10. The Positivstellensatz Basic semialgebraic sets Semialgebraic - PowerPoint PPT Presentation

10 - 1 The Positivstellensatz P. Parrilo and S. Lall 2006.06.07.01 10. The Positivstellensatz • Basic semialgebraic sets • Semialgebraic sets • Tarski-Seidenberg and quantifier elimination • Feasibility of semialgebraic sets • Real fields and inequalities • The real Nullstellensatz • The Positivstellensatz • Example: Farkas lemma • Hierarchy of certificates • Boolean minimization and the S-procedure • Exploiting structure

10 - 2 The Positivstellensatz P. Parrilo and S. Lall 2006.06.07.01 Basic Semialgebraic Sets The basic (closed) semialgebraic set defined by polynomials f 1 , . . . , f m is x ∈ R n | f i ( x ) ≥ 0 for all i = 1 , . . . , m � � Examples • The nonnegative orthant in R n • The cone of positive semidefinite matrices • Feasible set of an SDP; polyhedra and spectrahedra Properties • If S 1 , S 2 are basic closed semialgebraic sets, then so is S 1 ∩ S 2 ; i.e., the class is closed under intersection • Not closed under union or projection

10 - 3 The Positivstellensatz P. Parrilo and S. Lall 2006.06.07.01 Semialgebraic Sets Given the basic semialgebraic sets, we may generate other sets by set the- oretic operations; unions, intersections and complements. A set generated by a finite sequence of these operations on basic semial- gebraic sets is called a semialgebraic set . Some examples: • The set x ∈ R n | f ( x ) ∗ 0 � � S = is semialgebraic, where ∗ denotes <, ≤ , = , � = . • In particular every real variety is semialgebraic. • We can also generate the semialgebraic sets via Boolean logical oper- ations applied to polynomial equations and inequalities

10 - 4 The Positivstellensatz P. Parrilo and S. Lall 2006.06.07.01 Semialgebraic Sets Every semialgebraic set may be represented as either • an intersection of unions p i m � x ∈ R n | sign f ij ( x ) = a ij � � � S = where a ij ∈ {− 1 , 0 , 1 } i =1 j =1 • a finite union of sets of the form x ∈ R n | f i ( x ) > 0 , h j ( x ) = 0 for all i = 1 , . . . , m, j = 1 , . . . , p � � • in R , a finite union of points and open intervals Every closed semialgebraic set is a finite union of basic closed semialgebraic sets; i.e., sets of the form x ∈ R n | f i ( x ) ≥ 0 for all i = 1 , . . . , m � �

10 - 5 The Positivstellensatz P. Parrilo and S. Lall 2006.06.07.01 Properties of Semialgebraic Sets • If S 1 , S 2 are semialgebraic, so is S 1 ∪ S 2 and S 1 ∩ S 2 • The projection of a semialgebraic set is semialgebraic • The closure and interior of a semialgebraic sets are both semialgebraic • Some examples: Sets that are not Semialgebraic Some sets are not semialgebraic; for example ( x, y ) ∈ R 2 | y = e x � � • the graph ( x, y ) ∈ R 2 | y = ⌊ x ⌋ � � • the infinite staircase • the infinite grid Z n

10 - 6 The Positivstellensatz P. Parrilo and S. Lall 2006.06.07.01 Tarski-Seidenberg and Quantifier Elimination Tarski-Seidenberg theorem: if S ⊂ R n + p is semialgebraic, then so are x ∈ R n | ∃ y ∈ R p ( x, y ) ∈ S � � • (closure under projection) x ∈ R n | ∀ y ∈ R p ( x, y ) ∈ S � � • (complements and projections) i.e., quantifiers do not add any expressive power Cylindrical algebraic decomposition (CAD) may be used to compute the semialgebraic set resulting from quantifier elimination

10 - 7 The Positivstellensatz P. Parrilo and S. Lall 2006.06.07.01 Feasibility of Semialgebraic Sets Suppose S is a semialgebraic set; we’d like to solve the feasibility problem Is S non-empty? More specifically, suppose we have a semialgebraic set represented by poly- nomial inequalities and equations � x ∈ R n | f i ( x ) ≥ 0 , h j ( x ) = 0 for all i = 1 , . . . , m, j = 1 , . . . , p � S = • Important, non-trivial result: the feasibility problem is decidable . • But NP-hard (even for a single polynomial, as we have seen) • We would like to certify infeasibility

10 - 8 The Positivstellensatz P. Parrilo and S. Lall 2006.06.07.01 Certificates So Far • The Nullstellensatz: a necessary and sufficient condition for feasibility of complex varieties x ∈ C n | h i ( x ) = 0 ∀ i � � = ∅ ⇐ ⇒ − 1 ∈ ideal { h 1 , . . . , h m } • Valid inequalities: a sufficient condition for infeasibility of real basic semialgebraic sets x ∈ R n | f i ( x ) ≥ 0 ∀ i � � = ∅ ⇐ − 1 ∈ cone { f 1 , . . . , f m } = • Linear Programming: necessary and sufficient conditions via duality for real linear equations and inequalities

10 - 9 The Positivstellensatz P. Parrilo and S. Lall 2006.06.07.01 Certificates So Far Degree \ Field Complex Real Linear Range/Kernel Farkas Lemma Linear Algebra Linear Programming Polynomial Nullstellensatz ???? Bounded degree: LP ???? Groebner bases We’d like a method to construct certificates for • polynomial equations • over the real field

10 - 10 The Positivstellensatz P. Parrilo and S. Lall 2006.06.07.01 Real Fields and Inequalities If we can test feasibility of real equations then we can also test feasibility of real inequalities and inequations , because • inequalities: there exists x ∈ R such that f ( x ) ≥ 0 if and only if there exists ( x, y ) ∈ R 2 such that f ( x ) = y 2 • strict inequalities: there exists x such that f ( x ) > 0 if and only if there exists ( x, y ) ∈ R 2 such that y 2 f ( x ) = 1 • inequations: there exists x such that f ( x ) � = 0 if and only if there exists ( x, y ) ∈ R 2 such that yf ( x ) = 1 The underlying theory for real polynomials called real algebraic geometry

10 - 11 The Positivstellensatz P. Parrilo and S. Lall 2006.06.07.01 Real Varieties The real variety defined by polynomials h 1 , . . . , h m ∈ R [ x 1 , . . . , x n ] is x ∈ R n | h i ( x ) = 0 for all i = 1 , . . . , m � � V R { h 1 , . . . , h m } = We’d like to solve the feasibility problem; is V R { h 1 , . . . , h m } � = ∅ ? We know • Every polynomial in ideal { h 1 , . . . , h m } vanishes on the feasible set. • The (complex) Nullstellensatz: − 1 ∈ ideal { h 1 , . . . , h m } ⇒ V R { h 1 , . . . , h m } = ∅ = • But this condition is not necessary over the reals

10 - 12 The Positivstellensatz P. Parrilo and S. Lall 2006.06.07.01 The Real Nullstellensatz Recall Σ is the cone of polynomials representable as sums of squares . Suppose h 1 , . . . , h m ∈ R [ x 1 , . . . , x n ] . − 1 ∈ Σ + ideal { h 1 , . . . , h m } ⇐ ⇒ V R { h 1 , . . . , h m } = ∅ Equivalently, there is no x ∈ R n such that h i ( x ) = 0 for all i = 1 , . . . , m if and only if there exists t 1 , . . . , t m ∈ R [ x 1 , . . . , x n ] and s ∈ Σ such that − 1 = s + t 1 h 1 + · · · + t m h m

10 - 13 The Positivstellensatz P. Parrilo and S. Lall 2006.06.07.01 Example Suppose h ( x ) = x 2 + 1 . Then clearly V R { h } = ∅ We saw earlier that the complex Nullstellensatz cannot be used to prove emptyness of V R { h } But we have − 1 = s + th with s ( x ) = x 2 t ( x ) = − 1 and and so the real Nullstellensatz implies V R { h } = ∅ . The polynomial equation − 1 = s + th gives a certificate of infeasibility.

10 - 14 The Positivstellensatz P. Parrilo and S. Lall 2006.06.07.01 The Positivstellensatz We now turn to feasibility for basic semialgebraic sets , with primal problem Does there exist x ∈ R n such that f i ( x ) ≥ 0 for all i = 1 , . . . , m h j ( x ) = 0 for all j = 1 , . . . , p Call the feasible set S ; recall • every polynomial in cone { f 1 , . . . , f m } is nonnegative on S • every polynomial in ideal { h 1 , . . . , h p } is zero on S The Positivstellensatz (Stengle 1974) S = ∅ ⇐ ⇒ − 1 ∈ cone { f 1 , . . . , f m } + ideal { h 1 , . . . , h m }

10 - 15 The Positivstellensatz P. Parrilo and S. Lall 2006.06.07.01 Example 3 Consider the feasibility problem 2 ( x, y ) ∈ R 2 | f ( x, y ) ≥ 0 , h ( x, y ) = 0 1 � � S = 0 where −1 f ( x, y ) = x − y 2 + 3 −2 −3 h ( x, y ) = y + x 2 + 2 −4 −4 −3 −2 −1 0 1 2 By the P-satz, the primal is infeasible if and only if there exist polynomials s 1 , s 2 ∈ Σ and t ∈ R [ x, y ] such that − 1 = s 1 + s 2 f + th A certificate is given by � 2 + 6 � 2 , s 1 = 1 y + 3 x − 1 � � 3 + 2 s 2 = 2 , t = − 6 . 2 6

10 - 16 The Positivstellensatz P. Parrilo and S. Lall 2006.06.07.01 Explicit Formulation of the Positivstellensatz The primal problem is Does there exist x ∈ R n such that f i ( x ) ≥ 0 for all i = 1 , . . . , m h j ( x ) = 0 for all j = 1 , . . . , p The dual problem is Do there exist t i ∈ R [ x 1 , . . . , x n ] and s i , r ij , . . . ∈ Σ such that � � � − 1 = r ij f i f j + · · · h i t i + s 0 + s i f i + i i i � = j These are strong alternatives

10 - 17 The Positivstellensatz P. Parrilo and S. Lall 2006.06.07.01 Testing the Positivstellensatz Do there exist t i ∈ R [ x 1 , . . . , x n ] and s i , r ij , . . . ∈ Σ such that � � � − 1 = r ij f i f j + · · · t i h i + s 0 + s i f i + i i i � = j • This is a convex feasibility problem in t i , s i , r ij , . . . • To solve it, we need to choose a subset of the cone to search; i.e., the maximum degree of the above polynomial; then the problem is a semidefinite program • This gives a hierarchy of syntactically verifiable certificates • The validity of a certificate may be easily checked; e.g., linear algebra, random sampling • Unless NP=co-NP, the certificates cannot always be polynomially sized.