In computer science, the planar 3-satisfiability problem (abbreviated PLANAR 3SAT or PL3SAT) is an extension of the classical Boolean 3-satisfiability problem to a planar incidence graph. In other words, it asks whether the variables of a given Boolean formula—whose incidence graph consisting of variables and clauses can be embedded on a plane—can be consistently replaced by the values TRUE or FALSE in such a way that the formula evaluates to TRUE. If this is the case, the formula is called satisfiable. On the other hand, if no such assignment exists, the function expressed by the formula is FALSE for all possible variable assignments and the formula is unsatisfiable. For example, the formula "a AND NOT b" is satisfiable because one can find the values a = TRUE and b = FALSE, which make (a AND NOT b) = TRUE. In contrast, "a AND NOT a" is unsatisfiable. Like 3SAT, PLANAR-SAT is NP-complete, and is commonly used in reductions.
Definition Every 3SAT problem can be converted to an incidence graph in the following manner: For every variable v i {\displaystyle v_{i}} , the graph has one corresponding node v i {\displaystyle v_{i}} , and for every clause c j {\displaystyle c_{j}} , the graph has one corresponding node c j . {\displaystyle c_{j}.} An edge ( v i , c j ) {\displaystyle (v_{i},c_{j})} is created between variable v i {\displaystyle v_{i}} and clause c j {\displaystyle c_{j}} whenever v i {\displaystyle v_{i}} or ¬ v i {\displaystyle \lnot v_{i}} is in c j {\displaystyle c_{j}} . Positive and negative literals are distinguished using edge colorings. The formula is satisfiable if and only if there is a way to assign TRUE or FALSE to each variable node such that every clause node is connected to at least one TRUE by a positive edge or FALSE by a negative edge. 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. Planar 3SAT is a subset of 3SAT in which the incidence graph of the variables and 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 problems to be NP-hard.
Proof of NP-completeness
The following proof sketch follows the proof of D. Lichtenstein. Trivially, PLANAR 3SAT is in NP. It is thus sufficient to show that it is NP-hard via reduction from 3SAT. This proof makes use of the fact that ( ¬ a ∨ ¬ b ∨ c ) ∧ ( a ∨ ¬ c ) ∧ ( b ∨ ¬ c ) {\displaystyle (\lnot a\lor \lnot b\lor c)\land (a\lor \lnot c)\land (b\lor \lnot c)} is equivalent to ( a ∧ b ) ↔ c {\displaystyle (a\land b)\leftrightarrow c} and that ( a ∨ ¬ b ) ∧ ( ¬ a ∨ b ) {\displaystyle (a\lor \lnot b)\land (\lnot a\lor b)} is equivalent to a ↔ b {\displaystyle a\leftrightarrow b} . First, draw the incidence graph of the 3SAT formula. Since no two variables or clauses are connected, the resulting graph will be bipartite. Suppose the resulting graph is not planar. For every crossing of edges (a, c1) and (b, c2), introduce nine new variables a1, b1, α, β, γ, δ, ξ, a2, b2, and replace every crossing of edges with a crossover gadget shown in the diagram. It consists of the following new clauses:
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