v2009.01.01 - Convex Optimization
v2009.01.01 - Convex Optimization v2009.01.01 - Convex Optimization
526 CHAPTER 7. PROXIMITY PROBLEMS is a strictly convex quadratic in D ; 7.16 ∑ minimize d 2 ij − 2h ij d ij + h 2 ij D i,j (1289) subject to D ∈ EDM N Optimal solution D ⋆ is therefore unique, as expected, for this simple projection on the EDM cone equivalent to (1214). 7.3.1.1 Equivalent semidefinite program, Problem 3, convex case In the past, this convex problem was solved numerically by means of alternating projection. (Example 7.3.1.1.1) [130] [123] [159,1] We translate (1289) to an equivalent semidefinite program because we have a good solver: Assume the given measurement matrix H to be nonnegative and symmetric; 7.17 H = [h ij ] ∈ S N ∩ R N×N + (1262) We then propose: Problem (1289) is equivalent to the semidefinite program, for ∂ = ∆ [d 2 ij ] = D ◦D (1290) a matrix of distance-square squared, minimize − tr(V (∂ − 2H ◦D)V ) ∂ , D [ ] ∂ij d ij subject to ≽ 0 , N ≥ j > i = 1... N −1 d ij 1 (1291) D ∈ EDM N ∂ ∈ S N h 7.16 For nonzero Y ∈ S N h and some open interval of t∈R (3.2.3.0.2,D.2.3) d 2 dt 2 ‖(D + tY ) − H‖2 F = 2 trY T Y > 0 7.17 If that H given has negative entries, then the technique of solution presented here becomes invalid. Projection of H on K (1206) prior to application of this proposed technique, as explained in7.0.1, is incorrect.
7.3. THIRD PREVALENT PROBLEM: 527 where [ ∂ij d ij d ij 1 ] ≽ 0 ⇔ ∂ ij ≥ d 2 ij (1292) Symmetry of input H facilitates trace in the objective (B.4.2 no.20), while its nonnegativity causes ∂ ij →d 2 ij as optimization proceeds. 7.3.1.1.1 Example. Alternating projection on nearest EDM. By solving (1291) we confirm the result from an example given by Glunt, Hayden, et alii [130,6] who found an analytical solution to convex optimization problem (1287) for particular cardinality N = 3 by using the alternating projection method of von Neumann (E.10): ⎡ H = ⎣ 0 1 1 1 0 9 1 9 0 ⎤ ⎦ , D ⋆ = ⎡ ⎢ ⎣ 19 19 0 9 9 19 76 0 9 9 19 76 0 9 9 ⎤ ⎥ ⎦ (1293) The original problem (1287) of projecting H on the EDM cone is transformed to an equivalent iterative sequence of projections on the two convex cones (1155) from6.8.1.1. Using ordinary alternating projection, input H goes to D ⋆ with an accuracy of four decimal places in about 17 iterations. Affine dimension corresponding to this optimal solution is r = 1. Obviation of semidefinite programming’s computational expense is the principal advantage of this alternating projection technique. 7.3.1.2 Schur-form semidefinite program, Problem 3 convex case Semidefinite program (1291) can be reformulated by moving the objective function in minimize ‖D − H‖ 2 F D (1287) subject to D ∈ EDM N to the constraints. This makes an equivalent second-order cone program: for any measurement matrix H minimize t t∈R , D subject to ‖D − H‖ 2 F ≤ t (1294) D ∈ EDM N
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526 CHAPTER 7. PROXIMITY PROBLEMS<br />
is a strictly convex quadratic in D ; 7.16<br />
∑<br />
minimize d 2 ij − 2h ij d ij + h 2 ij<br />
D<br />
i,j<br />
(1289)<br />
subject to D ∈ EDM N<br />
Optimal solution D ⋆ is therefore unique, as expected, for this simple<br />
projection on the EDM cone equivalent to (1214).<br />
7.3.1.1 Equivalent semidefinite program, Problem 3, convex case<br />
In the past, this convex problem was solved numerically by means of<br />
alternating projection. (Example 7.3.1.1.1) [130] [123] [159,1] We translate<br />
(1289) to an equivalent semidefinite program because we have a good solver:<br />
Assume the given measurement matrix H to be nonnegative and<br />
symmetric; 7.17<br />
H = [h ij ] ∈ S N ∩ R N×N<br />
+ (1262)<br />
We then propose: Problem (1289) is equivalent to the semidefinite program,<br />
for<br />
∂ = ∆ [d 2 ij ] = D ◦D (1290)<br />
a matrix of distance-square squared,<br />
minimize − tr(V (∂ − 2H ◦D)V )<br />
∂ , D<br />
[ ]<br />
∂ij d ij<br />
subject to<br />
≽ 0 , N ≥ j > i = 1... N −1<br />
d ij 1<br />
(1291)<br />
D ∈ EDM N<br />
∂ ∈ S N h<br />
7.16 For nonzero Y ∈ S N h and some open interval of t∈R (3.2.3.0.2,D.2.3)<br />
d 2<br />
dt 2 ‖(D + tY ) − H‖2 F = 2 trY T Y > 0<br />
<br />
7.17 If that H given has negative entries, then the technique of solution presented here<br />
becomes invalid. Projection of H on K (1206) prior to application of this proposed<br />
technique, as explained in7.0.1, is incorrect.