By Raphael Pass, Abhi Shelat

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**Sample text**

C. Given two numbers a and b such that a ≥ b, Euclid’s algorithm computes the greatest common divisor of a and b, denoted gcd( a, b). It is not at all obvious how this value can be efficiently computed, without say, the factorization of both numbers. 6. Basic Computational Number Theory 43 notice that any divisor of a and b is also a divisor of b and a − b. The subtraction is easy to compute and the resulting pair (b, a − b) is a smaller instance of original gcd problem. The algorithm has since been updated to use a mod b in place of a − b to improve efficiency.

2: ExtendedEuclid( a, b) such that a > b > 0 1: 2: 3: 4: 5: 6: if a mod b = 0 then Return (0, 1) else ( x, y) ← ExtendedEuclid(b, a mod b) Return (y, x − y( a/b )) end if Note: by polynomial time we always mean polynomial in the size of the input, that is poly(log a + log b) Proof. 2 returns ( x, y) such that ax + by = gcd( a, b) = d via induction. First, let us argue that the procedure terminates in polynomial time. 3 If a > b ≥ 0 and a < 2n , then ExtendedEuclid( a, b) makes at most 2n recursive calls.

2 x+x x x + ( x − 1) ( x − 1) ··· x+1 1 Observe that X > 2x (since each term is greater than 2) and that the largest prime dividing X is at most 2x (since the largest numerator in the product is 2x). By these facts and unique factorization, we can write X= ∏ p νp ( X ) > 2 x p<2x 1 32 chapter 2. computational hardness where the product is over primes p less than 2x and νp ( X ) denotes the integral power of p in the factorization of X. Taking logs on both sides, we have ∑ νp ( X ) log p > x p<2x We now employ the following claim proven below.

### A Course in Cryptography by Raphael Pass, Abhi Shelat

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