Randomly splitting of a number in a sum format.
8 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
rohini more
le 27 Avr 2021
Commenté : rohini more
le 27 Avr 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 commentaires
John D'Errico
le 27 Avr 2021
Modifié(e) : John D'Errico
le 27 Avr 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
le 27 Avr 2021
Modifié(e) : John D'Errico
le 27 Avr 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.
Réponse acceptée
Bruno Luong
le 27 Avr 2021
Modifié(e) : Bruno Luong
le 27 Avr 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
12 commentaires
Plus de réponses (1)
Bruno Luong
le 27 Avr 2021
Modifié(e) : Bruno Luong
le 27 Avr 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)
Voir également
Catégories
En savoir plus sur Logical dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!