Planar graphs with no 5-cycles, 6-cycles or intersecting triangles are 3-colorable

1. Planar graphs with no 5-cycles, 6-cycles or intersecting triangles are 3-colorable Carl Yerger, Davidson College Clemson Mini-Conference 2012

2. A Historical Problem • Grotzsch’s Theorem (1959): Any planar graph that contains no 3-cycles is 3-colorable. • Thomassen found several short proofs of Grotzsch’s theorem as a consequence of some of his other results. • Are there other classes of graphs that exclude certain length cycles that are 3-colorable?

3. Steinberg’s Conjecture • Steinberg conjectured (1976) that any planar graph without 4- or 5- cycles is 3-colorable. • The conditions of Steinberg are necessary: • Erdos suggested a method of attack in 1991 by asking what is the minimum k such that if G excludes cycles of length 4 up to length k, G is 3-colorable?

4. Progress on Steinberg - Coloring • Results for k = 11, 9, 8, 7. The most recent result within the paradigm of Erdos is that of Borodin et al (2005) who show that k ≤ 7. • Proof Technique for k = 7: Show that every proper 3-coloring of the vertices of any single face of length 8 to 11 in a connected graph G that excludes cycles of length 4 through 7 extends to a proper 3-coloring of G. • The proof of this theorem uses discharging.

5. Discharging • Euler’s Formula: • Give all vertices charge • Give all faces charge except a precolored face, which is given • Goal: We want all vertices and faces to have nonnegative charge (positive charge for precolored face).

6. How do we distribute the charges?

7. Complications

8. One Forbidden Structure: A Tetrad • If G were a minimum counterexample and contained a tetrad, identify vertices x and t. Delete vertices . • If the resulting graph is 3-colorable, then the original graph is 3-colorable. • Excluding tetrads gives faces more charge after discharging, which allows us to prove that such a minimal counterexample does not exist.

9. The k = 7 proof • Consider a minimal counterexample. • Begin with a graph G having a 3-colored face, D. The aim is to prove structural properties. • No separating cycles of length at most 11. • G is 2-connected, vertices of degree 2 must be part of D. • No cycle of length at most 13 has a non-triangular chord, nor does D.

10. Forbidden Structures in the k = 7 Proof

11. Steinberg’s Conjecture and Higher Surfaces • The condition given by Erdos holds for arbitrary surfaces: Zhao showed that for every surface Σ, there exists some k such that if G is a graph embedded on Σ, and G has no cycles of length 4 through k, then G is 3-colorable. • Unfortunately, for surfaces with nonpositive Euler characteristic the bound for k is given by . • Euler characteristic of some simple surfaces are plane = 2, torus = 0, Klein bottle = 0, double torus = -2, etc. • Example: Double torus, k = 35.

12. A Topological Result • We prove that k = 10 with a linear size of minimal exceptions. • Theorem [Thomas, CY]: Let Σ be a surface of Euler genus g. If G is 4-critical and has no cycles of length four through ten, then

13. A similar construction • [Dvorak, Thomas] If G is a 4-critical, triangle free graph drawn on a torus, then

14. Tightness of this bound • On the plane, when cycles of length 4 through 9 are excluded, no 4-critical graphs exist by the result of Borodin et al. • On higher surfaces 4-critical graphs do exist since there are graphs with arbitrarily large girth and chromatic number. • The Hajos construction allows us to construct a sequence of critical graphs with size linear in the genus of G.

15. Intersecting Triangles • By including triangles spaced far apart, we can reduce the cycle restrictions dramatically. • 2010: Borodin and Glebov showed that every planar graph excluding 5-cycles and with minimum distance between triangles at least two is 3-colorable. • The (Strong) Bordeaux Coloring Conjecture: (Borodin and Raspaud) Can we reduce the restriction to only (adjacent) intersecting triangles and 5-cycles?

16. Eared cycles • Whalen showed that planar graphs with no cycles of length 4, 5, 6 or eared cycles of length 7 are 3-colorable.

17. Methodology • Whalen seeks a minimum counterexample and proves the customary pre-colored cycle analogue: • Theorem: Let G be a plane graph with outer cycle of length at most 11. Suppose G excludes cycles of length 4, 5, 6 and eared cycles of length 7. Suppose G does not contain a vertex adjacent to three vertices on the outer cycle. Then any proper 3-coloring of the outer cycle extends to a proper 3-coloring of G.

18. Coloring with an Independent Set and a Forest Can we partition the vertices of a graph into an independent set and a set that induces a forest?

19. How does this relate to Steinberg? Coloring with an independent set and a forest is stronger than 3-coloring. Does there exist a planar graph excluding 4-cycles and 5-cycles that is not colorable with an independent set and a forest? Suggests a possible extension of Steinberg’s conjecuture.

20. Borodin’s Technique • A common technique for solving problems related to Steinberg’s conjecture is to examine what happens inside a precolored cycle. • Colorings inside and outside the cycle do not affect each other. • What about for an independent set and a forest?

21. An extension of Grotzsch’s Theorem Kawarabayashi and Thomassen (2008) proved the following: Let G be a plane graph. Assume that every triangle of G has a vertex v which is on the outer face boundary and such that v is contained in no 4-cycle. Assume also that the distance between any two triangles is at least 5. Then G has chromatic number at most 3. This is proved with the help of a structural theorem in coloring with an independent set and a forest.

22. Our Theorem Theorem: Let G be a planar graph that excludes 5-cycles, 6-cycles and intersecting triangles. Then G is 3-colorable. Proof uses a discharging technique and strategies similar to the proof of Borodin and Glebov.

23. General Strategy Assume that there exists a pre-colored cycle of length 7 through 10 and other basic structural properties hold. Describe a set of discharging rules. Exclude various structures, including tetrads, most 4-cycles and some 7-cycles and 8-cycles. Show that every vertex and face has nonnegative charge, contradicting Euler’s formula.

24. The Discharging Rules, I

25. The Discharging Rules II

26. Excluding a Tetrad

27. Excluding most 4-cycles

28. An 8-cycle argument

29. An 8-cycle argument, continued

30. Reducible Configurations

31. More Reducible Configurations

32. To finish the proof, we must show each of the previous configurations is reducible. All remaining face combinations must have charge at least zero via the discharging rules alone. We get a contradiction via Euler’s formula. Wrap-up

33. Future directions • Cycles of length seven are the more challenging case for Steinberg. • No known counterexample for the Steinberg conditions for an independent set and a set that induces a forest. • Need another idea to push discharging to work for Steinberg’s conjecture. • Other types of coloring?

34. Thank you for your attention! Any questions?