An extremal graph problem with a transcendental solution

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1 An extremal graph problem with a trancendental olution Dhruv Mubayi Caroline Terry April 1, 017 Abtract We prove that the number of multigraph with vertex et {1,..., n} uch that every four vertice pan at mot nine edge i a n +o(n where a i trancendental (auming Schanuel conjecture from number theory. Thi i an eay conequence of the olution to a related problem about maximizing the product of the edge multiplicitie in certain multigraph, and appear to be the firt explicit (omewhat natural quetion in extremal graph theory whoe olution i trancendental. Thee reult may hed light on a quetion of Razborov who aked whether there are conjecture or theorem in extremal combinatoric which cannot be proved by a certain cla of finite method that include Cauchy-Schwarz argument. Our proof involve a novel application of Zykov ymmetrization applied to multigraph, a rather technical progreive induction, and a traightforward ue of hypergraph container. 1 Introduction All logarithm in thi paper are natural logarithm unle the bae i explicitly written. Given a et X and a poitive integer t, let ( X t = {Y X : Y = t}. A multigraph i a pair (V, w, where V i a et of vertice and w : ( V N = {0, 1,,...}. Definition 1. Given integer and q 0, a multigraph (V, w i an (, q-graph if for every X ( V we have xy ( X w(xy q. An (n,, q-graph i an (, q-graph with n vertice, and F (n,, q i the et of (n,, q-graph with vertex et := {1,..., n}. The main goal of thi paper i to prove that the maximum product of the edge multiplicitie over all (n, 4, 15-graph i γn +O(n (1 where γ = β + β(1 βlog 3 log and β = log 3 log 3 log. It i an eay exercie to how that both β and γ are trancendental by the Gelfond-Schneider theorem [9]. Uing (1, we will prove that F (n, 4, 9 = a n +o(n, ( where a = γ i alo trancendental (auming Schanuel conjecture in number theory. Department of Mathematic, Statitic, and Computer Science, Univerity of Illinoi at Chicago. upported in part by NSF Grant DMS ; mubayi@uic.edu Department of Mathematic, Univerity of Maryland, College Park; cterry@umd.edu Reearch 1

2 Due to the Erdő-Simonovit-Stone theorem [5, 6], many natural extremal graph problem involving edge denitie have rational olution, and their enumerative counterpart have algebraic olution. For example, the Erdő-Kleitman-Rothchild theorem [7] tate that the number of triangle-free graph on i n /4+o(n and 1/4 i algebraic ince 1/4 i rational. For hypergraph the ituation i more complicated, and the firt author and Talbot [14] proved that certain partite hypergraph Turán problem have irrational olution. Going further, the quetion of obtaining trancendental olution for natural extremal problem i an intriguing one. Thi wa perhap firt explicitly poed by Fox (ee [17] in the context of Turán denitie of hypergraph. Pikhurko [17] howed that the et of hypergraph Turán denitie i uncountable, thereby proving the exitence of trancendental one (ee alo [10], but hi lit of forbidden hypergraph wa infinite. When only finitely many hypergraph are forbidden, he obtained irrational denitie. To our knowledge, (1 and ( are the firt example of fairly natural extremal graph problem whoe anwer i given by (explicitly defined trancendental number (modulo Schanuel conjecture in the cae of (. Another area that (1 may hed light on i the general quetion of whether certain proof method uffice to olve problem in extremal combinatoric. The exploion of reult in extremal combinatoric uing Flag Algebra [18] in recent year ha put the potlight on uch quetion, and Razborov firt poed thi in (Quetion 1, [18]. A ignificant reult in thi direction i due to Hatami and Norine [11]. They prove that the related quetion (due in different form to Razborov, Lováz, and Lováz-Szegedy of whether every true linear inequality between homomorphim denitie can be proved uing a finite amount of manipulation with homomorphim denitie of finitely many graph i not decidable. While we will not attempt to tate (Quetion 1, [18] rigorouly here, it motivation i to undertand whether the finite method that are typically ued in combinatorial proof of extremal reult (formalized by Flag Algebra and the Cauchy-Schwarz calculu uffice for all extremal problem involving ubgraph denitie. Although we cannot ettle thi, one might peculate that thee finite Cauchy-Schwarz method may not be enough to obtain (1. In any event, (1 eem to be a good tet cae. Curiouly, our initial exploration into (1 were through Flag Algebra computation which gave the anwer to everal decimal place and motivated u to obtain harp reult, though our eventual proof of (1 ue no Flag Algebra machinery. Intead, it ue ome novel extenion of claical method in extremal graph theory, and we expect that thee idea will be ued to olve other related problem. A remarked earlier, ( i a fairly traightforward conequence of (1 and ince the expreion in (1 i obtained a a product (rather than um of number, it i eaier to obtain a trancendental number in thi way. However, we hould point out that an extremal example for (1 (and poibly all extremal example, though we were not able to how thi involve partitioning the vertex et into two part where one part ha ize approaching βn, and β i alo trancendental (ee Definition 3 and Theorem in the next ection. Thi might indicate the difficulty in proving (1 uing the ort of finite method dicued above. Finally, we would like to mention that the problem of aymptotic enumeration of (n,, q- graph i a natural extenion of the work on extremal problem related to (n,, q-graph by Bondy and Tuza in [4] and by Füredi and Kündgen in [8]. Further work in thi direction, including a ytematic invetigation of extremal, tability, and enumeration reult for a large cla of pair (, q, will appear in [16] (ee alo [15] for another example on multigraph. Alon [1] aked whether the trancendental behavior witneed here i an iolated cae. Although we believe that there are infinitely many uch example (ee Conjecture 1 in Section 7 we were not able to prove thi for any other pair (, q. The infinitely many pair for which we obtain precie extremal reult in [16] have either rational or integer denitie.

3 Reult Given a multigraph G = (V, w, define P (G = xy ( V w(xy and S(G = xy ( V w(xy. Definition. Suppoe and q 0 are integer. Define ( ex Π (n,, q = max{p (G : G F (n,, q}, and ex Π (, q = lim ex Π (n,, q ( n 1. n An (n,, q-graph G i product-extremal if P (G = ex Π (n,, q. The limit ex Π (, q (which we will how alway exit i called the aymptotic product denity. Our firt main reult i an enumeration theorem for (n,, q-graph in term of ex Π (, q + (. Theorem 1. Suppoe and q 0 are integer. If ex Π (, q + ( > 1, then ( ( n ( ( (1+o(1( n ex Π (, q + F (n,, q exπ, q +, and if ex Π (, q + ( 1, then F (n,, q o(n. Theorem 1 will be proved in Section 4 uing the hypergraph container method of [3, 19] along with a multigraph verion of the graph removal lemma. Theorem 1 reduce the problem of enumerating F (n, 4, 9 to computing ex Π (4, 15. Thi will be the focu of our remaining reult. Definition 3. Given n, let W (n be the et of multigraph G = (, w for which there i a partition L, R of uch that w(xy = 1 if xy ( ( L, w(xy = if xy R, and w(xy = 3 if (x, y L R. Notice that W (n F (n, 4, 15 for all n N. Straightforward calculu how that for G log 3 W (n, the product P (G i maximized when R βn, where β = log 3 log i a trancendental number. Thi might indicate the difficulty of obtaining thi extremal contruction uing a tandard induction argument. Given a family of hypergraph F, write P(F for the et of G F with P (G = max{p (G : G F}. Ue the horthand P(n,, q for P(F (n,, q. Theorem. For all ufficiently large n, P(W (n P(n, 4, 15. Conequently ex Π (n, 4, 15 = max P (G = +O(n G W (n γn and ex Π (4, 15 = γ, where γ = β / + β(1 β log 3 and β = log 3 log 3 log. For reference, β.73 and γ The reult below follow directly from Theorem 1 and. Theorem 3. F (n, 4, 9 = γn +o(n. Proof. Theorem implie ex Π (4, 15 = γ > 1. By Theorem 1, thi implie that ex Π (4, 15 (n F (n, 4, 9 exπ (4, 15 (1+o(1(n. Conequently, F (n, 4, 9 = γ(n +o(n = γn +o(n. 3

4 Recall that Schanuel conjecture from the 1960 (ee [1] tate the following: if z 1,..., z n are complex number which are linearly independent over Q, then Q(z 1,..., z n, e z 1,..., e zn ha trancendence degree at leat n over Q. A promied in the introduction and abtract, we now how that auming Schanuel conjecture, γ i trancendental. Oberve that thi implie ex Π (4, 15 = γ i alo trancendental over Q, auming Schanuel conjecture. Propoition 1. Auming Schanuel conjecture, γ i trancendental. Proof. Aume Schanuel Conjecture hold. It i well-known that Schanuel conjecture implie log and log 3 are algebraically independent over Q (ee for intance [0]. Oberve γ = f(log,log 3 g(log,log 3 where f(x, y = xy / + y (y x and g(x, y = x(y x. Note the coefficient of x 3 in f(x, y i 0 while in g(x, y it i 1. We now how log, log 3, γ log are linearly independent over Q. Suppoe toward a contradiction that thi i not the cae. Then there are non-zero rational p, q, r uch that f(log,log 3 g(log,log 3 p log + q log 3 + rγ log = 0. f(log,log 3 Replacing γ with, thi implie p log + q log 3 + r g(log,log 3 log = 0. By clearing the denominator of p, q, r and multiplying by g(log, log 3, we obtain that there are integer a, b, c, not all zero, uch that (a log + b log 3g(log, log 3 + cf(log, log 3 log = 0. Let p(x, y = (ax + byg(x, y + cf(x, yx. Oberve that p(x, y i a rational polynomial uch that p(log, log 3 = 0. Since the coefficient of x 3 i 1 in g(x, y and 0 in f(x, y, the coefficient of x 4 in p(x, y i a 0. Thu p(x, y ha at leat one non-zero coefficient, contradicting that log and log 3 are algebraically independent over Q. Thu log, log 3, γ log are linearly independent over Q, o Schanuel conjecture implie Q(log, log 3, γ log, γ ha trancendence degree at leat 3 over Q. Suppoe toward a contradiction that γ i not trancendental. Then log, log 3, γ log mut be algebraically independent over Q. Let h(x, y, z = zg(x, y xf(x, y. Then it i clear h(x, y, z ha non-zero coefficient, and h(log, log 3, γ log = (γ log g(log, log 3 (log f(log, log 3 = 0, f(log,log 3 where the econd equality ue the fact that γ = g(log,log 3. But thi implie log, log 3, γ log are algebraically dependent over Q, a contraction. Thu γ i trancendental. 3 General enumeration in term of aymptotic product denity In thi ection we prove Theorem 1, our general enumeration theorem for (n,, q-graph. We will ue a verion of the hypergraph container theorem (Balogh-Morri-Samotij [3], Saxton-Thomaon [19], a graph removal lemma for edge-colored graph, and Propoition below, which how ex Π (, q exit for all and q 0. Given G = (V, w and X V, let G[X] = (X, w ( X. Propoition. For all n and q 0, ex Π (, q exit and ex Π (n,, q ex Π (, q (n. If q (, then exπ (, q 1. Proof. Fix and q 0. Clearly, for all n, b n := (ex Π (n,, q ( n 0. We now how the b n are non-increaing. For n > and G F (n,, q, note ( 1/(n ( P (G = P (G[ \ {i}] b (n 1 1/(n n( n 1 n 1 = b /n n 1 = b (n n 1. i i 1 4

5 Therefore, for all G F (n,, q, P (G 1/(n bn 1, o b n b n 1 and lim n b n = ex Π (, q exit. The inequality ex Π (n,, q ex Π (, q (n follow becaue the bn are non-increaing. If q (, then for all n, the multigraph G = (, w, where w(xy = 1 for all xy (, i in F (n,, q. Thi how b n 1 for all n, o by definition, ex Π (, q 1. We now tate a verion of the hypergraph container theorem. Specifically, Theorem 4 below i a implified verion of Corollary 3.6 from [19]. We firt require ome notation. Given r, an r-uniform hypergraph i a pair H = (W, E where W i a et of vertice and E ( W r i a et of edge. Given C W, H[C] i the hypergraph (C, E ( C r. The average degree of H i d = r E / W, and e(h = E i the number of edge in H. Given a et X, X denote the power et of X. Definition 4. Suppoe H = (W, E i an r-uniform hypergraph with average degree d and fix τ > 0. For every σ W, x W, and j [r], et d(σ = {e E : σ e} and d (j (x = max{d(σ : x σ W, σ = j}. If d > 0, then for each j [r], define j = j (τ to atify the equation j τ j 1 nd = x W d (j (x and et (H, τ = (r 1 If d = 0, et (H, τ = 0. The function (H, τ i called the co-degree function. r j= (j 1 j. Theorem 4 (Corollary 3.6 from [19]. Fix 0 < ɛ, τ < 1. Suppoe H i an r-uniform hypergraph with vertex et W of ize N atifying (H, τ ɛ 1r!. Then there exit a poitive contant c = c(r and a collection C W uch that the following hold. (i For every independent et I in H, there i ome C C, uch that I C. (ii For all C C, we have e(h[c] ɛe(h. (iii log C c log(1/ɛnτ log(1/τ. Our next goal i to prove a verion of Theorem 4 for multigraph. Suppoe G = (V, w i a multigraph. For all xy ( V, we will refer to w(xy a the multiplicity of xy. The multiplicity of G i µ(g = max{w(xy : xy ( V }. Given another multigraph, G = (V, w, we ay that G i a ubmultigraph of G if V = V and for each xy ( V, w(xy w (xy. Definition 5. Suppoe and q 0 are integer. Set H(, q = {G = ([], w : µ(g q and S(G > q}, and g(, q = H(, q. If G = (V, w i a multigraph, let H(G,, q = {X ( V : G[X] = G ome G H(, q}. Oberve G = (V, w i an (, q-graph if and only if H(G,, q =. Suppoe n i an integer. We now give a procedure for defining a hypergraph H(n = (W, E. Set [0, q] = {0, 1,..., q}. The vertex et of H(n i W = {(f, u : f (, u [0, q]}. The idea i that each pair (f, u correpond to the choice the multiplicity of f i u. The edge et E of H(n conit of all et of the form {(f, w(f : f ( A }, where A and (A, w = G for ome G H(, q. For any σ W, define V (σ be the et of all i appearing in an element of σ, i.e., V (σ = { i : there i ome (f, u σ with i f }. 5

6 Oberve that for all e E, e = ( (, V (e =, and for all f V (e, there i exactly one u [0, q] uch that (f, u e. By definition, H(n i an ( ( -uniform hypergraph, W = (q + 1 n, and E = g(, q ( n. We now prove our multigraph verion of Theorem 4. Many of the computation in the proof are modeled on thoe ued to prove Corollary 3 in [13] Theorem 5. For every 0 < δ < 1 and integer, q 0, there i a contant c = c(, q, δ > 0 uch that the following hold. For all ufficiently large n, there i G a collection of multigraph of multiplicity at mot q and with vertex et uch that (i for every J F (n,, q, there i G G uch that J i a ubmultigraph of G, (ii for every G G, H(G,, q δ ( n, and (iii log G cn 1 4 log n. Proof. Clearly it uffice to how Theorem 5 hold for all 0 < δ < 1/. Fix 0 < δ < 1/ and integer, q 0. Let c 1 = c 1 ( ( be from Theorem 4, and et c = c(, q, δ = c 1 g(,q 4 log( δ (q + 1. Aume n i ufficiently large. We how there i a collection G of multigraph with multiplicity at mot q and vertex et uch that (i-(iii hold for thi c and δ. Let H := H(n be the ( -uniform hypergraph decribed above. In particular, H = (W, E where ( W = {(f, u : f, u [0, q]} and E = {{ (f, w(f : f ( A } : A and (A, w = G, for ome G H(, q}. Set ɛ = δ g(,q, τ = n 1 4, and N = W. We how the hypothee of Theorem 4 are atified by H with thi ɛ and τ. Since n i ufficiently large, 0 < τ < 1/. By definition of ɛ, 0 < ɛ δ < 1/. We mut now verify that (H, τ ɛ 1r!. We begin with bounding the j. Fix j ( and σ W with σ = j. We claim d(σ g(, qn 1 j. (3 If V (σ >, then becaue every e E atifie V (e =, we mut have d(σ = 0, o (3 hold. Similarly, if there are u v [0, q] and f ( V (σ uch that (f, u, (f, v σ, then becaue no e E can contain both (f, u and (f, v, we mut have d(σ = 0, o (3 hold. Aume now V (σ and for all f ( V (σ, there i at mot one u [0, q] uch that (f, u σ. Note thi implie σ ( V (σ. For each (f, u in σ, et w(f = u. Then w i a (poibly partial function from ( V (σ into [0, q]. Every edge e containing σ can be contructed a follow. Chooe an -element ubet X extending V (σ. There are ( n V (σ V (σ way to do thi. Extend w to a total function ( X [0, q] o that (X, w = G for ome G H(, q, and et e = {(f, w(f : f ( X }. There are at mot g(, q way to do thi. Thi how that ( n V (σ d(σ g(, q g(, qn V (σ g(, qn 1 j, (4 V (σ 6

7 where the econd inequality i becaue j = σ ( V (σ implie that V (σ j > 1 + j. Thu we have hown (3 hold for all σ W with σ = j. Therefore, for all x W, On the other hand, the average degree of H i ( ( g(, q n ( d = (q + 1 ( g(, q n n = q + 1 d (j (x g(, qn 1 j. (5 g(, q q + 1 ( n g(, q ( n n.1, q + 1 where the lat two inequalitie are becaue n i ufficiently large. Combining thi with (5 yield j = x W d(j (x dnτ j 1 1 Ng(, qn j dnτ j 1 = 1 g(, qn j dτ j 1 g(, qn 1 j +.1+(j = g(, qn 1.6 j+(j n 1.61 j+(j 1 1 4, (6 where the lat inequality i becaue n i large. Since j (, (j 1 <. Therefore, (j 1 j (j 1 j 4 4 (j 1 = j 1 j 4 = ( j 1 ( j = j 8 Combining thi with (6 we obtain that j n = n.14. Since thi bound hold for all j (, we have ( (H, τ = (( 1 j= (j 1 j n.14( ( (( 1 j= (j 1. (7 Since (( 1 ( j= (j 1 i a contant and n i ufficiently large, (7 implie that we have (H, τ ɛ/(1 (!, a deired. We have now verified the hypothee of Theorem 4 hold. Conequently, Theorem 4 implie there exit C W uch that the following hold. 1. For every independent et I in H, there i ome C C, uch that I C.. For all C C, we have e(h[c] ɛe(h. 3. log C c 1 log(1/ɛnτ log(1/τ. For each C C, define G C = (, w C where for each f (, w C (f = max{u : (f, u C}. Set G = {G C : C C}. We how thi G atifie (i-(iii of Theorem 5 for c and δ. Firt note that by contruction, every G C G ha multiplicity at mot q. We now how (i hold. Fix J = (, w F (n,, q and let I = {(f, w(f : f ( }. It i traightforward to verify that becaue J i an (, q-graph, I W i an independent et in H. By 1, there i C C uch that I C. By definition of G C, thi implie that for each f (, w(f w C (f. In other word, J i a ubmultigraph of G C. Thu G atifie (i. We now how part (ii. Fix G C = (, w C G. By, e(h[c] ɛe(h = δ ( n, where the equality i by definition of ɛ and becaue e(h = g(, q ( n. So it uffice to how that H(G C,, q e(h[c]. Given A H(G C,, q, define Θ(A = {(f, w C (f : f ( A }. We 7

8 how Θ i an injective map from H(G C,, q into E(H[C]. By definition of w C, Θ(A C, and by definition of E, Θ(A E. Thu Θ(A E(H[C]. Clearly for all A A H(G C,, q, Θ(A Θ(A. So Θ i an injection from H(G C,, q E(H[C], and H(G C,, q e(h[c], finihing our proof of (ii. For (iii, we mut compute an upper bound for log G. By definition of G, G C, o it uffice to bound log C. By 3 and the definition of N, τ, ɛ, and c, we have log C c 1 log ( 1 ( 1 Nτ log ɛ τ = c 1 log ( 1 ɛ (q + 1 ( n n 1 4 log(n 1 4 = c ( n Since ( n n, thi i at mot cn 1 4 log n. So log G cn 1 4 log n, a deired. n 1 4 log n. We now tate a generalization of the graph removal lemma, Lemma 1 below. Since the argument i merely an adjutment of the argument for graph, uing a multi-colored verion of Szemerédi regularity lemma (ee [], we omit the proof. Given two multigraph G = (V, w and G = (V, w, et (G, G = { xy ( V : w(xy w (xy }. We ay G and G are δ-cloe if (G, G δn, otherwie they are δ-far. Lemma 1. Fix integer and q 0. For all 0 < ν < 1, there i 0 < δ < 1 uch that for all ufficiently large n, the following hold. If G = (, w atifie µ(g q and H(G,, q δ ( n, then G i ν-cloe to ome G in F (n,, q. Given G = (V, w, let G + = (V, w + where w + (x, y = w(x, y + 1 for each xy ( V. Oberve that for any finite multigraph G, the number of ubmultigraph of G i P (G +, and if G F (n,, q, then G + F (n,, q + (. The following uperaturation type reult i a conequence of Lemma 1 and Propoition. Lemma. Suppoe, q 0. For all ɛ > 0 there i a δ > 0 uch that for all ufficiently large n, the following hold. Suppoe G = (, w ha µ(g q and atifie H(G,, q δ ( n. Then P (G + ex Π (, q + ( (1+ɛ( n if exπ (, q + ( > 1 and P (G + ɛ(n if exπ (, q + ( = 1. Proof. By Propoition, ex Π (, q + ( exit and i at leat 1. Fix ɛ > 0 and et { ex Π (, q + ( B = if exπ (, q > 1 if ex Π (, q = 1. Set ν = ɛ/(6 log B (q + 1. Apply Lemma 1 to ν to obtain δ. Aume n i ufficiently large, and G = (, w ha µ(g q and atifie H(G,, q δ ( n. Then Lemma 1 implie there i H = (, w F (n,, q uch that G and H are ν-cloe. Oberve (G, H = (G +, H +. Then P (G + = ( ( (w(xy + 1 = (w (xy + 1 (w(xy + 1 xy ( xy ( \ (G,H ( = P (H + xy (G,H xy (G,H w(xy + 1 w. (8 (xy + 1 Since max{µ(g, µ(h} q, we have that for all xy (, 1 w(xy + 1, w (xy + 1 q + 1, and w(xy+1 therefore, w (xy+1 q+1 1 = q. By aumption, (G, H νn 3ν ( n. Oberve that becaue H + F (n,, q + (, we have P (H + ex Π (n,, q + (. Combining thee fact with (8 yield the following. ( P (G + P (H + (q + 1 (G,H P (H + (q + 1 3ν(n exπ (n,, q + (q + 1 3ν(n. (9 8

9 Becaue n i ufficiently large an by definition of ex Π (, q+ (, we may aume that exπ (n,, q+ ( i at mot ex Π (, q + ( ( n B ɛ( n /. Combining thi with (9 and the definition of ν yield that ( ( n ( ( P (G + ex Π (, q + B ɛ( n ( n / (q + 1 3ν(n = exπ, q + B ɛ( n. When ex Π (, q + ( > 1, thi ay P (G + ex Π (, q + ( (1+ɛ( n, and when exπ (, q + ( = 1, thi ay P (G + ɛ(n. Proof of Theorem 1. Suppoe and q 0. Fix ɛ > 0. We how that for ufficiently large n, ex Π (, q + ( ( n ( F (n,, q exπ, q + ( (1+ɛ( n if ex Π (, q + ( 1 and F (n,, q ɛ( n if exπ (, q + ( = 1. We firt prove the upper bound. Set { ex Π (, q + ( B = if exπ (, q > 1 if ex Π (, q = 1. Apply Lemma to ɛ/ to obtain δ and apply Theorem 5 to δ for and q to obtain c. Aume n i ufficiently large. By Theorem 5, there i a collection G of multigraph of multiplicity at mot q and with vertex et uch that (i for every J F (n,, q, there i G G uch that J i a full ubmultigraph of G, (ii for every G G, H(G,, q δ ( n, and (iii log G cn 1 4 log n. By Lemma and (ii, for every G G, P (G + ex Π (, q + ( ( n B ɛ( n /. By (i, every element of F (n,, q can be contructed a follow. Chooe G G. By (iii, there are at mot cn 1 log n choice. Since n i ufficiently large, we may aume that cn 1 4 log(n B ɛ(n /. Chooe a ubmultigraph of G. There are P (G + ex Π (, q + ( ( n B ɛ( n / choice. Combining thee bound yield that F (n,, q ex Π (, q + ( ( n B ɛ( n. In other word, if ex Π (, q + ( > 1, then F (n,, q exπ (, q + ( (1+ɛ( n and if exπ (, q + ( = 1, then F (n,, q ɛ(n. We only have left to how that in the cae where exπ (, q + ( > 1, F (n,, q ex Π (, q + ( ( n. Chooe any product-extremal G0 = (, w 0 F (n,, q + (. Oberve that by aumption and Propoition, P (G 0 = ex Π (n,, q + ( exπ (, q + ( ( n > 1. Therefore, G 0 contain no edge of multiplicity 0, o we can define a multigraph G = (, w atifying w(xy = w 0 (xy 1 for all xy (. By contruction, G + = G 0, o G 0 F (n,, q + ( implie G F (n,, q. Then F (n,, q i at leat the number of ubmultigraph of G, which i P (G + = P (G 0 = ex Π (n,, q + ( exπ (, q + ( ( n. 9

10 4 Extremal reult for (n, 4, 15-graph: a two-tep reduction In thi ection we reduce Theorem to two tepping-tone theorem, Theorem 6 and 7, below. The main idea i that Theorem relie on undertanding the tructure of (4, 15-graph which are product-extremal ubject to certain contraint. Given a et F of multigraph, recall that P(F = {G F : P (G P (G for all G F} and P(n, 4, 15 = P(F (n, 4, 15. Definition 6. Given n N, define F 3 (n, 4, 15 = {G F (n, 4, 15 : µ(g 3} and D(n = F 3 (n, 4, 15 F (n, 3, 8. Theorem 6. For all ufficiently large n, P(n, 4, 15 = P(D(n. Theorem 7. For all ufficiently large n, P(D(n P(W (n. Thee two theorem will be proved in Section 6 and 5 repectively. We ue the ret of thi ection to prove Theorem, given Theorem 6 and 7. Given G = (, w W (n, let L(G and R(G denote the part in the partition of uch that w(xy = 1 if and only if xy ( L(G. Recall the definition of γ from Theorem. Lemma 3. For all G P(W (n, we have P (G = γn +O(n. Proof. Let G = (, w W (n. Set h(y = (y 3 y(n y and oberve that if L(G = n y and R(G = y, then P (G = h(y. Thu it uffice to how that max y h(y = γn +O(n. Baic calculu how that h(y ha a global maximum at τ = βn (log /(( log 3 log, log 3 where β = log 3 log i a in Theorem. Thi implie max y N h(y = max{h( τ, h( τ }. It i traightforward to check max{h( τ, h( τ } = max{h( βn,, h( βn }. By definition of γ and h, thi implie max y h(y = γn +O(n. Proof of Theorem. Fix n ufficiently large and G 1 P(W (n. By Theorem 7, there i ome G P(D(n P(W (n. Since G 1 and G are both in P(W (n, P (G 1 = P (G. Our aumption and Theorem 6 imply G P(D(n = P(n, 4, 15, o P (G = ex Π (n, 4, 15. Combining thee fact yield P (G 1 = P (G = ex Π (n, 4, 15, o G 1 P(n, 4, 15. Thi how P(W (n P(n, 4, 15. Since G 1 P(n, 4, 15 P(W (n, Lemma 3 implie ex Π (n, 4, 15 = P (G 1 = γn +O(n. By definition, ex Π (4, 15 = γ. 5 Proof of Theorem 7 The goal of thi ection i to prove Theorem 7. It will require many reduction and lemma. The general trategy i to how we can find element in P(D(n with increaingly nice propertie, until we can how there i one in W (n. The proof method can be viewed a a generalization of Zykov-ymmetrization to multigraph, where we ucceively replace and duplicate vertice if they do not have certain deirable propertie. We now give ome definition which will be ued in thi ection. Given a multigraph G = (V, w, let G be the binary relation on V defined by x G y if and only if w(xy = 1 and for all z V \ {x, y}, w(xz = w(yz. Propoition 3. Suppoe G = (V, w i a multigraph. Then G form an equivalence relation on V. Moreover, if Ṽ = {V 1,..., V t } i the et of equivalence clae of V under G, then for each i j, there i w ij {1,, 3} uch that for all (x, y V i V j, w(xy = w ij. 10

11 The proof i traightforward and left to the reader. Given G = (V, w and i, j, k N, an (i, j, k- triangle in G i a et {x, y, z} ( V 3 uch that {w(xy, w(yz, w(xz} = {i, j, k}. Say that G omit (i, j, k-triangle if there i no (i, j, k-triangle in G. Definition 7. A multigraph G i neat if µ(g 3 and G omit (i, j, k-triangle, for each (i, j, k {(1, 1,, (1, 1, 3, (1,, 3}. In neat multigraph, G i epecially well behaved. Propoition 4. Suppoe G = (V, w i a neat multigraph. Then for all x, y V, x G y if and only if w(xy = 1. Moreover, if Ṽ = {V 1,..., V t } i the et of equivalence clae of V under G, then for each i j, there i w ij {, 3} uch that for all (x, y V i V j, w(xy = w ij. Again, the proof i traightforward and left to the reader. 5.1 Finding a neat multigraph in P(D(n Definition 8. Suppoe n 1. Define C(n to be the et of neat multigraph in D(n, that i, C(n = {G D(n : G omit (i, j, k-triangle, for each (i, j, k {(1, 1,, (1, 1, 3, (1,, 3}}. Oberve that for all n, W (n C(n D(n. The goal of thi ubection i to prove Lemma 5, which ay the for all n, there i a product-extremal element of D(n which i alo in C(n. We begin with ome notation. Suppoe G = (V, w i a multigraph. Given x y V, define G xy = (V, w to be the multigraph uch that G xy [V \ {x, y}] = G[V \ {x, y}], w (xy = 1, and for all u V \ {x, y}, w (xu = w(yu. The idea i that G xy i obtained from G by making the vertex x look like the vertex y. Oberve that if G = G xy, then x G y. Given y V, et p(y = x V \{y} w(xy. We will ue the following equation for any xy ( V. p(y P (G xy = P (G. (10 p(xw(xy Lemma 4. Suppoe n 1, G D(n, and uv (. Then Guv D(n. Proof. Fix G = (, w D(n and let G := G uv = (, w. We how G D(n. Given X, let S(X = xy ( X w(xy and S (X = xy ( X w (xy. By definition of G uv and becaue G D(n, µ(g 3. We now check that G F (n, 4, 15. Suppoe X ( 4. If u / X, then S (X = S(X 15. If X {u, v} = {u}, then S (X = S((X \ {u} {v} 15. So aume {u, v} X, ay X = {u, v, z, z }. Becaue G F (n, 3, 8 and by definition of G uv, we have that S ({v, z, z } = S({v, z, z } 8. Combining thi with the fact that w (uv = 1 and µ(g 3 yield S (X = S ({v, x, y} + w (uv + w (ux + w (uy = 15. We now verify that G F (n, 3, 8. Suppoe X ( 3. If u / X, then S (X = S(X 8. If X {u, v} = {u}, then S (X = S((X \ {u} {v} 8. So aume {u, v} X, ay X = {u, v, z}. Becaue µ(g 3, S (X w (uv = = 7 8. Conequently, G F 3 (n, 4, 15 F (n, 3, 8 = D(n. 11

12 Corollary 1. Suppoe n 1, G = (, w P(D(n, and uv (. If w(uv = 1, then G uv P(D(n. Proof. By Lemma 4, G vu D(n. Combining thi with (10 and the fact that G P(D(n yield that P (G vu = p(u p(v P (G P (G. Thu p(u p(v. The ymmetric argument how p(v p(u, o p(u = p(v. Therefore, P (G uv = p(v p(u P (G = P (G, which implie G uv P(D(n. Given G = (V, w, et t G = {xy ( V : x G y}. We now prove the main reult of thi ubection. Lemma 5. For all n 1, P(D(n C(n. Conequently, P(C(n P(D(n. Proof. Fix G = (, w P(D(n uch that t G i maximal among the element of P(D(n. Let V 1,..., V k be the G -clae of G, enumerated o that V 1... V k. Suppoe there i i < j and (x, y V i V j uch that w(xy = 1. Let G = (, w = G xy. By Corollary 1, G P(D(n. It i traightforward to check that the G clae of G are V 1,..., V k, where V i = V i\{x}, V j = V j {x}, and V l = V l for all l [k] \ {i, j}. Conequently, t G = t G + V j V i + 1 > t G, contradicting the maximality of t G. Thu, for all i j and (x, y V i V j, w(xy 1. By Propoition 3, we mut have that for all i j, either w(xy = for all (x, y V i V j, or w(xy = 3 for all (x, y V i V j. Thi implie that G omit (i, j, k-triangle for each (i, j, k {(1, 1,, (1, 1, 3, (1,, 3}. Thu G P(D(n C(n. Combining thi with C(n D(n yield that P(C(n P(D(n. 5. Acyclic multigraph We ay two multigraph G = (V, w and G = (V, w are iomorphic, denoted G = G, if there i a bijection f : V V uch that w(xy = w (f(xf(y, for all xy ( V. We ay that G = (V, w contain a copy of G if there i X V uch that G[X] = G. Definition 9. Given t 3, define C t (3, to be the multigraph ([t], w uch that w(1 = w(3 =... = w((t 1t = w(t1 = 3, and w(ij = for all other pair i j. For n 1, et NC(n (NC= no cycle to be the et of G C(n which do not contain a copy of C t (3, for any t 3. We will how in the next ubection that for large n, all product-extremal element of C(n are in NC(n. However, we mut firt how that we can find product-extremal element of NC(n which are nice, and thi i the goal of thi ubection. In particular we will how that for all n 1, there i a product-extremal element of NC(n which i alo in W (n. We begin with ome notation and definition. If G contain a copy of C t (3,, we will write C t (3, G, and if not, we will write C t (3, G. A vertex-weighted graph i a triple (V, E, f where (V, E i graph and f : V N >0. We now give a way of aociating a vertex-weighted graph to a neat multigraph. Suppoe G = (V, w i a neat multigraph, Ṽ = {V 1,..., V t } i the et of equivalence clae of V under G, and for each i j, w ij {, 3} i from Propoition 4. Define the vertex-weighted graph aociated to G and G to be G = (Ṽ, Ẽ, f where Ẽ = {V iv j ( Ṽ : wij = 3} and f(v i = V i for all i [t]. We will ue the notation G to denote thi vertex-weight function f, and we will drop the upercript when G i clear from context. If H = (V, E i a graph and X V, then let H[X] = (X, E ( X. Lemma 6. Suppoe n 1 and G i a neat multigraph with vertex et. Then G NC(n if and only if G i a foret. 1

13 Proof. Suppoe G i not a foret. Then there i X = {V i1,..., V ik } Ṽ uch that G[X] i a cycle of length k 3. Chooe ome y j V ij for each 1 i k and let Y = {y 1,..., y k }. Then by definition of G, we mut have G[Y ] = C k (3,. Thu G / NC(n. On the other hand, uppoe G / NC(n. Then becaue G i neat, we mut have that either G / F (n, 4, 15 or C t (3, G for ome t 3. Suppoe G / F (n, 4, 15. Then there i ome Y ( 4 uch that S G (Y > 15. Since µ(g 3, thi implie that either (i {w(xy : xy ( Y } = {3, 3, 3, 3,, } or (ii {w(xy : xy ( Y } = {3, 3, 3, 3, 3, j}, ome j {1,, 3}. Let X be the et of equivalence clae interecting Y, that i X = {V i Ṽ : Y V i }. In Cae (i, becaue Y pan no edge of multiplicity 1 in G, the element of Y mut be in pairwie ditinct equivalence clae under G. Thu in G, X = 4 and X pan exactly 4 edge. Thi implie G[X] i either a 4-cycle or contain a triangle. In Cae (ii, if j = 1, then X = 3 and G[X] i a triangle. If j 1, then X = 4 and pan at leat 5 edge. Thi implie G[X] contain a triangle. Therefore, if G / F (n, 4, 15, then G i not a foret. Suppoe now C t (3, G, for ome t 3. Then if X i uch that G[X] = C t (3,, G[X] i a cycle, o conequently G i not a foret. Definition 10. Given a vertex-weighted graph G = (Ṽ, E,, et f π ( G = 3 U V U V. UV E Note that we have P (G = f π ( G for all G C(n. UV (Ṽ \E Two vertex-weighted graph G 1 = (V 1, E 1, f 1 and G = (V, E, f, are iomorphic, denoted G 1 = G, if there i a graph iomorphim g : V 1 V uch that for all v V 1, f 1 (v = f (g(v. Lemma 7. Suppoe n 1 and H = (Ṽ, E, i a vertex-weighted foret uch that V Ṽ V = n. Then there i a multigraph G NC(n uch that G i iomorphic to H. Proof. Let Ṽ = {V 1,..., V t } and for each i, let x i = V i. Since t i=1 x i = n, it i clear there exit a partition P 1,..., P t of uch that for each i [t], P i = x i. Fix uch a partition P 1,..., P t. Define G = (, w a follow. For each xy (, et 1 if xy ( P i for ome i [t] w(xy = 3 if xy E(P i, P j for ome i j uch that V i V j E if xy E(P i, P j for ome i j uch that V i V j / E. By contruction, G i a neat multigraph and G i iomorphic to H. Becaue H = G i a foret, Lemma 6 implie G NC(n. Given a vertex-weighted graph, H = (Ṽ, E, and V Ṽ, let dh (V to denote the degree of V in the graph (Ṽ, E. Given a graph (Ṽ, E and dijoint ubet X, Ỹ of Ṽ, let E( X = E ( X and E( X, Ỹ = E {XY : X X, Y Ỹ }. Lemma 8. Suppoe H = (Ṽ, E, i a vertex-weighted foret uch that (Ṽ, E i not a tar. Then there i a vertex-weighted graph H = (Ṽ, E, uch that (Ṽ, E i a tar, and f π (H f π (H. Moreover, if f π (H = f π (H, then V = W where V i the center of the tar (Ṽ, E and W Ṽ i ome vertex ditinct from V. 13

14 Proof. Let H = (Ṽ, E, be a vertex-weighted foret. Fix V Ṽ with V = max{ X : X Ṽ }. We now define a equence H 0, H 1,..., H k, where for each i, H i = (Ṽ, E i,. Step 0: Let X be the et of iolated point in H. If X = et H 0 = H and go to the next tep. If X, let E 0 = E {V X : X X} and H 0 = (Ṽ, E 0,. Clearly (Ṽ, E 0 i till a foret, ince any cycle mut contain a new edge, i.e. an edge of the form V X, ome X X. But d H 0 (X = 1 for all X X implie no X X can be contained in a cycle in H 0. Further, note ( 3 X f π (H 0 = f π (H X V X > f π (H. If H 0 i a tar, end the contruction and let k = 0, otherwie go to the next tep. Step i + 1: Suppoe by induction we have defined H 0,..., H i uch that (Ṽ, E i i foret but not a tar and contain no iolated point. Since (Ṽ, E i i not a tar, it i in particular, not a tar with center V. Thi implie the et Ỹi := Ṽ \ ({V } dh i (V. We how there i Y Ỹi uch that d H i (Y = 1. Since there are no iolated point in (Ṽ, E i, every Y Ỹi ha d H i (Y 1. Suppoe toward a contradiction that every Y Ỹi had d H i (Y. Chooe a maximal equence of point Y = (Y 1,..., Y u from Ỹi with the property that Y 1 Y,..., Y u 1 Y u E i. Since Y 1 and Y u have degree at leat two in (Ṽ, E i and becaue (Ṽ, E i i a foret, there are Z 1, Z u Ṽ \ Y uch that Y 1 Z 1, Y u Z u E i. Since Y 1, Y u Y i, Z 1, Z u V and ince Y wa maximal, Z 1, Z u / Ỹi. Thu Z 1, Z u Ṽ \ (Ỹi {V } which implie V Z 1, V Z u E i. Thi yield that V, Z 1, Y 1,..., Y u, Z u, V i a cycle in (Ṽ, E i, a contradiction. Thu there exit Y Ỹi uch that d H i (Y = 1. Fix uch a Y Ỹi and let W be the unique neighbor of Y in (Ṽ, E i. Define E i+1 = (E i \ {Y W } {V Y }. and let H i+1 = (Ṽ, E i+1,. We firt check (Ṽ, E i+1 i a foret. Since (Ṽ, E i i a foret, any cycle in (Ṽ, E i+1 will contain V Y. However, d H i+1 (Y = 1, o Y cannot be contained in a cycle. Note ( 3 Y ( V W f π (H i+1 = f π (H i 3 V Y Y W Y W V Y = f π (H i fπ (H i, where the inequality hold becaue V W by choice of V. Further, note that the inequality i trict unle V = W. Clearly thi proce mut end after ome 0 k < Ṽ tep. If k = 0, then H 0 = H k i a tar and f π (H k > f π (H. If k 1, then the reulting H k = (Ṽ, E k, will have the property that (Ṽ, E k i a tar with center V. Since k 1, one of the following hold. f π (H 1 > f π (H 0, o f π (H k > f π (H, or f π (H 0 = f π (H 1 and at tep 1, we found a vertex W V with V = W. Lemma 9. Suppoe n 1, G NC(n, and G = (Ṽ, E, i the vertex-weighted graph aociated to G and G. Suppoe (Ṽ, E i a tar with center V and there i W Ṽ \ {V } uch that W > 1. Then G / P(NC(n. Proof. Let Ṽ = (Ṽ \ {W } {W 1, W } and E = (E \ {V W } {V W 1, V W }, where W 1, W are new vertice. Let H = (Ṽ, E, where the vertex-weight function i defined by U = U for all U Ṽ \ {W }, W 1 = W 1, and W = 1. By definition of H, U Ṽ U = U Ṽ U = n. Since H i obtained from G by plitting the degree one vertex W into W 1 and W, and G i a foret, H i alo a foret. Thu H atifie the hypothee of Lemma 7, o there i an G NC(n 14

15 uch that G i iomorphic to H. Thi and Definition 10 implie f π (H = f π ( G = P (G. Let Z = Ṽ \ {V, W }. Then ( ( f π (H = 3 U V U U W 1 U + W U 3 V W 1 + V W W 1 W U Z ( = U Z UU ( Z U Z ( 3 U V U U UU ( Z = f π ( G W 1 f π ( G. U Z W U 3 V W W 1 So G NC(n and P (G = f π (H > f π ( G = P (G imply G / P(NC(n. We now prove the main reult of thi ubection. Lemma 10. For all n 1, P(NC(n W (n. Conequently, P(W (n P(NC(n. Proof. If n = 1, thi i trivial. If n = then the only element in P(NC(n i the G which conit of a ingle edge with multiplicity 3. Clearly thi G alo in W (n. Aume now n 3 and let G = (, w P(NC(n. Suppoe firt that G contain no edge of multiplicity 1. Then G = (, E where E = {xy ( : w(xy = 3}. By Lemma 6, G i a foret. It i a well known fact that becaue G i a foret with n vertice, E n 1. Therefore, we have that P (G = 3 E (n E 3 n 1 (n n+1. Let G = (, w be uch that w (1i = 3 for all i n and w (xy = for all other edge. Then G NC(n and P (G = 3 n 1 (n n+1 P (G. Since G P(NC(n, thi implie G P(NC(n a well. By definition, G W (n, o we are done. Aume now G contain ome xy with w(xy = 1. Conider now the vertex-weighted graph G = (Ṽ, E, aociated to G and G. Suppoe (Ṽ, E i a tar with center V. If W = 1 for all W Ṽ \ {V }, then G W (n and we are done. If there i W Ṽ \ {V } uch that W > 1, then Lemma 9 implie G / P(NC(n, a contradiction. Suppoe now (Ṽ, E i not a tar. Then Lemma 8 implie there i a vertex-weighted graph H = (Ṽ, E, uch that (Ṽ, E i a tar and f π (H f π ( G. Since (Ṽ, E i a tar, it i a foret. Since (Ṽ, E, i the vertex-weighted graph aociated to G and G, U Ṽ U = n. Thu H atifie the hypothee of Lemma 7, o there i G NC(n uch that G = H. Thu P (G = f π (H f π ( G, where the equality hold by Definition 10. Suppoe f π (H > f π ( G. Then P (G = f π (H > f π ( G = P (G, contradicting that G P(NC(n. Thu we mut have f π (H = f π ( G. By Lemma 8, thi only happen if there i ome W V Ṽ uch that V = W, where V i the center of the tar (Ṽ, Ẽ. Note that becaue G contain ome xy with w(xy = 1, there i ome vertex U Ṽ uch that U > 1. If U V, then U Ṽ \ {V } and U > 1. If U = V, then W Ṽ \ {V } and W = V = U > 1. In either cae Lemma 9 implie that G / P(NC(n. Since P (G = f π ( G = f π (H = P (G, thi implie G / P(NC(n, a contradiction. Thu we have hown that for all n 1, P(NC(n W (n. Since W (n NC(n, thi implie P(W (n P(NC(n. 5.3 Getting rid of cycle and proving Theorem 7 In thi ubection we prove Lemma 17, which how that for large n, all product-extremal element of C(n are in NC(n. We will then prove Theorem 7 at the end of thi ubection. Our proof ue an argument that i eentially a progreive induction. 15

16 Note that if G C(n, then C 3 (3, G (ince C(n F (n, 3, 8 and C 4 (3, G (ince S(C 4 (3, = 16. So to how ome G C(n i in NC(n, we only need to how C t (3, G for t 5. Given G = (V, w, X V, and z V \ X, et P G z (X = x X w(xz. Lemma 11. Let 5 t n and G = (, w C(n. Suppoe C t (3, G, and for all 5 t < t, C t (3, G. If X ( t i uch that G[X] = Ct (3,, then for all z \ X either 1. {x X : w(zx = 3} 1 and P G z (X 3 t 1 or. {x X : w(zx = 3} and P G z (X 3 t 3 < 3 t 1. Proof. Let X = {x 1,..., x t } where w(x i x i+1 = w(x 1 x t = 3 for each i [t 1] and w(x i x j = for all other pair ij ( [t]. Since G C(n, C3 (3,, C 4 (3, G. Combining thi with our aumption, we have that for all 3 t < t, C t (3, G. We will ue throughout that µ(g 3 (ince G C(n. Fix z \ X and let Z = {x X : w(zx = 3}. If Z 1, then clearly 1 hold. So aume Z and i 1 <... < i l are uch that Z = {x i1,..., x il }. Without lo of generality, aume i 1 = 1. Set Given (x, y I, let d(x, y = I = {(x ij, x ij+1 : 1 j l 1} {(x i1, x il }. { i j+1 i j if (x, y = (x ij, x ij+1 ome 1 j l 1 t i l + 1 if (x, y = (x i1, x il. Note that becaue C 3 (3, G, d(x, y t for all (x, y I. Suppoe firt that there i ome (u, v I uch that d(u, v = t. Then ince d(x, y for all (x, y I we mut have that I = 1 and either (u, v = (x i1, x il = (x 1, x t 1 or (u, v = (x i1, x il = (x 1, x 3. Without lo of generality, aume (u, v = (x 1, x 3. Then we mut have that w(zx 1 ince otherwie G[{z, x 1, x, x 3 }] = C 4 (3,, a contradiction. Thi how that P G z (X 3 1 t 3 < 3 t 1. Suppoe now that for all (x, y I, d(x, y t 3. Given (x, y I, ay an element x k i between x and y if either (x, y = (x ij, x ij+1 and i j < k < i j+1 or (x, y = (x i1, x il and i l < k. Then for each (x, y I, there mut be a x k between x and y uch that w(zx k 1, ince otherwie {z, x, y} {u : u i between x and y} i a copy of C d(x,y+ (3, in G, a contradiction ince d(x, y + < t. Thi implie there are at leat l element u in X \ Z uch that w(zu 1, o P G z (X 3 l t l 3 t 4 < 3 t 1. Given n, t N et } f(n, t = min { ( βt + βt c 3 βt (1 βt +c (1 βt + βt (n t c : c { β(n t, β(n t }. Definition 11. Suppoe t n and X ( t. Define GX = (, w to be the following multigraph, where Y = \ X. Chooe any A P(W (n t and B W (t o that R(B = βt and L(B = (1 βt. Define w on ( ( Y X to make GX [Y ] = A and G X [X] = B. Define w on the remaining pair of vertice in the obviou way o that G X W (n. Lemma 1. Suppoe t n and X ( t. Then for any A P(W (n t, P (G X P (A f(n, t. 16

17 Proof. Set Y = \ X and let G X = (, w. Let B W (t and A P(W (n t be a in the definition of G X o that G X [X] = B and G X [Y ] = A. Let L A, R A and L B, R B be the partition of Y and X repectively uch that w(xy = 1 for all xy ( L A ( LB. By choice of B, LB = (1 βt and R B = βt. Let c = R A. By definition, L A = n t c, and ince A P(W (n t, c { β(n t, β(n t } (by the proof of Lemma 3. Combining thee obervation with the definition of f(n, t implie ( R B + R A R B 3 L B R B + R B L A + L B R A = ( βt + βt c 3 βt (1 βt +c (1 βt + βt (n t c f(n, t. Combining thi with the definition of G X, we have P (G X = P (A ( R B + R A R B 3 L B R B + R B L A + L B R A P (Af(n, t. Since P (A = P (A for all A P(W (n t, thi finihe the proof. Definition 1. Given n, t N, let h(n, t = 3 n (t +t(n t n. Lemma 13. Let 5 t n, G C(n, and ν > 0. Suppoe X ( t, G[X] = Ct (3,, and there i ome A P(W (n t uch that P (G[ \ X] νp (A. Then P (G ν((h(n, t/f(n, tp (G X. Proof. Let Y = \ X. Becaue G[X] = C t (3,, P (G = P (G[Y ]3 t (t t z Y P G z (X. By Lemma 11, for each z Y, P G z (X 3 t 1. Thi implie P (G P (G[Y ]3 t (t t ( 3 t 1 n t = P (G[Y ]3 n (t +t(n t n = P (G[Y ]h(n, t. (11 By aumption, P (G[Y ] νp (A, o (11 implie P (G νp (Ah(n, t. Combining thi with Lemma 1 yield P (G νp (Ah(n, t = νp (Af(n, t(h(n, t/f(n, t νp (G X (h(n, t/f(n, t. The following will be proved in the Appendix. Lemma 14. There are γ > 0 and 5 < K M 1 uch that the following hold. 1. For all K t n, h(n, t < f(n, t.. For all 5 t K and n M 1, h(n, t < γn f(n, t. Lemma 15. Let K be from Lemma 14. Then for all K t n, the following hold. If G C(n, C t (3, G, and C t (3, G for all t < t, then for all G 1 P(W (n, P (G < P (G 1. Proof. Let t K and n = t + i. We proceed by induction on i. Suppoe firt i = 0. Fix G C(n uch that C t (3, G and C t (3, G for all t < t. Then n = t implie G = C t (3, and o P (G = 3 t (t t = h(t, t. Let H W (n have R(H { βn, βn } and L(H = V (H R(H. Then by definition of f(n, t, P (H = ( R(H 3 L(H R(H = f(t, t > h(t, t = P (G, where the inequality i by part (1 of Lemma 14. Since H W (n, thi implie P (G < P (G 1 for all G 1 P(W (n. Suppoe now that i > 0. Aume by induction that the concluion of Lemma 15 hold for all K t 0 n 0 where n 0 = t 0 + j and 0 j < i. Fix G C(n uch that C t (3, G and C t (3, G for all t < t. Let X ( t be uch that G[X] = Ct (3,. 17

18 Claim 1. For any A P(W (n t, P (G[ \ X] P (A. Proof. Note that C t (3, G[ \ X] for all 3 t < t. We have two cae. (1 If C t (3, G[ \ X] for all t t, then G[ \ X] i iomorphic to ome D NC(n t. By Lemma 10, for any A P(W (n t, P (G[ \ X] = P (D P (A. ( If C t (3, G[ \ X] for ome t t, then fix t 0 the mallet uch t, and et n 0 = n t. Our aumption imply t 0 t K and t 0 \ X = n t = i, o n 0 = n t = t 0 + (n t t 0 = t 0 + i t 0 = t 0 + j, where 0 i t 0 = j < i. Note G[ \ X] i iomorphic to ome D C(n t = C(n 0. Then we have that K t 0 n 0, n 0 = t 0 + j, 0 j < i, and D C(n 0 atifie C t0 (3, D and C t (3, D for all t < t 0. By our induction hypothei, for any A P(W (n 0 = P(W (n t, P (G[ \ X] = P (D < P (A. Claim 1 and Lemma 13 with ν = 1 imply P (G (h(n, t/f(n, tp (G X. Since K t n, Lemma 14 part (1 implie h(n, t/f(n, t < 1, o thi how P (G < P (G X. Since G X W (n, we have P (G < P (G X P (G 1 for all G 1 P(W (n. Lemma 16. Let M 1 and K be a in Lemma 14. There i M uch that for all 5 t K and n M 1 + K, the following hold. If G C(n, C t (3, G, and C t (3, G for all t < t, then for all G 1 P(W (n, P (G M (h(n, t/f(n, tp (G 1. Proof. Set M = M 1 + K. Chooe M ufficiently large o that for all 5 t K and t n M, ex Π (n, 4, 15 M (h(n, t/f(n, t. We how the concluion of Lemma 16 hold for all n M by induction. Suppoe firt n = M. Fix 5 t K and G C(n uch that C t (3, G and C t (3, G for all t < t. Then by our choice of M, P (G ex Π (n, 4, 15 M (h(n, t/f(n, t M (h(n, t/f(n, tp (G 1, for all G 1 P(W (n. Suppoe now n > M. Aume by induction the concluion of Lemma 16 hold for all 5 t 0 K and M n 0 < n. Fix 5 t K and G C(n uch that C t (3, G and C t (3, G for all t < t. Let X ( t be uch that G[X] = Ct (3, and et n 0 = n t. Claim. For any A P(W (n 0, P (G[ \ X] M P (A, Proof. Fix A P(W (n 0. Note that C t (3, G[ \ X] for all t < t and n 0 M 1 K (ince n t M K = M 1. We will ue the following obervation. For all 5 t 0 K, h(n 0, t 0 f(n 0, t 0 γn 0 < 1. (1 Thi hold by Lemma 14 part ( and the fact that n 0 M 1. Suppoe firt n 0 < M. Then G[\X] i iomorphic to ome D F (n 0, 4, 15 and n 0 K, o by our choice of M, P (G[ \ X] = P (D ex Π (n 0, 4, 15 M h(n 0, K f(n 0, K P (A M P (A, where the lat inequality i by (1. Aume now n 0 M. We have two cae. (1 If C t (3, G[ \ X] for all t t, then G[ \ X] i iomorphic to ome D NC(n 0. By Lemma 10, P (G[ \ X] = P (D P (A M P (A. 18

19 ( If C t (3, G[ \ X] for ome t t, chooe t 0 the mallet uch t, and let D C(n 0 be uch that G[ \ X] = D. Suppoe firt t 0 K. Then we have 5 t t 0 K, M n 0 < n, D C(n 0, C t0 (3, D, and C t (3, D for all t < t 0. Therefore our induction hypothei implie the concluion of Lemma 16 hold for D, n 0, t 0. In other word, ince A P(W (n 0, P (G[ \ X] = P (D M h(n 0, t 0 f(n 0, t 0 P (A M P (A, where the lat inequality i by (1. Suppoe finally that t 0 > K. Then K t 0 n 0, D C(n 0, C t0 (3, D, and C t (3, D for all t < t 0. Thu we have by Lemma 15 that P (G[ \ X] = P (D < P (A M P (A. Claim and Lemma 13 with ν = M imply P (G M (h(n, t/f(n, tp (G X. Since G X i in W (n, we have that P (G M (h(n, t/f(n, tp (G X M (h(n, t/f(n, tp (G 1, for all G 1 P(W (n. We can now prove that for large n, all product-extremal element of C(n are in NC(n. Lemma 17. For all ufficiently large n, P(C(n NC(n. Conequently, P(C(n = P(NC(n. Proof. Let γ, K, and M 1 be a in Lemma 14 and let M be a in Lemma 16. Chooe M M 1 + K ufficiently large o that M γn < 1 for all n M. Suppoe n > M and G / NC(n. We how G / P(C(n. Clearly if G / C(n we are done, o aume G C(n. Since W (n C(n, it uffice to how there i G 1 W (n uch that P (G 1 > P (G. Since G / NC(n, there i 5 t n uch that C t (3, G and for all t < t, C t (3, G. If t K, then Lemma 15 implie that for any G 1 P(W (n, P (G < P (G 1. If 5 t < K, then Lemma 16 implie that for any G 1 P(W (n, P (G M (h(n, t/f(n, tp (G 1 M γn P (G 1, where the econd inequality i becaue of Lemma 14 part (. By our choice of M, thi implie that for all G 1 W (n, P (G < P (G 1. Thi how P(C(n NC(n. Since NC(n C(n, thi implie P(C(n = P(NC(n. We can prove the main reult of thi ection, Theorem 7. Proof of Theorem 7. Aume n ufficiently large. By Lemma 5, we can chooe ome G in P(D(n C(n = P(C(n. By Lemma 17, P(C(n = P(NC(n, o G P(NC(n. By Lemma 10, there i ome G P(NC(n W (n = P(W (n. Since G and G are both in P(NC(n, P (G = P (G. Since G P(D(n and W (n D(n, thi implie that G P(D(n. Thu we have hown G P(D(n P(W (n. 6 Proof of Theorem 6 In thi ection we prove Theorem 6. We will need the following computational lemma, which i proved in the appendix. Given n, t, let k(n, t = 15 t (t +t(n t t. Lemma 18. There i M uch that for all n M and t n, k(n, t < f(n, t. The following can be checked eaily by hand and i left to the reader. Lemma 19. Suppoe a, b, and c are non-negative integer. If a + b 4, then a b. If a + b + c 6, then a b c 3. 19

20 Proof of Theorem 6. Let n be ufficiently large. It uffice to how P(n, 4, 15 D(n. Suppoe toward a contradiction there i G = (, w P(F (n, 4, 15\D(n. Given X and z \X, let S(X = S(G[X] and S z (X = x X w(xz. If G / F (n, 3, 8, let D 1,..., D k be a maximal collection of pairwie dijoint element of ( 3 uch that S(Di 9 for each i, and et D = k i=1 D i. If G F (n, 3, 8, et D =. If µ(g[\d] > 3, chooe e 1,..., e m a maximal collection of pairwie dijoint element of ( \D uch that S(ei 4 for each i and et C = m i=1 e i. If µ(g[ \ D] 3, et C =. Let X = D C and l = X = 3k + m. Note that by aumption X i nonempty, o we mut have l. We now make a few obervation. If D, then for each D i and z \ D i, S z (D i S(D i {z} S(D i 15 9 = 6 = 3, which implie by Lemma 19 that P G z (D i 3. By maximality of the collection D 1,..., D k, G[ \ D] i a (3, 8-graph. Thu if C, then for each i and z \ (D e i, S z (e i S(e i {z} = 4 =, which implie by Lemma 19 that P G z (e i. Since µ(g 15, for each D i and e j, P (D i 15 3 and P (e j 15. Let Y = \ X and write P (Y for P (G[Y ]. Our obervation imply that P (G i at mot ( k ( m P (Y P (D i P (e i (l +l(n l l+m P (Y 15 l m (l +l(n l l+m P (Y k(n, l. (13 i=1 i=1 Note that G[Y ] i iomorphic to an element of D(n l. Let n 0 be uch that Lemma 17 hold for all n > n 0. We partition the argument into two cae. Cae 1. n l n 0. In thi cae we can ue the crude bound P (G < (l 15 l m+( n 0 ln 0 < (l +4l+n 0 +ln 0 < ex Π (n, 4, 15 where the lat inequality hold ince we may aume that n i much larger than n 0 and l > n n 0. Thi contradict the fact that G P(n, 4, 15. Cae. n l > n 0. In thi cae may apply Lemma 17 to G[Y ] a Y = n l > n 0. Fix A P(W (n l. By Lemma 5, Lemma 17, and Lemma 10, P(W (n l P(D(n l, which implie that P (Y P (A. Combining thi with Lemma 1 yield P (G X P (Af(n, l P (Y f(n, l. Thi, along with the bound on P (G in (13, implie P (G P (Y k(n, l k(n, l = P (G X P (Y f(n, l f(n, l < 1, where the lat inequality i by choice of M and Lemma 18. So P (G < P (G X, a contradiction. 7 Concluding Remark The argument ued to prove Theorem can be adapted to prove a verion for um. If G = (V, w, let S(G = xy ( V w(xy. Given integer and q 0, et ex Σ (n,, q = max{s(g : G F (n,, q}. 0