Post #2058024
2026-04-19 12:32 UTC
@tnt @trcwm (yes, that's clearly a verb) from the input node to the first netlist entry connnected to it ("first" might mean: of type A, because that has no other rings to cross).
You can at every variable node assume it's closed in your favor. Unless, of course, it's been marked uncloseable by a previous routing step in the current attempt to find a shortest path. Keep a copy of the distance-from-origin-marked graph around, but for the next step, just use the current assignments of variable
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@funkylab@mastodon.social 2026-04-19 12:38
@tnt @trcwm nodes. Dijkstra guarantees you found a shortest route here, because the planar nature of your routing sub-grids doesn't allow for "I can take a shortcut if I take more turns". Then, next component also directly connected to input. Once all components of a type of the same depth in the netlist have been routed, check for same-ring (i.e., concentric) interconnects, and whether their length can be optimized by rotating "pizza slices" of your already embedded graph/slice permutations.