fill matrix with all options of successive, increasing numbers 1-5

I want to construct a matrix, m x n, filled with the numbers 1 till 5. The numbers have to be successive, and have to be increasing. One column of the matrix could for example look like.

A =1 1 2 2 2 2 3 4 5 5 5 5

or

A = 1 2 3 3 3 4 5 5 5 5 5 5

I want to construct a matrix with all possible options. The first column of matrix A therefore should look like this:

A(:,1) = 1 1 1 1 1 1 1 1 2 3 4 5

and the last one like this

A(:,end)= 1 2 3 4 5 5 5 5 5 5 5

In my case, the matrix will have a length of 96, instead of the above example where the length is 12. Could you help me?

[Adam Danz]

> In my case, the matrix will have a length of 96, instead of the above example where the length is 12.

This part isn't clear. By 'length' do you mean width? What would the first and last rows look like? Is the max value still 5 or is it 96?

[Rik]

Can you provide the code you already used? It is often easier to fix existing code than to write new code.

[Sjoukje de Lange]

I do not have a code yet, I'm already stuck in how to approach this. Sorry!

[Sjoukje de Lange]

@Adam Danz: The first row (or column, does not matter) should look like this: [1 1 1 1 (... there will be 92 ones in total) 2 3 4 5] and the last row (or column) like this: [1 2 3 4 5 5 5 5 (... 92 fives in total)]. So it has to become a huge matrix m x n, m being 96 and n being the total amount of possibilities to form a row (or column) of 96 values satisfying the conditions (only increasing or equal, successive). Does this make it a bit clearer?

[Adam Danz]

I think you mean the mxn matrix will by 96 columns (n), and probably tens of thousands of rows, if not more. What are you plans with this matrix? How will you use it?

[Sjoukje de Lange]

yes correct. I need to divide an area of 96 kilometer in 5 parts, in the most optimal way (optimal being defined by the values of certain parameters). I need this matrix to define all possible options to divide it. Unfortunatly something like cluster analysis to divide it won't be possible with my data since there are multiple (in)dependend variables, so we came up with this solution. Now the question is: how to tell the computer... :)

[Adam Danz]

Why do you need repeated values, then?

There are 61,124,064 ways to select 5 items out of 96 and that's without repetition. With the repetitions you're looking at billions.

Addendum: the number above includes indicies that increase and decrease. If you're only interested in increading indicies, that number will be reduced.

[Sjoukje de Lange]

Hmmm that might be a but extreme yeah... I need repeating values because I want to label each parameter at each kilometer with number 1-5.

Answer 1

[Bruno Luong]

p = 5;
n = 12;

j = nchoosek(2:n,p-1);
m = size(j,1); % == nchoosek(n-1,p-1) == 330 and not 96
i = repmat((1:m)',1,p-1);
A = cumsum(accumarray([i(:) j(:)],1,[m n]),2)+1

[Bruno Luong]

p=5;
n = 41;

tic
i=nchoosek(2:n,p-1);
m=size(i,1);
j=repmat((1:m)',1,p-1);
A=cumsum(accumarray([j(:) i(:)],1,[m n]),2)+1;
toc

Elapsed time is 0.033608 seconds.

[Sjoukje de Lange]

will try tomorrow (it's damn late here now... :) ), will let you know! thanks for helping!

[Adam Danz]

Nice one (+1)

When n==96, A has 3,183,545 rows and takes less than 3 sec using the same machine as above.

[Sjoukje de Lange]

neat! Amazing!