6. (6 points) Let A[1..n] be an array containing n distinct integers sorted in i

Question

6. (6 points) Let A[1..n] be an array containing n distinct integers sorted in increasing order, where n 1. The following algorithm from page 364 of the text determines if a query integer x is in A; it returns the position of x in A if x is present and returns zero otherwise. The initial call is Binary-Search(1, n, x).

1. Binary-Search(i,j,x) // Search subarray A[i..j] for x, 1 <= i <= j <= n

2. m <- floor((i+j)/2)

3. if (x = A[m]) then return(m)

4. else if (x < A[m] and i < m) then return(Binary-Search(i,m-1,x))

5. else if (x > A[m] and j > m) then return(Binary-Search(m+1,j,x))

6. else return(0)

7. end

Let P(s) be the assertion that when presented with a subarray A[i..j] of size s = j i + 1, Binary-Search(i, j, x) returns the position of x in A[i..j] if x is present in the subarray and returns zero otherwise, 1 i j n. Use Strong Induction to prove that P(s) is true for s 1. Similar to recursive linear search (discussed in class) consider separately the cases where x is present/not present in the subarray being searched (A[i..j] here).

The question requires using Strong Induction so please have an answer using strong induction

Explanation / Answer

We need to prove the only thing that the given algorithm returns the index of number if numberarray, or 0 if numberarray. We will now prove it by induction, where n=rightleft+1. We have to show that at each step it preserves the invariant.

CASE-1 : When number exists in the given sorted array

Base case : if n=1, since this is the only number, it is returned as the searched number exists in the array.

Inductive step :
Generally speaking, it doesn't matter whether the number is on left side or right side. All that matters is that the algorithm should do next iteration on a strictly smaller sub array.
   1. If number=array[middle]. This works correctly.
   2. If number<array[middle], then since it occurs in array atleast once, and the array is sorted, it must occur at a position between array[left] to array[middle-1]. Therfore, the invariant is preserved if we recur on array[left....(middle1)]
   3. If numberarray[middle], then since it occurs in array atleast once, it must occur at middle or later position. And therefore, the invariant is preserved if we recur on array[middle..rigth]

CASE-2 : When number doesn't exists in the given sorted array
Similar argument follows


some useful links
http://www.cs.cornell.edu/courses/cs211/2006sp/Lectures/L06-Induction/binary_search.html

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