v2009.01.01 - Convex Optimization
v2009.01.01 - Convex Optimization v2009.01.01 - Convex Optimization
314 CHAPTER 4. SEMIDEFINITE PROGRAMMING constraints in (711). (confer (707)) When a permutation A of vector B exists, number of iterations can be as small as 1. But this combinatorial Procrustes problem can be made even more challenging when vector A has repeated entries. solution via cardinality constraint Now the idea is to force solution at a vertex of permutation polyhedron (91) by finding a solution of desired sparsity. Because permutation matrix X is n-sparse by assumption, this combinatorial Procrustes problem may instead be formulated as a compressed sensing problem with convex iteration on cardinality of vectorized X (4.5.1): given nonzero vectors A, B minimize X∈R n×n ‖A − XB‖ 1 + w〈X , Y 〉 subject to X T 1 = 1 X1 = 1 X ≥ 0 (714) where direction vector Y is an optimal solution to minimize Y ∈R n×n 〈X ⋆ , Y 〉 subject to 0 ≤ Y ≤ 1 1 T Y 1 = n 2 − n (467) each a linear program. In this circumstance, use of closed-form solution for direction vector Y is discouraged. When vector A is a permutation of B , both linear programs have objectives that converge to 0. When vectors A and B are permutations and no entries of A are repeated, optimal solution X ⋆ can be found as soon as the first iteration. In any case, X ⋆ = Ξ is a permutation matrix. 4.6.0.0.4 Exercise. Combinatorial Procrustes constraints. Assume that the objective of semidefinite program (711) is 0 at optimality. Prove that the constraints in program (711) are necessary and sufficient to produce a permutation matrix as optimal solution. Alternatively and equivalently, prove those constraints necessary and sufficient to optimally produce a nonnegative orthogonal matrix.
4.6. CARDINALITY AND RANK CONSTRAINT EXAMPLES 315 4.6.0.0.5 Example. Tractable polynomial constraint. The ability to handle rank constraints makes polynomial constraints (generally nonconvex) transformable to convex constraints. All optimization problems having polynomial objective and polynomial constraints can be reformulated as a semidefinite program with a rank-1 constraint. [247] Suppose we require 3 + 2x − xy ≤ 0 (715) Assign ⎡ ⎤ x [x y 1] G = ⎣ y ⎦ = 1 [ X z z T 1 ] ∆ = ⎡ ⎤ x 2 xy x ⎣ xy y 2 y ⎦∈ S 3 (716) x y 1 The polynomial constraint (715) is equivalent to the constraint set (B.1.0.2) tr(GA) ≤ 0 [ X z G = z T 1 rankG = 1 ] (≽ 0) (717) in symmetric variable matrix X ∈ S 2 and variable vector z ∈ R 2 where ⎡ ⎤ 0 − 1 1 2 A = ⎣ − 1 0 0 ⎦ (718) 2 1 0 3 Then the method of convex iteration from4.4.1 is applied to implement the rank constraint. 4.6.0.0.6 Example. Boolean vector feasible to Ax ≼ b. (confer4.2.3.1.1) Now we consider solution to a discrete problem whose only known analytical method of solution is combinatorial in complexity: given A∈ R M×N and b∈ R M find x ∈ R N subject to Ax ≼ b δ(xx T ) = 1 (719) This nonconvex problem demands a Boolean solution [x i = ±1, i=1... N ].
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4.6. CARDINALITY AND RANK CONSTRAINT EXAMPLES 315<br />
4.6.0.0.5 Example. Tractable polynomial constraint.<br />
The ability to handle rank constraints makes polynomial constraints<br />
(generally nonconvex) transformable to convex constraints. All optimization<br />
problems having polynomial objective and polynomial constraints can be<br />
reformulated as a semidefinite program with a rank-1 constraint. [247]<br />
Suppose we require<br />
3 + 2x − xy ≤ 0 (715)<br />
Assign<br />
⎡ ⎤<br />
x [x y 1]<br />
G = ⎣ y ⎦ =<br />
1<br />
[ X z<br />
z T 1<br />
]<br />
∆<br />
=<br />
⎡ ⎤<br />
x 2 xy x<br />
⎣ xy y 2 y ⎦∈ S 3 (716)<br />
x y 1<br />
The polynomial constraint (715) is equivalent to the constraint set (B.1.0.2)<br />
tr(GA) ≤ 0<br />
[ X z<br />
G =<br />
z T 1<br />
rankG = 1<br />
]<br />
(≽ 0)<br />
(717)<br />
in symmetric variable matrix X ∈ S 2 and variable vector z ∈ R 2 where<br />
⎡ ⎤<br />
0 − 1 1<br />
2<br />
A = ⎣ − 1 0 0 ⎦ (718)<br />
2<br />
1 0 3<br />
Then the method of convex iteration from4.4.1 is applied to implement the<br />
rank constraint.<br />
<br />
4.6.0.0.6 Example. Boolean vector feasible to Ax ≼ b. (confer4.2.3.1.1)<br />
Now we consider solution to a discrete problem whose only known analytical<br />
method of solution is combinatorial in complexity: given A∈ R M×N and<br />
b∈ R M find x ∈ R N<br />
subject to Ax ≼ b<br />
δ(xx T ) = 1<br />
(719)<br />
This nonconvex problem demands a Boolean solution [x i = ±1, i=1... N ].