On a Conjecture of Erdos, Gallai, and Tuza - U.I.U.C. Math
On a Conjecture of Erd˝os, Gallai, and Tuza Gregory J. Puleo Coordinated Science Lab, UIUC
UIUC Graph Theory Seminar September 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 .
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 (Tur´an 1941) If G is an n-vertex graph with no copy of Kr , then G has at most r −2 n2 r −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 (Erd˝os 1965) Any graph G can be made bipartite by removing at most |E (G )| /2 edges.
How hard is it to make a graph bipartite? Theorem (Erd˝os 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.
How hard is it to make a graph bipartite? Theorem (Erd˝os 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 expected number of deleted edges is |E (G )| /2.
How hard is it to make a graph bipartite? Theorem (Erd˝os 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 expected number of deleted edges is |E (G )| /2.
Corollary If G has n vertices, then G can be made bipartite by deleting at most n2 /4 edges. The worst-case graph is Kn .
Hitting sets and triangle-independent sets
Definition X ⊂ E (G ) is a hitting set if it contains at least one edge from each triangle of G . A ⊂ E (G ) is triangle-independent if it contains at most one edge from each triangle of G .
Hitting sets and triangle-independent sets
Definition X ⊂ E (G ) is a hitting set if it contains at least one edge from each triangle of G . A ⊂ E (G ) is triangle-independent if it contains at most one edge from 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 .
Hitting sets and triangle-independent sets
Definition X ⊂ E (G ) is a hitting set if it contains at least one edge from each triangle of G . A ⊂ E (G ) is triangle-independent if it contains at most one edge from 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 Erd˝os bound implies: τ1 (G ) ≤ n2 /4. Mantel’s Theorem implies: α1 (G ) ≤ n2 /4.
The Erd˝os–Gallai–Tuza Conjecture Conjecture (Erd˝os–Gallai–Tuza 1996) If G is an n-vertex graph, then α1 (G ) + τ1 (G ) ≤ n2 /4. The conjecture is sharp, if true:
The Erd˝os–Gallai–Tuza Conjecture Conjecture (Erd˝os–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:
The Erd˝os–Gallai–Tuza Conjecture Conjecture (Erd˝os–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 ) =
n 2
− n2 /4 = n2 /4 − n/2,
The Erd˝os–Gallai–Tuza Conjecture Conjecture (Erd˝os–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 ) =
n 2
− n2 /4 = n2 /4 − n/2,
α1 (Kn ) = n/2,
The Erd˝os–Gallai–Tuza Conjecture Conjecture (Erd˝os–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 ) =
n 2
− n2 /4 = n2 /4 − n/2,
α1 (Kn ) = n/2, Thus α1 (Kn ) + τ1 (Kn ) = n2 /4.
The Erd˝os–Gallai–Tuza Conjecture Conjecture (Erd˝os–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 ) =
n 2
− 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 :
The Erd˝os–Gallai–Tuza Conjecture Conjecture (Erd˝os–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 ) =
n 2
− 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,
The Erd˝os–Gallai–Tuza Conjecture Conjecture (Erd˝os–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 ) =
n 2
− 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,
The Erd˝os–Gallai–Tuza Conjecture Conjecture (Erd˝os–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 ) =
n 2
− 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.
The Erd˝os–Gallai–Tuza Conjecture Conjecture (Erd˝os–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 ) =
n 2
− 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 .
From τ1 to τB Definition τ1 (G ) is the smallest size of a hitting set in G .
Equivalent Definition τ1 (G ) is the smallest size of an edge set X such that G − X is triangle-free.
From τ1 to τB Definition τ1 (G ) is the smallest size of a hitting set in G .
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.
From τ1 to τB Definition τ1 (G ) is the smallest size of a hitting set in G .
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.
From τ1 to τB Definition τ1 (G ) is the smallest size of a hitting set in G .
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.
Two approaches to the conjecture Conjecture (Erd˝os–Gallai–Tuza 1996) If G is an n-vertex graph, then α1 (G ) + τ1 (G ) ≤ n2 /4. 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.
Two approaches to the conjecture Conjecture (Erd˝os–Gallai–Tuza 1996) If G is an n-vertex graph, then α1 (G ) + τ1 (G ) ≤ n2 /4. 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.
Two approaches to the conjecture Conjecture (Erd˝os–Gallai–Tuza 1996) If G is an n-vertex graph, then α1 (G ) + τ1 (G ) ≤ n2 /4. 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 K4− , then α1 (G ) + τB (G ) ≤ n2 /4. In the rest of the talk, we prove these and other results.
K4−
Two Lemmas Let b(G ) = max{|B| : G [B] is bipartite}.
Lemma u
v
If A is triangle-independent, then for any uv ∈ A, the induced subgraph G [NA (u) ∪ NA (v )] is bipartite with dA (u) + dA (v ) vertices.
Two Lemmas Let b(G ) = max{|B| : G [B] is bipartite}.
Lemma u
v
If A is triangle-independent, then for any uv ∈ A, the induced subgraph G [NA (u) ∪ NA (v )] is bipartite with dA (u) + dA (v ) vertices. Thus, dA (u) + dA (v ) ≤ b(G ).
Two Lemmas Let b(G ) = max{|B| : G [B] is bipartite}.
Lemma u
v
If A is triangle-independent, then for any uv ∈ A, the induced subgraph G [NA (u) ∪ NA (v )] is bipartite with dA (u) + dA (v ) vertices. Thus, dA (u) + dA (v ) ≤ b(G ).
Lemma If A is triangle-independent, then |A| ≤
nb(G ) 4.
Two Lemmas Let b(G ) = max{|B| : G [B] is bipartite}.
Lemma u
v
If A is triangle-independent, then for any uv ∈ A, the induced subgraph G [NA (u) ∪ NA (v )] is bipartite with dA (u) + dA (v ) vertices. Thus, dA (u) + dA (v ) ≤ b(G ).
Lemma 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 ).
Two Lemmas Let b(G ) = max{|B| : G [B] is bipartite}.
Lemma u
v
If A is triangle-independent, then for any uv ∈ A, the induced subgraph G [NA (u) ∪ NA (v )] is bipartite with dA (u) + dA (v ) vertices. Thus, dA (u) + dA (v ) ≤ b(G ).
Lemma 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 ). P Summing over all edges in A, we get v dA (v )2 ≤ |A| b(G ).
Two Lemmas Let b(G ) = max{|B| : G [B] is bipartite}.
Lemma u
v
If A is triangle-independent, then for any uv ∈ A, the induced subgraph G [NA (u) ∪ NA (v )] is bipartite with dA (u) + dA (v ) vertices. Thus, dA (u) + dA (v ) ≤ b(G ).
Lemma 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 ). P Summing over all edges in A, we get v dA (v )2 ≤ |A| b(G ). 2 P Cauchy-Schwarz gives v dA (v )2 ≥ 4|A| n , and the conclusion follows.
α1 (G ) + τB (G ) ≤ 5n2 /16 Lemma (Erd˝os–Faudree–Pach–Spencer 1988) If G is a graph, and B is avertex set such that G [B] is bipartite, then 1 τB (G ) ≤ 2 E (G ) \ B2 .
α1 (G ) + τB (G ) ≤ 5n2 /16 Lemma (Erd˝os–Faudree–Pach–Spencer 1988) If G is a graph, and B is avertex set such that G [B] is bipartite, then 1 τB (G ) ≤ 2 E (G ) \ B2 .
Theorem For any n-vertex graph G , α1 (G ) + τB (G ) ≤ 5n2 /16.
α1 (G ) + τB (G ) ≤ 5n2 /16 Lemma (Erd˝os–Faudree–Pach–Spencer 1988) If G is a graph, and B is avertex set such that G [B] is bipartite, then 1 τB (G ) ≤ 2 E (G ) \ B2 .
Theorem For any n-vertex graph G , α1 (G ) + τB (G ) ≤ 5n2 /16.
Proof. EFPS yields τB (G ) ≤
1 2
h n 2
−
i
b(G ) 2
≤
n2 4
−
b(G )2 4 .
α1 (G ) + τB (G ) ≤ 5n2 /16 Lemma (Erd˝os–Faudree–Pach–Spencer 1988) If G is a graph, and B is avertex set such that G [B] is bipartite, then 1 τB (G ) ≤ 2 E (G ) \ B2 .
Theorem For any n-vertex graph G , α1 (G ) + τB (G ) ≤ 5n2 /16.
Proof. EFPS yields τB (G ) ≤
1 2
h n
Thus, α1 (G ) + τB (G ) ≤
−
2 nb(G ) 4
i
b(G ) 2
+
n2 4
−
b(G )2 n2 4 − 4 . 2 b(G ) 4 .
≤
α1 (G ) + τB (G ) ≤ 5n2 /16 Lemma (Erd˝os–Faudree–Pach–Spencer 1988) If G is a graph, and B is avertex set such that G [B] is bipartite, then 1 τB (G ) ≤ 2 E (G ) \ B2 .
Theorem For any n-vertex graph G , α1 (G ) + τB (G ) ≤ 5n2 /16.
Proof. EFPS yields τB (G ) ≤
1 2
h n
−
2 nb(G ) 4
i
b(G ) 2 2
b(G )2 n2 4 − 4 . 2 b(G ) 4 .
≤
+ n4 − Thus, α1 (G ) + τB (G ) ≤ The right side is maximized when b(G ) = n/2, yielding α1 (G ) + τB (G ) ≤ 5n2 /16.
Two versions of the conjecture Conjecture (Erd˝os–Gallai–Tuza 1996) If G is an n-vertex graph, then α1 (G ) + τ1 (G ) ≤ n2 /4.
Two versions of the conjecture Conjecture (Erd˝os–Gallai–Tuza 1996 ...sort of) If G is an n-vertex graph, then α1 (G ) + τ1 (G ) ≤ n2 /4.
Two versions of the conjecture Conjecture (Erd˝os–Gallai–Tuza 1996 ...sort of) If G is an n-vertex graph, then α1 (G ) + τ1 (G ) ≤ n2 /4.
Definition A graph is triangular if each of its edges lies in some triangle.
Conjecture (Erd˝os–Gallai–Tuza 1996) If G is an n-vertex triangular graph, then α1 (G ) + τ1 (G ) ≤ n2 /4. Later formulations of the conjecture, by both Erd˝ os and Tuza, silently dropped the “triangular” condition.
Two versions of the conjecture Conjecture (Erd˝os–Gallai–Tuza 1996 ...sort of) If G is an n-vertex graph, then α1 (G ) + τ1 (G ) ≤ n2 /4.
Definition A graph is triangular if each of its edges lies in some triangle.
Conjecture (Erd˝os–Gallai–Tuza 1996) If G is an n-vertex triangular graph, then α1 (G ) + τ1 (G ) ≤ n2 /4. Later formulations of the conjecture, by both Erd˝ os and Tuza, silently dropped the “triangular” condition.
Question (Grinberg 2012) Are these two forms of the conjecture equivalent?
Two versions of the conjecture Conjecture (Erd˝os–Gallai–Tuza 1996 ...sort of) If G is an n-vertex graph, then α1 (G ) + τ1 (G ) ≤ n2 /4.
Definition A graph is triangular if each of its edges lies in some triangle.
Conjecture (Erd˝os–Gallai–Tuza 1996) If G is an n-vertex triangular graph, then α1 (G ) + τ1 (G ) ≤ n2 /4. Later formulations of the conjecture, by both Erd˝ os and Tuza, silently dropped the “triangular” condition.
Question (Grinberg 2012) Are these two forms of the conjecture equivalent? Answer (P. 2014+): YES.
Warmup: α1 (G ) + τ1 (G ) ≤ |E (G )|
Theorem (Erd˝os–Gallai–Tuza 1996) In any graph G , α1 (G ) + τ1 (G ) ≤ |E (G )|.
Warmup: α1 (G ) + τ1 (G ) ≤ |E (G )|
Theorem (Erd˝os–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|.
Warmup: α1 (G ) + τ1 (G ) ≤ |E (G )|
Theorem (Erd˝os–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.
Warmup: α1 (G ) + τ1 (G ) ≤ |E (G )|
Theorem (Erd˝os–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 least two edges of X .
Warmup: α1 (G ) + τ1 (G ) ≤ |E (G )|
Theorem (Erd˝os–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 least two edges of X . Thus, X is a hitting set (and then some!), so that τ1 (G ) ≤ |X |.
Warmup: α1 (G ) + τ1 (G ) ≤ |E (G )|
Theorem (Erd˝os–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 least two 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] .
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] .
S
Proof. Let A be any triangle-independent set in G .
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] .
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.
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] .
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].
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] .
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.
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] .
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.
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] .
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+) For any S ⊂ V (G ), α1 (G ) + τ1 (G ) ≤ α1 (G [S]) + τ1 (G [S]) + α1 (G [S]) + τ1 (G [S]) + [S, S] .
Minimum Counterexamples have Dense 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] .
Theorem 2 If G is a vertex-minimal graph such that α 1 (G ) + τ1 (G ) > n /4, then for all proper nonempty S ⊂ V (G ), we have [S, S] > |S| (n − |S|)/2.
Minimum Counterexamples have Dense 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] .
Theorem 2 If G is a vertex-minimal graph such that α 1 (G ) + τ1 (G ) > n /4, then for all 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]
Minimum Counterexamples have Dense 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] .
Theorem 2 If G is a vertex-minimal graph such that α 1 (G ) + τ1 (G ) > n /4, then for all 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.
Minimum Degree in Minimum Counterexamples Theorem 2 If G is a vertex-minimal graph such that α 1 (G ) + τ1 (G ) > n /4, then for all proper nonempty S ⊂ V (G ), we have [S, S] > |S| (n − |S|)/2.
Improvement If G is a vertex-minimal counterexample, then δ(G ) > n/2.
Minimum Degree in Minimum Counterexamples Theorem 2 If G is a vertex-minimal graph such that α 1 (G ) + τ1 (G ) > n /4, then for all proper nonempty S ⊂ V (G ), we have [S, S] > |S| (n − |S|)/2.
Improvement If G is a vertex-minimal counterexample, then δ(G ) > n/2. Thus, G is triangular.
Minimum Degree in Minimum Counterexamples Theorem 2 If G is a vertex-minimal graph such that α 1 (G ) + τ1 (G ) > n /4, then for all proper nonempty S ⊂ V (G ), we have [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 , then α1 (G ) + τ1 (G ) ≤ n2 /4 for all graphs G . That is, the two forms of the conjecture are equivalent.
Dense Cuts for τB Lemma For any triangle-independent A and any S ⊂ V (G ), α1 (G ) + τB (G ) ≤ α1 (G [S]) + τ B (G [S]) + α1 (G [S]) + τB (G [S])+ 1 2 [S, S] + A ∩ [S, S] .
Dense Cuts for τB Lemma For any triangle-independent A and any S ⊂ V (G ), α1 (G ) + τB (G ) ≤ α1 (G [S]) + τ B (G [S]) + α1 (G [S]) + τB (G [S])+ 1 2 [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 them yielding a bipartition of V (G ).
S
S
Dense Cuts for τB Lemma For any triangle-independent A and any S ⊂ V (G ), α1 (G ) + τB (G ) ≤ α1 (G [S]) + τ B (G [S]) + α1 (G [S]) + τB (G [S])+ 1 2 [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 them yielding a bipartition of V (G ).
S
S
Dense Cuts for τB Lemma For any triangle-independent A and any S ⊂ V (G ), α1 (G ) + τB (G ) ≤ α1 (G [S]) + τ B (G [S]) + α1 (G [S]) + τB (G [S])+ 1 2 [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 them yielding a bipartition of V (G ).
S
S
Clique Cuts Corollary If G is a vertex-minimal graph with α1 (G ) + τB (G ) > n2 /4 and A is triangle-independent, then for any nonempty proper S ⊂ V (G ), we have 1 2 [S, S] + A ∩ [S, S] > |S| (n − |S|)/2.
Clique Cuts Corollary If G is a vertex-minimal graph with α1 (G ) + τB (G ) > n2 /4 and A is triangle-independent, then for any nonempty proper S ⊂ V (G ), we have 1 2 [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 is triangle-independent, and if S is a clique, then [S, S] > (|S| − 2)(n − |S|).
Clique Cuts Corollary If G is a vertex-minimal graph with α1 (G ) + τB (G ) > n2 /4 and A is triangle-independent, then for any nonempty proper S ⊂ V (G ), we have 1 2 [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 is triangle-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|.
Clique Cuts Corollary If G is a vertex-minimal graph with α1 (G ) + τB (G ) > n2 /4 and A is triangle-independent, then for any nonempty proper S ⊂ V (G ), we have 1 2 [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 is triangle-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|.
can’t happen:
S
v
K4− -free Graphs Theorem If G is K4− -free, then α1 (G ) + τB (G ) ≤ n2 /4.
K4− -free Graphs Theorem If G is K4− -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.
K4− -free Graphs Theorem If G is K4− -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 K4− -free graphs with α1 (G ) + τB (G ) > n2 /4, take G vertex-minimal. Note that G is vertex-minimal among all graphs.
K4− -free Graphs Theorem If G is K4− -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 K4− -free graphs with α1 (G ) + τB (G ) > n2 /4, take G vertex-minimal. Note that G is vertex-minimal among all graphs. Let S be a maximum clique, so that |S| ≥ 3.
S
K4− -free Graphs Theorem If G is K4− -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 K4− -free graphs with α1 (G ) + τB (G ) > n2 /4, take G vertex-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.
S
v
K4− -free Graphs Theorem If G is K4− -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 K4− -free graphs with α1 (G ) + τB (G ) > n2 /4, take G vertex-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 K4− . Contradiction!
S
v
α1 (G ) + τB (G ) ≤ n2 /4 for triangle-free G Lemma (Erd˝os–Faudree–Pach–Spencer 1988) If G is an n-vertex triangle-free graph with m edges, then τB (G ) ≤ m −
4m2 . n2
α1 (G ) + τB (G ) ≤ n2 /4 for triangle-free G Lemma (Erd˝os–Faudree–Pach–Spencer 1988) If G is an n-vertex triangle-free graph with m edges, then τB (G ) ≤ m −
4m2 . n2
Corollary If G is triangle-free, then α1 (G ) + τB (G ) ≤ n2 /4.
Proof. The lemma yields α1 (G ) + τB (G ) ≤ 2m −
4m2 . n2
α1 (G ) + τB (G ) ≤ n2 /4 for triangle-free G Lemma (Erd˝os–Faudree–Pach–Spencer 1988) If G is an n-vertex triangle-free graph with m edges, then τB (G ) ≤ m −
4m2 . n2
Corollary If G is triangle-free, then α1 (G ) + τB (G ) ≤ n2 /4.
Proof. The lemma yields α1 (G ) + τB (G ) ≤ 2m −
4m2 . n2
The upper bound is maximized when m = n2 /4, yielding α1 (G ) + τB (G ) ≤ n2 /4.
Acknowledgments
This research was performed while I was a graduate student at the UIUC math department.
Acknowledgments
This research was performed while I was a graduate student at the UIUC math department. Thank you for coming!
FIN
UIUC Graph Theory Seminar September 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 .
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 (Tur´an 1941) If G is an n-vertex graph with no copy of Kr , then G has at most r −2 n2 r −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 (Erd˝os 1965) Any graph G can be made bipartite by removing at most |E (G )| /2 edges.
How hard is it to make a graph bipartite? Theorem (Erd˝os 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.
How hard is it to make a graph bipartite? Theorem (Erd˝os 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 expected number of deleted edges is |E (G )| /2.
How hard is it to make a graph bipartite? Theorem (Erd˝os 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 expected number of deleted edges is |E (G )| /2.
Corollary If G has n vertices, then G can be made bipartite by deleting at most n2 /4 edges. The worst-case graph is Kn .
Hitting sets and triangle-independent sets
Definition X ⊂ E (G ) is a hitting set if it contains at least one edge from each triangle of G . A ⊂ E (G ) is triangle-independent if it contains at most one edge from each triangle of G .
Hitting sets and triangle-independent sets
Definition X ⊂ E (G ) is a hitting set if it contains at least one edge from each triangle of G . A ⊂ E (G ) is triangle-independent if it contains at most one edge from 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 .
Hitting sets and triangle-independent sets
Definition X ⊂ E (G ) is a hitting set if it contains at least one edge from each triangle of G . A ⊂ E (G ) is triangle-independent if it contains at most one edge from 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 Erd˝os bound implies: τ1 (G ) ≤ n2 /4. Mantel’s Theorem implies: α1 (G ) ≤ n2 /4.
The Erd˝os–Gallai–Tuza Conjecture Conjecture (Erd˝os–Gallai–Tuza 1996) If G is an n-vertex graph, then α1 (G ) + τ1 (G ) ≤ n2 /4. The conjecture is sharp, if true:
The Erd˝os–Gallai–Tuza Conjecture Conjecture (Erd˝os–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:
The Erd˝os–Gallai–Tuza Conjecture Conjecture (Erd˝os–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 ) =
n 2
− n2 /4 = n2 /4 − n/2,
The Erd˝os–Gallai–Tuza Conjecture Conjecture (Erd˝os–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 ) =
n 2
− n2 /4 = n2 /4 − n/2,
α1 (Kn ) = n/2,
The Erd˝os–Gallai–Tuza Conjecture Conjecture (Erd˝os–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 ) =
n 2
− n2 /4 = n2 /4 − n/2,
α1 (Kn ) = n/2, Thus α1 (Kn ) + τ1 (Kn ) = n2 /4.
The Erd˝os–Gallai–Tuza Conjecture Conjecture (Erd˝os–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 ) =
n 2
− 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 :
The Erd˝os–Gallai–Tuza Conjecture Conjecture (Erd˝os–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 ) =
n 2
− 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,
The Erd˝os–Gallai–Tuza Conjecture Conjecture (Erd˝os–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 ) =
n 2
− 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,
The Erd˝os–Gallai–Tuza Conjecture Conjecture (Erd˝os–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 ) =
n 2
− 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.
The Erd˝os–Gallai–Tuza Conjecture Conjecture (Erd˝os–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 ) =
n 2
− 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 .
From τ1 to τB Definition τ1 (G ) is the smallest size of a hitting set in G .
Equivalent Definition τ1 (G ) is the smallest size of an edge set X such that G − X is triangle-free.
From τ1 to τB Definition τ1 (G ) is the smallest size of a hitting set in G .
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.
From τ1 to τB Definition τ1 (G ) is the smallest size of a hitting set in G .
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.
From τ1 to τB Definition τ1 (G ) is the smallest size of a hitting set in G .
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.
Two approaches to the conjecture Conjecture (Erd˝os–Gallai–Tuza 1996) If G is an n-vertex graph, then α1 (G ) + τ1 (G ) ≤ n2 /4. 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.
Two approaches to the conjecture Conjecture (Erd˝os–Gallai–Tuza 1996) If G is an n-vertex graph, then α1 (G ) + τ1 (G ) ≤ n2 /4. 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.
Two approaches to the conjecture Conjecture (Erd˝os–Gallai–Tuza 1996) If G is an n-vertex graph, then α1 (G ) + τ1 (G ) ≤ n2 /4. 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 K4− , then α1 (G ) + τB (G ) ≤ n2 /4. In the rest of the talk, we prove these and other results.
K4−
Two Lemmas Let b(G ) = max{|B| : G [B] is bipartite}.
Lemma u
v
If A is triangle-independent, then for any uv ∈ A, the induced subgraph G [NA (u) ∪ NA (v )] is bipartite with dA (u) + dA (v ) vertices.
Two Lemmas Let b(G ) = max{|B| : G [B] is bipartite}.
Lemma u
v
If A is triangle-independent, then for any uv ∈ A, the induced subgraph G [NA (u) ∪ NA (v )] is bipartite with dA (u) + dA (v ) vertices. Thus, dA (u) + dA (v ) ≤ b(G ).
Two Lemmas Let b(G ) = max{|B| : G [B] is bipartite}.
Lemma u
v
If A is triangle-independent, then for any uv ∈ A, the induced subgraph G [NA (u) ∪ NA (v )] is bipartite with dA (u) + dA (v ) vertices. Thus, dA (u) + dA (v ) ≤ b(G ).
Lemma If A is triangle-independent, then |A| ≤
nb(G ) 4.
Two Lemmas Let b(G ) = max{|B| : G [B] is bipartite}.
Lemma u
v
If A is triangle-independent, then for any uv ∈ A, the induced subgraph G [NA (u) ∪ NA (v )] is bipartite with dA (u) + dA (v ) vertices. Thus, dA (u) + dA (v ) ≤ b(G ).
Lemma 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 ).
Two Lemmas Let b(G ) = max{|B| : G [B] is bipartite}.
Lemma u
v
If A is triangle-independent, then for any uv ∈ A, the induced subgraph G [NA (u) ∪ NA (v )] is bipartite with dA (u) + dA (v ) vertices. Thus, dA (u) + dA (v ) ≤ b(G ).
Lemma 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 ). P Summing over all edges in A, we get v dA (v )2 ≤ |A| b(G ).
Two Lemmas Let b(G ) = max{|B| : G [B] is bipartite}.
Lemma u
v
If A is triangle-independent, then for any uv ∈ A, the induced subgraph G [NA (u) ∪ NA (v )] is bipartite with dA (u) + dA (v ) vertices. Thus, dA (u) + dA (v ) ≤ b(G ).
Lemma 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 ). P Summing over all edges in A, we get v dA (v )2 ≤ |A| b(G ). 2 P Cauchy-Schwarz gives v dA (v )2 ≥ 4|A| n , and the conclusion follows.
α1 (G ) + τB (G ) ≤ 5n2 /16 Lemma (Erd˝os–Faudree–Pach–Spencer 1988) If G is a graph, and B is avertex set such that G [B] is bipartite, then 1 τB (G ) ≤ 2 E (G ) \ B2 .
α1 (G ) + τB (G ) ≤ 5n2 /16 Lemma (Erd˝os–Faudree–Pach–Spencer 1988) If G is a graph, and B is avertex set such that G [B] is bipartite, then 1 τB (G ) ≤ 2 E (G ) \ B2 .
Theorem For any n-vertex graph G , α1 (G ) + τB (G ) ≤ 5n2 /16.
α1 (G ) + τB (G ) ≤ 5n2 /16 Lemma (Erd˝os–Faudree–Pach–Spencer 1988) If G is a graph, and B is avertex set such that G [B] is bipartite, then 1 τB (G ) ≤ 2 E (G ) \ B2 .
Theorem For any n-vertex graph G , α1 (G ) + τB (G ) ≤ 5n2 /16.
Proof. EFPS yields τB (G ) ≤
1 2
h n 2
−
i
b(G ) 2
≤
n2 4
−
b(G )2 4 .
α1 (G ) + τB (G ) ≤ 5n2 /16 Lemma (Erd˝os–Faudree–Pach–Spencer 1988) If G is a graph, and B is avertex set such that G [B] is bipartite, then 1 τB (G ) ≤ 2 E (G ) \ B2 .
Theorem For any n-vertex graph G , α1 (G ) + τB (G ) ≤ 5n2 /16.
Proof. EFPS yields τB (G ) ≤
1 2
h n
Thus, α1 (G ) + τB (G ) ≤
−
2 nb(G ) 4
i
b(G ) 2
+
n2 4
−
b(G )2 n2 4 − 4 . 2 b(G ) 4 .
≤
α1 (G ) + τB (G ) ≤ 5n2 /16 Lemma (Erd˝os–Faudree–Pach–Spencer 1988) If G is a graph, and B is avertex set such that G [B] is bipartite, then 1 τB (G ) ≤ 2 E (G ) \ B2 .
Theorem For any n-vertex graph G , α1 (G ) + τB (G ) ≤ 5n2 /16.
Proof. EFPS yields τB (G ) ≤
1 2
h n
−
2 nb(G ) 4
i
b(G ) 2 2
b(G )2 n2 4 − 4 . 2 b(G ) 4 .
≤
+ n4 − Thus, α1 (G ) + τB (G ) ≤ The right side is maximized when b(G ) = n/2, yielding α1 (G ) + τB (G ) ≤ 5n2 /16.
Two versions of the conjecture Conjecture (Erd˝os–Gallai–Tuza 1996) If G is an n-vertex graph, then α1 (G ) + τ1 (G ) ≤ n2 /4.
Two versions of the conjecture Conjecture (Erd˝os–Gallai–Tuza 1996 ...sort of) If G is an n-vertex graph, then α1 (G ) + τ1 (G ) ≤ n2 /4.
Two versions of the conjecture Conjecture (Erd˝os–Gallai–Tuza 1996 ...sort of) If G is an n-vertex graph, then α1 (G ) + τ1 (G ) ≤ n2 /4.
Definition A graph is triangular if each of its edges lies in some triangle.
Conjecture (Erd˝os–Gallai–Tuza 1996) If G is an n-vertex triangular graph, then α1 (G ) + τ1 (G ) ≤ n2 /4. Later formulations of the conjecture, by both Erd˝ os and Tuza, silently dropped the “triangular” condition.
Two versions of the conjecture Conjecture (Erd˝os–Gallai–Tuza 1996 ...sort of) If G is an n-vertex graph, then α1 (G ) + τ1 (G ) ≤ n2 /4.
Definition A graph is triangular if each of its edges lies in some triangle.
Conjecture (Erd˝os–Gallai–Tuza 1996) If G is an n-vertex triangular graph, then α1 (G ) + τ1 (G ) ≤ n2 /4. Later formulations of the conjecture, by both Erd˝ os and Tuza, silently dropped the “triangular” condition.
Question (Grinberg 2012) Are these two forms of the conjecture equivalent?
Two versions of the conjecture Conjecture (Erd˝os–Gallai–Tuza 1996 ...sort of) If G is an n-vertex graph, then α1 (G ) + τ1 (G ) ≤ n2 /4.
Definition A graph is triangular if each of its edges lies in some triangle.
Conjecture (Erd˝os–Gallai–Tuza 1996) If G is an n-vertex triangular graph, then α1 (G ) + τ1 (G ) ≤ n2 /4. Later formulations of the conjecture, by both Erd˝ os and Tuza, silently dropped the “triangular” condition.
Question (Grinberg 2012) Are these two forms of the conjecture equivalent? Answer (P. 2014+): YES.
Warmup: α1 (G ) + τ1 (G ) ≤ |E (G )|
Theorem (Erd˝os–Gallai–Tuza 1996) In any graph G , α1 (G ) + τ1 (G ) ≤ |E (G )|.
Warmup: α1 (G ) + τ1 (G ) ≤ |E (G )|
Theorem (Erd˝os–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|.
Warmup: α1 (G ) + τ1 (G ) ≤ |E (G )|
Theorem (Erd˝os–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.
Warmup: α1 (G ) + τ1 (G ) ≤ |E (G )|
Theorem (Erd˝os–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 least two edges of X .
Warmup: α1 (G ) + τ1 (G ) ≤ |E (G )|
Theorem (Erd˝os–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 least two edges of X . Thus, X is a hitting set (and then some!), so that τ1 (G ) ≤ |X |.
Warmup: α1 (G ) + τ1 (G ) ≤ |E (G )|
Theorem (Erd˝os–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 least two 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] .
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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] .
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Proof. Let A be any triangle-independent set in G .
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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] .
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Proof. Let A be any triangle-independent set in G . Let X1 , X2 be smallest hitting sets in G [S], G [S] respectively.
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] .
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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].
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] .
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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.
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] .
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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.
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] .
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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+) For any S ⊂ V (G ), α1 (G ) + τ1 (G ) ≤ α1 (G [S]) + τ1 (G [S]) + α1 (G [S]) + τ1 (G [S]) + [S, S] .
Minimum Counterexamples have Dense 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] .
Theorem 2 If G is a vertex-minimal graph such that α 1 (G ) + τ1 (G ) > n /4, then for all proper nonempty S ⊂ V (G ), we have [S, S] > |S| (n − |S|)/2.
Minimum Counterexamples have Dense 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] .
Theorem 2 If G is a vertex-minimal graph such that α 1 (G ) + τ1 (G ) > n /4, then for all 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]
Minimum Counterexamples have Dense 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] .
Theorem 2 If G is a vertex-minimal graph such that α 1 (G ) + τ1 (G ) > n /4, then for all 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.
Minimum Degree in Minimum Counterexamples Theorem 2 If G is a vertex-minimal graph such that α 1 (G ) + τ1 (G ) > n /4, then for all proper nonempty S ⊂ V (G ), we have [S, S] > |S| (n − |S|)/2.
Improvement If G is a vertex-minimal counterexample, then δ(G ) > n/2.
Minimum Degree in Minimum Counterexamples Theorem 2 If G is a vertex-minimal graph such that α 1 (G ) + τ1 (G ) > n /4, then for all proper nonempty S ⊂ V (G ), we have [S, S] > |S| (n − |S|)/2.
Improvement If G is a vertex-minimal counterexample, then δ(G ) > n/2. Thus, G is triangular.
Minimum Degree in Minimum Counterexamples Theorem 2 If G is a vertex-minimal graph such that α 1 (G ) + τ1 (G ) > n /4, then for all proper nonempty S ⊂ V (G ), we have [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 , then α1 (G ) + τ1 (G ) ≤ n2 /4 for all graphs G . That is, the two forms of the conjecture are equivalent.
Dense Cuts for τB Lemma For any triangle-independent A and any S ⊂ V (G ), α1 (G ) + τB (G ) ≤ α1 (G [S]) + τ B (G [S]) + α1 (G [S]) + τB (G [S])+ 1 2 [S, S] + A ∩ [S, S] .
Dense Cuts for τB Lemma For any triangle-independent A and any S ⊂ V (G ), α1 (G ) + τB (G ) ≤ α1 (G [S]) + τ B (G [S]) + α1 (G [S]) + τB (G [S])+ 1 2 [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 them yielding a bipartition of V (G ).
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Dense Cuts for τB Lemma For any triangle-independent A and any S ⊂ V (G ), α1 (G ) + τB (G ) ≤ α1 (G [S]) + τ B (G [S]) + α1 (G [S]) + τB (G [S])+ 1 2 [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 them yielding a bipartition of V (G ).
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Dense Cuts for τB Lemma For any triangle-independent A and any S ⊂ V (G ), α1 (G ) + τB (G ) ≤ α1 (G [S]) + τ B (G [S]) + α1 (G [S]) + τB (G [S])+ 1 2 [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 them yielding a bipartition of V (G ).
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Clique Cuts Corollary If G is a vertex-minimal graph with α1 (G ) + τB (G ) > n2 /4 and A is triangle-independent, then for any nonempty proper S ⊂ V (G ), we have 1 2 [S, S] + A ∩ [S, S] > |S| (n − |S|)/2.
Clique Cuts Corollary If G is a vertex-minimal graph with α1 (G ) + τB (G ) > n2 /4 and A is triangle-independent, then for any nonempty proper S ⊂ V (G ), we have 1 2 [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 is triangle-independent, and if S is a clique, then [S, S] > (|S| − 2)(n − |S|).
Clique Cuts Corollary If G is a vertex-minimal graph with α1 (G ) + τB (G ) > n2 /4 and A is triangle-independent, then for any nonempty proper S ⊂ V (G ), we have 1 2 [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 is triangle-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|.
Clique Cuts Corollary If G is a vertex-minimal graph with α1 (G ) + τB (G ) > n2 /4 and A is triangle-independent, then for any nonempty proper S ⊂ V (G ), we have 1 2 [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 is triangle-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|.
can’t happen:
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K4− -free Graphs Theorem If G is K4− -free, then α1 (G ) + τB (G ) ≤ n2 /4.
K4− -free Graphs Theorem If G is K4− -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.
K4− -free Graphs Theorem If G is K4− -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 K4− -free graphs with α1 (G ) + τB (G ) > n2 /4, take G vertex-minimal. Note that G is vertex-minimal among all graphs.
K4− -free Graphs Theorem If G is K4− -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 K4− -free graphs with α1 (G ) + τB (G ) > n2 /4, take G vertex-minimal. Note that G is vertex-minimal among all graphs. Let S be a maximum clique, so that |S| ≥ 3.
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K4− -free Graphs Theorem If G is K4− -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 K4− -free graphs with α1 (G ) + τB (G ) > n2 /4, take G vertex-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.
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K4− -free Graphs Theorem If G is K4− -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 K4− -free graphs with α1 (G ) + τB (G ) > n2 /4, take G vertex-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 K4− . Contradiction!
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α1 (G ) + τB (G ) ≤ n2 /4 for triangle-free G Lemma (Erd˝os–Faudree–Pach–Spencer 1988) If G is an n-vertex triangle-free graph with m edges, then τB (G ) ≤ m −
4m2 . n2
α1 (G ) + τB (G ) ≤ n2 /4 for triangle-free G Lemma (Erd˝os–Faudree–Pach–Spencer 1988) If G is an n-vertex triangle-free graph with m edges, then τB (G ) ≤ m −
4m2 . n2
Corollary If G is triangle-free, then α1 (G ) + τB (G ) ≤ n2 /4.
Proof. The lemma yields α1 (G ) + τB (G ) ≤ 2m −
4m2 . n2
α1 (G ) + τB (G ) ≤ n2 /4 for triangle-free G Lemma (Erd˝os–Faudree–Pach–Spencer 1988) If G is an n-vertex triangle-free graph with m edges, then τB (G ) ≤ m −
4m2 . n2
Corollary If G is triangle-free, then α1 (G ) + τB (G ) ≤ n2 /4.
Proof. The lemma yields α1 (G ) + τB (G ) ≤ 2m −
4m2 . n2
The upper bound is maximized when m = n2 /4, yielding α1 (G ) + τB (G ) ≤ n2 /4.
Acknowledgments
This research was performed while I was a graduate student at the UIUC math department.
Acknowledgments
This research was performed while I was a graduate student at the UIUC math department. Thank you for coming!
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