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Finding Largest Small Polygons with Gloptipoly

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Finding Largest Small Polygons with Gloptipoly Finding largest small polygons with GloptiPoly Didier Henrion1,2,3, Frédéric Messine4 June 17, 2018 Abstract A small polygon is a convex polygon of unit diameter. We are interested in small polygons which have the largest area for a given number of vertices n. Many instances are already solved in the lit- erature, namely for all odd n, and for n = 4, 6 and 8. Thus, for even n ≥ 10, instances of this problem remain open. Finding those largest small polygons can be formulated as nonconvex quadratic pro- gramming problems which can challenge state-of-the-art global opti- mization algorithms. We show that a recently developed technique for global polynomial optimization, based on a semidefinite programming approach to the generalized problem of moments and implemented in the public-domain Matlab package GloptiPoly, can successfully find largest small polygons for n = 10 and n = 12. Therefore this signif- icantly improves existing results in the domain. When coupled with accurate convex conic solvers, GloptiPoly can provide numerical guar- antees of global optimality, as well as rigorous guarantees relying on interval arithmetic. Keywords: Extremal convex polygons, global optimization, nonconvex quadratic programming, semidefinite programming arXiv:1103.4456v1 [math.OC] 23 Mar 2011 1CNRS; LAAS; 7 avenue du colonel Roche, F-31077 Toulouse; France. [email protected] 2Université de Toulouse; UPS, INSA, INP, ISAE; UT1, UTM, LAAS; F-31077 Toulouse; France 3Faculty of Electrical Engineering, Czech Technical University in Prague, Technická 2, CZ-16626 Prague, Czech Republic 4ENSEEIHT-IRIT, 2 rue Charles Camichel, BP 7122, F-31071 Toulouse; France. [email protected] 1 1 Introduction The problem of finding the largest small polygons was first studied by Rein- hardt in 1922 [16]. He solved the problem by proving that the solution corresponds to the regular small polygons but only when the number of ver- tices n is odd. He also solved the case n = 4 by proving that a square with diagonal length equal to 1 is a solution. However, there exists an infinity of other solutions (it is just necessary that the two diagonals intersect with a right angle). The hexagonal case n = 6 was solved numerically by Graham in 1975 [7]. Indeed, he studied possible structures that the optimal solu- tion must have. He introduced the diameter graph of a polygon which is defined by the same vertices as the polygon and by edges if and only if the corresponding two vertices of the edge are at distance one. Using a result due to Woodall [17], he proved that the diameter graph of the largest small polygons must be connected, yielding 10 distinct possible configurations for n =6. Discarding 9 of these 10 possibilities (by using standard geometrical reasonings plus the fact that all the candidates must have an area greater than the regular small hexagon), he determined the only possible diameter graph configuration which can provide a better solution than the regular one. He solved this last case numerically, yielding the largest small hexagon. The name of this corresponding optimal hexagon is Graham’s little hexagon. Following the same principle, Audet et al. in 2002 found the largest small octagon [4]. The case n = 8 is much more complicated than the case n = 6 because it generates 31 possible configurations and just a few of them can be easily discarded by geometrical reasonings. Furthermore, for the remaining cases, Audet et al. had to solve difficult global optimization problems with 10 variables and about 20 constraints. In [4], these problems are formulated as quadratic programs with quadratic constraints. For solving this program, Audet et al. used a global solver named QP [1]. Notice that optimal solutions for n = 6 and n = 8 are not the regular polygons [4, 7]. At the Toulouse Global Optimization Workshop TOGO 2010, the corresponding optimal oc- tagon was named Hansen’s little octagon. The following cases n = 10, 12,... were open. However in 1975, Graham proposed a conjecture which is the following: when n is even and n ≥ 4, the largest small polygon must have a diameter graph with a cycle with n − 1 vertices and with an additional edge attached to a vertex of the cycle; this is true for n =4, 6 and also n =8, see Figure 1 for an example of this optimal structure for the octagon. Therefore, this yields only one possible diameter graph configuration that must have the optimal shape. In 2007, Foster and Szabo proved Graham’s conjecture [6]. Thus to solve the following open cases n ≥ 10 and n even, it is just necessary to solve one global optimization problem defined by the configuration of the 2 diameter graph with a cycle with n − 1 vertices and an additional pending edge. In order to have an overview of these and other related subjects about polygons, refer to [2, 3]. In this paper, using the Global Optimization software GloptiPoly [9], we solve the two open cases n = 10 and n = 12. Invariance of the quadratic programming problem under a group of permutation suggests that small polygons have a symmetry axis. Exploiting this symmetry allows to reduce signficantly the size of optimization problems to be solved, even though cur- rently we are not able to prove that this is without loss of generality. In Section 2, the general quadratic formulation is presented. Then, we show that the quadratic problem is invariant under a group of permutations, which suggests an important reduction of the size of the quadratic programs. In Section 3 GloptiPoly is presented and an example of its use is given to solve the case of the octagon yielding a rigorous certificate of the global optimum found in [4]. In Section 4, we present the solutions for the largest small decagon (n = 10) and dodecagon (n = 12), and conjecture the solutions for the tetradecagon (n = 14) and the hexadecagon (n = 16). Then, we conclude in Section 5. 2 Nonconvex Quadratic Optimization Problems As mentioned above, for even n ≥ 4, finding the largest small polygon with n vertices amounts to solving only one global optimization problem [6], namely a nonconvex quadratic programming problem. This formulation was previ- ously introduced by Audet et al. in [4] for the octagon case. On Figure 1, we recall the nomenclature and configuration corresponding to the octagon (n =8). 1 max A8 = x1 + {(x2 + x3 − 4x1)y1 + (3x1 − 2x3 + x5)y2 x,y 2 +(3x1 − 2x2 + x4)y3 +(x3 − 2x1)y4 +(x2 − 2x1)y5} 2 s.t. kvi − vjk ≤ 1 i, j =1,..., 8 2 kv2 − v6k =1 (1) x2 + y2 =1 i =1,..., 5 i i yi ≥ 0 i =1,..., 5 ≤ x ≤ 1 0 1 2 0 ≤ xi ≤ 1 i =2,..., 5. Without loss of generality we can insert the additional constraint x2 ≥ x3 which eliminates a symmetry axis. In program (1), all the constraints are 3 v4 = (0, 1) • v3 = (−x1,y1) • • v5 = (x1,y1) v2 = (−x1 + x3 − x5, v6 = (x1 − x2 + x4, y1 − y3 + y5) • • y1 − y2 + y4) • • v1 = (x1 − x2,y1 − y2) v7 = (x3 − x1,y1 − y3) • v8 = (0, 0) Figure 1: Case of n =8 vertices. Definition of variables following Graham’s conjecture. quadratic. The quadratic objective function corresponds to the computation of the area of the octagon following Graham’s diameter graph configuration. To generalize this quadratic formulation, we have to define two vertices n vn = (0, 0) and v 2 = (0, 1) and n − 2 other vertices vi with the help of the following variables: k k i i uk = x1 + (−1) x2i,y1 + (−1) y2i , i=1 i=1 ! k X k X i+1 i wk = (−1) x2i+1, (−1) y2i+1 , i=0 i=0 ! n X X with k = 0, 1,..., 2 − 2. On Figure 2, we represent the four first vertices, namely (0, 0), (−x1,y1), (0, 1), (x1,y1). Note that we have an axis of symme- try from the line passing through (0, 0) and (0, 1), i.e. this symmetry provides the maximal area for this quadrilateral polygon and its corresponding value is equal to x1. Thus, from vertex u0 = (x1,y1), we construct iteratively all the vertices ui following a path in the graph of diameter and we do the same from w0 =(−x1,y1) for the vertices wi, see Figure 2. The main idea using this notation is that it yields: k+1 k+1 2 2 kuk+1 − ukk = (−1) x2(k+1), (−1) y2(k+1) = x2(k+1) + y2(k+1) =1. 2 2 We obtain by the same way that kwk+1 − wkk = x + y = 1. 2(k+1)+1 2(k+1)+1 This allows to eliminate all the variables yi, reducing by half the size of the problem. 4 (0, 1) (0, 1) (0, 1) • • u0 w0 • (−x1,y1•) •(x1,y1) • • u2 w2 w3 u3 • • • • u1 • • u4 • w1 (0•, 0) (0•, 0) (0•, 0) Figure 2: Construction of the General Quadratic Formulation. Then, we rename the vertices by vi =(xi, yi): n−2 vi := u2i−1 i =1,..., ⌊ 4 ⌋ n−2 vn−i := w2i−1 i =1,..., ⌊ 4 ⌋ n−2 v n := w i =1, ··· , ⌈ ⌉ 2 −i 2(i−1) 4 n−2 v n u i , , 2 +i := 2(i−1) =1 ··· ⌈ 4 ⌉ The general quadratic program can be written as follows: n−2 1 max An = x1 + (yixi+1 − xiyi+1) x,y 2 i=1 n n i6= 2X−1,i6= 2 2 s.t. kvi − vjk ≤ 1 i, j =1,...,n 2 kv⌊ n ⌋ − v 3n k =1 (2) 4 ⌈ 4 ⌉ 2 2 xi + yi =1 i =1,...,n − 3 yi ≥ 0 i =1,...,n − 3 1 0 ≤ x1 ≤ 2 0 ≤ xi ≤ 1 i =1,...,n − 3 where xi and yi are linear functions respectively depending on xi and yi.
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