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2.6 Suppose that your algorithm to find the mean is (n), and that your algorithm to find the median is (n lg n), what will be the execution speed ratio between your algorithm for the mean and your algorithm for the median when the number of values is 1,000,000 mean : median mean : median mean : median It will be 20 :: 1,000,000 : 1,000,000 * lg 1,000,000 :: 1 : lg 1,000,000 :: 1 : 20 times faster to find the mean than the median.

, the smallest) element also to be in its correct location, so the array must be sorted The complexity function O(n2) means that, for large values of n, the number of loop iterations tends to be proportional to n2 That means that, if one large array is twice the size of another, it should take about four times as long to sort The inner j loop iterates n 1 times on the first iteration of the outside i loop, n 2 times on the second iteration of the i loop, n 3 times on the third iteration of the i loop, and so on.

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For example, when n = 7, there are six comparisons made on the first iteration of the i loop, five comparisons made on the second iteration of the i loop, four comparisons made on the third iteration of the i loop, and so forth, so the total number of comparisons is 6 + 5 + 4 + 3 + 2 + 1 = 21 In general, the total number of comparisons will be (n 1) + (n 2) + (n 3) + + 3 + 2 + 1 This sum is n(n 1)/2 (See Theorem A7 on page 323) For large values of n, that expression is nearly n2/2 which is proportional to n2..

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.

public static void sort(int[] a) { boolean sorted=false; for (int i = a.length; i > 0; i -= 2) { for (int j = 1; j < i; j++) { if (a[j-1] > a[j]) { swap(a,j-1,j); } } for (int j = i-2; j > 0; j--) { if (a[j-1] > a[j]) { swap(a, j-1, j); } } } }

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); }

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.

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.

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