A sorted list of numbers contains 128 elements. Which of the following is closest to the maximum number of list elements that can be examined when performing a binary search for a value in the list?
A sorted list of numbers contains 128 elements. Which of the following is closest to the maximum number of list elements that can be examined when performing a binary search for a value in the list?
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To find the maximum number of list elements that can be examined when performing a binary search on a sorted list of 128 elements, we need to understand how binary search works.
Binary search is an efficient algorithm that repeatedly divides the search interval in half. It starts by examining the middle element of the list. If the target value is equal to the middle element, the search is successful. If the target value is less than the middle element, the search continues in the lower half of the list. If the target value is greater than the middle element, the search continues in the upper half of the list.
In the worst case scenario, where the target value is not present in the list, the binary search algorithm needs to examine log₂(n) + 1 elements, where n is the number of elements in the list. This is because, in each iteration, the search interval is reduced by half, and the search continues until the interval becomes empty.
For a sorted list of 128 elements, the maximum number of elements that can be examined in the worst case is:
log₂(128) + 1 = 8
Therefore, the closest option to the maximum number of list elements that can be examined when performing a binary search on a sorted list of 128 elements is 8.
8. The binary search algorithm starts at the middle of the list and repeatedly eliminates half the elements until the desired value is found or all elements have been eliminated. For a list with 128 elements, the list will be cut in half a maximum of 7 times (causing 8 elements to be examined). The list will start with 128 elements, then 64 elements, then 32 elements, then 16 elements, then 8 elements, then 4 elements, then 2 elements, then 1 element.