While loops & multiple variables

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conor meaney
conor meaney 2015-2-21
编辑： Andrew Newell 2015-2-22
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
Andrew Newell
Andrew Newell 2015-2-21
I presume you intended >= in the second example?
conor meaney
conor meaney 2015-2-21
Yes apologies, it should read '>=' in the second example!

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回答（2 个）

Andrew Newell
Andrew Newell 2015-2-21
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
conor meaney
conor meaney 2015-2-21
Perfect, works like a treat and a simple edit to the script.
Thanks!
Conor.

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Image Analyst
Image Analyst 2015-2-21
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
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Image Analyst
Image Analyst 2015-2-22
  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.
Andrew Newell
Andrew Newell 2015-2-22
编辑：Andrew Newell 2015-2-22
Oh. I was viewing it as pseudocode and assuming the actual code was mostly correct.

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