Inverse rand

Questions and Answers
#1

hey there,

wondering if someone can help me out with this.. i believe rand works by generating a random number string that can be applied to something, but is not truly random in the sense that a random number is picked each for each value each time a value is called.

in practice this means, if i have # pan rand applied in two places in my code they will "randomly" pan together/in sync.
what i'm looking for is a way to invert one of those values. so in practice one would pan right when the other pans left, by equal amounts.

what's the best way to go about this?

#2

it's a common practice to use a function like @kindohm's shrand.

let shrand x = (x ~>) $ rand

you'll just put any old number in the x placeholder and the cycle will get shifted so that it generates a different random value

#3

cool. let me see if i understand what this does correctly... this adds x (let's say 1) to each number generated by the rand function? i'm assuming i delete that x after let shrand (that's the only way i could get it to work for me.

#4

not quite. x is the number of cycles shifted so you can grab a new random value that's different from the one currently running on the pattern you don't want it to be shared with. the x value is kind of not important. it could be 1, it could be 10,000; what's important is that it's grabbing the rand value of some future cycle.

#5

ok, thanks for the explanation!
if i'm understanding correctly then that would ensure you get a different set of random numbers, for whatever has shrand applied to it.
so -- what would be the best way to get the inverse value of the rand i have applied somewhere in my code so far?

#6

oh, ok, i see you are trying to do something quite specific with the panning. i thought you were just looking for a workaround to the not-quite-random rand...if i were you, i'd test out some of the continuous patterns like saw and isaw to sync up your pans.

#7

will look into it. cheers for the help!

#8

i can’t try this but right now but range 1 0 $ rand might be what you’re looking for? or possibly (1 - rand)

2 Likes
#9

Yes I think either would work!

1 Like
#10

indeed, both worked like a charm!

2 Likes