Better Answers to Real Questions
BBetter Answers to Real Questions
Marek Koˇsta and Thomas SturmMax-Planck-Institut f¨ur InformatikSaarbr¨ucken, Germany {mkosta,sturm}@mpi-inf.mpg.de
Andreas DolzmannLeibniz-Zentrum f¨ur InformatikSaarbr¨ucken, Germany [email protected]
January 21, 2015
Abstract
We consider existential problems over the reals. Extended quantifierelimination generalizes the concept of regular quantifier elimination byproviding in addition answers, which are descriptions of possible assign-ments for the quantified variables. Implementations of extended quantifierelimination via virtual substitution have been successfully applied to vari-ous problems in science and engineering. So far, the answers produced bythese implementations included infinitesimal and infinite numbers, whichare hard to interpret in practice. We introduce here a post-processingprocedure to convert, for fixed parameters, all answers into standard realnumbers. The relevance of our procedure is demonstrated by applicationof our implementation to various examples from the literature, where itsignificantly improves the quality of the results.
We consider existential problems over the reals. Extended quantifier eliminationgeneralizes the concept of regular quantifier elimination by providing in addi-tion answers, which are descriptions of possible assignments for the quantifiedvariables (Weispfenning, 1994b, 1997b). Implementations of extended quantifierelimination (Dolzmann and Sturm, 1997a, 1996) via virtual substitution (Loosand Weispfenning, 1993; Weispfenning, 1997a, 1994a, 1988; Dolzmann et al.,1999) have been successfully applied to various problems in science and en-gineering (Sturm and Weispfenning, 1997; Sturm, 1999b; Weispfenning, 2001;Sturm and Weber, 2008; Sturm et al., 2009; Errami et al., 2011; Weber et al.,2011; Errami et al., 2013; Sturm, 1999a).So far, the answers produced by these implementations included infinitesimaland infinite numbers, which are hard to interpret in practice. This has been ex-plicitly criticized in the literature, e.g., by Collard (2003). In the present article,we introduce a complete post-processing procedure to convert, for fixed values1 a r X i v : . [ c s . S C ] J a n f parameters, all answers into standard real numbers. We furthermore demon-strate the successful application of an implementation of our method withinRedlog (Dolzmann and Sturm, 1997a) to a number of extended quantifier elim-ination problems from the scientific literature including computational geom-etry (Sturm and Weispfenning, 1997), motion planning (Weispfenning, 2001),bifurcation analysis for models of genetic circuits and for mass action (Sturmand Weber, 2008; Sturm et al., 2009), and sizing of electrical networks (Sturm,1999a).The plan of the paper is as follows: In Section 2 we make ourselves familiarwith the concept of extended quantifier elimination. In Section 3 we give anintroduction of virtual substitution for extended quantifier elimination to theextent necessary to understand how nonstandard values enter the answers andwhat information is available for fixing them to standard values. Section 4 is thetechnical core; we describe and prove our method and illustrate it by discussingone example in detail. In Section 5 we revisit degree shifts , a successful heuristicsfor reducing the degree of quantified variables before their elimination. We re-interpret these degree shifts as quantifier eliminations by virtual substitution.This allows us in Section 6 to generalize our method to cover also possibledegree shifts during elimination. In Section 7 we revisit examples from thescientific literature where the application of extended quantifier elimination tovarious problems from planning, modeling, science, and engineering had yieldednonstandard answers. In all cases we can efficiently fix all nonstandard symbolsto standard values using our implementation of the method as it was introducedin Section 4 and generalized in Section 6. This significantly improves the qualityof the results from a practical point of view. Finally, in Section 8 we summarizeour results and discuss possible extensions of our method. For our purposes here, we restrict ourselves to existential problems ϕ ( u , . . . , u m ) = ∃ x n . . . ∃ x ψ ( x , . . . , x n , u , . . . , u m )in the Tarski language L = (0 , , + , − , · , ≤ , <, ≥ , >, =) interpreted in the Tarskialgebra ( R , , , + , − , · , ≤ , <, ≥ , >, =). As usual in algebraic model theory, thesymbol “=” and its interpretation as equality is part of first-order logic so thatit does not occur explicitly in the language L . Without loss of generality, ψ is an ∧ - ∨ -combination of atomic constraints, and we agree that all right handsides of the atomic constraints are 0. Extended quantifier elimination applied to ϕ yields an extended quantifierelimination result (EQR) β ( u ) x = e ( u ) . . . x n = e n ( u )... ... . . . ... β k ( u ) x = e k ( u ) . . . x n = e kn ( u ) . The conditions β i ( u ) are quantifier-free Tarski formulas such that R | = ϕ ←→ W ki =1 β i . In other words, W ki =1 β i is a regular quantifier elimination result for ϕ ,2nd extended quantifier elimination generalizes regular quantifier elimination.The answers e i ( u ) are terms in an extension language of the Tarski language.For a ∈ R m , if ϕ ( a ) holds, then at least one β i ( a ) holds, and so does ψ ( e i ( a ) , a ).We agree that “false” never occurs as a condition. If ϕ itself is equivalent to“false,” we possibly obtain the empty scheme [ ].As an example, consider the input formula ϕ = ∃ x ∃ yψ, ψ = ay + 3 x + 4 x ≤ a ∧ x ≥ a ≥ y. A possible extended quantifier elimination result for ϕ is given by a = 0 ∧ a + 3 ≥ y = − a − x = aa ≤ ∧ a − a − ≤ y = a x = √− a + 3 a + 4 − . From this extended quantifier elimination result we can derive a regular quan-tifier elimination result( a = 0 ∧ a + 3 ≥ ∨ ( a ≤ ∧ a − a − ≤ , which can be simplified to a ≥ ∨ a − a − ≤
0. Hence, ϕ holds if and onlyif a ≥ α , where α ≈ − . a − a − − . ≤ a and a = 0, while the second row covers α ≤ a ≤
0. Let usconsider some particular interpretations of a : • For a = 2, the condition in the first row holds and the correspondinganswers yield x = 2 and y = −
9. In fact, these three values satisfy ψ .The condition in the second row, in contrast, does not hold. If we plug a = 2 into the corresponding answers anyway, then we obtain a negativeargument for the square root, which cannot be interpreted over the reals. • For α < a = − . < − .
75, the condition in the second row holds andthe corresponding answers yield x = √ − and y = − . ψ . Now the condition in the first row does nothold. If we plug a = − . x = − . y = − . ψ . • For a = − .
5, both conditions hold and yield two different sets of val-ues satisfying ψ , viz. x = − . y = − . x = √ − , y = − . • For a = 0 only the condition in the second row holds, but the answers inthe first row happen to work as well. This shows that the conditions aresufficient but not necessary for the answers to be valid. Given ϕ ( u ) = ∃ xψ ( x, u ), we compute a finite elimination set E = (cid:8) . . . , (cid:0) γ ( u ) , e ( u ) (cid:1) , . . . , (cid:9) such that ∃ xψ ←→ _ ( γ,e ) ∈ E γ ∧ ψ [ x // e ] . (1)3n the elimination set E the e ( u ) are elimination terms substituted for thequantified variable x via a virtual substitution [ x // e ]. The γ are quantifier-freeTarski formulas serving as substitution guards . Equation (1) formally describesregular quantifier elimination of one quantifier ∃ x from ϕ . For the eliminationof several quantifiers, one assumes without loss of generality that the formula isprenex and processes the prenex quantifier block from the inside to the outside.We are now going to give an idea of the exact shape and computation of elim-ination sets that is sufficiently precise to understand our main contribution here.For a more thorough introduction into the theory of quantifier elimination byvirtual substitution we refer to the original publications by Weispfenning (1988,1994a, 1997a), Loos and Weispfenning (1993), and Dolzmann et al. (1999). InSection 3.1 we restrict ourselves to input formulas ϕ not containing any strictinequalities “ < ,” “ > ,” or “ =.” In that course it will become clear how exactly wederive extended quantifier elimination from the virtual substitution procedure.Later, in Section 3.2, we generalize to formulas containing also strict in-equalities. While with regular quantifier elimination the techniques used in thecourse of the generalization to strict inequalities remain completely transpar-ent to the user, with extended quantifier elimination they leave visible tracesin the answers by possibly introducing certain nonstandard elements, which donot have a straightforward interpretation over the reals. The purpose of thispaper is to convert these answers to real numbers, given fixed values for theparameters. Recall that our constraints are normalized such that their right hand sides are0. Assume that all occurrences of x in ϕ ( u ) = ∃ xψ ( x, u ) are at most quadratic.Consider fixed real interpretations u = a ∈ R m for all parameters. Then allconstraints in ψ ( x, a ) become univariate, and the set { c ∈ R | R | = ψ ( c, a ) } is a finite union of real intervals, where the interval endpoints are zeros of theunivariate left hand side polynomials.Our goal is to include at least one such interval endpoint into our eliminationset E : For each constraint f ( u ) x + f ( u ) x + f ( u ) %
0, with a weak relation % ∈ { = , ≤ , ≥} and discriminant ∆ = f − f f we add to E three pairs ( γ, e )as follows: (cid:18) f = 0 ∧ ∆ ≥ , − f + √ ∆2 f (cid:19) , (cid:18) f = 0 ∧ ∆ ≥ , − f − √ ∆2 f (cid:19) , (cid:18) f = 0 ∧ f = 0 , − f f (cid:19) . In order to obtain a quantifier-free equivalent for ϕ , such pairs have to be pluggedinto ϕ according to (1). To start with, note that the substitution guards γ makethe substitution terms e meaningful by ensuring that denominations are not zeroand arguments to square roots are not negative.Next, observe that our elimination terms e are not terms in the Tarski lan-guage as they contain division as well as root symbols. It is one central idea ofthe virtual substitution approach that the substitution operator does not map L -terms to L -terms but atomic L -formulas to quantifier-free L -formulas:[ x // t ] : atomic L -formulas → quantifier-free L -formulas4ote that when there is more than one quantifier, it is crucial to obtain L -formulas in (1) in order to be able to proceed.To give an impression of virtual substitution, we describe here the substitu-tion ( f = 0)[ x // g + g √ g g ] of a root expression g + g √ g g , g i ∈ Z [ u ]into an equation f = 0, where f ∈ Z [ u ][ x ] of arbitrary degree: It is easy to seethat there are g ∗ , g ∗ , g ∗ such that f (cid:18) g + g √ g g (cid:19) = g ∗ + g ∗ √ g g ∗ is again a root expression. Using this intermediate result, we transform g ∗ + g ∗ √ g g ∗ = 0 ˙= g ∗ + g ∗ √ g = 0˙= | g ∗ | = | g ∗ √ g | ∧ (cid:0) sgn( g ∗ ) = sgn( g ∗ ) ∨ sgn( g ∗ ) = sgn( g ∗ ) = 0 (cid:1) ˙= g ∗ − g ∗ g = 0 ∧ g ∗ g ∗ ≤ . Technical details and formal descriptions of virtual substitutions for all ourrelations have been given by Weispfenning (1997a).Let us now apply these ideas to extended quantifier elimination of severalexistential quantifiers via virtual substitution. Given ∃ x n . . . ∃ x ψ ( x , . . . , x n , u , . . . , u m )our intended result is a scheme as described in Section 2: β ( u ) x = e ( u ) . . . x n = e n ( u )... ... . . . ... β k ( u ) x = e k ( u ) . . . x n = e kn ( u ) . We successively apply (1) to x , . . . , x n using elimination sets E ( x , . . . , x n , u ),. . . , E n ( u ) to obtain β , . . . , β k as follows: _ ( γ n ,e n ) ∈ E n · · · _ ( γ ,e ) ∈ E γ n ∧ (cid:0) · · · ∧ (cid:0) γ ∧ ψ [ x // e ] (cid:1) · · · (cid:1) [ x n // e n ] | {z } β i ( u ) . (2)The index i of β i describes one choice (cid:0) ( γ n , e n ) , . . . , ( γ , e ) (cid:1) from the Cartesianproduct E n × · · · × E . In practice, the β i obtained this way undergo sophisti-cated simplification methods such as those described by Dolzmann and Sturm(1997b). Recall from the previous section that β i that become “false” are ig-nored. It is important to understand that for the computation of the β i we areusing exclusively virtual substitution.The corresponding e i ( u ), in contrast, are obtained from e n ( u ), e n − ( x n , u ),. . . , e ( x , . . . , x n , u ) via regular back-substitution of terms in an suitable exten-sion language of L : e i,n = e n , e i,n − = e n − [ x n /e i,n ] , e i, = e [ x /e i, ] . . . [ x n /e i,n ] . (3)5ote once more that the back-substitution possible creates objects like u + 3 p √ u − u − , which are not L -terms or even root expressions. It is thus suitable for obtain-ing the e i but not for the β i . Vice versa, virtual substitution requires atomicformulas as input. Thus it is suitable for obtaining the β i while for the e i itis not applicable at all. Virtual substitution and regular term substitution areindependent concepts, which complement each other. Let us return to the elimination of a single quantifier from ∃ xψ ( x, u ). Recallfrom the beginning of the previous Section 3.1 how we considered fixed inter-pretation u = a ∈ R m causing the set S = { c ∈ R | R | = ψ ( c, a ) } to be a finiteunion of intervals.When considering in addition strict inequalities, the intervals in S are pos-sibly open. Consequently, for a strict constraint ψ i ( x, u ) ˙= f i ( u ) x i + f i ( u ) x i + f i ( u ) % i , % i ∈ { <, >, = } , (4)contained in ψ ( x, u ) we cannot use the zeros z i ( u ) of the left hand side but needa point from inside the corresponding interval.In early versions of virtual substitution methods for the linear case, Weispfen-ning (1988) used arithmetic means ( z i + z j ) for all pairs ( ψ i , ψ j ) of strict con-straints. However, the size of the elimination set then grows quadratically inthe number of constraints, which turned out to be critical for the practical per-formance of the method (Burhenne, 1990; Loos and Weispfenning, 1993). Forthe quadratic case, observe that expressions12 − g i ± √ ∆ i g i + − g j ± p ∆ j g j ! are not root expression of the form g ∗ + g ∗ √ ∆ ∗ g ∗ , so that Weispfenning’s (1997a) virtual substitution rules sketched in the previousSection 3.1 cannot be used.The established approach for strict inequalities uses nonstandard extensionsof R : Let ε ∈ R ∗ ⊃ R be a positive infinitesimal number, i.e., 0 < ε < x for all0 < x ∈ R . Then for a strict constraint as defined in (4) we use four test points − f i ± √ ∆ i f i ± ε. As an optimization, it suffices to consider only upper bounds using − ε . Forsolution sets that are unbounded from above we have to add only one morepoint ∞ := 1 /ε ∈ R ∗ for the entire problem.6or the application of the elimination set as described in (1) and thus forthe computation of the β i with extended quantifier elimination, both ε and ∞ are equivalently translated into the Tarski language L via virtual substitution.For instance, let t be a standard term. Then( x + x − x − < x // t − ε ] ˙=( x + x − x − < ∨ ( x + x − x − ∧ (3 x + 2 x − > ∨ (3 x + 2 x − ∧ (6 x + 2 < ∨ (6 x + 2 = 0 ∧ > x // t ] . In a subsequent step, t can be virtually substituted for x as discussed in theprevious section. For understanding the principal idea, notice that 3 x + 2 x − x + 2, and 6 are the first, second, and third derivatives of x + x − x − ∞ we have, e.g.,( ax + bx + c < x // ∞ ] ˙= a < ∨ ( a = 0 ∧ b < ∨ ( a = 0 ∧ b = 0 ∧ c < . Again, precise definitions and proofs have been given by Weispfenning (1997a).When again performing regular back-substitution of terms on the side ofthe e i , nonstandard symbols cannot be removed but are propagated along theway. In the final result a single answer e i can even contain several of suchnonstandard symbols. For example, on input of ϕ = ∃ x ∃ yψ, ψ ˙= ay + 3 x + 4 x < ∧ x > y > a (5)we obtain a nonstandard extended quantifier elimination result: a + 4 < y = − a − ε − ε − x = − a − ε − a < ∧ a + 4 > y = − ε − ε x = − ε . (6)Given such answers containing nonstandard symbols, it is not hard to non-constructively prove from the elimination procedure that for fixed real interpre-tations of the parameter a there are positive real choices ε , ε ∈ R so thatthe answers satisfy ψ . Infinitesimals introduced at different stages of the elim-ination are indexed accordingly. It is noteworthy that they have to be chosendifferently in general: Fixing a = − ε has to be chosen different from ε because otherwise we would obtain y = 2 x ,which together with a = − ay + 3 x + 4 x < Proposition 1 (No Standard Answers for Unfixed Parameters) . Consider theformula ϕ = ∃ x ( a < x <
1) and the nonstandard extended quantifier eliminationresult (cid:2) a < x = 1 − ε (cid:3) . There is no standard choice e ε ∈ R such that (cid:2) a < x = 1 − e ε (cid:3) is anextended quantifier elimination result for ϕ as well.7 roof. Assume for a contradiction that e ε ∈ R is a suitable choice. Then bydefinition of extended quantifier elimination it follows for all a ∈ ] −∞ ,
1[ that a < − e ε <
1, in particular e ε >
0. On the other hand, for a = 1 − e ε wehave a ∈ ] −∞ ,
1[ and 1 − e ε < a <
1, a contradiction.
Given an extended quantifier elimination result and prescribed values for all pa-rameters, our goal is to compute answers containing only standard real numbers.For instance, given (5) and (6), and fixing a = − h true y = − x = − i . From the point-of-view of our method, it makes no difference whether theparameters are fixed after extended quantifier elimination or in advance. For thesake of a concise description, we are thus going to restrict to existential decisionproblems from now on. Recall that if the regular quantifier elimination resultis “false,” then the extended quantifier elimination result is [ ], i.e., empty. Ifthe result is “true,” then we assume for simplicity that the extended quantifierelimination result contains only one row, like (cid:2) true x = e , . . . x n = e ,n (cid:3) . (7)Recall from (3) in Section 3.1 that here the e , , . . . , e ,n have been obtainedfrom elimination terms e , . . . , e n via regular back-substitution. Our methodis going to use not the back-substituted answers but these original eliminationterms. In addition, we are going to use the substitution guards γ , . . . , γ n substituted with those elimination terms in (2). Hence the input for our methodis not an EQR like in (7) but a pre-EQR as follows: (cid:20) true x = e ( x , . . . , x n ) . . . x n − = e n − ( x n ) x n = e n () γ ( x , . . . , x n ) . . . γ n − ( x n ) γ n () (cid:21) . Lemma 2 (Semantics of Virtual Substitution) . Let ϕ ( x , . . . , x n ) be a Tarskiformula, and let a , . . . , a n ∈ R .(i) Assume that R | = ϕ [ x // x − ε ]( a , . . . , a n ). Then there is e ε ∈ R , 0 < e ε ,such that for all e ε ∈ R , 0 < e ε < e ε , we have R | = ϕ ( a − e ε, a , . . . , a n ).(ii) Assume that R | = ϕ [ x // ∞ ]( a , . . . , a n ). Then there is e a ∈ R , such thatfor all e a ∈ R , e a < e a , we have R | = ϕ ( e a , a , . . . , a n ). In particular,the set T = { a ∈ R | < a and R | = ϕ ( a, a , . . . , a n ) } is unbounded fromabove. Proof.
Consider L ε = L ∪{ ε, ∞} . Using the compactness theorem for first-orderlogic, there is a real closed field R ∗ where the interpretation of ε is a positiveinfinitesimal and ∞ = ε − . The L -restriction of R ∗ is a proper extension field of R and, by the Tarski principle, elementary equivalent to R . Formally, R ∗ | L ⊃ R and R ∗ | L ∼ = R . 8i) Using R ∼ = R ∗ | L and expanding to R ∗ it follows from R | = ϕ [ x // x − ε ]( a , . . . , a n ) that also R ∗ | = ϕ [ x // x − ε ]( a , . . . , a n ). It has been dis-cussed by Loos and Weispfenning (1993) and Weispfenning (1997a) thatfor our a , . . . , a n ∈ R the virtual substitution of ε has the followingproperty: R ∗ | = ϕ [ x // x − ε ]( a , . . . , a n ) ←→ ϕ [ x /x − ε ]( a , . . . , a n ) . It follows that R ∗ | = ϕ [ x /x − ε ]( a , . . . , a n ). Let n ∈ N \ { } . Then wecan conclude R ∗ | = ψ [ x /ε ]( a , . . . , a n ), where ψ = (0 < x ∧ nx < ∧ ϕ )[ x /x − x ] . Now we can generalize R ∗ | = ∃ x ψ ( a , . . . , a n ). Since ε does not oc-cur anymore, we restrict from R ∗ to R ∗ | L and then use the elemen-tary equivalence to obtain R | = ∃ x ψ ( a , . . . , a n ). We have just shownthat for any n ∈ N \ { } there exists a ∈ R , 0 < a < n , such that R | = ϕ ( a − a , a , . . . , a n ). It follows that inf S = 0, where S = { a ∈ R | < a and R | = ϕ ( a − a, a , . . . , a n ) } ⊆ R . On the other hand, S is a semialgebraic set and thus a finite union ofintervals and points. Hence there is e ε ∈ R , 0 < e ε , such that ]0 , e ε [ ⊆ S .(ii) The argument is essentially the same as in (i) above: We conclude that R ∗ | = ϕ [ x // ∞ ]( a , . . . , a n ). According to Loos and Weispfenning (1993)and Weispfenning (1997a) we know that R ∗ | = ϕ [ x // ∞ ]( a , . . . , a n ) ←→ ϕ [ x / ∞ ]( a , . . . , a n )so that for n ∈ N we can conclude R ∗ | = ψ [ x / ∞ ]( a , . . . , a n ), where ψ = ( n < x ∧ ϕ )[ x /x ] . Again, we generalize, restrict, and apply elementary equivalence to obtain R | = ∃ x ψ ( a , . . . , a n ). We thus know that for any n ∈ N there exists a ∈ R , n < a , such that R | = ϕ ( a , a , . . . , a n ). It follows that the set T is unbounded from above. On the other hand, T is a semialgebraic setand thus a finite union of intervals and points. Hence there is e a ∈ R , suchthat ] e a , ∞ [ ⊆ T . Lemma 3.
Consider a quantifier-free Tarski formula ψ ( x , . . . , x n ). Assumethat for each i ∈ { , . . . , n } we have a root expression e e i = a i + b i √ c i d i with a i , b i , c i , d i ∈ Z [ x i +1 , . . . , x n ]. Assume furthermore that R | = ψ −→ c i ≥ ∧ d i = 0,and let α i ∈ R be the interpretation of e i = e e i [ x i +1 / g e i +1 ] . . . [ x n / f e n ]. Then { α ∈ R | R | = ψ [ x // e e ] . . . [ x n // f e n ]( α ) } = { α ∈ R | R | = ψ ( α, α , . . . , α n ) } . Proof.
Consider L = L ∪{√ , − } and the L -expansion R of R where √ and − have the usual semantics if defined and an arbitrary but fixed value otherwise.Let ν = ( f % f ∈ Z [ x, u , . . . , u m ], % ∈ {≤ , <, ≥ , >, = , = } , and let e = a + b √ cd , where a , b , c , d ∈ Z [ u , . . . , u m ]. Let s ∈ R m such that c ( s ) (cid:62) d ( s ) = 0. Then according to Weispfenning (1997a) the following holds for s ∈ R m : R | = ( ν [ x // e ] ←→ ν [ x/e ])( s ) . For our formula ψ and all α ∈ R we obtain R | = ( ψ [ x // e e ] . . . [ x n // f e n ] ←→ ψ [ x / e e ] . . . [ x n / f e n ] ←→ ψ )( α, α , . . . , α n ) . It follows that R | = ( ψ [ x // e e ] . . . [ x n // f e n ] ←→ ψ )( α, α , . . . , α n ) for all α ∈ R . Lemma 4 (Commutation of Virtual Substitutions) . Consider a quantifier-freeTarski formula ψ ( x , . . . , x n ). Let e = a + b √ c d , with a , b , c , d ∈ Z , c (cid:62) d = 0. Furthermore, let i ∈ { , . . . , n } , let e i = a i + b i √ c i d i with a i , b i , c i , d i ∈ Z [ x i +1 , . . . , x n ], and let γ i ( x i +1 , . . . , x n ) be a quantifier-free formula suchthat R | = γ i −→ c i (cid:62) ∧ d i = 0. Then R | = ( γ i ∧ ψ )[ x i // e i ][ x // e ] ←→ ( γ i ∧ ψ )[ x // e ][ x i // e i ] . Proof.
Let L , R be as in the proof of Lemma 3. Recall the equivalence R | =( ν [ x // e ] ←→ ν [ x/e ])( s ) and observe that, on the other hand, if R | = ( c i < s )or R | = ( d i = 0)( s ) for s ∈ R n − i , then R | = ( γ i ∧ ψ ←→ false)( s ). It follows that R | = ( γ i ∧ ψ )[ x i // e i ][ x // e ] ←→ ( γ i ∧ ψ )[ x i /e i ][ x /e ] ←→ ( γ i ∧ ψ )[ x /e ][ x i /e i ] ←→ ( γ i ∧ ψ )[ x // e ][ x i // e i ] . Since virtual substitution eliminates all occurrences of √ and − , we can con-clude that R | = ( γ i ∧ ψ )[ x i // e i ][ x // e ] ←→ ( γ i ∧ ψ )[ x // e ][ x i // e i ]. Theorem 5 (Computation of Standard Answers) . Consider a closed Tarskiformula ϕ = ∃ x n . . . ∃ x ψ ( x , . . . , x n ) . Assume that (cid:20) true x = e . . . x n = e n γ . . . γ n (cid:21) is a pre-EQR for ϕ such that each e i is of one of the following forms:(a) a root expression a + b √ cd , where a , b , c , d ∈ Z [ x i +1 , . . . , x n ] ,(b) ∞ ,(c) x i +1 − ε .Then we can compute root expressions e e , . . . , f e n meeting the specification (a)above and e γ , . . . , f γ n such that the following is a pre-EQR for ϕ as well: (cid:20) true x = e e . . . x n = f e n e γ . . . f γ n (cid:21) . (8)10 roof. For the sake of the proof, we are going to show that in addition to therequired e e , . . . , f e n and e γ , . . . , f γ n we can compute real algebraic numbers α , . . . , α n for the values of e e , . . . , f e n after back-substitution. We representthese real algebraic numbers as pairs of univariate defining polynomials andopen isolating intervals with rational bounds. Given k ∈ { , . . . , n } , it sufficesto show that from (cid:20) true x = e . . . x k = e k x k +1 = g e k +1 . . . x n = f e n γ . . . γ k (cid:93) γ k +1 . . . f γ n (cid:21) and α k +1 , . . . , α n we can compute suitable e e k , f γ k , and α k . Define ϕ k ( x k , . . . , x n ) = ( γ k − ∧ · · · ∧ γ ∧ ψ )[ x // e ] . . . [ x k − // e k − ] ,ϕ k +1 ( x k +1 , . . . , x n ) = ( γ k ∧ ϕ k )[ x k // e k ] , and observe that (cid:20) true x k = e k x k +1 = g e k +1 . . . x n = f e n γ k (cid:93) γ k +1 . . . f γ n (cid:21) (9)and (cid:20) true x k +1 = g e k +1 . . . x n = f e n (cid:93) γ k +1 . . . f γ n (cid:21) (10)are pre-EQRs for ∃ x n . . . ∃ x k ϕ k and ∃ x n . . . ∃ x k +1 ϕ k +1 , respectively. On thebasis of these definitions it is sufficient for our proof to compute suitable e e k , f γ k ,and α k such that (cid:20) true x k = e e k x k +1 = g e k +1 . . . x n = f e n f γ k (cid:93) γ k +1 . . . f γ n (cid:21) is a pre-EQR for ∃ x n . . . ∃ x k ϕ k as well. We define furthermore ξ ( x k , . . . , x n ) = f γ n ∧ · · · ∧ (cid:93) γ k +1 ∧ ϕ k ,ξ ( x k ) = ξ [ x k +1 // g e k +1 ] . . . [ x n // f e n ] . Lemma 3 applied to the quantifier-free formula ξ , the root expressions g e k +1 ,. . . , f e n , and the real algebraic numbers α k +1 , . . . , α n yields { r ∈ R | R | = ξ ( r ) } = { r ∈ R | R | = ξ ( r, α k +1 , . . . , α n ) } . (11)We distinguish three cases depending on the type of e k .(a) We have e k = a k + b k √ c k d k , and γ k equals d k = 0 ∧ c k (cid:62)
0. We set e e k = e k and f γ k = γ k . Since (10) is a pre-EQR for ∃ x n . . . ∃ x k +1 ϕ k +1 and α k +1 , . . . , α n correspond to the values of g e k +1 , . . . , f e n after back-substitution, we have R | = ϕ k +1 ( α k +1 , . . . , α n ). It follows that in particular R | = γ k ( α k +1 , . . . , α n )and furthermore R | = d k ( α k +1 , . . . , α n ) = 0 and R | = c k ( α k +1 , . . . , α n ) (cid:62) α k = e e k ( α k +1 , . . . , α k ) from a k , b k , c k , d k , α k +1 ,. . . , α n . 11b) We have e k = ∞ , and γ k is “true.” Since (9) is a pre-EQR for formula ∃ x n . . . ∃ x k ϕ k we have R | = ξ [ x k // ∞ ][ x k +1 // g e k +1 ] . . . [ x n // f e n ] . Using the fact that α k +1 , . . . , α n are real algebraic numbers correspondingto the values of g e k +1 , . . . , f e n after back-substitution we conclude that R | = ξ [ x k // ∞ ]( α k +1 , . . . , α n ) . Lemma 2(ii) now guarantees that the set { r ∈ R | R | = ξ ( r, α k +1 , . . . , α n ) } is unbounded from above. Thus by (11) the set { r ∈ R | R | = ξ ( r ) } isunbounded from above as well. Using well-known bounds (Akritas, 2009)on the roots of the univariate polynomials contained in ξ , we compute asufficiently large pq ∈ Q satisfying ξ . We set e e k = p +0 √ q and constructa corresponding real algebraic number α k , and we set f γ k = true. Then R | = ( f γ k ∧ ξ )[ x k // e e k ], and n − k applications of Lemma 4 yield R | = ( f γ n ∧ · · · ∧ f γ k ∧ ϕ k )[ x k // e e k ][ x k +1 // g e k +1 ] . . . [ x n // f e n ] . (c) We have e k = x k +1 − ε , and γ k is “true.” Similarly to case (b), we observethat (9) is a pre-EQR for ∃ x n . . . ∃ x k ϕ k and obtain R | = ξ [ x k // x k +1 − ε ][ x k +1 // g e k +1 ] . . . [ x n // f e n ] , and conclude that R | = ξ [ x k // x k +1 − ε ]( α k +1 , . . . , α n ). Lemma 2(i) nowguarantees the existence of some e ε ∈ R , 0 < e ε , such that R | = ξ ( α k +1 − e ε, α k +1 , . . . , α n ) for 0 < e ε < e ε . By (11) it follows that R | = ξ ( α k +1 − e ε ) for all 0 < e ε < e ε . Therefore, afterfinitely many refinements of the isolating interval (cid:3) pq , u (cid:2) of α k +1 we obtain R | = ξ (cid:0) pq (cid:1) . We set e e k = p +0 √ q and construct a corresponding real algebraicnumber, and we set f γ k = true. Exactly as in case (b), R | = ( f γ k ∧ ξ )[ x k // e e k ],and n − k applications of Lemma 4 yield R | = ( f γ n ∧ · · · ∧ f γ k ∧ ϕ k )[ x k // e e k ][ x k +1 // g e k +1 ] . . . [ x n // f e n ] . Note that instead of using the lower bound pq one can heuristically try andfind a satisfying integer.A careful inspection of our proof reveals that in all cases f γ k = γ k . However,this is going to change in Corollary 7, which generalizes our theorem.Notice that the constructive proof of Theorem 5 suggests to recompute theintermediate quantifier elimination results ϕ k . In practice, there are argumentsfor saving these ϕ k during the quantifier elimination run. Consider, e.g., thefollowing common optimization: Whenever some ϕ k heuristically simplifies to adisjunction ϕ k, ∨ · · · ∨ ϕ k,s , then the virtual substitution procedure would treateach ϕ k,j separately, i.e., like originating from several elimination set elements.In general, in the course of the application of Theorem 5 such transformationscannot be reconstructed exclusively from the pre-EQR.12o illustrate the theorem we revisit our example given in (5) on page 7 forthe choice a = −
2. In that case ψ in (5) specializes to ψ ˙= − y + 3 x + 4 x < ∧ x > y > − . For our theorem we have to consider the following pre-EQR corresponding tothe specialization of the EQR (6) to a = − (cid:20) true y = x − ε x = h − ε h = 0true true true (cid:21) . (12)Notice the introduction of an artificial variable h to meet the requirement of thetheorem that infinitesimals occur only in expressions of the form x i = x i +1 − ε .To apply the theorem to the pre-EQR (12), we consider ϕ ( y, x, h ) ˙= − y + 3 x + 4 x < ∧ x > y > − ,ϕ ( x, h ) ˙= ϕ [ y // x − ε ]˙= x + 2 > ∧ x + 2 x < ,ϕ ( h ) ˙= ϕ [ x // h − ε ]˙= h + 2 > ∧ ( h = 0 ∨ h + 2 h < . As in the theorem, we proceed from the right to the left, i.e., our first step isfixing h and computing a respective algebraic number α h . Since h = 0, we arein case (a). Now (cid:20) true h = 0true (cid:21) is a pre-EQR for ∃ hϕ . Therefore, α h is the root of the polynomial h in theinterval ] − , α h is the rational number 0.We continue with x = h − ε , which is case (c). Now (cid:20) true x = h − ε h = 0true true (cid:21) is a pre-EQR for ∃ x ∃ hϕ . Lemma 2(i) ensures that there exists e ε ∈ R , 0 < e ε such that for all e ε ∈ R , 0 < e ε < e ε , we have R | = ϕ ( α h − e ε, α h ). Refining α h wecompute that R | = ϕ ( α x , α h ), where α x is the root of 32 x + 1 in the interval] − , [, i.e., α x is the rational number − .Finally consider y = x − ε , which is again case (c). Now (cid:20) true y = x − ε x = − h = 0true true true (cid:21) is a pre-EQR for ∃ y ∃ x ∃ hϕ . Lemma 2(i) ensures that there exists e ε ∈ R , 0 < e ε such that for all e ε ∈ R , 0 < e ε < e ε , we have R | = ϕ ( α x − e ε, α x , α h ). Refining α x we compute that R | = ϕ ( α y , α x , α h ), where α y is the root of 256 y + 9 in theinterval ] − , [, i.e., α y is the rational number − . To conclude we statethat (cid:20) true y = − x = − h = 0true true true (cid:21)
13s a pre-EQR for ∃ y ∃ x ∃ hϕ , which does not contain any nonstandard symbols.Since h does not occur in ϕ = ψ , (cid:20) true y = − x = − true true (cid:21) is a pre-EQR for ∃ y ∃ xψ .Finally, note that all quantified variables have to be present in a pre-EQRbefore Theorem 5 can be applied. This has a consequence, which we illustrateon an example. Consider a valid sentence ∃ x ∃ a ( x < a ). An extended quantifierelimination result containing a nonstandard symbol for this formula is (cid:2) true a = ∞ (cid:3) . Since x does not occur in this result, it can be “freely chosen,” i.e., the result isindependent of the value of x . Put another way, this means that (cid:2) true x = t a = ∞ (cid:3) is an extended quantifier elimination result for any standard term t as well. Thisdegree of freedom disappears when computing standard answers in the followingsense: The term t has to be fixed before the computation of standard answers.Fixing t = 2 and using Theorem 5 we obtain a standard EQR (cid:2) true x = 2 a = 3 (cid:3) . The computed standard answer for a depends on this choice of t , i.e., it ispossibly invalid for other choices. Fixing for example x = 4, we see that a = 3is not admissible anymore, because substituting these terms into ( x < a ) yields“false.” To compute a standard term for a when x = 4, we have to start with (cid:2) true x = 4 a = ∞ (cid:3) , and apply Theorem 5 again. We have already discussed that the feasibility of the virtual substitution methodstrongly depends on the degrees of the quantified variables. Among the heuris-tics for decreasing these degrees there is an observation, which was essentiallymade already by Weispfenning (1997a), and which was refined and named de-gree shift by Dolzmann et al. (1998). The following lemma restates the resultby Dolzmann et al.:
Lemma 6 (Degree Shift) . Consider a quantifier-free Tarski formula ψ . Let g be the GCD of all exponents of x in ψ . We divide all exponents of x in ψ by g yielding ψ . If g is odd, we have ∃ xψ ←→ ∃ xψ , if g is even we have ∃ xψ ←→ ∃ x ( x ≥ ∧ ψ ). For g > x in ψ . In orderto obtain larger GCDs and hence a better degree reduction, we may in advance“adjust” the degree n > x in polynomials of the form x n f , where x doesnot occur in f as follows: In equations and disequations, n may be equivalentlyreplaced by any m >
0. In ordering inequalities we may choose any m > n . 14e now want to reanalyze this result as a special case of virtual substitution.For this, we have to slightly generalize the framework by introducing shadowquantifiers . Recall that we are considering existential problems of the form ϕ ( u , . . . , u m ) = ∃ x n . . . ∃ x ψ ( x , . . . , x n , u , . . . , u m ) . As a first step we now switch to the equivalent problemˆ ϕ ( u , . . . , u m ) = ∃ ˆ x n ∃ x n . . . ∃ ˆ x ∃ x ψ ( x , . . . , x n , u , . . . , u m ) , where { ˆ x , . . . , ˆ x n } ∩ { x , . . . , x n , u , . . . , u m } = ∅ . That is, the variables ˆ x i donot occur in ψ . Consequently, proceeding with the elimination as discussed inSection 3, each shadow quantifier ∃ ˆ x i imposes a trivial elimination problem.Strictly following the virtual substitution framework, one would not sim-ply drop those quantifiers ∃ ˆ x i but eliminate them via trivial elimination setslike { (true , } . Notice that one cannot use ∅ as an elimination set here be-cause W ∅ = false. Furthermore, from the point of view of extended quantifierelimination the use of { (true , } formally provides answers also for ˆ x , . . . , ˆ x n .Consider now w.l.o.g. the elimination of ∃ x , and assume that we are in thesituation of Lemma 6, where g > x in ψ . We use an elimination set that depends on the parity of g : E = n(cid:16) γ, g p ˆ x (cid:17)o , where γ is x ≥ g is even and “true” otherwise. Of course, we have to definea suitable virtual substitution of g √ ˆ x for x within ψ : k X j =1 a j x j % !" x // g p ˆ x = k X j =1 a j ˆ x (cid:4) max( j,g ) g (cid:5) % ! , where a , . . . , a k ∈ Z [ x , . . . , x n , u , . . . , u m ] and % ∈ { = , ≤ , <, ≥ , >, = } . Thefloor function is applied to make the definition complete; with our eliminationsets we will always have divisibility g | max( j, g ). The max operator takes careof possible degree adjustments made for the computation of E .Observe that in contrast to the elimination sets studied so far we introducehere a variable ˆ x which was not present in ψ before. That variable is bound byshadow quantifier ∃ ˆ x . Intuitively, for the elimination of ∃ ˆ x ∃ x we switch fromone hard plus one trivial elimination step to two nontrivial elimination steps.The termination of quantifier elimination with shadow quantifiers followsfrom the termination of the underlying quantifier elimination method plus thefact that there are only finitely many shadow quantifiers for each regular quan-tifier.To keep the notation simple, we will in the sequel not formally introduceshadow quantifiers for all quantifiers. Instead, our procedure will silently assumetheir presence whenever it performs a degree shift. In the corresponding pre-EQR this can be recognized by assignments of the form x i = g √ ˆ x i , which cannotcome into existence otherwise. In this section we generalize Theorem 5 to admit more general pre-EQRs asinput. Furthermore, we discuss heuristics for obtaining rational numbers oreven integers instead of root expressions in our standard answers.15 orollary 7 (Generalized Computation of Standard Answers) . Consider aclosed Tarski formula ϕ = ∃ x n . . . ∃ x ψ ( x , . . . , x n ) . Assume that the follow-ing is a pre-EQR for ϕ : (cid:20) true x = e . . . x n = e n γ . . . γ n (cid:21) such that each e i is of one of the following forms:(a) a + b √ cd , where a , b , c , d ∈ Z [ x i +1 , . . . , x n ] ,(b) a + b √ cd ± ε , where a , b , c , d ∈ Z [ x i +1 , . . . , x n ] ,(c) ±∞ ,(d) g √ x i +1 , where g ∈ N \ { } ,where as usual “ ± ” denotes “ + ” or “ − .” Then we can compute root expressions e e , . . . , f e n each meeting either the specification (a) or the specification (d) aboveand e γ , . . . , f γ n such that the following is a pre-EQR for ϕ as well: (cid:20) true x = e e . . . x n = f e n e γ . . . f γ n (cid:21) . Proof.
From a theoretical point of view, the treatment of e i = a + b √ cd − ε canbe reduced to Theorem 5 via the introduction of artificial variables as we didfor our example in (12) above. From a practical point of view, it is not hard tosee how to algorithmically treat such expressions directly. Notice that then ourcorrected answer e e i will generally be a rational number. As already mentionedin the proof of Theorem 5(c), one might heuristically even find an integer. Inboth cases the possibly non-trivial guard γ i has to be replaced by e γ i = true.Next, the treatment of a + b √ cd + ε and −∞ in analogy to a + b √ cd − ε and ∞ ,respectively, is straightforward.Finally, having obtained algebraic numbers for x i +1 , . . . , x n , one can com-pute an algebraic number also for g √ x i +1 with g ∈ N \ { } .The proofs for both Theorem 5 and Corollary 7 are constructive. Recall thatthe ordering of the variables within the given pre-EQR is such that quantifierelimination has taken place from the left to the right, while the construction ofthe standard answers proceeds from the right to the left.Consider the computation of e e k for some e k . Here, the quantifier eliminationdirection mentioned above has played an important role in our proofs: Although e k +1 , . . . , e n have been replaced with g e k +1 , . . . , f e n , the expression e k is stillvalid. Taking that idea a bit further, we may replace e k with any valid expressionwithout affecting the validity of either e , . . . , e k − or g e k +1 , . . . , f e n .In fact, it is sometimes possible to convert a root expression e k into a rationalnumber or even an integer as follows: Before processing e k , we check whetherchanging it to one of e k ± ε yields a valid pre-EQR for ϕ as well. This can bedone by means of the virtual substitution( f γ n ∧ · · · ∧ (cid:93) γ k +1 ∧ γ k ∧ · · · ∧ γ ∧ ψ )[ x // e ] . . . [ x k − // e k − ][ x k // e k ± ε ][ x k +1 // g e k +1 ] . . . [ x n // f e n ] .
16n the positive case, we process e k ± ε instead of e k . In terms of the proofsof Theorem 5 and Corollary 7 this leads to the cases (c) and (b), respectively,where we generally obtain a rational solutions e e k and heuristically even integers.Finally, it is quite helpful in general to recognize rational numbers amongall occurring real algebraic numbers. This holds in particular for the final α n ,. . . , α , as they correspond to the values of the back-substituted f e n , . . . , e e ,which may be complicated nested root expressions. For this one can use thefollowing lemma. Lemma 8 (Rational Algebraic Numbers) . Consider a real algebraic number α = (cid:0) a n x n + · · · + a , ] l, u [ (cid:1) , where a , . . . , a n ∈ Z , a > l , u ∈ Q , l > (cid:3) a u , a l (cid:2) ∩ Z = { z } . Then α ∈ Q if and only if α = a z . Proof.
Let α ∈ Q . From l > α >
0, say α = pq , where p , q ∈ Z , p > q >
0. This admits the following factorization: q · n X i =0 a i x i = ( qx − p ) · n − X i =0 a i x i . It follows that p | a , say pp = a , and we obtain α = pp qp = a qp . On the otherhand, l < a qp < u , which is equivalent to a u < qp < a l , and it follows that qp = z . Together α = a qp = a z . The converse implication is obvious.The lemma can be straightforwardly generalized to arbitrary intervals ] l, u [. We have implemented our method in Redlog, which is a part of the computer al-gebra system Reduce. Reduce is freely available under a modified BSD license. Technically, our implementation is an extension of Redlog’s extended quantifierelimination rlqea by a switch rlqestdans , which toggles the computation ofstandard answers.In the following subsections we are going to revisit a number of applicationsof extended quantifier elimination that have been documented in the scientificliterature. In each case we are going to briefly explain the underlying prob-lem, recompute the solutions with nonstandard answers, and finally computesolutions with standard answers using our approach as described in this article.Since Redlog is very actively developed and improved, and the consideredapplications date back up to more than 15 years, the nonstandard answersobtained here partly differ from those reported in the literature. Of course, insuch cases both variants are correct.All computations have been carried out with the CSL version of Reduce,revision 2465, using 4 GB RAM on a 2.4 GHz Intel Xeon E5-4640 running64 bit Debian Linux 7.3.
Besides many standard problems from computational geometry, Sturm andWeispfenning (1997) consider in their Example 10 the reconstruction of a cuboid http://reduce-algebra.sourceforge.net x -axis with alens of focal length 5.The answers obtained by extended quantifier elimination is going to describevectors e , e , e ∈ R generating the cuboid together with a vector v ∈ R describing its translation from the origin. The input formula, which contains inaddition points i ∈ R on the camera sensor, contains 15 quantifiers: ∃ e ∃ e ∃ e ∃ v ∀ i (cid:0) ( ι ←→ π ) ∧ ∃ k (59 k v = (100 , , k + 295)) (cid:1) . The formula ι ( e , e , e , v , i ), which has been obtained by regular quantifierelimination earlier, generically describes that a point i lies in the image of acuboid generated by e , e , e , and translated by v . The formula π ( i ) is aquantifier-free description of one concrete image. The remaining part of theinput formula fixes i = (cid:0) , (cid:1) to be the image of the origin of the cuboid.Extended quantifier elimination yields “true” if and only if π is a picture ofa cuboid at all. In the positive case, the answers will provide suitable vectors e , e , e , and v .For π as considered by Sturm and Weispfenning (1997) in Example 10,the extended quantifier elimination yields “true” together with the followingnonstandard answers: e = (cid:18) ∞ , ∞ , ∞ (cid:19) , e = (cid:18) ∞ , ∞ , − ∞ (cid:19) , e = (cid:18) − ∞ , ∞ , ∞ (cid:19) , v = (cid:18) ∞ , ∞ , ∞ + 204 (cid:19) . Our method fixes ∞ = 1, which yields the following standard answers: e = (cid:18) , , (cid:19) , e = (cid:18) , , − (cid:19) , e = (cid:18) − , , (cid:19) , v = (cid:18) , , (cid:19) . The entire computation takes 189 s, of which the computation of the standardanswers takes less than 1 ms.
Weispfenning (2001) has studied motion planning problems in dimension two.Both the object to be moved and the free space between given obstacles aresemilinear sets. Extended quantifier elimination is used to decide whether ageometrical object can be moved from an initial to a final destination in at most n moves, where the trajectory of each move is a line segment. In the positivecase, the answers describe the coordinates u , . . . , u n ∈ R of the object aftereach of the n moves. Accordingly, the input formulas contain 2 n variables inthe prenex existential block.We have applied our answer correction to three of the examples discussedby Weispfenning (2001). For the concrete input formulas and pictures of thescenery we refer to that publication. 18xample u u u time6.4 (5 − ε , − ε ) (cid:0) − ε , − ε +12 (cid:1) (cid:0) , (cid:1) , ε ) (3 − ε ,
6) (7 ,
6) 9.5 s6.9 (0 , ε ) (3 − ε ,
6) 0.28 sTable 1: Summary of nonstandard answers and computation times for motionplanning examples considered by Weispfenning (2001).Example ε u u u (cid:0) , (cid:1) (cid:0) , (cid:1) (cid:0) , (cid:1) ,
4) (2 ,
6) (7 , ,
4) (2 , ε was less than 1 ms.For Example 6.4, we obtain the following nonstandard answers: u = (5 − ε , − ε ) , u = (cid:18) − ε , − ε + 12 (cid:19) , u = (cid:18) , (cid:19) . Our method fixes ε = , which yields the following standard answers: u = (cid:18) , (cid:19) , u = (cid:18) , (cid:19) , u = (cid:18) , (cid:19) . The entire computation takes 60 ms, of which the computation of the standardanswers takes less than 1 ms. Tables 1 and 2 summarize these results along withthe two other examples.
Recently, symbolic methods for the identification of Hopf bifurcations in vectorfields arising from biological networks or chemical reaction networks have re-ceived considerable attention in the literature (Sturm and Weber, 2008; Sturmet al., 2009; Errami et al., 2011; Weber et al., 2011; Errami et al., 2013). Givena polynomial vector field, El Kahoui and Weber (2000) introduced a method,which automatically generates first-order Tarski formulas describing the exis-tence of a Hopf bifurcation in terms of the parameters. Then real quantifierelimination is applied to obtain corresponding necessary and sufficient condi-tions. For efficiency reasons, one often existentially quantifies all parametersand applies extended quantifier elimination. In the positive case, the answersprovide one set of parameter values giving rise to a Hopf bifurcation.Based on models introduced by Boulier et al. (2007), Sturm and Weber(2008) and Sturm et al. (2009) used the approach sketched above to automat-ically derive the existence of Hopf bifurcations for the gene regulatory network19ontrolling the circadian clock of a certain unicellular green alga. The inputformula is ∃ (0 < v ∧ < v ∧ < v ∧ < ϑ ∧ < γ ∧ < µ ∧ < δ ∧ < α ∧ ϑ · ( γ − v − v v ) = 0 ∧ λ v + γ µ − v = 0 ∧ α ( γ − v − v v ) + δ ( v − v ) = 0 ∧ < ϑδ + ϑ v δ + 9 λ ϑ v v δ ∧ ϑ v α v + 162 ϑα v v + 162 α v v δ + ϑ + 2 ϑ v δ + ϑ v δ + ϑ v + 2 ϑδ + 81 α v v ϑδ + 81 α v v ϑδ + δ + ϑδ + ϑ δ + ϑ + 2 ϑ v + ϑ v + 6561 α v v + 2 ϑ v δ + δ + 81 α v v + ϑ v δ − λ ϑ v v δ = 0) , for which we obtain the following nonstandard answers: γ = √∞ ∞ +16 √∞ ∞ +8 √∞ ∞ ∞ ∞ +1458 ∞ ∞ +729 ∞ ∞ − ∞ ∞ ∞ − ∞ ∞ ∞ +9 ∞ ∞ ,µ = ∞ ∞ +1458 ∞ ∞ +729 ∞ ∞ − ∞ ∞ ∞ − ∞ ∞ ∞ + ∞ ∞ − ∞ ∞ ∞ +16 ∞ ∞ +8 ∞ ,ϑ = − ∞ ∞ − ∞ ∞ − ∞ ∞ − ∞ ∞ − ∞ +4374 ∞ ∞ ∞ +13122 ∞ ∞ ∞ +13122 ∞ ∞ ∞ +4374 ∞ ∞ − ∞ ∞ ∞ − ∞ ∞ + ∞ ∞ ∞ +32805 ∞ ∞ +65610 ∞ ∞ +65610 ∞ ∞ +32805 ∞ ∞ +6561 ∞ − ∞ ∞ ∞ − ∞ ∞ ∞ − ∞ ∞ ∞ − ∞ ∞ ∞ − ∞ ∞ +3078 ∞ ∞ ∞ +9234 ∞ ∞ ∞ +9234 ∞ ∞ ∞ +3078 ∞ ∞ − ∞ ∞ ∞ − ∞ ∞ ∞ − ∞ ∞ + ∞ ∞ + ∞ , v = √∞ ∞ +8 √∞ ∞ ∞ ∞ +1458 ∞ ∞ +729 ∞ ∞ − ∞ ∞ ∞ − ∞ ∞ ∞ +9 ∞ ∞ , v = √∞ , v = √∞ , α = ∞ , δ = 1 , λ = ∞ . Our method fixes ∞ = 1, ∞ = 9, and ∞ = 87481, which yields thefollowing standard solution: γ = 69984800 · √ , µ = 382716252169984800 , ϑ = 765291726176056937210 , v = 6998480 · √ , v = √ , v = √ ,α = 1 , δ = 1 , λ = 87481 . The entire computation takes 370 ms, of which the computation of the standardanswers takes 140 ms.
We are now going to discuss another example about Hopf bifurcation. Thistime, the considered system is a chemical reaction system, viz. the famous andwell-studied phosphofructokinase reaction. It has been firstly analyzed withsymbolic methods by Gatermann et al. (2005, Example 2.1). We adopt here thefirst-order formulation discussed by Sturm and Weber (2008) and Sturm et al.(2009) following the approach sketched in the previous subsection.20e obtain nonstandard answers of the following form: k = K ( ∞ , ∞ , ∞ , ∞ , ε ) , k = ∞ , k = ∞ ,k = K ( ∞ , ∞ , ∞ , ∞ , ∞ , ε ) , k = ∞ , k = ∞ ,k = K ( ∞ , ∞ , ∞ , ∞ , ∞ , ε ) , v = ∞ ∞ ∞ , v = V ( ∞ , ∞ , ∞ , ∞ , ∞ , ε ) , v = ∞ . The nonstandard terms K , . . . , K , V are so large that we cannot explicitlydisplay them here. To give an idea, K would fill more than 16 pages in thisdocument.Our method fixes ∞ = ∞ = ∞ = ∞ = 1, ∞ = 20, and ε = 2( √ − k = 3 , k = 1 , k = 1 ,k = √ , k = 20 , k = 1 ,k = −√ , v = 1 , v = −√ , v = 1 . The entire computation takes 13.2 s, of which the computation of the standardanswers takes 0.1 s.
Sturm (1999a, Section 5) has applied generic quantifier elimination to the sizingof a BJT amplifier. Description of a circuit is given as a set of operating pointequations E and a set of AC conditions E . For the concrete equations werefer to the mentioned publication. The system E ∧ E has to be solved w.r.t.the main variables M = { r , . . . , r , c } in terms of parameter variables P = { v cc , a high , a low , p, z in , z out } . Fixing values of the parameters to v cc = 3 , a high = 3 , a low = 2 , p = 12 , z in = 5 , z out = 5 , the answers contain one nonstandard term: r = 44570583955180672 , r = 44570583952590336 , r = − ,r = − , r = ∞ , r = 282999424999804520000 ,r = 5 , r = 2550959560533708675520836792295619328 , c = 64758413371175185 . Our method fixes ∞ = 1, which yields a standard answer for r . The entirecomputation takes less than 2 ms, of which the computation of the standardanswer takes less than 1 ms. 21 .6 A Linear Feasibility Example Korovin et al. (2009, Section 9) have considered a small linear existential prob-lem to demonstrate the difference between their conflict resolution method andthe Fourier–Motzkin elimination method. The following are nonstandard an-swers for that problem computed by Redlog: x = − , x = 1 − ε , x = −
14 + 13 ε ,x = − − ε + 65 ε , x = −
30 + 26 ε . Our method fixes ε = and ε = , which yields the following standardanswers: x = − , x = 0 , x = − , x = − , x = − . The entire computation takes 3 ms, of which the computation of the standardanswers takes less than 1 ms.
We have introduced extended quantifier elimination as a general concept, andfocused on virtual substitution as one possible method for its realization. Suc-cessful applications of extended quantifier elimination via virtual substitutionhave been documented in the literature over the past two decades. One problemthere was that the answers obtained via virtual substitution in general containnonstandard symbols, which can be hard to interpret. For fixed parametersthe present work resolves this issue by providing a complete post-processingmethod for fixing all answers to standard real numbers. We have implementedour method, and applied it to various extended quantifier elimination problemsfrom the literature. In these experiments we have generally obtained standardanswers that are meaningful in terms of the modeled problems. In most casesour post-processing method does not significantly contribute to the overall com-putation time. It is noteworthy that our method is compatible with our recentwork on combining virtual substitution with learning techniques (Korovin et al.,2014).Recall from our discussion in Section 6 on finding integer and rational an-swers that there is often a considerable degree of freedom in the choice of stan-dard answers. In future this can be further exploited in many interesting ways:For instance, using extended quantifier elimination methods as a theory solver inthe context of Satisfiability Modulo Theory (SMT) solving (Nieuwenhuis et al.,2006), in particular when combining several theories in a Nelson–Oppen (1979)style, one is specifically interested in avoiding identical answers for different vari-ables. Alternatively, one can try to identify certain variables, which might beinteresting in certain contexts. As another option, there is only a small step toautomatically generating for a given pre-EQR code for an interactive procedurethat suggests ranges and finds possible choices for certain variables in coopera-tion with the user. In cooperation projects with researchers from the sciences we22ave had the experience that those researchers often have a surprisingly preciseidea about reasonable choices for certain variables.A theoretically way more challenging step would be the generalization of ourmethod to the parametric case. Recall that Proposition 1 has shown that it isnot possible in general to determine real standard values for infinitesimals andinfinities without fixing values for the parameters beforehand. Nevertheless,it might well be possible to devise on the basis of our work here a completemethod for symbolically replacing nonstandard symbols with standard terms.In the example in Proposition 1 the infinitesimal ε could be replaced e.g. with − a yielding a standard extended quantifier elimination result. Acknowledgments
This research was supported in part by the German Transregional CollaborativeResearch Center SFB/TR 14 AVACS and by the DFG/ANR Programme BlancProject STU 483/2-1 SMArT.
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