3-SAT → Simple-polygon visibility graph recognition Complexity open
Unsolved
Why this question matters
Visibility recognition is a central inverse problem in computational geometry: reconstruct geometry from pairwise visibility information.
The cited recognition question supplies neither the polygon boundary order nor holes. Results for variants with that information do not settle the graph-only problem.
The reduction to construct
A source instance is an explicitly encoded Boolean formula with at most three literals per clause. Its outputs are satisfying Boolean assignments, or NO-SOLUTION when the formula is unsatisfiable.
The target is a graph alone, with no boundary order or holes supplied. A positive output must encode a simple polygon whose vertex visibility graph is exactly the input. The precise finite coordinate model remains to be fixed.
Construct deterministic polynomial-time maps F and G: F sends every legal source instance to a legal target instance, and G(x,y) is a valid source output for every valid output y of F(x). Preserve the stated threshold, domain and promises. The requested complexity conclusion is NP-hardness for the stated target problem.
Boundary-order freedom and finite coordinate encodings must be controlled while excluding unintended visibility edges.
What a complete answer must include
Give explicit construction and recovery algorithms, a general proof for every legal input and every valid target output, and polynomial runtime and encoding-size bounds. Specify finite output encodings and handle NO-SOLUTION outputs when applicable. Tests compare recovered source outputs with independent source solutions.
References
Literature checked 2026-09-16. This summarizes the archived literature search on the date above. Unpublished, unindexed and overlooked work remains outside coverage; no new novelty assessment was performed.