aa r X i v : . [ m a t h . C O ] A ug A sharp threshold phenomenon in string graphs
Istv´an Tomon ∗ Abstract
We prove that for every ǫ > δ > C be acollection of n curves in the plane such that there are at most ( − ǫ ) n pairs of curves { α, β } in C having a nonempty intersection. Then C contains two disjoint subsets A and B such that |A| = |B| ≥ δn , and every α ∈ A is disjoint from every β ∈ B .On the other hand, for every positive integer n there exists a collection C of n curves in theplane such that there at most ( + ǫ ) n pairs of curves { α, β } having a nonempty intersection,but if A , B ⊂ C are such that |A| = |B| and α ∩ β = ∅ for every ( α, β ) ∈ A × B , then |A| = |B| = O ( ǫ log n ). The intersection graph of family of sets C is the graph whose vertices are identified with the elementsof C , and two vertices are joined by an edge if the corresponding elements of C have a nonemptyintersection. A curve in the plane is the image of a continuous function φ : [0 , → R , and a stringgraph is the intersection graph of a family of curves.Combinatorial and computational properties of string graphs are extensively studied, both froma theoretical and a practical point of view. The concept of string graphs was introduced by Benzer[1] in order to study topological properties of genetic structures. Later, Sinden [17] considered stringgraphs to model printed circuits, and he proved the already non-trivial statement that not everygraph is a string graph.A separator in a graph G is a subset S of the vertices of G such that every connected componentof G − S has size at most | V ( G ) | . A classical result of Lipton and Tarjan [11] is that every planargraph on n vertices contains a separator of size O ( √ n ). Building on this result, Fox and Pach [4]proved that if C is a family of curves and m is the total number of crossings between the elementsof C , then the intersection graph of G contains a separator of size O ( √ m ). They conjectured thestrengthening of their result that the same conclusion holds if m denotes the number of intersectingpairs of curves in C . Almost settling this conjecture, Matouˇsek [12] proved that every string graph G with m edges contains a separator of size O ( √ m log m ). Recently, the conjecture of Fox and Pachwas confirmed by Lee [10]. Theorem 1. ([10]) If G is a string graph with m edges, then G contains a separator of size O ( √ m ) . ∗ ´Ecole Polytechnique F´ed´erale de Lausanne, Research partially supported by Swiss National Science Foundationgrants no. 200020-162884 and 200021-175977. e-mail : [email protected] C C C p p p p p p Figure 1: Consider four pairwise touching circles C , C , C , C that touch at the six points p ij for1 ≤ i < j ≤
4. For each vertex in V i , we can define a convex set that only slightly deviates fromthe circle C i . It is possible to define these convex sets such that if x ∈ V i and y ∈ V j for some1 ≤ i < j ≤
4, then the corresponding convex sets intersect in the small neighborhood of p ij if xy ∈ E ( G ), and these convex sets are disjoint otherwise.Fox and Pach [4, 6] gave several applications of the existence of small separators in string graphs.A bi-clique in a graph G is a pair of disjoint subsets of vertices ( A, B ) such that | A | = | B | and ab ∈ E ( G ) for every a ∈ A and b ∈ B , and a size of a bi-clique ( A, B ) is | A | . An immediateconsequence of Theorem 1 is that for every 0 < δ < there exists c > G is a stringgraph with n vertices and at most cn edges, then the complement of G contains a bi-clique of sizeat least δn . Pach and Tomon [14] proved that if restrict our attention to x -monotone curves (curvessuch that every vertical line intersects them in at most one point), then there is a sharp threshold forthe edge density when linear sized bi-cliques start to appear in the complement of G . More precisely,they proved that for every ǫ > δ > G is the intersection graph of nx -monotone curves and | E ( G ) | ≤ ( − ǫ ) n , then G contains a bi-clique of size at least δn . On theother hand, they showed that there exists a family of n convex sets whose intersection graph G hasat most ( + ǫ ) n edges, but the size of the largest bi-clique in G is O ( ǫ log n ). Indeed, the existenceof such families follows from the following result of Pach and T´oth [15]: if G is a graph whose vertexset can be partitioned into four parts V , V , V , V such that V i spans a clique for i = 1 , , ,
4, then G can be realized as the intersection graph of convex sets. See Figure 1 for a brief explanation. Butthen, a standard probabilistic construction shows that there exist such graphs G with n vertices, atmost ( + ǫ ) n edges such that the size of the largest bi-clique in G is O ( ǫ log n ).Pach and Tomon [14] also conjectured that their result holds without the assumption that thecurves are x -monotone. Our main result is the proof of this conjecture. Convex sets can be approximated arbitrarily closely by x -monotone curves, so intersection graphs of convex setsare also intersection graphs of x -monotone curves. heorem 2. For every ǫ > there exists δ > such that the following holds. For every positiveinteger n if G is a string graph with n vertices and at most ( − ǫ ) n edges, then there exist twodisjoint sets A, B ⊂ V ( G ) such that | A | = | B | ≥ δn and there are no edges between A and B . The problem of finding large bi-cliques in intersection graphs and in their complements gained alot of interest recently. Fox, Pach and T´oth [7] proved that if G is the intersection graph of n convexsets, then either G or its complement contains a bi-clique of size Ω( n ). Also, if G is the intersectiongraph of n x -monotone curves, then either G contains a bi-clique of size Ω( n log n ), or G contains abi-clique of size Ω( n ), and these bounds are best possible. Pach and Tomon [14] gave a differentproof of the latter result using ordered graphs. Fox, Pach and T´oth [8] also proved that for every k ∈ N there exists c k > G is the intersection graph of n curves, where any two of thecurves cross at most k times, then either G or its complement contains a bi-clique of size at least c k n .In the case there are no restriction on the curves, Fox and Pach [5] proved that every dense stringgraph contains a dense incomparability graph as subgraph. By results of Fox [3] and Fox, Pach andT´oth [7], this implies that every dense string graph contains a bi-clique of size Ω( n log n ). Combiningthis with the result of Lee [10], we get that if G is a string graph, either G contains a bi-clique ofsize Ω( n log n ), or G contains a bi-clique of size Ω( n ); again, these bounds are best possible. Let usremark that incomparability graphs exhibit a similar threshold phenomenon as Theorem 2. Indeed,in case G is an incomparability graph with n vertices and at most ( − ǫ ) n edges, then G containsa bi-clique of size Ω( ǫn ), but there are incomparability graphs G with at most ( + ǫ ) n edges suchthat the size of the largest bi-clique in G is O ( ǫ log n ), see [7].Finally, the existence of linear sized bi-cliques in the complement of not too dense string graphsalso follows from a recent graph theoretic result of Chudnovsky, Scott, Seymour and Spirkl [2]. Theorem 3. ([2]) Let H be a graph. Then there exists ǫ > such that if G is a graph with n verticesand at most ǫn edges, and G does not contain a subdivision of H as an induced subgraph, then G contains a bi-clique of size at least ǫn . Indeed, if G is a string graph, then G does not contain a subdivision of K as an induced subgraph(we discuss this in more detail below). In order to prove Theorem 2, we shall also consider graphswhich avoid certain weaker notions of subdivisions of K as induced subgraph.Our paper is organized as follows. In Section 2, we prove Theorem 2. In Section 3, we concludeour paper with some open problems and remarks. But first, let us agree on some terminology. Given two curves α and β , a crossing between α and β is a point z ∈ α ∩ β such that α passes to theother side of β at z . Given a collection of curves C , we assume that if two curves α, β ∈ C intersect atsome point z , then z is a crossing between α and β . Indeed, by a small perturbation any intersectionpoint can be turned into a crossing without changing the intersection graph. Also, we assume thatthere are a finite number of crossings between any two curves. Indeed, by a result of Schaefer andˇStefankoviˇc [16], every string graph can be realized as the intersection graph of such a collection ofcurves. 3or a positive integer t , K t denotes the complete graph on t vertices. Let G be a graph. Thecomplement of G is denoted by G . If v ∈ V ( G ), then N ( v ) = { w ∈ V ( G ) : vw ∈ E ( G ) } is theneighborhood of v . Also, if U is a subset of the vertices, then G [ U ] is the subgraph of G induced by U . Moreover, if U and V are disjoint subsets of V ( G ), then E ( U, V ) is the set of edges in G withone endpoint in U and one endpoint in V . In this section, we prove Theorem 2. First, we shall reduce Theorem 2 to a completely graph theoreticstatement. In order to do this, we make use of the following immediate consequence of Theorem 1.
Lemma 4.
There exists a constant
C > such that for every positive integer n , if G is a stringgraph with n vertices and at most Cn edges, then G contains a bi-clique of size at least n . Note that the constant has no significance in this lemma, any positive constant would serveour purposes. Therefore, instead of citing Theorem 1, we could have cited Theorem 3 as well to getsomewhat weaker version of this lemma. In the rest of the paper, C denotes the constant describedby Lemma 4. This lemma tells us that in order to prove Theorem 2, it is enough to consider densegraphs. But for dense graphs, we can use powerful tools such as the Regularity lemma.Let us introduce some of the main notions used in this section. Let G be a graph with n vertices.If 0 < δ < , say that G is δ -full if for every A, B ⊂ V ( G ) satisfying | A | , | B | ≥ δn there existsan edge between A and B . The density of G is d ( G ) = | E ( G ) | n . Say that G is ( α, β ) -dense if everyinduced subgraph of G with at least αn vertices has density at least β . By Lemma 4, if G is a stringgraph that is δ -full, then G is (4 δ, C )-dense.If H is a graph, a k -subdivision of an edge xy ∈ E ( H ) is the following operation: we replace theedge xy by a path x = x , x , . . . , x k +1 = y , where { x , . . . , x k } is disjoint from V ( H ). A subdivisionof an edge is a k -subdivision for some k ≥
1. A graph H ′ is a k -subdivision (or subdivision) of H if we get H ′ from H by k -subdividing (or subdividing) every edge of H . Also, H ′ is a partialsubdivision of H if we get H ′ from H by subdividing some (possibly zero) edges of H . If H ′ is apartial subdivision of H , we refer to the vertices of H in H ′ as branch-vertices , and we refer to theother vertices as side-vertices .If H ′ is a k -subdivision (or subdivision) of H , let H ′′ be a graph we get from H ′ by addingedges between some pairs of side-vertices that belong to the subdivisions of neighboring edges of H . Then H ′′ is called a weak- k -subdivision (or weak-subdivision) of H . More precisely, H ′′ is a weak- k -subdivision (or weak-subdivision) of H if there exists a k -subdivision (or subdivision) H ′ of H such that V ( H ′′ ) = V ( H ′ ), E ( H ′ ) ⊂ E ( H ′′ ) and for every uv ∈ E ( H ′′ ) \ E ( H ′ ), u and v areside-vertices of H ′ , and if u belongs to the subdivision of the edge xy ∈ E ( H ) and v belongs to thesubdivision of x ′ y ′ ∈ E ( H ), then xy = x ′ y ′ , and xy and x ′ y ′ share an endpoint. See Figure 2 for anexample.Slightly generalizing Lemma 3.2 in [15] and Lemma 11 in [13], we show that string graphs do notcontain weak-subdivisions of K . Lemma 5. If G is a string graph, then G does not contain a weak-subdivision of K as an inducedsubgraph. BC D E
Figure 2: A weak-subdivision of K , where A, B, C, D, E are the branch-vertices. The edges that arepart of the weak-subdivision but not the subdivision are red.
Proof.
It is enough to show that if H ′ is a weak-subdivision of K , then H ′ is not a string graph.Suppose that H ′ is a string graph and let C be a collection of curves realizing H ′ . Let v , . . . , v bethe branch-vertices of H ′ and let C i ∈ C be the curve corresponding to v i for i = 1 , . . . ,
5. Let x i bean arbitrary point of C i . For every 1 ≤ i < j ≤
5, there exists a path connecting v i and v j in H ′ .Let X i,j be the union of the curves in C corresponding to the vertices of this path. Then X i,j is aconnected set in the plane containing x i and x j , so there exists a curve γ i,j ⊂ X i,j with endpoints x i and x j . Note that as H ′ is a weak-subdivision, X i,j ∩ X i ′ ,j ′ = ∅ if and only if { i, j } ∩ { i ′ , j ′ } 6 = ∅ .Hence, ∪ ≤ i For every ǫ > there exists δ > such that the following holds. Let G be a graph with n vertices and at most ( − ǫ ) n edges such that G is (4 δ, C ) -dense and δ -full. Then G contains aweak-subdivision of K . In the rest of this section, we prove this theorem. Let us briefly outline the proof. With the helpof the Regularity lemma, we partition our graph G into a constant number of sets such that thebipartite graph induced by most pairs of these sets is random like. The notion of regularity and theRegularity lemma is described in Section 2.1. Given a partial subdivision H of K , we show that an5 i x k x j z γ γ γ γ x i x k x j Figure 3: Uncrossing two neighboring edges.induced weak-subdivision of H can be found in G if we can find | V ( H ) | parts in the partition of G satisfying certain properties. This argument is presented in Section 2.2. Then, we finish our proofin Section 2.3 by showing that a regular partition of G must contain | V ( H ) | parts with the desiredproperties. In this section, we define the notion of regularity and state the Regularity lemma.If G is a graph and A, B ⊂ V ( G ), let d ( A, B ) = | E ( A, B ) || A || B | . For λ > 0, a pair of subsets ( A, B ) of V ( G ) is λ -regular , if for every A ′ ⊂ A and B ′ ⊂ B satisfying | A ′ | ≥ λ | A | and | B ′ | ≥ λ | B | , we have | d ( A, B ) − d ( A ′ , B ′ ) | ≤ λ. A λ -regular partition of the graph G is a partition V ( G ) = V ∪ · · · ∪ V k such that • | V | , . . . , | V k | ∈ {⌊ nk ⌋ , ⌈ nk ⌉} , • all but at most λk of the pairs ( V i , V j ) is λ -regular. Regularity Lemma. For every λ > and positive integer m there exists a positive integer M suchthat the following holds. Let G be a graph, then G has a λ -regular partition into k parts, where m ≤ k ≤ M . Given a graph G with λ -regular partition ( V , . . . , V k ), the reduced graph of this partition is theedge-weighted graph ( R, w ), where w : E ( R ) → [0 , • The vertex set of R is [ k ], • i and j are joined by an edge if ( V i , V j ) is λ -regular, • if ij ∈ E ( R ), then w ( ij ) = d ( V i , V j ). 6 .2 Embedding Let ( R, w ) be a complete edge-weighted graph, and let ǫ > 0. An edge xy ∈ E ( R ) is ǫ -thin if w ( xy ) ≤ λ , and xy is ǫ -fat if w ( xy ) ≥ − ǫ . Let H be a graph with | V ( R ) | vertices. Say that( R, w ) is ( H, ǫ ) -admissible if there exists a bijection b : V ( H ) → V ( R ) and a total ordering ≺ of thevertices of H such that • no edge of R is ǫ -fat, • if xy ∈ E ( H ) such that x ≺ y and b ( x ) b ( y ) is ǫ -thin, then w ( b ( x ) b ( z )) + w ( b ( y ) b ( z )) < − ǫ holds for every z ∈ V ( H ) \ { x, y } satisfying x ≺ z .Say that b and ≺ witness that ( R, w ) is ( H, ǫ )-admissible if they satisfy the properties above.This section is devoted to the proof of the following embedding lemma. Lemma 7. Let H be a graph with h vertices, let k be a positive integer and let α, β, δ, λ, ǫ > be realnumbers satisfying ǫ < β , λ < β h ( ǫ ) h , δ < k ( ǫ ) h , and α < k ( ǫ ) h . Let G be an ( α, β ) -dense, δ -full graph with a λ -regular partition ( V , . . . , V k ) and corresponding reduced graph ( R, w ) . If ( R, w ) contains an ( H, ǫ ) -admissible subgraph, then G contains a weak-2-subdivision of H as an inducedsubgraph.Proof. Let N = ⌊ nk ⌋ and for simplicity, assume that | V | = · · · = | V k | = N . Let the vertex set of H be { v , . . . , v h } , and let R ′ be ( H, ǫ )-admissible subgraph of R . Without loss of generality, supposethat { , . . . , h } is the vertex set of R ′ . Also, the bijection b and ordering ≺ witnessing that R ′ is( H, ǫ )-admissible are defined as follows: b ( v i ) = i for i = 1 , . . . , h , and v ≺ · · · ≺ v h .Let H ′ be the 2-subdivision of H , and for every edge v i v j ∈ E ( H ), let s i,j and s j,i be the twoside-vertices in H ′ v i v j , where s i,j is joined to v i , and s j,i is joined to v j,i .We embed the vertices of H ′ in G such that • for i = 1 , . . . , h , the image of v i is some vertex v ′ i ∈ V i , and for v i v j ∈ E ( H ), the image of s i,j is some vertex s ′ i,j ∈ V i , • the image of each edge of H ′ is an edge, • the image of each non-edge of H ′ is a non-edge, with the exception that s ′ i,j and s ′ i,j ′ might bejoined by an edge if v i v j , v i v j ′ ∈ E ( H ).Clearly, such an embedding induces a weak-2-subdivision of H in G , so our task is reduced to showingthat such an embedding exists. See Figure 4 for an illustration.Given a vertex v ∈ V i for some i ∈ [ h ] and h − U j ⊂ V j for j ∈ [ h ] \ { i } , say that v isaverage with respect to ( U j ) j ∈ [ h ] \{ i } if (cid:12)(cid:12)(cid:12)(cid:12) | N ( v ) ∩ U j || U j | − w ( ij ) (cid:12)(cid:12)(cid:12)(cid:12) < λ for j ∈ [ h ] \ { i } . The following useful claim is an easy consequence of the regularity condition. Claim 8. Let i ∈ [ h ] , and for j ∈ [ h ] \ { i } , let U j ⊂ V j such that | U j | ≥ λN . Then the number ofvertices in V i which are not average with respect to ( U j ) j ∈ [ h ] \{ i } is at most hλN . V V V v ′ v ′ v ′ v ′ s ′ , s ′ , s ′ , s ′ , s ′ , s ′ , s ′ , s ′ , s ′ , s ′ , s ′ , s ′ , Figure 4: An embedding of a weak-2-subdivision of K . The edges that are part of theweak-subdivision but not the subdivision are red. Proof. Let W ⊂ V i be the set of vertices which are not average with respect to ( U j ) j ∈ [ h ] \{ i } . If | W | ≥ hλN , then there exists j ∈ [ h ] \ { i } and W ′ ⊂ W such that | W ′ | ≥ λN , and either | N ( v ) ∩ U j || U j | > w ( ij ) + λ for every v ∈ W ′ or | N ( v ) ∩ U j || U j | < w ( ij ) − λ for every v ∈ W ′ . In the firstcase d ( W ′ , U j ) > d ( V i , V j ) + λ , in the second case d ( W ′ , U j ) < d ( V i , V j ) − λ , both contradicting that( V i , V j ) is a λ -regular pair.For i = 1 , . . . , h , set U i := V i . We embed the vertices of H ′ in G step-by-step while updating thesets U , . . . , U h at the end of each step. During this procedure, say that an index i ∈ [ h ] is active ,if either v ′ i or s ′ i,j is not defined yet for some v i v j ∈ E ( H ). During our procedure, the followingproperties are satisfied: • at each step, we embed either one branch-vertex or two side-vertices connected by an edge, • when a set is updated, it is replaced with one of its subsets, • at the end of step s , | U i | ≥ ( ǫ ) s N for i ∈ [ h ], • at the end of each step, if v ∈ V ( H ′ ) is already embedded in G and its image is v ′ ∈ V i , thenfor every active index j ∈ [ h ] \ i we have N ( v ′ ) ∩ U j = ∅ .In the first h steps, we embed the vertices v , . . . , v h one-by-one. More precisely, for s = 1 , . . . , h , atstep s we define the image v ′ s ∈ V s of v s and update the sets U , . . . , U h such that in addition to theproperties above, the following condition is also satisfied: • at the end of step s , we have U s ⊂ N ( v ′ s ). 8e proceed at step s as follows. In the beginning of step s , we have | U i | ≥ ( ǫ ) s − N for i ∈ [ h ].As G is ( α, β )-dense and | U s | ≥ ( ǫ ) h N ≥ αn , we have d ( G [ U s ]) ≥ β . Let W be the set of verticesin U s whose degree in G [ U s ] is at least β | U s | . Then β | U s | ≤ E ( G [ U s ]) ≤ 12 ( | W || U s | + β | U s | ) , so | W | ≥ β | U s | ≥ β ( ǫ ) h N . As | W | > hλN , there exists w ∈ W such that w is average withrespect to ( U i ) i ∈ [ h ] \{ s } . Note that for i ∈ [ h ] \ { s } , we have | U i \ N ( w ) | = | U i | − | U i ∩ N ( w ) | ≥ | U i | − | U i | ( w ( ij ) + λ ) ≥ | U i | ( ǫ − λ ) > ǫ | U i | > (cid:16) ǫ (cid:17) s N, where the second inequality holds noting that ij is not ǫ -fat. Also, | U s ∩ N ( w ) | ≥ β | U s | ≥ ǫ | U s | > (cid:16) ǫ (cid:17) s N. Set v ′ s = w , and make the following updates: U i := U i \ N ( w ) for i ∈ [ h ] \{ s } and U s := N ( w ) ∩ U s .This finishes step s .Let f , . . . , f l be an enumeration of the edges of H such that if 1 ≤ i < j ≤ l , and f i = v a v b , f j = v a ′ v b ′ , where a < b , a ′ < b ′ , then a ≤ a ′ . Now for s = 1 , . . . , l , we embed the two side-verticeson the 2-subdivision of the edge f s in step h + s .We proceed at step h + s as follows. In the beginning of the step, we have | U i | ≥ (cid:0) ǫ (cid:1) h + s ) − N .Let f s = v a v b , where a < b . Note that by the ordering of the edges, the active indices at this stepare i ∈ { a, a + 1 , . . . , h } . Consider two cases. Case 1. w ( ab ) ≥ ǫ .As | U a | > ( ǫ ) h N > hλN , there exists a vertex u ∈ U a such that u is average with respectto ( U i ) i ∈ [ h ] \{ a } . For i ∈ [ h ] \ { a } , let U ′ i = U i \ N ( u ), and let U ′ a = U a . Then, by similarcalculations as before, we have | U ′ i | ≥ ǫ | U i | . Also, let W = U b ∩ N ( u ), then | W | = | U b ∩ N ( u ) | ≥ | U b | ( w ( ab ) − λ ) ≥ | U b | ( ǫ − λ ) ≥ ǫ | U b | . But then | W | > hλN , so there exists w ∈ W such that w is average with respect to ( U ′ i ) i ∈ [ h ] \{ b } .But then for i ∈ [ h ] \ { b } , by the same calculations as before, we have | U ′ i \ N ( w ) | ≥ ǫ | U ′ i | ≥ (cid:16) ǫ (cid:17) | U i | ≥ (cid:16) ǫ (cid:17) h + s ) N. Set s ′ a,b = u , s ′ b,a = w , and make the following updates: U i := U ′ i \ N ( w ) for i ∈ [ h ] \ { b } and U b := U ′ b . This finishes step h + s . Case 2. w ( ab ) < ǫ .Let W a be the set of elements in U a that are average with respect to ( U i ) i ∈ [ h ] \{ a } , and let W b be the set of elements in U b that are average with respect to ( U i ) i ∈ [ h ] \{ b } . Then | W a | ≥| U a | − hλN ≥ | U a | , and similarly | W b | ≥ | U b | . But then | W a | , | W b | > ( ǫ ) h N > δn and G is δ -full, so there exists an edge between W a and W b . Let w a ∈ W a and w b ∈ W b be the9ndpoints of such an edge. For every i ∈ { a + 1 , . . . , h } \ { b } , let U ′ i = U i \ ( N ( w a ) ∪ N ( w b )).Then | U ′ i | = | U i |−| N ( w a ) ∪ N ( w b ) | ≥ | U i |−| N ( w a ) |−| N ( w b ) | ≥ | U i |− ( w ( ai )+ λ ) | U i |− ( w ( bi )+ λ ) | U i | . Note that we have w ( ai ) + w ( bi ) ≤ − ǫ as R ′ is ( H, ǫ )-admissible and a < i . Hence, | U ′ i | ≥ | U i | ( ǫ − λ ) ≥ ǫ | U i | ≥ (cid:16) ǫ (cid:17) h + s ) N. Set s ′ a,b = w a , s ′ b,a = w b , and make the following updates: U i := U ′ i if i ∈ { a + 1 , . . . , h } \ { b } , U a := U a \ N ( w b ), U b := U b \ N ( w a ), and do not update the sets U , . . . , U a − . This finishesstep h + s .It is easy to check that the end of the procedure we get the desired embedding of H ′ in G . Given an edge weighted graph ( R, w ), the weight of R is defined as w ( R ) = P f ∈ E ( R ) w ( f ). The aimof this section is to prove the following lemma. Lemma 9. Let H be the family of partial subdivisions of K with at most vertices. For every ǫ > there exist k = k ( ǫ ) ∈ N + and ǫ = ǫ ( ǫ ) > such that the following holds for every k ≥ k . Let ( R, w ) be an edge weighted graph with k vertices, at least (1 − ǫ ) k edges such that w ( R ) ≤ ( − ǫ ) k . Then there exists H ∈ H such that R contains an ( H, ǫ ) -admissible subgraph. Given the edge weighted graph ( R, w ), let R be the complete graph on V ( R ), and define theweight function w : E ( R ) → [0 , 1] as follows. For every f ∈ E ( R ), let w ( f ) = max( w ( f ) + ǫ , 1) if f ∈ E ( R ) and w ( f ) > ǫ , f ∈ E ( R ) and w ( f ) ≤ ǫ , f E ( R ) . Then it is easy to see that for every graph H , an ( H, R , w ) is( H, ǫ )-admissible in ( R, w ). Also, if | E ( R ) | ≥ (1 − ǫ ) | V ( R ) | , then | w ( R ) − w ( R ) | < ǫ | V ( R ) | .Therefore, in order to prove Lemma 9, it is enough to prove the following lemma. Lemma 10. Let H be the family of partial subdivisions of K with at most vertices. For every ǫ > there exist k = k ( ǫ ) ∈ N + such that the following holds for every k ≥ k . Let ( R, w ) be acomplete edge weighted graph with k vertices such that w ( R ) ≤ ( − ǫ ) k . Then there exists H ∈ H such that R contains an ( H, -admissible subgraph.Proof of Lemma 9 assuming Lemma 10. Let ǫ = ǫ and let k = k ( ǫ ) be the constant given byLemma 10. We show that the choice k = k and ǫ = ǫ suffices. Let ( R, w ) be an edge weightedgraph with k ≥ k vertices, at least (1 − ǫ ) k edges such that w ( R ) ≤ ( − ǫ ) k , and define thecomplete edge weighted graph ( R , w ) as above. Then w ( R ) ≤ w ( R ) + 2 ǫ k ≤ ( − ǫ ) k . Hence,( R , w ) contains an ( H, H ∈ H . But then this subgraph is also( H, ǫ )-admissible in ( R, w ). 10s a reminder, a complete subgraph R ′ of the complete edge-weighted graph ( R, w ) is( H, H if | V ( R ′ ) | = | V ( H ) | , and there exist a bijection b : V ( H ) → V ( R ′ )and a total ordering ≺ of the vertices of H such that • R ′ contains no edge of weight 1, • if xy ∈ E ( H ) such that x ≺ y and w ( b ( x ) b ( y )) = 0, then w ( b ( x ) b ( y )) + w ( b ( x ) b ( z )) < z ∈ V ( H ) \ { x, y } satisfying x ≺ y .In what comes, fix a positive integer k and a finite family of graphs H . Let ( R, w ) be a complete edge-weighted graph on k vertices that does not contain an ( H, H ∈ H . Also, among every such weighted graph, choose the one that minimizes the weight w ( R ),and among the ones that minimize the weight, it minimizes the size of the set F w = { w ( f ) : f ∈ E ( R ) } \ (cid:26) , , (cid:27) . Note that the set of weight functions w : E ( R ) → [0 , 1] for which ( R, w ) does not contain an( H, H ∈ H is a compact subset of R E ( R ) , so w ( R ) attains its minimumon this set. We show that ( R, w ) must have a simple structure, reminiscent of the extremal graphsgiven by the classical Tur´an’s theorem [19] not containing a clique of given size. To show this, weproceed by following a similar train of thoughts as one of the traditional proofs of Tur´an’s theorem. Proposition 11. F w = ∅ . Proof. Say that a triple of vertices ( x, y, z ) in ( R, w ) is dangerous if w ( xy ) = 0 and w ( xz )+ w ( yz ) ≥ D w be the set of dangerous triples. We show that if F w is nonempty, then we can replacethe weight function w with w ′ such that w ′ ( R ) ≤ w ( R ), | F w ′ | < | F w | and D w ⊂ D w ′ . But whether asubgraph of ( R, w ) is ( H, R, w ) isnot ( H, H ∈ H , then neither is ( R, w ′ ).For r ∈ [0 , F ( r ) = { f ∈ E ( R ) : w ( f ) = r } . Suppose that F w is nonempty and let r ∈ F w .Consider two cases. Case 1. − r F w . If r < , let q be the largest real number smaller than r such that q ∈ F w or 1 − q ∈ F w ; if there exists no such q , let q = 0. If r > , let q be the largest real numbersmaller than r but not less than such that q ∈ F w or 1 − q ∈ F w ; if there exists no such q , let q = . Define the new weight function w ′ as w ′ ( f ) = ( q if f ∈ F ( r ) ,w ( f ) if f ∈ E ( R ) \ F ( r ) . Then w ′ ( R ) = w ( R ) − ( r − q ) | F ( r ) | < w ( R ), F w ′ = F w \ { r } and D w ⊂ D w ′ . Case 2. − r ∈ F w . Without loss of generality, we can suppose that | F ( r ) | ≥ | F (1 − r ) | . Again, if r < , let q be the largest real number smaller than r such that q ∈ F w or 1 − q ∈ F w ; if thereexists no such q , let q = 0. If r > , let q be the largest real number smaller than r but not11ess than such that q ∈ F w or 1 − q ∈ F w ; if there exists no such q , let q = . Define the newweight function w ′ as w ′ ( f ) = q if f ∈ F ( r ) , − q if f ∈ F (1 − r ) ,w ( f ) if f ∈ E ( R ) \ ( F ( r ) ∪ F (1 − r )) . Then w ′ ( R ) = w ( R ) + ( q − r )( | F ( r ) | − | F (1 − r ) | ) ≤ w ( R ), F w ′ = F w \ { r, − r } and D w ⊂ D w ′ . Proposition 12. Among the complete edge weighted graphs ( R, w ) on k vertices, where w : E ( R ) →{ , , } and ( R, w ) has no ( H, -admissible subgraph for any H ∈ H , there is one with minimumweight w ( R ) satisfying the following properties. The vertex set of R can be partitioned into someparts X , . . . , X s such that for i ∈ [ s ] , every edge in X i has weight , and for every ≤ i < j ≤ s ,either every edge between X i and X j has weight , or every edge between X i and X j has weight .Proof. Let ( R, w ) be a complete edge weighted graph satisfying the desired conditions of theproposition. For x ∈ V ( R ), let d w ( x ) = P y ∈ V ( R ) \{ x } w ( xy ). For x, y ∈ V ( R ), let x ∼ y if w ( xy ) = 1.We show that ∼ is an equivalence relation. Suppose not, then there exists x, y, z ∈ V ( R ) such that x ∼ y , y ∼ z but x z . Consider two cases. Case 1. Either d w ( y ) > d w ( x ) or d w ( y ) > d w ( z ).Suppose that d w ( y ) > d w ( x ), the other case being similar. Define the new weight function w ′ ( f ) = ( w ( xu ) if f = yu for some u ∈ V ( R ) ,w ( f ) otherwise.Then w ′ ( R ) = w ( R ) + d w ( x ) − d w ( y ) < w ( R ) and ( R, w ′ ) does not contain an ( H, H ∈ H . Indeed, ( R, w ) and ( R ′ , w ) differ only in the edges containing y . Hence, if ( R, w ′ ) contains an ( H, R ′ , then y ∈ V ( R ′ ). But as w ( xy ) = 1, we must have x V ( R ′ ). The neighborhoods of x and y are identical in ( R, w ′ ), sothen R ′ \ { y } ∪ { x } is also ( H, R, w ) and ( R, w ′ ), contradiction. Case 2. d w ( y ) ≤ d w ( x ) and d w ( y ) ≤ d w ( z ).Define the new weight function w ′ ( f ) = ( w ( yu ) if f = xu or f = zu for some u ∈ V ( R ) ,w ( f ) otherwise.Then w ′ ( R ) = w ( R ) + (2 d w ( y ) − − ( d w ( x ) + d w ( z ) − w ( xz )) < w ( R ). Also, by a similarargument as in the previous case, ( R, w ′ ) does not contain an ( H, H ∈ H w ( R ), so ∼ is an equivalence relation. Let X , . . . , X s bethe equivalence classes of ∼ .Now consider the following operation on weight functions. Let w : E ( R ) → R + be a weightfunction such that w ( xy ) = 1 for every x, y ∈ X i , i = 1 , . . . , s . Define the weight function C i w on E ( R ) as follows. Let x ∈ X i be a vertex such that d w ( x ) is minimal among the vertices in X i . Thenfor f ∈ E ( R ), let C i w ( f ) = ( w ( xz ) if f = yz for some y ∈ X i , u ∈ V ( R ) ,w ( f ) otherwise.Then C i w ( R ) = w ( R ) + | X i | d w ( x ) − P y ∈ X i d w ( y ) ≤ w ( R ). Also, for every graph H ,if ( R, w ) does not have an ( H, R, C i w ) also does not contain an( H, w ′ = C s C s − . . . C w . Then w ′ ( R ) ≤ w ( R ) and ( R, w ′ ) does not contain an ( H, H ∈ H . Moreover, for any 1 ≤ i < j ≤ s , either every edge between X i and X j hasweight 0, or every edge between X i and X j has weight . Indeed, it is easy to prove by inductionon l that if w l = C l C l − . . . C w , then for every pair ( i, u ), where 1 ≤ i ≤ l and u ∈ X j for some j ∈ [ s ] \ { i } , either for every v ∈ X i we have w l ( uv ) = 0, or for every v ∈ X i we have w l ( uv ) = .But then w ′ = w s has the desired properties.In what comes, suppose that ( R, w ) has the form described in Proposition 12. Define the vertex -weighted graph ( Q, φ ), where φ : V ( Q ) → [0 , X , . . . , X s be the partitionof V ( R ) given by Proposition 12. Then the vertex set of Q is [ s ], the weight of a ∈ [ s ] is φ ( a ) = | X a | k ,and ab is an edge of Q if every edge of R between X a and X b has weight . Note that X a ∈ [ s ] φ ( a ) = 1 . We define the weight of the graph Q in the following unconventional way: φ ( Q ) = X a ∈ [ s ] φ ( a ) + X ab ∈ E ( Q ) φ ( a ) φ ( b ) . Note that with this definition, we have w ( R ) = X xy ∈ E ( R ) w ( xy ) = s X i =1 (cid:18) | X i | (cid:19) + X ab ∈ E ( Q ) | X a || X b | = k (cid:18) φ ( Q ) − k (cid:19) . (1)Let H be a graph and let Q ′ be an induced subgraph of Q on | V ( H ) | vertices. Say that Q ′ is H -admissible if there exists a bijection b : V ( H ) → V ( Q ′ ) and a total ordering ≺ of the vertices of H such that • if xy ∈ E ( H ) such that x ≺ y and b ( x ) b ( y ) E ( Q ′ ), then for every z ∈ V ( H ) \ { x, y } satisfying x ≺ z , either b ( x ) b ( z ) E ( Q ′ ) or b ( y ) b ( z ) E ( Q ′ )13ay that b and ≺ witness that Q ′ is H -admissible if they satisfy the properties above. Note that Q contains an H -admissible subgraph if and only if ( R, w ) contains an ( H, Q does not contain an H -admissible subgraph for any H ∈ H . The following proposition isfinishing touch in the proof of Lemma 10. Proposition 13. Let H be the family of partial subdivisions of K with at most 8 vertices. If ( Q, φ ) has no H -admissible subgraph for any H ∈ H , then φ ( Q ) ≥ . Proof. In order to make the proof of this proposition more transparent, we collect a few simpleobservations about Q . Observation 1. Any subset of 5 vertices of Q contains two neighboring edges and two disjoint edges.Proof. Let A be a 5 element subset of V ( Q ). If Q [ A ] does not contain 2 neighboring edges, then anyordering ≺ of the vertices of K and any bijection b : V ( K ) → A witness that Q [ A ] is K -admissible.Now suppose that A does not contain two disjoint edges. Then either Q [ A ] is the union of atriangle and two isolated vertices, or Q [ A ] is the union of a star and some isolated vertices. In thefirst case, any ordering ≺ of the vertices of K and any bijection b : V ( K ) → A witness that Q [ A ]is K -admissible. In the second case, let 1 ≺ ≺ ≺ ≺ K , and let v ∈ A bethe unique vertex with degree at least two. Then any bijection b : V ( K ) → A satisfying b (1) = v with the ordering ≺ witness that Q [ A ] is K -admissible. Observation 2. V ( Q ) does not contain two vertices a, b and a set A ⊂ V ( Q ) \ { a, b } such that | A | = 3 , and either1. there are no edges between { a, b } and A ,2. there are no edges between a and A , and b is joined to every vertex of A ,3. ab is an edge, and a and b are joined to every element of A .Proof. Let the vertices of K be 1 ≺ ≺ ≺ ≺ 5. Suppose that Q contains a, b, A with one ofthe described properties, and let A = { x , x , x } . If Q [ A ] contains exactly one non-edge, supposethat it is x x . Define the bijection b : V ( K ) → A ∪ { a, b } as follows. Let b (1) = b, b (2) = a, b (3) = x , b (4) = x , b (5) = x . Then b and ≺ witness that Q [ A ∪ { a, b } ] is K -admissible. Observation 3. Let v ∈ V ( Q ) and A ⊂ V ( Q ) \ { v } such that | A | = 4 and either v is joined to everyelement of A , or v is joined to no element of A . Then Q [ A ] is either a cycle of length 4 or a path oflength 3.Proof. It is easy to check that Q [ A ] is K -admissible, unless Q [ A ] is a cycle of length 4 or a path oflength 3. Let 1 ≺ ≺ ≺ K and let b : K → A be a bijection such that ≺ and b witness that Q [ A ] is K -admissible. Then extending ≺ to V ( K ) as 5 ≺ ≺ ≺ ≺ b as b (5) = v , b and ≺ witness that Q [ A ∪ { v } ] is K -admissible. Observation 4. If s ≥ , the complement of Q has maximum degree . roof. Assume that there exists a ∈ V ( Q ) and B ⊂ V ( Q ) \ { v } such that | B | = 5 and there are noedges between a and B . Let b ∈ V ( Q ) \ ( B ∪ { a } ). Then there exists A ⊂ B such that | A | = 3 andeither b is joined to every element of A , or there are no edges between b and A . This contradictseither 1. or 2. in Observation 2.Now we show that Q cannot have many vertices. Claim 14. s < Proof. Suppose that s = 8. By Observation 4, the minimum degree of Q is at least 3. First, supposethat there exists v ∈ V ( Q ) of degree 3, and let A be the set of 4 vertices not joined to v . Then byObservation 3, Q [ A ] is either a path of length 3, or a cycle of length 4. In both cases, there are twonon-edges ab in Q [ A ] for which there exists c ∈ A \ { a, b } such that ac, bc ∈ E ( Q ). Fix one of thesenon-edges ab , and let B = ( V ( Q ) \ ( A ∪ { v } )) ∪ { a, b } . Then | B | = 5, and as the minimum degree of Q is 3, both a and b has degree at least 1 in Q [ B ].We claim that there exists a path between a and b in Q [ B ]. Indeed, suppose not and let C bethe connected component of Q [ B ] containing a . Then | C | ≥ | B \ C | ≥ a and b hasdegree at least 1. But then {| C | , | B \ C |} = { , } , and there are no edges between B and C , whichcontradicts 1. in Observation 2. See Figure 5.Let a = x , x , . . . , x p +1 = b be the vertices of a path connecting a and b in B , and set D = A ∪ { v, x , . . . , x p } . Also, let K ( p )5 be the partial subdivision of K in which one edge is p -subdivided.We show that D is K ( p )5 -admissible. Let 1 , , , , K ( p )5 , and let r , . . . , r p be the vertices p -subdividing the edge 23. Let c, d be the two vertices in A \ { a, b } . Consider thevertex ordering on K ( p )5 defined as r ≺ · · · ≺ r p ≺ ≺ ≺ ≺ ≺ 5, and define the bijection b as b ( r i ) = x i for i = 1 , . . . , p , b (1) = v, b (2) = a, b (3) = b, b (4) = c, b (5) = d . Then ≺ and b witness that Q [ D ] is K ( p )5 -admissible.Now we can suppose that every vertex of Q has degree at least 4. Let v ∈ V ( Q ) be an arbitraryvertex and let A ⊂ N ( v ) such that | A | = 4. Again, by Observation 3, we have that Q [ A ] is either acycle of length 4 or a path of length 3. Let B = ( V ( Q ) \ ( A ∪ { v } )) ∪ { a, b } . Then | B | = 5, and asthe minimum degree of Q is 4, both a and b have degree at least 1 in Q [ B ]. But then we can repeatthe exact same argument as before to find a K ( p )5 -admissible subgraph for some p ≤ 3. This finishesthe proof of the claim.Therefore, in order to finish the proof of Proposition 13, it is enough to consider the cases s = 1 , . . . , 7. The following simple inequality will help us to deal with these few cases. Claim 15. Let ≤ a ≤ b ≤ c ≤ d . Then ad + bc ≤ ac + bd ≤ ab + cd. Proof. The first inequality is equivalent to ( b − a )( d − c ) ≥ 0, while the second inequality is equivalentto ( c − a )( d − b ) ≥ φ (1) ≤ φ (2) ≤ · · · ≤ φ ( s ). Now let us consider thedifferent possible values of s . For every value of s , we show that φ ( Q ) can be lower bounded by thesum of the squares of 4 numbers whose sum is equal to 1. But then φ ( Q ) ≥ .15 ac bd x x x B Figure 5: An illustration for the proof of Claim 14. s ≤ In this case, we simply have φ ( Q ) ≥ s X i =1 φ ( i ) ≥ s ≥ . s = If Q is not K -admissible, then Q has at least two edges by Observation 1. But then φ ( Q ) ≥ X i =1 φ ( i ) + 2 φ (1) φ (2) = ( φ (1) + φ (2)) + φ (3) + φ (4) + φ (5) ≥ . s = By Observation 1, there are 3 vertices a, b, c ∈ { , , , , } such that ab, ac ∈ E ( Q ). Butthen φ ( a ) φ ( b ) + φ ( a ) φ ( c ) ≥ φ (2) φ (3) + φ (2) φ (4). Also, there are two disjoint edges a b , a b in Q [[6] \ { a } ]. But then by Claim 15, we have φ ( a ) φ ( b ) + φ ( a ) φ ( b ) ≥ φ (1) φ (4) + φ (2) φ (3).As ab, ac, a b , a b are four distinct edges of Q , we can write φ ( Q ) ≥ X i =1 φ ( i ) + φ (2) φ (3) + φ (2) φ (4) + φ (1) φ (4) + φ (2) φ (3) ≥ ( φ (1) + φ (4)) + ( φ (2) + φ (3)) + φ (5) + φ (6) ≥ . s = The analysis of this case is quite tedious, so we postpone it until the Appendix.Now we are ready to conclude the proof of Lemma 10, and therefore the proof of Lemma 9. Proof of Lemma 10. The choice k = ⌈ ǫ ⌉ + 1 suffices. Indeed, we proved that if ( R, w ) does notcontain an ( H, H ∈ H , then φ ( Q ) ≥ . But by (1), we have w ( R ) = k (cid:18) φ ( Q ) − k (cid:19) > (cid:18) − ǫ (cid:19) k , finishing the proof. 16 .4 Proof of Theorem 6 In this section, we put everything together to finish the proof of Theorem 6. Proof of Theorem 6. Let H be the family of partial subdivisions of K with at most 8 vertices. Asa reminder, C is the constant given by Lemma 4.Let ǫ = ǫ , ǫ = min { C , ǫ ( ǫ ) } and k = k ( ǫ ), where ǫ ( ǫ ) and k ( ǫ ) are the constants given byLemma 9. Let β = C , and λ = β h ( ǫ ) . By the Regularity lemma, there exists M = M ( k , λ ) suchthat G has a λ -regular partition ( V , . . . , V k ), where k ≤ k ≤ M . Let ( R, w ) be the correspondingreduced graph. Then | E ( R ) | ≥ (cid:0) k (cid:1) − λk > (1 − ǫ ) k . Also, n k w ( R ) ≤ | E ( G ) | ≤ (cid:18) − ǫ (cid:19) n , so w ( R ) ≤ ( − ǫ ) k . But then by Lemma 9, ( R, w ) contains an ( H, ǫ )-admissible subgraph forsome H ∈ H . Let h = | V ( H ) | ≤ δ = M ( ǫ ) and α = 4 δ < M ( ǫ ) . Then the parameters h, α, β, δ, λ, ǫ satisfy theconditions of Lemma 7. Hence, if G is ( α, β )-dense and δ -full, then G contains a weak-2-subdivisionof H as an induced subgraph. Note that a weak-subdivision of a partial subdivision of K mightnot be a weak-subdivision of K , but it contains a weak-subdivision of K as an induced subgraph.Thus, G contains a weak-subdivision of K as an induced subgraph. It follows from the combination of Theorem 3 and Theorem 6 that if G is a graph with n vertices, atmost ( − ǫ ) n edges such that G does not contain a weak-subdivision of K as an induced subgraph,then G contains a linear sized bi-clique. It would be interesting to decide whether this statementcan be generalized for graphs with no induced weak-subdivision of the complete graph K t . Conjecture 16. For every ǫ > and integer t ≥ , there exists δ > such that the following holds.If G is a graph with n vertices and at most ( t − − ǫ ) n edges such that G does not contain an inducedweak-subdivision of K t , then G contains a bi-clique of size at least δn . Note that if this conjecture is true, it is sharp. Indeed, if G is a graph whose vertex set canbe partitioned into t − V , . . . , V t − such that V i spans a clique for i = 1 , . . . , t − 1, then G does not contain a weak-subdivision of K t . But using standard probabilistic techniques, it is easy toconstruct such graphs G with less than ( t − + ǫ ) n edges such that the size of the largest bi-cliquein G is O ( ǫ log n ). Moreover, a positive answer to Conjecture 16 would have similar implications asTheorem 2 for intersection graphs of curves defined on certain surfaces. We omit the details.In order the prove Conjecture 16, it would be enough to prove the following generalization ofProposition 13, as this is the only part of our proof that does not apply to general families of graphs H . Conjecture 17. Let H be the family of partial subdivisions of K t . If ( Q, φ ) has no H -admissiblesubgraph for any H ∈ H , then φ ( Q ) ≥ t − . 17e can prove the following slightly weaker version of Conjecture 16, by proving a slightly weakerversion of Conjecture 17. Theorem 18. For every integer t ≥ , there exists δ > such that the following holds. If G is a graph with n vertices and at most t − · n edges such that G does not contain an inducedweak-subdivision of K t , then G contains a bi-clique of size at least δn .Sketch proof. It is enough to show that there exists ǫ > Q, φ ) has no K t -admissiblesubgraph, then φ ( Q ) ≥ t − + ǫ .Let ( Q, φ ) be the vertex weighted graph on vertex set [ s ] with no K t -admissible subgraph. Then s < R ( t ), where R ( t ) is the usual Ramsey number, that is, R ( t ) is the smallest integer N such thatevery graph on N vertices contains either a clique or independent set of size t . Indeed, if s ≥ R ( t ),then Q contains either a clique or independent set of size t , but both are K t -admissible.Now we show that if φ ( Q ) ≥ s + t − , then Q contains an independent set of size t , which isimpossible. But then setting ǫ = R ( t ) finishes the proof.By a similar Tur´an type argument as in Proposition 12, we can show that among the vertexweighted graphs ( Q, φ ) with vertex set [ s ], fixed weight function φ , and no independent set of size t , the one that minimizes φ ( Q ) has the following form. The vertex set of Q can be partitioned into t − V , . . . , V t − such that V i induces a clique in Q for i = 1 , . . . , t − 1. But then φ ( Q ) = s X i =1 φ ( i ) + t − X i =1 X a,b ∈ V i ,a
References [1] S. Benzer, On the topology of the genetic fine structure, Proc. Nat. Acad. Sci. (1959):1607–1620.[2] M. Chudnovsky, A. Scott, P. Seymour, S. Spirkl, Sparse graphs without linear anticomplete pairs, arXiv:1804.01060 (2018)[3] J. Fox, A bipartite analogue of Dilworth’s theorem, Order (2006): 197–209.184] J. Fox, J. Pach, A separator theorem for string graphs and its applications, Combin. Probab.Comput. (3) (2010): 371–390.[5] J. Fox, J. Pach, String graphs and incomparability graphs, Advances in Mathematics (2012):1381–1401.[6] J. Fox, J. Pach, Applications of a new separator theorem for string graphs, Combin. Probab.Comput. (1) (2014): 66–74.[7] J. Fox, J. Pach, C. D. T´oth, Tur´an-type results for partial orders and intersection graphs of convexsets, Israel Journal of Mathematics (2010): 29–50.[8] J. Fox, J. Pach, Cs. T´oth, Intersection patterns of curves, J. Lond. Math. Soc. (2011): 389–406.[9] G. H. Hardy, J. E. Littlewood, G. P´olya, Inequalities, Cambridge University Press (1952): Section10.2, Theorem 368.[10] J. R. Lee, Separators in region intersection graphs, in: 8th Innovations in Theoretical Comp.Sci. Conf. (ITCS 2017), LIPIcs (2017): 1–8.[11] J. Lipton, R. E. Tarjan, A separator theorem for planar graphs, SIAM J. Appl. Math. (2)(1979): 177–189.[12] J. Matouˇsek, Near-optimal separators in string graphs, Combinatorics, Probability & Computing (1) (2014): 135–139.[13] J. Pach, B. A. Reed, Y. Yuditsky, Almost all string graphs are intersection graphs of planeconvex sets , in: 34th Symposium on Computational Geometry (SoCG 2018): 68:1–14. Discrete& Computational Geometry, to appear.[14] J. Pach, I. Tomon, Ordered graphs and large bi-cliques in intersection graphs of curves, to appearin European Journal of Combinatorics (2019)[15] J. Pach, G. T´oth. How many ways can one draw a graph?, Combinatorica (2006): 559–576.[16] M. Schaefer, D. ˇStefankoviˇc, Decidability of string graphs, J. Comput. System Sci. (2004):319–334.[17] F. W. Sinden, Topology of thin film RC-circuits, Bell System Technological Journal (1966):1639–1662.[18] E. Szemer´edi, Regular partitions of graphs, in Proc. Colloque Inter. CNRS (J.-C. Bermond, J.-C.Fournier, M. Las Vergnas, D. Sotteau, eds.) (1978): 399–401.[19] P. Tur´an, On an extremal problem in graph theory, Matematikai ´es Fizikai Lapok (in Hungarian) (1941): 436–452. 19 ppendix - Proof of Proposition 13, s = 7 Suppose that s = 7. First, we show that if Q contains either a cycle of length 6 or 7, we have φ ( Q ) ≥ . Indeed, let a < · · · < a r be the vertices of this cycle, where r ∈ { , } , and let π ∈ S r bea permutation such that a π (1) a π (2) , a π (2) a π (2) , . . . , a π ( r ) a π (1) are the edges of this cycle. Then r X i =1 φ ( a π ( i ) ) φ ( a π ( i +1) ) ≥ r X i =1 φ ( a i ) φ ( a r +1 − i ) ≥ φ (1) φ (6) + 2 φ (2) φ (5) + 2 φ (3) φ (4) , where the first inequality is the consequence of the Rearrangement inequality [9]. Hence, φ ( Q ) ≥ X i =1 φ ( i ) + 2 φ (1) φ (6) + 2 φ (2) φ (5) + 2 φ (3) φ (4) ≥ ( φ (1) + φ (6)) + ( φ (2) + φ (5)) + ( φ (3) + φ (4)) + φ (7) ≥ . By Observation 4, every vertex of Q has degree at least 2. Suppose that Q has a vertex a ofdegree 2, and let b , b be the vertices joined to a , and let C = { c , c , c , c } be the rest of thevertices. By Observation 3, Q [ C ] is either a cycle of length 4 or path of length 3. Case 1. C is a cycle of length 4. Without loss of generality, let φ ( c ) ≤ φ ( c ) ≤ φ ( c ) ≤ φ ( c ) and φ ( b ) ≤ φ ( b ). Then φ ( Q ) = X i =1 φ ( i ) + X xy ∈ E ( Q ) φ ( x ) φ ( y ) ≥ X i =1 φ ( i ) + 2 φ ( c ) φ ( c ) + 2 φ ( c ) φ ( c ) + 2 φ ( a ) φ ( b )= ( φ ( c ) + φ ( c )) + ( φ ( c ) + φ ( c )) + ( φ ( a ) + φ ( b )) + φ ( b ) ≥ , where the first inequality holds by two applications of Claim 15. Case 2. C is a path of length 3. Without loss of generality, let the edges of this cycle be c c , c c , c c . As c and c both have degree at least 2, there is an edge from both c and c to { b , b } . Without loss of generality, suppose that c b is an edge. If c b is an edge, then Q contains a cycle of length 7, namely c c c c b ab , so we are done. Hence, we may assumethat c b is an edge, and c b and c b are non-edges. But then both b c and b c are edges,because if b c is a non-edge, say, then there are no edges between { a, b } and { c , c , c } ,contradicting Observation 2. But then Q contains cycle of length 6, namely c b c c b c , sowe are done. See Figure 6. 20 c c c c b b ac c c c b b Figure 6: An illustration for Case 2. Left is the subcase where c b is an edge, right is the subcasewhere c b is not an edge.Therefore, we have φ ( Q ) ≥ if Q contains a vertex with degree 2. Hence, we can suppose thatevery vertex has degree at least 3 in Q . As 7 is odd, Q has at least one vertex a of even degree, sothe degree of a is either 4 or 6.First, suppose that a has degree 4, and let B = { b , b , b , b } be the neighbors of a , and c , c be the rest of the vertices. By Observation 3, Q [ B ] is either a cycle of length 4, or a path of length3. In both cases Q [ B ∪ { a } ] contains a cycle C of length 5. If c c is not an edge, then there are atleast three edges between c and C , so there are two consecutive vertices of C joined to c . But then C ∪ { c } contains a cycle of length 6, so we are done. Therefore, we can suppose the c c is an edge.Then there are at least 2 edges between c i and C for i = 1 , 2. In this case, we can find two disjointedges between { c , c } and C , which also implies the existence of a cycle of length at least 6 in C .The only remaining case is when the degree of a is 6. Let b ∈ V ( Q ) \ { a } , and let A be theneighborhood of b in V ( Q ) \ { a } . If | A | ≥ 3, then we get a contradiction by 3. in Observation 2, andif | A | ≤ 2, then | V ( Q ) \ ( A ∪ { a } ) | ≥| ≥