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public static void sort(int[] a) { for (int i = 0; i < a.length-1; i++) { int j=i; for (int k = i+1; k < a.length; k++) { if (a[k] < a[j]) { j = k; } } swap(a, i, j); } } public static void sort(int[] a) { sort(a, a.length); } private static void sort(int[] a, int n) { if (n < 2) { return; } int j = 0; for (int k = 1; k < n; k++) { if (a[k] > a[j]) { j = k; } } swap(a, n-1, j); sort(a, n-1); }
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The last loop invariant proves correctness. So, like the proof for the bubble sort, we need only verify the loop invariants. On the first iteration of the main loop (step 1), a[i] is the last element in the array, so the index k of the inner loop runs through every element after a[0]. The value of the index j begins at 0 and then changes each time k finds a larger element. Since j is always reset to the index of the larger element, a[j] will be the largest element of the array when the inner loop finishes. This verifies the first loop invariant. On each successive iteration of the outer loop, the index k runs through the remaining unsorted segment of the array, so for the same reason, a[j] will be the largest element of that remaining segment when the inner loop finishes. This verifies that the first loop invariant is true on every iteration of the outer loop. Since swap(a,i,j) simply interchanges a[i] with a[j], the second loop invariant follows from the first. The third loop invariant follows from the second and by mathematical induction. During the first iteration of the main loop, the inner loop finds a[j] to be the largest element in the array. The swap(a,i,j) puts that largest element in the last location a[i], so a[i] must be >= all the a[j]. Prior to the ith iteration of the main loop, we have by the inductive hypothesis that the subarray a[i+1..n) is sorted and all the values in the subarray a[0..i] are smaller than a[i+1]. Then after the ith iteration, a[i] is one of those smaller elements, so a[i] a[i+1] ... a[n-1]. Again, the proof is essentially the same as that for the corresponding theorem for the bubble sort. On the first iteration of the outer i loop, the inner j loop iterates n 1 times. On the second, it iterates n 2 times. This progression continues, for a total of (n 1) + (n 2) + + 2 + 1 = n(n 1)/2.
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public static void sort(int[] a, int[] ndx) { for (int i = 1; i < a.length; i++) { int ndxi = ndx[i], j; for (j = i; j > 0 && a[ndx[j-1]] > a[ndxi]; j--) { ndx[j] = ndx[j-1]; } ndx[j] = ndxi; } }
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public static void sort(int[] a) { sort(a, a.length); } public static void sort(int[] a, int n) { if (n < 2) { return; } sort(a, n-1); int temp = a[n-1], j; for (j = n-1; j > 0 && a[j-1] > temp; j--) { a[j] = a[j-1]; } a[j] = temp; }
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On the first iteration of the main loop, a[1] is compared with a[0] and interchanged if necessary. So a[1] after the first iteration. If we assume that the loop invariant is true prior to some kth iteration, then it must also be true after that iteration because during it a[k] is inserted between the elements that are less than or equal to it and those that are greater. It follows by mathematical induction that the loop invariant is true for all k.
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The proof is similar to that for the corresponding theorems for the bubble sort and the selection sort. On the first iteration of the outer i loop, the inner j loop iterates once. On the second, it iterates once or twice, depending upon whether a[1] > a[2]. On the third iteration, the inner j loop iterates at most three times, again depending upon how many of the elements on the left of a[3] are greater than a[3]. This pattern continues, so that on the kth iteration of the outer loop, the inner loop iterates at most k times. Thus the maximum total number of iterations is 1 + 2 + + (n 1) = n(n 1)/2. In this case, the inner loop will iterate only once for each iteration of the outer loop. So the total number of iterations of the inner loop is: 1 + 1 + 1 + +1 + 1 = n 1. For the quick sort pivoting on the last element, the only changes needed are in the partition() method:
private static int partition(int[] a, int p, int q) { int pivot = a[q-1], i = p-1, j = q-1; while (i < j) { while (i < j && a[++i] <= pivot) ; // empty loop if (i < j) { a[j] = a[i]; } while (j > i && a[--j] >= pivot) ; // empty loop if (j > i) { a[i] = a[j]; } } a[j] = pivot; return j; }
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