# Randomly splitting of a number in a sum format.

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rohini more 2021년 4월 27일
댓글: rohini more 2021년 4월 27일
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.
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John D'Errico 2021년 4월 27일
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.

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### 채택된 답변

Bruno Luong 2021년 4월 27일
편집: Bruno Luong 2021년 4월 27일
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
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rohini more 2021년 4월 27일
Now its working . Thank you very much.

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### 추가 답변(1개)

Bruno Luong 2021년 4월 27일
편집: Bruno Luong 2021년 4월 27일
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]}
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rohini more 2021년 4월 27일
Thanks for giving your valuable time for solving my question. It really means a lot.
Thank you very much.

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