aa r X i v : . [ m a t h . C O ] M a y A note on a conjecture of Gy´arf´as
Ryan R. Martin ∗ Abstract
This note proves that, given one member, T , of a particular familyof radius-three trees, every radius-two, triangle-free graph, G , with largeenough chromatic number contains an induced copy of T . A ground-breaking theorem by Erd˝os [1] states that for any positive integers χ and g , there exists a graph with chromatic number at least χ and girth at least g . This has an important corollary. Let H be a fixed graph which contains acycle and let χ be a fixed positive integer. Then there exists a G such that χ ( G ) > χ and G does not contain H as a subgraph.Gy´arf´as [2] and Sumner [9] independently conjectured the following: Conjecture 1.1.
For every integer k and tree T there is an integer f ( k, T ) such that every G with ω ( G ) ≤ k and χ ( G ) ≥ f ( k, T ) contains an induced copy of T . Of course, an acyclic graph need not be a tree. But, Conjecture 1.1 is thesame if we replace T , by F where F is a forest. Suppose F = T + · · · + T p where each T i is a tree, then we can see by induction on both k and p that f ( k, F ) ≤ p + | V ( F ) | f ( k − , F ) + max ≤ i ≤ p { f ( k, T i ) } . A similar proof is given in [4]. Thus, it is sufficient to prove Conjecture 1.1 fortrees, as stated. ∗ Department of Mathematics, Iowa State University, Ames, IA 50011. The author partiallysupported by the Clay Mathematics Institute. email: [email protected] T Figure 2: A radius three tree covered in [7].
The first major progress on this problem came from Gy´arf´as, Szemer´edi andTuza [3] who proved the case when k = 3 and T is either a radius two treeor a so-called “mop.” A mop is a graph which is path with a star at the end.Kierstead and Penrice [4] proved the conjecture for k = 3 and when T is thegraph in Figure 1.The breakthrough for k > T is a radius two tree and k is anypositive integer. This result contains the one in [3]. Furthermore, Kierstead andZhu [7] prove the conjecture true for a certain class of radius three trees. Thesetrees are those with all vertices adjacent to the root having degree 2 or less. Agood example of such a tree is in Figure 2. The paper [7] contains the resultin [4].Scott [8] proved the following theorem: Theorem 1.2 (Scott) . For every integer k and tree T there is an integer f ( k, T ) such that every G with ω ( G ) ≤ k and χ ( G ) ≥ f ( k, T ) contains a subdivision of T as an induced subgraph. Theorem 1.2 results in an easy corollary:
Corollary 1.3 (Scott) . Conjecture 1.1 is true if T is a subdivision of a starand k is any positive integer. Kierstead and Rodl [6] discuss why Conjecture 1.1 does not generalize wellto directed graphs. 2igure 3: T (4 , T (5 , , In order to prove the theorem, we must define some specific trees. In general, let T ( a, b ) denote the radius two tree in which the root has a children and each ofthose children itself has exactly b children. (Thus, T ( a, b ) has 1+ a + ab vertices.)In particular, T ( t,
2) is the radius two tree for which the root has t children andeach neighbor of the root has 2 children. Figure 3 gives a drawing of T (4 , T ( t, ,
1) be the radius three tree in which the root has t children, eachneighbor of the root has 2 children, each vertex at distance two from the roothas 1 child and each vertex at distance three from the root is a leaf. Figure 4gives a drawing of T (5 , , Theorem 2.1.
Let t be a positive integer. There exists a function f , such thatif G is a radius two graph with no triangles and χ ( G ) > f ( t ) , then G must have T ( t, , as an induced subgraph. Proof.
We will let r be the root of G and let S = S ( r,
1) be the neighbors of r and S = S ( r,
2) be the second neighborhood of r . We will try to create a T ( t, ,
1) with a root r vertex by vertex. We look for a v ∈ S with the propertythat there exist w a , w a ∈ N S ( v ) as well as x a ∈ N S ( w a ) \ N S ( w b ) = ∅ and x b ∈ N S ( w b ) \ N S ( w a ) = ∅ such that x a x b . So, clearly, { v , w a , w b , x a , x b } induce the tree T (2 , G to create G : { v , w a , w b , x a , x b } ∪ N S ( v ) ∪ N ( w a ) ∪ N ( w b ) ∪ N ( x a ) ∪ N ( x b ) . Since G has no triangles, the graph induced by these vertices has chromaticnumber at most 4. Thus, χ ( G ) ≥ χ ( G ) − One such coloring is (1) N S ( w a ) ∪ N S ( x a ), (2) N S ( w b ) ∪ N S ( x b ), (3) N S ( v ) and(4) { v , x a , x b } .
3e continue to find v , . . . , v s from each of G , . . . , G s in the same mannerwith s < t so that G has an induced T ( s, ,
1) rooted at r . We also have a G s +1 so that χ ( G s +1 ) ≥ χ ( G ) − s . If we can continue this process to thepoint that s = t , we have our T ( t, ,
1) rooted at r . So, let us suppose that theprocess stops for some s < t . From this point forward, S will actually denote S ∩ V ( G s +1 ) and S will denote S ∩ V ( G s +1 ).Furthermore, in the graph G s +1 , each vertex v ∈ S has the followingproperty: For any w a , w b ∈ N ( v ), the pair( N S ( w a ) \ N S ( w b ) , N S ( w b ) \ N S ( w a ))induces a complete bipartite graph. If this were not the case, then we could findthe x a and x b that we need.Consider this property in reverse. Let v ∈ S and z , z ∈ S \ N S ( v ). Thenthe two sets N S ( v ) ∩ N ( z ) and N S ( v ) ∩ N ( z ) have the property that oneis inside the other or they are disjoint. As a result, N S ( v ) has two nonemptysubsets such that any z ∈ S \ N S ( v ) has the property that N S ( v ) ∩ N ( z )contains either one subset or the other.So, for each v ∈ S , there exists some (not necessarily unique and not neces-sarily distinct) pair of vertices, w a ( v ) , w b ( v ) ∈ N S ( v ) such that for all z ∈ S ,if z is adjacent to some member of N S ( v ) then either z ∼ w a ( v ) or z ∼ w b ( v )or both.For every v ∈ S , find such vertices and label them, arbitrarily as w a ( v ) or w b ( v ), recognizing that a vertex can have many labels. Now form the graph H ∗ induced by vertices from among those labelled as some w a ( v ) or w b ( v ). Find aminimal induced subgraph H so that if h ∗ ∈ V ( H ∗ ), then there exists h ∈ V ( H )such that N S ( h ∗ ) ⊆ N S ( h ).We have a series of claims that end the proof: Claim 1. χ ( H ) = χ ( S ). Proof of Claim 1.
Since H is a subgraph of S , χ ( H ) ≤ χ ( S ). If we properlycolor H with χ ( H ) colors, then we can extend this to a coloring of S . We dothis by giving z ∈ S the same color as that of some h ∈ V ( H ) with the propertythat N S ( z ) ⊆ N S ( h ).This is possible first because there must be some h ∗ = w A ( v ) or h ∗ = w B ( v )in H ∗ with N S ( z ) ⊆ N S ( h ∗ ). Further, there is an h such that N S ( h ∗ ) ⊆ N S ( h ). So, N S ( z ) ⊆ N S ( h ). Now suppose z and z are given the same colorbut are adjacent. Let h and h be the vertices in H whose neighborhoodsdominate those of z and z , respectively and whose colors z and z inherit.Because z ∼ z , h ∼ z and h ∼ z . But then it must also be the case that4 (2,8)z(1,1) z(1) z’ z(2) Figure 5: T (2 ,
8) with some vertices labelled h ∼ h . Thus, h and h cannot receive the same color, a contradiction. (cid:4) Claim 2. H induces a T (2 t + 1 , Proof of Claim 2.
Because S is an independent set, χ ( S ) ≥ χ ( G s +1 ) − χ ( G ), hence χ ( G s +1 ), is large, Claim 1 ensures that χ ( H ) is large.Claim 2 results from [3], because T (2 t + 1 ,
8) is a radius-two tree. (cid:4)
Let the tree T , guaranteed by Claim 2, have root z ′ , its children be labelled z (1) , . . . , z (2 t + 1) and the children of each z ( i ) be labelled z ( i, , . . . , z ( i, Claim 3. If v ∈ S is adjacent to z ( i, j ), then v cannot be adjacent to anyother vertices of T except one other vertex z ( i, j ′ ) or z ′ . Proof of Claim 3. If v ∈ S is adjacent to, say, z (1 , v z ( i, j ) if i = 1. This is because N S ( w A ( v )) △ N S ( w B ( v )) induces a complete bipartitegraph which would imply an edge between z (1) and z ( i ).It can be shown, for similar reasons, that if v ∼ z (1 , v z ( i ) forany i = 1. Also, v z (1) because G is triangle-free. (cid:4) Claim 4.
We may assume that there is a v ∈ S that is adjacent to (withoutloss of generality) z (1 ,
1) as well as z ′ . Proof of Claim 4.
We prove this by contradiction. Applying Claim 3 to everyleaf of T , we see that since Claim 4 is not true, then for i = 1 , . . . , t + 1, wecan find a set of 4 vertices of the form z ( i, j ) and 4 vertices from S so thatthey induce a perfect matching. Furthermore, the 4(2 t + 1) vertices from S areeach adjacent to no other vertices of T , because of Claim 3. Hence, we have ourinduced T ( t, , (cid:4) Because our definition of H guaranteed that vertices had neighborhoods thatwere not nested, there must be some z ′′ ∈ S that is adjacent to z (1 ,
1) but not z ′ . Call this vertex z ′′ . Claim 5.
For any z ( i, j ) with i = 1 and any v ∈ S adjacent to z ( i, j ), v cannotbe adjacent to both z ′ and z ′′ . Proof of Claim 5.
We again proceed by contradiction, supposing that v ∼ z ( i, j ) , z ′ , z ′′ . There is, without loss of generality, w a ( v ) ∈ N S ( v ) such that5 S ( z ′′ ) ⊆ N S ( w a ( v )). Thus, either N S ( z ′ ) ⊆ N S ( w a ( v )) or N S ( z ( i, j )) ⊆ N S ( w a ( v )). But if w a ( v ) were deleted from H ∗ to form H , either z ′ or z ( i, j )would have been deleted as well.Therefore, either w a ( v ) = z ′ or w a ( v ) = z ( i, j ). So, N S ( z ′′ ) ⊆ N S ( z ′ ) or N S ( z ′′ ) ⊆ N S ( z ( i, j )). We can conclude that either z ′ ∼ z (1 ,
1) or z ( i, j ) ∼ z (1 , T is an induced subtree. (cid:4) Claim 6.
For all i = 1, z ′′ is adjacent to z ( i ) but no vertex z ( i, j ). Proof of Claim 6.
Note that z (2) , . . . , z (2 t + 1) are adjacent to z ′ but not z (1 , N S ( z ′ ) △ N S ( z (1 , z ′′ must be adjacent to z (2) , . . . , z (2 t + 1). Because G is triangle-free, z ′′ cannot be adjacent to any vertex of the form z ( i, j ) where i = 1. (cid:4) Now we construct the tree we need. For each z ( i, j ), i = 1, find a vertex v ( i, j ) ∈ S to which z ( i, j ) is adjacent. According to Claim 3, no v ( i, j ) vertexcan be adjacent to any vertex of V ( T ) \ { z ′ } and, according to Claim 5, it isadjacent to at most one of { z ′ , z ′′ } .For each i ∈ { , . . . , t + 1 } , the majority of { v ( i, , . . . , v ( i, } have that v ( i, j ) is either nonadjacent to z ′ or nonadjacent to z ′′ . Without loss of gener-ality, we conclude that z ′ has the property that, for i = 2 , . . . , t + 1, the vertices v ( i, , . . . , v ( i,
4) fail to be adjacent to z ′ .Since any vertex of S can be adjacent to at most two vertices of H , thenfor i = 2 , . . . , t + 1, |{ v ( i, , . . . , v ( i, }| ≥
2. Therefore, we assume that foreach i ∈ { , . . . , t + 1 } , v ( i,
1) and v ( i,
2) are distinct. But now the vertex set { z ′ } ∪ t +1 [ i =2 ( { z ( i ) , z ( i, , z ( i, , v ( i, , v ( i, } )induces T ( t, , (cid:4) Very many thanks to an anonymous referee who identified a significant error inthe first version of this paper.
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