CS70 Midterm 2 Solutions - Computing with Polynomials, Exams of Discrete Mathematics

Download CS70 Midterm 2 Solutions - Computing with Polynomials and more Exams Discrete Mathematics in PDF only on Docsity! cs70, fall 2001 midterm 2 solutions professor wagner Problem #1 (18 pts.) Short-answer questions (a) C(n,i)/2^n (b) 0.3 <= Pr[E union F] <= 0.5. 0.3 occurs if E is a subset of F, and 0.5 occurs if E and F are disjoint. (c) 2002, since each such string is of the form [k]1[2000-k]0 (where the quantity in brackets indicates the number of times to repeat) for some k in {0, 1, ..., 2001}, and there are 2002 such k. (d) All of them, since 2z = n+1 = 1 (mod n), so z-inverse = 2 (mod n). (Alternate explanation: gcd((n+1)/2,n) = gcd((n+1)/2,(n-1)/2) = gcd(1,(n-1)/2)=1 by Euclidean algorithm, so gcd(z,n) = 1, so z has an inverse mod n.) Problem #2 (12 pts.) Digit sums Consider an arbitrary natural number n. Write n in decimal: n = Ak*10^k + ... + A1*10 + A0 Then f(n) = Ak^3 + ... + A1^3 + A0^3 [by defn of f] = Ak + ... + A1 + A0 = Ak*10^k + ... + A1+10 + A0 [since 10^j = 1 (mod 3) for all j] = n (mod 3) n was arbitrary, so this must hold for all natural numbers. Problem #3 (12 pts.) Computing with polynomials (a) O(d^2(lg p)^2) There are O(d^2) cross-terms, and each requires O((lg p)^2) work to do a modular multiplication and addition. (b) O(d(lg p)^2) Computing the sequence 1, u, u^2, ..., u^d mod p takes d multiplications, then multiplying by Ai mod p takes d+1 multiplications, plus d more additions. (c) 1. Let v = f(u) mod p using part (b). 2. Return v^2 mod p Takes O(d(lg p)^2 + (lg p)^2) = O(d(lg p)^2) time. cs70, midterm 2 solutions, fall 2001 cs70, fall 2001course, semester/year midterm 2 solutionsexam # professor wagnerprofessor (e.g., Professor J. Wawrzynek)1