How to create a checkerboard matrix without inbuilt function.

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

8 Kommentare

im trying to do it using zero(16) and plussing 1's in certain coordinates, or i could use meshgrid too mesh them together, not sure how i want to do it.
@Riley Smith: so try both of them. What is stopping you from trying things out? MATLAB does not explode if your first attempt does not work.
This is one of those problems where there is a vast number of different ways to try it. Many that give the right answer and many that give the wrong answer, but command line testing is free. My first attempt yielded the wrong answer as I forgot that the value at the end of a column is equal to that at the start of the next column. But a little tweak soon fixes those problems and experimenting is fun and good for learning.
If you get stuck, or just want to see how others have done this:
Also, the criterion restricting the use of builtins makes it harde to answer this. It is very difficult to define what counts as a builtin and what doesn't, as that quickly devolves into not being able to use Matlab at all (+ is a builtin as well).
Cedric
Cedric am 12 Sep. 2017
Bearbeitet: Cedric am 12 Sep. 2017
You can also use
abs(sin(...)) or abs(cos(...))
with the proper arguments.
Using the sin function for this would be a sin ...
Cedric
Cedric am 28 Nov. 2017
Bearbeitet: Cedric am 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

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 Akzeptierte Antwort

Joseph Cheng
Joseph Cheng am 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

Weitere Antworten (4)

Sangam K
Sangam K am 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
For an even sized checkerboard:
rows = 6;
cols = 4;
normal = repmat(eye(2,'logical'),[rows/2 cols/2]);
% or
inverted = repmat(~eye(2,'logical'),[rows/2 cols/2]);
Mendi
Mendi am 5 Sep. 2020
It is more natural to use modulus on meshgrid:
[iX,iY] = meshgrid(1:8,1:8);
Mask=mod(iX+iY,2);

2 Kommentare

One liner variant
mod((1:8)+(1:8)',2)
Yeap. The one liner is better.

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Bruno Luong
Bruno Luong am 5 Sep. 2020
Bearbeitet: Bruno Luong am 5 Sep. 2020
Two more methods
toeplitz(mod(0:7,2))
or for even size
kron(ones(4),[0 1;1 0])

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