How can I skip a piece of code in recursion?

Question

def bin_to_decimal(inp, degree=0, result=0):
    arr = []
    for i in inp:
        arr.append(i)
    if not arr:
        return result
    else:
        x = arr.pop(-1)
        result += int(x)**degree
        return bin_to_decimal(inp, degree+1, result)


bin_to_decimal('1001')

Is it possible to skip this part of the function, when I call recursion?

arr = []
for i in inp:
    arr.append(i)

Answer 1

You can use:

def bin_to_decimal(inp, degree=0, result=0, skip=False):
    if skip:
        arr = []
        for i in inp:
            arr.append(i)
    if not arr:
        return result
    else:
        x = arr.pop(-1)
        result += int(x)**degree
        # Flag True to skip
        return bin_to_decimal(inp, degree+1, result, True)

But you will now have an unreferenced variable arr

• along with a few more errors

I suggest you have a look at some other methods that you can use:

Built-in:

>>> int('1001',2)
9

Iteration:

def bin_to_decimal(inp):
    x = 0
    for i in inp:
        x = x*2 + int(i)
    return x

Answer 2

if whatever: ... is possible in recursion as well and can skip things inside your function - but there is no need:

You do not need to add an implicit carry (result=0) to your function signature if you do recursion the right way:

def bin_to_decimal(inp, degree=0):
    if len(inp) == 1:  # base case
        return  2**degree*int(inp[0])

    # "12345" => "1234" into remainder and "5" into x by decomposition
    *remainder, x = inp # decompose

    # calc the value of x and recurse with remainder and 1 degree higher
    return 2**degree*int(x)+bin_to_decimal(remainder, degree+1)
    
for n in range(17):
    b = bin(n)[2:]  # this function converts int to bin with a 0b prefix
    print(b, bin_to_decimal(b), sep=" ==> ")
 

Output:

0 ==> 0
1 ==> 1
10 ==> 2
11 ==> 3
100 ==> 4
101 ==> 5
110 ==> 6
111 ==> 7
1000 ==> 8
1001 ==> 9
1010 ==> 10
1011 ==> 11
1100 ==> 12
1101 ==> 13
1110 ==> 14
1111 ==> 15
10000 ==> 16

Recursion is not needed here, you can unroll it:

def bin_to_decimal(inp):
    num = 0
    degree = 0
    for digit in inp[::-1]: # reversed
        num += int(digit) * 2**degree
        degree += 1
    return num    

or simpy use num = int("0100101", 2) to convert it.

Answer 3

If you want a "closest to your code as possible" solution, you can use:

def bin_to_decimal(inp, degree=0, result=0):
    arr = list(inp) # this does what your loop does ... just shorter
    # list("abc") => ["a","b","c"]
    # list(["a","b","c"]) => ["a","b","c"] where it is not needed

    if not arr:
        return result
    else:
        x = arr.pop(-1)
        result += int(x)**degree

        # use arr here, not inp
        return bin_to_decimal(arr, degree+1, result)

but using a carry like you do is bad style regarding recursion.