How to get this while cycle?
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Hi, I'm trying to get a while cycle which checks convergence every iteration.
Within the cycle, I'm solving a system which returns each iteration 3 types of solutions: a (vector), B, C (matrices).
Subscript "0" is related to the (l-1) iteraton, the other refers to the (l) iteration.
I wrote this code:
while (abs(a_0-a) > 1e-6) & (abs(B_0-B) > 1e-6) & (abs(C_0-C) > 1e-6)
....
end
It starts, but it do only one iteration, I don't know why.
It should iterate until ALL residues are lower than 1e-6.
Probably it's wrong, maybe due to parentheses or logical operators. I've already tried to change them, but in the other cases it doesn't work.
What should I do?
Thank you for your help!!
采纳的回答
Fabio Freschi
2021-8-16
if B and C are matrices, make them vectors using :
while any(abs(a_0-a) > 1e-6) || any(abs(B_0(:)-B(:)) > 1e-6) || any(abs(C_0(:)-C(:)) > 1e-6)
更多回答（1 个）
Walter Roberson
2021-8-16
while (abs(a_0-a) > 1e-6) || (abs(B_0-B) > 1e-6) || (abs(C_0-C) > 1e-6)
5 个评论
Elia Paini
2021-8-16
编辑：Elia Paini
2021-8-16
Walter Roberson
2021-8-16
while any(abs(a_0-a) > 1e-6) || any(abs(B_0-B) > 1e-6) || any(abs(C_0-C) > 1e-6)
Elia Paini
2021-8-16
Walter Roberson
2021-8-16
while any(abs(a_0-a) > 1e-6, 'all') || any(abs(B_0-B) > 1e-6, 'all') || any(abs(C_0-C) > 1e-6, 'all')
The implication is that at least one of your subtractions is giving back a matrix that is 2 or more dimensions.
Please check that a_0 and a, and B_0 and B, and C_0 and C, have the same orientation if they are vectors. For example,
(1:3).' - (1:3)
is not the vector [0 0 0]
Elia Paini
2021-8-16
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