## How to create a checkerboard matrix without inbuilt function.

### Riley Smith (view profile)

on 12 Sep 2017
Latest activity Commented on by Nitika Gupta

on 2 Jun 2019

### Joseph Cheng (view profile)

I just want to write this matrix, but want to do it using for loops
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0

Cedric Wannaz

### Cedric Wannaz (view profile)

on 12 Sep 2017
You can also use
abs(sin(...)) or abs(cos(...))
with the proper arguments.
James Tursa

### James Tursa (view profile)

on 12 Sep 2017
Using the sin function for this would be a sin ...
Cedric Wannaz

### Cedric Wannaz (view profile)

on 28 Nov 2017
I had missed this comment ... ;-)
It's one of the rare cases where using a cos would be a sin too ..
0.5-(-1).^((1:n)+(1:n).')/2
or
a = 1 : n ;
a + a.' == 1

### Joseph Cheng (view profile)

on 12 Sep 2017

There are much easier ways to do this than nested for loops, reshape is readily available. if you must do it with for loops you want to start off with the template
for Rind=1:Nrows
for Cind=1:Ncols
Checkerboard(Rind,Cind) = ___;%what condition of Rind and Cind gives you the checker board pattern.
%start with the first row what makes a 1 appear when Rind==1 and Cind=a number.
%then think of the sigificant difference between Rind==1 and Rind==2.....
end
end

### ksam K (view profile)

on 25 Nov 2017

%Hope this helps you.
function a = checkerboard(n)
a = zeros(n);
for i = 1:n
for j = 1:n
if (i == j)
a (i, j) = 0;
elseif (mod(j, 2) == 0) && (mod(i,2) == 0)
a(i,j) = 0;
elseif (mod(j, 2) == 0) || (mod(i,2) == 0)
a(i,j) = 1;
end
end
end
end

Nitika Gupta

on 2 Jun 2019
thanks