CODEWITHSUNDEEP
pythonpython-3.xrecursion
Dongho Han
Dongho Han

Reputation: 97

How could I convert Binary into Decimal using recursive way?

I basically finished everything.. I just can't find a way to multiply each integer by 2^0, 2^1 ... and so on Here is my code 

def BinaryToDecimal(binaryString):
    if len(binaryString) == 0:
        return 0
    else:
        return int(binaryString[-1:])*(2**(4-len(binaryString))) + BinaryToDecimal(binaryString[:len(binaryString)-1])

If I have input '1000', I return last digit of the string and do recursion by eliminating the last digit so, '1000' --> '100' --> '10' and so on

The problem here is that I just cannot find a way to multiply last digit by its corresponding power of 2. Any thought of how to get 0,1,2,3 when length of binary string is 4,3,2,1 ?

Upvotes: 1

Views: 7129

Answers (2)

PM 2Ring
PM 2Ring

Reputation: 55489

You're making this more complicated than it needs to be. No power calculation is required.

def binary_to_decimal(bstring):
    if not bstring:
        return 0
    return binary_to_decimal(bstring[:-1]) * 2 + int(bstring[-1])

# Test

for i in range(16):
    b = format(i, 'b')
    n = binary_to_decimal(b)
    print('{:2} {:4} -> {}'.format(i, b, n))

output

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

This function also handles leading zeros correctly:

for i in range(16):
    b = format(i, '05b')
    n = binary_to_decimal(b)
    print('{:2} {} -> {}'.format(i, b, n))

output

 0 00000 -> 0
 1 00001 -> 1
 2 00010 -> 2
 3 00011 -> 3
 4 00100 -> 4
 5 00101 -> 5
 6 00110 -> 6
 7 00111 -> 7
 8 01000 -> 8
 9 01001 -> 9
10 01010 -> 10
11 01011 -> 11
12 01100 -> 12
13 01101 -> 13
14 01110 -> 14
15 01111 -> 15

Actually, we don't even need that int call, if we assume we only get passed valid strings:

def binary_to_decimal(bstring):
    if not bstring:
        return 0
    return binary_to_decimal(bstring[:-1]) * 2 + (bstring[-1] == '1')

Upvotes: 3

freakish
freakish

Reputation: 56527

If you are trying to recursively parse a number (as an exercise) then have a look at this:

def parse_binary(binary_string, power):
    if len(binary_string) == 0:
        return 0
    last_digit = int(binary_string[-1]) * (2**power)
    return last_digit + parse_binary(binary_string[:-1], power+1)

def BinaryToDecimal(binary_string):
    return parse_binary(binary_string, 0)

But if you want to do this efficiently then simply call int(binary_string, 2).

Upvotes: 0

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