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3. (Invariant: If x is in the sequence s, then it must be in the subsequence ss.) 4. Let si be the middle element of ss. 5. If si = x, return its index i . 6. If si < x, repeat steps 2 7 on the subsequence that lies above si . 7. Repeat steps 2 7 on the subsequence of ss that lies below si . It is implemented in Example 2.5. EXAMPLE 2.5 The Binary Search
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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 output is the same as in Example 2.4.
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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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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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2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 What is the difference between a component and an element of an array What does it mean to say that Java does not allow multidimensional arrays What is an ArrayIndexOutOfBoundsException exception, and how does its use distinguish Java from other languages such as C and C++ What types are valid for array indexes What s wrong with this definition:
Arrays arrays = new Arrays();
What is the simplest way to print an array of objects If the binary search is so much faster than the sequential search, why would the latter ever be used What happens if the sequential search is applied to an element that occurs more than once in the array What happens if the binary search is applied to an element that occurs more than once in the array
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