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Two graph triangles sharing a vertex $B$ force their two opposite edges to use the same two colours in a graph colouring with three colours. A palette graph triangle $T,F,B$ and another graph triangle $u,\bar{u},B$ therefore force the literal pair to have opposite Boolean colours. This gadget is used in a reduction from 3-SAT to the three-colourability problem.

A Boolean formula is in conjunctive normal form when it is a conjunction of Boolean clauses. It is true precisely when every clause is true; an empty conjunction is true. 3-SAT restricts each clause to at most three Boolean literals.

The decision problem is in NP: an assignment is a certificate, and counting its satisfied clause occurrences is polynomial in the input size.

Reduce 3-SAT to it. For each of the $m$ original clauses, use the seven-clause gadget for MAX-2SAT with its three literals in place of $a,b,c$ and with a fresh auxiliary variable. Keep all ten clause occurrences per gadget, including any repeated occurrences across gadgets. Set the target to $k=7m$.

If the original formula is satisfiable, choose each auxiliary value as in part (a), giving seven satisfied clauses per gadget. Conversely, no gadget can exceed seven. If an assignment satisfies at least $7m$ clauses in total, every gadget must reach seven, so every original clause is satisfied by the original-variable assignment. The construction has $10m$ clauses and $m$ auxiliary variables, so is polynomial. Consequently the decision version of MAX-2SAT is NP-complete.

The three-colourability problem belongs to NP: a colour assignment is a polynomial-length complexity certificate, and all edges can be checked in $O(|V|+|E|)$ time.

For NP-hardness, reduce 3-SAT to the three-colourability problem. Start with a palette graph triangle with vertices $T,F,B$. Its colours are necessarily distinct, and name them by these vertices. For each Boolean variable $u$, add vertices $u,\bar{u}$, join them to each other, and join each to $B$. The graph triangle $B,u,\bar{u}$ is the forcing component of the Boolean-pair colouring gadget, so the two literal vertices receive opposite colours $T,F$.

For each Boolean clause, attach a fresh copy of the three-colour clause gadget, identify its $t$ with the palette vertex $T$, and identify its three input vertices with the Boolean clause's Boolean literal vertices. Copies share only palette or literal vertices. A Boolean clause with one or two Boolean literals is padded to three by repeating a literal; an empty Boolean clause can be mapped immediately to the uncolourable graph $K_4$.

Any graph colouring with three colours gives a truth assignment by reading colour $T$ as true. The three-colour clause gadget forbids three false inputs, so every Boolean clause is satisfied. Conversely, any satisfying truth assignment colours the literal pairs, and the extension property proved above independently colours the fresh auxiliary vertices of each clause. Thus the entire graph is three-colourable exactly when the formula is satisfiable.

There are $3+2n+5m$ vertices for $n$ variables and $m$ nonempty padded clauses, and $O(n+m)$ edges. The construction is a polynomial-time many-one reduction. Using the permitted NP-completeness of 3-SAT, we obtain

A three-literal disjunction can be represented by ten unit or two-literal clause occurrences with one fresh auxiliary variable. If the number $r$ of true input literals is zero, the best auxiliary choice satisfies six; for $r=1,2,3$, the best count is seven. Applying separate gadgets to all clauses of a 3-SAT instance gives target seven times the original clause count, proving NP-completeness of the decision form of MAX-2SAT. The local counting argument is essential: no gadget may exceed seven and compensate for an unsatisfied input clause.

This decision problem asks whether a finite undirected graph has a graph colouring with three colours. It is NP-complete: a colouring certifies membership in NP, while 3-SAT reduces using Boolean-pair colouring gadgets and three-colour clause gadgets.