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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:

i + V _

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; }

(d )

27 A sort routine which is easy to program is the bubble-sort The program simply scans all of the elements to be sorted repeatedly On each pass, the program compares each element with the one next to it, and reorders the two, if they are in inverse order For instance, to sort the following list: 67314 Bubble-sort starts by comparing 6 and 7 They are in the correct order, so it then compares 7 and 3 They are in inverse order, so bubble-sort exchanges 7 and 3, and then compares 7 and 1 The numbers 7 & 1 are in reverse order, so bubble-sort swaps them, and then compares 7 & 4 Once again, the order is incorrect, so it swaps 7 & 4.

14.25 14.26

Fig. 1-1

The passive and active circuit elements introduced in the previous section are all available in SPICE modeling; however, the manner of node speci cation and the voltage and current sense or direction are clari ed for each element by Fig. 1-2. The universal ground node is assigned the number 0. Otherwise, the node numbers n1 (positive node) and n2 (negative node) are positive integers

partition() method: private static int partition(int[] a, int p, int q) { int m = (p + q)/2; m = indexOfMedian(a, p, m, q-1); swap(a, p, m); // The rest is the same as lines 20-32 in Example 14.9 on page 264 }

End of scan 1: 63147 Scanning left to right again results in: 31467 Scanning left to right again results in a correct ordering: 13467 Write pseudo code for the bubble-sort.

Fig. 1-2

This requires a method for locating the index of three array elements:

CHAP. 1]

private static int indexOfMedian(int[] a, int i, int j, int k) { // Returns the index of the median of {a[i], a[j], a[k]} if (a[i] <= a[j] && a[j] <= a[k]) return j; if (a[i] <= a[k] && a[k] <= a[j]) return k; if (a[j] <= a[i] && a[i] <= a[k]) return i; if (a[j] <= a[k] && a[k] <= a[i]) return k; if (a[k] <= a[i] && a[i] <= a[j]) return i; return j; }

bubbleSort( list ) length <-- length of list // look through the whole list to find // mis-ordered pairs, and continue until // we get through the whole list without // swapping any pairs do { // start at the left end of the list: index = 1 swapped_pair <-- false index <-- 1 while index <= length - 1 { // if this pair is mis-ordered, swap them if list[index] > list[index + 1] { swap( list[index], list[index + 1] ) swapped_pair = true // increment the index to look at the // next pair up index <-- index + 1 } } // quit only when we have gone through the whole // list and found it unnecessary to swap any pair } while( swapped = true ) end 28 What is the bubble-sort ( n2 ) 2.