A089676 a(n) is the maximal size of a set S of points in {0,1}^n in real n-dimensional Euclidean space such that every angle determined by three points in S is acute.

1, 2, 2, 4, 5, 6, 8, 9, 10, 16, 17

Offset: 0

Author

David Bevan, Jan 06 2004

Keywords

• nonn
• more
• nice

Comments

Consider the 2^n points {0,1}^n in real Euclidean space. Then a(n) = maximal size of a subset S of these 2^n points such that there is no triple of points P,Q,R in S which subtends a right angle. That is, we are not allowed to have P-Q perpendicular to R-Q.

There is an existence proof due to Erdős and Füredi that exponentially large subsets S exist: see for example Theorem 2.3 of Noga Alon's survey "Probabilistic Methods in Extremal Finite Set Theory". This was improved by Bevan and later by Ackerman and Ben-Zwi.

As explained by Erdős-Füredi, these sets of points are equivalent to set systems none of which is sandwiched between the intersection and the union of two others. In turn these are equivalent to so-called (2,1)-separating systems. As far as I know the best construction is the one in my Israel J. Math. 2013 paper. It uses algebraic geometry and coding techniques, and it implies a lower bound on a(n) of roughly 11^{3n/50}. This is better than Erdős-Füredi, Bevan, or Ackerman-Ben-Zwi, which are all about (4/3)^{n/2}. It is remarkable that this construction beats the probabilistic method (and note that it implies that for large n, naive computer search will have exponentially small chance to find optimal configurations). I should also add that Erdős-Füredi claimed (without proof) a lower bound of 2^{n/4} which, if correct, would be even better (but also non-constructive). - Hugues Randriambololona, Apr 08 2016

For a(10)=17 a combinatorial search algorithm shows that a cubic acute 10-set with 18 vertices is not possible. For a complete enumeration of sets with maximal size, see A289972. - Fausto A. C. Cariboni, Jul 17 2017

The best known lower bounds for a(11-15) are 24, 32, 33, 64 and 128. a(11-14) were found by D. Kamenetsky, while a(15) was found by D. Kamenetsky and V. Chubenko (see attached file). Lower bounds for n > 15 have been found by V. Harangi (see Table 3 in his paper). - Dmitry Kamenetsky, May 18 2018 and Jun 05 2018

Examples

			a(3) = 4: {000, 011, 101, 110}.
a(4) = 5: {0011, 0101, 0110, 1000, 1111}.
The following sets are given by Bevan (2006), who also shows they are optimal:
a(5) = 6:
  0 0 0 0 0
  0 0 0 1 1
  0 0 1 0 1
  0 1 0 0 1
  1 0 0 0 1
  1 1 1 1 0
a(6) = 8:
  0 0 0 0 0 0
  0 0 0 1 1 1
  0 1 1 0 0 1
  0 1 1 1 1 0
  1 0 1 0 1 0
  1 0 1 1 0 1
  1 1 0 0 1 1
  1 1 0 1 0 0
a(7) = 9:
  0 0 0 0 0 0 0
  0 0 0 0 0 1 1
  0 0 0 1 1 0 1
  0 1 1 0 0 0 1
  0 1 1 1 1 1 0
  1 0 1 0 1 0 1
  1 0 1 1 0 1 0
  1 1 0 0 1 1 0
  1 1 0 1 0 0 1
a(8) = 10:
  0 0 0 0 0 0 0 0
  0 0 0 0 0 0 1 1
  0 0 0 0 0 1 0 1
  0 0 0 1 1 0 0 1
  0 1 1 0 0 0 0 1
  0 1 1 1 1 1 1 0
  1 0 1 0 1 0 0 1
  1 0 1 1 0 1 1 0
  1 1 0 0 1 1 1 0
  1 1 0 1 0 0 0 1
For a(9) = 16 Bevan uses the construction in his Theorem 4.2, which shows that a(3k) >= a(k)^2 for all k, and then a computer search shows that this is optimal for k = 3. Let v0,v1,v2,v3 denote the four vectors for a(3). Then to get a(9)=16 use the vectors { v_i v_j v_{j-i mod 4}, 0 <= i,j <= 3 }. - _N. J. A. Sloane_, Mar 30 2016
a(10) = 17, from Hugues Randriambololona, Apr 08 2016:
  0 0 0 0 0 0 0 0 0 0
  0 0 0 0 1 1 0 1 1 1
  1 1 1 0 1 1 1 0 0 1
  0 0 1 0 0 1 1 1 0 1
  1 1 0 0 0 1 0 1 0 1
  1 1 0 0 1 0 1 1 1 1
  0 1 0 1 1 0 0 1 0 0
  1 1 1 0 0 0 0 0 1 1
  1 0 0 1 1 1 1 0 1 1
  1 0 1 0 1 1 0 0 1 0
  0 1 0 1 0 1 1 0 0 0
  1 0 0 0 0 1 1 1 1 0
  1 0 1 0 1 0 1 1 0 0
  1 0 1 1 0 0 1 1 1 1
  1 1 1 1 0 1 0 1 1 0
  0 1 1 0 1 1 1 1 1 0
  1 0 1 1 1 1 0 1 0 1
a(11) >= 23, from _Dmitry Kamenetsky_, Jan 05 2018:
  0 1 1 0 0 0 1 0 1 1 1
  0 0 0 0 1 0 0 0 0 0 1
  0 0 1 1 0 0 1 0 0 0 1
  0 1 1 0 1 0 1 1 1 0 0
  0 0 1 0 1 1 0 0 1 1 1
  1 1 0 1 1 0 0 0 1 1 0
  1 0 0 1 1 0 0 1 0 1 1
  1 0 0 1 0 1 1 0 1 1 0
  1 1 1 1 1 1 0 0 0 0 0
  1 0 0 1 1 1 1 1 1 0 1
  1 0 1 0 0 0 0 0 0 0 0
  1 0 0 0 0 1 0 1 1 1 1
  0 1 0 0 1 0 1 0 0 1 0
  1 1 0 0 0 0 0 0 1 0 1
  1 1 1 0 0 1 1 1 0 1 1
  1 1 0 0 0 1 1 0 0 0 0
  0 0 0 1 0 0 1 1 0 1 0
  1 0 1 1 1 1 1 0 0 1 1
  1 1 1 1 0 0 0 1 0 0 1
  0 1 1 1 0 0 0 1 1 1 0
  0 1 1 0 1 1 0 1 0 1 0
  0 0 0 0 0 1 0 1 0 0 0
  0 1 0 1 0 1 0 0 0 0 1
		

Links

• Eyal Ackerman and Oren Ben-Zwi, On sets of points that determine only acute angles, European Journal of Combinatorics 30 (2009) 908-910.
• N. Alon, Probabilistic Methods in Extremal Finite Set Theory
• D. Bevan, Sets of Points Determining Only Acute Angles and Some Related Coloring Problems, Electronic J. of Combinatorics, 13(1), 2006, #R12.
• L. Danzer and B. Grünbaum, Uber zwei Probleme bezüglich konvexer Körper von P. Erdős und von K. L. Klee, Math. Zeitschrift 79 (1962) 95-99.
• P. Erdős and Z. Füredi, The greatest angle among n points in the d-dimensional Euclidean space, Annals of Discrete Math. 17 (1983) 275-283.
• Viktor Harangi, Acute Sets In Euclidean Spaces, Journal on Discrete Mathematics, volume 25, issue 3, (2011) 1212-1229.
• Dmitry Kamenetsky, Lower bounds and their solutions for a(11-15)
• Math StackExchange, Tight Acute Sets
• Hugues Randriambololona, (2,1)-Separating systems beyond the probabilistic bound, Israel Journal of Mathematics, 195 (2013), 171-186; DOI: 10.1007/s11856-012-0126-9.
• Hugues Randriambololona, (2,1)-Separating systems beyond the probabilistic bound, arXiv:1010.5764 [math.CO], 2010-2012.

Crossrefs

Cf. A289972.

Formula

If k <= m <= n, a(k+2m) >= a(k)a(m), a(k+2m+3n) >= a(k)a(m)a(n).

a(n) >= 2*floor((sqrt(6)/9)(2/sqrt(3))^n), which is approximately 0.544*1.155^n.

Extensions

Edited by N. J. A. Sloane, Mar 29 2016 and Apr 08 2016

a(10) from Fausto A. C. Cariboni, Jul 17 2017