# Random matrix with no repeats in rows and columns.

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Francisco Boudagh on 25 Nov 2021
Edited: Image Analyst on 26 Nov 2021
So I have generated a random matrix with dimensions 10x15 with values between 2 and 18 with step length 2.
Simple code:
candidates = 2:2:18;
idx = randi(numel(candidates), 10, 15);
M = candidates(idx)
So my questions is how do I do to never get a value above, under or next to the same value. As the ones I have marked with red in the picture.
the cyclist on 25 Nov 2021
@Francisco Boudagh, I was just giving a single counter-example to dpb's concern that a valid solution might not even exist. I was not trying to solve your question.

Jan on 25 Nov 2021
Edited: Jan on 26 Nov 2021
Create the matrix elementwise. Choose a random element from the available elements. Exclude elements on top and left of the element to be set:
nx = 15;
ny = 10;
pool = 1:9; % Choose values 1:9, multiply by 2 finally
M = zeros(ny, nx);
for ix = 1:nx
for iy = 1:ny
% Exclude elements on the left and on the top from pool:
exclude = [M(gt0(iy - 1), ix), M(iy, gt0(ix - 1))];
choose = pool;
choose(exclude) = [];
% Nicer but slower in one line:
% choose = setdiff(pool, [M(gt0(iy - 1), ix), M(iy, gt0(ix - 1))]);
% Choose a random element from the remaining list:
M(iy, ix) = choose(randi(numel(choose)));
end
end
M = M * 2
ans = 10×15
6 16 14 8 14 16 14 18 14 6 8 16 18 10 18 18 10 18 4 6 4 18 10 6 4 12 6 8 14 6 2 4 14 18 16 6 16 8 4 10 18 12 14 8 16 4 6 12 8 2 10 6 18 6 14 8 6 18 16 12 8 12 4 2 18 16 10 14 2 10 2 10 6 2 14 10 14 18 14 2 4 16 4 10 6 12 6 12 14 6 8 10 12 2 8 6 12 18 14 18 6 18 8 4 18 14 18 8 6 2 8 14 2 8 6 14 16 18 12 6 16 14 18 12 18 4 2 12 10 18 6 12 16 10 12 6 8 16 6 8 18 12 8 4 8 16 6 18 6 8
function k = gt0(k)
% Reply [] as index, if it is <= 0
if k <= 0, k = []; end
end
I've experimented with cumsum(cumsum(randi([1, 9], 10, 15), 1), 2): There are no repetitions. Then limit the values by rem(). This was not successful. In one dimension it is working, but not in 2.
Image Analyst on 26 Nov 2021
Latin Squares are a very important concept in the "Design of Experiments" (DOX).
Actually there is Latin Square code in the File Exchange:

the cyclist on 26 Nov 2021
Edited: the cyclist on 26 Nov 2021
I believe this does what you want.
% Set the random number seed, for reproducibility
rng default
% The pool of random numbers to choose from
pool = 2:2:18;
% The size of the array
M = 10;
N = 15;
% Preallocate the output array
out = zeros(M,N);
% Fill the upper-left element from the pool
out(1,1) = randsample(pool,1);
% Fill the first column, never repeating the just-above value
for i1 = 2:M
out(i1,1) = randsample(setxor(pool,out(i1-1,1)),1);
end
% Fill the first row, never repeating the just-left value
for j1 = 2:N
out(1,j1) = randsample(setxor(pool,out(1,j1-1)),1);
end
% Fill the rest of the array, never repeating the just-above and just-left values
for ii = 2:M
for jj = 2:N
out(ii,jj) = randsample(setxor(pool,[out(ii-1,jj),out(ii,jj-1)]),1);
end
end
out
out = 10×15
16 12 16 12 14 12 14 6 2 14 18 8 16 4 10 4 14 4 14 2 18 10 16 10 18 10 4 14 16 18 14 4 12 6 14 16 14 6 2 10 18 10 2 6 8 6 8 4 14 8 4 6 12 4 2 4 16 4 8 10 18 4 6 18 10 16 8 6 8 6 10 18 10 6 8 4 16 4 2 18 10 4 16 2 14 12 8 16 12 6 18 12 18 10 8 12 16 12 16 6 18 12 4 10 16 14 4 6 8 10 18 8 4 12 14 4 8 10 2 14 8 18 10 4 18 10 12 14 6 2 18 2 8 4 2 16 8 16 18 12 2 10 18 12 16 4 18 10 18 10
I would caution you, though, that I have not thought carefully about the random properties of this array. (For example, does each element have equal probability for each value, etc., that might be important to you.) I think it's probably OK.
Also, if you do not have the Statistics and Machine Learning Toolbox, you will not have the randsample function, and therefore will need to replicate that function's behavior.
the cyclist on 26 Nov 2021
:-)

Image Analyst on 25 Nov 2021
Just construct a Latin square for n=9, then multiply it by 2