On the Existence of a Point Subset with 3 or 7 Interior Points

An interior point of a planar point set P is a point of P such that it is not on the boundary of the convex hull of P. For each integer k ≥ 3, let h(k) be the smallest integer such that every finite planar point set P with no three collinear points and with at least h(k) interior points has a subset Q whose the interior of the convex hull of Q contains exactly k or k + 4 interior points of P. In this paper, we show that h(3) = 7. Mathematics Subject Classification: 52C10


Introduction
In this paper, we focus on finite planar point sets in general position; that is, no three points are collinear. In 1935, Erdős and Szekeres [1] posed a problem: for any integer ≥ 3, determine the smallest positive integer ( ) such that any finite point set of at least ( ) points has a subset of points whose convex hull contains exactly vertices. In 1961, they [2] showed that ( ) ≥ 2 −2 + 1 for all integer ≥ 3 and then conjectured that ( ) ≥ 2 −2 + 1 for all integer ≥ 3.
In 2001, Avis et al. [6] proved that 3 is the smallest positive integer such that any finite point set of at least 3 interior points has a subset for which the interior of the convex hull of the set contains exactly 3 or 4 points in the set . Moreover, they [11] also proved that 7 is the smallest positive integer such that any finite point set of at least 7 interior points has a subset for which the interior of the convex hull of the set contains exactly 4 or 5 points in the set . In 2009, Wei and Ding [12] showed that any planar point set with 3 vertices and 9 interior points has a subset with 5 or 6 interior points of the set .
In 2010, Wei et al. [13] proved that 8 is the smallest positive integer such that any finite point set of at least 8 interior points has a subset for which the interior of the convex hull of the set contains exactly 3 or 5 points in the set .
In 2012, Sroysang [14] proved that 7 is the smallest positive integer such that any finite point set of at least 7 interior points has a subset for which the interior of the convex hull of the set contains exactly 3 or 7 points in the set .
In this paper, we pose an interior point problem: for any integer ≥ 3, determine the smallest positive integer ℎ( ) such that any finite point set of at least ℎ( ) points has a subset for which the interior of the convex hull of the set contains exactly or + 3 points in the set . We show that ℎ(3) = 8; that is, 8 is the smallest positive integer such that any finite point set of at least 8 interior points has a subset for which the interior of the convex hull of the set contains exactly 3 or 6 points in the set .

Preliminaries
In this section, we list propositions and notations about the set , where is a finite planar point set such that no three points are collinear.
An interior point of the set is a point of the set such that it is not on the boundary of the convex hull of the set .
We denote notations as follows:   An edge of the set is an edge in ( ). A subset of the set is called a -int subset if * ( ) = .
Proposition 1 (see [8]). 9 is the smallest integer such that any finite point set of at least 9 interior points has a subset for which the interior of the convex hull of the set contains exactly 3 points in the set .
For any positive integer ≥ 3, we let ℎ( ) be the smallest integer such that every planar point set with no three collinear points and with at least ℎ( ) interior points has a subset for which the interior of the convex hull of the set contains exactly or + 3 points of the set .

Main Results
In this section, we will show that 8 is the smallest positive integer such that any finite point set of at least 8 interior points has a subset for which the interior of the convex hull of the set contains exactly 3 or 6 points in the set .   Proof. Let be a finite planar point set such that ( ) ≥ 9.
By Proposition 1, there is a subset of the set such that * ( ) = 3. Then * ( ) ∈ {3, 6}. Hence, ℎ(3) ≤ 9. Proof. This suffices to show the existence of a deficient point set of type (3,7,3,6). We construct a deficient point set of type (3, 7, 3, 6) as shown in Figure 1. Hence, ℎ(3) ≥ 8. Proof. Suppose that each subset of a planar point set is not a 3-int subset. In [8], we have only three different configurations of the type (3, 8, 3, 3) as shown in Figures 2, 3, and 4. However, each configuration has a subset for which the interior of the convex hull of the set contains exactly 6 points of the set . Hence, the set has a 3-int or 6-int subset. Proof. Let ( ) = { , , , and } be such that vertices , , , are put into counterclockwise positions, respectively (see Figure 5).  Suppose that each subset of the planar point set is not a 6-int subset. Then the sets Δ , Δ , Δ , and Δ are not 6-int subsets.
If Δ is a 2-int subset, then the set Δ is a 6-int subset. Then the set Δ is not a 2-int subset. Similarly, the sets Δ , Δ , and Δ are not 2-int subsets. Let = {Δ , Δ , Δ , Δ }. To show that the set has a 3-int subset, we divide into seven cases.

Case 1.
There is an element in the set such that the set is a 3-int subset.
In this case, the set has a 3-int subset.

Case 2.
There is an element in the set such that the set is a 5-int subset. Without loss of generality, we assume that = Δ . Then the set Δ is a 3-int subset. Thus, the set has a 3-int subset.

Case 3.
There is an element in the set such that the set is a 7-int subset.
Without loss of generality, we assume that = Δ . Then the set Δ is a 1-int subset. If the set has a 3-int subset, then the set has a 3-int subset. Assume that the set is a deficient point set of type (3,7,3,3). By Proposition 2, there is a subset of the set (Δ ) with * ( ) = 2 such that the edge is an edge of the set . Let = ∪ Δ . Then * ( ) = 3. Thus, the set has a 3-int subset.

Case 4.
There is an element in the set such that the set is a 1-int subset.
Without loss of generality, we assume that = Δ . Then the set Δ is a 7-int subset. Similar to Case 3, the set has a 3-int subset.

Case 5.
There is an element in the set such that the set is an 8-int subset.
By Lemma 5, the set has a 3-int subset.

Case 6.
There is an element in the set such that the set is a 0-int subset. Without loss of generality, we assume that = Δ . Then the set Δ is an 8-int subset. By Lemma 5, the set has a 3-int subset.
If one of them has a 3-int subset, then the set has a 3-int subset. Assume that they are deficient point sets without a 3int subset. If the edge of the set Δ is of type 2, then we obtain that * ( \ { }) = 6. It follows that, the edge of the set Δ is of type 0 or type 1. If the edge of the set Δ is of type 0, then the edge of the set Δ is of type 3, so the set has a 3-int subset. Next, we will assume that the edge of the set Δ is of type 1. Similarly, it suffices to assume that the edge of the set Δ is only of type 1, the edge of the set Δ is only of type 1, and the edge of the set Δ is only of type 1. Hence, we obtain only one possible configuration as shown in Figure 6.
However, there is a subset of such that * ( ) = 3, as shown in Figure 7. Thus, the set has a 3-int subset. Therefore, the set has a 3-int or 6-int subset. This proof is completed.  Proof. Let V( ) = and ( ) = {V 1 , V 2 , . . . , V } be such that vertices V 1 , V 2 , . . . , V are put into counterclockwise positions, respectively (see in Figure 8). Suppose that each subset of the set is not a 6-int subset. Then the set ΔV 1 V V +1 is not a 6-int subset for all .
To show that the set has a 3-int subset, we divide into six cases.

Case 1.
There is an element in the set such that the set is a 3-int subset.
In this case, the set has a 3-int subset.

Case 2.
There is an element in the set such that the set is a 7-int subset. Then the set ΔV 1 V V +1 is a 1-int subset for some . Without loss of generality, we assume = ΔV 1 V V +1 for some > . If the set has a 3-int subset, then the set has a 3-int subset. Assume that the set is a deficient point set of type (3,7,3,3). By Proposition 2, there is a subset of the set ( ) with * ( ) = 2 such that the edge V 1 V is an edge of the set . Let = ∪ ΔV 1 V V +1 . Then * ( ) = 3. Thus, the set has a 3-int subset. Case 3. There is an element in the set such that the set is a 5-int subset.
We divide into three subcases.

Subcase 3.1.
There is an element in the set such that the set is a 3-int subset.
In this subcase, the set has a 3-int subset.

Subcase 3.2.
There exist elements , , and in the set such that the sets , , and are 1-int subsets. It follows that Without loss of generality, we assume that | − | = min{| − |, | − |, | − |}. Then the set ∪ is a 6-int subset. This is impossible.

Subcase 3.3.
There exist elements , in the set such that the set is a 1-int subset and the set is a 2-int subset.

Case 4.
There is an element in the set such that the set is a 4-int subset.
We divide into five subcases.
There is an element in the set such that the set is a 3-int subset. In this subcase, the set has a 3-int subset.

Subcase 4.2.
There exist elements , , , and in the set such that the sets , , , and are 1-int subsets. It follows that , ∈ {2, 3, . . . , }. Without loss of generality, we can assume that < < < . If < , then the set ∪ ∪ is a 3-int subset. If < , then the set ∪ ∪ is a 3-int subset. Thus, the set has a 3-int subset if < or < . Next, we will show that the statement " < < " is impossible. We suppose that < < . Then the set ∪ ∪ is a 6-int subset which is a contradiction. It follows that = ΔV 1 V V +1 , = ΔV 1 V V +1 and = ΔV 1 V V +1 where , , ∈ {2, 3, . . . , }. Without loss of generality, we assume | − | < | − |. Then the set ∪ is a 6-int subset. This is impossible. Subcase 4.4. There exist elements , , and in the set such that the sets and are 1-int subsets and the set is a 2-int subset.
It follows that Without loss of generality, we can assume that < . Let = max{ , } and = min{ , }. If < , then the set ∪ ∪ is a 6-int subset. If < , then the set ∪ is a 6-int subset. Thus, we obtain that < < . Then the set ∪ ΔV 1 V V +1 is a 3-int subset. Hence, the set has a 3-int subset.

Subcase 4.5.
There is an element in the set \ { } such that the set is a 4-int subset.
Without loss of generality, we assume that Then the sets and are not 6-int subsets. If the set is an 8-int subset or the set is an 8-int subset then, by Lemma 6, the set has a 3-int subset. If the set is a 7-int subset, then the set ΔV 1 V V is a 3-int subset, so the set has a 3-int subset. If the set is a 7-int subset, then the set ΔV 1 V +1 V +1 is a 3-int subset, so the set has a 3-int subset. If the set is a 5-int subset, then the set ΔV V +1 V is a 3-int subset, so the set has a 3-int subset. If the set is a 5-int subset, then the set ΔV +1 V V +1 is a 3-int subset, so the set has a 3-int subset. Next, we assume that the sets and are 4-int subsets. Then ΔV 1 V V +1 is a 0-int subset. Then the set {V , V +1 , V , V +1 } is an 8-int subset. By Lemma 6, the set has a 3-int subset.

Case 5.
There is an element in the set such that the set is an 8-int subset.
By Lemma 5, the set has a 3-int subset.
Case 6. We have * ( ) ≤ 2 for all ∈ . If there exist elements , , , and in the set such that the sets , , , and are 2-int subsets where the sets , , , and put into anticlockwise positions, then the set ∪ ∪ is a 6-int subset. Thus, we obtain that there is an element in the set such that it is a 1-int subset. It is easy to see that has a 3-int subset. Therefore, the set has a 3-int or 6-int subset. This proof is completed. Theorem 8. One has ℎ(3) = 8.

Conclusion and Discussion
In [6], 3 is the smallest positive integer such that any finite point set of at least 3 interior points has a subset for which the interior of the convex hull of the set contains exactly 3 or 4 points in the set .
In [13], 7 is the smallest positive integer such that any finite point set of at least 7 interior points has a subset for which the interior of the convex hull of the set contains exactly 3 or 5 points in the set .
In this paper, 8 is the smallest positive integer such that any finite point set of at least 8 interior points has a subset for which the interior of the convex hull of the set contains exactly 3 or 6 points in the set .
In [14], 7 is the smallest positive integer such that any finite point set of at least 7 interior points has a subset for which the interior of the convex hull of the set contains exactly 3 or 7 points in the set .
In [15], 8 is the smallest positive integer such that any finite point set of at least 8 interior points has a subset for which the interior of the convex hull of the set contains exactly 3 or 8 points in the set . Moreover, 9 is the smallest positive integer such that any finite point set of at least 9 interior points has a subset for which the interior of the convex hull of the set contains exactly 3 or points in the set , where ≥ 9.
For any positive integer ≥ 4, we let ℎ * ( ) be the smallest integer such that every planar point set with no three collinear points and with at least ℎ * ( ) interior points has a subset for which the interior of the convex hull of the set contains exactly 3 or points of the set . (3)

Conflict of Interests
The author declares that there is no conflict of interests regarding the publication of this paper.