Binary Search – Is this correct? and how to get position of search term

I am a beginner to programming. I have three questions about my binary search code I wrote (in Python):
(1) When I compared my code to the ones I found on the net, I see that everyone uses low and high values parameters. Wondering if mine can have errors that I am not seeing or is it just that the high/low parameters make the code more readable? I thought mine follows the recursive way of thinking (at least for me).

(2) edit: This part has since been answered.—>> I can’t get it to return True or False for the answer. Is it because the True/False is getting returned to the recursive function and not the print? How can I get it to return True, or False answer?

(3) I can’t figure out how to get the position of the search term in the list (of course, not using index function). I thought I could somehow use the “mid” variable which was in the code, but couldn’t. Can you give me some ideas how to get its position in the list?.

def binsearch(i, arr):

    mid = ((len(arr)-1)//2)

    # If we can't divide the list any further and it is not i , return false    
    if mid == 0:
        if i != arr[mid]:
            # print ("False")
            return False

    # else recursively search if i is at the mid point

    if i == arr[mid]:
        # print ("True")
        return True
    elif i < arr[mid]:
        arr = arr[0:mid+1]
        binsearch(i,arr)
    else:
        arr = arr[mid+1:]      
        binsearch(i,arr)   
i = 19
#1=1
#i=16
#i=17         
arr =[1,2,3,4,5,10,12,13,14,15,17,19]
print(binsearch(i, arr))

Thanks

Arun

Answers:

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Method 1

The problem here is that you’re not returning the value on the recursive cases. Refactor your code to this and it should work.

    if i == arr[mid]:
        return True
    elif i < arr[mid]:
        arr = arr[0:mid+1]
    else:
        arr = arr[mid+1:]

    return binsearch(i, arr)

Method 2

Answer to part 2: you need to return binsearch(i, arr) in the elif and else statements.

Edit:

Answering part 3 as well, you can add the current index as a third variable of binsearch and update it each recursion:

def binsearch(i, arr, index=0):

    mid = len(arr)//2
    index += mid

    # If the item is found return True and the index
    if i == arr[mid]:
        return True, index

    # i not in list
    elif not mid:
        return False

    # else recursively search if i is at the mid point
    elif i < arr[mid]:
        arr = arr[:mid]
        index -= mid

    else:
        arr = arr[mid:]

    return binsearch(i,arr, index)   
    
arr =[1,2,3,4,5,10,12,13,14,15,17,19]
for i in range(20):
    print(i, binsearch(i, arr))

Output:

0 False
1 (True, 0)
2 (True, 1)
3 (True, 2)
4 (True, 3)
5 (True, 4)
6 False
7 False
8 False
9 False
10 (True, 5)
11 False
12 (True, 6)
13 (True, 7)
14 (True, 8)
15 (True, 9)
16 False
17 (True, 10)
18 False
19 (True, 11)

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