How do I run two while loops independent of each other?

I have a simple VI. I have loop 1 with and LED and a stop button and I have loop 2 with an LED and stop button. They are both in while loops. When I run the VI,if I stop loop 1, I can not turn it back on. If I stop loop 2, it will not stop. I've attached a file with my VI.
Attachments:
2 Loops.vi ‏9 KB

bebesbabe,
This looks like a learning exercise, so I will not present a complete solution.
1. It is not polite to post to the Forum a VI which will not stop except by pressing the Abort button. The Abort button should only be used to stop a VI if something is wrong with the program during development. Add a stop button to the outer loop.
2. A loop with no delay will try to run as fast as possible, consuming all the CPU resources it can get. Since these loops are for user interfaces, delays of 100 to 200 ms are appropriate. Use the Wait (ms) function.
3. Your main problem is understanding dataflow. In LabVIEW any node may execute only when all of its inputs have been given data and it will not complete its execution until all nodes inside it have completed. Your VI does not have external inputs so waiting for data is not a factor. Each loop is considered a node. So the outer loop will iterate only after BOTH of the inner loops have stopped. On the next iteration both inner loops will be started again. Of course if the stop buttons are still pressed, the will stop after one iteration, which will seem almost instantaneous. But because of dataflow as I have described, your program cannot really work the way you want. You cannot restart one loop until both have stopped. Put an indicator on the "i" terminal of the outer loop.
4. You may wish to consider changing the mechanical action on your stop buttons to Latch when Released. Read the help to learn about mechanical actions.
The way the instructions on the front panel read, it is not clear what the intended solution is. What control or user action or program action is required to turn the loops back on? Changing the stop button? Automatically after a delay? A separate On switch? How is the entire program supposed to stop?
Lynn

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Proposal
In the vein of the Perl tradition, I'd like to see if there's any interest in solving short programming puzzles in LabVIEW. Anyone can post a problem and define the rules for solving it.
Here's a beginning/intermediate question to whet your "palette".
Description
With one user-action, how would you stop two while loops such that the ratio of their iterations is the always the same? One concrete way to visualize this situation is in taking measurements: two instruments use the same timing source to take measurements, but second divides the clock down so that it is an integer factor slower than the first. For example, if the slower instrument is four times slower, then at the end of the VI, if the slower instrument takes 100 measurements, the faster instrument has taken 400.
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Norbert_B wrote:
a) You work with an integer factor of speed. While this matches many application areas, there are still cases where the factor has to be floating point (e.g. 2.5). Additionally, the factor might "invert", so the master loop runs slower than the slave (factor 0.5 for instance).
It sounds like you're about to post another code golf question. If you think the problem scope is too small, you're welcome to fashion one to your taste. Thanks for volunteering :-D
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b) This example does not take into account that the hardware has to be synchronized with more accuracy than the software supplies (PXI?). If hardware synchronization is required, there are two possible ways:
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Hrm, maybe puzzles aren't really your thing after all. You are correct: I did not define every constraint that can lead to a provable best closed-form solution, but that does not prevent people from stating their assumptions and forming creative, well-reasoned, and still correct designs. I overlooked this reading of the description and neglected to add a statement that said solvers were free to add their own conditions. I will do so in the fugure :-)
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c) You don't state anything about "what happens with the data". Is this time consuming? Does it introduce additional delay in loop iterations? Is it OK to have data loss?
I don't know -- you decide, you disclose, and you solve :-)
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