# hw 2 - 1536 is even so it is 6 4824 is even so it is 1 It...

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1. O(n) Polynomial 2. 1, 2, 3, 5, 7, 10, 13, 15, 19 -------- 1 2 3 4 5 6 7 8, 9 given input x sum = 1 while (x > 1) sum *= x x-- return sum log(N!) = log(1) + log(2) + … log(N) upper bound = O(n 2 ) log(N!) <= log(N N ) = N log N = 2 n n. log N = n. size of number x = log x N/2 log (N/2) = O(N log N) 3. Given input x int square = 1; for (int i = 1; i < x; i++) { square = i*i; if (square > x) return false; else if (square == x) return true; } O(log x) since it’ll exit if the square is > the number 4a. 1 because it has even number of exponents. 4b. Using repeated squaring 2, 4, 8, 16, 32, 64, 128 (2 64 * 2 32 * 2 16 * 2 8 * 2 4 * 2 1 ) % (2 7 – 1) 2 5^3 4c. either ends in 4 or 6, must end in 9 or 1. Only way to be divisible by 35 is 4 – 9, or 6 – 1.
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Unformatted text preview: 1536 is even, so it is 6. 4824 is even, so it is 1. It is divisible by 5 for sure. Because 35 = 5*7, that means that 4*6 must have remainder 1, and 4824 and 1536 are divisible by 24. Therefore, 1-1 (mod 35) = 0. Yes. 4d. 4, 21, none, 14 5. gcd(Fn+1, Fn) = 1 1, 1, 2, 3, 5, 8, 13, 21, 34, 55 … Because F n is the sum of F n-1 + F n-2 , and if we are comparing F n-1 and F n , that means Base case: the gcd(1, 1) = 1, gcd(1, 2) = 1. 6. Input x, y. X*y/GCD(x,y) = LCM. Finding GCD = O(n 3 ) (this algorithm works) 2 7 – 1 1, 1, 2, 3, 5, 8, 13 Using euclid’s formula...
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## This note was uploaded on 01/09/2012 for the course CSE 101 taught by Professor Staff during the Spring '08 term at UCSD.

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