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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42

For x = -

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Algorithm 14.10 The Radix Sort (Precondition: s = {s0 . . . sn 1} is a sequence of n integers or character strings with radix r and width w.) (Postcondition: The sequence s is sorted numerically or lexicographically.) 1. Repeat step 2 for d = 0 up to w 1. 2. Apply a stable sorting algorithm to the sequence s, sorting only on digit number d. A sorting algorithm is said to be stable if it preserves the relative order of elements with equal keys. For example, the insertion sort is stable, but the heap sort is not. EXAMPLE 14.12 Sorting ISBNs with the radix sort

& we have, in sign-magnitude form,

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public static void sort(int[] a) { // POSTCONDITION: a[0] <= a[1] <= ... <= a[a.length-1]; int min = min(a); int max = max(a); int n = a.length; Bucket[] bucket = new Bucket[n]; // step 1 for (int j=0; j<n; j++) { bucket[j] = new Bucket(); } for (int i=0; i<n; i++) { // step 2 int j = n*(a[i] - min)/(max + 1 - min); bucket[j].add(a[i]); // step 3 } int i=0; for (int j=0; j<n; j++) { Bucket bj=bucket[j]; bj.sort(); // step 4 for (int k=0; k<bj.size(); k++) { // step 5 a[i++] = bj.get(k); } } } private static int min(int[] a) { int min = a[0]; for (int ai: a) { if (ai < min) { min = ai; } } return min; } private static int max(int[] a) { int max = a[0]; for (int ai: a) { if (ai > max) { max = ai; } } return max; }

Figure 14.5 shows a sequence of 12 ISBNs and the first four iterations of the radix sort applied to it.

x = 0.001 11 = 1.1 1000

x = 0.001 1 1

This program requires the implementation of this interface:

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Note how the stability is needed to conserve the work done by previous iterations. For example, after the first iteration, 8838650527 precedes 0830636528 because 7 < 8. Both of these keys have the same value 2 in their second least significant digit (digit number d = 1). So on the second iteration, which sorts only on digit number 1, these two keys evaluate as being equal. But they should retain their previous relative order because 27 < 28. Stability guarantees that they do. The columns that have been processed are shaded. So after the third iteration, the right-most 3-digit subsequences are sorted: 109 < 13X < 373 < 453. (Note that X stands for the value 10. So 13X numerically means 130 + 10 = 140.)

+ 0.00001 = 1.1 1001

IMPLEMENTATION O F DISCRETE-TIME SYSTEMS Consider the following implementation of a second-order filter:

public interface Bucket { public void add(int x); public int get(int k); public int size(); } // appends x to end of bucket // returns element k from bucket // returns the number of elements

CHAP. 14]

( I . sinwo) are quantized to 4 bits, draw the set of allowable pole locations in the z-plane. For what types of filters would this filter structure be preferred over a direct form structure'

(a) This filter structure is called the coupled form realization. The system function for this filter may be found as follows. The difference equations relating .r(n), v ( n ) .and v(n) are u(n) = x(n) - r sin(q)y(n - 1 )

For example:

This method assumes that the constants RADIX has WIDTH have been defined. For example, for arrays of ints:

+ r cos(wIJ)v(ri- I )

y (n) = r sin(wo)v(n - I ) + r cos(w,,)v(n- I ) Taking the z-transform of the first equation. we have

43 44 45 46 47

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

-.. -

- r cos q , j r sin q,

private static class Bucket extends java.util.ArrayList<Integer> { void sort() { java.util.Arrays.sort(this.toArray()); } }

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