By R. Miller, L. Boxer

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**Sample text**

Then prove by substitution. 1. Upper bound: Guess: T (n) ≤ dn lg n. Substitution: T (n) ≤ T (n/3) + T (2n/3) + cn ≤ d(n/3) lg(n/3) + d(2n/3) lg(2n/3) + cn = (d(n/3) lg n − d(n/3) lg 3) + (d(2n/3) lg n − d(2n/3) lg(3/2)) + cn = dn lg n − d((n/3) lg 3 + (2n/3) lg(3/2)) + cn = dn lg n − d((n/3) lg 3 + (2n/3) lg 3 − (2n/3) lg 2) + cn = dn lg n − dn(lg 3 − 2/3) + cn ≤ dn lg n if −dn(lg 3 − 2/3) + cn ≤ 0 , c . d ≥ lg 3 − 2/3 Therefore, T (n) = O(n lg n). , d) are different. 2. Lower bound: Guess: T (n) ≥ dn lg n.

We have T (n) ≥ c(n − 2) lg(n − 2) + d(n − 2) + 2 lg n = cn lg(n − 2) − 2c lg(n − 2) + dn − 2d + 2 lg n > cn lg(n − 2) − 2c lg n + dn − 2d + 2 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 inequality (1) above) = cn lg n − cn − 2(c − 1) lg n + dn − 2d ≥ cn lg n , if −cn−2(c−1) lg n+dn−2d ≥ 0 or, equivalently, dn ≥ cn+2(c−1) lg n+2d. Pick any constant c > 1/2, and then pick any constant d such that d ≥ 2(2c − 1) . ) Then d/2 ≥ 2c − 1 = c + (c − 1) , and adding d/2 to both sides, we have d ≥ c + (c − 1) + d/2 .

1. 2. 3. 4. The following justiÞcations explain some of the rankings: 1. en = 2n (e/2)n = ω(n2n ), since (e/2)n = ω(n). 2. (lg n)! = ω(n 3) by taking logs: lg(lg n)! = approximation, lg(n3 ) = 3 lg n. lg lg n = ω(3). (lg n lg lg n) by Stirling’s Solutions for Chapter 3: Growth of Functions 3-11 √ √ √ √ 3. ( 2)lg n = ω 2 2 lg n by taking logs: lg( 2)lg n = (1/2) lg n, lg 2 2 lg n = 2 lg n. (1/2) lg n = ω( 2 lg n). √ √ 4. 2 2 lg n = ω(lg2 n) by taking logs: lg 2 2 lg n = 2 lg n, lg lg2 n = 2 lg lg n.