On Some Numbers Related to the Erdös-Szekeres Theorem

A crossing family of segments is a collection of segments each pair of which crosses. Given positive integers and , a grid is the union of two pairwise-disjoint collections of segments (with and members, respectively) such that each segment in the first collection crosses all members of the other. Let j


Introduction
For each positive integer let 3 n    g n  be the least integer such that every planar set of  g n points in general position contains the vertices of some convex n-gon.This number was introduced by Erdös and Szekeres in 1935 (see [1] and [2]) who established the bounds and conjectured that the lower bound is in fact an equality.The values and are easy, and several proofs of the fact have been given.However, no other values have been computed exactly and the upper bound given by Erdös and Szekeres stood until recently as the best known.A seqence of 1998 papers by Chung and Graham [3], Kleitman and Pachter [4], and finally Tóth and Valtr [5] improved the above-mentioned bound to Morris and Soltan [6] provide an excellent survey of related results.
Given the apparent difficulty of determining values of

 
g n we might well seek weakened notions of these numbers.For example, let be a combinatorial property satisfied by the vertex set of a convex -gon.It might be interesting to ask how large a set   n  n X must be to guarantee the existence of a subset of X having property   n  .We will consider such generalizations where the property is a specified intersection behavior of some subset of the diagonals to a convex -gon.n We will say that a set 2 X   generates a collection of segments if each segment in has its endpoints in   X .Also, we will say that a collection  of segments is a crossing family if each pair of segments in crosses (intersects at a point that is not endpoint to either segment).Now if then the vertex set of a convex -gon clearly generates a crossing family of size .Define be the least integer such that any planar set of points in general position generates a  -grid.We again have the easy inequality , but a result of Nielsen and Sabo [8] implies the linear upper bound for these numbers.It appears at least superficially, then, that grids are easier to find than are crossing families, and that both are easier to find than are convex polygons.But while the progression from

 
g n to to represents a geometric and computational simplification, none of these can be said to be simple.A look at the six-point cases illustrates the situation well.

 
The value of is not known.It is conjectured that , but (0.1) gives only .
  The value of is not known, but we will prove in this paper that    .However, this is the largest case for which the exact value of is known.

 
,2 8 The purpose of this paper is to establish the facts mentioned in items (B) (see Section 1) and (C) (see Section 2).Bounds for some of the larger cases of these numbers will be given in a subsequent paper.The methods we use are not complicated, but some imagination is required to find an approach that reduces the number of cases to a manageable level.

An Improved Bound for c(3)
The bound (0.2) gives only -of course (from (0.1)) is better.We are able here to improve this substantially to .We would be quite surprised if the actual value of is not closer to the lower bound, but reducing the upper bound appears to be very difficult.

 
We begin by developing some notation that will be useful in the main proof.Let X be a finite planar set in general position.Now let A and B be vertices of the convex hull of X admitting parallel supporting lines.We may assume these supporting lines touch the convex hull of X only at points A and B so that the points of  1).
 For each i such that i X is defined let i F be the set consisting of those points in  lying on the boundary of the convex hull of i X , together with the points A and B .Furthermore, We think of i F as being a "convex fence" separating and W .
i i More generally, we will say that a sequence of points , and  every segment joining a point of V to a point of W crosses one of the f − 1 segments and W is the of points of X outside that fence.
Given positive integers , and we will say that ,  fence consisting of f points for some v v   and w w   .The sets described above yield a sequence of properties , , , , , for each i and es AB.Then X j the segment i j PQ cross generates a crossing family of three segments including AB .
Clearly it is enough to show that is a conve x quadrilateral for some , , , i j k m , and this it is enough to show that line sses segment k m Q Q and line   misses segm , , P P s con , , Q Q then it must be the case that there is a m Q meeting exactly two of the lines , , P C P (see Figure 2).segments by Now X generates no crossing family of three assumption, so by Lemma 1.1 there cannot be three points of W on the same side  of line for some 2 f  gener egments.Proof.Assume to reach a ates a crossing family of three s contradiction that a set X has pr for some 2 f  but rossing family of th generates no c ree segments n by Lemma 1.2 X also has property We leave it to the reader to verify that this set fails to generate any crossing family of three segments.Now let X be any set of 16 points in general e will show that X generates a crossing family of three segments.The construction given at the outset of this section established a sequence of propert ies for X .Consider the property   p ing family for X.Thus, we may assume 1 2 0 r r   and , , P P P d the other consists of the two points in with point A. Thus, we may assume 2 S contains no more than one point of X.
This show 2 toget S s that by discarding up to two points ( an P ) inside and one point (in 2 S ) outside the fence ABC we can eliminate all segme s (joining an inside an outside point) that cross AB.A mirror image of this same argument can be done for eliminating segments that cross CD.In this way we conclude that we can discard four points inside ABCD and two points outside ABCD and eliminate all segments except those that cross BC.So, BC is a   3,3 -fence for the remaining subset of X.As before, this fficient to guarantee that X generates the desired crossing family.
Case 2:  If R 4 contains these points then CE is a   3,5 -fence where the three points on one side consist of the point in R 2 together with A and B. If the points lie in R 5 then similarly AD is a   3,5 -fence.Both of the above cases, then, lead to a cr amily of ossing f three segments generated by X.The only remaining case to consider is 2 6 r  (and 1 3 4 5 0 r r r r     ).
Here, ABDE is a   6, 6 e for X ( C with the points ou the fence).Note that Lemma 1.2 would allow us to conclude that either X generates a crossing family of three segments or else its property -fenc e where we include tsid .Consider the convex hull of the five points up the fence ABCDE .The diagonals of this pentagon determine terior regions.If a point of X lies in any of the eleven in  ).Unfortunately, that is not sufficient under our lemm s of X in n int as to give the desired conclusion.Instead, we will need to be more careful in reducing the fence.
For our first step in the reduction, order the point region 2 R as 1 P through 6 P radially from B as in Figure 9.We may ssume the gion 1 S in that figure contains no more than one point of X, since otherwise BE is a   3,3 -fence (with two points from 1 S and A on one nd   , , P P P on the other).en discarding any point 1 ng with 1 P and 2 P , we may eliminate all segments (join g a po inside a alo re side a Th in S i ABDE to a point outside) that cross AB ; all this while leaving at least four inside points and at least five outside points.
The second part of the reduction is accomplished by ordering the remaining points of X in R 2 as Q 1 through Q 4 radially from D as in Figure 10.We may assume the region labeled as S 2 in that figure contains no more than 2 points of X, else DE is a  

An Exact Value for #(1,2)
nown for any of the ppose that  n t X must generate a crossing family of three segments, so the theorem is proved.
Here we prove the only exact value k six-point configuration numbers mentioned in the introduction.The following well-known fact will prove useful in the analysis.
Lemma 2.1: , , , , , , , , n , X A B P P P Q Q   Q and j the segment i j PQ crosses AB.T     half of Figu e that R is terior to the triangle APQ as in the right half of Figure re 13).Thus, we may also assum in 13.We now consider some subcases.Recall that there is a point Z of set X lying outside of the convex hull of X  .The segment RZ must meet either BC or one or both of the diagonals of ABCD.
Subcase 3A.Suppose first that RZ meets BC.Note that it must also meet one of the sides of triangle APQ, an D. Then these diagonals together w ular, then, RZ meets BD.But BD d that this side is disjoint from BC. Thus X generates a (1,2)-grid in this case.
Subcase 3B.Next suppose that RZ meets exactly one of the diagonals of ABC ith RZ form a (1,2)-grid.Subcase 3C.Finally, suppose that RZ meets both diagonals of ABCD.In partic meets both AP and AQ, and RZ must miss one of these.So, this case also yields a generated (1,2)-grid.
Case 4. The only remaining case is to assume that the convex hull of X  is a triangle, say ABC, with the four remaining points of X  interior to this triangle.If ZABC is a convex quadrilateral then by separating B from the remaining sev points we reduce to one of the earlier cases (see Figure 14).
Thus we may assume that C is interior to triangle ZAB (so that the convex hull of en X is a triangle).We may clearly assume that a similar configuration results if any of the three vertices of this triangle are separated from the other seven points.In this case the set X must be as pictured in Figure 15: the convex hull of X is a triangle     of each of the triangles and , again yielding a (1,2)-grid.
All cases have now been considered and the proof is complete.
We will show below that
a set generates a crossing family of three segments if it contains a subset with a   3,3 -fence of two points or a   5, 5 -fence of three .This fact will be used repeatedly in our next proof.eight points depicted in Figure 3. (A few lines are shown to indicate relative positioning of the points.)

4 
In this case, X has property 7 4 5 / / .As in Figure4, let the fence be ABCD an the that X then generates a crossing family of three segments (two of which are AC and BD).So, if r 1 > 0 then we may assume either 3

Figure 4 .
Figure 4.The general situation for Case reliminary lemmas would guarantee the desired cross-

4 R 4 R
he ts B and D) AC is a   3,3 -fe e.On t other hand, if 4 6 r  then ACD is  ,5 -fence.So (again using our lem as) we may assume 1,  , and the supporting line through A to the e ll of conv x hu X ) contains at most two poin of ts X , else AB is a   3 -fence (with the latter set of th points being   , P P D ).But then BCA is a the points outside consist of the (at least t oints outside ABCD not lying in 1 S together with D and the point of X in region .By our preliminary lemmas, X then generates a ssing family of three segments.S in Figure 6 (bounded by AB contain  ,  , and the supporting line).If this region o points of X then BD is a   3,3 -fence where one set of three points is 3

Figure 6 . 5 R ure 8 .
Figure 6.The arrangement for Subcase 1B.ve regions bounded by segments of the pentagon or in fi the central region bounded by all the diagonals, then X generates a crossing family of three segments as sho in Figure 7. Thus, we m wn ay assume that the six points of X inside the fence ABCDE lie in the five regions labeled 1 R through 5 R ure 8.As before, let i r denote t number o oints of in Fig p he f X in region i R . If points of X lie in each of two jac ad ent regions from among R 1 through R 5 then it is easy to construct a crossing family of three segments (two are diagonals of ABCDE and the other joins the points in question). If 2 4 0 r r   then ACE is a   6,5 -fence-more tha guarantee the desired crossing family.Putting together these two observations, we may now n enough to assume that 2 0 r  , 1 3 0 r r   , either 4 0 r  or 5 0 r  , and (of course) 2 4 5 6 r r r    . If 2 2 5 r  

Figure 7 .
Figure 7. Crossing families in easy instances of Case 2.

Figure 9 .
Figure 9.The first step in reducing the fence for Case 2.

Figure 10 .
Figure 10.The second step in reducing the fence for Case 2.each of which crosses AB .Proof.Let π be the permutation of such ngths of the segm is m   1, 2, , n  ents n P Q  that the sum of the le   n

Figure 14 .
Figure 14.Reducing an instance of Case 4 to a previous case.

Figure 15 .
Figure 15.The difficult part of Case 4.
be the least integer such that any planar set of   168 this bound might be improved to a linear bound-a question that remains open at present.
Consequently, we may assume that P and Q lie in the "inner pentago determined by the diagonals.We may also assume that Q is interior to the triangle PAB.But then QD crosses both the diagonal CE and either PA or PB, giving us a (1,2)-grid (see Figure12).