Using Representation Theorems for Proving

Using Representation Theorems for Proving Polynomials Non-negative Salvador Lucas Departamento de Sistemas Inform´ aticos y Computaci´ on (DSIC) Universidad Polit´ ecnica de Valencia, Spain http://users.dsic.upv.es/~slucas/

1

Salvador Lucas (UPV)

Using Representation Theorems

AISC’14, December 11, 2014

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Motivation

Automatic tools for program analysis

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Motivation


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Motivation


Example Consider the following Term Rewriting System (TRS) R: fact(0) 0×y x +0 p(s(x))

→ → → →

s(0) 0 x x

fact(s(x)) → s(x) × fact(p(s(x)))) s(x) × y → (x × y ) + y x + s(y ) → s(x + y )

together with the following dependency pairs [AG00] associated to R: FACT (s(x)) FACT (s(x)) s(x) ×] y

→ → →

s(x) ×] fact(p(s(x)))) P(s(x)) (x × y ) +] y

FACT (s(x)) s(x) ×] y x +] s(y )

→ → →

FACT (p(s(x))) x ×] y x +] y

During the proof of termination, we have to show that p(s(x)) & x and FACT (s(x)) = FACT (p(s(x))) for some (appropriate) quasi-ordering & and well-founded ordering = (on terms). 4




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Motivation


Abstraction. Polynomial interpretations: k-ary symbols f are given polynomial functions [f ] : Ak → A (for some numeric domain A) defined by some polynomial [f ] ∈ A[X1 , . . . , Xk ]. Terms are interpreted inductively.

Example The following polynomial interpretation (where A = [0, +∞)) [p](x) =

1 2x

[s](x) = 2x +

1 2

[FACT ](x) = 2x

can be used to prove the previous requirements (for all x ∈ A): x [p(s(x))] = x + 41 ≥ [FACT (s(x))] = 4x + 1 > 2x +

3 4

= [x] = [FACT (p(s(x)))]

Existing methods for proving positive (semi)definiteness of polynomials can be used to certify the proof! 5




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Motivation


Types Cond AC Built-in Strat Loop Rec

⇓ Modeling

Certificate

⇑

Proof Jumps Tr. Pred. Dep. Pairs Modules

⇓ Symbolic constraints s=t Abstraction

s=t

⇓

s&t

Vn

i=1 ci

Certificate

⇒ s ./ t

· · · PL

· · · MT

⇑

M ⊕N

· · · SC

⇑ Certificate

Ps,t > 0 Ps,t = 0· · · Pol

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Motivation



⇓ Modeling

Certificate

⇑


⇓ Symbolic constraints s=t Abstraction

s=t

⇓

s&t

Vn

i=1 ci

Certificate

⇒ s ./ t

· · · PL

· · · MT

⇑

M ⊕N

· · · SC

⇑ Certificate

Ps,t > 0 Ps,t = 0· · · Pol But, how do we obtain the ‘nice’ polynomials used above? 6




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Motivation


Automation. Parametric polynomials: monomial coefficients are parameters whose value is given by means of constraint solving processes.

Example (1) With the following parametric polynomial interpretation [p](x) = p1 x + p0

[s](x) = s1 x + s0

[FACT ](x) = F1 x + F0

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Motivation




[s](x) = s1 x + s0

[FACT ](x) = F1 x + F0

(2) the expressions of the symbolic constraints are interpreted as follows: [p(s(x))] = p1 s1 x + p1 s0 + p0 ≥ x = [x] [FACT (s(x))] = F1 s1 x + F1 s0 + F0 > F1 p1 s1 x + F1 p1 s0 + F1 p0 + F0 = [FACT (p(s(x)))]

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Motivation




[s](x) = s1 x + s0

[FACT ](x) = F1 x + F0


and (3) then written as ∃∀-sentences (of, e.g., Tarski’s FOLRCF): ∃pi , si ∈ C ∀x ∈ A, (p1 s1 − 1)x + p1 s0 + p0 ∃Fi , pi , si ∈ C ∀x ∈ A, (F1 s1 − F1 p1 s1 )x + F1 s0 + F0 − F1 p1 s0 − F1 p0 − F0

≥ >

0 0

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Motivation




[s](x) = s1 x + s0

[FACT ](x) = F1 x + F0



≥ >

0 0

which (4) if A, C ⊆ [0, +∞) are equivalent to the existential constraints: p1 s1 − 1 ≥ 0

p1 s0 + p0 ≥ 0

F1 s1 − F1 p1 s1 ≥ 0

F1 s0 + F0 − F1 p1 s0 − F1 p0 − F0 > 0 7




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Motivation




[s](x) = s1 x + s0

[FACT ](x) = F1 x + F0



≥ >

0 0

which (4) if A, C ⊆ [0, +∞) are equivalent to the existential constraints: p1 s1 − 1 ≥ 0

p1 s0 + p0 ≥ 0

F1 s1 − F1 p1 s1 ≥ 0

F1 s0 + F0 − F1 p1 s0 − F1 p0 − F0 > 0

that (5) can be solved: p1 = 21 , p0 = 0, s1 = 2, s0 = 12 , F1 = 2, F0 = 0. Salvador Lucas (UPV)



7

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Motivation



⇓ Modeling

Certificate

⇑


⇓ Symbolic constraints s=t

s=t

s&t

Vn

Certificate

i=1 ci

⇒ s ./ t

⇓ Abstraction Ps,t > 0 Ps,t = 0· · · Pol

⇒

· · · MT

⇑

M ⊕N

Certificate

CSP SMT

· · · PL

· · · SC

⇑ · · · CS

Resolution 8




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Summary

Goals and plan of the talk

Our goal: 1

Consider methods for testing P(S)Dness of more general polynomials (e.g., nonlinear, possibly including negative coefficients, i.e., C 6⊆ [0, +∞)) and devise how to use them in our setting

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Summary


Our goal: 1


2

Pay attention to the use of parametric polynomials

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Summary


Our goal: 1


2


3

We contribute with a new technique

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Summary


Our goal: 1


2


3


4

Take into account the cost due to constraint solving processes

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Summary


Our goal: 1


2


3


4

Take into account the cost due to constraint solving processes

Summary: 1

A parametric vector basis for positive polynomials in one variable

2

Representation theorems

3

An SMT approach to solve polynomial constraints

4

Combined cost analysis

5

Conclusions 9




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Parametric basis for positive polynomials in one variable

Motivation

Observation In termination tools, currently implemented tests of Psd([0, +∞)) actually check (or impose, in the parametric case) that the coordinates [P]Sn ∈ Rn+1 of P are non-negative (w.r.t. the standard vector basis Sn = (1, x, . . . , x n ) for univariate polynomials P ∈ Pn of degree at most n), see [CMTU06,HJ98]. This suffices for linear polynomials, but it is not satisfactory in many cases.

Example The polynomial Q(x) = x 3 − 4x 2 + 6x + 1 is Psd([0, +∞)). However, it does not fit the previous criterion.

Idea: If we change the vector basis B = (v0 , v1 , . . . , vd ), then some polynomials P with negative coefficients could get non-negative coordinates [P]B . If the vi are Psd([0, +∞)), then P is Psd([0, +∞)) as well. 10




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Parametric basis for positive polynomials in one variable

Main result

For n ∈ N, the sequence Pn of parametric univariate polynomials: Pi (x) =

Q 2i

Pi (x) = x

j=1 (x

Q

i−1 2

− γij )2

j=1 (x

− γij )2

if i is even if i is odd

where 2 ≤ i ≤ n and γij are parameters is a vector basis for Pn , the set of univariate polynomials of degree at most n.

Theorem A polynomial P ∈ R[X ] of degree n is Psd([0, +∞)) (Pd([0, +∞))) if and only if [P]Pn ≥ ~0 (resp. [P]Pn > ~0) for some assignment of values γij ≥ 0 to the parameters in Pn .

Example Consider Q(x) = x 3 − 4x 2 + 6x + 1 and B = {1, x, x 2 , x(x − 2)2 }. Since [Q]B = (1, 2, 0, 1)T ≥ ~0, we conclude that Q is Psd([0, +∞)). Moreover, since the first coordinate is 1 > 0, Q is actually Pd([0, +∞)). 11




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Hilbert

P ∈ R[X1 , . . . , Xm ] is Psd([0, +∞)m ) iff H(X1 , . . . , Xm ) = P(X12 , . . . , Xm2 ) is Psd(Rm ). We have the following result.

Proposition (Hilbert) If P ∈ R[X ] is Psd(R), then P is a sum of two squares of polynomials. We use this result (and the previous observation) in the opposite direction: if P ∈ R[X ] can be seen as a sum of two squares (sos), then P is Psd(R). Consider H(x) = Q(x 2 ) = x 6 − 4x 4 + 6x 2 + 1 = f1 (x) + f2 (x) where fi (x) = (ai x 3 + bi x 2 + ci x + di )2 for i = 1, 2. Then, H(x) should match P2

i=1

ai2 x 6 + 2ai bi x 5 + (bi2 + 2ai ci )x 4 + 2(bi ci + ai di )x 3 + (2bi di + ci2 )x 2 + 2ci di x + di2

which amounts at solving the following equalities: P2

ai2 i + ai di i=1 bi c P 2 2 i=1 di

P2

i=1

= = =

1 0 1

P2

ai bi 2 i=1 2bi di + ci

P2

i=1

= =

0 6

P2

i=1

bi2 + 2ai ci P2 i=1 ci di

= =

−4 0

A solution (with irrational numbers) is obtained by using Mathematica. 12




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P´ olya and Szeg¨ o

Proposition (Pólya-Szegö) If P is Psd([0, +∞)), then there are sos polynomials f , g such that P(x) = f (x) + xg (x) and deg(f ), deg(xg ) ≤ deg(P). For our running example Q, we have (every univariate sos polynomial f can be written as a sum of two sos): Q(x) = x 3 − 4x 2 + 6x + 1 = f1 (x) + f2 (x) + x(g1 (x) + g2 (x)) where fi (x) = (ai x + bi )2 and gi (x) = (ci x + di )2 for i = 1, 2. Then, Q(x) = (c12 + c22 )x 3 + (a12 + a22 + 2c1 d1 + 2c2 d2 )x 2 + (2a1 b1 + 2a2 b2 + d12 + d22 )x + b12 + b22

We solve now the inequalities 1 ≥ c12 + c22

−4 ≥ a12 + a22 + 2c1 d1 + 2c2 d2

6 ≥ 2a1 b1 + 2a2 b2 + d12 + d22

1 ≥ b12 + b22

to obtain the following solution with integer coefficients only: a1 = 0

a2 = 0


b1 = 1

b2 = 0

c1 = 1

c2 = 0


d1 = −2

d2 = 1


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Karlin and Studden

Theorem (Karlin and Studden) Let P2m be a polynomial of degree 2m for some m ≥ 0 with leading coefficient a2m > 0. If P2m > 0 on [0, +∞), then there exists a unique representation m m Y Y 2 P2m (X ) = a2m (X − αj ) + βX (X − γj )2 j=1

j=2

where β > 0 and 0 = γ1 < α1 < γ2 < · · · < γm < αm < ∞. Similarly, if P2m+1 is a polynomial of degree 2m + 1 for some m ≥ 0, with leading coefficient a2m+1 > 0 and P2m+1 > 0 on [0, +∞), then there exists a unique representation m+1 m Y Y P2m+1 (X ) = a2m+1 X (X − αj )2 + β (X − γj )2 j=2

j=1

where β > 0 and 0 = α1 < γ1 < α2 < γ2 < · · · < γm < αm+1 < ∞. This representation cannot be used to prove that P is Pd([0, +∞)) by matching: P = (x − 1)2 matches it, but it is not Pd([0, +∞)). Salvador Lucas (UPV)



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Karlin and Studden

If we just require αj , β, γj ≥ 0, then if P matches Karlin and Studden’s representation, we can conclude that P is Psd([0, +∞)).

Example Since the degree of Q is odd, we let KQ (x) = x(x − α2 )2 + β(x − γ1 )2 = x 3 + (β − 2α2 )x 2 + (α22 − 2βγ1 )x + βγ12 Thus, we have the following constraints: −4 ≥ β − 2α2

1≥0

6 ≥ α22 − 2βγ1

1 ≥ βγ12

The assignment α2 = 94 , β = 41 , and γ1 =

1 2

β≥0

γ1 ≥ 0

α2 ≥ 0

solves the system.

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Gram matrices

Theorem (Gram matrices) Let P be a polynomial of degree 2m and z(X ) be the vector of all monomials X α such that |α| ≤ m. Then, P is a sum of squares in R[X ] if and only if there exists a real, symmetric, psd matrix B such that P = z(X )T Bz(X ). The problem of proving that P is SOS is translated into the problem of proving whether a matrix B is positive semidefinite. This can be done (in practice) in several ways: 1

Compute the characteristic polynomial of B and require that all its roots are non-negative [PW98]. Powers and W¨ ormann show that this 2 can be translated into a constraint solving problem involving d 2−d new parameters λi .

2

Use semidefinite programming techniques [PP08]. 16




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First quantitative comparison

In the following table, V (n) is the number of auxiliary variables which are introduced by a method and I (n) is the number of obtained (in)equalities. Method: V (n): I (n):

Hilbert 2n + 2 2n + 1

P&S 2n + 2 n+1

K&S n+1 2n + 1

Gram

Vector

(n+1)(n+2) 2

3n+2

n+

n2 −n%2 4 2 1 + n −n%2 4

Note that Hilbert yields equations.

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P&S 2n + 2 n+1

K&S n+1 2n + 1

Gram

Vector

(n+1)(n+2) 2

3n+2

n+

n2 −n%2 4 2 1 + n −n%2 4


Preliminary Analysis The most critical aspect is the number of variables V (n) which are introduced. In this sense, we have the following: 1

For proving PSDness on [0, +∞), Vector is the best choice for 1 ≤ n ≤ 9 and P&S (2) is the best choice for n ≥ 10.

2

For proving PDness on [0, +∞), Vector is the best choice for 1 ≤ n ≤ 5 and K&S is the best choice for n ≥ 6.

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P&S 2n + 2 n+1

K&S n+1 2n + 1

Gram

Vector

(n+1)(n+2) 2

3n+2

n+

n2 −n%2 4 2 1 + n −n%2 4


Preliminary Analysis The most critical aspect is the number of variables V (n) which are introduced. In this sense, we have the following: 1

For proving PSDness on [0, +∞), Vector is the best choice for 1 ≤ n ≤ 9 and P&S (2) is the best choice for n ≥ 10.

2

For proving PDness on [0, +∞), Vector is the best choice for 1 ≤ n ≤ 5 and K&S is the best choice for n ≥ 6. For a more precise cost evaluation we need to know how polynomial inequalities will be handled! 17




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Transformation rules

Solve polynomial constraints C (e.g., P ≥ 0 or P > 0, where P is written as a sum of monomials with the corresponding coefficients) by transformation into a formula of the linear arithmetic. Highly efficient Satisfiability Modulo Theories (SMT) techniques are then used to find a solution. First P ≥ 0 is transformed into `P ≥ 0: `P is obtained by replacing the nonlinear monomials M in P by new variables xM ; new atoms xM = M are added and subsequently transformed by the following linearization rules

Definition (BLO+ 12) Let C be a pure non-linear constraint and D be a finite set. The transformation rules are the following (where v is a variable): V p L1: C ∧ x = v p =⇒ C ∧ a∈D V (v = a → x = a ), pif p > 1 p L2: C ∧ x = v · w =⇒ C ∧ Va∈D (v = a → x = a · w ) L3: C ∧ x = v p · M =⇒ C ∧ a∈D (v = a → x = ap · xM ) ∧ xM = M if M is not linear and v does not occur in M 18




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Quantitative analysis

Example For instance, for 1 ≥ c12 + c22 , 1 ≥ c12 + c22 ;L0 1 ≥ xc 2 + xc 2 ∧ xc 2 = c12 ∧ xc 2 = c22 1 2 2 V1 V ;L1 1 ≥ xc 2 + xc 2 ∧ d∈D c1 = d → xc 2 = d 2 ∧ d∈D c2 = d → xc 2 = d 2 1

2

1

2

we obtain 1 + 2|D| linear formulas and 2 new variables are required. Let VP (n) be the number of parameters obtained after (1) matching a given representation and (2) the preprocessing step for linearization. Method Hilbert P&S K&S Vector

n=1 10 6 2 0

n=2 18 10 4 2

n=3 28 20 7 6

n=4 40 28 13 28

n=5 54 36 20 96

n = 10 154 94 328 758086

n = 20 504 284 78741 2.48 · 1016

n = 100 10504 5404 9.6 · 1023 6.

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Conclusions

1

2

3

4

We have considered several representation theorems in connection with the characterization of positive (semi)definiteness of univariate polynomials on [0, +∞). We focused on those results which are well-suited to deal with parametric polynomials as required by termination tools We have shown that P(S)Dness of univariate polynomials P of degree at most n can be proved by finding an instance B of a vector basis Pn such that its coordinates [P]B are non-negative The quantitative analysis of the methods shows that Polya & Szegö’s representation theorem yields the most appropriate treatment for polynomials of degree n > 6.

Future work 1

Implementation of Polya & Szeg¨ o’s approach as part of the tool mu-term

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Conclusions

1

2

3

4


Future work 1


2

Representation theorems for finite domains 20




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Conclusions

1

2

3

4


Future work 1


2

Representation theorems for finite domains

3

Use with matrix interpretations 20




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References

T. Arts and J. Giesl. Termination of Term Rewriting Using Dependency Pairs. Theoretical Computer Science, 236(1–2):133–178, 2000. C. Borralleras, S. Lucas, A. Oliveras, E. Rodr´ıguez, and A. Rubio. SAT Modulo Linear Arithmetic for Solving Polynomial Constraints. Journal of Automated Reasoning 48:107-131, 2012. E. Contejean, C. Marché, A.-P. Tom´ as, and X. Urbain. Mechanically proving termination using polynomial interpretations. Journal of Automated Reasoning, 32(4):315-355, 2006. H. Hong and D. Jakuˇs. Testing Positiveness of Polynomials. Journal of Automated Reasoning 21:23-38, 1998. H. Peyrl and P.A. Parrilo. Computing sums of squares decompositions with rational coefficients. Theoretical Computer Science 409:269-281, 2008. V. Powers and T. W¨ ormann. An algorithm for sums of squares of real polynomials. Journal of Pure and Applied Algebra127:99-104, 1998. Salvador Lucas (UPV)


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Using Representation Theorems for Proving

Using Representation Theorems for Proving

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