Studia Scientiarum Mathematicarum Hungarica 41 (2), (2004)

Transcription

1 Studia Scientiarum Mathematicarum Hungarica 4 (), (004) PLANAR POINT SETS WITH A SMALL NUMBER OF EMPTY CONVEX POLYGONS I. BÁRÁNY and P. VALTR Communicated by G. Fejes Tóth Abstract A subset A of a finite set P of points in the plane is called an empty polygon, if each point of A is a vertex of the convex hull of A and the convex hull of A contains no other points of P. We construct a set of n points in general position in the plane with only.6n empty triangles,.94n empty quadrilaterals,.0n empty pentagons, and 0.n empty hexagons.. Introduction Results. We say that a set P of points in the plane is in general position if it contains no three points on a line. Let P be a finite set of points in general position in the plane. We call a subset A of k points in P an empty k-gon if the convex hull of A is a k-gon containing no point of P \ A. Let g k (n) be the minimum number of empty k-gons in a set of n points in general position in the plane. Horton [0] proved that g k (n) = 0 for any k 7 and any n N. The following bounds on g k (n), k = 3, 4, 5, 6, have been known: n O(n log n) g 3 (n) n = n, n O(n) g 4 (n) n =.3... n, 000 Mathematics Subject Classification. Primary 5C0. Key words and phrases. Planar point configuration, empty convex polygon, empty hexagon. Research by I. Bárány was partially supported by Hungarian Science Foundation Grants T 0345 and T Research by P. Valtr was supported by project LN00A056 of The Ministry of Education of the Czech Republic /$ 0 c 004 Akadémiai Kiadó, Budapest

2 44 I. BÁRÁNY and P. VALTR n 4 6 g 5 (n) n =.8... n, g 6 (n) n = n. The upper bounds have been shown in [5], improving previous bounds of [, 3, 3]. The lower bound on g 3 (n) has been shown in [3], the lower bound on g 4 (n) by Bárány (see [3]) and by Dumitrescu [5], and the lower bound on g 5 (n) in [4]. In this paper we give the following improved upper bounds: Theorem. ( g 3 (n) α 6 ) 3 β p n + o(n ) = n + o(n ), ( g 4 (n) α 6β + 6 ) 3 γ p n + o(n ) = n + o(n ), ( g 5 (n) ) 3 α 6β γ p n + o(n ) = n + o(n ), ( 93 g 6 (n) α 6 3 β + 6 ) 3 γ p n + o(n ) = n + o(n ), where p = 3 π = , α = β = γ = z=3 odd z=3 odd z=3 odd z = π 8 = , z log = , z z 4 log = z Our construction seems to be the final one of the type developed in [3, 5], and is, perhaps, the best possible up to the additive o(n )-factor. Several exciting questions remain open. The most interesting is whether g 6 (n) > 0

3 PLANAR POINT SETS 45 for sufficiently large n (e.g. see [7]). In other words, is it true that if P is a finite set of points in general position in the plane with P large enough, then P contains an empty hexagon. Another question is whether g 3 (n) ( + ε)n holds for large enough n for some fixed ε > 0. This would be the case if one could show that g 5 (n) > εn for some fixed ε > 0. These questions have turned out to be more difficult than expected: for instance the innocent looking g 6 (n) > 0 has been a challenge for more than 30 years now.. The construction Our construction giving the upper bounds in Theorem is a set obtained from the grid n n by a little perturbation (due to monotonicity of g k (n) it suffices to prove Theorem when n is a square of an integer). Throughout the rest of the paper, n is a square of an integer, and Λ is the grid {,,..., n} {,,..., n}. The perturbed set will be denoted by Λ. The construction of Λ uses so-called Horton sets [] which generalize a construction of Horton [0] giving g 7 (n) = 0 for any n. Horton sets. Let H be a finite set of points in general position in the plane such that no two points have the same x-coordinate, and let h 0, h,..., h m be the points of H listed by increasing x-coordinate. We say that a subset H H lies far below a subset H H (and H lies far above H ), if the entire set H lies above every line trough a pair of points of H and the entire set H lies below every line trough a pair of points of H. For 0 i < j, we define a subset H i,j of H as the set of points h k with k i (mod j). The set H is called a Horton set if, for every j =, 4, 8, 6,... and every integer i with 0 i < j/, the set H i,j lies far below or far above H i+j/,j. It was shown in [] that if H is Horton, then also each H i,j, 0 i < j, is Horton. Obviously, if H is Horton then also each contiguous segment of H (i.e., a set of points h k of H with k 0 k k ) is Horton. Construction of Λ. Set m := n and ε := /(0m). We construct an auxiliary random Horton set H = H(ε) of size m + as follows. We choose randomly and independently for each i, j, 0 i < j/, j = l m, the mutual position of the sets H i,j, H i+j/,j (whose union is the set H i,j/ ): the set H i,j will lie with probability / far above H i+j/,j and with probability / far below H i+j/,j. For a given choice of mutual positions, we define H as the set of points h k = (k, ε log m l= ±(m + ) l ), k = 0,..., m, where the choice of + or at (m + ) l corresponds to the choice of mutual position of those sets H i, l, H i+ l, l whose union H i,l contains h k (we take +(m + ) l in the sum if h k lies in that of the sets H i, l,

4 46 I. BÁRÁNY and P. VALTR H i+ l, l which lies far above the other of these sets; otherwise we take (m + ) l ). The x-coordinates of the points of H = H(ε) are 0,,..., m and the y-coordinates lie in the interval ( ε, ε). For ε > 0, we consider another, analogously defined random Horton set H = H (ε ) of size m +. Further, we consider the set H = H (ε ) obtained from H = H (ε ) by the interchange of the axes, i.e. H = T (H ), where T : (x, y) (y, x). We define Λ as the Minkowski sum of the sets H = H(ε) and H = H (ε ), where ε = ε (m) > 0 is sufficiently small compared to ε = /(0m) (e.g., ε = / ( 0m(m + ) +log m) will do). The set Λ approximates Λ. For a point X in Λ, we denote by X the corresponding point of Λ. We usually use letters I, J, K, L, R, S, T to denote points in Λ. We denote their coordinates by I = ( i, y(i) ), J = ( j, y(j) ), etc. (We use such a notation since we mostly work with the first coordinate). It follows from the choice of ε, ε that for any three points I, J, K Λ the following holds: Observation. (i) If I, J, K Λ are not collinear, then the triples I, J, K and I, J, K have the same orientation. (ii) If I, J, K Λ lie on a non-vertical common line, then the orientation of the triple I, J, K is equal to the orientation of the triple h i, h j, h k of points of H. (iii) If I, J, K Λ lie on a vertical common line, then the orientation of the triple I, J, K is determined by the orientation of the corresponding triple of points of H. It follows from Observation that the points of Λ corresponding to the intersection of a non-vertical line with Λ form a set having the same order type as a contiguous part of some set H i,j, 0 i < j. Consequently, such points form a Horton set (see Claim 3.0 in []). Observation 3. The points of Λ corresponding to the intersection of a non-vertical line with Λ form a random Horton set G. That is, randomly and independently for each i, j, 0 i < j/, j = l m, the set G i,j lies with probability / far above G i+j/,j and with probability / far below G i+j/,j. Notation. The lattice is the usual lattice of points in the plane with integer coordinates. A lattice point is a point of the lattice. We say that a line is a lattice line, if it contains infinitely many lattice points. For a non-vertical lattice line l, we denote by l + (resp. l ) the closest lattice line above (below) l and parallel to l. A lattice segment is a segment connecting two lattice points. We say that a lattice segment is s-prime, if it contains s + lattice points (including its endpoints). If a lattice segment is -prime (i.e., its relative interior contains no lattice points), then we call it a prime segment. Otherwise we call it a non-prime segment. If I I... I k is an empty k-gon in Λ, then we also say that I I... I k is an empty k-gon (it may be degenerate).

5 PLANAR POINT SETS 47 For an empty k-gon P = L L... L k with all vertices in Λ, we define the base of P as the segment L v L w connecting the vertex L v having the smallest x-coordinate with the vertex L w having the largest x-coordinate. If P has more vertices with the smallest x-coordinate, then we choose for L v that one with the smallest y-coordinate. Similarly, if P has more vertices with the largest x-coordinate, then we choose for L w that one with the largest y-coordinate. If the base of an empty polygon P is prime, non-prime, or s-prime, then we say that P is prime, non-prime, or s-prime, respectively. We say that a (possibly degenerate) polygon P with all vertices in Λ is a t-line polygon, if t is the least number such that the vertices of P lie on t neighboring parallel lattice lines. 3. Structure of the proof We note first that Λ contains no empty 7-gon. This was proved in [3]: the reason is that Λ is built from Horton sets. For each k = 3, 4, 5, 6, we distinguish five types of empty k-gons and estimate the expected number of empty k-gons for each of them separately. Here are the five types of empty k-gons: 3-line k-gons, -line prime k-gons, -line non-prime k-gons, -line s -prime k-gons (s N), -line r-prime k-gons (r s ). It follows from Observation 4 below that every empty polygon in Λ is -, -, or 3-line. Thus, the above five types embrace all empty polygons in Λ. Observation 4 ([]). If the convex hull of a subset S of Λ has no lattice point in the interior, then S lies either on one line, or on two parallel lines with no lattice point strictly between them, or on the perimeter of a lattice triangle with exactly one lattice point in the relative interior of each side. Next, let P be a finite point set, of n points, say, in the plane in general position. Consider the complex, C, of empty convex polygons in P. C is clearly a simplicial complex. Let f k (P ) be its f-vector (k =,,... ), that is, f k (P ) is the number of empty convex k-gons in P. Clearly f (P ) = n, and f (P ) = ( n ). It is proved by Edelman and Rainer [6] that C is contractible. Then it satisfies the Euler equation: f (P ) f (P ) + f 3 (P ) f 4 (P ) =.

6 48 I. BÁRÁNY and P. VALTR There is another linear relation satisfied by the f-vector: it is shown by Ahrens et al. [] that f (P ) f (P ) + 3f 3 (P ) 4f 4 (P ) = P int conv P. These two linear relations are very useful in our construction since there f (Λ ) = n, f (Λ ) = ( n ) and fk (Λ ) = 0 when k > 6. So out of the remaining four quantities f i (Λ ), i = 3, 4, 5, 6, only have to be determined. Our choice is to compute f 3 and f 6, which means that out of the 0 entries of the following table, we only compute 0. Empty: [ (3/π )n ] triangles quadrilaterals pentagons hexagons 3-line /4 /8 /8 /4 -line prime 0/3 9/7 6/7 0/ -line non-prime /3 54/49 4/49 8/47 -line s -prime 4/9 9/49 4/49 4/44 other -line α 3 β 8α 6β+ 3 γ 6 3 α 6β+ 3 3 γ 4 6 3α 3 β+ 6 3 γ Each entry in the table must be multiplied by (3/π )n to obtain a -approximation of the correspoding quantity. E.g., the entry 0/3 in the second row and first column means that Λ contains (0/3) (3/π )n + o(n ) -line prime triangles. It is easily verified that the 0 entries in the first and last column and the above equations on the f-vector give Theorem. In fact we have computed all entries of the above table. The method is to fix the base, IJ of the k-gon in question, then compute the expectation of the empty k-gons with base IJ, and then sum for all possible bases. This is fairly straightforward although lengthy in all five cases except the -line prime k-gons where we need a more detailed analysis. 4. Auxiliary statements In this section we collect several simple facts (and prove some of them) that will be needed later. Most of them are quite easy. We say that a segment I J, i j, in Λ is open up if K lies below the line I J for any lattice point K in the relative interior of IJ. Similarly, we say that a segment I J, i j, is open down if K lies above the line I J for any lattice point K in the relative interior of IJ. If I J is open up or down, then we also say that the segment IJ is open up or down, respectively. Clearly, each prime segment is open up and down, and each -prime segment is open either up or down. Here is a more general lemma:

7 PLANAR POINT SETS 49 Lemma 5. Let IJ be an s-prime segment in Λ. Then: (i) If s is a power of, then the segment IJ is open up (down, respectively) with probability /s. (ii) If s is not a power of, then the segment IJ is open neither up nor down. Observation 6. If I J K is an empty triangle in Λ, i j, and K lies strictly above the line IJ, then IJ is open up. Analogously, if I J K is an empty triangle in Λ, i j, and K lies strictly below the line IJ, then IJ is open down. Let f(n), g(n) be two real functions defined for any n = m, m N. We write f(n) g(n) (and say that f(n) equals g(n)), if f(m ) lim m g(m ) =. We denote the set of prime segments in the n n grid Λ by P, and its size by p n = P. It is well-known (see for instance [9]) that p n 6 ( ) n π 3 π n. Lemma 7. (i) For any r, the number of r-prime segments in Λ is pn. r (ii) For any r and n, the number of r-prime segments in Λ is at most 8n. r Proof. We first suppose that { n is divisible by r. Consider the mapping f : Λ,..., n } { } ( r,..., n r defined by f(i) = i ) r, y(i) r for I Λ. Each r-prime segment is mapped to a lattice segment of the same direction and /r of its original length. Thus, each r-prime segment { is mapped } { to a prime } segment. Moreover, each prime segment KL in,...,,..., is the image of exactly r r-prime segments in Λ, n r n r namely it is the image of the r-prime segments ( r K + (α, β), r L + (α, β) ), where α, β {0,,..., r }. It follows that if n is divisible by r then Λ determines r p n/r r- prime segments. This yields (i): for any n = m, Λ determines at least r p n/r p n and at most r p r n/r p n r-prime segments. r If r n then Λ determines no r-prime segments and (ii) clearly holds. Otherwise Λ determines at most r p n/r r ( n/r ) r (4n/r ) =

8 50 I. BÁRÁNY and P. VALTR 8n r r-prime segments, as required in (ii). Lemma 8. Let H, H be two Horton sets, let H lie far below H, and let H = {h 0,..., h z }, H = {h 0,..., h z }. Further, let P H H be the vertex set of an empty polygon in H H, and let P H and P H. Then P H 3 and P H 3. Moreover, if P H =3 then P H = {h i, h i+j, h j }, where j i is a power of. Analogously, if P H = 3 then, h l }, where l k is a power of. P H = {h k, h k+l Let H be a Horton set with vertices denoted as usual. Then we say that a segment h i h j, j > i, in H is open down, if all points h k, i < k < j, lie above it. Similarly, we say that h i h j is open up, if all points h k, i < k < j, lie below it. Lemma 9. (i) Any Horton set of size s determines s+ (s + ) open down segments. (ii) If H = {h 0,..., h s } is a Horton set of size s, where the points are listed according to the increasing x-coordinate, then H determines s (s + ) open down segments h i h j with j > i +. (iii) In (i) and (ii), open down can be replaced by open up. Proof. We proceed by induction on s. The lemma clearly holds for s = 0,. Suppose now that H = {h 0, h,..., h s } is a Horton set of size s, s. Let H be the lower of the sets H 0,, H,. By the inductive assumption, H determines s (s + ) open down segments. The set H determines the following two types of open down segments: (T) s segments h i h i+, (T) s (s + ) open down segments determined by H. Thus, H determines ( s ) + ( s (s + ) ) = s+ (s + ) open down segments. This gives (i). The open down segments h i h j with j > i + are just the segments of type (T). This gives (ii). (iii) follows from the symmetry. Observation 0. For each s N, let f s (n), g s (n) be two functions satisfying f s (n) g s (n). Moreover, suppose that for each ε > 0 there is a t N such that, for any n N, s=t+ f s (n) εn and s=t+ g s (n) εn. Then f s (n) g s (n) + o(n ). s= s=

9 PLANAR POINT SETS line triangles and hexagons 5.. The parity of the coordinates of lattice prime segments Here we estimate the number of -prime segments IJ, I, J Λ, such that j i is even. The standard method from [9] showing that p n 3 n gives π easily the following. Lemma. The number of prime segments IJ with j i even is Lemma. (i) Λ determines pn p n 3. -prime segments IJ with j i even. (ii) Λ determines p n -prime segments IJ with j i 0 even. Proof. Certainly, it suffices to prove the lemma for { n even. } { } Consider the mapping f : Λ,..., n,..., n defined by ( f(i) = i ), y(i), as in the proof of Lemma 7 (for r = ). Each -prime segment { IJ in } Λ is { mapped to } a prime segment, and each prime segment KL in,...,,..., is the image of exactly 4 -prime segments in n n Λ. Moreover, for a -prime segment IJ in Λ, j i is even if and only if l k ( = j i ) is even, where k, l are the x-coordinates of the points K = f(i), L = f(j), respectively. Thus, by Lemma, Λ determines 4 pn/4 3 p n -prime segments IJ with j i even. This gives (i). Since there are only O(n) -prime segments IJ with j i = 0, (ii) follows from (i) and from p n = Θ(n ) line triangles Let IJK be an IJ-triangle with all three sides -prime and no lattice point in the interior. We now find the probability that IJK is empty. Set R = I+J, S = I+K, T = J+K (see Fig. ). If j i is odd, then (exactly) one of the numbers k i, j k is also odd. Without loss of generality, let k i be odd. Then i j k r s (mod ). Consequently, R and S lie either both below or both above the segments

10 5 I. BÁRÁNY and P. VALTR Fig. The points R, S, T. I J and I K, respectively. Thus, (exactly) one of the points R, S lies inside the triangle I J K. We conclude that IJK is not an empty triangle in this case. Suppose now that j i is even. If both numbers k i, j k are even, then in the triangle IRS the y-components of a side are of the same parity, and then the midpoint of this side is a lattice point. Consequently, one of the sides of the original triangle IJK is not -prime. Thus j i is even and both k i, j k are odd. Consequently, i j k r s t (mod ). With probability /, both points S, T lie inside the triangle I J K. Independently and also with probability /, the point R lies inside the triangle I J K. Thus, if j i is even then I J K is empty with probability /4. For a -prime segment IJ P with j i > 0 even, there are exactly two lattice points K, i k j, such that IJK is a 3-line triangle. One such placement of K is on the line (IJ + ) + (in which case the points S, T are the two points on IJ + satisfying i s < t < j), the other placement of K is on the line (IJ ) (in which case the points S, T are the two points on IJ satisfying i < s < t j), see Fig.. It now follows from Lemma (ii) that Fig. Two possible placements of K. the expected number of empty 3-line triangles is 4 pn = pn 4.

11 PLANAR POINT SETS line hexagons Each empty 3-line triangle IJK corresponds to the empty 3-line hexagon I I+J J J+K K K+I, and vice versa. Thus, the number of empty 3-line hexagons equals the number of empty 3-line triangles. 6. -line prime triangles and hexagons 6.. Some lattice properties Given a non-vertical prime segment IJ P, there is a unique K (IJ) + with i k < j. We let q + (IJ) denote this lattice point K. Assume J I = (m, t) with 0 < t < m and let K I = (x, y). Then ym + x( t) = as one can readily check. Thus x is the inverse of t (mod m). We will use a theorem of Balog and Deshoulliers [] saying that x is uniformly distributed in [0, m). Theorem 3 ([]). Assume m is a positive integer. Then for any α (0, ], and any η > 0, the number of pairs (t, x) with t {,..., m} and x {,..., αm } where xt (mod m) is αϕ(m) + O(m /+η ) where the implied constant depends at most on η. Actually, the original result of Balog and Deshoulliers is more general and is stated in a slightly different form. For r N, we define a subset P r of P as the set of non-vertical prime segments IJ P such that the x-coordinate of q + (IJ) lies in the interval [i, i + j i r ). Lemma 4. (i) For any r, (ii) For any r, n, P r P r, P r 0 r n. Proof. (i) is a direct corollary of Theorem 3. To prove (ii), suppose that I Λ and that t {,,..., n }. The number of lattice points K Λ, k > i, with t K I < t + is approximately πt certainly smaller than 0t (say). Now, let K be one of

12 54 I. BÁRÁNY and P. VALTR these points. If IK is non-prime, then there is no lattice point J with K = q + (IJ). Otherwise the lattice points J with K = q + (IJ) lie on the lattice half-line IK { (x, y) R : x k } (see Fig. 3). The half-line Fig. 3 The lattice points J with K = q + (IJ). IK { (x, y) R : x k } contains at most J Λ. n K I n t lattice points It follows that for each I and t {,,..., n } there are at n t most 0t = 00n lattice points J with t q + (IJ) I < t +. If IJ P r then q + (IJ) I < k n. It follows that for each I there are at most r j n r 0 00n r n t= lattice points J Λ with IJ P r. Consequently, P r 0 r n. We denote the lattice points on the open halfline Iq + (IJ) by K = q + (IJ), K, K 3,..., so that K t = I + t ( q + (IJ) I ) for each t N. See Fig. 4. Similarly, we denote the lattice points on the open halfline J(J ( q + (IJ) I ) ) by L, L, L 3,..., so that L t = J t ( q + (IJ) I ) for each t N. We remark that P r is the set of segments IJ P such that i k r < j, where k r is the x-coordinate of the point K r = I + r( q + (IJ) I ). For IJ P and s 0, we define two events E + s = E + s (IJ) and E s = E s (IJ) as follows: E + s = E + s (IJ): the segment IK s is open down,

13 PLANAR POINT SETS 55 Fig. 4 The lattice points K i, L i. E s = E s (IJ): the segment JL s is open up. Clearly, K s lies in some empty -line IJ-polygons if and only if IJ P s and E + s is satisfied. Similarly, L s lies in some empty -line IJ-polygons if and only if IJ P s and E s is satisfied. The following observation follows from Lemma 5(i): Observation 5. For any s 0 and IJ P s, Prob (E + s ) = s, Prob (E s ) = s. and E s are almost inde- The following lemma shows that the events E + s pendent if IJ is taken uniformly from P r. Lemma 6. Let r N and 0 s, s r. Then Prob (E + s E s ) p n r. s+s IJ P r Proof. If s = 0 then, by Observation 5 and by Lemma 4(i), IJ P r Prob (E + s E s ) = Prob (E s ) p n r, s IJ P r as required. Analogously, the lemma holds also for s = 0. We further suppose that s, s. Let IJ P r and let K = K = q + (IJ). Not all three numbers i + j, i + k, j + k are even, since in that case one of the points I+J, I+K, J+K

14 56 I. BÁRÁNY and P. VALTR (corresponding to an even of the numbers y(i) + y(j), y(i) + y(k), y(j) + y(k)) would be a lattice point. Consequently, by a parity argument, exactly one of the numbers i + j, i + k, j + k is even. By Lemma, there are P 3 segments IJ P with i + j even. Consequently, by Theorem 3, there are P r 3 segments IJ P r with i + j even. By Lemma, there are P 3 segments IJ P with i + j odd, which is the same as j i odd. Thus, by symmetry, there are P 3 segments IJ P with k i even and also P 3 segments IJ P with k i odd. We want to use now Theorem 3 with J I = (m, t) and x [0, m r ), with the extra condition that x = k i is even (resp. odd). When x is even and lies in [0, m r ) then x/ is an integer in [0, m r ) for which (t)(x/) (mod m) and t runs through the reduced residue classes (mod m). The number of such pairs (t, x/) is then r ϕ(m) + O(m /+η ), which implies that there are Pr 3 segments IJ P r with j i odd and k i even. Then the complementary set of segments with j i odd and k i odd is also of size P r 3. Let IJ P r. If i + k is even (and i + j, j + k are odd), then the x- coordinates of I, K, K,... have the other parity than the x-coordinates of J, L, L,.... Consequently, the events E + s and E s are independent and Prob (E + s E s ) = Prob (E + s ) Prob (E s ) = s+s in this case. If i + j is even, then the x-coordinates of I, K, K 4,..., J, L, L 4,... have the other parity than the x-coordinates of K, K 3,..., L, L 3,.... Consequently, either K lies below the line I K s or L lies above the line J L s. Thus, Prob (E + s E s ) = 0 in this case (provided s, s ). If j + k is even, then the x-coordinates of K, K 3,..., J, L, L 4,... have the other parity than the x-coordinates of I, K, K 4,..., L, L 3,.... The following two conditions are necessary for E + s E s : C : {I, K, K 4,... } lies far below {K, K 3,... }, C : {L, L 3,... } lies far below {J, L, L 4,... }. Clearly, C is satisfied if and only if C is satisfied. Thus, Prob (C C ) = Prob (C ) = Prob (C ) =. Suppose that C C is satisfied. Then E + s is satisfied if and only if I K s is open down in the (random) Horton set {I, K, K 4,..., K s}, i.e., with

15 PLANAR POINT SETS 57 probability. Analogously, E s s is satisfied if and only if J L is open s up in the (random) Horton set {J, L, L 4,..., L }, i.e., with probability s. Moreover, s E+ s and E s are independent (provided j + k is even and C C is satisfied), since the x-coordinates of I, K, K 4,..., K s have the other parity than the x-coordinates of J, L, L 4,..., L s. Thus, if j + k is even then Prob (E + s E s ) = Prob (C C ) Prob (E + s C C ) Prob (E s C C ) = = Altogether, s s s+s. IJ P r Prob (E + s E s ) P r 3 s+s + P r P r 3 s+s = p n r s+s line prime triangles is The expected number of empty -line IJ-triangles with IJ P r \ P r+ = = IJ P r\p r+ IJ P r \P r+ ( r Prob (E + s ) + s=0 ( r s=0 r s + (4 r ) P r \ P r+, s s =0 r s =0 ) ) Prob (E s ) and thus the expected number of empty -line IJ-triangles with IJ P = (P r \ P r+ ) is r= () r= (4 r ) P r \ P r+.

16 58 I. BÁRÁNY and P. VALTR By Lemma 4(i), () P r \ P r+ p n r+. For every ε > 0, there is a t N such that, by Lemma 4(ii), the sum of the terms in () with r t + can be bounded from above by r=t+ 4 0 r n = 80 t n < εn. Thus, by Observation 0 and by (), (3) () = r= ( 4 r ) pn r+ ( ) p n 3 = 5 3 p n. Similarly, define P r as the set of non-vertical prime segments IJ P such that the x-coordinate of q + (IJ) lies in the interval [j j i, j). An analogue of the above proof shows that the expected number of empty r -line IJ-triangles with IJ P is also (4) 5 3 p n. Consequently, the expected number of empty -line prime triangles is the sum of (3) and (4), that is, 0 3 p n.

17 PLANAR POINT SETS line prime hexagons We first estimate the number of empty -line IJ-hexagons with IJ P r \ P r+. Each of them is of form JK t K t/ IL t L t /. Thus, their expected number is IJ P r \P r+ r s= r r Prob (E + s E s ) = s = s= r s = IJ P r \P r+ Prob (E + s E s ). It follows that the expected number of empty -line IJ-hexagons with IJ P = (P r \ P r+ ) is r= (5) r= r s= r s = IJ P r \P r+ Prob (E + s E s ). Lemma 6 and the inclusion P r+ P r imply that (6) r s= r s = IJ P r\p r+ Prob (E + s E s ) r s= r s = ( pn r p ) ( n r+ = ) p n s+s r r+. For every ε > 0, there is a t N such that, by Lemma 4(ii), the terms in (5) with r t + can be bounded from above by r=t+ r s= r s = IJ P r\p r+ r=t+ r P r \ P r+ r=t+ r 0 r n < εn. (7) Thus, by Observation 0 and by (6), (5) = r= r= ( r ) p n r+ ( r+ r + ) 3r+ p n

18 60 I. BÁRÁNY and P. VALTR = ( 3 + ) p n 4 = 5 p n. A similar argument as at the end of Paragraph 6. shows that the expected number of empty -line prime hexagons is twice as much as (7), i.e., 0 p n. 7. -line non-prime triangles and hexagons 7.. -line non-prime triangles If there is an empty -line non-prime IJ-triangle, then IJ must be open up or down and thus IJ must be s -prime for some s N. Let s N and let IJ P be a non-vertical s -prime segment with j > i. The line IJ + contains s points K Λ with i k < j (unless j = i +, y(i) = n). Each of these points determines an empty -line IJ-triangle IJK if and only if IJ is open up, i.e., with probability. Thus, the expected number of empty -line IJ-triangles with one vertex on the line IJ + s is equal to s =. By symmetry, the expected number of empty -line IJtriangles with one vertex on the line IJ is also (unless j = i +, y(i) = ). s Thus, the expected number of empty -line non-prime triangles is (8) s= (IJ is s -prime). The appears in (8) since the expected number of empty -line IJtriangles is 0 for vertical s -prime segments IJ P and is smaller than for s -prime segments IJ P, j = i +, with y(i) = n or y(j) =. It follows from Lemma 7(ii) that the first sum in (8) satisfies the assumptions of Observation 0, and thus (8) can be estimated by s= p n 4 s = 3 p n.

19 PLANAR POINT SETS line non-prime hexagons If there is an empty -line non-prime IJ-hexagon, then IJ must be open up or down and thus IJ must be s -prime for some s N. Let s N and let IJ be a non-vertical s -prime segment with j > i. The line IJ + contains s points K with i k < j (unless j = i +, y(i) = n), forming a (random) Horton set, which we denote by H. By Lemma 9(ii), H determines s (s + ) open down segments KK with K+K H. Each of these segments determines an empty -line IJ-hexagon I I+J JK K +K K if and only if IJ is open up, i.e., with probability. Thus, the expected number of empty -line IJ-hexagons with all vertices s on the lines IJ and IJ + is equal to s (s+) = s+ s (unless j = i +, y(i) = n). Altogether, the expected number s of two-line non-prime hexagons is (9) s= (IJ is s -prime) ( s + ) s. The appears in (9) for analogous reasons as in (8). It follows from Lemma 7(ii) that the first sum in (9) satisfies the assumptions of Observation 0, and thus (9) can be estimated by s= ( p n 4 s s + ) s ( = ) p n 49 = 8 47 p n. 8. -line s -prime triangles and hexagons For k N and s 0, we define V k (s) as the expected number of those empty k-gons in a random Horton set H of size s +, which contain both the leftmost point and the rightmost point of H. Lemma 7. For any s 0, V 3 (s) = s, V 6 (s) = s 4 + s + s.

20 6 I. BÁRÁNY and P. VALTR Proof. Let h 0, h,..., h s be the points of a Horton set H listed according to the increasing x-coordinate. For i = 0,..., s, we define a i -element subset H(i) of H by H(i) = H s i, s i = { h j H : j s i (mod s i ) }. Observe that H \ {h 0, h s} is a disjoint union of the sets H(i), i =,..., s and that each H(i) = H s i,s i lies far above or far below the set H 0, s = {h 0, h s} H 0, s i. In particular, each H(i) lies either below or above the line h 0 h s. We distinguish s combinatorial cases C, C,..., C s defined for i =,,..., s by C i : The line h 0 h s separates H(0) H() H(i ) from H(i). The remaining case C s is defined by C s : The whole set H(0) H(s ) lies on one side of the line h 0 h s. Clearly,, for i =,..., s, i Prob (C i ) =, for i = s. s The triangle h 0 h s h s is always empty. Moreover, in case C i ( i < s) there are i empty triangles h 0 h sp, p H(i). It is easy to see that there are no other empty triangles with the two vertices h 0, h s. Thus, s s V 3 (s) = + Prob (C i ) i = + = s. i= It remains to compute V 6 (s). Without loss of generality, let H(0) = {h s } lie under the line h 0 h s. By Lemma 9(ii), in case C i ( i < s) there are i (i + ) empty hexagons h 0 h s h sh v h v+w h w corresponding to the i (i + ) open down segments h w h v, v > w +, in H(i). By Lemma 8, there are no other empty hexagons with the two vertices h 0, h s. Thus, V 6 (s) = s i= i= i ( i (i + ) ) = s 4 + s + s. Lemma 8. Let k 3. If V k (s) = O(s), then the expected number of empty -line s -prime k-gons (s N) in Λ is s= V k (s) 4 s p n.

21 PLANAR POINT SETS 63 Proof. Let k 3. The expected number of empty -line s -prime k- gons (s N) is (0) s= (IJ is s -prime) V k (s). We may apply Observation 0, since, by Lemma 7(ii), for any ε > 0 and for any sufficiently large t = t(ε), s=t+ (IJ is s -prime) V k (s) s=t+ s=t+ 8n 4 s V k(s) ( s ) O 4 s n < εn. The lemma now follows from (0), Observation 0, and Lemma 7(i). We are ready to estimate the number of empty -line s -prime triangles and hexagons. By Lemmas 7 and 8, the expected number of -line s - prime triangles is s= s 4 s p n = 4 9 p n, and the expected number of empty -line s -prime hexagons is s= s 4 + s+ ( 4 s 4 s p n = ) p n = p n. 9. -line r-prime triangles and hexagons (r s ) For k N and odd z 3, we define W k (z) as the expected number of those empty k-gons in a random Horton set H of size z +, which contain both the leftmost point and the rightmost point of H. Lemma 9. For any odd z 3, where ω = log z. W 3 (z) = 4 4 ω, W 6(z) = 4 ω ω,

22 64 I. BÁRÁNY and P. VALTR Proof. Let z 3 be odd and let H = {h 0,..., h z } be a Horton set with vertices listed according to the increasing x-coordinate. For i =,,..., ω = log z, we put K i = h i and L i = h z i. Clearly, only the points h 0, h z, K i, L i ( i ω) may be vertices of empty polygons with the two vertices h 0, h z. Without loss of generality, let {h 0, K, K,..., K ω } H 0, = {h 0, h,..., h z } lie far below {L ω, L ω,..., L, h z } H, = {h, h 3,..., h z }. By Lemma 5, for any i =,..., ω, the segment h 0 K i is open up in H 0, with probability i. Analogously, L i h z is open down in H, also with probability i. Thus, each of the triangles h 0 K i h z and h 0 L i h z is empty with probability i, and W 3 (z) = ω i= i = 4 4 ω. Any two empty triangles h 0 K i h z and h 0 L j h z (i, j ) give rise to an empty hexagon h 0 K i K i h z L j L j. Thus, W 6 (z) = ω i= ω j= i ( ω j = i= ) ( i = ) ω = 4 ω ω. Observation 0. For any odd z 3 and any k, s N, the expected number of empty k-gons in a random Horton set H of size s z + containing both the leftmost point and the rightmost point of H is equal to W k (z). Proof. We denote the points of H as above. The set H 0, s = {h 0, h s,..., h s z} is a random Horton set of size z +. Its convex hull contains no other points of H. Thus, H 0, s determines, in expectation, W k (z) empty k- gons with the two vertices h 0, h s z. There are no other empty k-gons with the two vertices h 0, h s z, since the interior of every triangle h 0 h s zh i, h i H \ H 0, s, contains one of the points h s, h s (z ). Here is an analogue of Lemma 8: Lemma. Let k N. If W k (z) = O(), then the expected number of empty -line r-prime k-gons (r s ) in Λ is 4 3 z=3 odd W k (z) z p n.

23 PLANAR POINT SETS 65 Proof. Let k N. By Observation 0, the expected number of empty -line s -prime k-gons is () z=3 odd s=0 (IJ is s z-prime) W k (z). It follows from Lemma 7 and from two applications of Observation 0 that () can be estimated by z=3 odd s=0 p n 4 s z W k(z) = 4 3 z=3 odd W k (z) z p n. We are ready to estimate the number of empty -line r-prime triangles and hexagons (r s ). By Lemmas 9 and, the expected number of -line r-prime triangles (r s ) is 4 3 z=3 odd 4 4 ( 6 ω z p n = 3 α 6 ) 3 β p n, and the expected number of empty -line r-prime hexagons (r s ) is 4 3 z=3 odd ( ω 4 4 ω z p n = 3 α 6 3 β + 6 ) 3 γ p n. REFERENCES [] Ahrens, C., Gordon, G. and McMahon, E. W., Convexity and the beta invariant, Discrete Comput. Geom. (999), MR 000j:5007 [] Balog, A. and Deshouillers, J-M., On some convex lattice polytopes, in: Number theory in progress, Volume : Elementary and analytic number theory (K. Győry et al., eds.), de Gruyter, Berlin 999, MR 000f:083 [3] Bárány, I. and Füredi, Z., Empty simplices in Euclidean space, Canadian Math. Bull. 30 (987), MR 89g:5004 [4] Bárány, I. and Károlyi, Gy., Problems and results around the Erdős Szekeres convex polygon theorem, Lect. Notes Comput. Sci. 098 (00), [5] Dumitrescu, A., Planar sets with few empty convex polygons, Stud. Sci. Math. Hung. 36 (000), MR 00f:5037 [6] Edelman, P. and Reiner, V., Counting the interior of a point configuration, Discrete Comput. Geom. 3 (000), 3. MR 000i:508 [7] Erdős, P., On some problems of elementary and combinatorial geometry, Ann. Mat. Pura. Appl. (4) 03 (975), MR 54#3

24 66 I. BÁRÁNY and P. VALTR: PLANAR POINT SETS [8] Erdős, P. and Szekeres, Gy., A combinatorial problem in geometry, Compositio Math. (935), [9] Hardy, G. H. and Wright, E. M., An introduction to the Theory of Numbers, 5th ed., Clarendon Press (Oxford) 979. MR 8i:000 [0] Horton, J. D., Sets with no empty convex 7-gons, Canadian Math. Bull. 6 (983), MR 85f:5007 [] Katchalski, M. and Meir, A., On empty triangles determined by points in the plane, Acta. Math. Hungar. 5 (988), MR 89f:50 [] Valtr, P., Convex independent sets and 7-holes in restricted planar point sets, Discrete Comput. Geom. 7 (99), MR 93e:5037 [3] Valtr, P., On the minimum number of empty polygons in planar point sets, Studia Sci. Math. Hung. 30 (995), MR 96e:509 (Received March, 003) ALFRÉD RÉNYI INSTITUTE OF MATHEMATICS HUNGARIAN ACADEMY OF SCIENCES H-364 BUDAPEST, P.O.B. 7 HUNGARY DEPARTMENT OF MATHEMATICS UNIVERSITY COLLEGE LONDON GOWER STREET LONDON WCE 6BT ENGLAND barany@math-inst.hu INSTITUTE FOR THEORETICAL COMPUTER SCIENCE AND DEPARTMENT OF APPLIED MATHEMATICS CHARLES UNIVERSITY MALOSTRANSKÉ NÁM PRAHA CZECH REPUBLIC valtr@kam.mff.cuni.cz