-1...a1a0)b

Example: (5739)10=5·103+7·102+3·101+9·100

Ref: http://en.wikipedia.org/wiki/Radix

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Representations of Integers

Commonly used bases: 2, 4, 8, 10, 12, 16.For 10<b≤36, the letters “A” to “Z” are used to

designate the decimal values 10 to 35.In particular, for base 16 (“hexadecimal”)

A=10, B=11, C=12, D=13, E=14, F=15 Examples:

(231)4=2·42+3·4+1=(45)10

(276)8=2·82+7·8+6=(190)10

(2D)16=2·16+13=(45)10

(AB)16=10·16+11=(171)10

(1AB)16=1·162+10·16+11=(427) 10

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Computing Base ExpansionsConverting from base b to base 10:

1. Using the powers of the base b (5134)b = 5·b3+1·b2+3·b1+4·b0

2. Avoiding using the powers of the base b (5134)b = b·(b·(b·5 + 1) + 3) + 4

Why? To perform fewer multiplications [also for the “square and multiply” exponentiation algorithm on slide 21]

Converting between bases where one base is a power of the other is very easy (e.g., 2 and 8, 2 and 16), because we can do it block-by-block. For example: (0110 1001 0001 1101)2 = (6 9 1 D)16

General procedure for computing base b expansion of integer n:procedure base-b-expansion (n: positive integer)q:=n; k:=0;repeat

ak := q mod b; q:= q div b; k := k+1;until q=0;{ the base b expansion of n is (akak-1...a1a0)b }

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Arithmetic with Base Expansions

(1001 0001)2+(0011 1101)2 =?

(421)8+(75)8 =?

(A1)16+(3D)16 =?

(100 0011)2 ·(10 1010)2 =?

(342)8-(173)8=?

Referenceshttp://en.wikipedia.org/wiki/Tom_Lehrerhttp://members.aol.com/quentncree/lehrer/newmath.htmhttp://www.youtube.com/watch?v=a81YvrV7Vv8http://www.youtube.com/watch?v=tx5KDyvlG3Q

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Modular Exponentiation:“Square and Multiply” Algorithm

modular exponentiation (b: integer; a,m: positive integers) {computes ba (mod m)}

Let a = (anan-1...a1a0)2; Let x := 1 mod m; Let k := n;repeat

if ak = 1 then x := x·b (mod m) (1)x := x2 (mod m);k := k-1

until k<0; {x equals ba mod m when the loop terminates}

Why does it work? First do the (base-2)→(base-10) conversion on exponent a.Example: n=3, a=(a3a2a1a0)2 = 2·(2·(2·a3+a2)+a1)+a0

Note that we can replace the whole line (1) by the following: x := x·bak

Note also that if x=be then x·bak = be+ak. Also, if x=be then x2= b2·e.Now look at the values of x computed in the above loop:(k,x) = initially (3,1), then (2,b2·a3), then (1,b2·(2·a3+a2)), then (0,b2·(2·(2·a3+a2)+a1), and

finally (-1,b2·(2·(2·(2·a3+a2)+a1)+a0), so the output is correct!