Conditional Random number generation
18 visualizzazioni (ultimi 30 giorni)
Mostra commenti meno recenti
wang syr
il 2 Mar 2017
Commentato: Bruno Luong
il 12 Ago 2023
Hello there, For example; If I want to generate 5 random integer numbers with a sum of 20, how can I do that?
" ... example = ceil(10*rand(100, 5)) ... "
Risposta accettata
Roger Stafford
il 4 Mar 2017
function R = randfixedsumint(m,n,S);
% This generates an m by n array R. Each row will sum to S, and
% all elements are all non-negative integers. The probabilities
% of each possible set of row elements are all equal.
% RAS - Mar. 4, 2017
if ceil(m)~=m|ceil(n)~=n|ceil(S)~=S|m<1|n<1|S<0
error('Improper arguments')
else
P = ones(S+1,n);
for in = n-1:-1:1
P(:,in) = cumsum(P(:,in+1));
end
R = zeros(m,n);
for im = 1:m
s = S;
for in = 1:n
R(im,in) = sum(P(s+1,in)*rand<=P(1:s,in));
s = s-R(im,in);
end
end
end
return
6 Commenti
Yu Takahashi
il 9 Feb 2021
Modificato: Walter Roberson
il 10 Feb 2021
Wondering whether it is possible to specify the max and min of the devided value? i.e., something like what you kindly provided in the randfixedsum function, thanks!
Ref
Bruno Luong
il 12 Ago 2023
@Walter Roberson "Wondering whether it is possible to specify the max and min of the devided value?"
Più risposte (2)
Walter Roberson
il 2 Mar 2017
9 Commenti
Walter Roberson
il 4 Mar 2017
Ah. I don't think I know how to implement your suggestion, though, at least not without generating all of the possible choices that sum to 20 and then picking one at random.
John D'Errico's https://www.mathworks.com/matlabcentral/fileexchange/12009-partitions-of-an-integer can calculate all of the possible partitions; a question is whether we can avoid having to take that step.
Walter Roberson
il 4 Mar 2017
https://en.wikipedia.org/wiki/Partition_(number_theory)#Restricted_part_size_or_number_of_parts talks about restricted partitioning briefly, and ties it to change making problems, which does indeed sound equivalent to the approach I was taking. Those are in turn tied to knapsack problems.
Bruno Luong
il 10 Ago 2020
m = 5;
n = 3;
s = 10;
This will generate uniform distribution with sum criteria
% generate non-negative integer random (m x n) array row-sum to s
[~,r] = maxk(rand(m,s+n-1),n-1,2);
z = zeros(m,1);
r = diff([z, sort(r,2), (s+n)+z],1,2)-1;
1 Commento
Bimal Ghimire
il 4 Ott 2020
While generating conditional random numbers, how can we generate random numbers that has a limit of some maximum value and have certain specified sum value?
Vedere anche
Categorie
Scopri di più su Random Number Generation in Help Center e File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!