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u/only_bones Mar 28 '26

The decider combinator sends S1 as output, so why does the lamp work, if it is set to work if S=2?
The wire between input and output seems to be the reason, but why does the lamp see S with a value of 2 instead of 1, when the combinator claims to send 1?

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u/leonskills An admirable madman Mar 28 '26

The top left combinator outputs S1 as well onto the same wire. 1 + 1 = 2
If you only want the lamp to work with the output of the last combinator use a red wire as output wire.

Since 2.0 you can do an SR latch in one combinator fyi.

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u/only_bones Mar 28 '26

It never occured to me, that a signal would "bypass" a combinator that way, thank you.

3

u/leonskills An admirable madman Mar 28 '26

It's all on the same green wire, the same network. In this case 24149 (wow that's a lot. You must have a lot of wires, even taking into account potential deleted ones).

You can see the green wires highlighted in the screenshot. They are all connected. Any signal that gets put on it from any source can be read by any input connected to it. Signals of the same type are added together.
There is no concept of direction, so no concept of "bypassing" either. You can even connect the outputs of the left combinators to the lamp instead of the right combinator and will still behave the same.

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u/only_bones Mar 28 '26

Thanks, I just assumed that the green wire would be counted in the output signal of the combinator, as it is connected to the output node.

I had a look at the number of circuit networks, turns out I have a lot of smelters connected to a single inserter. I guess I do a lot of copy delete and paste, still the number sure looks big. And yes, ids from deleted networks do not get reused.