Does anyone know an algorithm or for finding the output of a belt after a arbitrary network of belt splitters?
Specifically:
Given
1) some number of input belts, each with some constant number resources coming in.
2) some network of belts and splitters, including splitters feeding back into themselves.
3) some number of output belts
4) no maximum belt throughput (unless this makes the math harder instead of easier).
5) single lane belts
How do I calculate the output rate of every output belt?
This is a graph theory question, but I don't know the right search terms to look for an answer.
Factorio inspired math problem: belts and splitters
Re: Factorio inspired math problem: belts and splitters
This problem is called Markov chain.King Mir wrote: Mon May 25, 2020 3:26 am2) some network of belts and splitters, including splitters feeding back into themselves.
There are many ways to solve it, here is a Javascript solution: https://kilin.clas.kitasato-u.ac.jp/sof ... arkov.html
Re: Factorio inspired math problem: belts and splitters
Yep, that's exactly the search term I'm looking for. Thanks for the prompt response! And I thought asking here was a long-shot.
Re: Factorio inspired math problem: belts and splitters?
Read up on Belt Balancers, as it deals with pretty much all the math you're concerned with.
https://wiki.factorio.com/Balancer_mechanics
https://en.wikipedia.org/wiki/Clos_netw ... n_.3D_2.29
(Fractal)
viewtopic.php?p=343062#p343062
( 63462 )
https://docs.google.com/document/d/1FsT ... VDioM/edit
[moderated by boskid. clone topic from imposter with the same first post was merged into this topic and first post of imposter was removed. This was the only remaining post]
https://wiki.factorio.com/Balancer_mechanics
https://en.wikipedia.org/wiki/Clos_netw ... n_.3D_2.29
(Fractal)
viewtopic.php?p=343062#p343062
( 63462 )
https://docs.google.com/document/d/1FsT ... VDioM/edit
[moderated by boskid. clone topic from imposter with the same first post was merged into this topic and first post of imposter was removed. This was the only remaining post]