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Integers

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

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Integers

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  1. Integers • Number Theory = Properties of Integers • (For this part, assume all values are integers.) • “a|b” = “a divides b” =  nZ (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. UCI ICS/Math 6D

  2. Division Algorithm There is a unique. There is one and only one. • Thrm: • If a,dZ  d>0, then ! q,rZ (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”) Functions on pairs (a,d) UCI ICS/Math 6D

  3. Congruent ... Modulo • For a, b, m integers with m>0, we say“a is congruent to b modulo m,” writtena  b (mod m), iff m | (a-b) • Thrm: For a, b, m integers with m>0, a  b (mod m) iff kZ a=b+km • Thrm: For a, b, m integers with m>0, a  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) and c d (mod m), thena+c  b+d (mod m) and ac bd (mod m). UCI ICS/Math 6D

  4. 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 7 • n1=0 ; n2=4 ; n3=2 ; n4=3 ; n5=6 ; n6=1 ; n7=0 ; … UCI ICS/Math 6D

  5. Applications of Congruences (cont) • Example: (a,c,m)=(3,4,7), i.e. nk+1 =(3nk+4) mod 7 • n1=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:Z26Z26, provides an encryption algorithm: • Examples: E(p) = (3n+13) mod 26 • E(p) = (15n+7) mod 26 UCI ICS/Math 6D

  6. 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.) UCI ICS/Math 6D

  7. Factorization into Primes • e.g. • 420 = 42·10 = 6·7·2·5 = 2·3·7·2·5 = 2·2·3·5·7 • 17 is prime • Sieve of Eratosthenes • Thrm: 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…) UCI ICS/Math 6D

  8. Prime Facts • Thrm: There are infinitely many primes. • Equivalently: There is no largest prime. • Prime Number Theorem: • If H(n)=|{k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. UCI ICS/Math 6D

  9. 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 thatp+2 is also prime. • http://en.wikipedia.org/wiki/Twin_prime_conjecture UCI ICS/Math 6D

  10. 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 UCI ICS/Math 6D

  11. 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. UCI ICS/Math 6D

  12. Finding gcd’s and lcm’s • Method 1: Factor each number into primes • a=p1j1·p2j2·...·pnjn, b=p1k1·p2k2·...·pnkn. • Then • gcd(a,b)=p1min(j1,k1)·p2min(j2,k2)·...·pnmin(jn,kn). • lcm(a,b)=p1max(j1,k1)·p2max(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 UCI ICS/Math 6D

  13. 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 UCI ICS/Math 6D

  14. 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+138 • 171=1·138+33 • 138=4·33+6 • 33=5·6+3 • 6=2·3+0 UCI ICS/Math 6D

  15. Greatest Common DivisorRepresented as Linear Combination of a & b: • Thrm: If a and b are integers, not both 0, then  s,tZ 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 UCI ICS/Math 6D

  16. Extended Euclidean Algorithm: Example • 309=1·171+138 • 171=1·138+33 • 138=4·33+6 • 33=5·6+3 • 6=2·3+0 • gcd(309,171) • = gcd(171,138) • = gcd(138,33) • = gcd(33,6) • = gcd(6,3) • = gcd(3,0) = 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) 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] UCI ICS/Math 6D

  17. 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-1bk-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 UCI ICS/Math 6D

  18. 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 UCI ICS/Math 6D

  19. Computing Base Expansions • Converting from base b to base 10: • Using the powers of the base b (5134)b = 5·b3+1·b2+3·b1+4·b0 • 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 } UCI ICS/Math 6D

  20. 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=? • References • http://en.wikipedia.org/wiki/Tom_Lehrer • http://members.aol.com/quentncree/lehrer/newmath.htm • http://www.youtube.com/watch?v=a81YvrV7Vv8 • http://www.youtube.com/watch?v=tx5KDyvlG3Q UCI ICS/Math 6D

  21. 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! UCI ICS/Math 6D

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