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.
How to create a checkerboard matrix without inbuilt function.
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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
採用された回答
Joseph Cheng
2017 年 9 月 12 日
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
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その他の回答 (4 件)
ksam K
2017 年 11 月 25 日
%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
Jan Siegmund
2020 年 5 月 21 日
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]);
0 件のコメント
Mendi
2020 年 9 月 5 日
It is more natural to use modulus on meshgrid:
[iX,iY] = meshgrid(1:8,1:8);
Mask=mod(iX+iY,2);
2 件のコメント
Bruno Luong
2020 年 9 月 5 日
編集済み: Bruno Luong
2020 年 9 月 5 日
Two more methods
toeplitz(mod(0:7,2))
or for even size
kron(ones(4),[0 1;1 0])
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