Ngày 9 tháng 12 năm Discrete Mathematics Lecture-15

Transcription

1 Discrete Mathematics Lecture-15 Ngày 9 tháng 12 năm 2011

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6 ex a 1 mod b (gcd(a,b) = 1)

7 ex a 1 mod b (gcd(a,b) = 1) Returns an integer c < b such that a c mod b = 1.

8 ex a 1 mod b (gcd(a,b) = 1) Returns an integer c < b such that a c mod b = 1. How many digits does the integer n have?

9 ex a 1 mod b (gcd(a,b) = 1) Returns an integer c < b such that a c mod b = 1. How many digits does the integer n have? len(str(n))

10 ex a 1 mod b (gcd(a,b) = 1) Returns an integer c < b such that a c mod b = 1. How many digits does the integer n have? len(str(n)) factorial(n)

11 ex a 1 mod b (gcd(a,b) = 1) Returns an integer c < b such that a c mod b = 1. How many digits does the integer n have? len(str(n)) factorial(n) Returns n!

12 ex a 1 mod b (gcd(a,b) = 1) Returns an integer c < b such that a c mod b = 1. How many digits does the integer n have? len(str(n)) factorial(n) Returns n! is_prime(n)

13 ex a 1 mod b (gcd(a,b) = 1) Returns an integer c < b such that a c mod b = 1. How many digits does the integer n have? len(str(n)) factorial(n) Returns n! is_prime(n) Returns True or False

14 ex a 1 mod b (gcd(a,b) = 1) Returns an integer c < b such that a c mod b = 1. How many digits does the integer n have? len(str(n)) factorial(n) Returns n! is_prime(n) Returns True or False pow(a,b,c)

15 ex a 1 mod b (gcd(a,b) = 1) Returns an integer c < b such that a c mod b = 1. How many digits does the integer n have? len(str(n)) factorial(n) Returns n! is_prime(n) Returns True or False pow(a,b,c) Returns a b mod c.

16 Some theorems we shall be using Fermat s Theorem: If p is prime and 0 < a < p then a p 1 mod p = 1.

17 Some theorems we shall be using Fermat s Theorem: If p is prime and 0 < a < p then a p 1 mod p = 1. Chinese Remainder Theorem: Given [a 1, a 2,..., a k ] pairwise relatively prime integers and integers [m 1, m 2,..., m k ], m i < a i then there is a unique integer M < k i=1 m i such that M mod a i = m i

18 Some theorems we shall be using Fermat s Theorem: If p is prime and 0 < a < p then a p 1 mod p = 1. Chinese Remainder Theorem: Given [a 1, a 2,..., a k ] pairwise relatively prime integers and integers [m 1, m 2,..., m k ], m i < a i then there is a unique integer M < k i=1 m i such that M mod a i = m i Euler s Theorem: Recall: φ(n) = {a 1 a < n}, gcd(n, a) = 1}. If gcd(n, a) = 1 then a φ(n) mod n = 1.

19 Some theorems we shall be using Fermat s Theorem: If p is prime and 0 < a < p then a p 1 mod p = 1. Chinese Remainder Theorem: Given [a 1, a 2,..., a k ] pairwise relatively prime integers and integers [m 1, m 2,..., m k ], m i < a i then there is a unique integer M < k i=1 m i such that M mod a i = m i Euler s Theorem: Recall: φ(n) = {a 1 a < n}, gcd(n, a) = 1}. If gcd(n, a) = 1 then a φ(n) mod n = 1. Wallis Theorem (p 1)! mod p = 1 if and only if p is prime.

20 Some theorems we shall be using Fermat s Theorem: If p is prime and 0 < a < p then a p 1 mod p = 1. Chinese Remainder Theorem: Given [a 1, a 2,..., a k ] pairwise relatively prime integers and integers [m 1, m 2,..., m k ], m i < a i then there is a unique integer M < k i=1 m i such that M mod a i = m i Euler s Theorem: Recall: φ(n) = {a 1 a < n}, gcd(n, a) = 1}. If gcd(n, a) = 1 then a φ(n) mod n = 1. Wallis Theorem (p 1)! mod p = 1 if and only if p is prime. Primitive Roots: The finite field GF(q) has primitive roots ( {a k, 0 k q 2} = GF (q)).