Planar SAT
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 formula_1, the graph has one corresponding node formula_1, and for every clause formula_3, the graph has  one corresponding node formula_4 An edge formula_5 is created between variable formula_1 and clause formula_3 whenever formula_1 or formula_9 is in formula_3. 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 formula_11 is equivalent to formula_12 and that formula_13 is equivalent to formula_14. 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", "c") and ("b", "c"), introduce nine new variables "a", "b", "α", "β", "γ", "δ", "ξ", "a", "b", and replace every crossing of edges with a crossover gadget shown in the diagram. It consists of the following new clauses: formula_15 If the edge ("a", "c") is  inverted in the original graph, ("a", "c") should be inverted in the crossover gadget. Similarly if the edge ("b", "c") is inverted in the original, ("b", "c") should be inverted. One can easily show that these clauses are satisfiable if and only if formula_16 and formula_17. This algorithm shows that it is possible to convert each crossing into its planar equivalent using only a constant amount of new additions. Since the number of crossings is polynomial in terms of the number of clauses and variables, the reduction is polynomial. Reductions. Logic puzzles. Reduction from Planar SAT is a commonly used  method in NP-completeness proofs of logic puzzles. Examples of these include Fillomino, Nurikabe, Shakashaka, Tatamibari, and Tentai Show. These proofs involve constructing gadgets that can simulate wires carrying signals (Boolean values), input and output gates, signal splitters, NOT gates and AND (or OR) gates in order to represent the planar embedding of any Boolean circuit. Since the circuits are planar, crossover of wires do not need to be considered. Flat folding of fixed-angle chains. This is the problem of deciding whether a polygonal chain with fixed edge lengths and angles has a planar configuration without crossings. It has  been proven to be strongly NP-hard via a reduction from planar monotone rectilinear 3SAT. Minimum edge-length partition. This is the problem of partitioning a polygon into simpler polygons such that the total length of all edges used in the partition is as small as possible. When the figure is a rectilinear polygon and it should be partitioned into rectangles, and the polygon is hole-free, then the problem is polynomial. But if it contains holes (even degenerate holes—single points), the problem is NP-hard, by reduction from Planar SAT. The same holds if the figure is any  polygon and it should be partitioned into convex figures. A related problem is minimum-weight triangulation - finding a triangulation of minimal total edge length. The decision version of this problem is proven to be NP-complete via a reduction from a variant of Planar 1-in-3SAT.