Randomly splitting of a number in a sum format.

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rohini more
rohini more 2021 年 4 月 27 日
コメント済み: rohini more 2021 年 4 月 27 日
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.
  2 件のコメント
John D'Errico
John D'Errico 2021 年 4 月 27 日
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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採用された回答

Bruno Luong
Bruno Luong 2021 年 4 月 27 日
編集済み: Bruno Luong 2021 年 4 月 27 日
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
  12 件のコメント
rohini more
rohini more 2021 年 4 月 27 日
Now its working . Thank you very much.

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その他の回答 (1 件)

Bruno Luong
Bruno Luong 2021 年 4 月 27 日
編集済み: Bruno Luong 2021 年 4 月 27 日
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]}
  1 件のコメント
rohini more
rohini more 2021 年 4 月 27 日
Thanks for giving your valuable time for solving my question. It really means a lot.
Thank you very much.

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