University of Tokyo: Advanced Algorithms Summer Lecture 4 13 May

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1 University of Tokyo: Advanced Algorithms Summer 2010 Lecture 4 13 May Lecturer: François Le Gall Scribe: Martsinkevich Tatiana Definition: A prime number is an integer p > 1 whose only divisors are 1 and p. For example: 2, 3, 5... Theorem 4.1 (Fundamental Theorem of Arithmetic). Any positive integer n > 1 can be decomposed as n = p 1 a1 p 2 a2.. p k a k, where pi are distinct primes and a i are positive integers. Moreover, the decomposition is unique. There exist two fundamental problems which are: PRIME: given integer n determine if it is a prime number. FACTORING: given n compute its factorization. The first problem is considered to be easy while second - hard. There is a deterministic algorithm solving PRIME in time polynomial in log n, but it s slow. This algorithm is presented in the paper Prime is in P by Agrawal-Kayal-Saxena, There exists faster and more practical randomized algorithms for PRIME. For FAC- TORING there is no good algorithm as yet: the best known algorithm has time complexity exponential in log n. 4.1 Euclid s Algorithms Notation: we write b a if b divides a, i.e. there exists an integer c so that a = bc. Definition: The greatest common divisor of two integers n and m is the largest positive d such that d m and d n. Then we write d = gcd(m, n). If d = 1 we say that m and n are coprime. e.g.: gcd(12, 30) = 6 gcd(m, n) can be read from the factorizations of m and n. For example, if m = and n = , then gcd(m, n) = Euclid s algorithm computes gcd(m, n) in time O(log m log n). 4-1

2 e.g.: let s compute gcd(1547, 560) Steps: 1. Divide the largest of the two numbers by the second number. 2. Take the number by which the division was performed and divide it by the remainder just obtained. 3. Repeat Step 2 until getting zero remainder. 4. gcd is the last non-zero remainder = = = = = = The last non-zero remainder is 7 = gcd(1547, 560) = 7 Theorem 4.2. For any integers m and n there exist integers u and v such that gcd(n, m) = un + vm. Moreover u and v can be found in time O(log m logn) by the extended Euclid s algorithm. Let s study example that demonstrates how the algorithm works. We reverse the computation done for the previous example of gcd(1547, 560): 7 = = 28 1 ( ) = = ( ) = = ( ) = = ( ) = }{{} 21 u 1547 }{{} 58 v Computation in Z n Let n be a positive integer. Z n denotes the set of integers {0, 1, 2..n 1}. The function modn maps an integer a to the integer b Z n such that n (b a). e.g.: 11 mod 3 = 2 2 mod 3 = 1 7 mod 5 = 3 Z n with addition modn is an abelian group. 4-2

3 Addition and substitution in Z n can be done in time O(log n). For example, if a,b Z n, then { a + b if a + b < n a + b mod n = a + b n if a + b n Multiplication can be done in time O((log n) 2 ). Multiplicative inverse Theorem 4.3. Let a be an element of Z n. Then ax mod n = 1 has a solution x if and only if gcd(a, n) = 1. The solution is unique in Z n and can be found in time O((log n) 2 ). Proof (Existence): Suppose there exists a solution x 0 so that ax 0 mod n = 1. Then: ax 0 + cn = 1 for some c = gcd(a, n) 1 = gcd(a, n) = 1. Suppose gcd(a, n) = 1. Then there exist integers u and v such that au + nv = 1. Then au mod n = 1, and x = u is a solution. Notice that also shows that a solution can be computed using the extended Euclid algorithm. Proof (Unicity): Suppose there exists x 0 x 1 Z n so that ax 0 mod n = 1 and ax 1 mod n = 1. Then: a(x 0 x 1 ) mod n = 0 = n a(x 0 x 1 ) = n (x 0 x 1 ) = x 0 x 1 = 0 (since n < x 0 x 1 < n) = x 0 = x 1. Conclusion: each element a Z n with gcd(a, n) = 1 has a multiplicative inverse a 1 such that aa 1 mod n = 1. For any positive integer, define Z n = {a Z n gcd(a, n) = 1}. Then Z n is a multiplicative group (with multiplication modulo n), and Z p, when p is a prime, is a field (with addition and multiplication modulo p). Power algorithm 4-3

4 Goal: evaluate a e mod n. If we compute each a, a 2, a 3.., this will take time O(e(log n) 2 ), which is not efficient. But there is an observation which we can use: { a e a a e 1 if e is odd = a (e/2)2 if e is even Here is an efficient algorithm for computing powers. POWER(a, e, n) [e 0, n 2, a Z n ] if e = 0 then return(1); else if e is odd then t POWER(a, e 1, n); return(at mod n); else t POWER(a, e/2, n); return(t 2 mod n); Theorem 4.4. POWER(a, e, n) returns a e mod n in time O(log e (log n) 2 ) e.g.: e = 13 P OW ER(a, 13, n) a a 12 = a 13 P OW ER(a, 12, n) (a 6 ) 2 = a 12 P OW ER(a, 6, n) (a 3 ) 2 = a 6 P OW ER(a, 3, n) a a 2 = a 3 P OW ER(a, 2, n) a 2 P OW ER(a, 1, n) 1 a = a P OW ER(a, 0, n) Euler Theorem Definition: for any positive integer n the Euler function ϕ(n) is defined as ϕ(n) = {1 b n gcd(n, b) = 1}. e.g.: ϕ(1) = 1 if p is a prime, then ϕ(p) = p 1 if p is a prime and α 1, then ϕ(p α ) = p α p α 1 4-4

5 Property: if n = p a 1 1 p a p a k k, where the p i s are distinct prime, then ϕ(n) = ϕ(p a 1 1 ) ϕ(p a 2 2 ) ϕ(p a k k ). In other words, if we know the factorization of n, we can compute ϕ(n). Theorem 4.5 (Euler Theorem). For all positive integers n and all x Z n x ϕ(n) mod n = 1. Proof: Suppose n > 1. Then ϕ(n) = Z n. For any finite group G and any g G we have g G = 1 (we have it from elementary group theory: Lagrange theorem). Special case of the Euler Theorem: if n = p prime then, for any x Z p, x p 1 mod p = 1. This is called Fermat s Little Theorem and it will be used for checking if p is prime. 4-5