Turán Numbers of Vertex-disjoint Cliques in r- Partite Graphs

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Univesity of Wyoming Wyoming Scholas Repositoy Honos Theses AY 16/17 Undegaduate Honos Theses Sping 5-1-017 Tuán Numbes of Vetex-disjoint Cliques in - Patite Gaphs Anna Schenfisch Univesity of

1 Univesity of Wyoming Wyoming Scholas Repositoy Honos Theses AY 16/17 Undegaduate Honos Theses Sping Tuán Numbes of Vetex-disjoint Cliques in - Patite Gaphs Anna Schenfisch Univesity of Wyoming, aschenfi@uwyo.edu Follow this and additional woks at: Pat of the Discete Mathematics and Combinatoics Commons Recommended Citation Schenfisch, Anna, "Tuán Numbes of Vetex-disjoint Cliques in -Patite Gaphs" (017). Honos Theses AY 16/ This Honos Thesis is bought to you fo fee and open access by the Undegaduate Honos Theses at Wyoming Scholas Repositoy. It has been accepted fo inclusion in Honos Theses AY 16/17 by an authoized administato of Wyoming Scholas Repositoy. Fo moe infomation, please contact scholcom@uwyo.edu.

2 Final Honos Poject Tuán numbes of vetex-disjoint cliques in -patite gaphs Anna Schenfisch May 1, Abstact In a boad sense, gaph theoy has always been pesent in civilization. Gaph theoy is the math of connections at a paty, who knows each othe? How many handshakes will each peson in a meeting have to give befoe shaking hands with eveyone? What is the best way to oute taffic though a city s netwok of oads? Extemal gaph theoy is a banch that deals with counting items (called vetices) and connections between two items (called edges) and detemining the maximum/minimum numbe of chaacteistics needed to satisfy a cetain popety. The specific topic of this pape is Tuán numbes, a topic of extemal gaph theoy that attempts to detemine the maximum numbe of edges a gaph may have without a specified patten emeging. Fo two gaphs, G and H, the Tuán numbe is denoted ex(g, H), and is the maximum numbe of edges in a subgaph of G that contains no copy of H. We wee able to find and pove a peviously unknown Tuán numbe fo a cetain patten in a cetain gaph. To be pecise, we found the Tuán numbe of copies of vetex-disjoint cliques in -patite gaphs (pat sizes n 1,..., n ). That is, ex(k n1,n,...,n, kk ) = n i n j n 1 n + n (k 1) 1 i<j This pape will descibe the motivation and histoy of extemal gaph theoy, discuss definitions and concepts elated to the eseach that was done, go though the poof of ou theoem, and finally discuss possible futue eseach as well as geneal open questions in the field. Note that much of this pape was adapted fom a pevious pape by the autho and othe contibutos [1]. Key concepts The following glossay may seve as an intoduction to those new to the field and as a eminde fo those moe familia with it. We encouage the eade to efe back to this section as a efeence to both concepts and notation. Gaph: A gaph G is a pai of sets G = (V, E), whee V is a fixed set of vetices, and the edge set E is a set of pais of distinct elements fom V. We often wite V as V (G) and E as E(G) (Note: all gaphs used in this pape have this definition; that is, they ae simple and undiected). An example of a gaph. The nodes epesent vetices and the lines epesent edges. 1

3 Subgaph: Let G be a gaph. A subgaph H of G is a pai of sets H = (V, E ) whee V V and E E, which is itself a gaph. If H is a subgaph of G, we wite H G. Cycle: A gaph G is called a cycle if the gaph is an altenating sequence of vetices and edges G = v 1 e 1 v e...v n 1 e n 1 v n e n v n+1, whee e i = v i v i+1, and if i < j n then v i v j, and v n+1 = v 1. A cycle of length n is typically denoted C n. C 5 Complete gaph: A gaph G = (V, E) is complete if fo evey pai x y in V we have xy E. If E(G) = n, this gaph is denoted K n. A subgaph that is a complete gaph is called a clique of size n. K 5, o clique of size 5 -Patite gaph: A gaph G is -patite if V (G) can be patitioned into sets, V 1 V V such that if x and y ae both in the same V i, then xy / E(G). A 3-patite (commonly called tipatite) gaph. Hee, each colo indicates to which pat a vetex might belong. Complete -patite gaph: A gaph G that is -patite is called complete -patite if evey pai of vetices among diffeent vetex sets in the patition (commonly called pats) ae adjacent (connected by an edge). If V i = m i, then this gaph is denoted K m1,m,...,m. A complete bipatite gaph, K 3, Join: The join of two disjoint gaphs G and H occus when all vetieces in G ae connected to all vetices in H. The join of G and H is notated G + H. Weight: Fo the puposes of this pape, we define the weight of a set of veticies as the numbe of edges in the subgaph containing exactly these vetecies. Fo a set of vetices S, the weight of S is denoted as w(s).

4 Tuán numbe: The Tuán numbe of a pai of gaphs S and G with S G is the maximum numbe of edges that a subgaph of G may have and still contain no copy of S. This numbe may also be called the extemal numbe, and is denoted ex(g, S). Induced Subgaph: A vetex induced subgaph is a gaph consisting of a specified set of vetices along with any edges between these specific vetices. The gaph induced by the vetex set S on a gaph G is denoted G[S]. 3 Motivation and histoy The study of gaph theoy is a faily ecent development, at least in the ways it is defined and eseached today. Aguably the fist embyonic instance of a gaph theoy question was posed in 1736, when the Swiss mathematician Leonad Eule published a pape called The Seven Bidges of Königsbeg. In this pape, Eule poved it was not possible to walk though Königsbeg cossing each of the city s seven bidges only once. Eule noticed that as the tavele went fom one bidge to anothe, the choice of oads between the bidges was ielevant. With ou moden teminology, this is like saying we do not need to pay attention to the shape of an edge between two veticies - all we need to know is that it connects them. Nealy two centuies late, we finally have ou fist majo extemal poblem esult made by Mantel in 1907 []. He found that if a gaph with n veticies (denoted G n ) does not contain a clique of size 3 (denoted K 3 ) then the numbe of edges in G n is no geate than n /, o [ ] n e(g n ) 4 This esult is much moe in line with the type of thinking we see today in gaph theoy. Howeve, Mantel s esult was lagely fogotten. It wasn t until yeas late that gaph theoy was again picked up, this time by a man named Paul Edős. Edős is consideed by many to be one of the fathes of the field. His influence is fa eaching - his wok influences gaph theoists today in how to think about questions, and even what type of questions to ask. One of Edős well known theoems (published in 1938 and colloquially called the C 4 -Theoem) is a good example of how gaph theoy uses techniques and has applications outside of pue gaph theoy alone. Pat of the theoem states: 1. Assume that n 1 < < n k ae positive integes such that n i n j n h n l unless i = h and j = l o i = l and j = h. What is the maximum numbe of such integes in [1, n]? Denote this maximum by B(n). In a late a pape, Edős would find the uppe bound ( ) n 3/4 B(n) π(n) + O log 3/ n whee π(n) is the numbe of pimes in [, n] Note that the last tem on the ight side epesents an eo of n ode 3/4 log 3/ n ; in othe wods, this bound may deviate fom eality by at most some facto of n 3/4 log 3/ n. At fist eading, this question seems to lie puely in the ealm of numbe theoy. Howeve, Edős was able to tansfom the question into a gaph theoy one by the following lemma: Lemma 1. Evey intege a [1, n] can be witten as a = bd : b B, d D whee D is the set of integes in [1, n /3 ], IP is the set of pimes in (n /3, n] and B = D IP. If we let A be a set that satisfies the condition in 1. Then we can epesent evey a A as a i = b j (i)d j (i). We assume that b j > d j. If we ceate a bipatite gaph G(B, D) by joining b j to d j fo each b j d j = a A. In a gaph theoetical sense then, each a j is an edge between its associated vetices. Note that if we wee to find a 4-cycle (b 1 d 1 b d ) in G(B, D), then a 1 = b 1 d 1 and a = b 1 d, a 3 = b d and a 4 = b d 1 meaning that a 1 a 3 = a a 4, which contadicts ou assumption. Theefoe, the efamed question becomes 3

5 . Given a bipatite gaph G(X, Y ) with m and n vetices in its colo classes, what is the maximum numbe of edges G(X, Y ) may have without containing any C 4? Edős poved the following theoem: Theoem 1. If C 4 G(X, Y ), X = Y = k, then e(g(x, Y )) 3k 3/ In yet anothe boowing fom a diffeent field, Edős tuned to geomety, boowing a lemma fom anothe mathematician, Eszte Klein: Lemma. Given p(p + 1) + 1 elements, (fo some pime p) we can constuct p(p + 1) + 1 combinations, taken (p + 1) at a time (meaning (p + 1)-tuples) having no two elements in common. Although we have glossed ove the details of the poof, the example of Edős C 4 -Theoem demonstates how gaph theoy has connections to othe fields (hee, we see connections to numbe theoy and finite geomety). We see the concepts of gaph theoy being applied to a multitude of othe fields, such as aithmetic poblems (fo example in Sidon numbes [4]), topology (in places such as the Edős Stone Theoem [5]), and compute science (Kuatowski s wok in netwoking [6]). Ou theoem deals with Tuán numbes, which fist appeas to be a bit moe abstact. Paul Tuán, like Edős, was a Hungaian mathematician inteested in gaph theoy. He was also highly influential, especially in extemal poblems, o poblems that seek to find a maximum o minimum numbe of something (such as edges, coloings, etc.) within a cetain constaint. Edős is even quoted as saying, Tuán initiated the field of extemal gaph theoy. [7] Tuán s most famous theoem is about the numbe of edges a gaph of n vetices may have without containing a clique of size p. Below, we use the notation T n,p 1, which is a Tuán gaph on n vetices with p 1 classes (meaning that it has the maximum possible numbe of edges without containing a clique of size p). Theoem. If G n contains no K p, then e(g n ) e(t n,p 1 ). In case of equality, G n = T n,p 1. While this theoem is fo quite specific gaphs, it opened the doo to moe geneal questions about host gaphs not containing specific subgaphs. Ou theoem is a descendant of this idea. Like Tuán, we have chosen a specific host gaph (an -patite gaph with pat sizes n 1,..., n ) and a specific fobidden subgaph (k vetex-disjoint copies of K ). Two pevious papes should be mentioned which may seve as helpful pedecessos to ou theoem. Moon s theoem [8] also has a fobidden gaph of a numbe of vetex disjoint copies of a specific fom. In 009, Chen, Li, and Tu [9] found the extemal numbe fo k vetex disjoint matchings in bipatite gaphs - ou theoem extends these esults. 4 Theoem Theoem 3. (Main Theoem) Fo any integes 1 k n 1 n, ex(k n1,n,...,n, kk ) = n i n j n 1 n + n (k 1). 1 i<j The appoach to the poof will fall into two main sections. Fist, we will show the lowe bound, o that indeed, the extemal numbe is at least the numbe we claim. Then we will show the uppe bound, o that the extemal numbe is at most the numbe we claim. By showing the extemal numbe is simultaneously at least and at most the value we claim, we will be able to then say that the extemal numbe is exactly equal to the value we claim. Fo the lowe bound, we conside the gaph ((n 1 (k 1))K 1 K k 1, n ) + K n3,...,n. In othe wods, we have (n 1 (k 1)) copies of K 1 (i.e., (n 1 (k 1) isolated vetices) and the complete bipatite gaph 4

6 between pats of size n and k 1. To this, we join the complete -patite gaph between the pats of size n 3 though n. Clealy, this is a subgaph of K n1,...n. Also note it has the equied numbe of edges. Finally, note that to have kk, we would clealy need k vetices fom each pat involved in the kk. Howeve, in ou gaph, we only have n 1 (k 1) vetices contibuting fom the pat of size n 1, and can theefoe not have kk. Theefoe, since ou gaph is a subgaph of K n1,...n, has the equied numbe of edges, but still does not contain a copy of kk, it seves as a poof of the lowe bound. The uppe bound will be a bit moe of an in-depth pocess. We conside two cases: n = n and n < n. In the fome case, the poof is by induction on n 1 + k. In the latte case, the poof is by induction on the total numbe of vetices in the host gaph. Fo ease of notation, we define h k (n 1, n,..., n ) = 1 i<j n i n j n 1 n + n (k 1). We begin with the case whee n = n. In ode to use induction, we need two base cases, which ae established in Lemmas 3 and 4. Given two disjoint subsets of the vetex set, A, B V (G), define AB as the gaph fomed by the set of edges in G incident to (connected to) a vetex in A and a vetex in B. Also fo ease of notation, given an -patite gaph G with pats V 1,..., V, we let R(G, ) = {{v 1,..., v } V (G) : v i V i fo all i []}. That is, R(G, ) is the set of all -tuples of vetices with exactly one vetex fom each pat. We will utilize R(G, ) thoughout to facilitate the counting of edges. Fo some S R(G, ), define w(s) as the numbe of edges in the subgaph of G induced by S, that is w(s) = E(G[S]. Note that fo S R(G, ), an edge v i v j V i V j is counted in w(s) if and only if {v i, v j } S. Theefoe, summing ove all S R(G, ), w(s) = E(V i V j ) n l. (1) Lemma 3. Fo 1 n 1 n, S R(G,) i i<j l i,j ex(k n1,n,...,n, K ) = h 1 (n 1, n,..., n ) (Note that the following lemma establishes the base case fo the induction on k, since we ae looking fo a single K ). Poof. Suppose G K n1,n,...,n does not contain a copy of K. Then, fo all S R(G, ), we would have w(s) ( ( ) 1 (Notice that this is because ) is the minimum weight of a K, so the weight must be at least one fewe to guaantee no K ). This means that fo the sum ove all S R(G, ), we would have (( ) ) w(s) < 1 n 1 n 1. () S R(G,) 5

7 Subtacting () fom (1) yields Theefoe, 0 j= E(V 1 V j ) n + = n 1 n 3 E(G) + i,j 1 j= E(V i V j ) n 1 n 3 (( ) E(V 1 V j ) n 3 (n n 1 ) (( ) 1 n 1 n 1 ( ( ) ) 1 n 1 n 3 E(G) + E (G) n n 3 (n n 1 ) ( ) (( ) ) 1 = n E(G) n 1 (n n 1 ) 1 n 1 n 1 (( ) ) = n E(G) ( )n 1 n 1 1 n 1 n 1. ) ) 1 n 1 n 1 (( ) ) 1 n 1 n 1 ( ) 1 E(G) n 1 n ( 1) + n n 1 n = h 1 (n 1, n,..., n ). We have shown that, although G does not contain a single K (as it was defined), the numbe of edges in G is less than o equal to ou theoized extemal numbe, meaning that the maximum numbe of edges G may have is ou theoized extemal numbe. Since the maximum numbe of edges a gaph may have without any appeaance of a paticula subgaph, this shows that ex(k n1,n,...,n, K ) = h 1 (n 1, n,..., n ), as desied. Now we will establish the second base case, whee we ae looking fo the same numbe of K s as n 1. As in the Lemma 3, we have all equal pat sizes, except fo n 1, which may be smalle. Lemma 4. Fo 1 n 1 n, ex(k n1,n,...,n, n 1 K = h n1 (n 1, n,..., n ). Poof. This lemma will be poved by induction on n 1. The base case of n 1 = 1 was shown fo all positive integes n by the Lemma 3. Assume the statement is tue fo n 1 < n 1 whee n 1. Now suppose towads a contadiction that G K n1,n,...,n contains moe than h n1 (n 1, n,..., n ) edges but does not contain a copy of n 1 K. Also note that simply by the way they ae defined, h n1 (n 1, n,..., n ) is geate than o equal to h 1 (n 1, n,..., n ) (the inequality is not stict hee because the lemma was defined fo 1 n 1 n, i.e., not with stict inequalities). Symbolically, this is expessed as E(G) > h n1 (n 1, n,..., n ) h 1 (n 1, n,..., n ) Since thee ae moe edges in G than the numbe of edges we have aleady established contains a K, we know that G contains a copy of K. Now let S R(G, ) such that G[S] = K (i.e., the subgaph of G induced by S is K ). Then, if we examine the gaph made by emoving S fom G, we see E(G\S) h n1 1(n 1 1, n 1,..., n 1), othewise G\S would contain a copy of (n 1 1)K, and this togethe with S is a copy of n 1 K in G (which we ae assuming cannot happen). Theefoe, E(G) E(G\S) > h n1 (n 1, n,..., n ) h n1 1(n 1 1, n 1,..., n 1) ( ) = ( 1)(n 1 + ( )n ) + ( 1)n 1. Anothe way of thinking about the above value is the numbe of edges in K n1,n,...,n that have a vetex in S. This implies that all edges in the host gaph containing a vetex in S ae pesent in G. Note that this is 6

8 tue fo evey S such that G[S] = K in G. Let u i V i and u j V J with i j, if eithe u i o u j is in S, then the edge u i u j E(G). Othewise, fo v i S V i, let S = (X\{v i }) {u i }. S induces a copy of K in G and theefoe u i u j E(G). Hence G = K n1,n,...,n, and thus contains n 1 K, a contadiction. Now that we have ou two necessay base cases, we ae eady to pove the main theoem. The poof is split up into two cases: n = n and n < < n. Poof. Case 1. Assuming n 1 = n, we poceed by induction on n 1 + k. The base case of k = 1 was shown to be tue fo all positive integes n 1 in Lemma 3. Now assume the statement is tue fo the paametes n 1, k such that n 1 +k < n 1 +k fo n 1 > k. Also assume that G K n1,n,...,n does not contain a copy of kk. Fist we will obtain a lowe bound on the numbe of copies of K in G. Note that we ae not equiing the copies of K to be vetex disjoint. Suppose that thee ae exactly q such copies of K in G, then ( ) (( ) ) w(s) q + (n 1 n 1 q) 1. Recall that S R(G,R) S R(G,R) w(s) = j= E(V 1 V j ) n + i,j 1 E(V i V j ) n 1 n 3. (this is just a ewoding of equation (1), made specific to ou G whee n = n ). Again using the same q as befoe, this gives q E(V 1 V j ) n + (( ) ) E(V i V j ) n 1 n 3 n 1 n 1 1. (3) j= i,j 1 We will use equation (3) to get an uppe bound on E(G) by counting S R(G,) j= S R(G,) i,j 1 E(G\S). An edge v i v j V i V j is counted in E(G\S) if and only if v i S and v j S, hence E(G\S) = E(V 1 V j ) (n 1 1)(n 1)n + E(V i V j ) (n 1) n 1 n 3. (4) Using equations (3) and (4), we now have (( ) E(G\S) + (q + n 1 n 1 S R(G,) ) 1 (n ) 1 j= + i,j 1 E(V 1 V j ) (n 1 1)(n 1)n E(V i V j ) (n 1) n 1 n 3 E(G) (n 1)n n 1. (5) Now fo S R(G, ), suppose G[S] is a copy of K. Then E(G\S) h k 1 (n 1, n 1,..., n 1), else by induction G\S contains a copy of (k 1)K, and so this togethe with S yields a copy of kk in G. If G[S] is not complete, then since G\S does not contain a copy of kk, induction gives E(G\S) h k (n 1 1, n 1,..., n 1). Hence E(G\S) q (h k 1 (n 1, n 1,..., n 1)) S R(G,) + (n 1 n 1 q) (h k (n 1, n 1,..., n 1)) = q(1 n ) + n 1 n 1 (h k (n 1, n 1,..., n 1)) 7

9 and thus, using equation (5) we have E(G) (n 1)n n 1 n 1 n 1 ( h k (n 1, n 1,..., n 1) + (( ) ) ) 1 (n 1). Theefoe E(G) n ( h k (n 1, n 1,..., n 1) + n 1 = h k (n 1, n,..., n ). (( ) ) ) 1 (n 1) Case. Assume n < n. We poceed by induction on the numbe of total vetices. The base case of n 1 = n is tue fo all positive integes k by case 1. Now assume the statement holds fo all paametes n 1,..., n such that i=1 n i < i=1 n i. Suppose that G K n1,...,n does not contain a copy of kk. Let v V. The gaph G\{v } does not contain a copy of kk, has fewe vetices than G, and n n 1. Theefoe, E(G) = E(G\{v }) + d(v ) = ex(k n1,...,n 1, kk ) + d(v ) = h k (n 1,...n 1) + d(v ) = 1 i<j 1 i<j 1 n i n j n i n 1 n + n)(k 1 + d(v ) i=1 = h k (n 1, n,..., n ). n i n j n 1 n + n (k 1) 5 Futue eseach and othe open questions The main theoem elies on the fact that both K and K n1,n,...,n ae -patite. Cetainly the host gaph must have moe pats than the size of the fobidden clique - in othe wods, it must be l-patite fo l to have K as a subgaph. An inteesting genealization would be to calculate ex(k n1,n,...,n l, kk ) fo <. In [10], De Silva, Heysse, and Young poved that ( t ex(k n1,n,...,n l, kk ) = (k 1) n i ), howeve the Tuán numbe is open fo 3. The gaph does not contain kk 3, hence i= ((n 1 + n k + 1)K 1 K k1,n3 ) + n 4 K 1 ex(k n1,n,n 3,n 4, kk 3 ) (n 1 + n + n 3 )n 4 + (k 1)n 3 This constuction can be easily genealized to -patite gaphs, but it is not clea that this is an extemal constuction. Refeences [1] J. De Silva, K. Heysse, A. Kapilow, A. Schenfisch, and M. Young: Tuán numbes of vetex-disjoint cliques in -patite gaphs, pepint. 8

10 [] W. Mantel: Poblem 8, Wiskundige Opgaven, 10 (1907) [3] P. Edős: On sequences of integes no one of which divides the poduct of two othes and elated poblems, Mitt. Fosch. Institut Mat. und Mech. Tomsk (1938) [4] P. Edős: On sequences of integes no one of which divides the poduct of two othes, and some elated poblems, Mitt. Foschungsinst. Math. u. Mech. Tomsk (1938), 74-8 [5] P. Edős, A.H. Stone: On the stuctue of linea gaphs. Bulletin of the Ameican Mathematical Society. 5, 1 (1946), [6] Kuatowski, C: Su l opeation A de l analysis situs. Fund. Math. 3 (19), [7] P. Edős: Paul Tuán : His wok in gaph theoy, J. Gaph Theoy, 1 (1977) [8] J.W. Moon: On independent complete subgaphs in a gaph, Canad. J. Math., 0 (1968) also in: Intenational Congess of Math. Moscow, (1966), vol 13. [9] Chen, He, Xueliang Li, and Jianhua Tu. Complete Solution fo the Rainbow Numbes of Matchings. Discete Mathematics (009): Web. [10] J. De Silva, K. Heysse, M. Young, Rainbow numbe fo matchings in patite gaphs, pepint. 9

Math 301: The Erdős-Stone-Simonovitz Theorem and Extremal Numbers for Bipartite Graphs

Math 301: The Erdős-Stone-Simonovitz Theorem and Extremal Numbers for Bipartite Graphs Math 30: The Edős-Stone-Simonovitz Theoem and Extemal Numbes fo Bipatite Gaphs May Radcliffe The Edős-Stone-Simonovitz Theoem Recall, in class we poved Tuán s Gaph Theoem, namely Theoem Tuán s Theoem Let