no repeating numbers in a matrix

I'm trying to find a method to shorten the code below. It take a lot of time to make it for more numbers, so I want to find something shorter and easier.

function y= example

m1=zeros(1,3);

m1(1,1)=ceil(6*rand);

m1(1,2)=ceil(6*rand);

m1(1,3)=ceil(6*rand);

m2=zeros(1,3);

m2(1,1)=ceil(6*rand);

m2(1,2)=ceil(6*rand);

m2(1,3)=ceil(6*rand);

while m1(1,2)==m1(1,1); m1(1,2)=ceil(6*rand); end

while m1(1,3)==m1(1,2)||m1(1,3)==m1(1,1); m1(1,3)=ceil(6*rand); end

while m2(1,1)==m1(1,3)||m2(1,1)==m1(1,2)||m2(1,1)==m1(1,1); m2(1,1)=ceil(6*rand); end

while m2(1,2)==m2(1,1)||m2(1,2)==m1(1,3)||m2(1,2)==m1(1,2)||m2(1,2)==m1(1,1); m2(1,2)=ceil(6*rand); end

while m2(1,3)==m2(1,2)||m2(1,3)==m2(1,1)||m2(1,3)==m1(1,3)||m2(1,3)==m1(1,2)||m2(1,3)==m1(1,1); m2(1,3)=ceil(6*rand); end

disp(['m1 is :', num2str(m1)]) disp(['m2 is :', num2str(m2)])

end

Answer 1

[Stephen23]

Your approach is extremely inefficient. Just use randperm:

>> V = randperm(6);
>> m1 = V(1:3)
m1 =
   4   2   5
>> m2 = V(4:6)
m2 =
   6   1   3

Answer 2

[Pawel Jastrzebski]

% 'mNumber'  - number of 'm's'
% 'elements' - number of elements in each 'm'

mNumber  = 2;
elements = 3;

M                     = zeros(mNumber,elements)
M(1:mNumber*elements) = ceil(6*rand(1,mNumber*elements))

% I don't get your 'While' loop. 
% But if you want to carry out operation that involves
% i.e the 'current element' in the row and the 'next one'
% Use vectors.
% i.e check if next element is the same as the current one

mNumberNEW  = 5;
elementsNEW = 8;

Mnew                           = zeros(mNumberNEW,elementsNEW)
Mnew(1:mNumberNEW*elementsNEW) = ceil(6*rand(1,mNumberNEW*elementsNEW))

currentElement = 1:elementsNEW-1
nextElement    = 2:elementsNEW

Mcomparison = Mnew(:,nextElement)== Mnew(:,currentElement)

[Stephen23]

This is rather strange:

M                     = zeros(mNumber,elements)
M(1:mNumber*elements) = ceil(6*rand(1,mNumber*elements))

It would be much simpler to just call randi:

M = randi(6,mNumber,elements)

[Pawel Jastrzebski]

True. I didn't have time to properly loop up the doc for 'rand'/'randi' and left the code at the first thing that came to my head that worked.