A note on the no-three-in-line problem on a torus
Zofia Stȩpień, Aleksander Misiak, Alicja Szymaszkiewicz, Lucjan Szymaszkiewicz, Maciej Zwierzchowski
aa r X i v : . [ m a t h . C O ] S e p A NOTE ON THE NO-THREE-IN-LINE PROBLEM ON A TORUS
ALEKSANDER MISIAK, ZOFIA STE¸ PIE ´N, ALICJA SZYMASZKIEWICZ,LUCJAN SZYMASZKIEWICZ, AND MACIEJ ZWIERZCHOWSKI
Abstract.
In this paper we show that at most 2 gcd( m, n ) points can beplaced with no three in a line on an m × n discrete torus. In the situationwhen gcd( m, n ) is a prime, we completely solve the problem. Introduction
The no-three-in-line-problem [2] asks for the maximum number of points thatcan be placed in the n × n grid with no three points collinear. This question hasbeen widely studied, but is still not resolved.The obvious upper bound is 2 n since one can put at most two points in eachrow. This bound is attained for many small cases, for details see [4] and [5]. In [7]the authors give a probabilistic argument to support the conjecture that for a large n this limit is unattainable.As a lower bound, Erd¨os’ construction (see [3]) shows that for p prime one canselect p points with no three collinear. In [8] it is shown, that for p prime one canselect 3( p −
1) points from a 2 p × p grid with no three collinear.In the literature we can find some extensions of the no-three-in-line problem (see[6], [9]). This paper is generalization of [6], where authors analyze the no-three-in-line-problem on the discrete torus. This modified problem is still interesting.Let m and n be positive integers greater than 1. By a discrete torus T m × n wemean { , , . . . , m − } × { , , . . . , n − } .Four integers a, b, u, v with gcd( u, v ) = 1 correspond to the line { ( a + uk, b + vk ) : k ∈ Z } on Z × Z . The condition gcd( u, v ) = 1 ensures that each pair P, Q of distinct pointsin Z × Z belongs to exactly one line. For instance, the points O = (0 , , P = (2 , { ( k, k ) : k ∈ Z } .We define lines on T m × n to be images of lines in the Z × Z under the projection π m,n : Z × Z → T m × n defined as follows π m,n ( a, b ) := ( a mod m, b mod n ) . By x mod y we mean the smallest non-negative remainder when x is divided by y . We say that a set X ⊂ T m × n satisfies the no-three-in-line condition if there areno three collinear points in X . Let τ ( T m × n ) denote the size of the largest set X satisfying the no-three-in-line condition.In our paper we will prove the following theorems. Theorem 1.1.
We have τ ( T m × n ) ≤ m, n ) . Theorem 1.2.
We have
Mathematics Subject Classification.
Key words and phrases.
Discrete torus; No-three-in-line problem; Chinese RemainderTheorem.Corresponding author: Z. St¸epie´n; e-mail: [email protected]; Tel/fax:+48914494826. For gcd( m, n ) = 1 , τ ( T m × n ) = 2 . (2) For gcd( m, n ) = 2 , τ ( T m × n ) = 4 . (3) Let gcd( m, n ) = p be an odd prime. (a) If gcd( pm, n ) = p or gcd( m, pn ) = p , then τ ( T m × n ) = 2 p. (b) If gcd( pm, n ) = p and gcd( m, pn ) = p , then τ ( T m × n ) = p + 1 . The Theorem 1.2(1) was proved in [6] by some algebraic argument. The Theorem1.2(2) is a generalized version of Proposition 2.1 from [6]. Similarly, Theorem1.2(3a) and Theorem 1.2(3b) are generalizations of Theorem 2.7 and Theorem 2.9from [6], respectively.2.
Proofs of Theorem 1.1 and Theorem 1.2(1)
One of the main tools used in this paper is the Chinese Remainder Theorem.
Theorem 2.1 (Chinese Remainder Theorem) . Two simultaneous congruences x ≡ a (mod m ) ,x ≡ b (mod n ) are solvable if and only if a ≡ b (mod gcd( m, n )) . Moreover, the solution is uniquemodulo lcm( m, n ) . Let us define the family L = { L s : s ∈ { , , . . . , gcd( m, n ) − }} of lines on T m × n , where L s = { π m,n ( k, k − s ) ∈ T m × n : k ∈ Z } . Lemma 2.2 (see [10]) . Let a = ( a x , a y ) ∈ T m × n and d = ( a x − a y ) mod gcd( m, n ) .Then a ∈ L d . Moreover, we have L s ∩ L s = ∅ for s = s and s , s ∈{ , , . . . , gcd( m, n ) − } . Proof.
By Theorem 2.1 there exists k ∈ Z such that k ≡ a x (mod m ) ,k ≡ a y + d (mod n ) . Consequently, ( a x , a y ) = π m,n ( k, k − d ) ∈ L d . Suppose that L s ∩ L s = ∅ . Thismeans that there are k , k ∈ Z such that π m,n ( k , k − s ) = π m,n ( k , k − s ) . Inother words k − k is the solution of the following system k − k ≡ m ) ,k − k ≡ s − s (mod n ) . By Theorem 2.1 again, we see that s − s ≡ m, n )). Hence L s ∩ L s = ∅ for s = s and s , s ∈ { , , . . . , gcd( m, n ) − } . (cid:3) Proof of Theorem 1.1.
Let X ⊂ T m × n satisfy the no-three-in-line condition. ByLemma 2.2 for every a ∈ X there exists L ∈ L such that a ∈ L . Consequently, τ ( T m × n ) ≤ · |L| = 2 · gcd( m, n ). (cid:3) Proof of Theorem 1.2(1).
Obviously it is always true that τ ( T m × n ) ≥
2. By Theo-rem 1.1 we get the statement. (cid:3) Proofs of Theorem 1.2(2) and Theorem 1.2(3a)
Let X = ( x , x ) ∈ Z × Z , Y = ( y , y ) ∈ Z × Z , Z = ( z , z ) ∈ Z × Z . Denoteby D ( X, Y, Z ) the following determinant (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x y z x y z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Recall the determinant criterion for checking whether points are in a line: emma 3.1. Three points
X, Y, Z ∈ Z × Z are in a line if and only if D ( X, Y, Z ) =0 . Now we prove the determinant criterion on a torus.
Lemma 3.2.
If three points a, b and c of T m × n are in a line, then D ( a, b, c ) ≡ m, n )) . Proof.
Suppose that three points a = ( a x , a y ) , b = ( b x , b y ) and c = ( c x , c y ) are ina line on T m × n . This means that there are A, B, C ∈ Z × Z such that π ( A ) = a,π ( B ) = b, π ( C ) = c and D ( A, B, C ) = 0 . More precisely A = ( a x + α x m, a y + α y n ) ,B = ( b x + β x m, b y + β y n ) ,C = ( c x + γ x m, c y + γ y n )for some α x , α y , β x , β y , γ x , γ y ∈ Z .We get0 = D ( A, B, C ) = D ( a, b, c ) + nD (( a x , α ) , ( b x , β y ) , ( c x , γ y ))+ mD (( α x , a y ) , ( β x , b y ) , ( γ x , c y )) + mnD (( α x , α y ) , ( β x , β y ) , ( γ x , γ y )) . Hence D ( a, b, c ) ≡ m, n )) . (cid:3) Proof of Theorem 1.2(2).
Let gcd( m, n ) = 2 . Let X = { (0 , , (0 , , (1 , , (1 , } ⊂ T m × n . It is easy to check that D ( a, b, c ) ≡ ± m, n )) for any a, b, c ∈ X. ByLemma 3.2, X satisfies the no-three-in-line condition. Thus τ ( T m × n ) ≥ . NowTheorem 1.1 finishes the statement. (cid:3)
Proof of Theorem 1.2(3a).
Let p = gcd( m, n ) be an odd prime. Assume withoutloss of generality that gcd( pm, n ) = p . Consequently m = pk and n = p l forsome positive integers k, l . Define X = { ( i, i p ) ∈ T m × n : i ∈ { , , . . . , p − }} and Y = { ( i, i p + 1) ∈ T m × n : i ∈ { , , . . . , p − }} . We will show that the set X ∪ Y of 2 p points satisfies the no-three-in-line condition.Take any three distinct points ( i, i p ) , ( j, j p ) , ( k, k p ) from X . We will showthat these points are not in a line on T m × n . To do this, we will show that threepoints A = ( i + α x m, i p + α y n ) , B = ( j + β x m, j p + β y n ) , C = ( k + γ x m, k p + γ y n ) , where α x , α y , β x , β y , γ x , γ y ∈ Z are not in a line on Z × Z . We get D ( A, B, C ) = D (( i, i p ) , ( j, j p ) , ( k, k p )) + nD (( i, α y ) , ( j, β y ) , ( k, γ y ))+ mD (( α x , i p ) , ( β x , j p ) , ( γ x , k p )) + mnD (( α x , α y ) , ( β x , β y ) , ( γ x , γ y ))= p · D (( i, i ) , ( j, j ) , ( k, k )) + p l · D (( i, α y ) , ( j, β y ) , ( k, γ y ))+ pk · p · D (( α x , i ) , ( β x , j ) , ( γ x , k )) + pk · p l · D (( α x , α y ) , ( β x , β y ) , ( γ x , γ y ))= p ( j − i )( k − i )( k − j ) + p M = 0 , since p ∤ ( j − i )( k − i )( k − j ) and M ∈ Z .In the same way it can be shown that any three points from Y are not in a lineon T m × n .Now take any two points ( i, i p ) , ( j, j p ) from X and ( k, k p + 1) from Y . Wehave D (( i, i p ) , ( j, j p ) , ( k, k p + 1)) ≡ j − i (mod gcd( m, n )) . By Lemma 3.2 these points are not in a line. The same argument works if we takeone point from X and any two points from Y . We showed that τ ( T m × n ) ≥ p. Now Theorem 1.1 gives the statement. (cid:3) . Proof of Theorem 1.2(3b)
Let p = gcd( m, n ) and ρ : T m × n → T p × p be the projection defined as follows ρ ( u, v ) = ( u mod p, v mod p ). Since p | m and p | n , the diagram Z × Z π p,p (cid:15) (cid:15) π m,n / / T m × nρ z z ✈✈✈✈✈✈✈✈✈ T p × p (1)commutes. Consequently, the image of every line on T m × n is a line on T p × p . Thisimmediately implies the following result. Lemma 4.1.
Let p = gcd( m, n ) . The following holds: (1) τ ( T m × n ) ≥ τ ( T p × p ) , (2) If the preimage of every line on T p × p is a line on T m × n , then τ ( T m × n ) = τ ( T p × p ) . Now, let us define the family P O = (cid:8) ℓ β : β ∈ { , , . . . , p − } (cid:9) ∪ { ℓ ∞ } of lineson T p × p passing through O = (0 , ℓ β = { π p,p ( k, βk ) ∈ T p × p : k ∈ Z } ,ℓ ∞ = { π p,p (0 , k ) ∈ T p × p : k ∈ Z } . The following results can also be found in [6].
Lemma 4.2.
Let p be an odd prime. Then T p × p = S ℓ ∈P O ℓ .Proof. Consider ( a x , a y ) ∈ T p × p . Suppose a x = 0. Since p is an odd prime, there is aunique β ∈ { , , . . . , p − } such that ( a x , a y ) ∈ ℓ β . If a x is zero, then ( a x , a y ) ∈ ℓ ∞ . Hence, T p × p ⊂ S ℓ ∈P O ℓ. The inclusion S ℓ ∈P O ℓ ⊂ T p × p is obvious. (cid:3) In the next two lemmas we will investigate the sets ρ − ( ℓ ) for ℓ ∈ P O . Lemma 4.3.
Let p be an odd prime. For every β ∈ { , , . . . , p − } the set ρ − ( ℓ β ) is a line on T m × n .Proof. First we claim that there is α ∈ Z such that α ≡ β (mod p ) and gcd( αm, n ) = p . Indeed, for instance, take α such that the following conditions are satisfied:(i) α = β + kp for some k ∈ Z , (ii) α is a prime, (iii) α is greater than n . Since p is an odd prime, the existence of such α is guaranteed by Dirichlet’s theorem onarithmetic progressions.Define L α = { π m,n ( k, αk ) : k ∈ Z } . Now, we will show that ρ − ( ℓ β ) = L α . Let( a x , a y ) ∈ ρ − ( ℓ β ). Then a y ≡ βa x (mod p ) and a y ≡ αa x (mod p ). By Theorem2.1 there exists k ∈ Z such that k ≡ αa x (mod αm ) ,k ≡ a y (mod n ) . It is easy to see that k = αk for k ∈ Z and we get αk ≡ αa x (mod αm ) ,αk ≡ a y (mod n ) . Hence k ≡ a x (mod m ) ,αk ≡ a y (mod n ) . This means that ( a x , a y ) ∈ L α . Since ρ ( L α ) ⊂ ℓ α = l β , we have L α ⊂ ρ − ( ℓ β ). Theproof is finished. Lemma 4.4.
Let p = gcd( m, n ) . The following holds: (1) If gcd( m, pn ) = p , then ρ − ( ℓ ∞ ) is a line in T m × n , (2) If gcd( pm, n ) = p , then ρ − ( ℓ ) is a line in T m × n .Proof. (1) Let L ∞ = { π m,n ( pk, k ) : k ∈ Z } . We will show that ρ − ( ℓ ∞ ) = L ∞ .Take ( a x , a y ) ∈ ρ − ( ℓ ∞ ). Hence a x ≡ p ). By Theorem 2.1 there exists k ∈ Z such that k ≡ a x (mod m ) ,k ≡ pa y (mod pn ) . It is easy to see that k = pk for some k ∈ Z and we get pk ≡ a x (mod m ) ,pk ≡ pa y (mod pn ) . Hence pk ≡ a x (mod m ) ,k ≡ a y (mod n )and a ∈ L ∞ . The inclusion L ∞ ⊂ ρ − ( ℓ ∞ ) is obvious. The proof is finished.(2) The proof is similar to (1). (cid:3) Theorem 4.5.
Let p = gcd( m, n ) be an odd prime such that gcd( pm, n ) = gcd( m, pn ) = p . Then we have τ ( T m × n ) = τ ( T p × p ) . Proof.
By Lemma 4.4 and 4.3 we get that ρ − ( ℓ ) is a line on T m × n for any ℓ ∈ P O .Hence the preimage of every line on T p × p is a line on T m × n . Lemma 4.1(2) finishesthe proof. (cid:3)
The following result can be found in [6]. Here we present the complete proof.
Theorem 4.6.
Let p be an odd prime. Then τ ( T p × p ) = p + 1 .Proof. Let p be an odd prime number. If p ≡ q to be somequadratic nonresidue modulo p . If p ≡ q to be some quadraticresidue modulo p . Define X = { ( x, y ) ∈ T p × p : x + q · y ≡ p ) } . Itis known that X has p + 1 points. See for example Theorem 10.5.1 in [1]. ByLagrange theorem for congruences, any line intersects X in at most two points.Hence the set X satisfies the no-three-in-line condition and τ ( T p × p ) ≥ p + 1 . Let X satisfy the no-three-in-line condition. We can assume that O ∈ X . ByLemma 4.2, every other point of the T p × p lies on one or the other of the p + 1 linespassing through O . Hence | X | ≤ p + 2 and we have τ ( T p × p ) ≤ p + 2.Assume that there exists a set Y with p + 2 points which satisfies the no-three-in-line condition. Take any line L in T p × p . We claim that either | L ∩ Y | = 0 or | L ∩ Y | = 2. Indeed, if L ∩ Y = { y } then L is the line passing through point y ∈ Y which does not pass by any other point of Y and consequently | Y | ≤ p + 1,a contradiction. Now fix any point z not in Y . Since each line through z containseither 0 or 2 points of Y , the number of points in Y is even, a contradiction withthe fact that p + 2 is odd. This means that the set Y does not exist. (cid:3) Proof of Theorem 1.2(3b).
Theorem 4.5 together with Theorem 4.6 gives the statement. (cid:3)
Acknowledgment.
We express our sincere thanks to the anonymous referee forhis/her valuable advice which resulted in an improvement of this article. eferences [1] B. C. Berndt, R. J. Evans, K. S. Williams, Gauss and Jacobi sums, Wiley, 1998.[2] H. E. Dudeney, Amusements in Mathematics, Nelson, Edinburgh 1917, pp. 94, 222.[3] P. Erd¨os. Appendix, in K.F. Roth, On a problem of Heilbronn, J. London Math. Soc. 26,198–204, 1951.[4] A. Flammenkamp, Progress in the no-three-in-line problem, J. Combin. Theory Ser. A, 60(2),305–311, 1992.[5] A. Flammenkamp, Progress in the no-three-in-line problem II, J. Combin. Theory Ser. A,81(1), 108–113, 1998.[6] J. Fowler, A. Groot, D. Pandya, B. Snapp, The no-three-in-line problem on a torus,arXiv:1203.6604v1.[7] R. K. Guy P. A. Kelly, The No-Three-Line Problem, Math. Bull. Vol. 11, pp. 527-531, 1968.[8] R. R. Hall, T. H. Jackson, A. Sudbery, and K. Wild, Some advances in the no-three-in-lineproblem, J. Combinatorial Theory Ser. A, 18:336–341, 1975.[9] A. Por, D.R. Wood, No-Three-in-Line-in-3D, Algorithmica 47(4), 481–488, 2007.[10] Z. St¸epie´n, L. Szymaszkiewicz, M. Zwierzchowski, The Cartesian product of cycles with small2-rainbow domination number, Journal of Combinatorial Optimization, doi:10.1007/s10878-013-9658-0. School of Mathematics, West Pomeranian University of Technology, al. Piast´ow48/49, 70-310 Szczecin, Poland
E-mail address : [email protected]
School of Mathematics, West Pomeranian University of Technology, al. Piast´ow48/49, 70-310 Szczecin, Poland
E-mail address : [email protected] School of Mathematics, West Pomeranian University of Technology, al. Piast´ow48/49, 70-310 Szczecin, Poland
E-mail address : [email protected] Institute of Mathematics, Szczecin University, Wielkopolska 15, 70-451 Szczecin,Poland
E-mail address : [email protected] School of Mathematics, West Pomeranian University of Technology, al. Piast´ow48/49, 70-310 Szczecin, Poland
E-mail address : [email protected]@zut.edu.pl