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.

[Andrew Newell]

I presume you intended >= in the second example?

[conor meaney]

Yes apologies, it should read '>=' in the second example!

Answer 1

[Andrew Newell]

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])

[conor meaney]

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

Thanks!

Conor.

Answer 2

[Image Analyst]

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

[conor meaney]

Thank you for your answer! I will be going with the one below merely for simplicity. Thank you again.

[Image Analyst]

OK, but be sure to fix the syntax errors with it. See my answer if you need help.

[Andrew Newell]

What syntax errors do you see?

[Image Analyst]

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]

Oh. I was viewing it as pseudocode and assuming the actual code was mostly correct.