The minimum number of triangular edges and a symmetrization for multiple graphs
arXiv:1411.0771v1 [math.CO] 4 Nov 2014
Zolt´an F¨ uredi
∗
Zeinab Maleki
†
This work was done while the authors visited the Department of Mathematics and Computer Science, Emory University, Atlanta, GA, USA.
Abstract We give an asymptotic formula for the minimum number of edges contained on triangles in a graph having n vertices and e edges. Our main tool is a generalization of Zykov’s symmetrization what we can apply for several graphs simultaneously.
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Graphs with few triangular edges
Erd˝ os, Faudree, and Rousseau [2] showed that a graph on n vertices and at least bn2 /4c+1 edges has at least 2bn/2c + 1 edges on triangles. To see that this result is sharp, just consider the graph obtained by adding one edge to the side of the complete bipartite graph Kdn/2e,bn/2c with dn/2e vertices. More generally, for a graph G let E(G, K3 ) denote the set of edges of G contained on triangles, Tr(G) := |E(G, K3 )|, Tr(n, e) := min{Tr(G) : |V (G)| = n, e(G) = e}. With this notation the above result of Erd˝os, Faudree, and Rousseau can be reformulated as Tr(n, bn2 /4c + 1) = 2bn/2c + 1.
(1)
Note that Tr(n, e) = 0 whenever e ≤ ex(n, K3 ), where ex(n, F ) denotes the Tur´ an number of the graph F , i.e., the maximum number of edges in an n-vertex graph not containing F as a subgraph. To avoid trivialities, we usually tacitly suppose that e > n2 /4 and n ≥ 5 (although it is not necessary in most cases). Obviously, Tr(4, 4) = 0 and Tr(4, e) = e for e ∈ {5, 6}. Given n ≥ 3, n2 ≥ e > n2 /4 integers we define a class of graphs, G1 (n, e), with many non-triangular edges as follows. Take a partition (A, B, C) of the n-element vertex set V with |A| ≥ 2. The class of graphs G1 (A, B, C) are graphs G with vertex set V containing a complete bipartite graph K(B, C), such ∗ Alfr´ ed R´ enyi Institute of Mathematics, 13–15 Re´ altanoda Street, 1053 Budapest, Hungary. E-mail:
[email protected]. Research supported in part by the Hungarian National Science Foundation OTKA 104343, by the Simons Foundation Collaboration Grant #317487, and by the European Research Council Advanced Investigators Grant 267195. † Department of Mathematical Sciences, Isfahan University of Technology, Isfahan 84156-83111, Iran. E-mail:
[email protected] 2010 Mathematics Subject Classifications: 05C35, 05C22, 05D99. Furedi˙Maleki˙Symm˙Method˙2014˙11˙03 Key Words: Tur´ an number, triangles, extremal graphs, symmetrization. November 5, 2014
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¨ redi and Maleki: Triangular edges and a symmetrization Fu
that B and C are independent sets, the vertices of C has neighbors only in B, and G[A] and G[A, B] are almost complete graphs, they span more than |A|−1 + |A||B| edges. Then the edges of G[B, C] are the 2 non-triangular edges.
a
c Kb,c
b Figure 1: A graph from G1 (A, B, C), where |A| = a, |B| = b, and |C| = c. Put a graph G ∈ G1 (A, B, C) into the class G1 (n, e) if it has e edges and it has the minimum number of triangular edges among these type of graphs, i.e., Tr(G) ≤ Tr(G0 ) holds for G0 ∈ G1 (A0 , B 0 , C 0 ) with |A0 | + |B 0 | + |C 0 | = n and e(G0 ) = e. Define g(n, e) := Tr(G) for some G ∈ G1 (n, e). We have a Tr(n, e) ≤ g(n, e) = min{e − bc : a + b + c = n, a, b, c ∈ N ∪ {0}, + ab + bc ≥ e}. (2) 2 We think one can extend the Erd˝ os, Faudree, Rousseau theorem as follows. Conjecture 1. Suppose that G is an n-vertex graph with e edges, such that e > n2 /4 and it has the minimum number of triangular edges, i.e., Tr(G) = Tr(n, e). Then G ∈ G1 (n, e). So we conjecture that Tr(n, e) = g(n, e). We develop a graph symmetrization method to prove a slightly weaker result. Our method is useful solving many similar problems [4]. Theorem 2. For e > n2 /4 we have g(n, e) − (3/2)n ≤ Tr(n, e) ≤ g(n, e).
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The symmetrization method
Given a Kp -free graph G and a pair of nonadjacent vertices u and v such that deg(u) ≤ deg(v). Replace all edges incident to u by new edges incident to u and N (v). This process is called Zykov’s symmetrization. He observed [6] that symmetrization does not increase the size of the largest clique and does not decrease the number of edges. Repeated applications of symmetrization lead to a (p − 1)-partite complete graph, thus Zykov gave a proof for Tur´ an’s theorem concerning ex(n, Kp ). This method does not seem to apply to determine Tr(n, e) because we need to increase simultaneously the number of edges and the number of non-triangular edges. Given a graph G with vertex set {v1 , . . . , vn } define a real polynomial X w(G, x) := {xi xj : vi vj ∈ E}. P Define a simplex Sn := {x ∈ Rn : ∀xi ≥ 0 and xi = 1}. Let w(G) := max{w(G, x) : x ∈ Sn }. Motzkin and Straus [5] provided a new proof of an asymptotic version of Tur´an’s theorem by observing a remarkable connection between the clique number, ω(G), and w(G). They proved that w(G) = (ω − 1)/(2ω). Their main tool was a continuous version of Zykov’s symmetrization.
¨ redi and Maleki: Triangular edges and a symmetrization Fu
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Theorem 3. (Motzkin and Strauss) [5] For any G, x ∈ S, there exists a y ∈ S such that w(G, x) ≤ w(G, y) and support(y) induces a complete subgraph. We generalize this so that it can be applied simultaneously for several graphs. Theorem 4. Let G1 , G2 , . . . , Gd be graphs on a common vertex set V := {v1 , . . . , vn }. Take any x ∈ Sn . Then there exists a subset K ⊆ V and a vector y ∈ Sn with support K such that w(Gi , x) ≤ w(Gi , y) for Sd every 1 ≤ i ≤ d, and α(H) ≤ d where the graph H is defined by V (H) := K and E(H) := i=1 E(Gi |K). To prove Theorem 4 we need the following lemma. Lemma 5. Suppose that a1 , . . . , ad ∈ Rd+1 . Then one can find a non-zero vector z ∈ Rd+1 such that P aTi z ≥ 0 for every 1 ≤ i ≤ d and the sum of its coordinates is 0, 1≤i≤d+1 zi = 0. Proof. Let j ∈ Rd+1 be the all 1 vector and define the matrix A as {a1 , . . . , ad , j}. If det(A) = 0, then there are non-trivial solutions of AT z = 0. If det(A) 6= 0 define a := (1, . . . , 1, 0)T ∈ Rd+1 . There is a unique solution z of AT z = a. This z is not the zero-vector and we are done. Proof of Theorem 4. Let y ∈ Sn be a vector such that, |support(y)| = min{|support(y0 )| : w(Gi , x) ≤ w(Gi , y0 ), ∀1 ≤ i ≤ d}. If {v1 , v2 , . . . , vd+1 } ⊆ support(y) is an independent set, then for any z = (z1 , . . . , zd+1 , 0, 0, . . . )T ∈ Rn , t ∈ R, and 1 ≤ i ≤ d we have w(Gi , y + tz) = w(Gi , y) + t(aTi z) for some ai ∈ Rd+1 . Here ai depends only on Gi and y, not from z and t. Apply Lemma 5 to obtain a non-zero vector z = (z1 , . . . , zd+1 , 0, 0, . . . )T with jT z = 0 and aTi z ≥ 0 for 1 ≤ i ≤ d. Choosing an appropriate t > 0 we can have a y + tz ∈ Sn such that support(y + tz) ⊆ support(y) − {vj } for some 1 ≤ j ≤ d + 1. This is a contradiction, so y has the desired property.
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Maximizing non-triangular edges in weighted graphs
Lemma 6. Let G1 and G2 be graphs on a common vertex set V := {v1 , . . . , vn }. Suppose that ∅ = 6 E(G2 ) ⊂ E(G1 ) and no edge in G2 can appear on a triangle of G1 . Take any x ∈ Sn . Then there exists a subset K ⊆ V and a vector y ∈ Sn with support K such that w(Gi , x) ≤ w(Gi , y) for i ∈ {1, 2}, α(H) ≤ 2 where the graph H is the subgraph of G1 induced by K. Moreover, H contains only a single edge e of G2 and H \ e is a complete graph. Proof. By Theorem 4, we know that there is a y ∈ Sn such that w(G1 , x) ≤ w(G1 , y), w(G2 , x) ≤ w(G2 , y) and α(H) ≤ 2. Let y = (y1 , . . . , yn ) be such a vector with |support(y)| is minimized. We claim that K := support(y) satisfies the other properties, too. First we show that the structure of G2 |K is rather simple, then we show that by finding an appropriate y0 one can further reduce K if G2 |K has two or more edges. Suppose that vk and vh ∈ K are nonadjacent vertices and ∂ ∂ ∂ ∂ w(G1 , y) ≥ w(G1 , y) and w(G2 , y) ≥ w(G2 , y). ∂yk ∂yh ∂yk ∂yh
(3)
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¨ redi and Maleki: Triangular edges and a symmetrization Fu
P P In other words, {y` : vk v` ∈ E(Gi |K)} ≥ {y` : vh v` ∈ E(Gi |K)} for i = 1, 2. Define the vector y0 by y0 := y + yh (ek − eh ) ∈ Sn , i.e., its k’th coordinate yk0 = yk + yh , its h’th coordinate is 0, and y`0 = y` otherwise. Then we have w(Gi , y) ≤ w(Gi , y0 ) for i ∈ {1, 2} and support(y0 ) = K \ {vh }, a contradiction. We conclude that the conditions of (3) do not hold. If w(G1 , y) ≤ 1/4 then choose a G2 -edge v1 v2 of H and define y0 = (1/2, 1/2, 0, . . . , 0). We obtain w(G2 , y) ≤ w(G1 , y) ≤ 1/4 = w(G1 , y0 ) = w(G2 , y0 ). This implies K = {1, 2} and we are done. So from now on, we suppose that w(G1 , y) > 1/4. Then the Motzkin-Straus theorem implies that the graph H is not triangle-free. Assume that v1 v2 and v1 v3 ∈ E(H) are G2 edges. We claim that deg(v1 ) = 2 and H \ {v1 , v2 , v3 } is a complete graph. In this section if we talk about ’edges’, ’degrees’ etc., then we always mean H-edges, degree in H, etc., except if it is otherwise stated. Observe that v2 and v3 are independent, otherwise the triangle v1 v2 v3 contains G2 edges. Suppose, on the contrary, that degH (v1 ) ≥ 3, v1 v4 ∈ E(H). Since the set {v2 , v3 , v4 } is not independent, it must contain some edge of H, other than the pair v2 v3 . However, if v3 v4 ∈ E(H) then again a triangle, v1 v3 v4 , contains a G2 edge (namely v1 v3 ). Finally, α(H) ≤ 2 implies that K \ {v1 } \ {N (v1 )} induces a complete graph. The above paragraph already implies that the structure of G2 edges is rather simple in H. Using (3) and some other techniques we eliminate the degree 2 case completely in three steps. Assume that v1 v2 and v1 v3 ∈ E(H) are G2 edges and no other G2 edge connects {v1 , v2 , v3 } to K \ {v1 , v2 , v3 }. Then ∂y∂ 2 w(G2 , y) = ∂y∂ 3 w(G2 , y) (namely, both are y1 ). Since v2 and v3 are independent the conditions of (3) hold, a contradiction. So we get that there is at least another G2 edge of H joined to v2 or v3 . Assume that v1 v2 , v1 v3 and v3 v4 ∈ E(H) are G2 edges. Suppose first that v2 v4 ∈ / E(H). Then the set A := {v1 , ..., v4 } only spans these three G2 edges, v1 and v3 are degree 2 vertices, and (K \ A) ∪ {vi } are complete graphs for i ∈ {2, 4}. Since H must contain triangles we have |K \ A| ≥ 2 and H does not contain further G2 edge. Suppose that y1 ≥ y3 . We obtain that ∂ ∂ w(G2 , y) = y1 ≥ w(G2 , y) = y3 , ∂y2 ∂y4 and
X X ∂ ∂ w(G1 , y) = y1 + y` ≥ w(G1 , y) = y3 + y` . ∂y2 ∂y4 `>4
`>4
This contradicts to (3), so we may assume that A contains the edge v2 v4 . Assume that v1 v2 , v1 v3 and v3 v4 ∈ E(H) are G2 edges and v2 v4 ∈ E(H). Then the set A := {v1 , ..., v4 } only spans these four edges, v1 and v3 are degree 2 vertices, and K \ {v1 , v3 } is a complete graph of size at least 3. H does not contain further G2 edges. We have ! X XX w(G1 , y) = (y1 + y4 )(y2 + y3 ) + (y2 + y4 ) y` + yi yj `>4
i>j>4
and w(G2 , y) = y1 y2 + y1 y3 + y3 y4 . Substitute y0 := y0 (t) = y + t(e1 + e2 − e3 − e4 ) into the above equations. Note that y0 ∈ Sn if t ∈ I := [max{−y1 , −y2 }, min{y3 , y4 }]. We get w(G1 , y0 ) = w(G1 , y) and w(G2 , y0 ) − w(G2 , y) = t2 + t(y2 − y4 ).
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¨ redi and Maleki: Triangular edges and a symmetrization Fu
The right hand side is a convex polynomial of t and it takes its maximum on I in one of the endpoints. Taking this optimal t we obtain that maxt∈I w(G2 , y0 ) > w(G2 , y) and |support(y0 )| < |support(y)|, a contradiction. This completes the proof that H has no adjacent G2 edges.
y2 + t y1 + t
y3
t y4
t
v1
v2
v3
v4
}
B1 = B2
A
B2 = B4
Figure 2: The cases when the G2 edges are on a C4
and
when G2 is a matching.
From now on, we may suppose that the G2 -edges of H are pairwise disjoint. We prove that H has no two disjoint G2 edges, which completes the proof of |E(G2 |K)| = 1. Suppose, on the contrary, that v1 v2 and v3 v4 are two disjoint G2 edges of H. We claim that either {v1 , v3 } and {v2 , v4 } or {v1 , v4 } and {v2 , v3 } are missing from E(H). Indeed, if, for example, v1 v3 is an edge, then the triple {v1 , v2 , v3 } must miss at least one pair, so we get {v2 , v3 } ∈ / E(H). Similarly, considering the triple {v1 , v3 , v4 } we get that {v1 , v4 } ∈ / E(H). From now on, we assume that v1 v4 and v2 v3 6∈ E(G1 ). Define A := {v1 , . . . , v4 } and Bi := {v ∈ K \ A : vvi ∈ E(H)} for 1 ≤ i ≤ 4. Since v1 v2 is a G2 edge we have B1 ∩ B2 = ∅. We claim that K = A ∪ B1 ∪ B2 , a partition of K. Indeed, if v5 6∈ (A ∪ B1 ) then v5 v4 ∈ E(H), otherwise {v5 , v1 , v4 } form an independent set. Then v5 v3 ∈ / E(H) otherwise {v5 , v3 , v4 } form a triangle. Then the set {v5 , v3 , v2 } contains two non-edges (v2 v3 and v5 v3 ) so we get v5 v2 is an edge, i.e., v5 ∈ B2 . This implies that A, B1 and B2 are indeed forming a partition of K. The same is true for A, B3 and B4 . We also obtained that v5 must belong to B4 , so B2 ⊆ B4 . Similarly, B1 ⊆ B3 . Since B3 and B4 are disjoint, we get B1 = B3 and B2 = B4 . We distinguish two cases. Suppose first that the pair v1 v3 is independent. Suppose that y2 ≥ y4 . Since no G2 -edge joins A to K \ A and B1 = N (v1 ) ∩ (K \ A) = N (v3 ) ∩ (K \ A) we obtain that ∂ ∂ w(G2 , y) = y2 ≥ w(G2 , y) = y4 , ∂y1 ∂y3 and
X X ∂ ∂ w(G1 , y) = y2 + y` ≥ w(G1 , y) = y4 + y` . ∂y1 ∂y3 y` ∈B1
y` ∈B1
This contradicts (3), so we may assume that A contains the edge v1 v3 . By symmetry, we may suppose that A contains the edge v2 v4 , too. We have X X XX w(G1 , y) = (y1 + y4 )(y2 + y3 ) + (y1 + y3 ) y` + (y2 + y4 ) y` + yi yj , y` ∈B1
and w(G2 , y) = y1 y2 + y3 y4 .
y` ∈B2
vi ,vj ∈A /
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¨ redi and Maleki: Triangular edges and a symmetrization Fu
Substitute y0 := y0 (t) = y + t(e1 + e2 − e3 − e4 ) into the above equations. Note that y0 ∈ Sn if t ∈ I := [max{−y1 , −y2 }, min{y3 , y4 }]. We get w(G1 , y0 ) = w(G1 , y) and w(G2 , y0 ) − w(G2 , y) = 2t2 + t(y1 + y2 − y3 − y4 ). The right hand side is convex, it takes its maximum on I in one of the endpoints. Taking this optimal t we obtain that maxt∈I w(G2 , y0 ) > w(G2 , y) and |support(y0 )| < |support(y)|, a contradiction. This completes the proof that H has a unique G2 edge. Finally, we claim that the vertices in H which are not adjacent to any G2 edge of H induce a clique. To see this, consider two such vertices v1 and v2 . We have ∂y∂ 1 w(G2 , y) = 0 = ∂y∂ 2 w(G2 , y) so the inequalities of (3) hold. Therefore v1 and v2 must be adjacent to avoid a contradiction.
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A continuous lower bound for the number of triangular edges
Let t(n, e) be a kind of real valued version of g(n, e) t(n, e) := min{e − bc : a + b + c = n, a, b, c ∈ R+ ,
1 2 a + ab + bc ≥ e}. 2
Obviously, t(n, e) ≤ g(n, e) for n2 /4 ≤ e ≤ n2 . However, the integer and the real valued functions are not too far from each other g(n, e) − (3/2)n ≤ t(n, e). (4) Indeed, suppose that (a, b, c) ∈ R3+ yields the optimal value, t(n, e) = e − bc. It is a straightforward calculation to show that (a0 , b0 , c0 ) := (da + 1e, dbe, n − a0 − b0 ) satisfy (2) and the difference of (e − b0 c0 ) and (e − bc) is at most (3/2)n. We cannot prove Conjecture 1 that g(n, e) ≤ Tr(n, e) (i.e., that they are equal), but as an application of Lemma 6 we will show that t(n, e) is a lower bound. Theorem 7. For e > n2 /4 we have t(n, e) ≤ Tr(n, e). Proof. Suppose that G1 is a graph with n vertices, e edges and minimum number of edges on triangles, i.e., G1 has Tr(n, e) triangle edges. Let G2 be the subgraph of G1 consisting of the edges not on any triangle of G1 . Consider the vector (1/n)j = (1/n, 1/n, . . . , 1/n) ∈ Rn . By Lemma 6 there exists a y = (y1 , . . . , yn ) ∈ Sn with support K such that G2 |K consists of a single edge, say v2 v3 . Moreover e − Tr(n, e) e = w(G1 , (1/n)j) ≤ w(G1 , y) and = w(G2 , (1/n)j) ≤ w(G2 , y) = y2 y3 . 2 n n2 P Assume that y2 ≥ y3 and define a := k6=2,3 yk n, b := y2 n, c := y3 n. We obtain that Tr(n, e) ≥ e − bc. Since v2 v3 is not on any triangle, N (v2 ) ∩ N (v3 ) = ∅. We get X X XX e ≤ w(G1 , y) = y2 y3 + y2 ( yk ) + y3 ( yk ) + yi yj 2 n yk ∈N (v2 ),k6=3
≤
yk ∈N (v3 ),k6=2
bc b a 1 a + × + ( )2 . n2 n n 2 n
This and the definition of t(n, e) give e − bc ≥ t(n, e). We are done.
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