To prove that 3-COLOR is NP-complete, we use a reduction from 3-CNF-SAT. Given a formula o of m clauses on n variables X1, X2, ..., Xn, we construct a graph G = (V, E) as follows. The set V consists of a vertex for each variable, a vertex for the negation of each variable, 5 vertices for each clause, and 3 special vertices: TRUE, FALSE, and RED. The edges of the graph are of two types: literal edges that are independent of the clauses and clause edges that depend on the clauses. The literal edges form a triangle on the special vertices and also form a triangle on Xi, X;, and RED for i = 1,2,...,n. d. Argue that in any 3-coloring c of a graph containing the literal edges, exactly one of a variable and its negation is colored c(TRUE) and the other is colored C(FALSE). Argue that for any truth assignment for 0, there exists a 3-coloring of the graph containing just the literal edges. The widget shown in Figure 34.20 helps to enforce the condition corresponding to a clause (x v y v z). Each clause requires a unique copy of the 5 vertices that are heavily shaded in the figure; they connect as shown to the literals of the clause and the special vertex TRUE. e. Argue that if each of x, y, and z is colored c(TRUE) or c(FALSE), then the widget is 3-colorable if and only if at least one of x, y, or z is colored c(TRUE). f. Complete the proof that 3-COLOR is NP-complete. TRUE Figure 34.20 The widget corresponding to a clause (x V y V z), used in Problem 34-3