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EXAMPLE 6.14 The Binary Search Algorithm
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This program tests a function that implements the Binary Search algorithm. It uses the same test driver that was used in Example 6.12 on page 133 to test the Linear Search algorithm: int index(int,int[],int); int main() { int a[] = { 22, 33, 44, 55, 66, 77, 88 }; cout << "index(44,a,7) = " << index(44,a,7) << endl; cout << "index(60,a,7) = " << index(60,a,7) << endl; } int index(int x, int a[], int n) { // PRECONDITION: a[0] <= a[1] <= ... <= a[n-1]; // binary search: int lo=0, hi=n-1, i; while (lo <= hi) { i = (lo + hi)/2; // the average of lo and hi if (a[i] == x) return i; if (a[i] < x) lo = i+1; // continue search in a[i+1..hi] else hi = i-1; // continue search in a[lo..i-1] } return n; // x was not found in a[0..n-1] } index(44,a,7) = 2 index(60,a,7) = 7 Note that the array is already sorted before the Binary Search is applied. That requirement is expressed in the PRECONDITION specified as a comment in the function s code. On each iteration of the while loop, the middle element a[i] of the subarray a[lo..hi] (i.e., all the elements from a[lo] to a[hi]) is examined. If it is not the target x, then the search continues either on the upper half a[i+1..hi] or on the lower half a[lo..i-1]. If (a[i] < x), then x could not be in the lower half (since the array is sorted into increasing order), so the lower half can be ignored and the search continued on only the upper half. Similarly, if the condition (a[i] < x) is false, then the search is continued on only the lower half. So on each iteration of the loop, the scope of the search is reduced by about 50%. The loop stops either when x is found at a[i] and the function returns, or when lo > hi. In that latter case, the subarray a[lo..hi] is empty, meaning that x was not found, so the function returns n. Here is a trace of the call index(44,a,7). When lo hi i a[i] x the loop begins, x = 44, n = 7, lo = 0, and hi = 6; the 0 6 3 55 > 44 middle element of the array a[0..6] is a[3] = 55 2 1 33 < 44 which is greater than x, so hi gets reset to i-1 = 2. On the second iteration, lo = 0 and hi = 2; the middle 2 2 44 == 44 element of the subarray a[0..2] is a[1] = 33 which is less than x, so lo gets reset to i+1 = 2. On the third iteration, lo = 2 and hi = 2; the middle element of the subarray a[2..2] is a[2] = 44 which is equal to x, so the function returns 2, indicating that the target x is was found at a[2]. Here is a trace of the call index(60,a,7). When lo hi i a[i] x the loop begins, x = 44, n = 7, lo = 0, and hi = 6; the 0 6 3 55 < 60 middle element of the array a[0..6] is a[3] = 55 4 5 77 > 60 which is less than x, so lo gets reset to i+1 = 4. On the second iteration, lo = 4 and hi = 6; the middle element 4 4 66 > 60 of the subarray a[4..6] is a[5] = 77 which is
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greater than x, so hi gets reset to i-1 = 4. On the third iteration, lo = 4 and hi = 4; the middle element of the subarray a[4..4] is a[4] = 66 which is greater than x, so hi gets reset to i-1 = 3. That terminates the loop, so the function returns 7, indicating that the target x was not found.
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The Binary Search algorithm is significantly different from the Linear Search algorithm. The most important distinction is that the Binary Search works only on sorted arrays. The benefit of that requirement is that the Binary Search is much faster than the Linear Search. For example, on an array of 100 elements, the Linear Search could take up to 100 iterations, but the Binary Search will not need more than 8 iterations, no matter what the target is. That is because the Binary Search runs in logarithmic time; i.e., the number of iterations cannot exceed lgn + 1, where n is the size of the array and lgn is the binary (base 2) logarithm of n. When n = 100, lgn + 1 = 7.64. Note that in Example 6.14, n = 7 elements, so lgn + 1 = 3.81; this means that no more than 3 iterations will ever be needed. A third distinction between the two algorithms is that the Linear Search returns the smallest index i for which a[i] == x. But the Binary Search is not specific: if there are multiple copies of x, you cannot be sure which one is located by the returned index. Since the Binary Search requires that the array be sorted, it is useful to have a separate function that tests that condition. EXAMPLE 6.15 Determining whether an Array is Sorted
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This program tests a boolean function that determines whether a given array is nondecreasing. bool isNondecreasing(int a[], int n); int main() { int a[] = { 22, 44, 66, 88, 44, 66, 55 }; cout << "isNondecreasing(a,4) = " << isNondecreasing(a,4) << endl; cout << "isNondecreasing(a,7) = " << isNondecreasing(a,7) << endl; } bool isNondecreasing(int a[], int n) { // returns true iff a[0] <= a[1] <= ... <= a[n-1]: for (int i=1; i<n; i++) if (a[i]<a[i-1]) return false; return true; } isNondecreasing(a,4) = 1 isNondecreasing(a,7) = 0 If the function finds any adjacent pair (a[i-1],a[i]) of elements that decrease (i.e., a[i]<a[i-1]), then it returns false. If that doesn t happen, then it returns true, meaning that the array is nondecreasing. Note that the boolean values true and false are printed as the integers 1 and 0; that is how they are stored in memory.
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If the precondition in Example 6.14 that the array be sorted is not true, the Binary search function search() will not work correctly. Such conditions can be checked automatically using the assert() function defined in the <cassert> header. This function takes a boolean argument. If the argument is false, the function terminates the program and reports the fact to the operating system. If the argument is true, the program continues unaffected.
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