Why does times(A,B) gives me negative values when A and B don't have any ???

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Benjamin
Benjamin 2025-6-19
编辑： Torsten 2025-6-19
Hello, i'm working with this code, which is a non-negative factorization :
function [G] = NN_Update_G(G_0,S,H,V,nb_iter,beta)
mustBePositive(G_0);
G=G_0;
if beta<1
gamma_beta=1/(1-beta);
elseif 1<=beta && beta <=2
gamma_beta = 1;
else
gamma_beta = 1/(beta-1);
end
G_old=G;
for k=1:nb_iter
mustBePositive(G_old);
product = (((S'*(((S*G_old*H).^(beta-2)).*V)*H'))./(S'*((S*G_old*H).^(beta-1))*H')).^gamma_beta;
mustBePositive(product);mustBePositive(G_old);
G = times(G_old,product);
mustBePositive(G);
G_old = G;
end
end
I don't know how but only the last mustBepostive(G) is triggered. It's the only one, which would mean that both G and product are positive beforehand.
I really hope someone can make some sense out of this because it's really bothering me not understanding what's the problem...
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Benjamin
Benjamin 2025-6-19
编辑：Benjamin 2025-6-19
Thanks for that, didn't know I could put parameters into a file like that (:
BUT, I tried but all the parameters (even alone) are too heavy to be included here /: (Apparently some files are attached so tell me if it works I guess. PS : it's just one matrice that's too heavy, so it's missing H. I'll try to find another way to send it)
for nb_iter I use 5 because it's in another loop but feel free to try anything
I use 0 for beta
Torsten
Torsten 2025-6-19
编辑：Torsten 2025-6-19
If you apply
load ("G_zero.mat")
G_0(G_0<=0)
G_0(G_0<0)
you will see that your G_0 has 0 negative elements, but 148 zero elements.

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采纳的回答

Steven Lord
Steven Lord 2025-6-19
Ran in:
Try using mustBeNonnegative instead of mustBePositive. This will detect if your calculation underflowed to 0.
x = realmin
x = 2.2251e-308
mustBePositive(x) % true so no error
y = x*x % underflow
y = 0
mustBeNonnegative(y) % true so no error
mustBePositive(y) % false
Value must be positive.

更多回答（1 个）

dpb
dpb 2025-6-19
移动：dpb 2025-6-19
Ran in:
mustBePositive(0)
Value must be positive.
shows that zero is not considered positive by mustBePositive. You don't give any klews as to what the magnitudes of inputs are, but one must surmise that one or more of the elements must have underflowed owing to a large exponent in the product expression.

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