v2010.10.26 - Convex Optimization

v2010.10.26 - Convex Optimization v2010.10.26 - Convex Optimization

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412 CHAPTER 5. EUCLIDEAN DISTANCE MATRIXx 3x 4x 5x 6Figure 119: Arbitrary hexagon in R 3 whose vertices are labelled clockwise.x 1x 2zero distance between unknowns is revealed. An optimal solution to (924)gravitates toward gradient discontinuities (D.2.1), as in Figure 72, whereasoptimal solution to (926) is less compact in the unknowns. 5.11 5.4.2.3.3 Example. Hexagon.Barvinok [25,2.6] poses a problem in geometric realizability of an arbitraryhexagon (Figure 119) having:1. prescribed (one-dimensional) face-lengths l2. prescribed angles ϕ between the three pairs of opposing faces3. a constraint on the sum of norm-square of each and every vertex x ;ten affine equality constraints in all on a Gram matrix G∈ S 6 (912). Let’srealize this as a convex feasibility problem (with constraints written in thesame order) also assuming 0 geometric center (911):findD∈S 6 h−V DV 1 2 ∈ S6subject to tr ( )D(e i e T j + e j e T i ) 1 2 = l2ij , j−1 = (i = 1... 6) mod 6tr ( − 1V DV (A 2 i + A T i ) 2) 1 = cos ϕi , i = 1, 2, 3tr(− 1 2 V DV ) = 1−V DV ≽ 0 (927)5.11 Optimal solution to (924) has mechanical interpretation in terms of interconnectingsprings with constant force when distance is nonzero; otherwise, 0 force. Problem (926) isinterpreted instead using linear springs.

5.4. EDM DEFINITION 413Figure 120: Sphere-packing illustration from [373, kissing number].Translucent balls illustrated all have the same diameter.where, for A i ∈ R 6×6 (919)A 1 = (e 1 − e 6 )(e 3 − e 4 ) T /(l 61 l 34 )A 2 = (e 2 − e 1 )(e 4 − e 5 ) T /(l 12 l 45 )A 3 = (e 3 − e 2 )(e 5 − e 6 ) T /(l 23 l 56 )(928)and where the first constraint on length-square l 2 ij can be equivalently writtenas a constraint on the Gram matrix −V DV 1 via (921). We show how to2numerically solve such a problem by alternating projection inE.10.2.1.1.Barvinok’s Proposition 2.9.3.0.1 asserts existence of a list, correspondingto Gram matrix G solving this feasibility problem, whose affine dimension(5.7.1.1) does not exceed 3 because the convex feasible set is bounded bythe third constraint tr(− 1 V DV ) = 1 (915).25.4.2.3.4 Example. Kissing number of sphere packing.Two nonoverlapping Euclidean balls are said to kiss if they touch. Anelementary geometrical problem can be posed: Given hyperspheres, eachhaving the same diameter 1, how many hyperspheres can simultaneouslykiss one central hypersphere? [394] The noncentral hyperspheres are allowed,but not required, to kiss.As posed, the problem seeks the maximal number of spheres K kissinga central sphere in a particular dimension. The total number of spheres isN = K + 1. In one dimension the answer to this kissing problem is 2. In twodimensions, 6. (Figure 7)

412 CHAPTER 5. EUCLIDEAN DISTANCE MATRIXx 3x 4x 5x 6Figure 119: Arbitrary hexagon in R 3 whose vertices are labelled clockwise.x 1x 2zero distance between unknowns is revealed. An optimal solution to (924)gravitates toward gradient discontinuities (D.2.1), as in Figure 72, whereasoptimal solution to (926) is less compact in the unknowns. 5.11 5.4.2.3.3 Example. Hexagon.Barvinok [25,2.6] poses a problem in geometric realizability of an arbitraryhexagon (Figure 119) having:1. prescribed (one-dimensional) face-lengths l2. prescribed angles ϕ between the three pairs of opposing faces3. a constraint on the sum of norm-square of each and every vertex x ;ten affine equality constraints in all on a Gram matrix G∈ S 6 (912). Let’srealize this as a convex feasibility problem (with constraints written in thesame order) also assuming 0 geometric center (911):findD∈S 6 h−V DV 1 2 ∈ S6subject to tr ( )D(e i e T j + e j e T i ) 1 2 = l2ij , j−1 = (i = 1... 6) mod 6tr ( − 1V DV (A 2 i + A T i ) 2) 1 = cos ϕi , i = 1, 2, 3tr(− 1 2 V DV ) = 1−V DV ≽ 0 (927)5.11 Optimal solution to (924) has mechanical interpretation in terms of interconnectingsprings with constant force when distance is nonzero; otherwise, 0 force. Problem (926) isinterpreted instead using linear springs.

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