12-23-2009 09:36 PM
Solved! Go to Solution.
12-23-2009 09:55 PM
12-24-2009 08:16 AM
12-24-2009 09:12 AM
I know the first example gives you 8 possible states. However, maybe you can insert some logic to get rid of the two states you want? If you add up the three booleans as distinctly different numbers you end up with only 4 states, and with that situation you won't know which boolean contributed to the final result (Example 2)!
If I understand you correctly, I think your best options is to use Example 1 in the VI attached and then use some logic to filter out the two cases you don't want. I'm not sure how else you'll get 6 cases from three booleans...
12-24-2009 10:53 AM
Thank you for the examples! Your example one is closest to what I am looking for.
Your logic for example one is as follows:(turning on/off one boolean at a time):
boolean option1,option2
boolean2 option3,option1
boolean3 option5,option1
The logic for my application will need to be:
boolean option0,option1
boolean2 option2,option3
boolean3 option4,option5
This leaves gives you 6 distinct states! 🙂
12-24-2009 10:57 AM
sorry, your logic is:
boolean option1,option2
boolean2 option1,option3
boolean3 option1,option5
I stand corrected...:-(
12-24-2009 10:58 AM
12-24-2009 11:01 AM
12-24-2009 11:16 AM
12-24-2009 11:18 AM