Cycles of 8


    Sep 30 2012 | 8:43 pm
    Hello,
    I really want to be able to count in cycles of 8. Currently I have a number box which shows what (absolute) bar I am on in an Ableton live track. I want to know what this means in a cycle of 8, so that the number box would start at 1 every 8 numbers. Therefore number 9 would equal 1, 10 would equal 2, 11 would equal three..........then 17 would equal 1, 18 would equal 2, etc. Is there a way this could be done?

    • Sep 30 2012 | 8:59 pm
      [%]
    • Sep 30 2012 | 9:25 pm
      Thank you so much. I have been thinking about this for days (!)
      One more question.
      When I have it set up, If I make the right operand 9, (so that that the result goes as high as 8), on the 9th, 17th, 25th, etc. count the remainder is 0. I would like this to start at 1.
      I am counting bars in cycles of 8, so I want the cycle to go 1, 2, 3, 4, 5, 6, 7, 8, (THEN) 1,2,3,4,5,6,7,8. Right now it goes
      1,2,3,4,5,6,7,8 THEN 0,1,2,3,4,5,6,7,8 etc. Anyway to do this.
      Thanks!
    • Sep 30 2012 | 9:27 pm
      [% 8] -> [+ 1]
    • Sep 30 2012 | 9:36 pm
      Also thank you for your help.
      But wont that add 1 to every number?
      All I need is that instead of the result equaling 0 when the incoming number is 8, or 16, I need it to actually say 8, then on the following number start at 1 again.
      Thank you
    • Sep 30 2012 | 10:01 pm
      expr ($i1-1)%8+1
    • Sep 30 2012 | 10:02 pm
      Okay, I'll spell it out for you. %9 will give you a set of 9 numbers (0...8) - you only want 8, so you need %8. This will give you a cycle of 0-7. If your first incoming number is 1, and not zero, you should subtract 1 on the input on your modulo - and then add it back at the end. [- 1] -> [%8] -> [+ 1]
      Given a series of numbers from 1...N, this will give you cycles of 1...8 mapped correctly. IE, the first bar (1) is now labelled 0, into modulo (the %), that stays 0, and then gets 1 added to it to become number 1. Your 7th bar is now a '6' before it gets to a modulo, stays 6 inside the modulo, and becomes 7 again on the output. The 9th bar is now '8' before it gets to modulo, which modulo will wrap down to '0' - and then you add 1.
      So yes, it will add 1 to all the numbers. That's what you want.
    • Sep 30 2012 | 10:02 pm
      let me join this chain of genius:
      [- 1] -> [% 8] -> [+ 1]
    • Sep 30 2012 | 10:03 pm
      WE'RE ALL MATHLETES HERE
    • Sep 30 2012 | 10:48 pm
      Seriously, thank you guys. I had to laugh at my own dumbass, but you helped a lot.
      I am clearly not as well versed as you are all in MAX so it is great that you guys take the time to help. @MuShoo @Andrzej I was really laughing hard, but you seriously taught me what's up.