Binary search infinite recursion - matlab

Question

I have written simple code in Matlab for binary Search. It's working properly if the searched item is in the array but goes into an infinite recursion loop if not.
I'm not sure where the problem lies.
function [] = BinarySearch(A,beg,last,item)
mid=floor((last+beg)/2);
if (beg)<=last
if item==A(mid)
fprintf('Item found at position %d \n',mid);
else
if(item<A(mid))
BinarySearch(A,beg,mid,item)
else
BinarySearch(A,mid,last,item)
end
end
else
fprintf('Item not found\n');
end

Answer 1

Imagine the really simple case that you only have 2 items in your list
A = [1 3]
and you call your BinarySearch on an item which would lie in the middle of the list. Look at the comments below, which follow how your function behaves...
BinarySearch(A, 1, 2, 2)
% mid = 1
% item ~= A(1): go to else
% item > A(1): go to else
% BinarySearch(A, 1, 2, 2)
% ... rinse and repeat
If your item was too small
BinarySearch(A, 1, 2, 0)
% mid = 1
% item ~= A(1): go to else
% item < A(1)
% BinarySearch(A, 1, 1, 0)
% mid = 1
% beg <= last (this is still true)
% item ~= A(1): go to else
% item < A(1)
% BinarySearch(A, 1, 1, 0)
% ... rinse and repeat
Similarly for an item which is larger than any in the list,
BinarySearch(A, 1, 2, 5)
% leads to BinarySearch(A, 1, 2, 5)
% ... repeat!!
You keep re-checking the same region, because your left (beg) and right (last) indices are allowed to both stay the same.
Let's re-implement the function, returning an actual value instead of just printing out the position to the console too. The comments directly relate to the numbered steps in the Wikipedia article for binary search, which looks similar in structure to what you've attempted:
function idx = BinarySearch(A, L, R, item)
%% BINARYSEARCH search for an item in the array A. Assumes that A is sorted ascending
% 1. Should be initially called using idx = BinarySearch(A, 1, n, item)
% where n is the number of elements in A, numel(A)
% 2. if L > R, the search terminates as unsuccessful
if L > R
idx = 0;
else
% 3. set m (position of middle element) to the floor of (L+R)/2
m = floor((L+R)/2);
% 4. if Am < item, set L to m+1 and go to 2.
if A(m) < item
L = m + 1; % YOU MISSED THIS STEP, CAUSING OVERLAPPING SEARCH REGIONS
idx = BinarySearch(A, L, R, item);
% 5. if Am > item, set R to m-1 and go to 2.
elseif A(m) > item
R = m - 1; % THE OTHER HALF OF THE STEP YOU MISSED
idx = BinarySearch(A, L, R, item);
% 6. Now Am = item, search is done, return index
else
idx = m;
end
end
end
Tests with A as before:
BinarySearch(A, 1, 2, 2); % returns 0: not found
BinarySearch(A, 1, 2, 0); % returns 0: not found
BinarySearch(A, 1, 2, 5); % returns 0: not found
BinarySearch(A, 1, 2, 3); % returns 2: 3 is at index 2
BinarySearch(A, 1, 2, 1); % returns 1: 1 is at index 1
Note that it may not be most efficient to implement this recursively. I'll leave it as an exercise, but this could be easily implemented using a while loop instead. The same logical structure would be used.

Answer 2

Try to set breakpoints to examine values where you think a problem might be.
You can use the while loop and the break command to avoid going into infinite loop.
For example try something like this:
function [index] = binarySearch(A, n, num)
left = 1;
right = n;
flag = 0;
while left <= right
mid = ceil((left + right) / 2);
if A(mid) == num
index = mid;
flag = 1;
break;
else if A(mid) > num
right = mid - 1;
else
left = mid + 1;
end
end
end
if flag == 0;
index = -1;
end
end