Gregory J. Puleo

On a Conjecture of Erdos, Gallai, and Tuza

Gregory J. Puleo

Coordinated Science Lab, UIUC

UIUC Graph Theory SeminarSeptember 23, 2014

How many edges in a triangle-free graph?

Definition

A graph is triangle-free if it has no subgraph isomorphic to K3.

Theorem (Mantel 1907)

If G is a triangle-free n-vertex graph, then G has at most n2/4 edges.Equality holds if and only if G = Kn/2,n/2.

Theorem (Turan 1941)

If G is an n-vertex graph with no copy of Kr , then G has at most(r−2r−1

2 edges.

Equality holds if and only if G = Kn/(r−1),··· ,n/(r−1).

How many edges in a triangle-free graph?

Definition

A graph is triangle-free if it has no subgraph isomorphic to K3.

Theorem (Mantel 1907)

If G is a triangle-free n-vertex graph, then G has at most n2/4 edges.Equality holds if and only if G = Kn/2,n/2.

Theorem (Turan 1941)

If G is an n-vertex graph with no copy of Kr , then G has at most(r−2r−1

2 edges.

Equality holds if and only if G = Kn/(r−1),··· ,n/(r−1).

How hard is it to make a graph bipartite?

Theorem (Erdos 1965)

Any graph G can be made bipartite by removing at most |E (G )| /2 edges.

Proof.

Randomly color each vertex red or blue, each with probability 1/2.Delete all monochromatic edges.

Each edge has probability 1/2 of being deleted, so the expectednumber of deleted edges is |E (G )| /2.

Corollary

If G has n vertices, then G can be made bipartite by deletingat most n2/4 edges. The worst-case graph is Kn.

Proof.

Corollary

Proof.

Corollary

Proof.

Corollary

Hitting sets and triangle-independent sets

Definition

X ⊂ E (G ) is a hitting set if it contains at least one edgefrom each triangle of G .

A ⊂ E (G ) is triangle-independent if it contains at most one edgefrom each triangle of G .

Definition

τ1(G ) is the smallest size of a hitting set in G .

α1(G ) is the largest size of a triangle-independent set in G .

The Erdos bound implies: τ1(G ) ≤ n2/4.Mantel’s Theorem implies: α1(G ) ≤ n2/4.

Definition

The Erdos–Gallai–Tuza Conjecture

Conjecture (Erdos–Gallai–Tuza 1996)

If G is an n-vertex graph, then α1(G ) + τ1(G ) ≤ n2/4.

The conjecture is sharp, if true:

Consider the complete graph Kn, where n is even:

τ1(Kn) =(n2

)− n2/4 = n2/4− n/2,

α1(Kn) = n/2,

Thus α1(Kn) + τ1(Kn) = n2/4.

Next, consider the complete bipartite graph Kn/2,n/2:

τ1(Kn/2,n/2) = 0,

α1(Kn/2,n/2) = n2/4,

Thus α1(Kn/2,n/2) + τ1(Kn/2,n/2) = n2/4.

More generally, the conjecture is sharp for Kr1,r1 ∨ · · · ∨ Krt ,rt .

τ1(Kn) =(n2

)− n2/4 = n2/4− n/2,

α1(Kn) = n/2,

Thus α1(Kn) + τ1(Kn) = n2/4.

τ1(Kn/2,n/2) = 0,

α1(Kn/2,n/2) = n2/4,

Thus α1(Kn/2,n/2) + τ1(Kn/2,n/2) = n2/4.

τ1(Kn) =(n2

)− n2/4 = n2/4− n/2,

α1(Kn) = n/2,

Thus α1(Kn) + τ1(Kn) = n2/4.

τ1(Kn/2,n/2) = 0,

α1(Kn/2,n/2) = n2/4,

Thus α1(Kn/2,n/2) + τ1(Kn/2,n/2) = n2/4.

τ1(Kn) =(n2

)− n2/4 = n2/4− n/2,

α1(Kn) = n/2,

Thus α1(Kn) + τ1(Kn) = n2/4.

τ1(Kn/2,n/2) = 0,

α1(Kn/2,n/2) = n2/4,

Thus α1(Kn/2,n/2) + τ1(Kn/2,n/2) = n2/4.

τ1(Kn) =(n2

)− n2/4 = n2/4− n/2,

α1(Kn) = n/2,

Thus α1(Kn) + τ1(Kn) = n2/4.

τ1(Kn/2,n/2) = 0,

α1(Kn/2,n/2) = n2/4,

Thus α1(Kn/2,n/2) + τ1(Kn/2,n/2) = n2/4.

τ1(Kn) =(n2

)− n2/4 = n2/4− n/2,

α1(Kn) = n/2,

Thus α1(Kn) + τ1(Kn) = n2/4.

τ1(Kn/2,n/2) = 0,

α1(Kn/2,n/2) = n2/4,

Thus α1(Kn/2,n/2) + τ1(Kn/2,n/2) = n2/4.

τ1(Kn) =(n2

)− n2/4 = n2/4− n/2,

α1(Kn) = n/2,

Thus α1(Kn) + τ1(Kn) = n2/4.

τ1(Kn/2,n/2) = 0,

α1(Kn/2,n/2) = n2/4,

Thus α1(Kn/2,n/2) + τ1(Kn/2,n/2) = n2/4.

τ1(Kn) =(n2

)− n2/4 = n2/4− n/2,

α1(Kn) = n/2,

Thus α1(Kn) + τ1(Kn) = n2/4.

τ1(Kn/2,n/2) = 0,

α1(Kn/2,n/2) = n2/4,

Thus α1(Kn/2,n/2) + τ1(Kn/2,n/2) = n2/4.

τ1(Kn) =(n2

)− n2/4 = n2/4− n/2,

α1(Kn) = n/2,

Thus α1(Kn) + τ1(Kn) = n2/4.

τ1(Kn/2,n/2) = 0,

α1(Kn/2,n/2) = n2/4,

Thus α1(Kn/2,n/2) + τ1(Kn/2,n/2) = n2/4.

τ1(Kn) =(n2

)− n2/4 = n2/4− n/2,

α1(Kn) = n/2,

Thus α1(Kn) + τ1(Kn) = n2/4.

τ1(Kn/2,n/2) = 0,

α1(Kn/2,n/2) = n2/4,

Thus α1(Kn/2,n/2) + τ1(Kn/2,n/2) = n2/4.

From τ1 to τB

Definition

Equivalent Definition

τ1(G ) is the smallest size of an edge set X such that G −X is triangle-free.

Definition

τB(G ) is the smallest size of an edge set X such that G − X is bipartite.

Observe that τ1(G ) ≤ τB(G ) ≤ n2/4.

Conjecture

If G is an n-vertex graph, then α1(G ) + τB(G ) ≤ n2/4.

From τ1 to τB

Definition

Conjecture

From τ1 to τB

Definition

Conjecture

From τ1 to τB

Definition

Conjecture

Two approaches to the conjecture

Two obvious ways to try to find partial results:

Find c > 1/4 such that α1(G ) + τ1(G ) ≤ cn2 for all G .Find a class F such that α1(G ) + τ1(G ) ≤ n2/4 for all G ∈ F .

Theorem (P. 2014+)

If G is an n-vertex graph, then α1(G ) + τB(G ) ≤ 5n2/16.

Theorem (P. 2014+)

If G has no induced K−4 , then α1(G ) + τB(G ) ≤ n2/4.

In the rest of the talk, we prove these and other results.K−4

Theorem (P. 2014+)

Two Lemmas

Let b(G ) = max{|B| : G [B] is bipartite}.

If A is triangle-independent, then for any uv ∈ A,the induced subgraph G [NA(u) ∪ NA(v)] isbipartite with dA(u) + dA(v) vertices.

Thus, dA(u) + dA(v) ≤ b(G ).

If A is triangle-independent, then |A| ≤ nb(G)4.

Proof.

For any uv ∈ A, the first lemma gives dA(u) + dA(v) ≤ b(G ).

Summing over all edges in A, we get∑

v dA(v)2 ≤ |A| b(G ).

Cauchy-Schwarz gives∑

v dA(v)2 ≥ 4|A|2n , and the conclusion follows.

Two Lemmas

If A is triangle-independent, then for any uv ∈ A,the induced subgraph G [NA(u) ∪ NA(v)] isbipartite with dA(u) + dA(v) vertices.Thus, dA(u) + dA(v) ≤ b(G ).

Proof.

v dA(v)2 ≤ |A| b(G ).

Two Lemmas

Proof.

v dA(v)2 ≤ |A| b(G ).

Two Lemmas

Proof.

v dA(v)2 ≤ |A| b(G ).

Two Lemmas

Proof.

v dA(v)2 ≤ |A| b(G ).

Two Lemmas

Proof.

v dA(v)2 ≤ |A| b(G ).

α1(G ) + τB(G ) ≤ 5n2/16

Lemma (Erdos–Faudree–Pach–Spencer 1988)

If G is a graph, and B is a vertex set such that G [B] is bipartite, then

τB(G ) ≤ 12

(∣∣∣E (G ) \(B2

)∣∣∣).

Theorem

For any n-vertex graph G , α1(G ) + τB(G ) ≤ 5n2/16.

Proof.

EFPS yields τB(G ) ≤ 12

)−(b(G)

)]≤ n2

4 −b(G)2

Thus, α1(G ) + τB(G ) ≤ nb(G)4 + n2

4 −b(G)2

4 .The right side is maximized when b(G ) = n/2, yieldingα1(G ) + τB(G ) ≤ 5n2/16.

α1(G ) + τB(G ) ≤ 5n2/16

τB(G ) ≤ 12

(∣∣∣E (G ) \(B2

)∣∣∣).Theorem

Proof.

)−(b(G)

)]≤ n2

4 −b(G)2

Thus, α1(G ) + τB(G ) ≤ nb(G)4 + n2

4 −b(G)2

α1(G ) + τB(G ) ≤ 5n2/16

τB(G ) ≤ 12

(∣∣∣E (G ) \(B2

)∣∣∣).Theorem

Proof.

)−(b(G)

)]≤ n2

4 −b(G)2

Thus, α1(G ) + τB(G ) ≤ nb(G)4 + n2

4 −b(G)2

α1(G ) + τB(G ) ≤ 5n2/16

τB(G ) ≤ 12

(∣∣∣E (G ) \(B2

)∣∣∣).Theorem

Proof.

)−(b(G)

)]≤ n2

4 −b(G)2

Thus, α1(G ) + τB(G ) ≤ nb(G)4 + n2

4 −b(G)2

The right side is maximized when b(G ) = n/2, yieldingα1(G ) + τB(G ) ≤ 5n2/16.

α1(G ) + τB(G ) ≤ 5n2/16

τB(G ) ≤ 12

(∣∣∣E (G ) \(B2

)∣∣∣).Theorem

Proof.

)−(b(G)

)]≤ n2

4 −b(G)2

Thus, α1(G ) + τB(G ) ≤ nb(G)4 + n2

4 −b(G)2

Two versions of the conjecture

Definition

A graph is triangular if each of its edges lies in some triangle.

If G is an n-vertex triangular graph, then α1(G ) + τ1(G ) ≤ n2/4.

Later formulations of the conjecture, by both Erdos and Tuza, silentlydropped the “triangular” condition.

Question (Grinberg 2012)

Are these two forms of the conjecture equivalent?

Answer (P. 2014+): YES.

Conjecture (Erdos–Gallai–Tuza 1996 ...sort of)

Definition

Warmup: α1(G ) + τ1(G ) ≤ |E (G )|

Theorem (Erdos–Gallai–Tuza 1996)

In any graph G , α1(G ) + τ1(G ) ≤ |E (G )|.

Proof.

Let A be a largest triangle-independent set, so that α1(G ) = |A|.Let X = E (G )− A.Any triangle of G contains at most one edge of A, thus contains at leasttwo edges of X .Thus, X is a hitting set (and then some!), so that τ1(G ) ≤ |X |.Therefore α1(G ) + τ1(G ) ≤ |A|+ |X | = |E (G )|.

This proof is global, dealing with all edges of G .We localize it, dealing only with the edges of some edge cut [S ,S ].

Warmup: α1(G ) + τ1(G ) ≤ |E (G )|

Proof.

Let A be a largest triangle-independent set, so that α1(G ) = |A|.

Let X = E (G )− A.Any triangle of G contains at most one edge of A, thus contains at leasttwo edges of X .Thus, X is a hitting set (and then some!), so that τ1(G ) ≤ |X |.Therefore α1(G ) + τ1(G ) ≤ |A|+ |X | = |E (G )|.

Warmup: α1(G ) + τ1(G ) ≤ |E (G )|

Proof.

Let A be a largest triangle-independent set, so that α1(G ) = |A|.Let X = E (G )− A.

Any triangle of G contains at most one edge of A, thus contains at leasttwo edges of X .Thus, X is a hitting set (and then some!), so that τ1(G ) ≤ |X |.Therefore α1(G ) + τ1(G ) ≤ |A|+ |X | = |E (G )|.

Warmup: α1(G ) + τ1(G ) ≤ |E (G )|

Proof.

Let A be a largest triangle-independent set, so that α1(G ) = |A|.Let X = E (G )− A.Any triangle of G contains at most one edge of A, thus contains at leasttwo edges of X .

Thus, X is a hitting set (and then some!), so that τ1(G ) ≤ |X |.Therefore α1(G ) + τ1(G ) ≤ |A|+ |X | = |E (G )|.

Warmup: α1(G ) + τ1(G ) ≤ |E (G )|

Proof.

Let A be a largest triangle-independent set, so that α1(G ) = |A|.Let X = E (G )− A.Any triangle of G contains at most one edge of A, thus contains at leasttwo edges of X .Thus, X is a hitting set (and then some!), so that τ1(G ) ≤ |X |.

Therefore α1(G ) + τ1(G ) ≤ |A|+ |X | = |E (G )|.

Warmup: α1(G ) + τ1(G ) ≤ |E (G )|

Proof.

Let A be a largest triangle-independent set, so that α1(G ) = |A|.Let X = E (G )− A.Any triangle of G contains at most one edge of A, thus contains at leasttwo edges of X .Thus, X is a hitting set (and then some!), so that τ1(G ) ≤ |X |.Therefore α1(G ) + τ1(G ) ≤ |A|+ |X | = |E (G )|.

This proof is global, dealing with all edges of G .We localize it, dealing only with the edges of some edge cut [S , S ].

Edge cuts

Lemma (P. 2014+)

For any S ⊂ V (G ),α1(G ) + τ1(G ) ≤ α1(G [S ]) + τ1(G [S ]) + α1(G [S ]) + τ1(G [S ]) +

∣∣[S , S ]∣∣ .

Proof.

Let A be any triangle-independent set in G .Let X1, X2 be smallest hitting sets in G [S ], G [S ] respectively.Let A1 = A ∩ E (G [S ]), A2 = A ∩ E (G [S ]), C = A ∩ [S ,S ].Have |A1|+ |X1| ≤ α1(G [S ]) + τ1(G [S ]) and likewise for S .X1 ∪ X2 hits all triangles except those with vertices in both S and S .Let X = X1 ∪ X2 ∪ ([S , S ]− C ); now X hits all triangles.

Edge cuts

Lemma (P. 2014+)

∣∣[S , S ]∣∣ .

Proof.

Let A be any triangle-independent set in G .

Let X1, X2 be smallest hitting sets in G [S ], G [S ] respectively.Let A1 = A ∩ E (G [S ]), A2 = A ∩ E (G [S ]), C = A ∩ [S ,S ].Have |A1|+ |X1| ≤ α1(G [S ]) + τ1(G [S ]) and likewise for S .X1 ∪ X2 hits all triangles except those with vertices in both S and S .Let X = X1 ∪ X2 ∪ ([S ,S ]− C ); now X hits all triangles.

Edge cuts

Lemma (P. 2014+)

∣∣[S , S ]∣∣ .

Proof.

Let A be any triangle-independent set in G .Let X1, X2 be smallest hitting sets in G [S ], G [S ] respectively.

Let A1 = A ∩ E (G [S ]), A2 = A ∩ E (G [S ]), C = A ∩ [S ,S ].Have |A1|+ |X1| ≤ α1(G [S ]) + τ1(G [S ]) and likewise for S .X1 ∪ X2 hits all triangles except those with vertices in both S and S .Let X = X1 ∪ X2 ∪ ([S ,S ]− C ); now X hits all triangles.

Edge cuts

Lemma (P. 2014+)

∣∣[S , S ]∣∣ .

Proof.

Let A be any triangle-independent set in G .Let X1, X2 be smallest hitting sets in G [S ], G [S ] respectively.Let A1 = A ∩ E (G [S ]), A2 = A ∩ E (G [S ]), C = A ∩ [S ,S ].

Have |A1|+ |X1| ≤ α1(G [S ]) + τ1(G [S ]) and likewise for S .X1 ∪ X2 hits all triangles except those with vertices in both S and S .Let X = X1 ∪ X2 ∪ ([S ,S ]− C ); now X hits all triangles.

Edge cuts

Lemma (P. 2014+)

∣∣[S , S ]∣∣ .

Proof.

Let A be any triangle-independent set in G .Let X1, X2 be smallest hitting sets in G [S ], G [S ] respectively.Let A1 = A ∩ E (G [S ]), A2 = A ∩ E (G [S ]), C = A ∩ [S ,S ].Have |A1|+ |X1| ≤ α1(G [S ]) + τ1(G [S ]) and likewise for S .

X1 ∪ X2 hits all triangles except those with vertices in both S and S .Let X = X1 ∪ X2 ∪ ([S ,S ]− C ); now X hits all triangles.

Edge cuts

Lemma (P. 2014+)

∣∣[S , S ]∣∣ .

Proof.

Let A be any triangle-independent set in G .Let X1, X2 be smallest hitting sets in G [S ], G [S ] respectively.Let A1 = A ∩ E (G [S ]), A2 = A ∩ E (G [S ]), C = A ∩ [S ,S ].Have |A1|+ |X1| ≤ α1(G [S ]) + τ1(G [S ]) and likewise for S .X1 ∪ X2 hits all triangles except those with vertices in both S and S .

Let X = X1 ∪ X2 ∪ ([S ,S ]− C ); now X hits all triangles.

Edge cuts

Lemma (P. 2014+)

∣∣[S , S ]∣∣ .

Proof.

Let A be any triangle-independent set in G .Let X1, X2 be smallest hitting sets in G [S ], G [S ] respectively.Let A1 = A ∩ E (G [S ]), A2 = A ∩ E (G [S ]), C = A ∩ [S ,S ].Have |A1|+ |X1| ≤ α1(G [S ]) + τ1(G [S ]) and likewise for S .X1 ∪ X2 hits all triangles except those with vertices in both S and S .Let X = X1 ∪ X2 ∪ ([S , S ]− C ); now X hits all triangles.

Minimum Counterexamples have Dense Cuts

Lemma (P. 2014+)

∣∣[S , S ]∣∣ .

Theorem

If G is a vertex-minimal graph such that α1(G ) + τ1(G ) > n2/4, then forall proper nonempty S ⊂ V (G ), we have

∣∣[S ,S ]∣∣ > |S | (n − |S |)/2.

Proof.

Lemma + minimality gives

α1(G ) + τ1(G ) ≤ |S |2 /4 + (n − |S |)2/4 +∣∣[S , S ]

∣∣

= n2/4− |S | (n − |S |)/2 +∣∣[S ,S ]

∣∣ .

Since α1(G ) + τ1(G ) > n2/4, the claim follows.

Lemma (P. 2014+)

∣∣[S , S ]∣∣ .

Theorem

∣∣[S ,S ]∣∣ > |S | (n − |S |)/2.

Proof.

α1(G ) + τ1(G ) ≤ |S |2 /4 + (n − |S |)2/4 +∣∣[S , S ]

∣∣

= n2/4− |S | (n − |S |)/2 +∣∣[S ,S ]

∣∣ .

Lemma (P. 2014+)

∣∣[S , S ]∣∣ .

Theorem

∣∣[S ,S ]∣∣ > |S | (n − |S |)/2.

Proof.

α1(G ) + τ1(G ) ≤ |S |2 /4 + (n − |S |)2/4 +∣∣[S , S ]

∣∣

= n2/4− |S | (n − |S |)/2 +∣∣[S , S ]

∣∣ .Since α1(G ) + τ1(G ) > n2/4, the claim follows.

Lemma (P. 2014+)

∣∣[S , S ]∣∣ .

Theorem

∣∣[S ,S ]∣∣ > |S | (n − |S |)/2.

Proof.

α1(G ) + τ1(G ) ≤ |S |2 /4 + (n − |S |)2/4 +∣∣[S , S ]

∣∣= n2/4− |S | (n − |S |)/2 +

∣∣[S , S ]∣∣ .

Minimum Degree in Minimum Counterexamples

Theorem

∣∣[S ,S ]∣∣ > |S | (n − |S |)/2.

Improvement

If G is a vertex-minimal counterexample, then δ(G ) > n/2.

Thus, G is triangular.

Corollary

If α1(G ) + τ1(G ) ≤ n2/4 for all triangular graphs G , thenα1(G ) + τ1(G ) ≤ n2/4 for all graphs G .

That is, the two forms of the conjecture are equivalent.

Theorem

∣∣[S ,S ]∣∣ > |S | (n − |S |)/2.

Improvement

If G is a vertex-minimal counterexample, then δ(G ) > n/2.Thus, G is triangular.

Corollary

Theorem

∣∣[S ,S ]∣∣ > |S | (n − |S |)/2.

Improvement

If G is a vertex-minimal counterexample, then δ(G ) > n/2.Thus, G is triangular.

Corollary

Dense Cuts for τB

For any triangle-independent A and any S ⊂ V (G ),α1(G ) + τB(G ) ≤ α1(G [S ]) + τB(G [S ]) + α1(G [S ]) + τB(G [S ])+

∣∣[S ,S ]∣∣+∣∣A ∩ [S ,S ]

∣∣ .

Proof.

Suffices to show that τB(G ) ≤ τB(G [S ]) + τB(G [S ]) + 12

∣∣[S ,S ]∣∣.

Given bipartitions of S and S , consider the two ways to combine themyielding a bipartition of V (G ).

Dense Cuts for τB

∣∣[S ,S ]∣∣+∣∣A ∩ [S ,S ]

∣∣ .Proof.

∣∣[S ,S ]∣∣.

Dense Cuts for τB

∣∣[S ,S ]∣∣+∣∣A ∩ [S ,S ]

∣∣ .Proof.

∣∣[S ,S ]∣∣.

Dense Cuts for τB

∣∣[S ,S ]∣∣+∣∣A ∩ [S ,S ]

∣∣ .Proof.

∣∣[S ,S ]∣∣.

Clique Cuts

Corollary

If G is a vertex-minimal graph with α1(G ) + τB(G ) > n2/4 and A istriangle-independent, then for any nonempty proper S ⊂ V (G ), we have12

∣∣[S , S ]∣∣+∣∣A ∩ [S , S ]

∣∣ > |S | (n − |S |)/2.

Corollary

If G is a vertex-minimal graph with α1(G ) + τB(G ) > n2/4, if A istriangle-independent, and if S is a clique, then

∣∣[S ,S ]∣∣ > (|S |− 2)(n−|S |).

Proof.

Since S is a clique, each vertex v ∈ S has at most one A-neighbor in S .Thus,

∣∣A ∩ [S ,S ]∣∣ ≤ n − |S |.

S vcan’t happen:

Clique Cuts

Corollary

∣∣[S , S ]∣∣+∣∣A ∩ [S , S ]

∣∣ > |S | (n − |S |)/2.

Corollary

∣∣[S , S ]∣∣ > (|S |− 2)(n−|S |).

Proof.

∣∣A ∩ [S ,S ]∣∣ ≤ n − |S |.

S vcan’t happen:

Clique Cuts

Corollary

∣∣[S , S ]∣∣+∣∣A ∩ [S , S ]

∣∣ > |S | (n − |S |)/2.

Corollary

∣∣[S , S ]∣∣ > (|S |− 2)(n−|S |).

Proof.

∣∣A ∩ [S , S ]∣∣ ≤ n − |S |.

S vcan’t happen:

Clique Cuts

Corollary

∣∣[S , S ]∣∣+∣∣A ∩ [S , S ]

∣∣ > |S | (n − |S |)/2.

Corollary

∣∣[S , S ]∣∣ > (|S |− 2)(n−|S |).

Proof.

∣∣A ∩ [S , S ]∣∣ ≤ n − |S |.

S vcan’t happen:

K−4 -free Graphs

Theorem

If G is K−4 -free, then α1(G ) + τB(G ) ≤ n2/4.

Proof (by contradiction).

First, show that α1(G ) + τB(G ) ≤ n2/4 when G is triangle-free.Among the K−4 -free graphs with α1(G ) + τB(G ) > n2/4, take Gvertex-minimal. Note that G is vertex-minimal among all graphs.Let S be a maximum clique, so that |S | ≥ 3.Corollary gives

∣∣[S ,S ]∣∣ > (|S | − 2)(n − |S |), so some v ∈ S has

dS(v) ≥ |S | − 1.Maximality forces dS(v) = |S | − 1, so this gives an induced K−4 .Contradiction!

K−4 -free Graphs

Theorem

First, show that α1(G ) + τB(G ) ≤ n2/4 when G is triangle-free.

Among the K−4 -free graphs with α1(G ) + τB(G ) > n2/4, take Gvertex-minimal. Note that G is vertex-minimal among all graphs.Let S be a maximum clique, so that |S | ≥ 3.Corollary gives

K−4 -free Graphs

Theorem

First, show that α1(G ) + τB(G ) ≤ n2/4 when G is triangle-free.Among the K−4 -free graphs with α1(G ) + τB(G ) > n2/4, take Gvertex-minimal. Note that G is vertex-minimal among all graphs.

Let S be a maximum clique, so that |S | ≥ 3.Corollary gives

K−4 -free Graphs

Theorem

First, show that α1(G ) + τB(G ) ≤ n2/4 when G is triangle-free.Among the K−4 -free graphs with α1(G ) + τB(G ) > n2/4, take Gvertex-minimal. Note that G is vertex-minimal among all graphs.Let S be a maximum clique, so that |S | ≥ 3.

Corollary gives∣∣[S ,S ]

∣∣ > (|S | − 2)(n − |S |), so some v ∈ S hasdS(v) ≥ |S | − 1.Maximality forces dS(v) = |S | − 1, so this gives an induced K−4 .Contradiction!

K−4 -free Graphs

Theorem

dS(v) ≥ |S | − 1.

Maximality forces dS(v) = |S | − 1, so this gives an induced K−4 .Contradiction!

K−4 -free Graphs

Theorem

α1(G ) + τB(G ) ≤ n2/4 for triangle-free G

If G is an n-vertex triangle-free graph with m edges, then

τB(G ) ≤ m − 4m2

Corollary

If G is triangle-free, then α1(G ) + τB(G ) ≤ n2/4.

Proof.

The lemma yields

α1(G ) + τB(G ) ≤ 2m − 4m2

The upper bound is maximized when m = n2/4, yielding

α1(G ) + τB(G ) ≤ n2/4.

τB(G ) ≤ m − 4m2

Corollary

Proof.

The lemma yields

α1(G ) + τB(G ) ≤ 2m − 4m2

α1(G ) + τB(G ) ≤ n2/4.

τB(G ) ≤ m − 4m2

Corollary

Proof.

The lemma yields

α1(G ) + τB(G ) ≤ 2m − 4m2

α1(G ) + τB(G ) ≤ n2/4.

Acknowledgments

This research wasperformed while I was agraduate student at theUIUC math department.

Thank you for coming!

Acknowledgments

This research wasperformed while I was agraduate student at theUIUC math department.

Thank you for coming!

Gregory J. Puleo

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