The Difference Between 5 5 Doubly Nonnegative and Completely Positive Matrices

Transcription

1 The Difference Between 5 5 Doubly Nonnegative and Completely Positive Matrices Sam Burer and Kurt M. Anstreicher University of Iowa Mirjam Dür TU Darmstadt IMA Hot Topics Workshop, November 2008

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3 Consider the cones of n n Completely Positive (CPP) and Doubly Nonnegative (DNN) matrices: C n := {X S n : X = NN T for some N 0}, D n := {X S n : X 0, X 0}. We are interested in the difference between C n and D n, particularly for n = 5.

4 Consider the cones of n n Completely Positive (CPP) and Doubly Nonnegative (DNN) matrices: C n := {X S n : X = NN T for some N 0}, D n := {X S n : X 0, X 0}. We are interested in the difference between C n and D n, particularly for n = 5. Why Bother?

5 Consider the cones of n n Completely Positive (CPP) and Doubly Nonnegative (DNN) matrices: C n := {X S n : X = NN T for some N 0}, D n := {X S n : X 0, X 0}. We are interested in the difference between C n and D n, particularly for n = 5. Why Bother? [Burer 2007] Indefinite quadratic programming, with or without binary variables, can be exactly formulated as a linear optimization problem over C n.

6 Consider the cones of n n Completely Positive (CPP) and Doubly Nonnegative (DNN) matrices: C n := {X S n : X = NN T for some N 0}, D n := {X S n : X 0, X 0}. We are interested in the difference between C n and D n, particularly for n = 5. Why Bother? [Burer 2007] Indefinite quadratic programming, with or without binary variables, can be exactly formulated as a linear optimization problem over C n. Optimization over D n is tractable, and clearly C n D n.

7 Consider the cones of n n Completely Positive (CPP) and Doubly Nonnegative (DNN) matrices: C n := {X S n : X = NN T for some N 0}, D n := {X S n : X 0, X 0}. We are interested in the difference between C n and D n, particularly for n = 5. Why Bother? [Burer 2007] Indefinite quadratic programming, with or without binary variables, can be exactly formulated as a linear optimization problem over C n. Optimization over D n is tractable, and clearly C n D n. Known that C n = D n for n 4, but not for n 5.

8 Extreme Rays of D 5 For a symmetric matrix X let G(X) be the undirected graph on vertices {1,..., n} induced by the nonzero off-diagonal elements of X. A complete characterization of extreme rays of D 5 based on rank(x) and G(X) is known. Theorem 1. [Hamilton-Jester and Li, 1996] Suppose X D 5. Then X Ext(D 5 ) if and only if one of the following holds:

9 Extreme Rays of D 5 For a symmetric matrix X let G(X) be the undirected graph on vertices {1,..., n} induced by the nonzero off-diagonal elements of X. A complete characterization of extreme rays of D 5 based on rank(x) and G(X) is known. Theorem 2. [Hamilton-Jester and Li, 1996] Suppose X D 5. Then X Ext(D 5 ) if and only if one of the following holds: (i) rank(x) = 1, in which case X C 5 ;

10 Extreme Rays of D 5 For a symmetric matrix X let G(X) be the undirected graph on vertices {1,..., n} induced by the nonzero off-diagonal elements of X. A complete characterization of extreme rays of D 5 based on rank(x) and G(X) is known. Theorem 3. [Hamilton-Jester and Li, 1996] Suppose X D 5. Then X Ext(D 5 ) if and only if one of the following holds: (i) rank(x) = 1, in which case X C 5 ; (ii) rank(x) = 3 and G(X) is a 5-cycle, in which case X C 5. P T XP =

11 We call X D n \ C n a bad matrix, and X Ext(D n ), X / C n an extremely bad matrix. Let E 5 be the set of extremely bad 5 5 matrices, and B 5 = Cone(E 5 ). It follows that D 5 = C 5 + B 5.

12 We call X D n \ C n a bad matrix, and X Ext(D n ), X / C n an extremely bad matrix. Let E 5 be the set of extremely bad 5 5 matrices, and B 5 = Cone(E 5 ). It follows that D 5 = C 5 + B 5. Can show that B 5 and C 5 have a nontrivial intersection.

13 We call X D n \ C n a bad matrix, and X Ext(D n ), X / C n an extremely bad matrix. Let E 5 be the set of extremely bad 5 5 matrices, and B 5 = Cone(E 5 ). It follows that D 5 = C 5 + B 5. Can show that B 5 and C 5 have a nontrivial intersection. By Caratheodory s Theorem, know that any X D 5 can be represented using at most 16 elements from Ext(D 5 ), any number of which could be in E 5.

14 We call X D n \ C n a bad matrix, and X Ext(D n ), X / C n an extremely bad matrix. Let E 5 be the set of extremely bad 5 5 matrices, and B 5 = Cone(E 5 ). It follows that D 5 = C 5 + B 5. Can show that B 5 and C 5 have a nontrivial intersection. By Caratheodory s Theorem, know that any X D 5 can be represented using at most 16 elements from Ext(D 5 ), any number of which could be in E 5. We prove that in fact D 5 = C 5 + E 5, that is, any X D 5 \ C 5 differs from a CPP matrix by a single extremely bad matrix.

15 CPP Reduction Definition 1. A matrix X D n is CPP-reducible if there are 0 Y C n and Z D n so that X = Y + Z. If no such Y, Z exist then X is CPP-irreducible.

16 CPP Reduction Definition 1. A matrix X D n is CPP-reducible if there are 0 Y C n and Z D n so that X = Y + Z. If no such Y, Z exist then X is CPP-irreducible. Theorem 4. Let X D n. Then X is CPP-reducible iff there exists a partition (I, J) of {1,..., n} such that I, X II > 0, and there is an f Range(X) with f 0, f J = 0, f I 0.

17 CPP Reduction Definition 1. A matrix X D n is CPP-reducible if there are 0 Y C n and Z D n so that X = Y + Z. If no such Y, Z exist then X is CPP-irreducible. Theorem 4. Let X D n. Then X is CPP-reducible iff there exists a partition (I, J) of {1,..., n} such that I, X II > 0, and there is an f Range(X) with f 0, f J = 0, f I 0. For any I, conditions in Theorem 4 can be checked using the LP max e T f s. t. Xw = f, e T f 1, (1) f 0, f J = 0. Let f be an optimal solution. If e T f = 1 then X is CPPreducible, and can take Y := ε f (f ) T and Z := X ε f (f ) T, where ε = argmax{ε > 0 : X εf (f ) T D n }.

18 Algorithm 1 CPP Reduction of a DNN Matrix Input: X D n 1: Set Y 0 := 0 and Z 0 := X. 2: for k = 0, 1, 2,... do 3: Find a partition (I, J) of {1,..., n} such that: (i) I, (ii) [Z k ] II > 0, (iii) the optimal value of (1) with X replaced by Z k is one. Let f k be an optimal solution of (1). If no such partition, set Y := Y k, Z := Z k and STOP. 4: Let ε k = argmax{ε > 0 : Z k εf k fk T D n}. 5: Set Y k+1 := Y k + ε k f k fk T and Z k+1 := Z k ε k f k fk T. 6: end for 7: Set Y := Y k+1 and Z := Z k+1. Output: Y C n and Z D n with X = Y + Z.

19 Theorem 5. Algorithm 1 terminates after no more than n + n(n + 1)/2 iterations.

20 Theorem 5. Algorithm 1 terminates after no more than n + n(n + 1)/2 iterations. Algorithm 1 is not polynomial-time; number of partitions that must be checked is exponential in n.

21 Theorem 5. Algorithm 1 terminates after no more than n + n(n + 1)/2 iterations. Algorithm 1 is not polynomial-time; number of partitions that must be checked is exponential in n. In decomposition X = Y + Z produced by Algorithm 1, Z D n is CPP-irreducible.

22 Theorem 5. Algorithm 1 terminates after no more than n + n(n + 1)/2 iterations. Algorithm 1 is not polynomial-time; number of partitions that must be checked is exponential in n. In decomposition X = Y + Z produced by Algorithm 1, Z D n is CPP-irreducible. If input to Algorithm 1 is X C n, output could be X = Y + Z where 0 Z D n. In particular Algorithm 1 does not determine if a given input X is CPP.

23 Theorem 5. Algorithm 1 terminates after no more than n + n(n + 1)/2 iterations. Algorithm 1 is not polynomial-time; number of partitions that must be checked is exponential in n. In decomposition X = Y + Z produced by Algorithm 1, Z D n is CPP-irreducible. If input to Algorithm 1 is X C n, output could be X = Y + Z where 0 Z D n. In particular Algorithm 1 does not determine if a given input X is CPP. Theorem 6. Let X D 5. Then X is CPP-irreducible if and only if X is extremely bad. Moreover, if X D 5 \ C 5 then there are Y C 5 and Z E 5 so that X = Y + Z.

24 Proof of Theorem 6 uses new explicit representation for matrices in E 5 : Theorem 7. X E 5 if only if there exists a permutation matrix P, a positive-diagonal matrix Λ and a 5 3 matrix R = r 21 r r r 21 (r 21 > 0, r 22 > 0) such that P T XP = ΛRR T Λ.

25 Separating an X E 5 from C 5 We also use Theorem 7 to construct a cut that separates a given X E 5 from C 5. Let H be the Horn matrix H := H is a copositive matrix which cannot be represented as the sum of a positive semidefinite and a nonnegative matrix, i.e., H C 5 \D 5.

26 Separating an X E 5 from C 5 We also use Theorem 7 to construct a cut that separates a given X E 5 from C 5. Let H be the Horn matrix H := H is a copositive matrix which cannot be represented as the sum of a positive semidefinite and a nonnegative matrix, i.e., H C 5 \D 5. Theorem 8. Let X E 5 with P T XP = ΛRR T Λ its representation provided by Theorem 7. Define w := (P T XP H) 1 e > 0 and K := P Diag(w)H Diag(w)P T. Then K Y 0 for all Y C 5, but K X < 0.

27 Application: Computing the maximum stable set in a graph. Let A be the adacency matrix of a graph G on n vertices, and let α be the maximum size of a stable set. It is known that α 1 = min {(I + A) X : E X = 1, X C n }. (2)

28 Application: Computing the maximum stable set in a graph. Let A be the adacency matrix of a graph G on n vertices, and let α be the maximum size of a stable set. It is known that α 1 = min {(I + A) X : E X = 1, X C n }. (2) Relaxing C n to D n results in a polynomial-time computable upper bound on α: (ϑ ) 1 = min {(I + A) X : E X = 1, X D n }. (3) The bound ϑ was first established (via a different derivation) by Schrijver as a strengthening of Lovász s ϑ number.

29 Application: Computing the maximum stable set in a graph. Let A be the adacency matrix of a graph G on n vertices, and let α be the maximum size of a stable set. It is known that α 1 = min {(I + A) X : E X = 1, X C n }. (2) Relaxing C n to D n results in a polynomial-time computable upper bound on α: (ϑ ) 1 = min {(I + A) X : E X = 1, X D n }. (3) The bound ϑ was first established (via a different derivation) by Schrijver as a strengthening of Lovász s ϑ number. For the case where G is a 5-cycle have α = 2 but ϑ = 5 > 2.

30 Application: Computing the maximum stable set in a graph. Let A be the adacency matrix of a graph G on n vertices, and let α be the maximum size of a stable set. It is known that α 1 = min {(I + A) X : E X = 1, X C n }. (2) Relaxing C n to D n results in a polynomial-time computable upper bound on α: (ϑ ) 1 = min {(I + A) X : E X = 1, X D n }. (3) The bound ϑ was first established (via a different derivation) by Schrijver as a strengthening of Lovász s ϑ number. For the case where G is a 5-cycle have α = 2 but ϑ = 5 > 2. Unique solution of problem in (3) is an X E 5. Adding cut from Theorem 8 and re-solving inc reases solution value from 1/ 5 to 1/2; i.e. gap between (3) and (2) is closed.

31 Open Problem 1: Separate arbitrary X D 5 \ C 5 from C 5.

32 Open Problem 1: Separate arbitrary X D 5 \ C 5 from C 5. Know that X D 5 \ C 5 has a decomposition X = Y + Z, Y C 5, Z E 5.

33 Open Problem 1: Separate arbitrary X D 5 \ C 5 from C 5. Know that X D 5 \ C 5 has a decomposition X = Y + Z, Y C 5, Z E 5. Separation procedure described above covers the case where Y = 0.

34 Open Problem 1: Separate arbitrary X D 5 \ C 5 from C 5. Know that X D 5 \ C 5 has a decomposition X = Y + Z, Y C 5, Z E 5. Separation procedure described above covers the case where Y = 0. If Y 0, separation cannot be based on representation X = Y + Z alone, since know that X C 5 may have representation of this form with Z 0.

35 Open Problem 1: Separate arbitrary X D 5 \ C 5 from C 5. Know that X D 5 \ C 5 has a decomposition X = Y + Z, Y C 5, Z E 5. Separation procedure described above covers the case where Y = 0. If Y 0, separation cannot be based on representation X = Y + Z alone, since know that X C 5 may have representation of this form with Z 0. Open Problem 2: Give complete inner description of C 5 give complete outer description of C 5.

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