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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..

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); }

This is the maximum TU she can receive by spending her total income of $10 on X and Y when Px = $2 and Py = $1 Thus, the consumer is in equilibrium by purchasing 2X and 6Y b To buy the third unit of X (at Px = $2), this consumer would have to give up the fth and sixth units of Y (at Py = $1) She would gain 11 utils by purchasing the third unit of X but lose 13 utils (7 + 6) by giving up her fth and sixth Y, with a net loss of 2 utils The consumer s TU would be only 79 utils if she purchased 3X and 4Y (compared with a TU.

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..