While loops & multiple variables

Hi There,
I've run into an issue whilst using a while loop in my script involving iterations. The loop must end once a certain number of variables in the script converges to a pre-defined error value/limiing criterion between the nth and n-1th iteration.
i.e.
INPUT
while abs(xn - xn-1) >= limiting criteron
SCRIPT HERE
end
OUTPUTS
That's fine for one variable and works as wanted.. as soon as the variable on the left become less than the limiting criterion, it ends.
What I wanted to do is the same idea but with multiple variables requiring this limiting criterion i.e. I want it to keep iterating so that multiple variables converge to less than this value. (assuming convergence exists).
My attempt is something along the lines of:
INPUTS
while [abs(xn - xn-1), abs(yn - yn-1), abs(zn-zn-1)] <= [limiting criterion, limiting criterion, limiting criterion]
SCRIPT HERE
end
OUTPUTS
What appears to be happening is as soon as any of the variables becomes less than this limiting criterion, the loop ends, but what I want to happen is it to keep iterating until all the variables are less than the limiting criterion, again assuming convergence exists.
Any ideas?
Thanks in advance
Conor.

2 个评论

I presume you intended >= in the second example?
Yes apologies, it should read '>=' in the second example!

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回答(2 个)

Your while condition returns a logical vector, and if any of its components is false the whole thing is false and the loop ends. What you need is
while any([abs(xn - xn-1), abs(yn - yn-1), abs(zn-zn-1)] >= [limiting criterion, limiting criterion, limiting criterion])

1 个评论

Perfect, works like a treat and a simple edit to the script.
Thanks!
Conor.

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Try
continueWithLoop = true;
while continueWithLoop
% Some code....
% Check if we can quit loop or need to continue
continueWithLoop = abs(x(n) - x(n-1)) > limitingCriterion || ...
abs(y(n) - y(n-1)) > limitingCriterion || ...
abs(z(n) - z(n-1)) > limitingCriterion
end

5 个评论

Thank you for your answer! I will be going with the one below merely for simplicity. Thank you again.
OK, but be sure to fix the syntax errors with it. See my answer if you need help.
What syntax errors do you see?
  1. xn should be x(n)
  2. xn-1 should be x(n-1)
  3. yn should be y(n)
  4. yn-1 should be y(n-1)
  5. zn should be z(n)
  6. zn-1 should be z(n-1)
  7. Each of the three instances of "limiting criterion" should be "limitingCriterion" since variables can't have spaces in their names.
Oh. I was viewing it as pseudocode and assuming the actual code was mostly correct.

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