Certificate of impossibility of Hilbert-Artin representation of - PowerPoint PPT Presentation

Certificate of impossibility of Hilbert-Artin representation of given degree for definite polynomials and functions Feng Guo Key Laboratory of Mathematics Mechanization Chinese Academy of Sciences, Beijing North Carolina State University Joint work with Erich Kaltofen and Lihong Zhi

2 Polynomial Optimization Problem Unconstrained polynomial minimization problem f ∗ def = inf { f ( ξ ) | ξ ∈ R n } where f ∈ R [ ¯ X ] = R [ X 1 ,..., X n ] . The problem is equivalent to compute f ∗ = sup { a ∈ R | f − a ≥ 0 on R n }

3 Sums of Squares (SOS) Relaxation • If f attains a minimum on R n f − f ∗ = SOS mod I Gradient Ideal Nie/Demmel/Sturmfels, Math. Program., 2005. • Otherwise f − f ∗ = SOS mod P • Gradient Tentacle Schweighofer, SIAM J. Optim. 2006. • Tangency Variety Hà/Pham, SIAM J. Optim. 2008. • Polar Variety Guo/Safey El Din/Zhi, ISSAC 2010. ⇒ Numerical Lower Bound Semidefinite Programming (SDP) =

4 Hilbert-Artin Representation of PSD polynomials Emil Artin’s 1927 Theorem (Hilbert’s 17th Problem): ∀ ξ 1 ,..., ξ n ∈ R : f ( ξ 1 ,..., ξ n ) ≥ 0 � X ) = ∑ l i = 1 u 2 i ∃ u i , v j ∈ R [ ¯ X ] : f ( ¯ ∑ l j = 1 v 2 j

4 Hilbert-Artin Representation of PSD polynomials Emil Artin’s 1927 Theorem (Hilbert’s 17th Problem): ∀ ξ 1 ,..., ξ n ∈ R : f ( ξ 1 ,..., ξ n ) ≥ 0 � X ) = ∑ l i = 1 u 2 i ∃ u i , v j ∈ R [ ¯ X ] : f ( ¯ ∑ l j = 1 v 2 j � ∃ e ≥ 0 , W [ 1 ] � 0 , W [ 2 ] � 0 , W [ 2 ] � = 0 : X ) T W [ 2 ] m e ( ¯ X ) T W [ 1 ] m d ( ¯ f ( ¯ X ) · ( m e ( ¯ X )) = m d ( ¯ X ) where d = e +( deg f ) / 2, m e ( ¯ X ) and m d ( ¯ X ) are vectors of terms. W � 0 ⇔ P T L T DLP , D diagonal, D i , i ≥ 0 (Cholesky)

4 Hilbert-Artin Representation of PSD polynomials Emil Artin’s 1927 Theorem (Hilbert’s 17th Problem): ∀ ξ 1 ,..., ξ n ∈ R : f ( ξ 1 ,..., ξ n ) ≥ 0 � ∑ l i = 1 u 2 i ∃ u i , v j ∈ R [ ¯ X ] : f ( ¯ X ) = ∑ l j = 1 v 2 j � ∃ e ≥ 0 , W [ 1 ] � 0 , W [ 2 ] � 0 , W [ 2 ] � = 0 : X ) T W [ 2 ] m e ( ¯ X ) T W [ 1 ] m d ( ¯ f ( ¯ X ) · ( m e ( ¯ X )) = m d ( ¯ X ) where d = e +( deg f ) / 2, m e ( ¯ X ) and m d ( ¯ X ) are vectors of terms. W � 0 ⇔ P T L T DLP , D diagonal, D i , i ≥ 0 (Cholesky)

4 Hilbert-Artin Representation of PSD polynomials Emil Artin’s 1927 Theorem (Hilbert’s 17th Problem): ∀ ξ 1 ,..., ξ n ∈ R : f ( ξ 1 ,..., ξ n ) ≥ 0 � X ) = ∑ l i = 1 u 2 i ∃ u i , v j ∈ R [ ¯ X ] : f ( ¯ ∑ l j = 1 v 2 j � ∃ e ≥ 0 , W [ 1 ] � 0 , W [ 2 ] � 0 , W [ 2 ] � = 0 : X ) T W [ 1 ] m d ( ¯ X ) T W [ 2 ] m e ( ¯ f ( ¯ X ) · ( m e ( ¯ X ) ) = m d ( ¯ X ) where d = e +( deg f ) / 2, m e ( ¯ X ) and m d ( ¯ X ) are vectors of terms. W � 0 ⇔ P T L T DLP , D diagonal, D i , i ≥ 0 (Cholesky)

5 Exact Certification of PSD Polynomial � � W [ 1 ] Denote W = , then the affine linear hyperplane is W [ 2 ] � � X ) T W [ 2 ] m e ( ¯ X ) T W [ 1 ] m d ( ¯ f ( ¯ X ) · ( m e ( ¯ X )) = m d ( ¯ L = X ) W W SDP W adjust Newton iteration W Newton recover an integer or rational matrix orthogonal exact projection � W L hard � W symmetric positive semidefinite matrices L easy References: "Easy Case" Peyrl, Parrilo ’07,’08; "Hard Case" Kaltofen, Li, Yang, Zhi ’08,’09

6 Our Result ⇒ ∄ W [ 1 ] � 0 , W [ 1 ] � 0 , W [ 2 ] � = 0 , s.t. If e is too small = X ) T W [ 2 ] m e ( ¯ X ) T W [ 1 ] m d ( ¯ f ( ¯ X ) · ( m e ( ¯ X )) = m d ( ¯ X ) SDP solver in Maltab can only numerically detect it! Notation: SOS / SOS 2 e = { ∑ u 2 i / ∑ v 2 j | u i , v j ∈ R [ ¯ X ] , deg v j ≤ e } . Our result: Given integer e ≥ 0, we give exact certification if f ( ¯ X ) / ∈ SOS / SOS 2 e . Remark: More general, we can certify f ( ¯ ∈ SOS / SOS 2 e with X ) / terms in v j are restricted in a given subset of m e .

7 Reduce to Semidefinite Programming ∈ SOS / SOS 2 e if and only if ∄ W [ 1 ] � 0 , W [ 2 ] � 0, s.t. f ( ¯ X ) / Tr W [ 2 ] = 1 , e W [ 2 ] m e ) = m T d W [ 1 ] m d , X ) · ( m T f ( ¯ where d = e +( deg f ) / 2.

7 Reduce to Semidefinite Programming ∈ SOS / SOS 2 e if and only if ∄ W [ 1 ] � 0 , W [ 2 ] � 0, s.t. f ( ¯ X ) / Tr W [ 2 ] = 1 , e W [ 2 ] m e ) = m T d W [ 1 ] m d , X ) · ( m T f ( ¯ where d = e +( deg f ) / 2. For symmetric matrices C , W , define inner product as C • W = � C , W � = ∑ i ∑ c i , j w i , j = Trace ( CW ) . j Let d W [ 1 ] m d = ∑ ( G [ α ] • W [ 1 ] ) ¯ X α , m T α and e W [ 2 ] m e ) = ∑ ( H [ β ] • W [ 2 ] ) ¯ X β . X ) · ( m T − f ( ¯ β

8 Infeasibility of Semidefinite Programming Let � � � � � � G [ α ] W [ 1 ] ∗ 0 , A [ α ] = W = , A = . H [ α ] W [ 2 ] ∗ I Tr W [ 2 ] = 1 , e W [ 2 ] m e ) = m T d W [ 1 ] m d , X ) · ( m T f ( ¯ ⇓  W ∈ S k × k − C • W sup            . .  . .  . .     ( P ) A [ α ] • W     0       , W � 0 s . t . =      . .  . .       . .    A • W 1

8 Infeasibility of Semidefinite Programming Let � � � � � � G [ α ] W [ 1 ] ∗ 0 , A [ α ] = W = , A = . H [ α ] W [ 2 ] ∗ I Tr W [ 2 ] = 1 , e W [ 2 ] m e ) = m T d W [ 1 ] m d , X ) · ( m T f ( ¯ ⇓  W ∈ S k × k − C • W sup            . .  . .  . .     ( P ) A [ α ] • W     0       , W � 0 s . t . =      . .  . .       . .    A • W 1 f ( ¯ ∈ SOS / SOS 2 e if and only if SDP ( P ) is infeasible. X ) /

9 Dual Problem The dual problem of SDP ( P ) :  s ∗ = ( y , s ) ∈ R m + 1 s inf   ( D ) s . t . M ( y , s ) � 0 where � � M d ( y ) M ( y , s ) = , M e (( − f ) y )+ sI 2 d ∈ R N n 2 d and m = # N n y = ( y α ) α ∈ N n 2 d , moment matrix M d ( y ) and localizing moment matrix M e (( − f ) y ) are indexed by N n 2 d . For α , β ∈ N n 2 d , M d ( y )[ α , β ] = y α + β M e (( − f ) y )[ α , β ] = − ∑ f γ y α + β + γ γ

10 Semidefinite Farkas’ Lemma Standard SDP: y ∈ R l b T y W ∈ S n × n − C • W sup inf s . t . A i • W = b i , i = 1 ··· l , l ∑ s . t . C + y i A i � 0 , W � 0 . i = 1 Lemma. [Jon Dattorro2005] If the set of vectors   A 1 • W   . .  , ∀ W � 0  . A l • W is closed, then primal problem is feasible if and only if ∀ y ∈ R l s.t. ∑ l i = 1 y i A i � 0 , we have b T y ≥ 0 .

11 Closedness of the Convex Cone Lemma. [Robinson73] The cone SOS 2 e is convex and closed. Lemma. In our case, the subset of vector space   . . .   A [ α ] • W     , ∀ W � 0 ,   . .   . A • W is closed. ⇒ Assumption in Farkas’ Lemma is satisfied! =

12 Main Contribution Theorem. The following are equivalent: ∈ SOS / SOS 2 e , 1. f / 2. ∃ feasible point ( y , s ) ∈ R m + 1 of ( D ) , s.t. s < 0 , 3. ∃ rational vector y ′ = ( y ′ α ) ∈ Q m , s.t., M d ( y ′ ) � 0 , M e ( fy ′ ) ≺ 0 . Proof: Semidefinite Farkas’ Lemma + Dual problem ( D ) is strictly feasible ⇓ ∃ rational vector y ′ = ( y ′ α ) ∈ Q m

13 Interpretation by Linear Forms on R [ ¯ X ] X ]) ∗ by Given y = ( y α ) ∈ R N n , define the linear form L y ∈ ( R [ ¯ L y ( f ) = y T CoeffVec ( f ) = ∑ for f = ∑ X α ∈ R [ ¯ f α ¯ y α f α , X ] . α α For u ( ¯ X ) , v ( ¯ X ) ∈ R [ ¯ X ] , we have L y ( u 2 ) = CoeffVec ( u ) T M ( y ) CoeffVec ( u ) L y ( fv 2 ) = CoeffVec ( v ) T M ( fy ) CoeffVec ( v )

14 Interpretation by Linear Forms on R [ ¯ X ] Theorem. The following are equivalent: ∈ SOS / SOS 2 e , 1. f / 2. ∃ y ′ ∈ Q m , s.t. ∀ v , u ∈ R [ ¯ X ] with deg v ≤ e, deg u ≤ e +( deg f ) / 2 , we have L y ′ ( u 2 ) ≥ 0 and L y ′ ( fv 2 ) < 0 . If f = ∑ u 2 i / ∑ v 2 j with deg v j ≤ e , then i ) = ∑ L y ′ ( fv 2 0 ≤ L y ′ ( ∑ u 2 j ) < 0 which is a contradiction.

15 Special Case: e = 0 X ) � = ∑ X ) 2 , u i ( ¯ ⇒ Certify f ( ¯ u i ( ¯ X ) ∈ R [ ¯ e = 0 = X ] i Theorem. The following are equivalent: X ) � = ∑ i u i ( ¯ 1. f ( ¯ X ) 2 , u i ( ¯ X ) ∈ R [ ¯ X ] , 2. ∃ y ′ ∈ Q m , s.t. ∀ u ∈ R [ ¯ X ] with deg u ≤ ( deg f ) / 2 . we have L y ′ ( u 2 ) ≥ 0 and L y ′ ( f ) < 0 . y ′ → separating hyperplane . Ahmadi and Parrilo 2009:

16 Exploit Sparsity in SOS X α , let X ) = ∑ α p α ¯ Theorem. [Reznick78] For a polynomial p ( ¯ C ( p ) be the convex hull of { α | p α � = 0 } . If f = ∑ i g 2 i then C ( g i ) ⊆ 1 2 C ( f ) . f ∈ SOS / SOS 2 e if and only if e W [ 2 ] m e ) = m T d W [ 1 ] m d . X ) · ( m T f ( ¯

16 Exploit Sparsity in SOS X α , let X ) = ∑ α p α ¯ Theorem. [Reznick78] For a polynomial p ( ¯ C ( p ) be the convex hull of { α | p α � = 0 } . If f = ∑ i g 2 i then C ( g i ) ⊆ 1 2 C ( f ) . f ∈ SOS / SOS 2 e if and only if e W [ 2 ] m e ) = m T d W [ 1 ] m d . X ) · ( m T f ( ¯ � e W [ 2 ] m e ) = m T G W [ 1 ] m G . X ) · ( m T f ( ¯ m G ⊆ m d ⇒ Sizes of the SDPs ( P ) and ( D ) decrease.