1 Snarks the Hunting of the Snark by Lewis Carroll by Vizing's Theorem, (As Well As by Shannon's Theorem), Every Degree
Total Page：16
File Type：pdf, Size：1020Kb
1 Snarks The Hunting of the Snark by Lewis Carroll By Vizing’s theorem, (as well as by Shannon’s theorem), every degree- three graph is 4-colorable. The problem of determining if χ′(G)=3 or 4 is NP-complete: I.J. Holyer, The NP-completeness of edge col- oring, SIAM J. Comput., 1981, Vol. 10, pp. 718-720. Tutte conjectured in “On the algebraic theory of graph colorings,” J. Combinatorial Theory, 1 (1966), pp. 15-50, that every 2-connected cubic graph not containing the Petersen graph is 3-edge-colorable. This would imply the 4-color-theorem, which was computer-proved. 1 Definition 1 A graph G = (V,E) is called k-edge-cyclically con- nected if the removal of any k−1 edges yields a graph with at least one connected components which is not a tree, i.e. k is smallest number such that the removal of some k edges disconnects the graph into trees. Definition 2 A snark is a 3-regular graph with the chromatic index 4 satisfying the additional condition of non-triviality: it is at least 5-edge-cyclically connected. Petersen Graph: smallest snark; The Flower snark 2 Why 5-edge-cyclically connected? Consider the construction illustrated below: the dot-edges and result- ing isolated vertices are removed from the two copies of the Petersen graphs; 4 new edges are added to the union. One can prove that the chromatic index of the new graph is 4. The reverse is also true: if G is a cubic, 4-edge-chromatic graph with 4 edges disconnecting G, then it is obtained from two 4-edge-chromatic cubic graphs by the operation Linking (below). 4-cyclically connected 4-edge-chromatic graphs 3 Lemma 1 Let G be a 3-edge-colorable graph with the maximum vertex degree 3 (not necessarily regular), and let φ be a 3-coloring of the edges of G. For i = 1, 2, 3, let ni denote the number of vertices that miss color i. Then n1 ≡ n2 ≡ n3(mod 2). Proof. If n is the number of vertices of G, and e1,e2,e3 are the number of edges colored 1,2,3, respectively, then ni = n − 2ei for i =1, 2, 3. The result follows. e edgese vertices i 2 i n = n e i + 2 i n vertices i 4 Operation Linking: Let G and H be two snarks, let xy ∈ E(G), x1 and x2 (resp. y1 and y2) be two more vertices adjacent to x (resp. y). Let u1u2 and v1v2 be two independent edges in H. Remove vertices x and y from G, and edges u1u2 and v1v2 from H. Add the following edges to the union G∪H: x1u1, x2u2, y1v1, y2v2. The resulting graph is 3-regular and 4-chromatic. u_1 u_1 x_1 x_1 u_2 x_2 x u_2 x_2 v_1 y_1 y v_1 y_1 y_2 v_2 v_2 y_2 Snark 1 Snark 2 New snark Proof. The resulting graph F is obviously 3-regular. If G − {x,y} (resp. H − u1v1 − u2v2) is 4-chromatic, then F is 4-chromatic. Let both G − {x,y} and H − u1v1 − u2v2 be 3-colorable. Then, for every 3-coloring of G − {x,y}, consider the colors α1,α2,α3,α4 missing at x1,x2,x3,x4, respectively. If α1 6= α2, then by Lemma 1, α3 6= α4, and {α1,α2} = {α3,α4}, implying that G is 3-edge- colorable. Thus, α1 = α2 and α3 = α4. (1) On the other hand, the same type of the consideration proves that for every 3-edge-coloring of H − u1v1 − u2v2, the colors β1, β2, β3, β4 missing at v1,v2,v3,v4, respectively, must satisfy β1 6= β2 and β3 6= β4. (2) . Since (1) and (2) are contradictory, F is 4-chromatic. 5.