How can I create a matrix of alternating 1s and 0s for any size matrix?

I am trying to script that will display a square matrix of alternating 1s and 0s.

ex: 1 0 1 0
    0 1 0 1
    1 0 1 0
    0 1 0 1

I have found that my solution works only for odd values of n, but not even values of n.

Here is what I have so far:

function y=exercise1(n)

%create a square matrix of size n x n

if rem(n,2)== 0
m = zeros(n,n);%displays a matrix of zeros
m(1:2:end,2) = 1 %extracts odd elements from column 2 and makes them a 1
m(2:2:end,1) = 1 %extracts even elements from column 1 and makes them 1

else 
    mod(n,n)
m = zeros(n,n);%displays a matrix of zero
m(1:2:end) = 1 %extracts all odd elements and makes them a one

end

[Dipesh Mudatkar]

function a = AlternateZeroOnes(n)
a=ones(n);
    if rem(n,2)
        a(2:2:numel(a))=0;
    else
        a(n+1,:)=1;
        a(:,n+1)=1;
        a(2:2:numel(a))=0;
        a(n+1,:)=[];
        a(:,n+1)=[];
    end
end

[Stephen23]

@Dipesh Mudatkar: expanding with an extra column is not required, as only adding the extra row makes any difference to the linear indexing. So you could just use this:

a(n+1,:)=1;
a(2:2:numel(a))=0;
a(n+1,:)=[];

To avoid the first resizing and moving of the array in memory you could easily create the array with the correct size in the first place:

function a = AlternateZeroOnes(n)
if rem(n,2)
    a = ones(n);
    a(2:2:numel(a)) = 0;
else
    a = ones(n+1,n);
    a(2:2:numel(a)) = 0;
    a(n+1,:)=[];
end
end

Unfortunately with this concept there is no way to avoid the second resize and move in memory.

Answer 1

[Stephen23]

You could use toeplitz for this:

>> toeplitz(mod(1:n,2))
ans =
   1     0     1     0
   0     1     0     1
   1     0     1     0
   0     1     0     1

[Stephen23]

And just for completeness...

If the rows~=columns:

>> toeplitz(mod(1:3,2),mod(1:5,2))
ans =
   1     0     1     0     1
   0     1     0     1     0
   1     0     1     0     1

and if the first value should be zero:

>> toeplitz(mod(0:2,2))
ans =
   0     1     0
   1     0     1
   0     1     0

[Jay]

This has helped a lot.

Thankyou.

Answer 2

[James Tursa]

1 - mod(bsxfun(@plus,(1:n)',1:n),2)