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{22, 33, 44, 55, 66, 77, 88, 99} search(a, 44): 2 search(a, 50): -1 search(a, 77): 5 search(a, 100): -1
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The search() method returns the index of the target x: search(a, 44) returns 2, because a[2] = 44; search(a, 77) returns 5, because a[5] = 77. The method returns 1 when the target is not in the array: search(a, 50) returns 1, because 50 is not in the array.
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The sequential search is correct. This means that it works. The following argument is a proof of that fact. If n = 0, then the sequence is empty and the loop does not execute at all. Only step 4 executes, immediately returning 1. This satisfies the postconditions: x cannot equal any of the elements because there aren t any. If n = 1, then the loop iterates only once, with i = 0. On that iteration, either s0 = x or s0 x. If s0 = x, then 0 is returned and the postcondition is satisfied. If s0 x, then the loop terminates, step 4 executes, and 1 is returned, and that satisfies the postcondition because the single element of the sequence is not equal to x. Suppose n > 1. We want to apply the First Principle of Mathematical Induction to deduce that the loop invariant must be true on every iteration of the loop. (See page 321.) That requires the verification of the invariant on the first iteration and the deduction of the invariant on iteration i from the corresponding invariant on iteration i 1. On the first iteration of the loop, i = 0, and the loop invariant in step 2 is true vacuously because the subsequence {s0...si 1} is empty. Then in step 3, either s0 = x or s0 x. If s0 = x, then 0 is returned and the postcondition is satisfied. If s0 x, then the loop continues on to a second iteration. Then i = 1, and the loop invariant in step 2 is again true because the subsequence {s0...si 1} = {s0} and s0 x. Suppose now that on iteration i 1, the loop invariant is true; that is, none of the elements in the subsequence {s0..si 1} is equal to x. If the loop continues on to the next iteration, then the condition si = x at step 3 was not true. Thus, si x. Therefore, none of the elements in the subsequence {s0..si } is equal to x, which is the loop invariant on the ith iteration The sequential search runs in O(n) time. This means that, on average, the running time is proportional to the number of elements in the array. So if everything else is the same, then applying the sequential search to an array twice as long will take about twice as long, on average. The following argument is a proof of that fact. If x is in the sequence, say at x = si with i < n, then the loop will iterate i times. In that case, the running time is proportional to i, which is O(n) since i < n. If x is not in the sequence, then the loop will iterate n times, making the running time proportional to n, which is O(n). THE BINARY SEARCH ALGORITHM The binary search is the standard algorithm for searching through a sorted sequence. It is much more efficient than the sequential search, but it does require that the elements be in order. It repeatedly divides the sequence in two, each time restricting the search to the half that would contain the element. You might use the binary search to look up a word in a dictionary. Here is the binary algorithm: (Precondition: s = {s0, s1, . . ., sn 1} is a sorted sequence of n values of the same type as x.) (Postcondition: either the index i is returned where si = x, or 1 is returned.) 1. Let ss be a subsequence of the sequence s, initially set equal to s. 2. If the subsequence ss is empty, return 1.
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