Suppose an array of length k is input to the while loop of the modified binary search algorithm. Show that after one iteration of the loop. if a [ m i d ] ≠ x , the input to the next iteration is an array of length at most ⌊ k / 2 ⌋ .

Show that after one iteration of the loop, if a[mid]≠x, the input to the next iteration is an array of length at most ⌊k/2⌋.

Given information:

Suppose an array of length k is input to the while loop of the modified binary search algorithm.

Proof:

Let the input to the while-loop of the modified binary search algorithm contain k elements, thus a[bot],a[bot+1],...,a[top] contains k elements.

By the number of elements in a list theorem, we also know that the number of integers from bot to top top is top − bot + 1 and thus a[bot],a[bot+1],...,a[top] contains top + bot − 1 element

top+bot−1=k

We then assign mid to the upper of the two middle indices of the array.

mid=⌈bot+top2⌉

If a[mid]≠x, then either a[mid]<x or a[mid]>x.

FIRST CASE: a[mid]<x

If a[mid]<x, then we assign mid − 1 to top and thus the input of the next iteration is then a[bot],a[bot+1],...,a[mid−1]. By the number of elements in a list theorem, we also know that the number of integers from bot to mid − 1 top is ( mid − 1) − bot + 1 = mid − bot and thus a[bot],a[bot+1],...,a[mid−1] contains mid − bot elements.

Length array next iteration = mid − bot

=⌈bot+top2⌉−bot

=⌈k+12⌉−bot                                              top+bot−1=k

≤⌈k+12⌉−1                                                            as bot≥1

=⌈k+12−1⌉

=⌈k+1−22⌉

=⌈k−12⌉

⌈k−12⌉=k2=⌈k2⌉=⌊k2⌋ when k even and thus k−1 odd

⌈k−12⌉=k−12=⌊k2⌋ when k odd and thus k−1 even

=⌊k2⌋

SECOND CASE: a[mid]>x

If a[mid]>x, then we assign mid + 1 to bot and thus the input of the next iteration is then a[mid+1],a[mid+2],