1 Overview
The intersection graph of a collection of sets is a graphs whose vertex set is and in which two sets in are connected by an edge if and only if they have nonempty intersection. A curve is a subset of the plane which is homeomorphic to the interval . The intersection graph of a finite collection of curves (“strings”) is called a string graph.
Ever since Benzer [Be59] introduced the notion in 1959, to explore the topology of genetic structures, string graphs have been intensively studied both for practical applications and theoretical interest. In 1966, studying electrical networks realizable by printed circuits, Sinden [Si66] considered the same constructs at Bell Labs. He proved that not every graph is a string graph, and raised the question whether the recognition of string graphs is decidable. The affirmative answer was given by Schaefer and Štefankovič [ScSt04] 38 years later. The difficulty of the problem is illustrated by an elegant construction of Kratochvíl and Matoušek [KrMa91], according to which there exists a string graph on vertices such that no matter how we realize it by curves, there are two curves that intersect at least times, for some . On the other hand, it was proved in [ScSt04] that every string graph on vertices and edges can be realized by polygonal curves, any pair of which intersect at most times, for some other constant . The problem of recognizing string graphs is NPcomplete [Kr91, ScSeSt03].
In spite of the fact that there is a wealth of results for various special classes of string graphs, understanding the structure of general string graphs has remained an elusive task. The aim of this paper is to show that almost all string graphs have a very simple structure. That is, the proportion of string graphs that possess this structure tends to as tends to infinity.
Given any graph property P and any , we denote by the set of all graphs with property P on the (labeled) vertex set . In particular, is the collection of all string graphs with the vertex set . We say that an element set is partitioned into parts of almost equal size if the sizes of any two parts differ by at most for some , provided that is sufficiently large.
As , the vertex set of almost every string graph can be partitioned into parts of almost equal size such that of them induce a clique in and the th one splits into two cliques with no edge running between them.
Every graph whose vertex set can be partitioned into parts such that of them induce a clique in and the th one splits into two cliques with no edge running between them, is a string graph.
Theorem 1 settles a conjecture of Janson and Uzzell from [JaU17], where a related weaker result was proved in terms of graphons.
We also prove that a typical string graph can be realized using relatively simple strings.
Let denote the set of all intersection graphs of families of labeled convex sets in the plane. For every pair , select a point in , provided that such a point exists. Replace each convex set by the polygonal curve obtained by connecting all points selected from by segments, in the order of increasing coordinate. Observe that any two such curves belonging to different s intersect at most times. The intersection graph of these curves (strings) is the same as the intersection graph of the original convex sets, showing that . Taking into account the construction of Kratochvíl and Matoušek [KrMa91] mentioned above, it easily follows that the sets and are not the same, provided that is sufficiently large.
There exist string graphs that cannot be obtained as intersection graphs of convex sets in the plane.
We call a graph canonical if its vertex set can be partitioned into parts such that of them induce a clique in and the th one splits into two cliques with no edge running between them. The set of canonical graphs on vertices is denoted by . Theorem 1 states . In fact, this is an immediate corollary of and the relation , formulated as
The vertices of every canonical graph can be represented by convex sets in the plane such that their intersection graph is .
The converse is not true. Every planar graph can be represented as the intersection graph of convex sets in the plane (Koebe [Ko36]). Since no planar graph contains a clique of size exceeding four, for no planar graph with vertices is canonical.
Almost all string graphs on labeled vertices are intersection graphs of convex sets in the plane.
See Figure 1 for a sketch of the containment relation of the families of graphs discussed above.
The rest of this paper is organized as follows. In Section 2, we recall the necessary tools from extremal graph theory, and adapt a partitioning technique of Alon, Balogh, Bollobás, and Morris [AlBBM11] to analyze string graphs; see Theorem 2. Since the modifications are not entirely straightforward, we include a sketch of the proof of Theorem 2 in the appendix. In Section 3, we collect some simple facts about string graphs and intersection graphs of plane convex sets, and combine them to prove Theorem 1. In Section 4, we strengthen Theorem 2 in two different ways and, hence, prove Theorem 1 modulo a small number of exceptional vertices. We wrap up the proof of Theorem 1 in Section 5.
2 The structure of typical graphs in an hereditary family
A graph property P is called hereditary if every induced subgraph of a graph with property P has property P, too. With no danger of confusion, we use the same notation P to denote a (hereditary) graph property and the family of all graphs that satisfy this property. Clearly, the properties that a graph is a string graph () or that is an intersection graph of plane convex sets () are hereditary. The same is true for the properties that contains no subgraph, resp., no induced subgraph isomorphic to a fixed graph .
It is a classic topic in extremal graph theory to investigate the typical structure of graphs in a specific hereditary family. This involves proving that almost all graphs in the family have a certain structural decomposition. This research is inextricably linked to the study of the growth rate of the function , also known as the speed of P, in two ways. Firstly, structural decompositions may give us bounds on the growth rate. Secondly, lower bounds on the growth rate help us to prove that the size of the exceptional family of graphs which fail to have a specific structural decomposition is negligible. In particular, we will both use a preliminary bound on the speed in proving our structural result about string graphs, and apply our theorem to improve the best known current bounds on the speed of the string graphs.
In a pioneering paper, Erdős, Kleitman, and Rothschild [ErKR76] approximately determined for every the speed of the property that the graph contains no clique of size . Erdős, Frankl, and Rödl [ErFR86] generalized this result as follows. Let be a fixed graph with chromatic number . Then every graph of vertices that does not contain as a (not necessarily induced) subgraph can be made partite by the deletion of edges. This implies that the speed of the property that the graph contains no subgraph isomorphic to is
(1) 
Prömel and Steger [PrS92a, PrS92b, PrS93] established an analogous theorem for graphs containing no induced subgraph isomorphic to . Throughout this paper, these graphs will be called free. To state their result, Prömel and Steger introduced the following key notion.
A graph is colorable for some if there is a coloring of the vertex set , in which the first color classes are cliques and the remaining color classes are independent sets. The coloring number of a hereditary graph property P is the largest integer for which there is an such that all colorable graphs have property P. Consequently, for any , there exists a colorable graph that does not have property P.
The work of Prömel and Steger was completed by Alekseev [Al93] and by Bollobás and Thomason [BoT95, BoT97], who proved that the speed of any hereditary graph property P satisfies
(2) 
The lower bound follows from the observation that for , there exists such that all colorable graphs have property P. In particular, contains all graphs whose vertex sets can be partitioned into cliques and independent sets, and the number of such graphs is equal to the righthand side of (2).
If we want to tighten the above estimates, another idea of Prömel and Steger
[PrS91] is instructive. They noticed that the vertex set of almost every free graph can be partitioned into a clique and an independent set, and no matter how we choose the edges between these two parts, we always obtain a free graph. Therefore, the speed of freeness is at most , which is much better than the general bound that follows from (2). Almost all free graphs permit similar “certifying partitions”. It is an interesting open problem to decide which hereditary families permit such partitions and what can be said about the inner structure of the subgraphs induced by the parts. This line of research was continued by Balogh, Bollobás, and Simonovits [BaBS04, BaBS09, BaBS11]. The strongest result in this direction was proved by Alon, Balogh, Bollobás, and Morris [AlBBM11], who proved that for almost every graph with a hereditary property P, one can delete a small fraction of the vertices in such a way that the rest can be partitioned into parts with a very simple inner structure. This allowed them to replace the bound (2) by a better one:This will be the starting point of our analysis of string graphs. As we shall see, in the case of string graphs, our results allow us to replace the in this bound by . See [BB11, KKOT15, RY17, ReSc17], for related results.
We need some notation. Following Alon et al., for any integer , define as a bipartite graph with vertex classes and , where a vertex in the first class is connected to a vertex in the second if and only if . We think of as a “universal” bipartite graph on vertices, because for every subset of the first class there is a vertex in the second class whose neighborhood is precisely this subset.
As usual, the neighborhood of a vertex of a graph is denoted by or, if there is no danger of confusion, simply by . For any disjoint subsets , let and denote the subgraph of induced by and the bipartite subgraph of consisting of all edges of running between and , respectively. The symmetric difference of two sets, and , is denoted by .
Let be a positive integer. A graph is said to contain if there are two disjoint subsets such that the bipartite subgraph induced by them is isomorphic to . Otherwise, with a slight abuse of terminology, we say that is free.
By slightly modifying the proof of the main result (Theorem 1) in [AlBBM11] and adapting it to string graphs, we obtain
For any sufficiently large positive integer and for any which is sufficiently small in terms of , there exist and a positive integer with the following properties.
The vertex set of almost every string graph can be partitioned into eight sets, , and a set of at most vertices such that

is free for every ;

; and

for every and there is such that
In other words, for the right choice of parameters, almost all string graphs have a partition into parts satisfying the following conditions. There is a set of sublinear size in the number of vertices such that deleting its elements, the subgraphs induced by the parts are U(k)free. Moreover, there is another set of at most constantly many vertices such that the neighborhood of every vertex with respect to the part it belongs to is similar to the neighbourhood of some vertex in . In Appendix A.1, we sketch the proof of this result, indicating the places where we slightly deviate from the original argument in [AlBBM11].
3 String graphs vs. intersection graphs of convex sets–Proof of Theorem 1
Instead of proving Theorem 1, we establish a somewhat more general result.
Given a planar graph with labeled vertices and positive integers , let denote the class of all graphs with vertices that can be obtained from by replacing every vertex with a clique of size , and adding any number of further edges between pairs of cliques that correspond to pairs of vertices with .
Then every element of is the intersection graph of a family of plane convex sets.
Proof.
Fix any graph . The vertices of can be represented by closed disks with disjoint interiors such that and are tangent to each other for some if and only if (Koebe, [Ko36]). In this case, let denote the point at which and touch each other. For any , let be the center of . Assume without loss of generality that the radius of every disk is at least .
has vertices denoted by , where and . In what follows, we assign to each vertex a finite set of points , and define to be the convex hull of . For every we include in all sets with , to make sure that for each , all sets have a point in common, therefore, the vertices that correspond to these sets induce a clique.
Let be the minimum of all angles at which the arc between two consecutive touching points and on the boundary of the same disc can be seen from its center, over all and over all and . Fix a small satisfying .
For every with , let be a circular arc of length on the boundary of , centered at the point . We select distinct points , each representing a different subset . A point will belong to the set if and only if . (Warning: Note that the roles of and are not interchangeable!)
If for some with , the intersection of the neighborhood of a vertex for any with the set is equal to , then we include the point in the set assigned to , see Figure 2 for a sketch. Hence, for every and , we have
In other words, the intersection graph of the sets assigned to the vertices of is isomorphic to .
It remains to verify that
Suppose that the intersection graph of the set of convex polygonal regions
differs from the intersection graph of
Assume first, for contradiction, that there exist with such that and are tangent to each other and contains a point for which
(5) 
Consider the unique point that belongs to , that is, we have
Draw a tangent line to the arc at point . See Figure 3. The polygon has two sides meeting at ; denote the infinite rays emanating from and containing these sides by and . These rays either pass through or intersect the boundary of in a small neighborhood of the point of tangency of with some other disk . Since was chosen to be much smaller than , we conclude that and lie entirely on the same side of where , the center of , is. On the other hand, all other points of , including the point satisfying (5) lie on the opposite side of , which is a contradiction.
Essentially the same argument and a little trigonometric computation show that for every and , the set is covered by the union of some small neighborhoods (of radius ) of the touching points between and the other disks . This, together with the assumption that the radius of every disk is at least (and, hence, is much larger than and ) implies that cannot intersect any polygon with , for which and are not tangent to each other. ∎
4 Strengthening Theorem 2
In this section, we strengthen Theorem 2 in two different ways. To avoid confusion, in the formulation of our new theorem, we use in place of and in place of . We will see that we can insist that the four parts of the partition have approximately the same size. Secondly, we can guarantee that , , and are cliques and induces the disjoint union of two cliques. More precisely, setting , we prove the following result, which is similar in flavour to a result in [ReSc17].
For every sufficiently small , there are with the following property. For almost every string graph on , there is a partition of into such that for some set of at most vertices the following conditions are satisfied:

, , and are cliques and induces the disjoint union of two cliques.

,

for every and every , there exists such that

for every , we have .
For the proof of Theorem 4 we need the following statement which is a slight generalization of Lemma 3.2 in [PaT06], and it can be established in precisely the same way, details are given in the appendix.
Let be a graph on the vertex set , where and every is connected by an edge to and . The graph may have some further edges connecting pairs of vertices with . Then is not a string graph.
For each of the following types of partition, there exists a nonstring graph whose vertex set can be partitioned in the specified way:
(a) stable (that is, independent) sets each of size at most 10;
(b) cliques each of size at most five and a vertex;
(c) cliques each of size at most five and a stable set of size ;
(d) cliques each of size at most five and a path with three vertices;
(e) cliques both of size at most five and graphs that can be obtained as the disjoint union of a point and a clique of size at most .
Proof of Theorem 4..
We choose sufficiently large and then sufficiently small in terms of . We choose such that Theorem 2 holds for this choice of and and so that is less than the of Lemma 4 for this choice of . We set and consider large enough to satisfy certain implicit inequalities below. We know that the subset of , consisting of those graphs for which there is a set of at most vertices and a partition into and satisfying (a),(b), and (c) set out in Theorem 2, contains almost every string graph. We call such a partition, certifying. We need to show that almost every graph in has a certifying partition for which we can repartition into so that (I),(II), and (IV) all hold (that (III) holds, is simply Theorem 2 (c) and ).
We prove this fact via a sequence of lemmas. In doing so, for a specific partition, we let be the number of pairs of vertices not lying together in some . The first lemma gives us a lower bound on , obtained by simply counting the number of graphs which permits a partition into four cliques all of size within one of . Its four line proof is given in the appendix.
.
The second gives us an upper bound on the number of choices for for graphs in for which is a certifying partition. It is Corollary 8 in [AlBBM11].
For every k, there is a positive such that for every sufficiently large , the number of free graphs with vertices is less than .
Next we prove:
The number of graphs in which have a certifying partition such that for some , is .
Proof.
The number of choices for a partition of into is at most . If this partition demonstrates that is free and is large, Lemma 4 tells us that there are only choices for . The number of choices for the edges out of each vertex of is . So, since is at most , we know there are at most choices for the edges out of . It follows that there are at most choices for our partition and the graphs over all in which can be certified using this partition. Furthermore,the number of graphs in permitting such a certifying choice is at most . Since, , it follows that almost every graph in has no certifying partition for which . The desired result follows. ∎
Setting , we have the following.
The number of graphs in which have a certifying partition for which there are distinct and such that both and contain disjoint independent sets of size 10 is .
Proof.
Consider a choice of certifying partition and induced subgraphs where . By Corollary 4(a), for any pair of independent sets of size , at least one of the choices of edges between the sets yields a bipartite nonstring graph. Thus, the number of choices for edges between the partitions which extend our choice to yield a graph in is at most . Since and , it follows that for almost every graph in , almost every certifying partition does not contain two distinct such and . ∎
Ramsey theory tells us that if a graph does not contain disjoint stable sets of size 10, it contains disjoint cliques of size 5. Combining applications of this fact to three of the , Corollary 11(c), and an argument similar to that used in the proof of Lemma 4 allows us to prove the following lemma. Details can be found in the appendix. The number of graphs in which have a certifying partition for which there is an such that does not contain disjoint cliques of size 5 is
With this lemma in hand, we can mimic the argument used in its proof to obtain the following two lemmas. In doing so, we apply Corollary 11 (c),(d), and (e).
The number of graphs in which have a certifying partition for which there is an such that contains disjoint sets of size three each inducing a stable set or a path is .
The number of graphs in which have a certifying partition for which there are two distinct such that contains disjoint sets of size four each inducing the disjoint union of a vertex and a triangle is .
Combining these lemmas, and possibly permuting indices, we see that almost every graph in has a certifying partition for which for every we have , no contains more than sets inducing a path of length three or a stable set of size three, and for every , does not contain disjoint sets inducing the disjoint union of a vertex and a triangle. For each such graph, we consider such a partition. For all , we let be the union of and a maximum family of disjoint sets in each inducing a path of length 3, a stable set of size three, or the disjoint union of a triangle and a vertex. We let be the union of and a maximum family of disjoint sets in each inducing a path of length three or a stable set of size three. We set . ∎
5 Completing the proof of Theorem 1
In this section, we prove our main result. By a great partition of we mean a partition of its vertex set into such that for , is a clique and is the disjoint union of two cliques. We call a graph great if it has a great partition and mediocre otherwise. Theorem 1 simply states that almost every string graph on is great.
Thus, we are trying to show that almost every string graph has a partition into sets satisfying Theorem 4 (I) with the sets empty. We choose so small that Theorem 4 holds and also satisfies certain inequalities implicitly given below. We apply Theorem 4 and obtain that for some positive and , for almost every graph in there is a partition of into satisfying (I), (II), (III), and (IV). Note that if we reduce the theorem remains true. We insist that is at most . We call such partitions good. We need to show that the number of mediocre string graphs on with a good partition is of smaller order than the number of great graphs on .
The following result tells us that the number of great graphs on is of the same order as the number of great partitions of graphs on .
Claim .
The ratio between the number of great partitions of graphs on and the number of graphs which permit such partitions is .
So, it is sufficient to show that the number of mediocre string graphs with a good partition on is of smaller order than the number of graphs with a great partition on . In doing so, we consider each partition separately. For every partition of we say that a good partition satisfying (I)(IV) with for every is good. We prove:
Claim .
For every partition of , the number of graphs which permit a great partition with for every is of larger order then the size of the set of mediocre string graphs which permit a good partition.
To complete the proof of Theorem 1 we need to show that our two claims hold.
Before doing so, we deviate momentarily and discuss the speed of the string graphs. Combining Theorem 1 and Claim 5, we see that the ratio of the size of over the number of ordered great partitions of graphs on is , so we need only count the latter. There are ordered partitions of into , and there are graphs for which this is a great partition, where, as before, is the number of pairs of vertices not lying together in some . This latter term is at most , which gives us the claimed upper bound on the speed of string graphs. Furthermore, a simple calculation of the ordered 4partitions of shows that there is an proportion where no two parts differ in size by more than one. This gives us the claimed lower bound.
We now prove our two claims. In proving both, we exploit the fact that if a string graph has a great partition and we fix the subgraph induced by the parts of the partition, then any choice we make for the edges between the sets will yield another string graph permitting the same great partition.
This fact implies that the edge arrangements between the partition elements of a graph permitting a particular great partition are chosen uniformly at random and, hence, are unlikely to lead to a graph permitting some other great partition. This allows us to prove Claim 5, which we do in the appendix.
Proof of Claim 5:.
Let be the number of pairs of vertices not contained in a partition element and note that there are exactly choices for for a graph for which is a great partition, and hence graphs for which is a great partition.
Our approach is to show that while there may be more choices for the for mediocre graphs for which is a good partition, for each such choice we have many fewer than choices for mediocre string graphs extending these subgraphs.
We note that by the definition of good, we need only consider partitions such that each has size .
Let and let be the projection of on the sets , that is, the disjoint union of the sets , and .
Now, (I) of Theorem 4 bounds the number of choices for by 1 if and if . Furthermore, (III) bounds the number of edges out of in terms of its size and (II) bounds its size. Putting this all together we obtain the following lemma. Its proof can be found in the appendix.
Let be a partition of , the number of possible projections on of graphs in is
For a mediocre graph in , we call a set versatile if for each with , there is clique in such that for all subsets of there are vertices of which are adjacent to all elements of and to none of .
The number of mediocre string graphs in such that for some there is a versatile subset of 3 vertices of inducing a path or a stable set of size three,is .
Proof.
To begin, we count the number of mediocre graphs which extend a given projection on where induces such a graph. We first expose the edges from to determine if is versatile and then count the number of choices for the remaining edges between the partition elements. If is versatile we choose cliques which show this is the case.
By Corollary 4 (c) or (d), there is a nonstring graph whose vertex set can be partitioned into 3 cliques of size at most five, and a graph isomorphic to the subgraph of the projection induced by . We label these three cliques as for and let be an isomorphism from to . For each vertex , let and be those vertices of whose neighbourhhod on is . Now, since for all in each , for each , we can choose cliques of size at most five such that there is bijection from to with for every .
If we choose our cliques in this way then for any set of three cliques there is a choice of edges between the cliques which would make the union of these three cliques with induce . Thus, there is one choice of edges between the cliques which cannot be used in any extension of to a string graph. Mimicking an earlier argument, this implies that the number of choices for edges between the partition elements which extend to a string graph is at most . By the bound in Lemma 5 on the number of possible projections, the desired result follows. ∎
Using Corollary 4 (e) in places of (c) & (d), we can ( and do in the appendix) prove an analogous result for sets of size 8 intersecting two partition elements. To state it we need a definition. A graph is extendible if there is some nonstring graph whose vertex set can be partitioned into two cliques of size five and a set inducing .
The number of mediocre string graphs in such that for some distinct and there are subsets of and of , both of size four, whose union is both versatile and induces an extendible graph is .
For every mediocre string graph in , we choose a maximum family of disjoint sets each of which is either (a) contained in some and induces one of a stable set of size three or a path of length three, or (b) contains exactly four vertices from each of two distinct partition elements and is extendible. For every such choice we count the number of elements of whose projection yields the given choice of .
Now, by the definition of a good partition, each contains a clique containing half the vertices of and hence at least vertices. Lemmas 5 and 5 imply that we can restrict our attention to graphs for which for any subset in , there is a subset of and a with disjoint from such that there are fewer than vertices of which are adjacent to all of and none of . This implies that the number of choices for the edges from to other partition elements is .
Every element of must intersect , so that . Set , and let . Note that for every , has more than vertices and is the disjoint union of two cliques. Given a choice of , the number of choices for projections on is less than . Mimicking the proof of Lemma 5, the number of choices for the vertices of , and the edges of from the vertices in which remain within the partition elements of is . Combining this with the result of the last paragraph yields:
There is a constant such that the number of mediocre string graphs in for which is .
So, we can restrict our attention to mediocre graphs which have a partition for which . Similar tradeoffs allow us to handle them. Full details are found in the Appendix. ∎
6 Acknowledgement
This research was carried out while all three authors were visiting IMPA in Rio de Janeiro. They would like to thank the institute for its generous support.
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Appendix A The Appendix
a.1 Sketch of the proof of Theorem 2
Proof.
We only need to prove this result for sufficiently small as it then follows for all . We will set to be for some which is required to be sufficently small. So, we can and do replace by in what follows. We essentially follow the [AlBBM11] proof of their Theorem 1 given in Section 7 of their paper. We note that our statement differs from their statement in the following ways (i) for us the hereditary family is the family of string graphs hence, as Pach and Toth proved , (ii) we allow to be any large enough integer rather than one fixed large integer, (ii) we allow to be arbitrarilly small as long as it is small enough in terms of ( and ), (iii) is chosen as a function of and , (iv) there is an integer which is chosen as a function of and such that there is a choice of at most vertices and a partition of into for which our property (c) holds, and (v) the sentence beginning Moreover is deleted. We will not reproduce the entire proof. We simply set out the very minor modifications these changes require.
We want to use the strengthening of their Lemma 23 obtained by replacing and such that in its statement with such that for any sufficiently small in terms of and , and (iii) replacing with in the definition of just before the statement of Lemma 23 by with for the of Lemma 18 or. Their proof of the lemma actually proves this strengthening, provided that (a) in the first paragraph we set out that is the of Lemma 18, (b) replace by in the definition of given on its fourth line, and (iii) delete if is sufficeintly large.
Now while following their (three paragraph) proof of their Theorem 1, we again replace by sufficiently small in terms of and , and insist and are sufficiently small in terms of both these parameters. Furthermore, we define to be the of Lemma 18. We also add and at the end of the second paragraph before for almost every.
Then we consider the adjustment and exceptional set they obtain and set . Now, as in their proof, consider a maximal bad set . By our strengthened version of Lemma 23 the size of is at most We set to be this . Now, (a) is their Theorem 1(b), (b) is their Theorem 1 (a) where is , and (c) follows immediately from the fact that is an adjustment and the definition of adjustment.
∎
a.2 The Proof of Lemma 4
Proof.
Suppose for contradiction that has a string representation. Continuously contract each of string curve representing to a point , without changing the intersection pattern of the curves. For every pair consider some nonself intersecting arc of the curve representing with endpoints and . These arcs define a drawing of , in which no two independent edges intersect. However, is not a planar graph, hence, by a well known theorem of Hanani and Tutte [Ch34], [Tu70], no such drawing exists. ∎
a.3 The Proof of Lemma 4
Proof.
For any partition of into four sets , each of size between and , there are at least string graphs on vertices in which the partition elements form cliques. We note these graphs are in with the empty and containing one vertex from each clique. So . ∎
a.4 The Proof of Lemma 4
Proof.
By Lemmas 4 and 4, it is enough to consider graphs in with respect to which every contains more than vertices and there are no two distinct such that and contain disjoint stable sets of size .
By Ramsey theorem, every set of vertices in any contains either a clique of size or stable sets of size . By our assumption does not contain disjoint cliques of size , therefore for large enough it must contain disjoint stable sets of size . Therefore, for all , does not contain disjoint stable sets of size , and, hence, it contains a set of cliques of size .
By Corollary 4, for the union of any one of the independent sets in and a clique of size five from each of the other , there is choice of edges between the partition element which extends these stable sets and cliques to a nonstring graph induced by the 25 vertices. Now, for some prime between and , we consider such unions given by, for each : (where addition is modulo ). We see that the number of choices for edges between the partition elements which gives a string graph, given the choices for the edges within is at most . The desired result follows. ∎
a.5 The Proof of Claim 5
Proof.
To prove our claim, we focus on graph–great partition pairs , that is, where the partition is a great partition of with the following property:
(P*)

any two vertices of in the same partition element which forms a clique, have at least common neighbours;

two vertices in different partition elements have fewer than common neighbours;

for every partition element and every vertex not in , forms a path of length three with two vertices of ; and

does not induce a clique.
Clearly, every great graph has at least six great partitions obtained by permuting the indices of the partition elements. We show now that (i) every graph on has at most six great partitions satisfying (P*), and (ii) almost every graph–great partition pair on satisfies (P*). These two statements prove our claim.
To prove (i), we assume that and are two great partitions of a graph , both of which satisfy property (P*). Clearly, (a) and (b) tell us that for , is contained in some . Now, (c) tells us that each such is, in fact, nonempty and equal to some . Hence, the set of partition elements is the same. Therefore, by (d), and (i) follows.
It remains to show (ii). For any (ordered) partition