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, 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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Listed in Table 9.2 are the side-lobe amplitudes of several windows along with the approximate transition width and stopband attenuation that results when the given window is used to design an N th-order low-pass filter.

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; }

w(n) =

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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..

Hanning'

14.25 14.26

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); } } } }

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Table 9-2 The Peak Side-Lobe Amplitude of Some Common Windows and the Approximate Transition Width and Stopband Attenuation of an Nth-Order Low-Pass Filter Designed Using the Given Window.

Window Side-Lobe Amplitude (dB)

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 }

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); }

- 13 -31 -41 -57

Stopband Attenuation (dB) -21 -44 -53 - 74

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

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.

EXAMPLE 9.3.1 specifications:

Suppose that we would like to design an FIR linear phase low-pass filter according to the following

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