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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

H (el") = HR(el") j Hr(el")

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Although we could find H,(ejw) using the discrete Hilbert transform, an easier approach is as follows. Because h(n) is causal, the wen and odd parts of h ( n )are

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TRANSFORM ANALYSIS OF SYSTEMS Therefore, the relationship between h,(n) and h,,(n) is as follows: h,,(n) = sgn(n) . h,(n)

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public static void sort(int[] a) { // POSTCONDITION: a[0] <= a[1] <= ... <= a[a.length-1]; int d = 1, j, n = a.length; // step 1 while (9*d < n) { // step 2 d = 3*d + 1; // step 3 } while (d > 0) { // step 4 for (int i = d; i < n; i++) { // step 5 int ai = a[i]; j = i; while (j >= d && a[j-d] > ai) { a[j] = a[j-d]; j -= d; } a[j] = ai; } d /= 3; // step 6 } }

sgn(n) =

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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The inverse DTFT of HR(eJW), which is the even part of h(n), is h,(n) = 6(n) Thus,

THE MERGE SORT The merge sort applies the divide-and-conquer strategy to sort a sequence. First it subdivides the sequence into subsequences of singletons. Then it successively merges the subsequences pairwise until a single sequence is re-formed. Each merge preserves order, so each merged subsequence is sorted. When the final merge is finished, the complete sequence is sorted. Although it can be implemented iteratively, the merge sort is naturally recursive: Split the sequence in two, sort each half, and then merge them back together preserving their order. The basis occurs when the subsequence contains only a single element.

+ 0. IS(n - 2) + 0.16(n + 2)

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

CHAP. 14]

+ 2)

and Hl(eJ"), the discrete-time Fourier transform of h,(n), is Hl(eJ") = -0.2sin(2o) Therefore, and ~ ) = H(eJ") = ~ ~ ( e jj Hl(eJW) I

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Here is a simple test driver for the factorial method:

+ 0.2e-*'"

h(n) = 6(n)

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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