By Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, Clifford Stein
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"Introduction to Algorithms, the 'bible' of the sphere, is a complete textbook overlaying the total spectrum of recent algorithms: from the quickest algorithms and information buildings to polynomial-time algorithms for probably intractable difficulties, from classical algorithms in graph conception to important algorithms for string matching, computational geometry, and quantity thought. The revised 3rd version significantly provides a bankruptcy on van Emde Boas bushes, essentially the most priceless info buildings, and on multithreaded algorithms, a subject matter of accelerating importance."--Daniel Spielman, division of laptop technology, Yale University
(Daniel Spielman )
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Extra resources for Introduction to Algorithms: Solutions Manual (2nd Edition)
Sample text
We have T (n) = T (n − 2) + 2 lg n ≤ c(n − 2) lg(n − 2) + 2 lg n ≤ c(n − 2) lg n + 2 lg n = (cn − 2c + 2) lg n Solutions for Chapter 4: Recurrences 4-15 = cn lg n + (2 − 2c) lg n ≤ cn lg n if c > 1 . Therefore, T (n) = O(n lg n). For the lower bound of T (n) = (n lg n), we’ll show that T (n) ≥ cn lg n + dn, for constants c, d > 0 to be chosen. We assume that n ≥ 4, which implies that 1. lg(n − 2) ≥ lg(n/2), 2. n/2 ≥ lg n, and 3. n/2 ≥ 2. ) 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.
Meaning that whenever you interview a candidate who is better than your current ofÞce assistant, you must Þre the current ofÞce assistant and hire the candidate. Since you must have someone hired at all times, you will always hire the Þrst candidate that you interview. Goal: Determine what the price of this strategy will be. 5-2 Lecture Notes for Chapter 5: Probabilistic Analysis and Randomized Algorithms Pseudocode to model this scenario: Assumes that the candidates are numbered 1 to n and that after interviewing each candidate, we can determine if it’s better than the current ofÞce assistant.
Upper bound: Guess: T (n) ≤ dn lg n for some positive constant d. We are given c in the recurrence, and we get to choose d as any positive constant. It’s OK for d to depend on c. Substitution: T (n) ≤ 2T (n/2) + cn n n + cn = 2 d lg 2 2 n = dn lg + cn 2 = dn lg n − dn + cn ≤ dn lg n if −dn + cn ≤ 0 , d ≥ c Therefore, T (n) = O(n lg n). 2. Lower bound: Write T (n) ≥ 2T (n/2) + cn for some positive constant c. Guess: T (n) ≥ dn lg n for some positive constant d. Substitution: T (n) ≥ 2T (n/2) + cn n n + cn = 2 d lg 2 2 n = dn lg + cn 2 = dn lg n − dn + cn ≥ dn lg n if −dn + cn ≥ 0 , d ≤ c 4-4 Lecture Notes for Chapter 4: Recurrences Therefore, T (n) = (n lg n).