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Theorem 14.3 The selection sort is correct. See the solution to Problem 14.19 on page 276 for a proof of this theorem. Theorem 14.4 The selection sort runs in O(n2) time. See the solution to Problem 14.20 on page 276 for a proof of this theorem. Note that even though the bubble sort and the selection sort have the same complexity function, the latter runs quite a bit faster. That fact is suggested by the two traces: The bubble sort made 18 swaps while the selection sort made only 7. The selection sort has the advantage of swapping elements that are far apart, so it makes one swap where the bubble sort could require several. (See Exercise 11.8.) THE INSERTION SORT Like the two previous algorithms, the insertion sort makes n 1 passes through a sequence of n elements. On each pass it inserts the next element into the subarray on its left, thereby leaving that subarray sorted. When the last element is inserted this way, the entire array is sorted. Algorithm 14.3 The Insertion 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 = 1 up to n 1. 2. Hold the element si in a temporary space. 3. Locate the least index j for which sj >= si . 4. Shift the subsequence {sj . . . si 1} up one position to {sj+1 . . . si}. 5. Copy the held value of si into sj. 6. (Invariant: the subsequence { s0..si} is sorted.) EXAMPLE 14.6 The Insertion Sort

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, sn 1} These are two of the three skip-3-subsequences Algorithm 144 The Shell Sort (Precondition: s = {s0 sn 1} is a sequence of n ordinal values) (Postcondition: The entire sequence s is sorted) 1 Set d = 1 2 Repeat step 3 until 9d > n 3 Set d = 3d + 1 4 Do steps 5 6 until d = 0 5 Apply the insertion sort to each of the d skip-d-subsequences of s 6 Set d = d/3 Suppose that s has n = 200 elements Then the loop at step 2 would iterate three times, increasing d from 1 to d = 4, 13, and 40 The first iteration of the loop at step 4 would apply the insertion sort to each of the 40 skip-40subsequences {s0, s40, s80, s120, s160}, {s1, s41, s81, s121, s161}, {s2, s42, s82, s122, s162}, .

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multiplications, and six additions. The cascade, on the other hand, requires one additional delay but only two multiplications and two additions.

public static void sort(int[] a) { // POSTCONDITION: a[0] <= a[1] <= ... <= a[a.length-1];

, {s39, s79, s119, s159, s199} Then step 6 would reduce d to 13, and then the second iteration of the loop at step 4 would apply the insertion sort to each of the thirteen skip-13-subsequences {s0, s13, s26, s39, s52, s65, , s194}, {s1, s14, s27, s40, s53, s66, , s195}, , {s12, s25, s38, s51, s64, s77, , s193} Then step 6 would reduce d to 4, and the third iteration of the loop at step 4 would apply the insertion sort to each of the four skip-4-subsequences {s0, s4, s8, s12, , s196}, {s1, s5, s9, s13, , s197}, {s2, s6, s10, s14, , s198}, and {s3, s7, s11, s15, , s199} Then step 6 would reduce d to 1, and, and the fourth.

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A DSP chip used in real-time signal processing applications has an instruction cycle time of 100 ns. One of the instructions in the instruction set, MACD, will fetch a value from data memory (input signal), fetch another data value from program memory (filter coefficient), multiply the two numbers together, add the product to the accumulator, and then move a number in data memory into the next memory location (this corresponds to a shift or delay of the data sequence). Thus, for an FIR filter of order N, to find the value

CHAP. 14]

of the output at time n , we need one instruction to read the new input value, x ( n ) , into the processor, we need (N 1) MACD instructions to evaluate the sum

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