A new progress on Weak Dirac conjecture

A NEW PROGRESS ON WEAK DIRAC CONJECTURE

arXiv:1607.08398v1 [math.CO] 28 Jul 2016

HOANG HA PHAM AND TIEN CUONG PHI Abstract. In 2014, Payne-Wood proved that every non-collinear set P of n points in n the Euclidean plane contains a point in at least lines determined by P. This is a 37 remarkable answer for the conjecture, which was proposed by Erd˝ os, that every nonn lines determined by P , for collinear set P of n points contains a point in at least c1 some constant c1 . In this article, we refine the result of Payne-Wood to give that every n + 2 lines determined by P non-collinear set P of n points contains a point in at least 26 . Moreover, we also discuss some relations on theorem Beck that every set P of n points 1 n(n − l) lines. with at most l collinear determines at least 61

1. Introduction Let P be a set of points in the Euclidean plane. A line that contains at least two points in P is said to be determined by P. In 1951, G. Dirac ([4]) made the following conjecture, which remains unsolved: Conjecture 1 (Strong Dirac Conjecture). Every non-collinear set P of n points in n the plane contains a point in at least −c0 of the lines determined by P , for some constant 2 c0 . In 2011, J. Akiyama, H. Ito, M. Kobayashi, and G. Nakamura ([2]) gave somes examples n to show that the bound would be tight. We note that if P is non-collinear and contains 2 n or more collinear points, then Dirac’s Conjecture holds.Thus we may assume that P 2 n does not contain collinear points, and n ≥ 5. 2 In 1961, P. Erd˝os ([5]) proposed the following weakened conjecture. Conjecture 2 (Weak Dirac Conjecture). Every non-collinear set P of n points conn tains a point in at least lines determined by P, for some constant c1 . c1 2010 Mathematics Subject Classification. 52C10, 52C30. Key words and phrases. Arrangement of points, Incident-line-number, Dirac conjecture, lines with few point. 1

2

HOANG HA PHAM AND TIEN CUONG PHI

In 1983, Beck ([3]) and Szemer´edi-Trotter ([18]) proved the Weak Dirac Conjecture for the case c1 but it is unspecied or very large. In 2014, Payne-Wood ([15]) proved the following theorem: n lines Theorem 1. Every non-collinear set P of n points contains a point in at least 37 determined by P . For the first purpose of this article, we would like to give a new progress for the Weak Dirac conjecture. In particular, we prove the following: n Main theorem 1. Every non-collinear set P of n points contains a point in at least +2 26 lines determined by P. Moreover, relate to work on the Weak Dirac Conjecture, Beck gave the number of lines determined by P. He proved the following theorem. Theorem 2. ([3]) Every set P of n points with at most l collinear determines at least c2 n(n − l) lines, for some constant c2 . In 2014, Payne - Wood also gave a remarkable improvement of Beck’s theorem by proving the following. Theorem 3. ([15]) Every set P of n points with at most l collinear determines at least 1 n(n − l) lines. 98 We note that the number 98 can be instead by 93. The details can be found in [14]. For the final purpose, we would like to give some results for the number of lines with few points from n points in plane. Then, we also give the following theorems. Main theorem 2. Every set P of n points with at most l collinear determines at least 1 n(n − l) lines. 61 Main theorem 3. Every set P of n points with at most l collinear determines at least 1 n(n − l) lines with at most 3 points. 122 2. Auxiliary Results We list here some known results which are very helpful for the proofs of the main theorems. The crossing number of a graph G, denoted by cr(G), is the minimum number of crossings

WEAK DIRAC CONJECTURE

3

in a drawing of G. The following version due to J. Pach, R. Radoiˇci´c, G. Tardos and G. T´oth [13] is the strongest to date. 103 n edges, Lemma 4. (Crossing lemma [13]). For every graph with n vertices and m ≥ 16 then 1024m3 cr(G) ≥ . 31827n2 We set E(H) to be the set of all edges of a graph H. The visibility graph G of a point set P has vertex set P , where vw ∈ E(G) whenever the line segment vw contains no other point in P (that is, v and w are consecutive on a line determined by P ). For i ≥ 2, an i − line is a line containing exactly i points in P. Let si be the number of i − lines. Let Gi be the spanning subgraph of the visibility graph of P consisting of all edges in j − lines P where j ≥ i. Note that since each i − line contributes i − 1 edges,|E(Gi )| = j≥i (j − 1)sj . We introduce some useful results: Theorem 5. (Hirzebruch’s Inequality [10]). Let P be a set of n points with at most n − 3 collinear. Then X 3 s2 + s3 ≥ n + (2i − 9)si . 4 i≥5 Theorem 6. (Szemer´edi-Trotter [18]). Let α and β be positive constants such that every graph H with n vertices and m ≥ αn edges satisfies cr(H) ≥

m3 βn2

Let P be a set of n points in the plane. Then a)

|E(Gi )| =

b)

X

X

(j − 1)sj ≤ max{αn,

j≥i

j≥i

sj ≤ max{

βn2 }, 2(i − 1)2

βn2 αn , }. i − 1 2(i − 1)3

3. A new progress on Weak Dirac’s conjecture In order to get the main theorem 1, we refine the method of Payne-Wood to find the largest number ε such that every set P of n non-collinear points in the plane at most εn + 2 collinear points, the arrangement of P has at least εn2 + 2n point- line incidents. We start by the following result.

4

HOANG HA PHAM AND TIEN CUONG PHI

Theorem 7. Let α and β be positive constants such that every graph G with n vertices m3 and m ≥ αn edges satisfies cr(G) ≥ . βn2 1 Fix two integers c ≥ 8, 0 ≤ q ≤ 3 and a real number ǫ ∈ (0; ), ǫn ≥ 2. Let h := 2 c(c − 2) . Then for every set P of n points in the plane with at most ǫn + q collinear 5c − 18 points, the arrangemnet of P has at least δn2 + rn point- line incident, where !! 1 β −18(c − 2) X i + 1 δ= 1 − ǫα − + , h+1 2 c3 (5c − 18) i≥c i3 r=

2h − 1 + α . h+1

Proof. Let J = {2; 3; ...; ⌊ǫn⌋+q} and assume that ǫn ≥ 2. Considering the visibility graph G of P and its subgraphs Gi as defined previously. Let k be the minimum integer such that |E(Gk )| ≤ αn. If there is no such k then let k = ⌊ǫn⌋ + q + 1. An integer i ∈ J is large if i ≥ k , and is small if i ≤ c. An integer in J that is neither small nor large is medium. Recall that an i-line is a line containing exactly i points in P . An i-pair is a pair of points in an i-line. A small pair is an i-pair for some small i. Define large pair, medium pair analogously. Let PS , PM and PL denote the number of small, medium and large pairs respectively. An i-incidence is an incidence between a point of P and an i-line. A small incidence is an i-incidence for some small i, and define medium, large incidences analogously. Let IS , IM and IL denote the number of small, medium and large incidences respectively and let I denote the total number of incidences. Since every si has i points incidence with its, then X I= isi = IS + IM + IL i∈J

n n Because P has no more than collinear points and n ≥ 5, thus ⌊ǫn⌋ + q ≤ ⌊ ⌋ ≤ n − 3. 2 2 Therefore, for n points of P has no more than n − 3 collinear points. Applying the Hirzebruch’s Inequality (Theorem 5), we have X 3 s2 + s3 ≥ n + (2i − 9)si . 4 i≥5

WEAK DIRAC CONJECTURE

5

Since h > 0 then, X 3 (2i − 9)si ≥ 0. hs2 + hs3 − hn − h 4 i≥5

PS =

c X

i 2

i=2

!

si

= s2 + 3s3 + 6s4 +

c X

i 2

i=5

!

si

! X i si − hn − h (2i − 9)si 2 i≥5  c  X h+1 h+4 3 9h i−1 isi .2s2 + .3s3 + .4h4 + − 2h + = 2 4 2 2 i i=5 c

X 3h ≤ (h + 1)s2 + ( + 3)s3 + 6s4 + 4 i=5

− h

k−1 X

(2i − 9)si − h

i=c+1

X

(2i − 9)si − hn

i≥k

c

X h+4 3 h+1 .2s2 + .3s3 + .4h4 + ≤ 2 4 2 i=5 − h

k−1 X

(2 −

i=c+1

Setting X := max{



9h i−1 − 2h + 2 i

isi

X 9 7 )isi − h (2 − )(i − 1)si − hn. c+1 c i≥k

h+1 h+4 3 i−1 9h ; ; ; max5≤i≤c ( − 2h + )} implies that, 2 4 2 2 i

PS ≤ XIS − h

k−1 X

(2 −

i=c+1

X 9 7 )isi − h (2 − )(i − 1)si − hn. c+1 c i≥k

i−1 9h − 2h + for 5 ≤ i ≤ c. 2 i h We have: γi” ≥ 0 ∀i ∈ (5, c) ⇒ γ(h, i)max = γ(h, 5) = 2 − 5 9h c−1 − 2h + for c ≥ 8. or γ(h, i)max = γ(h, c) = 2 c

Let γ(h, i) =



(3.1)

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HOANG HA PHAM AND TIEN CUONG PHI

Clearly, h(c) =

c(c − 2) has minimum 5c − 18 h+1 ≥ 2 h+1 ≥ 2 h+1 ≥ 2 h+1 = 2

value

24 when c = 8. Hence, 11

3 2 h+4 4 h 2− 5 c−1 9h − 2h + . 2 c

h+1 . 2 On the other hand, if i ∈ J is medium (c < i < k) then i is not large. Therefore, P P (j − 1)sj > αn. Because if (j − 1)sj ≤ αn then |E(Gi )| ≤ αn, contradict with

Thus, X =

j≥i

j≥i

minimum property of k. Using part (a) and (b) of the Szemerdi- Trotter theorem 6, X X X βn2 βn2 βn2 i jsj = (j − 1)sj + sj ≤ + = . (3.2) 2 3 3 2(i − 1) 2(i − 1) 2(i − 1) j≥i j≥i j≥i Given X as above, we have PM − XIM =

k−1 X

i=c+1

=

i 2

! ! si

−X

k−1 1 X (i − 1 − 2X)isi . 2 i=c+1

k−1 X

i=c+1

isi

!

Combining with (3.1), we get PS + PM ≤ XIS − hn + XIM

 k−1  18h 7 1 X i − 1 − 2X − 4h + isi − h(2 − )|E(Gk )|. + 2 i=c+1 c+1 c (3.3)

We define 18h c+1 c(c − 2) 18c(c − 2) = c−2−5 + 5c − 18 (c + 1)(5c − 18) −18(c − 2) = . (c + 1)(5c − 18)

Y = c − 5h − 2 +

WEAK DIRAC CONJECTURE

7

This implies −1 < Y < 0 with c ≥ 8. Thus we have,  k−1  k−1 1 X 18h 1 X T = i − 1 − 2X − 4h + isi = (i − c + Y )isi 2 i=c+1 c+1 2 i=c+1 ! ! k−1 X k−1 k−1 X X Y 1 jsj + isi = 2 i=c+1 j=i 2 i=c+1 ! ! k−1 X 1 X X Y ≤ jsj + isi . 2 i=c+1 j≥i 2 i≥c+1 Applying (3.2) and Y + 1 > 0, this yields 1 X βn2 i Y βn2 (c + 1) βn2 T ≤ + . = 2 i≥c+1 2(i − 1)3 2 2c3 4

Finally, we have

⌊εn⌋+q

PL −XIL =

X i=k

i 2

!

si −X

X

isi ≤

i≥k

c+1 Xi+1 Y 3 + c i3 i≥c

!

.

(3.4)

X ǫn + q X ǫn + q (i−1)si −X (i−1)si = ( −X)|E(Gk )|. 2 i≥k 2 i≥k (3.5)

Combining (3.3), (3.4), (3.5), we get PS + PM + PL ≤ X(IS + IM + IL ) − hn βn2 c + 1 X i + 1 1 7h + (Y 3 + ) + (ǫn + q − 2X − 4h + )|E(Gk )| 3 4 c i 2 c i≥c

1 βn2 c + 1 X i + 1 (Y 3 + ) + (ǫn − 2)|E(Gk )| ( by 1 ≤ q ≤ 3, c ≥ 8) ≤ XI − hn + 3 4 c i 2 i≥c

βn2 c + 1 X i + 1 1 (Y 3 + ) + (ǫn − 2)αn. 3 4 c i 2 i≥c ! n 1 On the other hand, we have PS + PM + PL = = (n2 − n). 2 2 Thus, we get 1 2 βn2 c + 1 X i + 1 ǫαn2 (n − n) ≤ XI − hn + (Y 3 + ) + − αn. 2 4 c i3 2 i≥c    β c+1 P i+1 2h − 1 + α 1 2 1 − ǫα − Y 3 + n + n. ⇒I≥ 3 2X 2 c 2X i≥c i ≤ XI − hn +

8

HOANG HA PHAM AND TIEN CUONG PHI

Since X =

h+1 −18(c − 2) and Y = 3 then, 2 c (5c − 18)

1 I≥ h+1

β 1 − ǫα − 2

= δn2 + rn.

−18(c − 2) X i + 1 + c3 (5c − 18) i≥c i3

!!

n2 +

2h − 1 + α n h+1

 Theorem 8. Every set P of n non-collinear points in the plane with at most n2 collinear points, the arrangement of P has at least + 2n point- line incidents. 26

n +2 26

n2 n < 1, then the arrangement of P is n2 − n > + 2n by n ≥ 5. 26 26 n n2 − n n2 Case 2. If 1 ≤ < 2, then I = 2s2 + 3s3 ≥ s2 + 3s3 = > + 2n by n ≥ 26. 26 2 26 1 n ≥ 2, then the assumptions of Theorem 7 satisfy with ǫ = , c = 46, q = 2. Case 3. If 26 26 We have n2 + 2n. I ≥ δn2 + rn ≥ 26 The proof of Theorem 8 is completed.  Proof. Case 1. If 0
. This shows that 26 1 is the best constant. 26

4. The lines with few points Theorem 9. Let α, β be positive constants such that every graph H with n vertices and m ≥ αn edges satisfies m3 . cr(H) ≥ βn2 Fix an integer c ≥ 29. Then for every set P of n points in the plane with at most l collinear points, the arrangement of P has at least  β 1− 2

c2 − 3c − 14 X 1 + 2c3 (c − 4) i2 i≥c

!

c2

(2c − 8)α 2c − 8 n2 − 2 ln. + 3c − 18 c + 3c − 18

lines with at most c points. Proof. We define the small, medium and large pairs and lines respectively as in the proof c2 − c − 2 , where c ≥ 29. Thus, h > 0. Using the Inequality of of Theorem 7. Set h = 4c − 16 Hirzebruch (Theorem 5), we have

s2 +

X 3h s3 − hn − h (2i − 9)si ≥ 0. 4 i≥5

10

HOANG HA PHAM AND TIEN CUONG PHI

Now we have,

PS =

c X 2

i 2

!

si

= s2 + 3s3 + 6s4 +

c X i=5

i 2

!

si

!  c X X i 3h + 3 s3 + 6s4 + si − hn − h (2i − 9)si ≤ (h + 1)s2 + 4 2 i=5 i≥5  c  X X 3 i(i − 1) ≤ (h + 1)s2 + (h + 4)s3 + 6s4 + − h(2i − 9) si − hn − h (2i − 9)si . 4 2 i=5 i≥c+1 

By c ≥ 29, it is easy to see that

i(i − 1) 3 − h(2i − 9))}, X := h + 1 = max{h + 1; (h + 4); 6; max5≤i≤c ( 4 2

and thus we get

PS ≤ XLS − hn − h

X

(2i − 9)si .

i≥c+1

For the medium index i, we use the Crossing Lemma 4 and part (a) of Theorem 6 to imply that

βn2 , (j − 1)sj ≤ 2(i − 1)2 j≥i

X

WEAK DIRAC CONJECTURE

11

thus we have

PS +PM − XLS ≤ −hn − h

X

(2i − 9)si +

i≥c+1

=

=

=

k−1 X

i=c+1

i 2

!

si

 k−1  X X i(i − 1) − h(2i − 9) si −hn − h (2i − 9)si + 2 i=c+1 i≥k X 1 −hn − h (2i − 9)si + 2 i≥k X 1 −hn − h (2i − 9)si + 2 i≥k

≤

−hn − h

=

−hn − h

X 1 (2i − 9)si + 2 i≥k X 1 (2i − 9)si + 2 i≥k

k−1 X

k−1 X k−1 X 4hi − 18h (c − )(i − 1)si + (j − 1)sj i−1 i=c+1 i=c+1 j=i

!

! k−1 X k−1 X 14h (c − 4h + )(i − 1)si + (j − 1)sj i−1 i=c+1 i=c+1 j=i ! k−1 k−1 X k−1 X X 14h (c − 4h + )(i − 1)si + (j − 1)sj c i=c+1 i=c+1 j=i ! k−1 k−1 X k−1 X 14h X (c − 4h + ) (i − 1)si + (j − 1)sj c i=c+1 i=c+1 j=i k−1 X

On the other hand, we have

c − 4h +

c2 − c − 2 7c2 − 7c − 14 c2 − 3c − 14 14h = c−4 + = c 4c − 16 c(2c − 8) 2c(c − 4) 14h > 0 ( by c ≥ 29). ⇒ c − 4h + c

So we get

PS + PM

X 1 − XLS ≤ −hn − h (2i − 9)si + 2 i≥k

≤ −hn − h

X (2i − 9)si + i≥k

k−1 X X c2 − 3c − 14 X (i − 1)si (i − 1)si + 2c(c − 4) i≥c+1 j=c+1 i≥j ! 2 X 1 βn2 c − 3c − 14 + . 2c3 (c − 4) i2 4 i≥c

!

12

HOANG HA PHAM AND TIEN CUONG PHI

Thus, we now have ! n − XLS = PS + PM + PL − XLS 2 ! ! l 2 X X i 1 βn c − 3c − 14 si (2i − 9)si + + + ≤ −hn − h 3 2 2c (c − 4) i 4 2 i≥c i=k i≥k !   c2 − 3c − 14 X 1 βn2 X i(i − 1) = −hn + + + − 2hi + 9h si 2 2c3 (c − 4) i 4 2 i≥c i≥k !   7h c2 − 3c − 14 X 1 βn2 X i (i − 1)si + + − 2h + = −hn + 2 2c3 (c − 4) i 4 2 i − 1 i≥c i≥k ! c2 − 3c − 14 X 1 βn2 l ≤ −hn + + + |E(Gk )| 3 2 2c (c − 4) i 4 2 i≥c ! c2 − 3c − 14 X 1 βn2 l ≤ −hn + + + αn. 2 2c3 (c − 4) i 4 2 i≥c X

2

So we get LS ≥



1 β − 2 4

c2 − 3c − 14 X 1 + 2c3 (c − 4) i2 i≥c

≥



1 β − 2 4

c2 − 3c − 14 X 1 + 2c3 (c − 4) i2 i≥c

!

  1 lα n n2 + h− − X 2 2 X

!

n2 lαn − . X 2X

c2 + 3c − 18 , we thus get 4c − 16 ! 2c − 8 (2c − 8)α c2 − 3c − 14 X 1 + n2 − 2 ln. 3 2 2 2c (c − 4) i c + 3c − 18 c + 3c − 18 i≥c

On the other hand, X = h + 1 =  β LS ≥ 1 − 2 Theorem 9 is proved.



For the case c = 36, we get the following. 1 Corollary 10. Every set P of n points with at most l collinear determines at least n2 − 39 1 ln lines with at most 36 points. 3 We now apply Theorem 9 to give the proof of Main theorem 2.

WEAK DIRAC CONJECTURE

13

Proof. We may assume that l is the size of the longest line. For some integer c ≥ 29, then by Theorem 9 we have L ≥ LS ≥ A(c)n2 − B(c)nl for some A(c) and B(c) evident in the theorem. Observe that, !  β c2 − 3c − 14 X 1 2c − 8 A(c) = 1− + 3 2 2 2 2c (c − 4) i c + 3c − 18 i≥c B(c) =

(2c − 8)α . c2 + 3c − 18

We note that, 2A ≥ ǫ 1 + 2B ǫ2 ǫ + Bǫ − ⇒A ≥ 2 2 ǫ ǫn + (B − )ǫn ⇒ An ≥ 2 2 ǫn ǫ ⇒ An ≥ + (B − )l 2 2 ǫn(n − l) ⇒ An2 − Bnl ≥ . 2 2A(c) to get a largest number ǫ. Now, set c = 44 we 1 + 2B(c) 1 1 . So we choose ǫ = to complete Main theorem 2.  get ǫ ≤ 30.2 30.5 We now get Main theorem 3 by using Main theorem 2 and the following observation: So we can find the maximum of

Theorem 11. ([15]) Let P be a set of n non-collinear points in a plane. Then at least half the lines determined by P contain at most 3 points. References [1] Martin Aigner and G¨ unter M. Ziegler, Proofs from The Book.Springer, 3rd edn., 2004. [2] Jin Akiyama, Hiro Ito, Midori Kobayashi, and Gisaku Nakamura, Arrangements of n points whose incident-line-numbers are at most n/2. Graphs Combin., 27(3):321-326, 2011. [3] J´ozsef Beck, On the lattice property of the plane and some problems of Dirac, Motzkin and Erdos in combinatorial geometry, Combinatorica, 3(3-4):281-297, 1983. [4] Gabriel A. Dirac,Collinearity properties of sets of points. Quart. J. Math., Oxford Ser. (2), 2:221-227, 1951. [5] Paul Erdos, Some unsolved problems. Magyar Tud. Akad. Mat. Kutat´ o Int. K¨ozl., 6:221-254, 1961. http://www.renyi.hu/∼ p erdos/1961-22.pdf.

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[6] Paul Erd˝ os and George Purdy, Some combinatorial problems in the plane. J. Combin.Theory Ser. A, 25(2):205-210, 1978. [7] Paul Erd˝ os and George Purdy, Two combinatorial problems in the plane. Discrete Comput. Geom., 13(3-4):441-443, 1995. [8] Paul Erd˝ os and Endre Szemer´edi, On sums and products of integers. In Studies in pure mathematics, pp. 213-218. Birkhauser, Basel, 1983. [9] Ben J. Green and Terence Tao, On sets defining few ordinary lines, Discrete Comput Geom., 50:409468, 2013. [10] Friedrich Hirzebruch, Singularities of algebraic surfaces and characteristic numbers. In The Lefschetz Centennial Conference, Part I, vol. 58 of Contemp. Math., pp. 141-155. Amer. Math. Soc., 1986. [11] Leroy M. Kelly and William O. J. Moser, On the number of ordinary lines determined by n points. Canad. J. Math., 10:210-219, 1958. ¨ [12] Eberhard Melchior, Uber Vielseite der projektiven Ebene. Deutsche Math., 5:461-475, 1941. [13] J´anos Pach, Radoˇs Radoiˇci´c, G´ abor Tardos and G´eza T´oth, Improving the cross-ing lemma by finding more crossings in sparse graphs. Discrete Comput. Geom., 36(4):527-552, 2006. [14] Michael S. PhD thesis,

Payne, Combinatorial geometry The University of Melbourne,

of point sets Department of

with collinearities. Mathematics and

Statistics,2014.http://www.ms.unimelb.edu.au/∼mspayne/MichaelPayneThesis.pdf. [15] Michael S. Payne and David R. Wood, Progress on Dirac’s Conjecture, The electronic journal of combinatorics, 21(2) 2-12, 2014. [16] George Purdy, A proof of a consequence of Diracs conjecture. Geom. Dedicata, 10(1-4):317-321, 1981. [17] L´ aszl´o A. Sz´ekely, Crossing numbers and hard Erdos problems in discrete geometry. Combin. Probab. Comput. 6(3):pp 353-358, 1997. [18] Endre Szemer´edi and William T. Trotter, Jr, Extremal problems in discrete geometry. Combinatorica, 3(3-4):381-392, 1983. Department of Mathematics, Hanoi National University of Education, 136 XuanThuy str., Hanoi, Vietnam E-mail address: [email protected]; [email protected]