How to do negative offset?

I need to switch between 24 possible inputs. One switch obviously is not enough.
So two switches (together the required 24 inputs) and 1 controller from 1 - 24.
Here I get lost. I want to use an IntSlider (obvious choice for controlling a switch), but I get a warning that my range (1-24) exceeds the max. Ok, it still works, although it might not be elegant. But then, for the second switch to get its selector input value, I need to Offset its values from the IntSlider with -12 (minus 12). I have checked remapping, but I can only apply positive Offsets. This can't be right. How do I apply negative Offsets?

I also tried with an RGBMath component, 'If'. That does not work either in the way I suspected it to be able to work. In this case I want the if to work as follows :

Input A is the IntSlider (range 1 - 24)
Input B is a fixed value of 12
Statement : if A > B (so 13 -24) then Output Switch 2 (13-24)
ELSE output Switch 1 (1-12).

I noticed that the inputs have to be from 0-1. Can I somehow switch to HDRcolors, so I can just use my numbers (like 12)? And can I output numbers (again also > 1)?

The Subtract component works alike. Only results between 0-1 seem acceptable. I understand that, in view of normal colors (not HDR). But is it not possible in FF to add or subtract 2 numbers < 1?

And is there somewhere more extended information about the RGBmath components than the Help? The otherwise excellent Help pages are very thin about these components.

This is exactly what I was looking for. Great, thanks Rachel. I will study how it works tomorrow, my brain has already gone to bed

Cheers, Beat

I have made a simple snippet to do all the permutations of 3 or 4 inputs (Colors in this case, but could be anything really). I made one for 3 colors, which gives 6 permutations, of which three are the circular permutations and one for 4 colors.

The number of permutations of n elements is as follows :
Circular permutations (original included) = n
General perms (circular + all the others) = n! (n factorial, which means n!= 1*2*3* ...n-1*n. So 3 elements give you 1*2*3 = 6 permutations. 4 Elements give you 4 circular perms and 24 (1*2*3*4) perms. 5 Elements gives you already 120 , 6 > 720 perms and rapidly rising. So you might understand that I stopped at 4 elements.

I have used the switch solution rachelduim kindly supplied.

Question : I'd like to make the 4 elements perm part into a group. As a group only has 1 output and I need 4 for the 4 colors, how to address that? The input side is ok, also with more than 1 settings controller, but the output is a problem.

## Perm 3 + 4 Group.ffxml

Thanks Rachel, I got it. Wonderful. I have created two new Blend Modes and with the eyedropper I could see that there are actually negative values where they should be. I will post them after some more checking and tweaking.

For others : you have to open up the permutation snippet to see whats going on, as there can be only one final output and my snippet gives you 3 or 4 outputs. So look inside