RamseyType Results for Geometric Graphs, I
Discrete Comput Geom
0 Courant Institute of Mathematical Sciences, New York University , New York, NY 10012 , USA
1 City College, CUNY , New York, NY 10031 , USA
2 Institute for Advanced Study , Princeton, NJ 08540 , USA
3 RamseyType Results for Geometric Graphs , I
For any 2coloring of the ¡n2¢ segments determined by n points in general position in the plane, at least one of the color classes contains a nonselfintersecting spanning tree. Under the same assumptions, we also prove that there exist b.n C 1/=3c pairwise disjoint segments of the same color, and this bound is tight. The above theorems were conjectured by Bialostocki and Dierker. Furthermore, improving an earlier result of Larman et al., we construct a family of m segments in the plane, which has no more than mlog 4= log 27 members that are either pairwise disjoint or pairwise crossing. Finally, we discuss some related problems and generalizations. ¤ G. Ka´rolyi was supported by NSF Grant DMS9304580, by the Alfred Sloan Foundation, and by DIMACS under NSF Grant STC9119999. J. Pach was supported by NSF Grant CCR9424398, PSCCUNY Research Award 663472, and OTKA4269. G. To´th was supported by OTKA4269 and OTKA14220.

A geometric graph is a graph drawn in the plane so that every vertex corresponds to a
point, and every edge is a closed straightline segment connecting two vertices but not
passing through a third. The ¡n2¢ segments determined by n points in the plane, no three
of which are collinear, form a complete geometric graph with n vertices (see [PA]). In
classical Ramsey theory, we want to find large monochromatic subgraphs in a complete
graph whose edges are colored with several colors [B3], [GRS]. Most questions of
this type can be generalized to complete geometric graphs, where the monochromatic
subgraphs are required to satisfy certain geometric conditions.
Our first two theorems settle two problems raised by Bialostocki and Dierker [BD2].
Theorem 1.1. If the edges of a finite complete geometric graph are colored by two
colors, there exists a nonselfintersecting spanning tree, all of whose edges are of the
same color.
Theorem 1.2. If the edges of a complete geometric graph with 3n ¡ 1 vertices are
colored by two colors, there exist n pairwise disjoint edges of the same color.
The analogues of Theorems 1.1 and 1.2 for abstract graphs, i.e., when the geometric
constraints are ignored, were noticed by Erdo˝s and Rado (see [EGP]) and Gerencse´r and
Gya´rfa´s [GG], respectively. In fact, Gerencse´r and Gya´rfa´s proved the stronger result
that for any 2coloring of the edges of a complete graph with 3n ¡ 1 vertices, there exists
a monochromatic path of length 2n ¡ 1. This statement, as well as Theorem 1.2, is best
possible, as is shown by the following example. Take the disjoint union of a complete
graph of n ¡ 1 vertices and a complete graph of 2n ¡ 1 vertices, all of whose edges
are red and blue, respectively, and color all edges between the two parts red. For many
interesting generalizations of these results, consult [EG], [EGP], [G2], [G1], [HKSS],
and [R].
Theorem 1.2 also has an “asymmetric” version.
Theorem 1.3. Let k and l be positive integers, n D k C l C maxfk; lg ¡ 1: If the edges
of a complete geometric graph with n vertices are colored by red and blue, one can find
either k disjoint red edges or l disjoint blue edges. This result cannot be improved.
Our Theorem 4.1 (see below) gives a nontrivial upper bound for the smallest positive
number R D R.n/ such that every complete geometric graph of R vertices whose edges
are colored by two colors contains a nonselfintersecting monochromatic path of length
n. We have been unable to determine the exact order of magnitude of R.n/.
Two segments are said to cross each other if they have an interior point in common. It
appears to be difficult to obtain any exact results analogous to Theorems 1.2 and 1.3 for
pairwise crossing edges. It follows from [AEGC] that if we color the edges of a complete
geometric graph of 12n2 vertices by two colors, one can always find n pairwise crossing
edges of the same color, but the assertion is probably true for much smaller graphs.
In [LMPT], the following question was discussed. What is the smallest positive
number r D r .n/ such that any family of r closed segments in general position in the plane has
n members that are either pairwise disjoint or pairwise crossing? This theorem improves
the lower bound r .n/ ¸ nlog 5= log 2 ¼ n2:322, obtained in [LMPT].
Theorem 1.4. For infinitely many n, there exists a family of nlog 27= log 4 ¼ n2:377
segments in general position in the plane, which has at most n members that are pairwise
disjoint and at most n members that are pairwise crossing.
2. Proofs of Theorems 1.1–1.3
Proof of Theorem 1.1. Let P D f p1; : : : ; png denote the vertex set of a complete
geometric graph Kn whose edges are colored with red and blue. Suppose without loss of
generality that no two vertices have the same x coordinate and that the vertices are listed
in increasing order of their x coordinates. The assertion is trivial for n · 2. Thus, we can
assume that n ¸ 3 and the theorem has already been proved for all complete geometric
graphs having fewer than n vertices.
We can also assume that all edges along the boundary of the convex hull of P are of the
same color (say, red). Indeed, if two consecutive edges of the convex hull have different
colors, then remove their common endpoint from Kn. By the induction hypothesis, the
remaining graph has a monochromatic nonselfintersecting spanning tree. Depending on
its color, this spanning tree can be completed to a monochromatic nonselfintersecting
spanning tree of Kn, by putting back one of the two previously deleted edges of the
convex hull of P.
For every i; 1 < i < n; let Kil and Kir denote the subgraphs of Kn induced by the
points f p1; : : : ; pi g and f pi ; : : : ; png, respectively. By the induction hypothesis, both Kil
and Kir have a monochromatic nonselfintersecting spanning tree, Til (resp. Tir ). We
can assume that these two trees have different colors, otherwise their union will meet the
requirements of the theorem. We can also assume that T2l is red and T2r is blue. Otherwise,
T2r would be red, and it could be completed to a nonselfintersecting red spanning tree
of Kn by the addition of any edge of the convex hull of P incident to p1. Similarly, we
can suppose that Tnl¡1 is blue and Tnr¡1 is red. Hence, there exists an i; 1 < i < n ¡ 1,
such that:
(a) Til is red and Tir is blue, and
(b) TilC1 is blue and TirC1 is red.
Connecting Til and TirC1 by any edge of the convex hull of P which intersects a vertical
line separating pi and piC1, we obtain a nonselfintersecting red spanning tree of Kn,
as required.
Proof of Theorem 1.2. Let P denote the vertex set of a complete geometric graph
K3n¡1, whose edges are colored with red and blue. Suppose for contradiction that K3n¡1
does not contain n pairwise disjoint edges of the same color. Since the theorem is trivial
for n D 1, we can assume that n ¸ 2 and that the statement has already been proved for
every complete geometric graph with 3k ¡ 1 vertices, where 1 · k < n.
An i element subset of P is called an i set if it can be obtained by intersecting P
with an open halfplane. It is easy to see that all i sets can be generated by the following
procedure [ELSS]: Take an oriented line ` passing through precisely one point p 2 P
and having i elements of P on its left side. Rotate ` around p in the clockwise direction
until it hits another point q 2 P, and then continue the rotation around q, etc. Whenever
` passes through only one element of P, the points lying on its left side form an i set.
By the induction hypothesis, the subgraph of K3n¡1 induced by any .3k ¡ 1/set
contains k disjoint edges of the same color (1 · k < n). If these edges are red (blue),
we say that the type of the corresponding .3k ¡ 1/set is red (blue). Note that a set may
have both types. Just like in the previous proof, we can suppose that all edges along the
boundary of the convex hull of P are of the same color (say, red). In other words, for
k D 1, the type of every .3k ¡ 1/set is supposed to be red.
Lemma 2.1.
(i) For any given k < n, all .3k ¡ 1/sets are of the same type.
(ii) For any k; l > 1 for which k C l D n, the .3k ¡ 1/sets and the .3l ¡ 1/sets
have opposite types.
(iii) For any k; l > 1 for which k Cl D n ¡1, either all .3k ¡1/sets or all .3l ¡1/sets
are of blue type.
To establish (i) and (ii), consider an oriented line ` passing through precisely one point
p 2 P and dividing P ¡ f pg into a .3k ¡ 1/set P¡.`/ and a .3l ¡ 1/set PC.`/, where
k C l D n. If P¡.`/ and PC.`/ had the same type, then P would contain k C l D n edges
of the same color, contradicting our assumption. Now rotate ` around p in the clockwise
direction until it hits another point q 2 P, and let `0 denote a line obtained by slightly
continuing the rotation around q. Notice that either P¡.`/ D P¡.`0/ or PC.`/ D PC.`0/:
Since P¡.`0/ and PC.`0/ have opposite types, we can conclude that P¡.`/ and P¡.`0/
are of the same type. Thus, (i) follows from the fact that any .3k ¡ 1/set can be reached
from P¡.`/ by repeating the above step a finite number of times.
To show (iii), fix a vertex p of the convex hull of P, and let p0 2 P be the next
vertex of the convex hull immediately after p in the clockwise order. Let p1 and p2
denote those elements of P, for which there are exactly 3k ¡ 1 points to the left of the
oriented line pp1 and exactly 3l ¡ 1 points to the right of pp2, where k C l D n ¡ 1.
That is, we have j P¡. pp1/j D 3k ¡ 1; j PC. pp2/j D 3l ¡ 1; and there is just one
point p3 2 P in the angular region p1 pp2; see Fig. 1. Assume now, for contradiction,
that all .3k ¡ 1/sets and all .3l ¡ 1/sets are of red type. In particular, the type of
P¡. pp1/ is red, which implies that the type of PC. pp2/ [ f p; p2; p3g must be blue;
otherwise, we could find k C .l C 1/ D n disjoint red edges. This, in turn, yields that
P¡. pp1/ [ f p1g ¡ f p0g is a .3k ¡ 1/element set that cannot contain k disjoint blue
edges. Thus, P¡. pp1/ [ f p1g ¡ f p0g has k disjoint red edges, PC. pp2/ has l disjoint red
edges, and by our assumption that every edge of the convex hull of P is red (including
pp0), we would obtain k C l C 1 D n pairwise disjoint red edges. This contradiction
proves the lemma.
Now we are in a position to complete the proof of Theorem 1.2. We distinguish
between two cases.
Case 1: n is even. Consider a line ` passing through precisely one point of P and dividing
the remaining points into two equal classes. Applying Lemma 2.1(i) with k D n=2, we
obtain that these classes are of the same type (say, red). Thus, there are n=2 pairwise
disjoint red edges on both sides of `, contradicting our assumption that there are no n
disjoint edges of the same color in K3n¡1.
Case 2: n is odd. Applying part (iii) of Lemma 2.1 with k D .n ¡ 1/=2, we obtain that
all [3.n ¡ 1/=2 ¡ 1]sets are of blue type. By part (ii), this implies that the type of all
[3.n C 1/=2 C 1]sets is red. Applying (iii) again, we find that all [3.n ¡ 3/=2 ¡ 1]sets
are of blue type. Proceeding like this, we conclude that the type of every 2set is blue. In
other words, every edge of the convex hull of P is blue, contradicting our assumption.
Proof of Theorem 1.3. Let R.k; l/ denote the smallest number R with the property that
in any complete geometric graph of R vertices, whose edges are colored with red and
blue, one can find either k disjoint red edges or l disjoint blue edges. It is enough to show
that R.k; l/ D 2k C l ¡ 1, for every l; 1 · l · k. This is trivial for l D 1, and, according
to Theorem 1.2, it also holds for l D k. To complete the proof, it is sufficient to verify
that R.k; l/ < R.k; l C 1/, for every l < k. Indeed, adding a new point p to any complete
geometric graph which contains neither k disjoint red edges nor l disjoint blue edges,
and connecting p to every other point by a blue edge, does not change the maximum
number of disjoint red edges, and the maximum number of disjoint blue edges can only
increase by one.
3. A Construction
The aim of this section is to prove Theorem 1.4 by a construction. A family of segments
is in general position if no three of their endpoints are collinear. Let S1 denote the family
of 27 segments depicted in Fig. 2. Clearly, S1 is in general position, and it can be checked
by an easy case analysis that S1 has no five pairwise crossing and no five pairwise disjoint
members.
Let S D fs1; : : : ; sng be a family of segments in general position in the plane. We say
that S can be flattened if for every " > 0 there are two disks of radius " at unit distance
from each other, and there is another family of segments S0 D fs10; : : : ; sn0g in general
position such that si0 and sj0 are disjoint if and only if si and sj are disjoint, and every si
connects two points belonging to different disks.
Lemma 3.1. Any system S of segments in general position, whose endpoints form the
vertex set of a convex polygon, can be flattened.
Proof. Let p1; : : : ; p2n denote the endpoints of the segments in counterclockwise order.
Notice that moving the endpoints to any convex curve does not effect the crossing
pattern of S, provided that the order of the endpoints remains unchanged. Let ppi0xD,
¡"=4i¡1; p"=2i¡1¢ ; 1 · i · 2n: Since all of these points are on the parabola y D
connecting the corresponding pairs by segments, we obtain a family S0, which has the
same crossing pattern as S.
It can be shown by easy calculation that if we have two disjoint segments pi0 p0j ; pk0 pl0 2
S0 for some i < k < l < j , then the slope of pi0 p0j is smaller than the slope of pk0 pl0:
Thus, extending all segments of S0 to the right until they hit the line x D 1 C "; does
not change the crossing pattern of the family. The lemma follows by applying an affine
transformation .x ; y/ ! .x ; ±y/ for some ± > 0; and moving the points into general
position.
Consider now the family S1 depicted in Fig. 2, and let " denote the minimum distance
between the endpoints. Replace every segment s 2 S1 by a suitably flattened copy of
S1, consisting of 27 segments whose endpoints are closer to the endpoints of s than "=2.
Replacing every member of the resulting family S2 by a (very) flattened copy of S1, we
obtain S3, etc. In this manner, for every k, we construct a family Sk of 27k segments in
general position, which has at most 4k pairwise disjoint members and at most 4k pairwise
crossing members. This completes the proof of Theorem 1.4.
4. Related Problems and Generalizations
Geometric Ramsey Numbers. Let G1; : : : ; Gk be not necessarily different classes of
geometric graphs. Let R .G1; : : : ; Gk / denote the smallest positive number R with the
property that any complete geometric graph of R vertices whose edges are colored with
k colors (1; : : : ; k, say) contains, for some i , an i colored subgraph belonging to Gi . If
G1 D ¢ ¢ ¢ D Gk D G, we write R.GI k/ instead of R .G1; : : : ; Gk /. If k D 2; for the sake
of simplicity, let R.G/ stand for R.GI 2/.
In Theorems 1.1 and 1.2, we determined R .G/ in the special case when G is the class
of all nonselfintersecting trees of n vertices and the class of all geometric graphs having
n disjoint edges, respectively. Theorem 1.3 gives the exact value of R .G1; G2/, when G1
and G2 denote the classes of all geometric graphs consisting of k disjoint edges and l
disjoint edges, respectively.
The proof of Theorem 1.2 can be easily generalized to give an upper bound for R .H/,
when H is, e.g., the class of all geometric graphs consisting of n pairwise disjoint triangles
[KPTV]. More generally, for any class of geometric graphs G and for any positive integer
n, let nG denote the class of all geometric graphs that can be obtained by taking the union
of n pairwise disjoint members of G, any two of which can be separated by a straight
line.
Theorem 4.1.
Let G be any class of geometric graphs, and let n be a power of 2. Then
R.nG/ · . R.G/ C 1/n ¡ 1:
In particular, if G is the class of triangles, we have R.G/ D 6. Moreover, in this case
Theorem 4.1 cannot be improved. We will return to these questions in a forthcoming
paper.
NonSelfIntersecting Paths. The length of a path is the number of its edges. Let Pn
denote the class of all nonselfintersecting paths of length n.
To give a nontrivial upper bound on R.Pk ; Pl /, we recall the following wellknown
(and very easy) lemma of Dilworth [D].
Lemma 4.2. Any partially ordered set of size kl C 1 either has a totally ordered subset
of size k C 1 or contains l C 1 pairwise incomparable elements.
Theorem 4.3. If the edges of a complete geometric graph of kl C 1 vertices are colored
by red and blue, one can find either a nonselfintersecting red path of length k or a
nonselfintersecting blue path of length l.
Proof. Let pi .0 · i · kl/ denote the vertices of a complete geometric graph. Suppose
that they are listed in increasing order of their x coordinates, which are all distinct.
Define a partial ordering of the vertices, as follows. Let pi < pj if i < j and there is
an x monotone red path connecting pi to pj . By Lemma 4.2, one can find either k C 1
elements that form a totally ordered subset Q ½ P , or l C 1 elements that are pairwise
incomparable. In the first case, there is an x monotone red path visiting every vertex of
Q. In the second case, there is an x monotone blue path of length l, because any two
incomparable elements are connected by a blue edge. Since an x monotone path cannot
intersect itself, the proof is complete.
Using the notation introduced above, Theorem 4.3 implies that R .Pn / D O .n2/, but
the best lower bound we are aware of is linear in n.
Constructive Vertex and Edge Ramsey Numbers. Given a class of geometric graphs
G, let Rv.G/ denote the smallest number R such that there exists a (complete)
geometric graph of R vertices with the property that for any 2coloring of its edges, it has
a monochromatic subgraph belonging to G. Similarly, let Re.G/ denote the minimum
number of edges of a geometric graph with this property. Rv.G/ and Re.G/ are called
the vertex and edge Ramsey number of G, respectively. Clearly, we have
Rv.G/ · R.G/;
µR.G/¶
Re.G/ · 2
:
(For abstract graphs, a similar notion is discussed in [EFRS] and [B1].)
It follows from the previous subsection that for Pn, the class of nonselfintersecting
paths of length n, Rv.Pn/ D O.n2/ and Re.Pn/ D O.n4/: It is not difficult to improve
the latter bound, as follows.
Proposition 4.4.
Re.Pn/ D O.n2/.
Proof. Construct a geometric graph G on the vertex set P D f.i; j /j0 · i; j · ng by
connecting every .i; j / to the points .i C 1; j /, .i; j C 1/, and .i C 1; j C 1/ (provided
that they belong to P). For any coloring of the edges of G with red and blue, color every
closed triangular face of G red (blue) if at least two of its sides are red (blue). Notice
that any two vertices of a red (blue) triangle can be joined in G by a red (blue) path of
length at most two. Thus, any two vertices belonging to the same connected component
of the union of the red (blue) triangles can be joined by a red (blue) path in G. The result
now follows from the fact that one can always find a pair of vertices lying on opposite
sides of the square f.x ; y/ 2 R2 j 0 · x ; y · ng, which belong to the same connected
component of the union of the red triangles or the union of the blue triangles. (See, e.g.,
page 85 in [B2] or the section about the game “Hex” in [BCG].)
Covering with NonSelfIntersecting Monochromatic Paths. Is it true that for every
k there exists an integer C .k/ such that the vertex set of every complete geometric
graph whose edges are colored by k colors can be covered by C .k/ nonselfintersecting
monochromatic paths? We cannot even decide the following weaker question for k D 2:
Does there exist a positive " such that every complete geometric graph G whose edges
are colored by k colors contains a nonselfintersecting monochromatic path of length
"jV .G/j?
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