Consider that the following lines of code:
appear two times, once for x < 0 and once for x > 0 (with the one for x < 0 being preceded by x =0-x;) That means those lines are executed in either case (x < 0 or x > 0), so you can separate them from the if x < 0 ... else ... end logic:
Now consider what happens if x > 2*y. Say, for instance, x = 10 and y = 3:
at that point the while condition will be checked again, this time with r = 7. It finds that the condition r > y (i.e., 7 > 3) is true, so the loop iterates again. The first line inside the loop is to update r:
but the update is based on x, which hasn't changed since the first time through, so you just get the same value of r (10-3=7) again. Then the while condition is checked again, nothing's changed, the loop iterates again, etc., etc., ad infinitum. The while loop's just going to run forever in this case.
Hopefully by considering that case, it's clear that the updated value of r needs to be based on the current value of r, not the initial value of r, which is x. That is, that line should be:
(just as q is incremented by 1 on the next line, r is decremented by y on this line). So that block should be this:
If you make that change the code should return a result in all cases where y > 0.
In case y < 0 (if your code has to support this case), you'll have to do something else because subtracting y from r will only increase the value of r. I'll leave that to you to think about, and also consider what should happen when y = 0 (something divided by zero).
And what about when x = 0 (zero divided by something)? As it is now the code would return a scalar z = 0 in that case. Maybe that's what you want, or maybe it should also return a quotient and remainder in that case (i.e., a 1x2 vector).
