Randomly splitting of a number in a sum format.

Suppose n=5

we can split this number like

5=3+2,4+1..... so on.

But I just want to select a only one sum but randomly and I would like to specify by using varible

like

5=1+3+1

then n_{1}=1, n_{2}=3, n_{3}=1.

How to implement matlab code for any value of n as per above method.

Please help me.

Thanks in advance.

[John D'Errico]

PLEASE STOP ASKING THE SAME QUESTION!

You have asked 6 questions so far on ANSWERS. 5 of them have been essentially duplicates. The rest of them had answers already, but I just closed the latest one, and will now close further duplicate questions by you. You have already gotten multiple answers to your question.

[John D'Errico]

No. There have been multiple questions about random partitions of a set. Learn how to find the set of all partitions, then choose one randomly. You cannot create variable names on the fly, and to the extent that you can do so, you SHOULD not. Instead, learn to use vectors.

Answer 1

[Bruno Luong]

n = 10;
for j=1:10
    r = n;
    i = 1;
    clear s
    while r > 0
        s(i) = ceil(r*rand);
        r = r-s(i);
        i = i+1;
    end
    disp(s)
end

     4     2     4

     8     1     1

     7     3

     4     2     2     1     1

     8     1     1

     7     1     1     1

     7     1     1     1

    10

     8     2

     1     9

[rohini more]

Thanks for your answer. It really means a lot.

But I have one doubt , how we can choose at a time only one random sum of given number.

How to assign a varible name.

Please help me.

Thanks

[Bruno Luong]

"But I have one doubt , how we can choose at a time only one random sum of given number."

Remove the for loop on j

"How to assign a varible name."

It's call s in my code.

[John D'Errico]

Note that things like

n_{1}=1

are invalid syntax in MATLAB.

[rohini more]

Yes sir.

at least we can mention n1=1 like that.

But really thanks a lot for your response.

[Bruno Luong]

EDIT: this post lost continuity since OP deletes his comment that requires "set" partition

Simply convert interger partiion to "set" partition

n = 10;

r = n;
i = 1;
clear h
while r > 0
    h(i) = ceil(r*rand);
    r = r-h(i);
    i = i+1;
end

H = mat2cell(1:n,1,h)

[rohini more]

Thank you very much.

[rohini more]

I would like to say thanks, for helping me. It really means a lot.

how to check that a or b is belong any colum of cell where a and b are any positive integer but less that n.

[Bruno Luong]

for x in (1:n) to check which x belong to

x = randi(n,1,1) % generate random x for testing

% h=cellfun('length',H)
loc = discretize(x,cumsum([1 h]));
% x belong to H(loc)

[rohini more]

Okay.

Thank you very much.

[rohini more]

I am able to generate the random number as per your guideline but unable to specify whether this belong to particular column of cell or not. Please give me guideline.

Thanks in advance.

[Bruno Luong]

Ops I have a typo in my code, can you try again.

[rohini more]

Now its working . Thank you very much.

Answer 2

[Bruno Luong]

This code will generate "uniform" partition distribution, in the sense that all possible partition has equal probability:

n = 10;

% This part is done once if n is fix
L = 1:n;
p = arrayfun(@(k) nchoosek(n-1,n-k), L);
e = [0, cumsum(p)];
e = e/e(end);

% This part must be repeated when an new random partition is requested
[~,k] = histc(rand,e);
h = diff([0 sort(randperm(n-1,n-k)) n]);
H = mat2cell(L, 1, h)

H = 1×2 cell array
    {[1 2 3 4 5]}    {[6 7 8 9 10]}

[rohini more]

Thanks for giving your valuable time for solving my question. It really means a lot.

Thank you very much.