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for (int i = 1; i < a.length; i++) { // int ai = a[i], j; // for (j = i; j > 0 && a[j-1] > ai; j--) { // a[j] = a[j-1]; // } a[j] = ai; // // INVARIANT: a[0] <= a[1] <= ... <= a[i]; } }
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Theorem 14.5 The insertion sort is correct. See the solution to Problem 14.23 on page 277 for a proof of this theorem. Theorem 14.6 The insertion sort runs in O(n2) time. See the solution to Problem 14.24 on page 277 for a proof of this theorem. Theorem 14.7 The insertion sort runs in O(n) time on a sorted sequence. See the solution to Problem 14.25 on page 283 for a proof of this theorem. THE SHELL SORT Theorem 14.7 suggests that if the sequence is nearly sorted, then the insertion sort will run nearly in O(n) time. That is true. The shell sort exploits that fact to obtain an algorithm that in general runs in better than O(n1.5) time. It applies the insertion sort repeatedly to skip subsequences such as {s0, s3, s6, s9, . . . , sn 2} and {s1, s4, s7, s10, . . . , sn 1}. These are two of the three skip-3-subsequences. Algorithm 14.4 The Shell Sort (Precondition: s = {s0 . . . sn 1} is a sequence of n ordinal values.) (Postcondition: The entire sequence s is sorted.) 1. Set d = 1. 2. Repeat step 3 until 9d > n. 3. Set d = 3d + 1. 4. Do steps 5 6 until d = 0. 5. Apply the insertion sort to each of the d skip-d-subsequences of s. 6. Set d = d/3. Suppose that s has n = 200 elements. Then the loop at step 2 would iterate three times, increasing d from 1 to d = 4, 13, and 40. The first iteration of the loop at step 4 would apply the insertion sort to each of the 40 skip-40subsequences {s0, s40, s80, s120, s160}, {s1, s41, s81, s121, s161}, {s2, s42, s82, s122, s162}, . . . , {s39, s79, s119, s159, s199}. Then step 6 would reduce d to 13, and then the second iteration of the loop at step 4 would apply the insertion sort to each of the thirteen skip-13-subsequences {s0, s13, s26, s39, s52, s65, . . . , s194}, {s1, s14, s27, s40, s53, s66, . . . , s195}, . . . , {s12, s25, s38, s51, s64, s77, . . . , s193}. Then step 6 would reduce d to 4, and the third iteration of the loop at step 4 would apply the insertion sort to each of the four skip-4-subsequences {s0, s4, s8, s12, . . ., s196}, {s1, s5, s9, s13, . . . , s197}, {s2, s6, s10, s14, . . . , s198}, and {s3, s7, s11, s15, . . . , s199}. Then step 6 would reduce d to 1, and, and the fourth
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iteration of the loop at step 4 would apply the insertion sort to the entire sequence. This entire process would apply the insertion sort 58 times: 40 times to subsequences of size n1 = 5, 13 times to subsequences of size n2 = 15, 4 times to subsequences of size n3 = 50, and once to the entire sequence of size n4 = n = 200. At first glance, the repeated use of the insertion sort within the shell sort would seem to take longer than simply applying the insertion sort directly just once to the entire sequence. Indeed, a direct calculation of the total number of comparisons, using the complexity function n2, yields 40(n12) + 13(n22) + 4(n32) + 1(n42) = 40(52) + 13(152) + 4(502) + 1(2002) = 53,925 which is quite a bit worse than the single n2 = 2002 = 40,000 But after the first iteration of step 4, the subsequent subsequences are nearly sorted. So the actual number of comparisons needed there is closer to n. Thus, the actual number of comparisons is more like 40(n12) + 13(n2) + 4(n3) + 1(n4) = 40(52) + 13(15) + 4(50) + 1(200) = 1,595 which is quite a bit better than 40,000. Theorem 14.8 The shell sort runs in O(n1.5) time. Note that, for n = 200, n1.5 = 2001.5 = 2,829, which is a lot better than n2 = 2002 = 40,000. EXAMPLE 14.7 The Shell Sort
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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 } }
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.
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