Python - Printing a binary Square Pattern with one argument

Question

I spend time working on a simple python function that allows for printing binary square patterns. Below is the code.

def print_binary_square(n):
    i = 0
    odd = True
    while(i<n):
        j=0
        while(j<n):
            if(odd):
                if (j % 2 == 0):
                    print(0, end='')
                else:
                    print(1,end='')
                j += 1
            else:
                j += 1
                if (j % 2 == 0):
                    print(0, end='')
                else:
                    print(1,end='')
        
        odd = not odd    
        i += 1
        print()
        

As you can see, the function takes one argument. It would print binary numbers in a square pattern. Each row starts with either 0 for odd-numbered or 1 for even-numbered row The above code worked, but I was wondering if there's I can improve for the code since the nested if-else seems redundant for me.

Thanks in advance!

Answer 1

If you look at the output as zero-indexed matrix, you can note that zero occurs at index (i, j) whenever i + j is even. Using this observation, you could write the following. It is shorter but not necessarily more legible.

n = 5

result = "\n".join(
    "".join("0" if (i + j) % 2 == 0 else "1" for i in range(n))
    for j in range(n)
    )

Answer 2

Here is a solution using itertools and a generator. Generators and itertools.cycle make it easy to repeat things without needing to count. The only trick here is to waste one element at each line in case n is even (as the last number repeats).

from itertools import cycle, islice

c = cycle(['0','1'])
n = 6
print('\n'.join(''.join(islice(c, n+1-n%2))[:n]
                for _ in range(n)))

Output:

010101
101010
010101
101010
010101
101010

This makes it very easy to generalize by just changing the input:

from itertools import cycle, islice

def pattern(l, n):
    c = cycle(l)
    r = 1-n%len(l)
    print ('\n'.join(''.join(islice(c, n+r))[:n]
                     for _ in range(n)))

Example:

>>> pattern(list('012'), 5)
0120
1201
2012
0120
1201

>>> pattern(list('01234'), 7)
012340
123401
234012
340123
401234
012340
123401