How to get all possible arrays by randomizing in each cells with four items (1x6 array)

Hi! I would like to know all possible arrays for 1x6 vector. Each cells may be 0 (or) x (or) y (or) z. Some can be repeated and some may exclude in an array.

Some of the results should be like below:

[x,y,x,0,z]; [x,x,x,x,z]; [y,0,0,0,0,z]; [z,0,y,0,0,x];......

Thank you so much.

Answer 1

[madhan ravi]

Use random indexing by using randi if you want the elements to repeat or randperm if you don’t want the elements to be repeated

c=cell(1,4); % preallocate
syms x y z  % define x y and z as you wish
a=[0,x,y,z];
for i =1:numel(c)
    c{i}=a(randi(numel(a),1,6));
end
celldisp(c)

Note : 1 X 6 array means you want to fill each cell ( totally 4 cells) with 6 elements.

[San May]

Yes Thank you so much... It works well.

[madhan ravi]

Anytime :)

[Image Analyst]

I'm not sure how using randi, or even randperm, will get you all possible 4^6 combinations.

[madhan ravi]

Yes your right , perhaps the OP wanted to just fill the cell arrays with those 4 elements since his quote "Yes Thank you so much... It works well." means he got what he was looking for .

Answer 2

[Image Analyst]

At first I though of using perms() on your list of numbers but since that doesn't allow more output numbers than input numbers (6 output with only 4 input), I thought that a 6-nested for loop would work. This code figures out all possible permutations of the 4 variables taken 6 per draw. You will of course have 4^6 possible unique combinations.

tic
x = 99;
y = 188;
z = 277;  % Whatever numbers you want...
listOfNumbers = [0, x, y, z]
c = zeros(4^6, 6); % Preallocate
% Generate all possible combinations.
row = 1;
for k1 = 1 : length(listOfNumbers)
for k2 = 1 : length(listOfNumbers)
for k3 = 1 : length(listOfNumbers)
for k4 = 1 : length(listOfNumbers)
for k5 = 1 : length(listOfNumbers)
for k6 = 1 : length(listOfNumbers)
	c(row, :) = [listOfNumbers(k1), ...
		listOfNumbers(k2), ...
		listOfNumbers(k3), ...
		listOfNumbers(k4), ...
		listOfNumbers(k5), ...
		listOfNumbers(k6)];
	row = row + 1;
end
end
end
end
end
end
toc

Time to run is a speedy 0.001 seconds (a millsecond), so don't be afraid of the for loops.

[San May]

wow incredible! Thank you so much.. It is so fast and got all possible arrays!

[madhan ravi]

+1 @sir Image Analyst , always precise!

Answer 3

[Stephen23]

A simple solution with ndgrid:

>> x = 99;
>> y = 188;
>> z = 277;
>> C = cell(1,6);
>> [C{:}] = ndgrid([0,x,y,z]);
>> C = cellfun(@(a)a(:),C,'uni',0);
>> M = [C{:}]
M =
     0     0     0     0     0     0
    99     0     0     0     0     0
   188     0     0     0     0     0
   277     0     0     0     0     0
     0    99     0     0     0     0
    99    99     0     0     0     0
   188    99     0     0     0     0
   277    99     0     0     0     0
     0   188     0     0     0     0
    99   188     0     0     0     0
    ... lots of lines here
    99    99   277   277   277   277
   188    99   277   277   277   277
   277    99   277   277   277   277
     0   188   277   277   277   277
    99   188   277   277   277   277
   188   188   277   277   277   277
   277   188   277   277   277   277
     0   277   277   277   277   277
    99   277   277   277   277   277
   188   277   277   277   277   277
   277   277   277   277   277   277
>>  size(M)
ans =
   4096      6