Planar SAT

A planar graph is a graph that can be drawn on the plane in a way such that no two of its edges cross each other. Every 3SAT problem can be converted to an incidence graph in the following manner: For every variable ${\displaystyle v_{i}}$, the graph has one corresponding ${\displaystyle v_{i}}$, and for every clause ${\displaystyle c_{j}}$, the graph has one corresponding node ${\displaystyle c_{j}.}$We create an edge ${\displaystyle (v_{i},c_{j})}$ between variable ${\displaystyle v_{i}}$and clause ${\displaystyle c_{j}}$ whenever ${\displaystyle v_{i}}$ or ${\displaystyle {\bar {v_{i}}}}$ is in ${\displaystyle c_{j}}$. We distinguish positive and negative literals using edge colorings.

Planar 3SAT is a subset of 3SAT in which the incidence graph of the variables and the clauses of a Boolean formula is planar. It is important because it is a restricted variant, and is still NP-complete. Many problems (for example games and puzzles) cannot represent non-planar graphs. Hence, Planar 3SAT provides a way to prove those games to be NP-hard.

The following proof sketch follows the proof of D. Lichtenstein.[1]

Trivially, PLANAR 3SAT is in NP. It is thus sufficient to show that it is NP-hard via reduction from 3SAT.

First, draw the incidence graph of the 3SAT formula. Since no two variables or clauses are connected, the resulting graph will be bipartite. The resulting graph may not be planar. Replace every crossing of edges with a crossover gadget shown here. However, the figure leads to a minor error—some clauses contain 4 variables and some contain only 2 variables so the premises of 3SAT are not followed exactly in the gadget shown. This glitch is easily fixable: For a clause that only contains 2 variables, either create parallel edges from one variable to the clause or create a separate false variable to include in the constraint.

For the 4 variable clause, borrow the reduction from 4SAT to 3SAT to create a gadget that involves introducing an extra variable set to false which represents whether the left or right side of the original clause contained the satisfying literal. This completes the reduction.[2]

• Planar 3SAT with a variable-cycle: Here, in addition to the incidence-graph being planar, the following graph should also be planar: there is a cycle going through all variables, each clause is either inside the cycle or outside it, and linked to its corresponding variables. This problem is NP-complete.[1]
  • However, if we further restrict the problem such that all clauses are inside the variable-cycle, or all clauses ar outside it, then the problem can be solved in polynomial time using dynamic programming.
• Planar 3SAT with literals: The bipartite incidence graph of the literals and clauses is planar too. This problem is NP-complete.[1]
• Planar rectilinear 3SAT: Each variable is a horizontal segment on the x-axis while each clause is a horizontal segment above/below the x-axis with vertical connections to the 3 variables it includes on the x-axis. The connections in each clause are either all-positive or all-negative. This problem is NP-complete.[3]
  • Planar monotone rectilinear 3SAT: This is a variant of planar rectilinear 3SAT where the clauses above the x-axis are all-positive and the clauses below the x-axis are all-negative. This problem is NP-complete.[4]
• Planar 1-in-3SAT: This is the planar equivalent of 1-in-3SAT. It is NP-complete.[5]
  • Planar positive rectinilear 1-in-3SAT: This is the planar equivalent of positive 1-in-3SAT. It is NP-complete.[6]
• Planar NAE 3SAT: This problem is planar equivalent of NAE 3SAT. Surprisingly, it can be solved in polynomial time. The proof is by reduction to planar maximum cut.[7]
• Planar circuit SAT: This is a variant of circuit SAT in which the circuit, computing the SAT formula, is a planar directed acyclic graph. Note that this is a different graph than the adjacency graph of the formula. This problem is NP-complete.[8]