필터 지우기
필터 지우기

How to get this while cycle?

조회 수: 1 (최근 30일)
Elia Paini
Elia Paini 2021년 8월 16일
댓글: Elia Paini 2021년 8월 16일
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!!

채택된 답변

Fabio Freschi
Fabio Freschi 2021년 8월 16일
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)
  댓글 수: 1
Elia Paini
Elia Paini 2021년 8월 16일
Great, it works! Thank you!

댓글을 달려면 로그인하십시오.

추가 답변 (1개)

Walter Roberson
Walter Roberson 2021년 8월 16일
while (abs(a_0-a) > 1e-6) || (abs(B_0-B) > 1e-6) || (abs(C_0-C) > 1e-6)
  댓글 수: 5
Walter Roberson
Walter Roberson 2021년 8월 16일
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
Elia Paini 2021년 8월 16일
I checked, the couples have the same dimension and orientation.
a_0 and a are (4x1), B_0 and B are (4x2), C_0 and C are (4x3)

댓글을 달려면 로그인하십시오.

카테고리

Help CenterFile Exchange에서 MATLAB Coder에 대해 자세히 알아보기

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by