ch0 3 - Solutions to Homework One 1. Exercise 0.1. (a) ;...

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Solutions to Homework One CSE 101 1. Exercise 0.1. (a) Θ; (b) O ; (c) Θ; (d) Θ; (l) O ; (m) O ; (n) Θ; (o) Ω; (p) O . 2. Exercise 0.3. (a) Base cases n = 6 and n = 7 are easy to check individually. Now assume the statement holds upto some value of n greater than 6; we need to extend it to n + 1. F n +1 = F n + F n 1 2 0 . 5 n + 2 0 . 5( n 1) (induction hypothesis) 2 · 2 0 . 5( n 1) = 2 0 . 5( n +1) . Note: for this problem (and some of the others below), we need two base cases. Do you see why? (b) We’ll show, again by induction, that F n 2 cn for c = log 2 ((1 + 5) / 2) 0 . 694. The base cases n = 0 , 1 are easy to check. As before, assume the statement is true upto some n > 0; we’ll show it for n + 1: F n +1 = F n + F n 1 2 cn + 2 c ( n 1) = 2 c ( n +1) · (2 c + 2 2 c ) , and the particular value of c chosen satisFes 2 c +2 2 c = 1. (We found c by solving this quadratic equation.) (c) Combined with (b). 3. Exercise 3.5. function reverse ( G ) Input: directed graph G = ( V,E ) in adjacency list format Output: the reverse of G , also in adjacency list format create G R = ( V,E R ) , with edge-set E R initially empty for each u V : for each ( u,w ) E : add edge ( w,u ) to E R return G R ±or each edge ( u,w ) E , adding ( w,u ) to E R takes O (1) time: the edge is simply added to the front of the adjacency list for node w . Thus the total running time is O ( | V
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ch0 3 - Solutions to Homework One 1. Exercise 0.1. (a) ;...

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