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public class TestBinarySearch { public static void main(String[] args) { int[] a = {22, 33, 44, 55, 66, 77, 88, 99}; ch02.ex02.DuplicatingArrays.print(a); System.out.println("search(a, 44): " + search(a, 44)); System.out.println("search(a, 50): " + search(a, 50)); System.out.println("search(a, 77): " + search(a, 77)); System.out.println("search(a, 100): " + search(a, 100)); } public static int search(int[] a, int x) { // POSTCONDITIONS: returns i; // if i >= 0, then a[i] == x; otherwise i == -1; int lo = 0; int hi = a.length; while (lo < hi) { // step 1 // INVARIANT: if a[j]==x then lo <= j < hi; // step 3 int i = (lo + hi)/2; // step 4 if (a[i] == x) { return i; // step 5 } else if (a[i] < x) { lo = i+1; // step 6 } else { hi = i; // step 7 } } return -1; // step 2 } }

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The system function of a continuous-time filter H a ( s ) of order N 2 2 may be expressed as a cascade of two lower-order systems:

Ha(s)= H,I(s)H,~(s) Therefore, a digital filter may either be designed by applying a transformation directly to H,(s) or by individually transforming H a l ( s )and H a z ( s )into H I ( z ) and H:!(z),respectively, and then realizing H ( z ) as the cascade: H ( z ) = Hi(z)Hz(z) ( a ) If H l ( z ) and H2(z) are designed from H O l ( s )and Ha2(s) using the impulse invariance technique, with the filter that is designed by using the impulse invaricompare the cascade H ( z ) = H I ( z ) H 2 ( z ) ance technique directly on H,(s).

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The binary search is correct. The loop invariant is true on the first iteration because the current subsequence is the same as the original sequence. On every other iteration, the current subsequence was defined in the preceding iteration to be the half of the previous subsequence that remained after omitting the half that did not contain x. So if x was in the original sequence, then it must be in the current subsequence. Thus the loop invariant is true on every iteration. On each iteration, either i is returned where si = x, or the subsequence is reduced by more than 50 percent. Since the original sequence has only a finite number of elements, the loop cannot continue indefinitely. Consequently, the algorithm terminates either by returning i from within the loop or at step 6 or step 7 where 1 is returned. If i is returned from within the loop, then si = x. Otherwise, the loop terminates when hi < lo; that is, when the subsequence is empty. In that case we know by the loop invariant that si is not in the original sequence.

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(a) Due to sampling, aliasing occurs when designing a digital filter using the impulse invariance method. Because the operations of sampling and convolution do not commute, the filter designed by using impulse invariance on H,(s) will not be the same as the filter designed by cascading the two filters that are designed using impulse invariance on H,,(s) and Ha2(s). In other words, if

where h(n) = h,(nT,), hl(n) = hal(nTx), h2(n) = ha2(nT,). As an example, consider the continuous-time and filter that has a system function tf,(s) = I I ( s + l)(s+2) s+ 1 I s+2

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On the other hand, writing H,(s) as a cascade of two first-order systems,

( b ) With the bilinear transformation. ( T , = 2 )

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The binary search runs in O(lgn) time. This means that, on average, the running time is proportional to the logarithm of the number of elements in the array. So if everything else is the same, if it takes an average of T milliseconds to run on an array of n elements, then will take an average of 2T milliseconds to run on an array of n 2 elements. For example, if it takes 3 ms to search 10,000 elements, then it should take about 6 ms to search 100,000,000 elements! The following argument is a proof of that fact. Each iteration of the loop searches a subarray that is less than half as long as the subarray on the previous iteration. Thus the total number of iterations is no more than the number of times that the length n can be divided by 2. That number is lgn. And the total running time is roughly proportional to the number of iterations that the loop makes. Review Questions

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