Start with a set cover problem, NP. OK. Translate it mechanically to an e-graph in polynomial time, sure, that instance of the e-graph is still unsolvable in polynomial time. It doesn't follow that all instances of e-graphs are unsolvable in polynomial time, only that some aren't.
Your e-graph is really heavily constrained by the ISA and optionally by the ABI. It's also constrained by the code it was created from which will tend to have its own constraints. Constraints can lead to faster solutions.
SSA construction is putting phi's in, if you're using them instead of block arguments. Destruction is taking phis out. Whether that's linear time or polynomial or non-polynomial depends on how optimal a job of eliding copies you choose to do.
>It doesn't follow that all instances of e-graphs are unsolvable in polynomial time, only that some aren't.
Yes and hardness results are about proving lower bounds on worst case complexity not best case or sometimes case or whatever-you-chose-as-meaningful case.
Specifically, e-graphs are used with respect to an ISA from which the cost model is derived. My intuition is that this constrains the structure sufficiently that polynomial solutions can be found. Probably involving a partial order calculated ahead of time.
For me, whether the result in the opening post is applicable to uses of e-graphs in compilers has a strong influence over whether I need to worry about it or not. I'm on the fence about e-graphs as a model. If rolling one into production means I'll occasionally hit exponential increases in running time as heuristics fall over, that's an argument against. If those cliffs are unreachable for a given ISA, they don't matter.
By rice's theorem, for any ISA that has a jmp or mov, there is no way to know/prove/predict this. So you're still not saying anything useful.
Also: if you are okay with heuristics that aren't guaranteed to output the optimal answer the whole point is moot
Are you like actually misunderstanding or are you trying to say something else - egraph is in np because it's equivalent to ilp and it's complete because you can reduce set cover to it not the other way around (the reduction always goes to the problem not from the problem in a completeness proof).
>Despite that, register allocation on a chordal graph can be done in linear time, and SSA form gives you a chordal graph.
This isn't saying much since SSA construction (phi elimination) isn't linear.