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Theorem 1411 The quick sort runs in O(n lgn) time in the best case The best case is when the sequence values are uniformly randomly distributed so that each call to the quick partition algorithm will result in balanced split of the sequence In that case, each recursive call to the quick sort algorithm divides the sequence into two subsequences of nearly equal length As with the binary search and the merge sort (Algorithm 145 on page 261), this repeated subdivision takes lgn steps to get down to size 1 subsequences, as illustrated in the diagram in Figure 142 on page 262 So there are O(lgn) calls made to the quick partition algorithm which runs in O(n) time, so the total running time for the quick sort algorithm is O(n lgn) Theorem 1412 The quick sort runs in O(n2) time in the worst case.

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pI = - 1

for (int i = n - 1; i > 0; i--) { swap(a, 0, i); heapify(a, 0, i); } }

(h) If r3= -I. follows:

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The worst case is when the sequence is already sorted (or sorted in reverse order) In that case, the quick partition algorithm will always select the last element (or the first element, if the sequence is sorted in reverse order), resulting in the most unbalanced split possible: One piece has n 2 elements, and the other piece has 1 element Repeated division of this type will occur O(n) times before both pieces get down to size 1 So there are O(n) calls made to the quick partition algorithm which runs in O(n) time, so the total running time for the quick sort algorithm is O(n2) Note that in the worst case, the quick sort reverts to the selection sort (Algorithm 142 on page 257) because each call to quick partition amounts to selecting the largest element from the subsequence passed to it So actually, Theorem 14.

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which are again on the unit circle. (c) If A,,(z) is a pth-order FIR filter with reflection coefficients r, where (r,I < I for j = 1.. . . . p lr,*l 1, = A,,(z) = A p - I ( ~ ) z - " A ~ - ~ ( ~ - ' ) f and it follows that the polynomial A,(z) is symmetric or antisymmetric, that is.

private static void heapify(int[] a, int i, int j) { int ai = a[i]; // step 1 48 while (2*i+1 < j) { 49 int k = 2*i + 1; 50 if (k + 1 < j && a[k+1] > a[k]) { 51 ++k; // a[k] is the larger child 52 } 53 if (ai >= a[k]) { 54 break; // step 3 55 } 56 a[i] = a[k]; // step 4 57 i = k; // step 5 58 } 59 a[i] = ai; // step 7 60 } The sort() function first converts the array so that its underlying complete binary tree is transformed into a heap. This is done by applying the heapify() function to each nontrivial subtree. The nontrivial

- I,

12 is a corollary to Theorem 144 on page 258 Theorem 1413 The quick sort runs in O(n lgn) time in the average case The proof of this fact is beyond the scope of this outline Theorem 1414 The quick sort is correct The invariant inside the while loop proof claims that all the elements to the left of a[i] are less than or equal to the pivot element and that all the elements to the right of a[j] are greater than or equal to the pivot.

Therefore, A,&) has (generalized) linear phase, which implies that all of the zeros of A,@) lie on the unit circle or in conjugate reciprocal p a n However, if Ir, 1 < I for j = 1.2, . . . p - I, the zeros must lie on the unit circle. The reason for this is as follows. The Schur-Cohn stability criterion states that none of the roots of A,(z) may lie outside the unit circle if ITj\ 5 1 for j = 1.2, . . . , p. Therefore, if A&) has generalized linear phase with no zeros outside the unit circle. then all of the zeros must be on the unit circle.

46 47

and determine the number of multiplications, additions, and delays required to implement the filter. Compare this structure to a direct form realization of H ( z ) in terms of the number of multiplies, adds, and delays.

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