How to do binary search in python (Recursion) , but not including max and min values of list?

Question

I have been ,wanting to do binary search (by recursion) , and I dont know why my code is not working... Can anyone pls, correct my code and tell the reason why it is not working??

As you can look in my code, i am slicing the string per recursion, and at last when the target is found i will return the position of the target..

def binary(n,target):
    n.sort()
    mid = (0 + (len(n)-1))//2
    if target == n[mid]:
        return mid
    elif target < n[mid]:
        return binary(n[:mid],target)
    elif target > n[mid]:
        return binary(n[mid:],target)

This is the error message i am recieving... RecursionError: maximum recursion depth exceeded while calling a Python object.

Print out the n and mid to see what is going on? Also, you're unnecessarily sorting your list (segment) many times. I'm assuming mid becomes 0. — L3viathan
– L3viathan, Commented Aug 14, 2019 at 7:08

Nir Alfasi · Accepted Answer · 2019-08-14 07:47:19Z

1

There are two problems, both of them are on the last line:

def binary(n, target):
    n.sort()
    mid = (0 + (len(n) - 1)) // 2
    if target == n[mid]:
        return mid
    elif target < n[mid]:
        return binary(n[:mid], target)
    elif target > n[mid]:
        return mid + 1 + binary(n[mid + 1:],target)
                ^^^^^^^^                ^

since you're slicing, and in order to provide the index in the original sorted list, we need to keep track of the indexes we "lose" while recursing
The complete to n[:mid] is n[mid+1:] - not n[mid:] because you already checked the target (in mid) and want to remove it from future iterations - this is, by the way, what causes the infinite loop!

Since we slice the list at mid+1 we need to add mid+1 before calling recursively, in order to preserve the index of the item on the right-side of the list:

[1,2,3,4,5]
     ^ say we slice here and get [4,5]
       we want to save the indexes so we'll add mid (2) + 1 since now in [4,5] the item 4 will get the index zero

Comment: by calling n.sort() upon every iteration we "lose" all the advantage of binary search since, even after the list is sorted, to re-sort it will take at least O(n). So if we need to sort first, we might as well just iterate the array until the item is found/not-found. Or, if we insist on sorting, do it only once and only then call recursively:

n.sort()
binary(n, 2)

where binary does't include the sorting anymore:

def binary(n, target):
    mid = (0 + (len(n) - 1)) // 2
    if target == n[mid]:
        return mid
    elif target < n[mid]:
        return binary(n[:mid], target)
    elif target > n[mid]:
        return mid + 1 + binary(n[mid + 1:], target)

edited Aug 14, 2019 at 7:47

answered Aug 14, 2019 at 7:19

Nir Alfasi

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2 Comments

Aryan Kumar Over a year ago

Yeah it worked...! Can you please tell me... Why we are doing mid + 1 + binary (n[mid + 1 :], target)??? I am unable to understand...

Nir Alfasi Over a year ago

@AryanKumar the explanation is in #2 above

L3viathan · Accepted Answer · 2019-08-14 07:22:17Z

Your midpoint calculation is too aggressive:

Consider n = [1, 2, 3] and target = 3. mid (len(n)-1//2) will then be 0, when surely it should be 1 or 2. The issue stems from you subtracting 1 from the length, and then dividing the result by 2.

Since you anyways do integer division, there is no need for subtracting one:

def binary(n,target):
    n.sort()
    mid = len(n)//2
    ...

There is also the issue of sorting the list (segment) every time binary is entered, which is unnecessary. You could write a wrapper function that sorts the list once and then calls the inner recursive function:

def binary(n, target):
    return _binary(sorted(n), target)

Then just rename your original function to _binary.

Finally, when cutting from the left, you'll need to make sure to keep track of the lost number that you need to add to the index:

elif target > n[mid]:
    return mid + binary(n[mid:],target)

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