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The payoff from this mean between the extremes is greater efficiency: O(n lg n) instead of O(n2) THE SPEED LIMIT FOR COMPARISON SORTS Theorem 1417 No sorting algorithm that rearranges the array by comparing its elements can have a worst-case complexity function better than O(n lg n) Consider the decision tree that covers all possible outcomes of the algorithm on an array of size n Since the algorithm rearranges the array by comparing its elements, each node in the decision tree represents a condition of the form (a[i] < a[j]) Each such condition has two possible outcomes (true or false), so the decision tree is a binary tree And since the tree must cover all possible arrangements, it must have at least n! leaves Therefore, by Corollary 113 on page 203, the height of the decision tree must be at least lg(n!).

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public static void sort(int[] a) { // POSTCONDITION: a[0] <= a[1] <= ... <= a[a.length-1]; int n = a.length; for (int i = n/2 - 1; i >= 0; i--) { // step 1 heapify(a, i, n); // step 2 }

Therefore,

H (z)=

In the worst case, the number of comparisons that the algorithm makes is the same as the height of the decision tree Therefore, the algorithm s worst-case complexity function must be O(lg(n!)) Now by Stirling s Formula (outlined on page 325), n! so log n! log 2n n -e.

CHAP. 14]

b(0)

In order for H ( z )to be stable, it is necessary and sufficient for the reflection coefficients TI and Tzto have a magnitude that is less than 1. In t e n s of the first two reflection coefficients, the denominator of H(z) is

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(Here, log means the binary logarithm log2.) Therefore, the algorithm s worst-case complexity function must be O(n log n). Theorem 14.17 applies only to comparison sorts. A comparison sort is an algorithm that sorts elements by comparing their values and then changes their relative positions according to the outcomes of those comparisons. All the sorting algorithms outlined previously are comparison sorts. In contrast, the following sorting algorithms are not comparisons sorts.

40 41 42 43 44 45

Because we require that 1 rl1 < I,

THE RADIX SORT The radix sort assumes that the sequence s element type is a lexicographic array of constant size, that is, either a character string type or an integer type. Let r be the array element s radix (e.g., r = 26 for ASCII character strings, r = 10 for decimal integers, r = 2 for bit strings), and let w be the constant width of the lexicographic array. For example, for U.S. Social Security numbers, d = 10 and w = 9. EXAMPLE 14.11 Sorting Books by Their ISBNs

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To find the lattice filter structure for this allpass filter, we use the step-down recursion to find the reflection coefficients corresponding to the denominator polynomial, A3(z)= I - 0.82-I First, we note that Az(z) =

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

r3= -0.5

Every book published since the 1970s has been assigned a unique international standard book number (ISBN). These are usually printed at the bottom of the back cover of the book. For example, the ISBN for this book is 0071476989. (ISBNs are usually hyphenated, like this: 0-07-147698-9, to distinguish the four separate fields that make up the code.) The last digit is a check digit, computed from the other nine digits. Since it can be any of the 10 numeric digits or the letter X, we have that the radix r = 11, while the number of digits d = 10.

12. Then, we find Az(z) as follows:

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