An Improved Lower Bound for. an Erdös-Szekeres-Type Problem. with Interior Points

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1 Applied Mathematical Sciences, Vol. 6, 2012, no. 70, An Improved Lower Bound for an Erdös-Szekeres-Type Problem with Interior Points Banyat Sroysang Department of Mathematics and Statistics, Faculty of Science and Technology, Thammasat University, Pathumthani Thailand Centre of Excellence in Mathematics, CHE, Si Ayutthaya RD.,Bangkok 10400, Thailand Abstract For each finite planar point set P with no three collinear points, an interior point of P is a point in P such that it is not on the boundary of the convex hull of P. For any integer k > 0, let g(k) be the smallest integer such that every finite planar point set P with no three collinear points and with at least g(k) interior points has a subset Q whose the interior of the convex hull of Q contains exactly k points of P. In this paper, we show that g(k) k 2 for all integer k 4. Mathematics Subject Classification: 52C10 Keywords: Interior point; finite planar set; convex hull; deficient point set 1 Introduction In 1935, Erdös and Szekeres [2] prove that, for any integer k 3, there exists a smallest positive integer f (k) such that any finite planar point set of at least f (k) points with no three collinear points has a subset of k points whose the convex hull contains exactly k vertices. They [3] presented the best known lower bound

2 3454 B. Sroysang on f (k) for all integer k 3 that f (k) 2 k Moreover, they conjecture for all integer k 3 that f (k) = 2 k In 2001, Avis, Hosono and Urabe [1] posed an similar problem, called an Erdös-Szekeres-type problem: for any positive integer k, determine the smallest positive integer g(k) such that any finite planar point set P of at least g(k) points with no three collinear points has a subset Q whose the interior of the convex hull of Q contains exactly k points in P. And then they showed that g(1) = 1, g(2) = 4, g(3) 8, and g(k) k + 2 for all k 4. In 2003, Fevens [4] presented the lower bound for all integer k 3 that g(k) 3k 1. In 2008, Wei and Ding [7] presented the better lower bound for all integer k 3 that g(k) 3k. In 2009, Wei and Ding [6] proved that g(3) = 9. In 2011, Sroysang [5] presented the new lower bound for all integer k 4 that g(k) 4k. We note that g(1) = 1 2, g(2) = 2 2 and g(3) = 3 2. And then we conjecture for all positive integer k that g (k) = k 2. In this paper, we present the best lower bound on g(k) for all integer k 4 that g(k) k 2. 2 Preliminaries An interior point of any finite planar point set P with no three collinear points is a point in P such that it is not on the boundary of the convex hull of P. For any finite planar point set P with no three collinear points, and for any subset Q of P, we denote the notations as follows: intch(p) is the interior of the convex hull of P, I(P) is the set of all interior points of P, i(p) is the number of elements of I(P), I * (Q) is the set I(P) intch(q), i * (Q) is the number of elements of I * (Q), and V(P) is the set P \ I(P). For any positive integer k, let g(k) be the smallest integer such that every finite planar point set P with no three collinear points and with at least g(k) interior points has a subset Q whose the interior of the convex hull of Q contains exactly k points of P. Thus, for each positive integer k, g(k) = min{ s : i(p) s Q P s.t. i * (Q) = k }. We recall two propositions about lower bounds as follows. Proposition 2.1 [7] For any integer k 3, g(k) 3k. Proposition 2.2 [5] For any integer k 4, g(k) 4k. For any finite planar point set P with no three collinear points, we say that P is

3 Erdös-Szekeres-type problem 3455 a deficient point set of type P(m, s, k) if v(p) = m, i(p) = s, and i * (Q) k for all subset Q of P. To show that g(k) k 2 for all integer k 4, by following Lemma, it suffices to prove the existence of a deficient point set of type P(3, k 2 1, k). Lemma 2.3 (Extension Lemma) [1] Every deficient point set P(m, s, k) can be extended to a deficient point set P(m + 1, s, k). For any three points x, y, z, we denote the notations as follows: C(x; y, z) is the cone formed by the two rays xy and xz, H(xy; z) is the half plane bounded by the line xy and containing z, H * (xy; z) is the half plane bounded by the line xy and not containing z, and Δxyz is the triangle with vertices x, y, z. 3 Main Results In [1, 6], we perceive that g(1) = 1 2, g(2) = 2 2 and g(3) = 3 2. Thus, we may conjecture that g (k) = k 2 for all positive integer k. In this section, we improve the lower bound on the number g(k) in Proposition 2.3. This new lower bound may be possible that it is the best lower bound on the number g(k). Theorem 3.1 For any integer k 4, g(k) k 2. Proof. Let k be an integer such that k 4. To show that g(k) k 2, for each integer m 3, we will seek a deficient point set of type P(m, k 2 1, k). By the Lemma 2.3, it suffices to construct a deficient point set P of type P(3, k 2 1, k). Let V(P) = { v 1, v 2, v 3 } be such that v 1, v 2, v 3 put into anticlockwise positions, respectively. Then v(p) = 3. We suppose that i(p) = k 2 1. Take I(P) = { a 1, a 2,, a k (k 1)} { d 1, d 2,, d k 1 }. Let a 0 = v 1 and a k (k 1) + 1 = v 3. Take P 0 = P \{v 2 } and V(P 0 ) = { a 0, a 1, a 2,, a k (k 1), a k (k 1) + 1 } where a 0, a 1, a 2,, a k (k 1), a k (k 1) + 1 put into anticlockwise positions, respectively. Let D = { d 1, d 2,, d k 1 }. Then I(P 0 ) = D and i * (P 0 ) = k 1. Next, we will locate d 1, d 2,, d k 1 as follows. For each i {1, 2,..., k 1}, d i C(v 2 ; a i (k 1), a i (k 1) + 1), d i H(a (i 1) (k 1) a i (k 1) + 2 ; v 2 ) H(a (i 1) (k 1) + 1 a i (k 1) + 3 ; v 2 ) H(a (i 1) (k 1) + k 2 a i (k 1) + k ; v 2 ), and d i H * (a (i 1) (k 1) a i (k 1) + 1 ; v 2 ) H * (a (i 1) (k 1) + 1 a i (k 1) + 2 ; v 2 ) H * (a (i 1) (k 1) + k 1 a i (k 1) + k ; v 2 ). For examples, the set P with k = 4 as shown in Fig. 1 and the set P with k = 5 as shown in Fig. 2.

4 3456 B. Sroysang The construction of P is valid. a 13 a 10 d 3 a 9 a 7 d 2 v 2 a 6 a 4 d 1 a 3 a 0 Fig. 1. A deficient point set P(3, 15, 4)

5 Erdös-Szekeres-type problem 3457 a 21 a 17 d 4 a 16 a 13 d 3 a 12 v 2 d 2 a 8 a 9 d 1 a 5 a 4 a 0 Fig. 2. A deficient point set P(3, 24, 5) Finally, we will show that i * (Q) k for all Q P. Let Q P. Now, we divide into three cases: (1) v 2 Q, (2) v 2 Q and D intch(q) = φ, and (3) v 2 Q and d i intch(q) for some i {1, 2,..., k 1}.

6 3458 B. Sroysang Case 1: v 2 Q. Then Q P \{v 2 } = P 0. Since i * (P 0 ) = k 1, it follows that i * (Q) k 1 < k. Case 2: v 2 Q and D intch(q) = φ. Then CH(Q) CH(Δv 2 a j a j+ k ) for some j {0, 1, 2,..., k 2 2k + 1}. Now, we note that I * (Q) I * (Δv 2 a j a j + k ) = { a j + 1, a j + 2,, a j + k 1 }. This implies that i * (Q) i * (Δv 2 a j a j+ k ) = k 1 < k. Case 3: v 2 Q and d i intch(q) for some i {1, 2,..., k 1}. Then CH(Δv 2 a j + (i 1)(k 1) a j + (i 1)(k 1) + k + 1 ) CH(Q) for some j = 0, 1,, n 2. Let T = Δv 2 a j + (i 1)(k 1) a j + (i 1)(k 1) + k + 1. Now, we note that I * (T ) I * (Q) and I * (T ) = { a j + (i 1)(k 1) + 1, a j + (i 1)(k 1) + 2,, a j + (i 1)(k 1) + k } { d i }. This implies that i * (Q) i * (T ) = k + 1 > k. Thus, we conclude by above three cases that i * (Q) k for all Q P. Hence, P is a deficient point set of type P(3, k 2 1, k). This proof is completed. 4 Conjecture We believe that the new lower bound in this paper is the best lower bound on the number g(k). And then we pose the following conjecture. Conjecture 4.1 For any integer k 4, g(k) = k 2. Acknowledgement. This research is supported by the Centre of Excellence in Mathematics, the Commission on Higher Education, Thailand. References [1] D. Avis, K. Hosono and M. Urabe, On the existence of a point subset with a specified number of interior points, Discrete Math., 241 (2001), [2] P. Erdös and G. Szekeres, A combinatorial problem in Geometry, Compos. Math., 2 (1935), [3] P. Erdös and G. Szekeres, On some extremum problems in elementary geometry, Ann. Univ. Sci. Budapest, 3 4 ( ), [4] T. Fevens, A note on point subset with a specified number of interior points, Discrete and Computational Geometry, Lecture Notes in Comput. Sci., 2866 (2003),

7 Erdös-Szekeres-type problem 3459 [5] B.Sroysang, A lower bound for Erdös-Szekeres-type problem with interior points, Int. J. Open Probl. Comput. Sci. Math., 4 (December 2011), [6] X. Wei and R. Ding, More on an Erdös-Szekeres-type problem for interior points, Discrete Comput. Geom., 42 (2009), [7] X. Wei and R. Ding, More on planar point subsets with a specified number of interior points, Math. Notes, 83 (2008), Received: March, 2012