How to create matrix with fixed sum in rows and fixed increment in elements

I would like to create a 231*3 matrix whose elements are all multiples of 0.05 and the sum of each row be equal to 1.

E.g.

0       0      1
0       0.05   0.95
0.05    0      0.95
0       0.1    0.90
0.1     0      0.90
0.05    0.05   0.90
......

[Zijin Xu]

Sorry about the ambiguity, actually I want this matrix to contain all the possible combinations of multiples of 0.05, without duplicated rows in the matrix.

Answer 1

[Stephen23]

Straightforward exhaustive search which does not generate any superfluous rows:

V = 0:5:100;
Z = nan(0,3);
for n1 = V
    for n2 = V
        for n3 = V
            if (n1+n2+n3)==100
                Z(end+1,:) = [n1,n2,n3];
            end
        end
    end
end
Z = Z/100;

This gives all 231 unique rows:

>> Z
Z =
 0.00000   0.00000   1.00000
 0.00000   0.05000   0.95000
 0.00000   0.10000   0.90000
 0.00000   0.15000   0.85000
 0.00000   0.20000   0.80000
 0.00000   0.25000   0.75000
 0.00000   0.30000   0.70000
 0.00000   0.35000   0.65000
 0.00000   0.40000   0.60000
 0.00000   0.45000   0.55000
 0.00000   0.50000   0.50000
 0.00000   0.55000   0.45000
 0.00000   0.60000   0.40000
 0.00000   0.65000   0.35000
 0.00000   0.70000   0.30000
... lots of rows here
 0.85000   0.05000   0.10000
 0.85000   0.10000   0.05000
 0.85000   0.15000   0.00000
 0.90000   0.00000   0.10000
 0.90000   0.05000   0.05000
 0.90000   0.10000   0.00000
 0.95000   0.00000   0.05000
 0.95000   0.05000   0.00000
 1.00000   0.00000   0.00000
>> size(Z)
ans =
   231     3
>> size(unique(Z,'rows'))
ans =
   231     3

[Zijin Xu]

Thanks a lot, Stephen, beautiful code!

[Rik]

As is often the case, it is a trade-off between memory and speed. My solution is slightly faster, but yours will consume less memory (especially for cases with smaller step sizes).

[Stephen23]

"My solution is slightly faster, but yours will consume less memory (especially for cases with smaller step sizes)."

The speed of my answer could be improved by preallocation and using a counter as the index:

Z = nan(231,3);
K = 0;
...
        K = K+1;
        Z(K,:) = [...];
...

I guess there is also some formula for getting "231", but it escapes me at this moment...

[Bruno Luong]

The formula is 231=nchoosek(20+3-1,3-1), you can see my code to see why.

Answer 2

[Bruno Luong]

s = 0.05;
n = round(1/s);
m = 3;
r = nchoosek(1:n+m-1,m-1);
z = zeros(size(r,1),1);
r = (diff([z, r, n+m+z],1,2)-1)/n

[Rik]

I don't know if it is relevant, but this method is not guaranteed to give unique rows.

[Bruno Luong]

I miss understood whereas OP wants random or all combinations. Now I fix it since the number of combinations is just 231.

[Zijin Xu]

Thank you guys, this now works perfectly well.

[Stephen23]

+1 Very nice.