- Email: [email protected]
Ramsey-type constructions for arrangements of segments
Electronic Notes in Discrete Mathematics 31 (2008) 265–269 www.elsevier.com/locate/endm
Ramsey-type constructions for arrangements of segments Jan Kynˇcl 1 Department of Applied Mathematics and Institute for Theoretical Computer Science (ITI) Charles University, Malostransk´e n´ am. 25 118 00 Prague, Czech Republic
Abstract Improving a result of K´ arolyi, Pach and T´ oth, we construct an arrangement of n segments in the plane without nlog 8/ log 169 pairwise crossing or pairwise disjoint segments. We use the recursive method based on flattenable arrangements which was established by Larman, Matouˇsek, Pach and T¨or˝ ocsik. We also discuss some related problems.
1
Introduction
An arrangement of segments is a finite set of compact straight-line segments in the plane in general position (i.e., no three end-points are collinear). We study the following Ramsey-type problem [3]: what is the largest number r(k) such that there exists an arrangement of r(k) segments with no k pairwise crossing or k pairwise disjoint segments? 1
Email: [email protected]. The author was partially supported by the Phenomena in High Dimensions project, in the framework of the European Community’s “Structuring the European Research Area” program. 1571-0653/$ – see front matter © 2008 Elsevier B.V. All rights reserved. doi:10.1016/j.endm.2008.06.054
266
J. Kynˇcl / Electronic Notes in Discrete Mathematics 31 (2008) 265–269
Larman et al. [3] proved that k 5 ≥ r(k) ≥ k log 5/ log 2 > k 2.3219 . The upper bound has remained unchanged since then. K´arolyi et al. [1] improved the lower bound to r(k) ≥ k log 27/ log 4 > k 2.3774 . We improve the construction for the lower bound even further and prove the following theorem. Theorem 1.1 For infinitely many positive integers k there exists an arrangement of k log 169/ log 8 > k 2.4669 segments with at most k pairwise crossing or k pairwise disjoint segments.
2
Construction
Both previous constructions for the lower bound [1,3] use the same approach. The starting configuration is an arrangement M0 of n0 segments with at most k0 pairwise crossing or pairwise disjoint segments. In the i-th step, an arrangement Mi of ni+1 segments is constructed from the arrangement Mi−1 by 0 replacing each of its segments by a flattened copy (a precise definition will follow) of M0 , which acts as a “thick segment”. Then two segments from different copies of M0 cross if and only if the two corresponding segments in Mi−1 cross. Our new arrangement Mi has then at most k0i+1 pairwise crossing or pairwise disjoint segments. This gives a lower bound r(k) ≥ k log n0 / log k0 for infinitely many values of k. We improve the construction by making a better starting arrangement. Unlike previous constructions, our basic pieces will be arrangements with different maximal numbers of pairwise crossing and pairwise disjoint segments. By putting them together, we obtain our starting arrangement M0 . A (k, l)-arrangement is an arrangement of segments with at most k pairwise crossing and at most l pairwise disjoint segments. An intersection graph G(M) of an arrangement M is a graph whose vertices are the segments of M and two vertices are joined by an edge if and only if the corresponding segments intersect. An arrangement M of segments is flattenable if for every ε > 0 there is an arrangement Mε with G(Mε ) = G(M) and two discs D1 , D2 of radius ε whose centers are at unit distance, such that each segment from Mε has one endpoint in D1 and the second endpoint in D2 . A flattened copy of M is the arrangement Mε with sufficiently small ε. The key result is the following lemma. Lemma 2.1
(i) There exists a flattenable (2, 4)-arrangement of 13 segments.
(ii) There exists a flattenable (4, 2)-arrangement of 13 segments.
267
J. Kynˇcl / Electronic Notes in Discrete Mathematics 31 (2008) 265–269
1
5
13
6 2
3
12
7 13
11
7 11
12 4
1 10
4 2
8 9
10
9
3
5 6 8
Fig. 1. A (4, 2)-arrangement of 13 segments (left) and a (2, 4)-arrangement of 13 segments (right).
Note that 13 is the largest possible number of segments for these two types of arrangements since every graph on more than 13 vertices contains either a clique of size 5 or an independent set of size 3. On the other hand, Cay(Z13 ; 1, 5), the Cayley graph of the cyclic group Z13 corresponding to the generators 1 and 5, has no clique of size 3 and no independent set of size 5. Proof. Both previous constructions used arrangements of segments with endpoints in convex position, that are flattenable by a relatively simple argument [1]. This is not the case of our construction. Hence, the flattening of our arrangements requires more complicated approach. For each sufficiently small ε > 0, we construct an arrangement Ma (ε) with the intersection graph Cay(Z13 ; 1, 5) and an arrangement Mb (ε) whose intersection graph is the complement of Cay(Z13 ; 1, 5). See Figure 1 for a non-flattened illustration. In the tables 1 and 2, we provide precise coordinates of the end-points of all the 13 segments, as a function of ε. Since the coordinates of all the left end-points converge to (0, 0) and the coordinates of all the right end-points converge to (1, 0), it remains to verify that for sufficiently small ε > 0, each of these two described arrangements has the desired intersection graph. This is a straightforward calculation, which can be done by a simple algorithm. 2 Proof of Theorem 1.1. Take a sufficiently flattened arrangement Ma (ε) and replace each of its segments by a copy of sufficiently flattened arrangement Mb (δ). Such a way we obtain our starting arrangement M0 of 169 segments
268
J. Kynˇcl / Electronic Notes in Discrete Mathematics 31 (2008) 265–269
left x
left y
right x
right y
1
−ε
0
1 − 2ε
2ε2 + 2ε6
2
ε2
ε − ε3
1 − ε2
ε3
3
0
ε4 + ε6
1
ε3 + 3ε4
4
0
ε4 − ε6
1 − 2ε
2ε2 − ε6
5
−ε + ε2
0
1 − 2ε2
2ε3 − 2ε4
6
−ε
2ε6
1−ε
2ε6
7
0
ε6
1
ε3 + 2ε4
8
0
ε
1 + ε3
0
9
0
ε
1 − 2ε2
2ε3 − ε4
10 −ε2 + 3ε3
3ε6
1 − 2ε
2ε2 + ε6
11 −ε2
ε6
1 − 2ε2
2ε3 − 3ε4
12 0
ε4
1
0
13 −ε
0
1+ε
0
Table 1 Arrangement Ma (ε).
with at most 8 pairwise crossing or pairwise disjoint segments. Then we proceed by the method described above. 2 In the full version of this paper, we show that not every arrangement can be flattened. A necessary condition for an arrangement to be flattenable is the existence of a line crossing all the segments, in a sufficiently flattened realization. We construct an intersection graph of segments which cannot be realized by an arrangement of segments crossing a common line. Moreover, we also construct an intersection graph of segments crossing a common line which cannot be realized by a flattenable arrangement. Finally, we construct an arrangement of c · k 2 log(k) segments with endpoints in a convex position, with no k pairwise crossing and no k pairwise disjoint segments. This bound is tight due to the result of Kostochka [2].
269
J. Kynˇcl / Electronic Notes in Discrete Mathematics 31 (2008) 265–269
left x
left y
right x
right y
1
ε
ε2 − ε3 + ε4 − 2ε5
1 + ε2
−ε4 + ε6
2
0
ε2 + 3ε5
1 − ε3
ε7
3
0
ε2 + 4ε5
1+ε
−ε3
4
0
2ε3
1 + 3ε4
−ε8
5
ε − ε 2 + ε3
ε2 − ε3 + ε4 − ε8
1+ε
−ε4
6
0
ε2 + ε5
1+ε
−ε3
7
0
ε2 + 5ε5
1 + 3ε4
−3ε7
8
ε − ε2 + ε3 + ε4 + 2ε5
ε2 − ε3 + ε4 + ε5 + ε6
1 + ε − ε4
−ε3
9
0
ε2
1+ε
−ε4
10
0
0
1 + 5ε3
0
11
0
ε2 + 2ε5
1 + 3ε4 − 2ε5
ε8
12
ε − ε3
ε3 − ε4
1+ε
−ε4
13
0
0
1
ε
Table 2 Arrangement Mb (ε).
Acknowledgement ˇ y for helpful and inspiring The author is grateful to G´eza T´oth and Jakub Cern´ discussions on this problem.
References [1] K´ arolyi, G., J. Pach and G. T´ oth, Ramsey-type results for geometric graphs I, Discrete Comput. Geom. 18 (1997), pp. 247–255, ACM Symposium on Computational Geometry (Philadelphia, PA, 1996). [2] Kostochka, A., Upper bounds on the chromatic number of graphs, Trudy Inst. Mat. (Novosibirsk) 10 (1988), pp. 204–226. [3] Larman, D., J. Matouˇsek, J. Pach and J. T¨ or˝ ocsik, A Ramsey-type result for convex sets, Bull. London Math. Soc. 26 (1994), pp. 132–136.
Ramsey-type constructions for arrangements of segments Jan Kynˇcl 1 Department of Applied Mathematics and Institute for Theoretical Computer Science (ITI) Charles University, Malostransk´e n´ am. 25 118 00 Prague, Czech Republic
Abstract Improving a result of K´ arolyi, Pach and T´ oth, we construct an arrangement of n segments in the plane without nlog 8/ log 169 pairwise crossing or pairwise disjoint segments. We use the recursive method based on flattenable arrangements which was established by Larman, Matouˇsek, Pach and T¨or˝ ocsik. We also discuss some related problems.
1
Introduction
An arrangement of segments is a finite set of compact straight-line segments in the plane in general position (i.e., no three end-points are collinear). We study the following Ramsey-type problem [3]: what is the largest number r(k) such that there exists an arrangement of r(k) segments with no k pairwise crossing or k pairwise disjoint segments? 1
Email: [email protected]. The author was partially supported by the Phenomena in High Dimensions project, in the framework of the European Community’s “Structuring the European Research Area” program. 1571-0653/$ – see front matter © 2008 Elsevier B.V. All rights reserved. doi:10.1016/j.endm.2008.06.054
266
J. Kynˇcl / Electronic Notes in Discrete Mathematics 31 (2008) 265–269
Larman et al. [3] proved that k 5 ≥ r(k) ≥ k log 5/ log 2 > k 2.3219 . The upper bound has remained unchanged since then. K´arolyi et al. [1] improved the lower bound to r(k) ≥ k log 27/ log 4 > k 2.3774 . We improve the construction for the lower bound even further and prove the following theorem. Theorem 1.1 For infinitely many positive integers k there exists an arrangement of k log 169/ log 8 > k 2.4669 segments with at most k pairwise crossing or k pairwise disjoint segments.
2
Construction
Both previous constructions for the lower bound [1,3] use the same approach. The starting configuration is an arrangement M0 of n0 segments with at most k0 pairwise crossing or pairwise disjoint segments. In the i-th step, an arrangement Mi of ni+1 segments is constructed from the arrangement Mi−1 by 0 replacing each of its segments by a flattened copy (a precise definition will follow) of M0 , which acts as a “thick segment”. Then two segments from different copies of M0 cross if and only if the two corresponding segments in Mi−1 cross. Our new arrangement Mi has then at most k0i+1 pairwise crossing or pairwise disjoint segments. This gives a lower bound r(k) ≥ k log n0 / log k0 for infinitely many values of k. We improve the construction by making a better starting arrangement. Unlike previous constructions, our basic pieces will be arrangements with different maximal numbers of pairwise crossing and pairwise disjoint segments. By putting them together, we obtain our starting arrangement M0 . A (k, l)-arrangement is an arrangement of segments with at most k pairwise crossing and at most l pairwise disjoint segments. An intersection graph G(M) of an arrangement M is a graph whose vertices are the segments of M and two vertices are joined by an edge if and only if the corresponding segments intersect. An arrangement M of segments is flattenable if for every ε > 0 there is an arrangement Mε with G(Mε ) = G(M) and two discs D1 , D2 of radius ε whose centers are at unit distance, such that each segment from Mε has one endpoint in D1 and the second endpoint in D2 . A flattened copy of M is the arrangement Mε with sufficiently small ε. The key result is the following lemma. Lemma 2.1
(i) There exists a flattenable (2, 4)-arrangement of 13 segments.
(ii) There exists a flattenable (4, 2)-arrangement of 13 segments.
267
J. Kynˇcl / Electronic Notes in Discrete Mathematics 31 (2008) 265–269
1
5
13
6 2
3
12
7 13
11
7 11
12 4
1 10
4 2
8 9
10
9
3
5 6 8
Fig. 1. A (4, 2)-arrangement of 13 segments (left) and a (2, 4)-arrangement of 13 segments (right).
Note that 13 is the largest possible number of segments for these two types of arrangements since every graph on more than 13 vertices contains either a clique of size 5 or an independent set of size 3. On the other hand, Cay(Z13 ; 1, 5), the Cayley graph of the cyclic group Z13 corresponding to the generators 1 and 5, has no clique of size 3 and no independent set of size 5. Proof. Both previous constructions used arrangements of segments with endpoints in convex position, that are flattenable by a relatively simple argument [1]. This is not the case of our construction. Hence, the flattening of our arrangements requires more complicated approach. For each sufficiently small ε > 0, we construct an arrangement Ma (ε) with the intersection graph Cay(Z13 ; 1, 5) and an arrangement Mb (ε) whose intersection graph is the complement of Cay(Z13 ; 1, 5). See Figure 1 for a non-flattened illustration. In the tables 1 and 2, we provide precise coordinates of the end-points of all the 13 segments, as a function of ε. Since the coordinates of all the left end-points converge to (0, 0) and the coordinates of all the right end-points converge to (1, 0), it remains to verify that for sufficiently small ε > 0, each of these two described arrangements has the desired intersection graph. This is a straightforward calculation, which can be done by a simple algorithm. 2 Proof of Theorem 1.1. Take a sufficiently flattened arrangement Ma (ε) and replace each of its segments by a copy of sufficiently flattened arrangement Mb (δ). Such a way we obtain our starting arrangement M0 of 169 segments
268
J. Kynˇcl / Electronic Notes in Discrete Mathematics 31 (2008) 265–269
left x
left y
right x
right y
1
−ε
0
1 − 2ε
2ε2 + 2ε6
2
ε2
ε − ε3
1 − ε2
ε3
3
0
ε4 + ε6
1
ε3 + 3ε4
4
0
ε4 − ε6
1 − 2ε
2ε2 − ε6
5
−ε + ε2
0
1 − 2ε2
2ε3 − 2ε4
6
−ε
2ε6
1−ε
2ε6
7
0
ε6
1
ε3 + 2ε4
8
0
ε
1 + ε3
0
9
0
ε
1 − 2ε2
2ε3 − ε4
10 −ε2 + 3ε3
3ε6
1 − 2ε
2ε2 + ε6
11 −ε2
ε6
1 − 2ε2
2ε3 − 3ε4
12 0
ε4
1
0
13 −ε
0
1+ε
0
Table 1 Arrangement Ma (ε).
with at most 8 pairwise crossing or pairwise disjoint segments. Then we proceed by the method described above. 2 In the full version of this paper, we show that not every arrangement can be flattened. A necessary condition for an arrangement to be flattenable is the existence of a line crossing all the segments, in a sufficiently flattened realization. We construct an intersection graph of segments which cannot be realized by an arrangement of segments crossing a common line. Moreover, we also construct an intersection graph of segments crossing a common line which cannot be realized by a flattenable arrangement. Finally, we construct an arrangement of c · k 2 log(k) segments with endpoints in a convex position, with no k pairwise crossing and no k pairwise disjoint segments. This bound is tight due to the result of Kostochka [2].
269
J. Kynˇcl / Electronic Notes in Discrete Mathematics 31 (2008) 265–269
left x
left y
right x
right y
1
ε
ε2 − ε3 + ε4 − 2ε5
1 + ε2
−ε4 + ε6
2
0
ε2 + 3ε5
1 − ε3
ε7
3
0
ε2 + 4ε5
1+ε
−ε3
4
0
2ε3
1 + 3ε4
−ε8
5
ε − ε 2 + ε3
ε2 − ε3 + ε4 − ε8
1+ε
−ε4
6
0
ε2 + ε5
1+ε
−ε3
7
0
ε2 + 5ε5
1 + 3ε4
−3ε7
8
ε − ε2 + ε3 + ε4 + 2ε5
ε2 − ε3 + ε4 + ε5 + ε6
1 + ε − ε4
−ε3
9
0
ε2
1+ε
−ε4
10
0
0
1 + 5ε3
0
11
0
ε2 + 2ε5
1 + 3ε4 − 2ε5
ε8
12
ε − ε3
ε3 − ε4
1+ε
−ε4
13
0
0
1
ε
Table 2 Arrangement Mb (ε).
Acknowledgement ˇ y for helpful and inspiring The author is grateful to G´eza T´oth and Jakub Cern´ discussions on this problem.
References [1] K´ arolyi, G., J. Pach and G. T´ oth, Ramsey-type results for geometric graphs I, Discrete Comput. Geom. 18 (1997), pp. 247–255, ACM Symposium on Computational Geometry (Philadelphia, PA, 1996). [2] Kostochka, A., Upper bounds on the chromatic number of graphs, Trudy Inst. Mat. (Novosibirsk) 10 (1988), pp. 204–226. [3] Larman, D., J. Matouˇsek, J. Pach and J. T¨ or˝ ocsik, A Ramsey-type result for convex sets, Bull. London Math. Soc. 26 (1994), pp. 132–136.