to Solution Exercise 8.1- 4
Solution: Let S be a sequence of n elements divided into n/k subsequences each of length k where all of the elements in any subsequence are larger than all of the elements of a preceding subsequence and smaller than all of the elements of a succeeding subsequence. Claim Any comparison-based sorting algorithm to sort S must take (nlgk) time in the worst case. Proof First notice that, as point out in the hint, we cannot prove the lower bound by multiplying together the lower bounds for sorting each subsequence. That would only prove that there is no faster algorithm that sorts the subsequences independently. This was not what we are asked to prove; we cannot introduce any extra assumptions. Now, consider the decision tree of height h for any comparison sort for S. Since the elements of each subsequence can be in any order, any of the k! permutations correspond to the final sorted order of a subsequence. And, since there are n/k such subsequences, each of which can be in any order, there are (k!)n/k permutations of S that could corresponding to the sorting of some input order.
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Solution to Exercise 8.1- 4 (cont ...)
Thus, any decision tree for sorting S must have at least (k!)n/k leaves. Since a binary tree of height h has no more than 2h leaves, we must have 2h (k!)n/k or h lg((k!)n/k). We therefore obtain h lg((k!)n/k = (n/k) lg(k!) (n/k) lg((k/2)k/2) = (n/2) lg(k/2). The third line comes from k! having k/2 largest terms being at least k/2 each. (We implicitly assume here that k is even, We could adjust with floors and ceilings if k were odd). Since there exists at least one path in any decision tree for sorting S that has length at least (n/2) lg(k/2), the worst-case running time of any comparisonbased sorting algorithm for S is (n lgk)
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Solution to Exercise 8.3-2
Insertion sort is stable. When inserting A[ j ] into the sorted sequence A[1..j 1], we do it the following way: compare A[ j ] to A[ i ], starting with i = j 1 and going down to i = 1. Continue as long as A[ j ] < A[ i ]. Merge sort as defined is stable, because when two elements compared are equal, the tie is broken by taking the elements from array L which keeps them in the original order. Heapsort and quicksort are not stable. One scheme that makes a sorting algorithm stable is to store the index of each element (the element's place in the original ordering) with the element. When comparing two elements, compare them by values their and break ties by their indices. Additional space requirements: For n elements, their indices are 1...n. Each can be written in lgn bits, so together they take O(n lgn) additional space. Additional time requirements: The worst case is when all elements are equal. The asymptotic time does not change because we add a constant amount of work to each comparison.
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Solution to Exercise 25.1-3
0 ... 0 ... (0) What does the matrix L I= 0 ... ... ... ... ... ... ... 0 used in the shortest-paths algorithms correspond to in the regular matrix multiplication? Answer: The matrix L(0) corresponds to the identity matrix 1 0 0 ... 0 0 1 0 ... 0 I= 0 0 1 ... 0 ... ... ... ... ... 0 0 0 ... 1 of regular matrix multiplication. Substitute 0 (the identity for +) for (the identity for min), and 1 (the identity for .) for 0 (the identity for +).
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Solution to Exercise 25.1- 5
The all-pairs shortest-paths algorithm in Section 25.1 computes L(n1) = W n1 = L(0) . W n1 ( 1 lijn - ) =(i , j ) and L(0) is the identity matrix. That is, the entry in the ith where row and jth column of the matrix "product" is the shortest-path distance from vertex i to vertex j, and row i of the product is the solution to the single-source shortest-paths problem for vertex i. Notice that in a matrix "product" C = A . B, the ith row of C is the ith row of A "multiplied" by B. Since all we want is the ith row of C, we never need more than the ith row of A. Thus the solution to the single-source shortest-paths from vertex i is Li(0) . Wn1, where Li(0) is the ith row of L(0) a vector whose ith entry is 0 and whose other entries are .
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Solution to Exercise 25.1- 5 (cont ...)
Doing the above "multiplications" starting from the left is essentially the same as the BELLMAN-FORD algorithm. The vector corresponds to the d values in BELLMAN-FORD the shortest-path estimates from the source to each vertex. The vector is initially 0 for the source and for all other vertices, the same as the values set up for d by INITIALIZE-SINGLE-SOURCE. Each "multiplication" of the current vector by W relaxes all edges just as BELLMAN-FORD does. That is, a distance estimate in the row, say the distance to v, is updated to the smaller estimate, if any, formed by adding some w(u, v) to the current estimate of the distance to u. The relaxation/multiplication is done n 1 times.
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