![4. (15 pts) Crabbe and Goyle are arguing about binary search. Goyle writes the following pseudocode on the board, which he claims implements a binary search for a target value v within input array A containing n elements. bSearch(A, v) f return binarySearch(A, 0, n, v) binarySearch (A, 1, r, v) f if 1r then return -1 p floor i r)/2) if Alp]v then return m if Am] < v thern return binarySearch(A, m+1, r, v) else return binarySearch (A, 1, m-1, v) (a) Help Crabbe determine whether this code performs a correct binary search. If it does, prove to Goyle that the algorithm is correct. If it is not, state the bug(s) give line(s) of code that are correct, and then prove to Goyle that your fixed algorithm is correct. (b) Goyle tells Crabbe that binary search is efficient because, at worst, it divides the remaining problem size in half at each step. In response Crabbe claims that nary search, which would divide the remaining array A into thirds at each step, would be even more efficient. Explain who is correct and why.](https://media.cheggcdn.com/media%2F579%2F5797bffa-aefc-4480-9aee-7edfb176763f%2Fimage)
Crabbe and Goyle are arguing about binary search. Goyle writes the following pseudocode on the board, which he claims implements a binary search for a target value v within input array A containing n elements. bSearch(A, v) { return binarySearch(A, 0, n, v) } binarySearch (A, 1, r, v) { if 1 > = r then return -1 p = floor ((1 + r)/2) if A[p] == v then return m if A[m]
(a)
the goyle algorithm has bugs :
the value used to store (l+r)/2 and a[m]==v should be m not p as in futher steps we used m to check the values and returned m
———————————–
the not found condition must be l>r then return -1 beacuse if we consider the example
no of elements 5
elements are 1 2 5 7 8
and key is 2
then the fisrt mid value is 2 then a[2] is 5 as 5 is greater than 2 binarySearch(A,0,2-1,v) is called
then mid is 0 as a[0] is 1 as 1 is less than 2 binarySearch(A,0+1,1,v) is called
now l is equal to r so it returns -1 but the element is present so the condition should be l>r
then mid is i=1 a[1]=2 as 2 is equal to 2 it returns the index of 2.
——————————————
(b)
Goyle is correct .
If we consider the worst -case binary search makes (log n to base 2)*2+1 searches where as trinary search makes
(log n to base 3)*4+1 searches which can be written as (2/(log 3 to base 2) )*(log n to base 2) whose value is greater than 1. so tinary search makes more comparisions hence binary search is the best of two
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