Constructing Fewer Open Cells by GCD Computation in CAD Projection Jingjun Han
Liyun Dai
Bican Xia
School of Mathematical Sciences & Beijing International Center for Mathematical Research Peking University
School of Mathematical Sciences & Beijing International Center for Mathematical Research Peking University
LMAM & School of Mathematical Sciences Peking University
[email protected]
[email protected]
arXiv:1401.4953v1 [cs.SC] 20 Jan 2014
ABSTRACT A new projection operator of cylindrical algebraic decomposition (CAD) is proposed. The new operator computes the intersection of projection factor sets produced by different CAD projection orders. In other words, it computes the gcd of projection polynomials in the same variables produced by different CAD projection orders. We prove that the new operator still guarantees obtaining at least one sample point from every connected component of the highest dimension. In general, the gcd computation produces smaller projection factor sets and thus fewer open cells. Some examples that are difficult to be solved by existing tools have been worked out efficiently by our program based on the new operator.
Categories and Subject Descriptors G.4 [Mathematics of computation]: Mathematical software — Algorithm design and analysis
General Terms Algorithms
Keywords CAD projection, positive semi-definite, polynomial.
1.
INTRODUCTION
The cylindrical algebraic decomposition (CAD) method and its application to quantifier elimination (QE) for elementary real algebra was first proposed by Collins [3, 2]. A key role in CAD algorithm is its projection operator. A well known improvement of CAD projection is Hong’s projection operator which is applicable for all cases of QE [8]. For many problems of QE, a smaller projection operator given by McCallum in [9, 10], with an improvement by Brown in [1], is more efficient.
ISSAC2014 ’2014 Kobe University, Japan
[email protected]
Strzebo´ nski proposed in [12] an algorithm called generic cylindrical algebraic decomposition (GCAD) for solving systems of strict polynomial inequalities, which made use of the so-called generic projection, the same projection operator as that in [1] proposed later. Rong proposed in [14] the open CAD method which is similar to GCAD. As a property of open CAD, in terms of the Brown projection, at least one sample point can be taken from every highest dimensional cell via the open CAD lifting phase. For traditional CAD projection operators, such as Brown’s projection operator, different projection orders may lead to a great difference in complexity. Thus, in order to reduce the projection scale, it is meaningful to study the relationship between those different projection orders. For related work, see for example [4]. The reason for such difference is mainly because the projection factors in the same variables produced by different projection orders may be different. For example, when we apply Brown’s projection operator Bp (see Definition 8) to any given polynomial f ∈ Z[x1 , ..., xn ], it is quite possible that Bp(f, [xn , xn−1 ]) 6= Bp(f, [xn−1 , xn ]). In this paper, we propose a new CAD projection operator Hp. The new operator computes the intersection of projection factor sets produced by different CAD projection orders. In other words, it computes the gcd of projection polynomials in the same variables produced by different CAD projection orders. In some sense, the polynomial in the projection factor sets of Hp is irrelevant to the projection orders. We prove that the new operator still guarantees obtaining at least one sample point from every connected component of the highest dimension. In general, the gcd computation produces smaller projection factor sets and thus fewer open cells. The structure of this paper is as follows. In Section 2, a simple example illustrates the main idea and steps of the new projection operator Hp. Section 3 introduces basic definitions, lemmas and concepts of CAD. In Section 4, the new projection operator Hp is defined and a new algorithm based on Hp is proposed. Our main result (Theorem 3) is proved. In Section 5, we prove that it is valid if we replace Bp with Hp in some steps of the projection phase of the simplified CAD projection Np we proposed recently for inequality proving in
[7]. Section 6 includes several examples which demonstrate the effectiveness of our algorithms. We conclude the paper in Section 7 with some discussions on our future work.
2.
MAIN IDEA
Let us show the comparison of our new operator and Brown’s projection operator on the following simple example. Formal description and proofs of our main results are given in subsequent sections.
Remark 2. Computing projection factor sets (polynomials) under different projection orders brings additional costs compared to traditional CAD projection operators. However, it has two gains. First, it produces fewer sample points (representing open cells) in many cases as shown in Example 1. Second, the most important thing is, if the number of variables is greater than 3, it may also reduce the scale of projection. Please see Definition 12, Algorithm 1, Remark 5 and Remark 6 for details.
Example 1. Let f = x − 2x y + 2x z + y − 2y z + z 4 + 2x2 + 2y 2 − 4z 2 − 4 ∈ Z[x, y, z].
For problems with many variables and/or special structures, the new method performs very well. Our program can solve problems of proving polynomial inequalities with more than 10 variables (even for polynomials with 17 variables!). Please see Section 6 for details.
We first compute an open CAD (see Definition 9) defined by f 6= 0 in R3 by Brown’s operator. Take the order z y x. Step 1, compute the projection polynomial (up to a nonzero constant)
3.
4
fz
2 2
2 2
4
2 2
∂ sqrf(f ), z) = Res(sqrf(f ), ∂z = (x4 − 2x2 y 2 + y 4 + 2x2 + 2y 2 − 4)(3x2 − y 2 − 4)2
where “Res” means the Sylvester resultant and “sqrf” means “squarefree” that is defined in Definition 5. Step 2, compute the projection polynomial (up to a nonzero constant) fzy
∂ = Res(sqrf(fz ), ∂y sqrf(fz ), y) 2 4 = (3x − 4)(x + 2x2 − 4)(4x2 − 5)2 (x − 1)8 (x + 1)8
which has 8 distinct real zeros. Step 3, by open CAD lifting under the order z y x and using the projection factor set {fzy , fz , f }, we will finally get 113 sample points of f 6= 0 in R3 .
PRELIMINARIES
If not specified, for a positive integer n, let xn be the set of variable {x1 , . . . , xn } and αn and βn denote the point (α1 , . . . , αn ) ∈ Rn and (β1 , . . . , βn ) ∈ Rn , respectively. Definition 1. Let f ∈ Z[xn ], denote by lc(f, xi ) and discrim(f, xi ) the leading coefficient and the discriminant of f with respect to (w.r.t.) xi , respectively. Definition 2. Let f ∈ Z[xn ], the set of real zeros of f is denoted by Zero(f ). Denote by Zero(L) or Zero(f1 , . . . , fm ) the common real zeros of L = {f1 , . . . , fm } ⊂ Z[xn ]. Definition 3. The level for f ∈ Z[xn ] is the biggest j such that (s.t.) deg(f, xj ) > 0 where deg(f, xj ) is the degree of f w.r.t. xj . For polynomial set L ⊆ Z[xn ], Li is the set of polynomials in L with level i.
Now, we compute a reduced open CAD (see Definition 13) defined by f in R3 by the new projection operator proposed in this paper. Step 1, take the order z y x and compute the projection polynomial fzy as above. Step 2, take another order y z x and we can similarly obtain a projection polynomial (up to a constant)
Definition 4. Let Pn be the symmetric permutation group of x1 , . . . , xn . Define Pn,i to be the subgroup of Pn , where any element σ of Pn,i fixes x1 , . . . , xi−1 , i.e., σ(xj ) = xj for j = 1, . . . , i − 1.
fyz = (3x2 − 4)2 (x4 + 2x2 − 4)(4x2 − 5)(6x2 − 7)8 .
h = al12j1 −1 · · · lt2jt −1 h1 2i1 · · · hm 2im ,
Step 3, compute 2
4
2
2
gcd(fyz , fzy ) = (3x − 4)(x + 2x − 4)(4x − 5) which has 6 distinct real zeros. Step 4, by open CAD lifting under the order z y x and using the projection factor set {gcd(fyz , fzy ), fz , f }, we will finally get 87 sample points of f 6= 0 in R3 .
Remark 1. The main result of this paper is Theorem 3 which guarantees that the new projection operator can obtain at least one sample point from every connected component of the highest dimension. Especially for Example 1, that means the intersection of the set of those 87 sample points and each connected component of f 6= 0 in R3 is not empty.
Definition 5. If h∈ Z[xn ] can be factorized in Z[xn ] as:
where a ∈ Z, t ≥ 0, m ≥ 0, lj (i = 1, . . . , t) and hi (i = 1, . . . , m) are pairwise different irreducible primitive polynomials with positive leading coefficient (under a suitable ordering) and positive degree in Z[xn ]. Define sqrf(h) = l1 · · · lt h1 · · · hm , sqrf 1 (h) = {li , i = 1, 2, . . . , t}, sqrf 2 (h) = {hi , i = 1, 2, . . . , m}. If h is a constant, let sqrf(h) = 1, sqrf 1 (h) = {1}, sqrf 2 (h) = {1}. In the following, we introduce some basic concepts and results of CAD. The reader is referred to [3, 8, 9, 10, 1, 14] for a detailed discussion on the properties of CAD and open CAD.
Definition 6. [3, 9] An n-variate polynomial f (xn−1 , xn ) over the reals is said to be delineable on a subset S (usually connected) of Rn−1 if (1) the portion of the real variety of f that lies in the cylinder S × R over S consists of the union of the graphs of some t ≥ 0 continuous functions θ1 < · · · < θt from S to R; and (2) there exist integers m1 , . . . , mt ≥ 1 s.t. for every a ∈ S, the multiplicity of the root θi (a) of f (a, xn ) (considered as a polynomial in xn alone) is mi . Definition 7. [3, 9] In the above definition, the θi are called the real root functions of f on S, the graphs of the θi are called the f -sections over S, and the regions between successive f -sections are called f -sectors. Theorem 1. [9, 10] Let f (xn , xn+1 ) be a polynomial in Z[xn , xn+1 ] of positive degree and discrim(f, xn+1 ) is a nonzero polynomial. Let S be a connected submanifold of Rn on which f is degree-invariant and does not vanish identically, and in which discrim(f, xn+1 ) is order-invariant. Then f is analytic delineable on S and is order-invariant in each f -section over S. Based on this theorem, McCallum proposed the projection operator MCproj, which consists of the discriminant of f and all coefficients of f . Theorem 2. [1] Let f (xn , xn+1 ) be a (n+1)-variate polynomial of positive degree in xn+1 such that discrim(f, xn+1 ) 6= 0. Let S be a connected submanifold of Rn in which discrim(f, xn+1 ) is order-invariant, the leading coefficient of f is sign-invariant, and such that f vanishes identically at no point in S. f is degree-invariant on S. Based on this theorem, Brown obtained a reduced McCallum projection in which only leading coefficients, discriminants and resultants appear. The Brown projection operator is defined as follows. Definition 8. [1] Given a polynomial f ∈ Z[xn ], if f is with level n, the Brown projection operator for f is Bp(f, [xn ]) = Res(sqrf(f ),
∂(sqrf(f )) , xn ). ∂xn
Otherwise Bp(f, [xn ]) = f . If L is a polynomial set with level n, then [ ∂(sqrf(f )) Bp(L, [xn ]) = {Res(sqrf(f ), , xn )} ∂xn f ∈L [ {Res(sqrf(f ), sqrf(g), xn )}. f,g∈L,f 6=g
Define Bp(f, [xn , xn−1 , . . . , xi ]) =Bp(Bp(f, [xn , xn−1 , . . . , xi+1 ]), [xi ]). Open CAD is a modified CAD construction algorithm, which was named in Rong Xiao’s Ph.D. thesis [14]. In fact, open
CAD is similar to the generic cylindrical algebraic decomposition (GCAD) proposed in [12] and was used in DISCOVERER [13] for real root classification. For convenience, we describe the framework of the open CAD here. Definition 9. (Open CAD) For a polynomial f (xn ) ∈ Z[xn ], an open CAD defined by f (xn ) is a set of sample points in Rn obtained through the following three phases: (1) Projection. Use the Brown projection operator on f (xn ), let F = {f, Bp(f, [xn ]), . . . , Bp(f, [xn , . . . , x2 ])}; (2) Base. Choose one point in each of the open intervals defined by the real roots of F 1 ; (3) Lifting. Substitute each sample point of Ri−1 for xi−1 in F i and then, by the same method as Base phase, choose sample points for F i (xi ).
4.
REDUCED OPEN CAD
Definition 10. (Open sample) A set of sample points T ⊂ Rk is said to be an open sample defined by f (xk ) ∈ Z[xk ] in Rk if it has the following property: for every open connected set U ⊂ Rk defined by f 6= 0, T ∩ U 6= ∅. Suppose g(xk ) is another polynomial. If T is an open sample defined by f (xk ) in Rk such that g(α) 6= 0 for any α ∈ T , then we denote the open sample by Tg6=0 .
Starting from an open sample of Rj (j < n), by a similar construction as the lifting phase of open CAD we can obtain a set of sample points of Rn . More concretely, we state the procedure as follows and will prove in this section that the output of the procedure is indeed an open sample of Rn under some conditions. Given two polynomial sets L1 = {fn (xn ), fn−1 (xn−1 ), . . . , fj (xj )} and L2 ={gn (xn ), gn−1 (xn−1 ), . . . , gj (xj )} and an open sample Tgj 6=0 defined by fj (xj ) in Rj , we compute a set of sample points in Rn by repeating the following step: substitute each sample point of Ri−1 for xi−1 in fi (j < i ≤ n) and then choose one point in each of the open intervals defined by the real roots of fi such that gi does not vanish at the point. Remark 3. In the following, for convenience, we call the above procedure OpenSP and the calling sequence for inputs L1 , L2 and T is OpenSP(L1 , L2 , T ).
Definition 11. (Open delineable) Let L1 = {fn (xn ), fn−1 (xn−1 ), . . . , fj (xj )},
(1)
L2 = {gn (xn ), gn−1 (xn−1 ), . . . , gj (xj )}
(2)
s
be two polynomial sets and S an open set of R (s ≤ j). The polynomial fn (xn ) is said to be open delineable on S w.r.t. L1 and L2 , if for any open sample Tgj 6=0 defined by fj (xj ) in Rj , any A = OpenSP(L1 , L2 , T ) and any T open connected setTU ⊂ Rn defined by fn 6= 0 with U (S × Rn−s ) 6= ∅, A U 6= ∅.
Open delineability has the following four properties. The first three can be derived easily from the definition of open delineable, we will omit the proof here.
Proposition 1. (open sample property) Let L1 , L2 be as in (1) and (2). If fn (xn ) is open delineable on every open connected set of fj (xj ) 6= 0 w.r.t. L1 and L2 , then for any open sample Tgj 6=0 defined by fj (xj ) in Rj , any A = OpenSP(L1 , L2 , T ) is an open sample defined by fn (xn ) in Rn .
Definition 12. Let f ∈ Z[x1 , . . . , xn ]. For m(1 ≤ m ≤ n), denote [y] = [y1 , . . . , ym ] where yi ∈ {x1 , . . . , xn } for 1 ≤ i ≤ m and yi 6= yj for i 6= j. For 1 ≤ i ≤ m, Hp(f, [y], yi ) and Hp(f, [y]) are defined recursively as follows. ˆ ), [yi ]), Hp(f, [y], yi ) = Bp(Hp(f, [y] i Hp(f, [y]) = gcd(Hp(f, [y], y1 ), . . . , Hp(f, [y], ym )), ˆ = [y1 , . . . , yi−1 , yi+1 , . . . , ym ] and Hp(f, [ ]) = f . where [y] i Define Hp(f, i) = {f, Hp(f, [xn ]), . . . , Hp(f, [xn , . . . , xi ])}, and
Proposition 2. (transitive property) Let L1 , L2 be as in (1) and (2). For a given open set S ⊂ Rs (s ≤ j), there exists k(j ≤ k ≤ n) such that fk (xk ) is open delineable on S w.r.t. {fk (xk ), . . . , fj (xj )} and {gk (xk ), . . . , gj (xj )}, and fn (xn ) is open delineable on every open connected set of fk (xk ) 6= 0 w.r.t. {fn (xn ), . . . , fk (xk )} and {gn (xn ), . . . , gk (xk )}, then fn (xn ) is open delineable on S w.r.t. L1 and L2 . Proposition 3. (nonempty intersection property) Let L1 , s L2 be as in (1) and T (2). For two open sets S1 and S2 of R (s ≤ j) with S1 S2 6= ∅, if fn (xn ) is open delineable on both SS 1 and S2 w.r.t. L1 and L2 , fn (xn ) is open delineable on S1 S2 w.r.t. L1 and L2 . Proposition 4. (union property) Let L1 , L2 be as in (1) and (2). For σ ∈ Pn,j+1 , denote yn = (y1 , y2 , . . . , yn ) = σ(xn ) and yi = (y1 , y2 , . . . , yi ). Let L01 ={fn (xn ), pn−1 (yn−1 ), . . . , pj (yj )} and L02 ={qn (xn ), qn−1 (yn−1 ), . . . , qj (yj )} where pi (yi ) and qi (yi ) are polynomials in i variables. For two open sets S1 and S2 of Rj , if (a) S1 (S2 ) is S in some open connected component of fj (xj ) 6= 0, (b) S1 S2 is in some open connected component of pj 6= 0, (c) fn (xn ) is open delineable on both S1 andSS2 w.r.t. L1 and L2 , (d) fn (xn ) is open delineable on S1 S2 w.r.t. L01 and L02 , and S (e) qj (yj ) vanishes atSno point in S1 S2 , then fn (xn ) is open delineable on S1 S2 w.r.t. L1 and L2 .
f H p(f, i) = {f, Hp(f, [xn ], xn ), . . . , Hp(f, [xn , . . . , xi ], xi )}. As a Corollary of Theorem 1 and Theorem 2, we have Proposition 5. Let f ∈ Z[xn ] be a squarefree polynomial with level n. f is open delineable on every open connected set defined by Bp(f, [xn ]) 6= 0 in Rn−1 w.r.t. Hp(f, n) f and H p(f, n). Definition 13. (Reduced open CAD) A reduced open CAD of f (xn ) w.r.t. [xn , . . . , xj ] is a set of sample points in Rn obtained through the following three phases: f 1. Projection. Compute Hp(f, j) and H p(f, j); 2. Base. Choose an open sample THp(f,[xn ,...,xj ],xj )6=0 defined by Hp(f, [xn , . . . , xj ]) in Rj−1 ; 3. Lifting. Substitute each sample point βi−1 in Ri−1 (i ≥ j) for xi−1 in Hp(f, [xn , . . . , xi+1 ]) and then choose one point in each of the open intervals defined by the real roots of Hp(f, [xn , . . . , xi+1 ])(βi−1 , xi ) and such that Hp(f, [xn , . . . , xi+1 ], xi+1 )(βi−1 , xi ) does not vanish at that point.
Proof. For point α1 ∈ S1 , α2 ∈ S2 , s.t. gj (αt )qj (αt ) 6= 0, t = 1, 2, let Tgtj 6=0 be any open sample defined by fj with Tgtj 6=0 ∩ St = {αt }. For any At = OpenSP(L1 , L2 , Tgtj 6=0 ), let Atαt = {βn | βn ∈ At , (β1 , ..., βj ) = αt }. Let Tq0tj 6=0 be any S open sample defined by pj with Tq0tj 6=0 ∩ (S1 S2 ) = {αt }. t t For any A0 = OpenSP(L01 , L02 , Tq0tj 6=0 ), let A0 αt = {βn | βn ∈ t A0 , (β1 , ..., βj ) = αt }.
Lemma 1. [7] Let f and g be coprime in Z[xn ]. For any connected open set U in Rn , the open set V = U \Zero(f, g) is also connected.
For T any open connected set UTdefined by fn1 6=T 0 with U (α1 × Rn−j ) 6= ∅, then A1α1 U 6= ∅Sand A0 α1 U 6= ∅. Since fn (xn ) is open delineable on S1 S2 w.r.t. L01 and T 2 T L02 , A0 α2 T U 6= ∅. This implies that U (S2 × Rn−j ) 6= ∅. 2 Thus, S Aα2 U 6= ∅. Therefore, fn (xn ) is open delineable S on (S1 S2 )\Zero(qj ) w.r.t. L1 andSL2 . Since 0 ∈ / qj (S1 S2 ), fn (xn ) is open delineable on S1 S2 w.r.t. L1 and L2 .
Theorem 3. Let j be an integer and 2 ≤ j ≤ n. For any given polynomial f (xn ) ∈ Z[xn ] and any open connected set U ⊂ Rj−1 of Hp(f, [xn , . . . , xj ]) 6= 0, let S = U \Zero({Hp(f, [xn , . . . , xj ], xt ) | t = j, . . . , n}). Then f (xn ) is open delineable on the open connected set S w.r.t. Hp(f, j) f and H p(f, j). As a result, a reduced open CAD of f (xn ) w.r.t. [xn , . . . , xj ] is an open sample defined by f (xn ).
Now, we define the new projection operator Hp.
Proof. First, by Lemma 1, S is open connected. We prove the theorem by induction on k = n − j. When k = 0,
The following Theorem is the main result of this paper, which shows that the reduced open CAD owns the property of open delineability.
it is obvious true from Proposition 5. Suppose the theorem is true for all polynomials g(xk ) ∈ Z[xk ] with k = 0, 1, . . . , n − i − 1. We now consider the case k = n − i. Let [z] = [xn , . . . , xi ]. For any given polynomial f (xn ) ∈ Z[xn ], let U ⊂ Ri−1 be an open connected set of Hp(f, [z]) 6= 0 and S = U \Zero({Hp(f, [z], xt ) | t = i, . . . , n}). For any point α ∈ S with Hp(f, [z], xi )(α) 6= 0, there exists an open connected set Sα ⊂ Ri−1 such that α ∈ Sα and 0∈ / Hp(f, [z], xi )(Sα ). By induction, Hp(f, [xn , . . . , xi+1 ]) is open delineable on Sα w.r.t. {Hp(f, [z])} and {Hp(f, [z], xi )}. By induction again and the transitive property of open delineable (Proposition 2), f is open delineable on Sα w.r.t. f Hp(f, i) and H p(f, i). For any point α ∈ S with Hp(f, [z], xi )(α) = 0, there exists an i0 such that n ≥ i0 ≥ i + 1 and Hp(f, [z], xi0 )(α) 6= 0. 0 Thus there exists an open connected set Sα of Ri−1 such 0 0 that α ∈ Sα and 0 ∈ / Hp(f, [z], xi0 )(Sα ). Let σ ∈ Pn,i with σ(xi ) = xi0 , it is easy to see that f (σ(xn )) is open delineable 0 f p(f (σ(xn )), i). For any on Sα w.r.t. Hp(f (σ(xn )), i) and H 0 β ∈ Sα with Hp(f, [z], xi )(β) 6= 0, it is clear that there exists 00 0 an open connected set Sα ⊂ Sα and f is open delineable on 00 f p(f, i). From union property of open Sα w.r.t. Hp(f, i) and H 0 delineable (Proposition 4), f is open delineable on Sα w.r.t. f Hp(f, i) and Hp(f, i). To summarize, the above discussion shows that for any point α ∈ S, there exists an open connected set Sα ⊂ S such that α ∈ Sα and f is open delineable on Sα w.r.t. Hp(f, i) and f H p(f, i). By the nonempty intersection property of open delineable (Proposition 3) and the fact that S is connected, f f (xn ) is open delineable on S w.r.t. Hp(f, i) and H p(f, i) as desired.
Therefore, Theorem 3 provides us many ways for designing various algorithms for computing open samples. Obviously, different choices of m and different orders of the m variables at each step of recursion will result in different algorithms. For example, we may set m = 2 and choose [xn , xn−1 ], [xn−2 , xn−3 ], etc. successively in each step. Because there are only two different orders for two variables, we compute the gcd of two projection polynomials under the two orders in each step. The following algorithm is based on this choice.
Algorithm 1 HpTwo Require: A polynomial f ∈ Z[xn ] of level n. Ensure: An open sample defined by f , i.e., a set of sample points which contains at least one point from each connected component of f 6= 0 in Rn 1: g := f ; 2: L1 := {}; 3: L2 := {}; 4: for i from n downto 2 do 5: if i ≥ 3 then S 6: L1 := L1 Hp(g, i − 1); S f 7: L2 := L2 H p(g, i − 1); 8: g := Hp(g, [xi , xi−1 ]); 9: i := i − 1; 10: else S 11: L1 := L1 Hp(g, i); S f 12: L2 := L2 H p(g, i); 13: g := Hp(g, [xi ]); 14: end if 15: end for 16: T :=An open sample TL1 6=0 in R defined by L11 ; 2 17: C:= OpenSP(L1 , L2 , T ); 18: return C.
Therefore, the theorem is true for k = n−j by induction.
Remark 4. If we do not require U to be open in Theorem 3, i.e., let U ⊂ Rj−1 be a connected submanifold, in which Hp(f, [xn , . . . , xj ]) is order-invariant. If S is connected, we may generalize the notation of open delineable and obtain similar results to Theorem 3 with similar discussion. That will make it possible to choose at least one sample point from every connected component of f = 0 by using Hp.
As an application of Theorem 3, for a given polynomial f (xn ), we could obtain a CAD based method to get an open sample defined by f . Roughly speaking, if we have already got an open sample defined by Hp(f, [xn , . . . , xj ]) in Rj−1 , according to Theorem 3, we could obtain an open sample defined by f . That process could be done recursively.
Remark 5. Roughly speaking, in the definition of Hp, we first choose m variables from {x1 , ..., xn }, compute all projection polynomials under all possible orders of those m variables, and then compute the gcd of all those projection polynomials. It is not difficult to see that if we modify the definition of Hp by choosing several (not all) orders of those m variables and computing the gcd of the projection polynomials under those orders, Theorem 3 is still valid.
Remark 6. If Hp(f, [xn , xn−1 )] 6= Bp(f, [xn , xn−1 ]) and n > 3, it is obvious that the scale of projection in Algorithm 1 is smaller than that of open CAD in Definition 9.
Remark 7. Let f (x1 ) and g(x1 ) be two univariate polynomials. Isolating the real roots of f (x1 ) and choosing one point from each open interval defined by the real roots such that g(x1 ) does not vanish at that point will give an open sample Tg6=0 in R defined by f .
5.
PROJECTION OPERATOR NP
In this section, we combined the idea of Hp and the simplified CAD projection operator Np we introduced previously in [7], to get a new algorithm for proving polynomial inequality. Definition 14. [7] Suppose f ∈ Z[xn ] is a polynomial of level n. Define Oc(f, xn ) = sqrf 1 (lc(f, xn )), Od(f, xn ) = sqrf 1 (discrim(f, xn )), Ec(f, xn ) = sqrf 2 (lc(f, xn )), Ed(f, xn ) = sqrf 2 (discrim(f, xn )), Ocd(f, xn ) = Oc(f, xn ) ∪ Od(f, xn ), Ecd(f, xn ) = Ec(f, xn ) ∪ Ed(f, xn ).
The secondary and principal parts of the projection operator Np are defined as Np1 (f, [xn ]) =Ocd(f, xn ), Y Np2 (f, [xn ]) ={
g}.
g∈Ecd(f,xn )\Ocd(f,xn )
If L is a set of polynomials of level n, define [ Np1 (L, [xn ]) = Ocd(g, xn ), g∈L
Np2 (L, [xn ]) =
[
{
Notice that the proof of Theorem 3 only uses the properties of open delineable (Propositions 1-4) and Proposition 5, and Proposition 6 is similar to Proposition 5. We can prove the following theorem by the same way of proving Theorem 3.
Y
h}.
g∈L h∈Ecd(g,xn )\Np1 (L,[xn ])
Based on the projection operator Np, we proposed an algorithm, Proineq, in [7] for proving polynomial inequalities. Algorithm Proineq takes a polynomial f (xn ) ∈ Z[xn ] as input, and returns whether or not f (xn ) ≥ 0 on Rn . The readers are referred to [7] for the details of Proineq. The projection operator Np is extended and defined in the next definition. Definition 15. Let f ∈ Z[x1 , . . . , xn ] with level n. Denote [y] = [y1 , . . . , ym ], for 1 ≤ m ≤ n, where yi ∈ {x1 , . . . , xn } for 1 ≤ i ≤ m and yi 6= yj for i 6= j. Define Y Np(f, [xi ]) = Np2 (f, [xi ]), Np(f, [xi ], xi ) = g. g∈Np1 (f,[xi ])
For m(m ≥ 2) and i(1 ≤ i ≤ m), Np(f, [y], yi ) and Np(f, [y]) are defined recursively as follows. ˆ ), yi ), Np(f, [y], yi ) = Bp(Np(f, [y] i Np(f, [y]) = gcd(Np(f, [y], y1 ), . . . , Np(f, [y], ym )), ˆ = [y1 , . . . , yi−1 , yi+1 , . . . , ym ]. Define where [y] i
Theorem 5. Let j be an integer and 2 ≤ j ≤ n. For any given polynomial f (xn ) ∈ Z[xn ], and any open connected set U of Np(f, [xn , . . . , xj ]) 6= 0 in Rj−1 , let S = U \Zero({Np(f, [xn , . . . , xj ], xt ) | t = j, . . . , n}). If the polyS nomials in n−j i=0 Np1 (f, [xn−i ]) are all semi-definite on U × Rn−j , f (xn ) is open delineable on S w.r.t. Np(f, j) and f N p(f, j). Theorem 5 and Proposition 2 provide us a new way to decide the non-negativity of a polynomial as stated in the next theorem. Theorem 6. Given a positive integer n. Let f ∈ Z[xn ] be a squarefree polynomial with level n and U a connected open set of Np(f, [xn , . . . , xj ]) 6= 0 in Rj−1 . Denote S = U \Zero({Np(f, [xn , . . . , xj ], xt ) | t = j, . . . , n}). The necessary and sufficient condition for f (xn ) to be positive semidefinite on U × Rn−j+1Sis the following two conditions hold. n−j (1)The polynomials in i=0 Np1 (f, [xn−i ]) are all semi-definite n−j on U × R . (2)There exists a point α ∈ S such that f (α, xj , . . . , xn ) is positive semi-definite on Rn−j+1 . Based on the above theorems, it is easy to design some different algorithms (depending on the choice of j) to prove polynomial inequality. For example, the heuristic algorithm PSD-HpTwo for deciding whether a polynomial is positive semi-definite, which we will introduce later, is based on Theorem 6 when j = n − 1 (Proposition 7).
Np(f, i) = {f, Np(f, [xn ]), . . . , Np(f, [xn , . . . , xi )]}, and f N p(f, i) = {f, Np(f, [xn ], xn ), . . . , Np(f, [xn , . . . , xi ], xi )}. Theorem 4. [7] Given a positive integer n ≥ 2. Let f ∈ Z[xn ] be a non-zero squarefree polynomial and U a connected component of Np(f, [xn ]) 6= 0 in Rn−1 . If the polynomials in Np1 (f, [xS n ]) are semi-definite on U , then f is delineable on V = U \ h∈Np (f,[xn ]) Zero(h). 1
Lemma 2. [7] Given a positive integer n ≥ 2. Let f ∈ Z[xn ] be a squarefree polynomial with level n and U a connected open set of Np(f, [xn ]) 6= 0 in Rn−1 . If f (xn ) is semi-definite on U × R, then the polynomials in Np1 (f, [xn ]) are all semi-definite on U .
Proposition 7. Given a positive integer n ≥ 3. Let f ∈ Z[xn ] be a squarefree polynomial with level n and U a connected open set of Np(f, [xn , xn−1 ]) 6= 0 in Rn−2 . Denote S = U \Zero(Np(f, [xn , xn−1 ], xn ), Np(f, [xn , xn−1 ], xn−1 )). The necessary and sufficient condition for f (xn ) to be positive semi-definite on U × R2 is the following two conditions hold. (1)The polynomials in either Np1 (f, [xn ]) or Np1 (f, [xn−1 ]) are semi-definite on U × R. (2)There exists a point α ∈ S such that f (α, xn−1 , xn ) is positive semi-definite on R2 .
Lemma 3. Given a polynomial f (xn , x) ∈ Z[xn , x], say f (xn , x) =
Now, we can rewritten Theorem 4 in another way. Proposition 6. Let f ∈ Z[xn ] be a squarefree polynomial with level n and U a connected component of Np(f, [xn ]) 6= 0 in Rn−1 . If the polynomials in Np1 (f, [xn ]) are semi-definite on U , then f is open delineable on U w.r.t. Np(f, n) and f N p(f, n).
l X
ci xi , cl 6≡ 0,
i=0
where ci is a polynomial in xn for each i = 0, . . . , l. Let f h (xn , x, y) =
l X
ci xi y l−i ,
i=0
then f ≥ 0 for all (xn , x) ∈ Rn+1 is equivalent to f h ≥ 0 for all (xn , x, y) ∈ Rn+2 .
Algorithm 2 PSD-HpTwo Require: An irreducible polynomial f ∈ Z[xn ]. Ensure: Whether or not ∀αn ∈ Rn , f (αn ) ≥ 0. 1: if n ≤ 2 then 2: if Proineq(f (xn ))=false then 3: return false 4: end if 5: else S 6: L1 := Np1 (f, [xn ]) Np1 (f, [xn−1 ]) 7: L2 := Np(f, [xn , xn−1 ]) 8: for g in L1 do 9: if PSD − HpTwo(g) =false then 10: return false 11: end if 12: end for 13: Cn−2 := A reduced open CAD of L2 w.r.t. [xn−2 , . . . , x2 ], which satisfies that Zero(Np(f, [xn , xn−1 ], xn ), Np(f, [xn , xn−1 ], xn−1 )) ∩Cn−2 = ∅. 14: if ∃αn−2 ∈ Cn−2 such that Proineq(f (αn−2 , xn−1 , xn ))=false then 15: return false 16: end if 17: end if 18: return true
Definition 16. Let Et(f, x) := sqrf 2 (lc(f h , x))
\
sqrf 2 (lc(f h , y)),
Ot(f, x) := sqrf 1 (lc(f h , x))
\
sqrf 1 (lc(f h , y)),
Etd(f, x) := Et(f, x)
[
Ed(f, x),
Otd(f, x) := Ot(f, x)
[
Od(f, x).
connected open set of HNp(f, [xn , . . . , xj ]) 6= 0. Denote S = U \Zero({HNp(f, [xn , . . . , xj ], xt ) | t = j, . . . , n}). The necessary and sufficient condition for f (xn ) to be positive semidefinite on U × Rn−j+1Sis the following two conditions hold. n−j HNp1 (f, [xn−i ]) are all semi-definite (1)The polynomials in i=0 n−j on U × R . (2)There exists a point α ∈ S such that f (α, xj , . . . , xn ) is positive semi-definite on Rn−j+1 .
6.
EXAMPLES
The Algorithm HpTwo has been implemented using Maple. In this section, we first report the numbers of sample points obtained by the algorithm HpTwo and open CAD, respectively, on one example in [12].
Example 6.1. [12] f = ax3 + (a + b + c)x2 + (a2 + b2 + c2 )x + a3 + b3 + c3 − 1. Under the order a ≺ b ≺ c ≺ x, an open CAD defined by f has 132 sample points, while an open sample obtained by the algorithm HpTwo has 15 sample points.
Example 6.2. For 100 random polynomials f (x, y, z) with degree 81 , Figure 1 shows the numbers of real roots of Bp(f, [z, y]), Bp(f, [y, z]) and Hp(f, [y, z]), respectively. It is clear that the number of real roots of Hp(f, [y, z]) is always less than those of Bp(f, [z, y]) and Bp(f, [y, z]).
The projection operator HNp is defined as HNp1 (f, [x]) = Otd(f, x), Y
HNp(f, [x]) = {
g}.
g∈Etd(f,x)\Otd(f,x)
If L is a set of polynomials of level n, define [ HNp1 (L, [xn ]) = Otd(g, xn ), g∈L
HNp(L, [xn ]) =
[
{
Figure 1: The number of real roots. Y
h}.
g∈L h∈Etd(g,xn )\Np1 (L,xn )
It is obvious that HNp(f, [x]) | Np(f, [x]) and HNp1 (f, [x]) | Np1 (f, [x]). Like Np, we can define HNp(f, [y]), HNp(f, [y], yi ) similarly. Notice that discrim(f h , x) = y l(l−1) discrim(f, x), and discrim(f h , y)= xl(l−1) discrim(f, x), the following theorem could be shown easily.
Algorithm PSD-HpTwo has been implemented using Maple. Now, we illustrate the performance of our program with several non-trivial examples. We report the timings of the program PSD-HpTwo, the program Proineq [7], and the program Raglib2 [5] on these examples. All computations were performed on a laptop with Inter Core2 2.10GHz CPU and 2.00GB RAM, Windows XP and Maple 15. 1
Theorem 7. Given a positive integer n. Let f ∈ Z[xn ] be a squarefree polynomial with level n and U ⊂ Rj−1 a
Generated by randpoly([x,y,z],degree=8) in Maple 15. RAGlib release 3.19.4 (Oct., 2012) implemented by Mohab Safey El Din. 2
For more examples, please visit the homepage3 of the first author. Example 6.3. [6] Prove that (
n X
x2i )2 − 4
n X
i=1
x2i x2i+1 ≥ 0,
i=1
8.
ACKNOWLEDGEMENTS
The authors would like to convey their gratitude to Hoon Hong who provided his valuable comments, advice and suggestion on this paper when he visited Peking University. The authors also would like to thank all members in the symbolic computation seminar of Peking University, including Xiaoxian Tang, Zhenghong Chen and Ting Gan, who together checked the proofs.
where xn+1 = x1 .
9. Hereafter “>4000” means either the running time is over 4000 seconds or the software is failure to get an answer. The timings in the table are in seconds. n 5 8 11 14 17
PSD-HpTwo 0.156 0.635 4.688 36.453 295.110
Raglib 6.984 177.750 3990.015 >4000 >4000
Proineq 0.297 >4000 >4000 >4000 >4000
Example 6.4. Prove that 3m+2 X
B(x3m+2 ) = (
x2i )2 − 2
i=1
3m+2 X i=1
x2i
m X
x2i+3j+1 ≥ 0,
j=1
where x3m+2+r = xr . When m = 1, it is equivalent to the case n = 5 of the last example. This form was once studied in [11]. 3m + 2 5 8 11
7.
PSD-HpTwo 0.156 1.000 7.781
Raglib 6.984 144.907 2989.516
Proineq 0.297 23.094 >4000
CONCLUSION
In this paper, we propose a new CAD projection operator Hp. The new operator computes the intersection of projection factor sets produced by different CAD projection orders. In other words, it computes the gcd of projection polynomials in the same variables produced by different CAD projection orders. In some sense, the polynomial in the projection factor sets of Hp is irrelevant to the projection orders. We prove that the new operator still guarantees obtaining at least one sample point from every connected component of the highest dimension. Some examples that are difficult to be solved by existing tools have been worked out efficiently by our program based on the new operator. It is not difficult to see that, if the input polynomial f (xn ) is symmetric, the new projection operator Hp cannot reduce the projection scale and the number of sample points. If f (xn ) is not symmetric, it is most likely that Hp produces smaller projection factor sets and thus fewer open cells. On the other hand, as stated in Remark 5, one has many choices based on Theorem 3 to design different algorithms for computing open samples defined by f (xn ). How to get better strategies of choosing variables and orders for concrete examples is our ongoing work. 3 https://sites.google.com/site/jingjunhan/home/ software
REFERENCES
[1] Brown C W. Improved projection for cylindrical algebraic decomposition. J.Symbolic Computation, 2001,32: 447–465. [2] Caviness B F, Johnson J R. Quantifier elimination and cylindrical algebraic decomposition. Springer Verlag, 1998. [3] Collins, G. E. Quantifier Elimination for the Elementary Theory of Real Closed Fields by Cylindrical Algebraic Decomposition. Lect. Notes Comput. Sci. 33, 134–183, 1975. [4] Dolzmann A, Seidl A, Sturm T. Efficient projection orders for CAD. Proceedings of the 2004 international symposium on Symbolic and algebraic computation. ACM, 2004: 111-118. [5] M. S. El Din, RAGLib (Real Algebraic Geometry Library), 2007. [6] Han J J: An Introduction to the Proving of Elementary Inequalities. Harbin: Harbin Institute of Technology Press, 2011, 49–50. (in Chinese) [7] Han J J, Jin Z, and Xia B C. Proving Inequalities and Solving Global Optimization Problems via Simplified CAD Projection, arXiv preprint arXiv:1205.1223, 2012. [8] Hong H. An improvement of the projection operator in cylindrical algebraic decomposition. In: Proceedings of the International Symposium on Symbolic and Algebraic Computation, 261–264, 1990. [9] McCallum, S. An improved projection operation for cylindrical algebraic decomposition of three-dimensional space. J.Symbolic Computation., 1988, 5:141–161. [10] McCallum, S. An improved projection operation for cylindrical algebraic decomposition. In Caviness, B.,Johnson, J. eds, Quantifier Elimination and Cylindrical Algebraic Decomposition, Texts and Monographs in Symbolic Computation. Vienna, Springer-Verlag, 242–268, 1998. [11] Parrilo, P.A., 2000. Structured semidefinite programs and semialgebraic geometry methods in robustness and optimization. Dissertation (Ph.D.), California Institute of Technology. [12] Strzebo´ nski, A. Solving systems of strict polynomial inequalities. J. Symbolic Comput. 29, 471–480, 2000. [13] Xia B, DISCOVERER: A tool for solving problems involving polynomial inequalities. In: Proceedings of ATCM’2000, 472–481, ATCM Inc., Blacksburg, USA, Dec, 2000. [14] Xiao R, Parametric Polynomial Systems Solving. Dissertation (Ph.D.), Peking University, 2009. (in Chinese)
