3-SAT → Depth-three unit 2-interval recognition Complexity open
Solution submitted Research history & proof on GitHub ↗
Why this question matters
Depth three is a boundary between simpler representations and known harder cases in intersection-graph recognition.
The cited representation results leave depth three between depth-two tractability and depth-four hardness.
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.
Given a simple graph, represent each vertex by two disjoint unit intervals so that intersection is exactly adjacency and no point lies in more than three intervals, or report NO-SOLUTION. A finite endpoint encoding is required.
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.
Every alternative geometric representation must decode consistently; merely drawing the intended representation proves only completeness.
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 2026-09-15. This summarizes the archived literature search on the date above. Unpublished, unindexed and overlooked work remains outside coverage; no new novelty assessment was performed.