By Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, Clifford Stein
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"Introduction to Algorithms, the 'bible' of the sector, is a entire textbook masking the total spectrum of recent algorithms: from the quickest algorithms and information buildings to polynomial-time algorithms for likely intractable difficulties, from classical algorithms in graph thought to important algorithms for string matching, computational geometry, and quantity conception. The revised 3rd version significantly provides a bankruptcy on van Emde Boas bushes, the most invaluable information buildings, and on multithreaded algorithms, an issue of accelerating importance."--Daniel Spielman, division of machine technology, Yale University
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Additional resources for Introduction to Algorithms: Solutions Manual (2nd Edition)
To start, let us deÞne a merge-inversion as a situation within the execution of merge sort in which the M ERGE procedure, after copying A[ p . q] to L and A[q + 1 . r] to R, has values x in L and y in R such that x > y. Consider an inversion (i, j ), and let x = A[i] and y = A[ j ], so that i < j and x > y. We claim that if we were to run merge sort, there would be exactly one mergeinversion involving x and y. To see why, observe that the only way in which array elements change their positions is within the M ERGE procedure.
For the upper bound, T (n) = O(n2 ), we use the inductive hypothesis that T (n) ≤ cn 2 for some constant c > 0. By a similar derivation, we get that Solutions for Chapter 4: Recurrences 4-11 T (n) ≤ cn 2 if −2cn + n + c ≤ 0 or, equivalently, n(1 − 2c) + c ≤ 0. This condition holds for c = 1 and n ≥ 1. Thus, T (n) = (n 2 ) and T (n) = O(n 2 ), so we conclude that T (n) = √ h. T (n) = T ( n) + 1 (n2 ). The easy way to do this is with a change of variables, as on page 66 of the text. Let m = lg n and S(m) = T (2m ).
N lg n), we conclude that T (n) = i. T (n) = T (n − 2) + 2 lg n We guess that T (n) = (n lg n). We show the upper bound of T (n) = O(n lg n) by means of the inductive hypothesis T (n) ≤ cn lg n for some constant c > 0. 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.