Randomly splitting of a number in a sum format.
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rohini more
el 27 de Abr. de 2021
Comentada: rohini more
el 27 de Abr. de 2021
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 comentarios
John D'Errico
el 27 de Abr. de 2021
Editada: John D'Errico
el 27 de Abr. de 2021
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
el 27 de Abr. de 2021
Editada: John D'Errico
el 27 de Abr. de 2021
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.
Respuesta aceptada
Bruno Luong
el 27 de Abr. de 2021
Editada: Bruno Luong
el 27 de Abr. de 2021
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
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Bruno Luong
el 27 de Abr. de 2021
Editada: Bruno Luong
el 27 de Abr. de 2021
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)
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