... lg(n − 2) − 2c lg(n − 2) + dn − 2d + lg n > cn lg(n − 2) − 2c lg n + dn − 2d + lg n (since − lg n < − lg(n − 2) ) = cn lg(n − 2) − 2( c − 1) lg n + dn − 2d ≥ cn lg(n /2) − 2( c − 1) lg n + dn − 2d (by ... 1, 2, 1, 3, 2, 1, 2, 3, 3, 1, 3, 2, probability 4 /27 5 /27 5 /27 5 /27 4 /27 4 /27 Solutions for Chapter 5: Probabilistic Analysis and Randomized Algorithms 5-13 Although these probabilities add to ... √ ( 2) lg n = ω 2 lg n by taking logs: lg( 2) lg n = (1 /2) lg n, lg 2 lg n = lg n (1 /2) lg n = ω( lg n) √ √ 2 lg n = ω(lg2 n) by taking logs: lg 2 lg n = lg n, lg lg2 n = lg lg n lg n = ω (2 lg...