checkerboard matrix (Problem Nr 4 from Cody)

I found this problem on Cody:

Given an integer n, make an n-by-n matrix made up of alternating ones and zeros as shown below. The a(1,1) should be 1.

Example:

 Input  n = 5
 Output a is [1 0 1 0 1
              0 1 0 1 0
              1 0 1 0 1
              0 1 0 1 0 
              1 0 1 0 1]

My solution seems logical but it doesn't work! can someone help me figure out what's messing.

Here is my solution:

function a = checkerboard(n)
 x = ones(n,n)
  k = 0;
  for i =1:n
      for j = 1:n
          if rem(k, 2) == 0
              x(i,j) = 0;
              k = k+1;
          end
      end
  end
  a = x;

end

Answer 1

[DGM]

I can see what you're going for, but you've got a couple issues.

The reason that you're not getting any pattern out is that the k=k+1 increment happens inside the conditional, where it should be happening in every iteration of the inner loop. That said, there's still a problem with using k like this. In this context, k is a linear index, so this solution will only produce a checkerboard if n is odd. otherwise, it'll produce stripes. The way you would fix this is to consider both i and j:

x = ones(n,n);
k = 0;
for i =1:n
	for j = 1:n
		if xor(rem(i, 2) == 1, rem(j, 2) == 1)
			x(i,j) = 0;
		end
	end
end
a = x;

That said, you should start to see how xor and checkerboards go hand in hand. Ideally, you shouldn't even need any loops.

% assuming nxn square output
n=5;
x=mod(1:n,2);
a=bsxfun(@xor,x,~x')

or

% generalized mxn output
m=4;
n=5;
x=mod(1:n,2);
y=mod(1:m,2);
a=bsxfun(@xor,x,~y')