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Combinators can't do anything with a specific wire.
It will "add" the negative value to the value already on the wire.
This is something relatively widely used, for things like sushi belts for example (belts that can have a wide variety of items on it and that loops around on itself).
I don't see where you would need to multiply or divide "anything (red wire)" and "anything (green wire)".
In case you didn't know, "anything" individually processes each signal (again, it doesn't do "per wire" but "per signal") and releases the result (usually on "anything" again, meaning the same signal).
For conditions, you also have "any" and "all" depending on if you want the condition to work if at least one fits or if all of them (signals that are not at 0) need to fit.
Specific examples of what you are trying to accomplish would help us help you.
You can do same operation on all incoming signals, regardless if by constant or by signal.
All combinators sum up same sifnal comming from red and green wire before doing operations on them. So incoming red A6 and green A3 loose their colour before calculation and become A9. So setting A*A will give 81, not 18.
Colours exist only as colour of wire carring signal not signal it self.
This way colours are way to not mix signals comming from two different sources in case they are send to more than one place.
Only way to do calculation on same signal comming from red and green wire is to change one of colours in to other signal, but that makes imposible to do "batch" calculation.
Taking example above only way to get result A18 of incomming red A6 and green A3 uis to change one of them in to other signal. If you will use other combinator to change A3 in to B3 then you can set B*A and set result as A to get A18.
That means you need to have 2 combinators for each pair of signals to do same operation on them.
Only sum and division can be done in "batch" as described by You. Thats done "automaticaly" when both signals meet in one combinator/building (for division you need to multipy one colour by -1).
No.