An extremal problem concerning graphs not containing and

Discrete Mathematics 303 (2005) 56 – 64 www.elsevier.com/locate/disc An extremal problem concerning graphs not containing Kt and Kt,t,t Ervin Győria,1 , Yoomi Rhob a Rényi Institute of Mathematics, Hungarian Academy of Sciences, Budapest, Hungary b Department of Computer Science and Engineering, Seoul National University, Seoul, Korea Received 30 August 2003; received in revised form 25 February 2004; accepted 8 December 2004 Available online 25 October 2005 Abstract Brualdi and Mellendorf raised the following problem as a modification of Turán’s theorem. Let t and n be positive integers with n
t
2. Determine the maximum number of edges of a graph of order n that contains neither Kt nor Kt,t as a subgraph. They solved it for n = 2t (see R.A. Brualdi, S. Mellendorf, Two extremal problems in graph theory, The Electron. J. Combin. 1 (1994) #R2). We consider the following similar modification of Turán’s theorem. Let t and n be positive integers with n
t
2. Determine the maximum number of edges of a graph of order n that contains neither Kt nor Kt,t,t as a subgraph. We solve it for n = 3t. © 2005 Elsevier B.V. All rights reserved. Keywords: Turán’s theorem; A complete graph; A complete tripartite graph; An independent set; A trisectable graph 1. Introduction One of the best known results in extremal graph theory is the following theorem of Turán. E-mail addresses: [email protected], [email protected] (E. Győri), [email protected] (Y. Rho). 1 Research is partially supported by OTKA Grants T34702 and AT48826. 0012-365X/$ - see front matter © 2005 Elsevier B.V. All rights reserved. doi:10.1016/j.disc.2004.12.017 E. Győri, Y. Rho / Discrete Mathematics 303 (2005) 56 – 64 57 Theorem 1. Let t and n be positive integers with n
t
2. Then the maximum number of edges of a graph of order n that does not contain a complete subgraph Kt of order t equals
n 2 − t−1
  ni i=1 2 (1) where n = n1 + · · · + nt−1 is a partition of n into t − 1 parts which are as equal as possible. Furthermore, the only graph of order n whose number of edges equals (1) that does not contain a complete subgraph Kt is the complete (t − 1)-partite graph Kn1 ,...,nt−1 with parts of sizes n1 , . . . , nt−1 , respectively. Denote this graph by Tt−1 (n) and its edge number by Tt−1 (n). Brualdi and Mellendorf raised the following extremal problem as a modification of a special case of Turán’s theorem (see [1]). Problem 1. Let t and n be positive integers with n
t
2. Determine the maximum number of edges of a graph of order n that contains neither the complete graph Kt nor the complete bipartite graph Kt,t as a subgraph. A solution of Problem 1 for n = 2t is given in [1]. We consider the following problem as a modification of the next special case of Turán’s theorem. Problem 2. Let t and n be positive integers with n
t
2. Determine the maximum number of edges of a graph of order n that contains neither the complete graph Kt nor the complete tripartite graph Kt,t,t as a subgraph. Before stating our theorems and a solution of Problem 2 for n = 3t, we introduce some notation. Let G and H be graphs, then their sum is the graph G + H obtained by taking disjoint copies of G and H and joining each vertex of G to each vertex of H by an edge. The union of G and H is the graph denoted by G ∪ H consisting  of disjoint copies of G and H. The join of G1 , G2 , . . . , Gk is a graph obtained from ki=1 Gi by adding k − 1 edges to make it connected. (Attention, the join is not a well-defined graph since we have many choices where to put the extra k − 1 edges. We use this notation when it does not count what particular graph is chosen.) Let m, k be positive integers in the following definitions. Let mG be the graph consisting of m disjoint copies of G. The complete t-partite graph with parts of size n1 , . . . , nt is denoted by Kn1 ,...,nt . We call a graph bisectable provided its vertices can be partitioned into two parts of equal size such that there are no edges between the two parts. We call a graph trisectable provided its vertices can be partitioned into three parts of equal size such that there are no edges between the three parts. A path of order m is denoted by Pm . It is an odd path if m is even. A tree of order m is denoted by Tm . (I.e., Tm does not denote a particular tree of m vertices, it can be any tree of m vertices.) A cycle of order m is denoted by Cm . A connected graph is unicyclic if it has a unique cycle, i.e. if its edge number is equal to its vertex number. 58 E. Győri, Y. Rho / Discrete Mathematics 303 (2005) 56 – 64 A unicyclic graph of order m is denoted by Um . The complement of a graph G is denoted by G. A graph with k more edges added to G is denoted by Gk+ . For simplicity, G1+ is denoted by G+ . A set of vertices of a graph is independent if no two vertices of it are joined by an edge. The independence number of G is the maximum cardinality of an independent set of G and is denoted by
(G). If m is odd, then Hm denotes a unicyclic graph of order m with
(Hm ) = (m − 1)/2. The graph G is 2-edge connected if at least two edges of G need to be removed to make it disconnected. When W is a vertex set of G, an induced subgraph G[W ] of G has vertex set W and all edges of G whose both endpoints are contained in W . For a vertex v of G, the degree of v which is denoted by d(v) is the number of vertices which are adjacent to v. G denotes the minimum degree of all the vertices of G. The following theorem of Brualdi and Mellendorf (see [1]) solves Problem 1 for n=2t. It is interesting that the order of magnitude of the edge number is different for t’s of different parities. Theorem 2. Let t be a positive integer with t
3. Then the maximum number of edges of a graph of order 2t that contains neither Kt nor Kt,t as a subgraph equals   2t − t − 4 if t is odd (2) 2 and equals   2t 3t − −1 2 2 if t is even. (3) If t is odd, then the only graphs of order 2t that contain neither Kt nor Kt,t as a subgraph and whose number of edges equals (2) are the graphs of the form K2,...,2,4 and K2,...,2 + V where V is the graph obtained from K3,3 by removing an edge, and the graphs K1,3,3,3 and K3,3 + P4 for t = 5. If t is even, then the only graphs of order 2t that contain neither Kt nor Kt,t as a subgraph and whose number of edges equals (3) are the graphs of the form K2,...,2 + Ha + Hb where a and b are odd integers with a + b = t + 2. The following theorem solves Problem 2 for n = 3t. It is very interesting again that the order of magnitude of the edge numbers are different for t’s divisible and not divisible by 3. Theorem 3. Let t be a positive integer with t
4. Then the maximum number of edges of a graph of order 3t that contains neither Kt nor Kt,t,t equals   3t − 3t − 7 if t ≡ 1 mod 3, (4) 2   3t − 3t − 6 if t ≡ 2 mod 3, (5) 2 and equals   3t t − 3t − − 3 if t ≡ 0 mod 3. 2 3 (6) E. Győri, Y. Rho / Discrete Mathematics 303 (2005) 56 – 64 59 If t ≡ 1 mod 3, then the only graphs of order 3t that contain neither Kt nor Kt,t,t as a subgraph and whose number of edges equals (4) are the graphs of the form K3,...,3,4,5 , K3,...,3,4 + A where A is a join of two K4 ’s, K3,4,4,4 + B where B is a join of two K3 ’s or K2,3,4,4,4,4 when t = 7. If t ≡ 2 mod 3, then the only graphs of order 3t that contain neither Kt nor Kt,t,t as a subgraph and whose number of edges equals (5) are the graphs of the form K3,...,3,4,4,4 . If t ≡ 0 mod 3, say t =3s, then the only 3t vertex graphs that contain neither Kt nor Kt,t,t as a subgraph and whose edge number equals (6) are of form K3,...,3 + F3r1 +1 + F3r2 +1 + Ft−3r1 −3r2 +1 for t
9 where the i vertex graphs Fi are joins of a K4 and appropriately many K3 ’s. We prove Theorem 3 in its equivalent complementary form stated in the next theorem. Theorem 4. Let t be a positive integer with t
4. Then the minimum number of edges of a graph of order 3t that contains neither an independent set of t vertices nor contained in Kt,t,t equals 3t + 7 if t ≡ 1 mod 3, (7) 3t + 6 if t ≡ 2 mod 3, (8) and equals 3t + t +3 3 if t ≡ 0 mod 3. (9) If t ≡ 1 mod 3, then the only non-trisectable graphs of order 3t that does not contain t independent vertices and whose edge number equals (7) are the graphs of the form (t − 3)K3 ∪ K4 ∪ K5 , (t − 4)K3 ∪ K4 ∪ A where A is a join of two K4 ’s, K3 ∪ 3K4 ∪ B where B is the join of two K3 ’s, and K2 ∪ K3 ∪ 4K4 when t = 7. If t ≡ 2 mod 3, then the only non-trisectable graphs of order 3t that does not contain t independent vertices and whose edge number equals (8) are the graphs of the form (t − 4)K3 ∪ 3K4 . If t ≡ 0 mod 3, say t = 3s then the only non-trisectable graphs of order 3t that does not contain t independent vertices and whose edge number equals (6) are of the following form: the union of 2s + 2 components, namely 2s − 1 K3 ’s, F3r1 +1 , F3r2 +1 and Ft−3r1 −3r2 +1 for t
9 where the graphs Fj are joins of a K4 and (j − 4)/3 triangles. 2. Proof of Theorem 4 Before the proof of Theorem 4, we prove some edge number estimates for connected graphs with given independence numbers. These lemmas will be used later in the proof of the main theorem. Lemma 5. Let G be a connected graph. Then e
2n − 2
− 1 where the equality holds only for graphs that are joins of odd cycles and paths. 60 E. Győri, Y. Rho / Discrete Mathematics 303 (2005) 56 – 64 Proof. The inequality of the lemma is proved in [2]. Let G be a graph where the equality of the lemma holds. Suppose that G contains an even cycle C. Then delete an edge of it and extend the remaining path to a spanning tree T. T has a bipartition with vertex sets A and B. Then
(G)
max(
(G[A]),
(G[B])). Adding one edge may decrease either
(G[A]) or
(G[B]) by one, which gives the theorem. However, adding the last edge of C decreases neither
(G[A]) nor
(G[B]), so then equality cannot hold. If G has two odd cycles sharing an edge or at least two vertices then G contains an even cycle. So if equality holds, then it consists of edge disjoint cycles joined by paths along a tree. (The length of the path can be 0 too.) Among them only the graphs of the form which consists of edge disjoint cycles and odd paths joined along a tree satisfy the inequality. Thus the theorem follows.  Lemma 6. Let G be a connected graph. Then e(G)
3n − 5
(G) − 1 where equality holds iff the graph is of the form of a tree replacing each vertex by either K3 or K4 , i.e. if G is a join of K3 ’s and K4 ’s. Furthermore: (i) If every maximum size independent set in G contains x then e(G)
3n − 5
(G) + 2 where equality holds iff G is a join of x, some K3 ’s and K4 ’s. Proof. We prove the statement by an induction on n. We will go back by 2, 3 or 4 vertices, so it is sufficient and easy to check it for 1, 2, 3, 4. (We do not have equality for n = 1, 2, 5.) Assume that the main statement of the lemma holds if v(G) < n, (i) holds if v(G) < n−3. First we prove (i). Let G be a connected graph with n − 3 vertices and with edge number e(G) = e and
(G) =
and suppose that the main statement of the lemma holds for graphs with less than n vertices. Suppose that every maximum size independent set contains x. Then add two vertices u, w and three edges xu, xw, uw to G to get the graph G0 . Since every
(G) element independent set in G contains x, so
(G0 ) =
and by the induction hypothesis, e(G0 )
3(n − 1) − 5
(G) − 1 where equality holds iff G0 is a join of K3 ’s and K4 ’s. It follows that e(G)
3(n − 3) − 5
(G) + 2 where equality holds iff G is a join of x, some K3 ’s and K4 ’s. Now we prove the main statement of the lemma. Let G be a connected graph with n vertices and with edge number e(G) = e and
(G) =
and suppose that the lemma holds for graphs with less than n vertices, including (i) for v(G)n − 3 and that the extremal graphs are only the graphs described in the lemma. If
n/5 then by Theorem 1 (not using connectivity) we have at least 2n > 3n − 5
− 1 = 2n − 1 edges. If
> n/5 then 3n − 5
< 2n − 1, so we may assume that G has a vertex of degree at most 3. We distinguish three cases according to the minimum degree G of G. Let  be G . Case 1:  = 1. Let d(v) = 1, x be the neighbor of v, and x1 , . . . , xk be the other neighbors of x. Then take G−v −x and add a minimal subset of the k −2 possibly missing edges x1 x2 , x1 x3 , . . . , x1 xk to the remaining graph what makes it connected. The resulting graph H is connected, has n − 2 vertices, at most e − 2 edges and
(H ) is at most
− 1. (If
, then adding v to the E. Győri, Y. Rho / Discrete Mathematics 303 (2005) 56 – 64 61 maximum independent set, we get an
+ 1 element independent set in G, a contradiction.) By the induction hypothesis, e is at least e(H ) + 2 which is at least 3(n − 2) − 5(
− 1) − 1 + 2 = 3n − 5
. Case 2:  = 2. Let d(v)=2, the neighbors of v are x and y, their further neighbors (not v, x or y)x1 , . . . , xk and y1 , . . . , yl , respectively. Case 2.1: x and y are adjacent. Then delete v, x and y from G and add a minimal subset of the possibly missing edges x1 x2 , x1 x3 , . . . , x1 xk , x1 y1 , . . . , x1 yl to it to make it connected. The resulting connected graph H has n − 3 vertices, at most e − 4 edges and
(H )is at most
− 1. (If not, then add v to get
+ 1 independent vertices in G, a contradiction.) By the induction hypothesis, e is at least e(H ) + 4 which is at least 3(n − 3) − 5(
− 1) − 1 + 4 = 3n − 5
− 1. If equality holds then H has exactly e − 4 edges, we had to add all the edges x1 x2 , x1 x3 , . . . , x1 xk , x1 y1 , . . . , x1 yl to it to make the graph connected and H is the join of K3 ’s and K4 ’s. But then G is a join of K3 ’s and K4 ’s too, we just “inserted” the triangle vxy. Case 2.2: x and y are not adjacent. Then delete v and x and add a minimal subset of the possibly missing edges yx 1 , . . . , yx k to make the graph connected. The resulting graph H has n − 2 vertices and at most e − 2 edges.
(H ) is at most
− 1 since if S is an independent set of
elements in H then if y is in S then S ∪ x, if y is not in S then S ∪ v is independent in G, a contradiction. By the induction hypothesis, e is at least e(H ) + 2 which is at least 3(n − 2) − 5(
− 1) − 1 + 2 = 3n − 5
. Case 3:  = 3. First we prove that we may assume that G is 2-edge-connected. Suppose that xy is a bridge in G and Gx and Gy are the n1 and n2 vertex components of G − xy containing x and y, respectively. Both Gx and Gy have at least n − 4 vertices because of the minimum degree condition. If S and T are maximum size independent sets in Gx and Gy , respectively, then S ∪ T is a maximum size independent set in G unless x ∈ S and y ∈ T , when deleting x or y we get an independent set in G. So
(G)
(Gx ) +
(Gy ) 
(G) + 1. Suppose that
(Gx )+
(Gy )=
(G)+1. Then every maximum size independent set in Gx and Gy contains x and y, respectively. Then e(Gx )
3n1 − 5
(Gx ) + 2, and e(Gy )
3n2 − 5
(Gy ) + 2 by (i). It follows that e(G)
3n − 5
(G). Now suppose that
(Gx ) +
(Gy ) =
(G). By the induction hypothesis, e(Gx )
3n1 − 5
(Gx ) − 1 and e(Gy )
3n2 − 5
(Gy ). It follows that e(G)
3n − 5
(G) − 1 and equality holds only if we had equalities in the used estimates, i.e. if Gx and Gy are joins of K3 ’s and K4 ’s and then G is the join of K3 ’s and K4 ’s, too. Let d(v) = 3, and let x, y, z denote the neighbors of v. Consider the graph G − v. It has at most two components because of the 2-edge-connectivity of G. Note that G − v is connected because if not then one of the edges vx, vy, vz is a bridge. Clearly,
(G − v) 
. If we have strict inequality then by the induction hypothesis, e(G − v)
3(n − 1) − 5
(G − v) − 1 > 3(n − 1) − 5
− 1 and so e(G) > 3n − 5
− 1. If
(G − v) =
then by the induction hypothesis e(G − v)
3(n − 1) − 5
(G − v) − 1 = 3(n − 1) − 5
− 1 and we have equality iff G − v is the join of K3 ’s and K4 ’s. If x, y, z belong to a K4 block xyzw of G − v then v and w can be extended to an
+ 1 element independent set in G going from the block xyzw on the branches of the constructing tree of 62 E. Győri, Y. Rho / Discrete Mathematics 303 (2005) 56 – 64 G − v since always possibly at most one vertex cannot be taken from the next block, the neighbor of the previous block. If x, y, z constitute a block (i.e. are not contained in a K4 ) then G is the join of xyzv and K3 ’s and K4 ’s. If x, y, z do not belong to the same block then we construct an
element independent set in G − v that does not contain any vertex from x, y, z which can be extended to an
+ 1 independent set in G by adding v, a contradiction. If two vertices out of x, y, z, say x and y are contained in the same block, then start with this block, otherwise we can start with any block. Take a vertex from the starting block, then take the neighboring blocks. Choose a vertex which is not a neighbor of the already settled block—one vertex forbidding—and different from x, y and z—at most one vertex forbidding, so it can be done because each block has at least three vertices and the procedure can be finished by taking the neighboring block of the settled blocks one by one along the tree structure of the join.  Proof of Theorem 4 for 3∤ t. Theorem 1 implies that if
(G)t − 2 then e(G)
36 + 3(t − 8) = 3t + 12 > 3t + 7. So, we may assume that
(G) = t − 1 and so the number of components is at most t −1. Applying Lemma 6 to each component, we get that e(G)
9t − 5(t − 1) − k where k is the number of components in which Lemma 6 holds with equality. Thus e(G)
3t + 6 and if equality holds then k = t − 1 and G has t − 1 complete graph components which should be K3 ’s and K4 ’s, namely t − 4 triangles and three K4 ’s. It is not trisectable if t ≡ 2 mod 3, so we are done in that case. If t ≡ 1 mod 3 then this graph is trisectable, so we get e(G)
3t + 7 and we may have equality only if k = t − 2. If G has t − 1 components then they are complete but one of them is not a K3 or K4 and in that component G0 we have e(G0 ) = 3v(G0 ) − 5
(G0 ) = 3v(G0 ) − 5. Then G0 is either a K5 and the components of G are this K5 , a K4 and t − 3 triangles or a K2 and the components are this K2 , four K4 ’s and t − 6 triangles, but the second one is only for t = 7, because otherwise G is trisectable. If G has t − 2 components then Lemma 6 holds with equality in each of them, and we have one non-complete component G0 with
= 2. By Lemma 6, G0 is the join of two K3 ’s, two K4 ’s or a K3 and a K4 , when the number of components K4 is 3,1,2, respectively, and the other components are triangles. However, the graph in the first or third case is trisectable and we get extremal graphs just in the second case.  Proof of Theorem 4 for 3|t. The following Lemma provides the proof of Theorem 4 for 3 | t. Lemma 7. Let G be a graph of order 3t where t | 3. Assume that G does not contain an independent set of size t and not trisectable. Then G has at least 10/3t + 3 edges. Proof. Let t = 3s. Apply Lemma 6 to each component Gi of G. Let ni , ei , and
i denote the order, the edge number and the independence number of the ith component. Then ei
3ni − 5
i − 1. Let k denote the number of components with equality in this estimate. Then adding up these inequalities, we obtain that e
9t − 5
− k
9t − 5t + 5 − k which is sufficient if k < 2s + 3, since
t − 1. So, we may assume that k
2s + 3. If we want to determine the extremal cases too then we still may assume that k
2s + 2. E. Győri, Y. Rho / Discrete Mathematics 303 (2005) 56 – 64 63 Similarly, apply Lemma 5 to each component of G. Then ei
2ni − 2
i − 1. Let l denote the number of components with equality in this estimate. Then adding up these inequalities, we obtain that e
6t − 2
− l which is sufficient if l < 2s. So, we may assume that l
2s and if we want to determine the extremal cases too then we still may assume that l
2s − 1. Before the further steps of the proof, we make some observations. 1. Note that if a component Gi has equality in the estimates of both Lemma 5 and 6 then ei = 3ni − 5
i − 1 = 2ni − 2
i − 1 and so ni = 3
i , i.e. 3 | ni . 2. If we have equality in Lemma 5 then n = 1 and if we have equality in Lemma 10 then n = 1, 2, 5. Note that if we have, say, x components with 3ni − 5
i − 1 edges where 3|ni − 2 then in these components the number of edges is (2ni − 2
i − 1) + (ni − 3
i ) where ni − 3
i is at least 2, so Lemma 5 not just not sharp but the gap is at least 2, so we must have not just 2s but 2s + x components where Lemma 5 holds with equality. Now, let us color the number ni blue if it has 2ni − 2
i − 1 edges and color it red if it has 3ni − 5
i − 1 edges. (A number may get two colors!) Now we prove a claim for these colored numbers which implies that the graph is trisectable. Claim. Suppose that ni (i ∈ I ) is a system of (not necessarily distinct) positive integers colored with red and blue such that  (a) i∈I ni = n = 9s; (b) every ni has 0,1 or 2 colors but if ni is red and blue at the same time then 3|ni ; (c) the number k of red numbers is at least 2s + 2, let x denote the number of red numbers congruent to 2 mod 3; (d) the number l of blue numbers is at least 2s − 1 + x; (e) ni = 1 if ni is blue, (f) ni = 1, 2, 5 if ni is red. Then  ni is trisectable, i.e., there is a partition I1 , I2 , I3 of I such  of integers  the system that i∈I1 ni = i∈I2 ni = i∈I3 ni = 3s unless we have 2s − 1 red and blue 3’s and three red numbers congruent to 1 mod 3 which case will be called extremal. Remark. Note that this claim finishes the proof of the main theorem. Proof of Claim. We prove the Claim by induction on s. We skip the case by case analysis proof if s 7. We do not get a trisection if s = 3 and the ni ’s are 4, 4, 4, 3, 3, 3, 3, 3. Because of the induction, we can get equality for bigger values of s too if we have 2s − 1 3’s (red and blue) and three red numbers 3r1 + 1, 3r2 + 1, 3r3 + 1 where ri
1 and r1 + r2 + r3 = s. Let m denote the number of ni ’s equal to 3. Then we have at least 2s − m + x − 1 blue numbers different from 3 which are at least 2 and at least 2s + 2 − m red numbers different from 3 which are at least 4. Thus, n
sum of the colored numbers
3m + 2(2s − m + x − 1) + 4(2s + 2 − m). This argument did not care the case when a number different from 3 has both colors but they are at least 6 = 2+4 by (b), so the estimate above is correct. So, we 64 E. Győri, Y. Rho / Discrete Mathematics 303 (2005) 56 – 64 get n = 9s
12s + 6 − 3m + 2x which implies that m
s + 2 + 2x/3. So, we have at least ten 3’s. If we have a colored ni0
6 such that ni0 is not a red 8 or we have an uncolored ni0
3 then delete two 3’s from among the ni ’s and replace ni0 by n′i0 = ni0 − 3 (which is a deletion too if we get 0) and color it with the possible color(s) of ni0 . It is easy to see that the new system satisfies the conditions (a)–(f) for s − 1, so it is either trisectable or extremal. But, it is easy to get a trisection of the original system by replacing n′i1 by ni0 (or put ni0 into any subsystem if it is 3 and add a 3 to both other subsystems. And if the system is extremal for s − 1 then the original system is extremal for s. So, we may assume that the red numbers are 3’s, 4’s and 8’s, the blue ones are 2’s, 3’s and 4’s and the uncolored ones are 1’s and 2’s. Now, we prove that there are twodisjoint subsystems S1 , S2 of the red numbers with  (disjoint) index sets I1 , I2 such that i∈I1 ni = i∈I2 ni = 3s unless we have an extremal system. First put the 8’s (8+8+8) into S1 and S2 so that the sum of each subsystem should be divisible by 3 (i.e. three by three) and should be at most 3s. Then put 8’s and 4’s (8 + 4) or just 4’s (4 + 4 + 4) into them keeping the same rules. Finally, try to put 3’s into them to get the sum 3s in each subsystem. If we can do it then we are done, the remaining numbers constitute the third subsystem. If we cannot get 3s in both subsystems then look at just the 8’s and 4’s in the subsystems. These 8’s and 4’s and all the 3’s add up to at most 6s − 3, so the number of these numbers are at most 2s − 1 and we have at least 3 numbers (8’s or 4’s) not put  into any Si . But then 8 + 4 or 4 + 4 + 4 can be put into some subsystem unless n , i∈I2 ni
3s −9 and we have at least ten 3’s so we can reach 3s in the subsystems i∈I1 i adding some 3’s to them in this case. So we may assume  have no 4’s outside the  that we n , subsystems. Then we have at least three 8’s there. Then i∈I2 ni
3s −21. If, say, i∈I1 i  i ∈ I1
3s − 9 then ten 3’s are still sufficient, so we may assume that the sum of the 8’s and 4’s is at most 3s −12 in both subsystems. If we have a 4 in one of the subsystems then we can replace a 4+4+4 or an 8+4 by 8+8+8, and we get a previous case. However, if s
8 then each subsystem must contain atleast three 8’s, so x
9 and m
s +2+2x/3
8+2+6 > 14 i.e. we can add the appropriate number of 3’s to the subsystems to get 3s.  References [1] R.A. Brualdi, S. Mellendorf, Two extremal problems in graph theory, The Electron. J. Combin. 1 (1994) #R2. [2] Y. Rho, An extremal problem concerning graphs not containing Kt and Kt,n−t , Discrete Math. 187 (1998) 195–209.