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L 1 L 2 L 3
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D----------------------------------------E
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C
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Sorry, this is intended to be the same picture as in the document I sent last evening. I have given labels to every vertex. The top and bottom vertex are actually the same vertex. The Li vertices are U and "U bar" vertices as described in the document. This is just the"gadget" for one clause. For each clause there is a similar gadget, and all have vertices R and C in common. They also share Li vertices when the clauses have a common variable. In any proper coloring of this gadget (3 colors or not) we may assume R has a color and specify it to be 3. We know C will receive a color - it can not be 3, but it can be any other color we want. So we will arbitrarily choose it to be 1. We also know each Li will have a color and it will also not be 3, but we can not say what or how many at this point. Lemma 1. If a gadget is 3-colorable, then at least one of L 1 , L 2 , or L 3 must have the same color as C, that is colored 1. Proof: Suppose not. Then, we may assume the color of L 1 , L 2 , and L 3 are all 2. Then, X and Y must be colored 1 and 3 - it doesn't matter which is which, and, hence, Z must be colored 2. Now, Z and L 3 are both colored 2. By the same argument, D and E must
be colored 1 and 3. Again, it doesn't matter which is which. Finally, C must be colored 2, contradicting the color of C being 1. Theorem 2. If G, the graph constructed from the given instance of 3-SAT, is 3- Colorable, then the 3-SAT instance is satisfiable. Proof: When G is 3-colorable, then each gadget is 3–colorable. By Lemma 1, each set of L 1 , L 2 , and L 3 have at least one variable colored the same as vertex C. Since C and R are common to all gadgets in G. We simply say the color of C is 1 = TRUE. Then, every gadget has a TRUE label (color) on at least one of its three Li vertices. Thus, in the instance of 3-SAT from which G was constructed, every clause has a TRUE variable, hence, a satisfying assignment. Theorem 3. If a 3-SAT instance is satisfiable, then the graph G, constructed as above, is 3-colorable. Proof: Suppose the instance of 3-SAT has a satisfying assignment. Then every clause has a true variable. Color the corresponding Ui and "Ui bar" vertices in G with TRUE or FALSE as determined by the satisfying assignments. Then, in each gadget, at least one of its Li vertices is (are) colored TRUE. We may arbitrarily color vertex C with TRUE and vertex R with color 3. Now we must show it is possible to properly color the remaining vertices of G with at most 3 colors (TRUE, FALSE, and 3). Since R, C and the U and "U bar" vertices are already given an assignment, we need only show the coloring for X, Y, Z, D, and E in the clauses. Further, by considering an arbitrary assignment, with at least one TRUE for the Li vertices, it is sufficient to show a 3- coloring for one clause, since their colorings are independent of each other. First, suppose either, or both, of L 1 and L 2 is (are) TRUE. Then, it is possible to color X and Y FALSE and 3, and it doesn't matter which, since either forces us to color Z TRUE. Now, Z and/or L 3 is TRUE. As before, it is possible to color D and E with FALSE and 3, which is consistent with C being colored TRUE. Thus, the gadget is 3-colored. Hence, all gadgets can be 3-colored, and G is 3-colored. Together, Theorems 2 and 3 give Theorem 4. G, as constructed, is 3-Colorable if and only if the 3-SAT instance is satisfiable. Corollary 5. Graph 3-Colorability is NP-Complete. Corollary 6. Graph k-Colorability is NP-Complete. Note: It is possible to make Lemma 1 "if and only if." That would simplify the proof of Theorem 3 a little. I had already proven Theorem 3 before I noticed that.
