Exiting out of loop when not found - Binary Search in Python

Question

I am making a binary search program to search a number in a list. Below is the current program. I require the program to find a the index of a number if it is found in the list, otherwise, it should return "not found".

But the code below does not exit out of the loop if the number is not found. How can I make it do so ?

def binary_search(n):
    l = [1,6,7,10,19,24,42,81]
    low = 0
    high = len(s)-1
    index = (high + low)/2
    while s[index] != n:
        if s[index] < n:
            low = index + 1
        else:
            high = index
        index = (high + low)/2

binary_search(43)

I have tried adding

if index == None:
    print "Not Found"
else:
    print index

before the function call, but as I said, the loop does not exit if it cannot find the number.

Have you tried using an if-statement inside it and then break when it matches those conditions? — user2961646
– user2961646, Commented Jun 24, 2015 at 2:57
Yes, I tried adding break, but it gives me the result as 5 even when its not found. I gives me the location just before the least greatest number. — adb16x
– adb16x, Commented Jun 24, 2015 at 2:59

salmanwahed · Accepted Answer · 2015-06-24 03:15:52Z

1

You need to break the while loop in case there is no element. You can add a condition like this:

while s[index] != n:
    if low>=high:
        return "Not Found"
    if  s[index] < n:
        low = index + 1 
    else:
        high = index
    index = (high + low)/2

Or change the condition for the while loop.

while low<high:
    if l[index] == n:
        print index
    if  l[index] < n:
        low = index + 1 
    else:
        high = index
    index = (high + low)/2

edited Jun 24, 2015 at 3:15

answered Jun 24, 2015 at 3:00

salmanwahed

9,6878 gold badges37 silver badges56 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

adb16x Over a year ago

Thanks! Your edit is great but how can I print not found now, since there is no condition?

salmanwahed Over a year ago

that is too easy. one way is not printing but returning from the loop using return index instead of print index and from the binary_search function return "Not Found". @WhiteFlameAB also why did you unaccepted the answer after accepting it?

Sufian Latif · Accepted Answer · 2015-06-24 02:51:06Z

1

It is because of the condition of the while loop. The loop goes on until it finds the element in the list. So when the element is not in the list, the loop won't break. You can debug the code by printing the values of high, low and index in each iteration.

To confirm that the element is not in the list, you can check another condition. When low >= high is true, it means that the search range is empty, thus the element is not present in the list.

answered Jun 24, 2015 at 2:51

Sufian Latif

13.4k3 gold badges36 silver badges71 bronze badges

Collectives™ on Stack Overflow

Exiting out of loop when not found - Binary Search in Python

2 Answers 2

2 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related