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(a) Draw a signal flowgraph for this system that requires the minimum number of multiplications. (b) If the input to this system is bounded with Ix(n)l 4 I for all n , what is the maximum value that the output. y(n ), can attain

3 4 5 6 7 8 9 10 11

(a) Because this system is a linear phase filter, it may be implemented with a network that has only four multiplies and six delays as shown in the figure below.

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Theorem 14.1 The Bubble Sort is correct. See the solution to Problem 14.14 on page 276 for a proof of this theorem. Theorem 14.2 The Bubble Sort runs in O(n2) time. See the solution to Problem 14.15 on page 276 for a proof of this theorem. THE SELECTION SORT The selection sort is similar to the bubble sort. It makes the n 1 passes through a sequence of n elements, each time moving the largest of the remaining unsorted elements into its correct position. But it is more efficient than the bubble sort because it doesn t move any elements in the process of finding the largest. It makes only one swap on each pass after it has found the largest. It is called the selection sort because on each pass it selects the largest of the remaining unsorted elements and puts it in its correct position. Algorithm 14.2 The Selection Sort (Precondition: s = {s0 . . . sn 1} is a sequence of n ordinal values.) (Postcondition: The entire sequence s is sorted.) 1. Do steps 2 4 for i = n 1 down to 1. 2. Locate the index m of the largest element among {s0 . . si} . 3. Swap si and sm. 4. (Invariants: the subsequence { si . . . sn 1} is sorted, and si = max{ s0 . . . si}.)

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With Ix(n)l < 1 for all n,

for (int i = 1; i < a.length; i++) { // int ai = a[i], j; // for (j = i; j > 0 && a[j-1] > ai; j--) { // a[j] = a[j-1]; // } a[j] = ai; // // INVARIANT: a[0] <= a[1] <= ... <= a[i]; } }

h(n) =

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

1 2 3 4

(c) For both of these implementations, determine the number of multiplications and additions required to compute each output value and the number of storage registers that are required.

which converges for Izl z 0. Thus, H ( z ) may be implemented as a cascade of an IIR system,

public static void sort(int[] a) { // POSTCONDITION: a[0] <= a[1] <= ... <= a[a.length-1]; for (int i = a.length-1; i > 0; i--) { // step 1 int m = 0; // step 2 for (int j = 1; j <= i; j++) { if (a[j] > a[m]) { m = j; } } // INVARIANT: a[m] >= a[j] for all j <= i; swap(a, i, m); // step 3 // INVARIANTS: a[j] <= a[i] for all j <= i; // a[i] <= a[i+1] <= ... <= a[a.length-1]; } }

Theorem 145 The insertion sort is correct See the solution to Problem 1423 on page 277 for a proof of this theorem Theorem 146 The insertion sort runs in O(n2) time See the solution to Problem 1424 on page 277 for a proof of this theorem Theorem 147 The insertion sort runs in O(n) time on a sorted sequence See the solution to Problem 1425 on page 283 for a proof of this theorem THE SHELL SORT Theorem 147 suggests that if the sequence is nearly sorted, then the insertion sort will run nearly in O(n) time That is true The shell sort exploits that fact to obtain an algorithm that in general runs in better than O(n15) time It applies the insertion sort repeatedly to skip subsequences such as {s0, s3, s6, s9, , sn 2} and {s1, s4, s7, s10, .

with an FIR system,

H2(z)= 1 - u7z-'

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