Asserting that a number is in binary

Question

Im working on my homework and encountered a problem. I had to string numbers in binary back to back. Lets say we have joined(1, 3) - 1 is the starting number and 3 says that Im gonna work with 2 more numbers (if i had joined(5, 4), id work with 5, 6, 7 and 8) 1 1 2 10 3 11 result 11011

def num_to_2(number):
    num2 = 0
    exp = 0
    while number >= 1:
        num2 += (number % 2) * (10 ** exp)
        number = number // 2
        exp += 1
    return num2


def num_lenght(number):
    if number == 0:
        return 0
    lenght = 0
    while number >= 1:
        number /= 10
        lenght += 1
    return lenght


def joined(start, count):
    result = 0
    end = start + count - 1
    for i in range(end, start - 1, -1):
        number = num_to_2(i)
        res_lenght = num_lenght(result)
        result += number * (10 ** res_lenght)
    return result


def main():
    assert joined(1, 3) == 0b11011
    assert joined(10, 4) == 0b1010101111001101
    assert joined(8, 5) == 0b10001001101010111100
    assert joined(99, 2) == 0b11000111100100
    assert joined(999, 3) == 0b111110011111111010001111101001
    assert joined(1111, 1) == 0b10001010111

The function works properly but it gives me results without the 0b prefix, which i need in order to pass the asserts. How do i add it there? and also: i cant use strings.

Thanks in advance!

Edit: i added the rest of my code so it makes more sense

Answer 1

I already solved my problem, what i had to do was "manually" convert the pseudo-binary number into base 10 number. this is my code now:

def convert_to_2(number):
    num2 = 0
    exp = 0
    while number >= 1:
        num2 += (number % 2) * (10 ** exp)
        number = number // 2
        exp += 1
    return num2


def num_lenght(number):
    if number == 0:
        return 0
    lenght = 0
    while number >= 1:
        number /= 10
        lenght += 1
    return lenght


def joined(start, count):
    result = 0
    end = start + count - 1
    for i in range(end, start - 1, -1):
        number = convert_to_2(i)
        res_lenght = num_lenght(result)
        result += number * (10 ** res_lenght)
    result = convert_to_10(result)
    return result


def convert_to_10(number):
    num10 = 0
    exp = 0
    while number >= 1:
        num10 += (number % 10) * (2 ** exp)
        number = number // 10
        exp += 1
    return num10


def main():
    assert joined(1, 3) == 0b11011
    assert joined(10, 4) == 0b1010101111001101
    assert joined(8, 5) == 0b10001001101010111100
    assert joined(99, 2) == 0b11000111100100
    assert joined(999, 3) == 0b111110011111111010001111101001
    assert joined(1111, 1) == 0b10001010111


if __name__ == "__main__":
    main()

thank you so much for your help(:

Answer 2

Well, it's unclear what your teachers mean with "don't use strings". I found this a very strange homework, so I did it, without strings.

I'm not sure what you are supposed to learn from it, but here it is:

def to_binary(number):
    b = 0
    while number >= 2**(b+1):
        b += 1
        
    while True:
        if number >= 2**b:
            yield 1
            number -= 2**b
        else:
            yield 0
        b -= 1
        if b < 0:
            return


def joined(start, count):
    result = 0
    for i in range(start, start + count):
        for digit in to_binary(i):
            result = (result << 1) + digit
    return result
    

def main():
    assert joined(1, 3) == 0b11011
    assert joined(10, 4) == 0b1010101111001101
    assert joined(8, 5) == 0b10001001101010111100
    assert joined(99, 2) == 0b11000111100100
    assert joined(999, 3) == 0b111110011111111010001111101001
    assert joined(1111, 1) == 0b10001010111

It implements a to_binary() function that will yield a sequence of integers. Then the join function will binary shift the result to the left, and add that one or zero.

No strings attached...