combinatorics and optimization 749 presentation

Havel’s conjecture

John M. Campbell

University of Waterloo

[email protected]

April 06 2015

John M. Campbell (Waterloo) April 06 2015 1 / 46

Overview

This presentation is divided into five sections, given below:

1. The history of Havel’s conjecture.

2. The girth 5 version of Havel’s conjecture.

3. The proof of Havel’s conjecture: some structural results involvingtightness of 4-faces.

4. The proof of Havel’s conjecture: some results on cylindrical grids andwinding numbers.

5. Conclusion.

Overview

5. Conclusion.

Overview

5. Conclusion.

Overview

5. Conclusion.

Overview

5. Conclusion.

Overview

5. Conclusion.

3-colorability of planar graphs

The four color theorem is one of the most famous theorems in the historyof mathematics, and the proof of the four color theorem given by KennethAppel and Wolfgang Haken in 1976 is one of the most famous proofs in thehistory of mathematics.

Theorem (The Four color theorem)Every planar graph is 4-colorable.

QuestionWhat planar graphs are 3-colorable?

DefinitionA decision problem is NP-complete when it is both in NP and NP-hard.

Theorem (Garey, Johnson, & Stockmeyer, 1976)It is NP-complete to decide if a given graph admits a 3-coloring.

Theorem (Dailey, 1980)It is NP-complete to decide if a given planar graph of degree 4 admits a3-coloring.

Grötzsch’s theorem

Grötzsch’s theorem is an important result in graph theory which was provenby Herbert Grötzsch in 1959.

Theorem (Grötzsch’s theorem)Every triangle-free planar graph is 3-colorable.

In 2003, Carsten Thomassen offered another proof of Grötzsch’s theorem.

Theorem (Thomassen, 2003)Every planar graph with girth at least five is 3-list-colorable.

Grötzsch’s theorem follows from the above theorem.

Theorem (Voigt, 1995)There exists a planar graph without 3-cycles which is not 3-choosable.

QuestionIs it possible to preserve 3-colorability while still somehow allowingtriangles?

Grötzsch’s theorem has lead to numerous generalizations involvingnon-triangle-free planar graphs.

Theorem (Aksionov, 1974)Every planar graph with at most three triangles is 3-colorable.

Havel considered generalizing Grötzsch’s theorem to planar graphs withtriangles “far apart”.

The distance between sets of vertices in a graph

Let G be a graph, and let X ,Y ⊆ V (G ).

Let d be the maximum integer such that every path with one end in X andthe other end in Y has length at least d . Then the sets X and Y are saidto be at distance d in G .

Two subgraphs are at distance d if their vertex-sets are at distance d .

Let dO(G ) denote the minimum distance between triangles in a planargraph G .

Let d2(G ) denote the minimum distance between cycles of length at most4 in a planar graph G .

Havel’s conjecture or Havel’s three color conjecture states that there existsa constant d such that a planar graph G with dO(G ) ≥ d is 3-colorable.

This conjecture was considered by was considered by Václav J. Havel asearly as 1969.

Havel proved that if such a constant d exists, then d ≥ 3.

Aksionov and Mel’nikov proved that if such a constant d exists, then d ≥ 4.

A major result suggesting the verity of Havel’s conjecture was made in2003:

Theorem (Borodin & Raspaud, 2003)Every planar graph G with dO(G ) ≥ 4 and no 5-cycles is 3-colorable.

Havel’s conjecture was proven in 2009 by Zdenek Dvořák, Daniel Král’ andRobin Thomas, in the paper Coloring planar graphs with triangles far apart.

Theorem (Dvořák, Král’ & Thomas, 2009)There exists an absolute constant d such that if G is a planar graph andevery two distinct triangles in G are at distance at least d, then G is3-colorable

The original proof of Havel’s conjecture given by Dvořák, Král’ andThomas is largely based upon the following result.

TheoremThere exists an absolute constant K such that the following holds. Let Gbe a planar graph with no separating cycles of length at most four, and letC be an induced facial cycle of G of length at most five or the nullsubgraph of G. Assume that there exists a 3-coloring of C that does notextend to a 3-coloring of G, but extends to every proper subgraph of Gthat includes C. Then ∑

f ∈F (G)|f |≥5

|f | ≤ Kt

where t is the number of triangles in G.

The above theorem is very powerful compared to the correspondinginequality ∑

|f | ≤ K (t + # of 4-cycles)

which was previously proven by Dvořák, Král’ and Thomas. This theorem isalso powerful in that it implies the so-called “girth five version of Havel’sconjecture”:

Girth five version of Havel’s conjecture: There exists a constant Dsuch that a planar graph G with d�(G ) ≥ D is 3-colorable.

The above theorem is also powerful in that it implies that if G has no4-faces, then triangles in G are far apart in the sense described above.

|f | ≤ K (t + # of 4-cycles)

But the above theorem involving the inequality∑|f | ≤ Kt

is not enough to prove Havel’s conjecture:

Main problem: 4-cycles. If G has no 4-cycles, then Havel’s conjectureholds. So 4-cycles are the “main problem” in this sense.

Key idea: 4-cycles have certain “tight” structure, and form a cylindricalgrid.

Let G be a plane graph, and let C0 be either the null graph or an inducedfacial cycle of length at most 5. For our purposes, C0 may be understoodto be the null graph. Let K be an integer such that K − 4 satisfies theconclusion of the above theorem, and let:

d = (2K + 7blog2(K + 8)c+ 28)(K + 1) + 1.

Induction on |V (G )| is used in this proof. Let G be as given above, let φ0be a 3-coloring of C0, and, by way of contradiction, assume that φ0 doesnot extend to a 3-coloring of G . Assume without loss of generality that φ0extends to every proper subgraph of G that includes C0. If G has at mostone triangle, then the theorem follows by a lemma due to Aksionov.Assume that G has at least two triangles. As an inductive hypothesis,assume that the theorem holds for all graphs with strictly less than |V (G )|vertices.

The above inductive hypothesis is used to prove the following claim.

If G has a separating cycle C of length at most five, then C has lengthexactly five and E (C ) includes an edge of a triangle of G .

Let S 6= ∅ denote the set of vertex-sets of all triangles in G . Now observethat 4-cycles are the “key structures” in this argument in the sense that ifG has no 4 faces, then Havel’s conjecture holds by the above theoreminvolving a sum of the form: ∑

f ∈F (G)|f |≥5

|f | ≤ Kt.

The above inductive hypothesis is used to prove the following claim.

If G has a separating cycle C of length at most five, then C has lengthexactly five and E (C ) includes an edge of a triangle of G .

Let S 6= ∅ denote the set of vertex-sets of all triangles in G . Now observethat 4-cycles are the “key structures” in this argument in the sense that ifG has no 4 faces, then Havel’s conjecture holds by the above theoreminvolving a sum of the form: ∑

f ∈F (G)|f |≥5

|f | ≤ Kt.

S-tightness

DefinitionLet G be a graph, let S ⊆ V (G ), and let C be a cycle in G . Then C isS-tight if C has length four and the vertices of C can be numbered v1, v2,v3, v4 in order such that for some integer t ≥ 0 the vertices v1, v2 are atdistance exactly t from S , and the vertices v3, v4 are at distance exactlyt + 1 from S .

S-tightness

For example, let G denote the graph illustrated below.

Let S be the vertex set of the unique triangle in G , and let C be theleftmost 4-cycle in G . It is clear that C is S tight, since the two rightmostvertices of C are at distance exactly 2 from S , and the two leftmostvertices of C are at distance exactly 3 from S .

The following claim is key in that is proves an important “tightness”property of 4-cycles with respect to triangles.

ClaimIf C is a cycle in G of length four at distance at most d − 1 from a setS ∈ S with |V (C ) ∩ V (C0)| ≤ 1 and C shares no edge with a triangle ofG , then either C is S-tight, or V (C )∩ S = ∅ and at least two vertices of Care adjacent to a vertex in S.

The following claim is key in that is proves an important “tightness”property of 4-cycles with respect to triangles.

ClaimIf C is a cycle in G of length four at distance at most d − 1 from a setS ∈ S with |V (C ) ∩ V (C0)| ≤ 1 and C shares no edge with a triangle ofG , then either C is S-tight, or V (C )∩ S = ∅ and at least two vertices of Care adjacent to a vertex in S.

So given a configuration such as the following, there are two possibleoutcomes, as given in the above claim.

Given such a configuration, where C shares no edge with a triangle of G ,then it is possible that C is S-tight as illustrated below.

Now suppose that it is not the case that C is S-tight. By the above claim,it thus follows that V (C ) and S are disjoint and at least two vertices of Care adjacent to a vertex in S , as illustrated below.

For our purposes it is not important to define the integer i0 given below,and for our purposes it is not important to define the quantity εK givenbelow. The following claim will be used to prove the existence of animportant “grid-like” structure.

S will henceforward denote the vertex-set of a fixed triangle in G .

ClaimIf v ∈ V (G ) is at distance i from S , where i0 ≤ i ≤ i0 + εK where εK is acertain quantity defined in terms of K then every face incident with v isS -tight and is not bounded by C0.

Consider a vertex situated in the region given by an inequality of the formi0 ≤ i ≤ i0 + εK .

By the above claim, we know that every face incident with this vertex mustbe S -tight as illustrated below:

The following two claims are also used to prove the existence of theaforementioned “grid-like” structure:

ClaimThere exists an equidistant cycle at distance i0 from S .

ClaimFor every i = 1, 2, . . . , d − 1, every equidistant cycle at distance i from Shas length at most K + 8.

Cylindrical grids

DefinitionLet r , s ∈ N, and let r ≥ 3. An r × s cylindrical grid is the Cartesianproduct of a cycle of length r and a path on s vertices.

For example, the cylindrical grid C52P5 is illustrated in two different waysbelow.

Image taken from Bounds on the Size of the Minimum Dominating Sets of Some Cylindrical Grid Graphs by

M. Nandi, S. Parui & A. Adhikari.John M. Campbell (Waterloo) April 06 2015 25 / 46

An r × (r + 5) cylindrical grid

By the previous three claims, together with a lemma given in Coloringplanar graphs with triangles far apart, we have that the graph G has asubgraph isomorphic to an r × (r + 5) cylindrical grid for some r ≤ K + 8.

We will use a heuristic argument to show how to construct such a gridbased on the above claims.

By the previous three claims, together with a lemma given in Coloringplanar graphs with triangles far apart, we have that the graph G has asubgraph isomorphic to an r × (r + 5) cylindrical grid for some r ≤ K + 8.

We will use a heuristic argument to show how to construct such a gridbased on the above claims.

Recall that a vertex situated in the region given by an inequality of theform i0 ≤ i ≤ i0 + εK yields a “grid” of S -tight 4-cycles. Suppose that thedistances from S of the vertices in this “grid” are as given below.

Recall that there exists an equidistant cycle at distance i0 from S , asillustrated below. But since faces incident with vertices situated in theregion given by the inequality i0 ≤ i ≤ i0 + εK are S -tight, we thus have aconfiguration such as that given below, a cylindircal grid of the formCr2P2.

By S -tightness of 4-cycles, we obtain the subgraph illustrated below, anr × 3 cylindrical grid, i.e. a subgraph isomorphic to Cr2P3.

Continuing in this manner, we obtain an r × (r + 5) cylindrical grid.

Winding number

DefinitionLet C = {v1, v2, . . . , vk} be a cycle in a graph G , where the vertices v1, v2,. . . , vk are listed in clockwise order, and let φ : V (C )→ {1, 2, 3} be a3-coloring of C . The winding number of φ on C is the number of indicesi ∈ [k] such that φ(vi ) = 1 and φ(vi+1) = 2 minus the number of indices isuch that φ(vi ) = 2 and φ(vi+1) = 1, where vk+1 means v1. The windingnumber of φ on C is denoted by wφ(C ).

Winding number

Consider the cycle C = {v1, v2, . . . , v17} illustrated below, where thevertices are listed in clockwise order as illustrated below.

Winding number

Let φ : V (C )→ {1, 2, 3} be the 3-coloring on C illustrated below, lettingvertices labeled 1 be colored red, letting vertices labeled 2 be colored green,and letting vertices labeled 3 be colored blue.

Winding number

(φ(v5), φ(v6)) = (φ(v7), φ(v8)) = (φ(v12), φ(v13)) = (φ(v17), φ(v1)) =(1, 2).

(φ(v6), φ(v7)) = (φ(v13), φ(v14)) = (φ(v16), φ(v17)) = (2, 1).

So the winding number of φ on C is 1.John M. Campbell (Waterloo) April 06 2015 34 / 46

Winding number

PropositionThe sum of the winding numbers of all the face boundaries of a graph G iszero.

For example, consider the 3-coloring on the graph G illustrated below.

Winding number

Now let C1, C2 and C3 respectively denote the face bounaries of Gillustrated below.

Clearly, wφ(C1) = 0, wφ(C2) = 1, and wφ(C3) = −1. We thus have thatwφ(C1) + wφ(C2) + wφ(C3) = 0.

Winding number

PropositionThe winding number of every 3-coloring on a cycle of length four is zero.

Recall that we showed that there exists an r × (r + 5) cylindrical grid in G .

Let D1, . . ., Dr+5 be the hoops of H as illustrated below.

Let ∆1 be the component of R2 − D2 containing D1 as illustrated below.

Let ∆r+5 be the component of R2 − Dr+4 containing Dr+5 as illustratedbelow.

Let G1 be obtained from G by deleting all vertices and edges drawn in theinterior of ∆r+5, and then adding edges into the face bounded by Dr+4 insuch a way that all faces contained in ∆r+5 are bounded by cycles oflength four, except possibly one, and if there is an exceptional face, then itis bounded by a cycle of length five. This is referred to as “thenear-quadrangulation property”.

Let Gr+5 be defined analogously.

By induction φ0 extends to a 3-coloring ψ1 of G1. Similarly, the graphGr+5 has a 3-coloring ψ2. If follows from the near quadrangulationproperty that |wψ1(D1)| ≤ 1 and |wψ2(Dr+5)| ≤ 1.

It shown using the previous propositions that there exists a 3-coloring of Gthat extends to φ0, using the coloring of D1 ∪ Dr+5 obtained by restrictingψ1 to D1 and restricting ψ2 to Dr+5. This is a contradiction.

Conclusion

We conclude with some open problems related to Havel’s conjecture.

ConjectureEvery 3-colorable planar graph is 4-choosable.

RemarkThe above conjecture implies the four color theorem.

ProblemFind a linear-time algorithm to 3-list-color planar graphs of girth 5.

RemarkThomassen’s Proof gives a quadratic algorithm.

Conclusion

We conclude with some open problems related to Havel’s conjecture.

ConjectureEvery 3-colorable planar graph is 4-choosable.

RemarkThe above conjecture implies the four color theorem.

ProblemFind a linear-time algorithm to 3-list-color planar graphs of girth 5.

RemarkThomassen’s Proof gives a quadratic algorithm.

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