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Algorithm 14.5 The Merge Sort (Precondition: s = {sp . . . sq 1} is a sequence of q p ordinal values.) (Postcondition: The entire sequence s is sorted.) 1. If q p > 1, do steps 2 5. 2. Split s into two subsequences, a = {sp . . . sm 1} and b = {sm . . . sq 1}, where m = (q p)/2. 3. Sort a. 4. Sort b. 5. Merge a and b back into s, preserving order. EXAMPLE 14.8 The Merge Sort
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public static void sort(int[] a) { // POSTCONDITION: a[0] <= a[1] <= ... <= a[a.length-1]; sort(a, 0, a.length); } private static void sort(int[] a, int p, int m, int q) { // PRECONDITIONS: 0 <= p <= q <= a.length; // POSTCONDITION: a[p..q) is sorted; if (q - p < 2) { // step 1 return; } int m = (p + q)/2; // step 2 sort(a, p, m); // step 3 sort(a, m, q); // step 4 merge(a, p, m, q); // step 5 }
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private static void merge(int[] a, int p, int m, int q) { // PRECONDITIONS: 0 <= p <= m < q <= a.length; 20 // a[p..m) is sorted; 21 // a[m..q) is sorted; 22 // POSTCONDITION: a[p..q) is sorted; 23 if (a[m-1] <= a[m]) { 24 return; 25 } 26 int i = p, j = m, k = 0; 27 int[] tmp = new int[q-p]; 28 while (i < m && j < q) { 29 // INVARIANT: tmp[0..k) is sorted 30 tmp[k++] = ( a[i]<=a[j] a[i++] : a[j++] ); 31 } 32 System.arraycopy(a, i, a, p+k, m-i); 33 System.arraycopy(tmp, 0, a, p, k); 34 } The main sort() method sorts the entire array by invoking the overloaded sort() method with
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parameters for the starting index k and the length n of the subarray. That three-parameter method sorts the subarray by sorting its left half and its right half separately and then merging them. The merge() method merges the two halves a[p..m) and a[m..q) into a temporary array, where m is the middle index m = p + n/2. The while loop copies one element on each iteration; it copies the smaller of the two elements a[i] and a[j]. The post increment operator automatically advances the index of the copied element. When all the elements of one half have been copied, the while loop stops and then all the elements are copied back into a[].
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Theorem 14.9 The merge sort runs in O(n lgn) time. In general, the merge sort works by repeatedly dividing the array in half until the pieces are singletons, and then it merges the pieces pairwise until a single piece remains. This is illustrated by the diagram in Figure 14.1. The number of iterations in the first part equals the number of times n can be halved: that is, lg n 1. In terms of the number and sizes of the pieces, the second part of the process reverses the first. So the second part also has lg n 1 steps. So the entire algorithm has O(lg n) steps. Each step compares all n elements. So the total number of comparisons is O(n lg n). Theorem 14.10 The merge sort is correct. The proof follows from the preconditions and postconditions given in the code. In the Figure 14.1 The merge sort main sort() method, the array is already sorted if its length is 0 or 1. Otherwise, the postcondition of the three-parameter sort() method guarantees that the array will be sorted after that method returns because the entire array is passed to it. That postcondition is the same as the postcondition of the merge() method, which is invoked last, so it remains to verify that the merge() method s postcondition will be true. The merge() method s postcondition follows from its loop invariant, because when that loop has finished, the tmp[] array is sorted and that is copied back into a[] in the same order. So it remains to verify the loop invariant for all k < q - p. Suppose the invariant is false for some k, that is, tmp[0..k) is not sorted. Then there must be some x and y in tmp[0..k), where x was copied into tmp[] before y but x > y. We may assume without loss of generality that x was copied from the left half of a[] and y from the right half, as Figure 14.2 The merge sort shown in Figure 14.2. Thus, x = a[r] and y = a[s] for some indexes r and s such that p r < i and m s < j. Now the two halves of a[] are each already separately sorted. Then for every element z in a[m..s], z a[s]. But a[s] = y < x. Therefore, every element z in a[m..s] must have been copied into tmp[] before x was, because this assignment
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tmp[k++] = ( a[i]<=a[j] a[i++] : a[j++] ); always copies the smaller element first. But that means that a[s] was copied into tmp[] before x. But a[s] = y, which was assumed to have been copied into tmp[] after x. This contradiction
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