Algorithms for computing strong μ-bases for rational tensor product surfaces

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Computer Aided Geometric Design www.elsevier.com/locate/cagd

Algorithms for computing strong product surfaces

μ-bases for rational tensor

Li-Yong Shen a,b,∗ , Ron Goldman c a b c

School of Mathematical Sciences, University of Chinese Academy of Sciences, 100049, Beijing, China Key Laboratory of Big Data Mining and Knowledge Management, CAS, 100190, Beijing, China Computer Science Department, Rice University, 6100 Main St., MS-132, Houston, TX 77005, USA

a r t i c l e

i n f o

Article history: Available online xxxx Keywords: Rational surface Base point Implicitization Moving plane Syzygy Strong μ-basis

a b s t r a c t Implicitizing rational surfaces is a fundamental computational task in Algorithmic Algebraic Geometry. Although the resultant of a μ-basis for a rational surface is guaranteed to contain the implicit equation of the surface as a factor, this resultant may also contain extraneous factors. Moreover, μ-bases for rational surfaces are, in general, notoriously difficult to compute. Here we develop fast algorithms to find μ-bases for rational tensor product surfaces whose resultants are guaranteed to be the implicit equation of the corresponding rational surface with no extraneous factors. We call these μ-bases strong μ-bases. Surfaces with strong μ-bases are relatively rare. We show how these strong μ-bases are related to the number of base points counting multiplicity of the corresponding surface parametrization. In addition, when the base points are simple, we provide tables of rational tensor product surfaces with strong μ-bases based on the bidegree of the rational surface and the number of base points of the parametrization. The bidegrees of the corresponding strong μ-bases are also listed in these tables. © 2017 Elsevier B.V. All rights reserved.

1. Introduction

μ-Bases are a powerful tool for representing and analyzing rational planar curves. μ-Bases for rational planar curves are bases for the syzygy module with respect to homogeneous parameterizations, the degrees of their elements are unique and sum to the degree of the parametrization, their cross product retrieves the homogeneous parametrization, and their resultant generates the implicit equation of the curve with no extraneous factors (Chen and Wang, 2002). Thus we can easily recover both the parametric equations and implicit equation of a rational planar curve from a μ-basis for the curve. We can also use μ-bases to locate and analyze the singularities of rational planar curves (Jia and Goldman, 2009). The notion of a μ-basis readily extends from rational curves to rational surfaces (Chen et al., 2005, 2001; Chen and Wang, 2003; Song and Goldman, 2009). Nevertheless, μ-bases for rational surfaces have many properties that are qualitatively different from the characteristic properties of μ-bases for rational curves. μ-Bases for rational surfaces are bases for the syzygy module only with respect to affine parameterizations rather than homogeneous parameterizations, the degrees of their elements are not unique and do not necessarily sum to the degree of the parametrization, and although their outer product retrieves the affine parametrization, the resultant of the μ-basis that generates the implicit equation of the surface may contain extraneous factors.

*

Corresponding author at: School of Mathematical Sciences, University of Chinese Academy of Sciences, 100049, Beijing, China. E-mail addresses: [email protected] (L.-Y. Shen), [email protected] (R. Goldman).

http://dx.doi.org/10.1016/j.cagd.2017.03.001 0167-8396/© 2017 Elsevier B.V. All rights reserved.

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Moreover, while there are simple, fast algorithms for computing μ-bases for rational planar curves (Chen and Wang, 2002), fast efficient algorithms for computing μ-bases for rational surfaces are known only for rational ruled surfaces (Chen and Wang, 2003; Shen, 2016), quadric surfaces (Chen et al., 2007; Wang et al., 2008), Steiner surfaces (Wang and Chen, 2012), surfaces of revolution (Shi and Goldman, 2012) and cyclides (Jia, 2014). Algorithms for computing μ-bases for general rational surfaces are neither simple nor fast (Deng et al., 2005). For rational surfaces a class of μ-bases more analogous to the notion of μ-bases for rational curves are the strong μ-bases (Shen and Goldman, 2017). Like μ-bases for rational curves, strong μ-bases for rational surfaces are bases for the syzygy module, the degrees of their elements sum to the degree of the parametrization, and their resultant generates the implicit equation of the surface with no extraneous factors. Based on numerical experiments, we conjecture as well that the degrees of the elements of a strong μ-basis are unique (see Section 5). The primary goal of this paper is to present fast algorithms for finding strong μ-bases for rational tensor product surfaces by solving a simple system of linear equations. We will also show how these exceptional μ-bases are related to the number of base points counting multiplicity of the corresponding surface parametrization. In addition, when the base points are simple, we provide some tables of rational tensor product surfaces with strong μ-bases based on the bidegree of the rational surface and the number of base points of the parametrization; the bidegrees of the corresponding strong μ-bases are also listed in these tables. Surfaces with strong μ-bases are relatively rare. We can use these tables to find strong μ-bases in the following manner: If we know the number of simple base points of a rational tensor product surface or the implicit degree of a rational tensor product surface (which is easy to compute – see Section 4.2), these tables predict whether or not the surface has a strong μ-basis. Moreover, if the surface does have a strong μ-basis, then these tables tell us the bidegrees of the elements of this μ-basis, so we can solve a simple system of linear equations to compute the elements of this μ-basis. The resultant of this strong μ-basis will give the implicit equation of the surface without any extraneous factors. We proceed in the following fashion. In Section 2, we provide the basic background, definitions, and notation for base points, moving planes (i.e. syzygies), strong μ-bases, and resultants that we shall use throughout the remainder of this paper. In Section 3 we investigate the existence of syzygies of different bidegrees for rational tensor product surfaces of fixed bidegrees. Section 4 contains our main results: fast algorithms for finding strong μ-bases for rational tensor product surfaces. Here we also provide tables of rational tensor product surfaces with strong μ-bases. Section 5 states a conjecture about the uniqueness of the bidegrees of the elements of a strong μ-basis for a rational tensor product surface. We conclude in Section 6 with a brief summary of our results along with a few details about our implementations. 2. Preliminaries: base points, moving planes, strong μ-bases and resultants A rational tensor product surface P of bidegree (m, n) can be represented by homogeneous parametric equations with bihomogeneous parameters

P((s, u ), (t , v )) = (a((s, u ), (t , v )), b((s, u ), (t , v )), c ((s, u ), (t , v )), d((s, u ), (t , v )))

(1)

where n m  

a((s, u ), (t , v )) =

ai , j (si um−i )(t j v n− j ), b((s, u ), (t , v )) =

i =0 j =0

c ((s, u ), (t , v )) =

m  n 

n m  

b i , j (si um−i )(t j v n− j ),

i =0 j =0

c i , j (si um−i )(t j v n− j ), d((s, u ), (t , v )) =

i =0 j =0

m  n 

di , j (si um−i )(t j v n− j )

i =0 j =0

are bihomogeneous polynomials in R[s, u ; t , v ] and gcd(a, b, c , d) = 1. The parametrization can also be written in matrix form P((s, u ), (t , v )) T = M C v T , where

⎛

a0,0 ⎜ b0,0 ⎜ MC = ⎝ c 0,0 d0,0

··· ··· ··· ···

··· ··· ··· ···

am,0 bm,0 cm,0 dm,0

a0,n b0,n c 0,n d0,n

··· ··· ··· ···

⎞

am,n bm,n ⎟ ⎟ cm,n ⎠ dm,n

is the coefficient matrix of (a, b, c , d) with respect to v = (um v n · · · sm v n · · · um t n · · · sm t n ). The affine form (i.e. u = 1, v = 1) of the surface is P(s, t ). We drop the trivial case where P(s, t ) defines a plane, i.e., we assume that a, b, c , d are linearly independent. We shall also assume that the rational surface is properly parametrized, i.e., the map

(s, t ) → is birational.

a(s, t ) b (s, t ) c (s, t )

,

,

d(s, t ) d(s, t ) d(s, t )



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2.1. Base points A base point of a rational surface P((s, u ), (t , v )) is a parameter pair ((s0 , u 0 ), (t 0 , v 0 )) such that P((s0 , u 0 ), (t 0 , v 0 )) = (0, 0, 0, 0). It is known that the bidegrees of the syzygies of a rational surface are related to the number and types of the base points of the surface (see Busé et al., 2003; Cox, 2004; Zheng et al., 2003). Here we review two typical types of the base points, triangular k-ple and rectangular k × l-ple base points (Zheng et al., 2003). Definition 1. (Zheng et al., 2003) A base point B P is a triangular k-ple point if there is a projective transformation sending B P to the origin such that the monomials of the transformed parametric polynomials do not include si um−i t j v n− j , i + j < k; furthermore, the coefficient matrix of the transformed parametric polynomials with respect to the degree k monomials has rank k + 1. Notice that here k ≤ 3. Definition 2. (Zheng et al., 2003) A point B P is a rectangular k × l-ple point if there is a projective transformation sending B P to the origin such that the monomials of the transformed parametric polynomials do not include si um−i t j v n− j , i < k; j < l; furthermore, if a generic linear combination of the transformed parametric polynomials is written as λ1 a((s, u ), (t , v )) + λ2 b((s, u ), (t , v )) + λ3 c ((s, u ), (t , v )) + λ4 d((s, u ), (t , v )) = sk P + t l Q , the coefficient matrix of polynomials P and Q has rank 2. In this paper, we assume that each base point of P(s, t ) is either a triangular or a rectangular base point. k+1 A triangular k-ple base point ((s0 , u 0 ), (t 0 , v 0 )) leads to 2 equations

∂ i + j P((s, u ), (t , v )) ((s0 , u 0 ), (t 0 , v 0 )) = (0, 0, 0, 0), 0 ≤ i + j ≤ k − 1. ∂ si ∂ t j

(2)

Similarly a rectangular k × l-ple base point ((s0 , u 0 ), (t 0 , v 0 )) leads to kl equations

∂ i + j P((s, u ), (t , v )) ((s0 , u 0 ), (t 0 , v 0 )) = (0, 0, 0, 0), 0 ≤ i ≤ k − 1, 0 ≤ j ≤ l − 1. ∂ si ∂ t j

(3)

The number of linear equations associated to a base point is called the degree of the base point. Suppose P((s, u ), (t , v )) has η base points. The formula for the degree of the implicit equation of P(s, t ) is.

Deg(P) = 2mn −

η 

di

(4)

i =1

where di is the multiplicity of the base point and di = k2i for a triangular ki -ple base point and di = ki li for a rectangular ki × li -ple base point (Zheng et al., 2003). A base point is called a simple base point if this base point is 1-ple (also 1 × 1-ple). Proposition 1. Let P((s, u ), (t , v )) be a parametrization with bidegree (m, n). The different simple base points of P((s, u ), (t , v )) cannot share the same (s, u ) parameter more than n times or the same (t , v ) parameter more than m times. Proof. Suppose that ((sk , uk ), (t 0 , v 0 )), k = 1, . . . , m + 1 are m + 1 different simple base points of P((s, u ), (t , v )). Consider

⎛ ⎞ m n   j n − j ⎝ f (s, u ) = a((s, u ), (t 0 , v 0 )) = ai , j (t 0 v 0 )⎠ (si um−i ). i =0

j =0

f (s, u ) is a homogeneous polynomial of degree m but has m + 1 different roots (sk , uk ), k = 1, . . . , m + 1, so a((s, u ), (t 0 , v 0 )) = f (s, u ) ≡ 0; hence t v 0 − t 0 v is a factor of a((s, u ), (t , v )). Similarly t v 0 − t 0 v is a factor of b, c , d. Therefore (t − t 0 )| gcd(a, b, c , d), which contradicts the assumption that gcd(a, b, c , d) = 1. Thus different simple base points of P((s, u ), (t , v )) cannot share the same (t , v ) parameter more than m times. The same proof works for the (s, u ) parameter. 2 Proposition 1 gives upper bounds for the number of simple base points sharing the same parameters in rational parametric surfaces. For a rational surface with ηt triangular ki -ple base points i = 1, . . . , ηt and ηr rectangular ki × li -ple base points i = ηt + 1, . . . , ηt + ηr , the total degree of the base points is

K=

ηt  ki + 1 i =1

2

+

η t +ηr i =ηt +1

ki li .

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Now consider the vector v = (um v n

⎛

···

sm v n

···

∂v ((s1 , u 1 ), (t 1 , v 1 )) ∂ s0 ∂ t 0

um t n

⎞

···

sm t n ) and the matrix M B of size K × (m + 1)(n + 1)

⎜ ⎟ ⎜ ⎟ .. ⎜ ⎟ . ⎜ ⎟ ⎜ ⎟ ∂ k 1 −1 v ⎜ ⎟ (( s , u ), ( t , v )) 1 1 1 1 k1 −1−i i ⎜ ⎟ ∂ s ∂t ⎜ ⎟ ⎜ ⎟ .. ⎜ ⎟ . ⎜ ⎟ ⎜ ⎟ ∂v ⎜ ⎟ (( s , u ), ( t , v )) η η η η 0 0 t t t t ∂ s ∂t ⎜ ⎟ ⎜ ⎟ . ⎜ ⎟ . ⎜ ⎟ . ⎜ ⎟ ⎜ ⎟ ∂ ηt −1 v ⎜ ⎟ kηt −1−i ((s ηt , u ηt ), (t ηt , v ηt )) i ∂ s ∂t ⎜ ⎟ MB = ⎜ ⎟. ∂v ⎜ ⎟ (( s , u ), ( t , v )) η + 1 η + 1 η + 1 η + 1 0 0 t t t t ∂ s ∂t ⎜ ⎟ ⎜ ⎟ . ⎜ ⎟ . . ⎜ ⎟ ⎜ k +l −2 ⎟ ⎜ ∂ ηt +1 ηt +1 v ⎟ ⎜ k −1 l −1 ((sηt +1 , u ηt +1 ), (t ηt +1 , v ηt +1 )) ⎟ ⎜ ∂ s ηt +1 ∂ t ηt +1 ⎟ ⎜ ⎟ ⎜ ⎟ .. ⎜ ⎟ . ⎜ ⎟ ⎜ ⎟ ∂ v ⎜ ⎟ (( s , u ), ( t , v )) η η η η ∂ s0 ∂ t 0 ⎜ ⎟ ⎜ ⎟ .. ⎜ ⎟ ⎜ ⎟ . ⎝ ⎠ ∂ kη +lη −2 v (( s , u ), ( t , v )) η η η η k η −1 l η −1 ∂s

(5)

∂t

The rank of M B is determined by the relationships between the base points. We say that the ηt triangular ki -ple base

η k +1 points i = 1, . . . , ηt and the ηr rectangular ki × li -ple base points i = ηt + 1, . . . , ηt + ηr with total degree K = i =t 1 i 2 +

ηt +ηt

k l are in general position if Rank( M B ) = K (also see Sederberg and Chen, 1995). The following proposition gives an i =ηt +1 i i upper bound for the number of the base points in general position.

Proposition 2. Let P((s, u ), (t , v )) with bidegree (m, n) define a non-planar rational surface. If P((s, u ), (t , v )) has ηt triangular ki -ple base points i = 1, . . . , ηt and has ηr rectangular ki × li -ple base points i = ηt + 1, . . . , ηt + ηr in general position, then the total

η k +1 η +η degree of the base points K ≡ i =t 1 i 2 + i =t η +r 1 ki li ≤ (m + 1)(n + 1) − 4. t

Proof. Let ((si , u i ), (t i , v i )), i = 1, . . . , ηt + ηr be base points of P((s, u ), (t , v )) in general position. By condition (2), the coefficient vector of a((s, u ), (t , v )) satisfies

M B · (a0,0 , a1,0 , . . . , am,0 , . . . , a0,n , . . . , am,n ) T = 0.

(6)

So too do the coefficient vectors of b, c , d. Since Rank( M B ) = K , the number of solutions of (6) is (m + 1)(n + 1) − K . If K > (m + 1)(n + 1) − 4 then the number of independent solutions is less than 4, in which case a, b, c , d must be linearly dependent and P((s, u ), (t , v )) degenerates to a plane. 2 Corollary 1. Let P((s, u ), (t , v )) with bidegree (m, n) define a non-planar rational surface. The number of simple base points of P((s, u ), (t , v )) in general position is at most (m + 1)(n + 1) − 4. Proof. By assumption the base points are 1-ple (also 1 × 1-ple). Hence

K=

ηt  ki + 1 i =1

2

+

η t +ηr i =ηt +1

ki li =

η t +ηr

1=η

i =1

is exactly the number of simple base points in general position. Therefore this result follows immediately from Proposition 2. 2 The base points introduce a system of linear constraints (2) for the coefficients of P(s, t ). There are (m + 1)(n + 1) − K freedoms for choosing these coefficients. For fixed base points these coefficients are generic if (m + 1)(n + 1) − K coefficients remain free.

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In the following discussions, we prefer to use the affine form P(s, t ) (that is, u = 1, v = 1) since μ-bases are defined only with respect to affine parametrizations. We shall use the bihomogeneous form P((s, u ), (t , v )) only when we are considering base points at infinity. 2.2. Moving planes and strong μ-bases A moving plane of bidegree (σ1 , σ2 ) has the implicit form

L (s, t , x, y , z, w ) ≡

σ  ( A i x + B i y + C i z + D i w )γi (s, t ) = 0

(7)

i =0

where A i , j , B i , j , C i , j , D i , j ∈ R and γi (s, t ), i = 1, . . . , σ are polynomials with bidegree at most (σ1 , σ2 ). We call γi (s, t ) blending functions of the moving plane. If the blending functions are si t j , i = 0, . . . , σ1 , j = 0, . . . , σ2 , then σ = (σ1 + 1)(σ2 + 1). For each parameter pair (s, t ), equation (7) is the implicit equation of a plane in R3 . A moving plane is also sometimes written in parametric form as the vector

L(s, t ) = ( A (s, t ), B (s, t ), C (s, t ), D (s, t )) by extracting the coefficients of (7) with respect to x, y , z, w. A moving plane L(s, t ) is said to follow the surface P(s, t ) if

L(s, t ) · P(s, t ) = A (s, t )a(s, t ) + B (s, t )b(s, t ) + C (s, t )c (s, t ) + D (s, t )d(s, t ) ≡ 0.

(8)

Thus L(s, t ) follows the surface P(s, t ) if L(s, t ) is a syzygy of P(s, t ). The syzygy module of P(s, t ) is known to be a free module with three generators (Chen et al., 2005). Three moving planes p(s, t ), q(s, t ) and r(s, t ) are said to form a μ-basis of P(s, t ) if p(s, t ), q(s, t ) and r(s, t ) are a basis for the syzygy module of P(s, t ). Chen et al. (2005) show that three syzygies p(s, t ), q(s, t ) and r(s, t ) are a μ-basis for the syzygy module of P(s, t ) if and only if

[p, q, r] = κ P, κ = 0

(9)

where [·] denotes the outer product defined by

    ⎞   ⎛  p2 p3 p4   p1 p3 p4   p1 p2 p4   p1 p2 p3          [p, q, r] = ⎝ q2 q3 q4  , −  q1 q3 q4  ,  q1 q2 q4  , −  q1 q2 q3 ⎠ .  r2 r3 r4   r1 r3 r4   r1 r2 r4   r1 r2 r3 

Introducing the homogeneous form of the tensor parameters, we have

[p((s, u ), (t , v )), q((s, u ), (t , v )), r((s, u ), (t , v ))] = κ (u , v )P((s, u ), (t , v )). By comparing the degrees of the homogeneous parameters, it follows that

(10)

κ (u , v ) = λu v for some constant λ = 0 where i

j

(i , j ) = bideg(p) + bideg(q) + bideg(r) − bideg(P). We can also retrieve the implicit equation of a rational surface P(s, t ) from a

(11)

μ-basis p, q, r for the surface. Let X =

(x, y , z, w ). Then, in general, Res(s,t ) (p · X, q · X, r · X) = F (x, y , z, w ) E (x, y , z, w ) where F (x, y , z, w ) = 0 is the implicit equation of P(s, t ) and E (x, y , z, w ) is an extraneous factor if DegX (Res(s,t ) (p · X, q · X, r · X)) > DegX ( F (x, y , z, w )) (Chen and Wang, 2003; Chen et al., 2005; Shi et al., 2012). Moreover there are some cases where Res(s,t ) (p · X, q · X, r · X) is identically zero (Shen and Goldman, 2017). We are interested to find those cases where this resultant is exactly equal to the implicit equation of the rational surface. Definition 3. A μ-basis p(s, t ), q(s, t ), r(s, t ) for a rational tensor product surface P(s, t ) is algebraically strong if E (x, y , z, w ) is a nonzero constant, or equivalently if

Deg(P) = DegX (Res(s,t ) (p · X, q · X, r · X))

(12)

where Deg(P) = the implicit degree of the rational surface P(s, t ). Definition 4. A μ-basis p(s, t ), q(s, t ), r(s, t ) for a rational tensor product surface P(s, t ) is parametrically strong if is a nonzero constant in (10), or equivalently if

bideg(p) + bideg(q) + bideg(r) = bideg(P).

κ (u , v ) = κ (13)

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Definition 5. A μ-basis p(s, t ), q(s, t ), r(s, t ) for a rational tensor product surface P (s, t ) is strong if the algebraically strong and parametrically strong. Example 1. Here we show that a parabolic dupin cyclide always has a strong parametrization with bidegree (2, 2) (see Jia, 2014)

μ-basis is both

μ-basis. A parabolic dupin cyclide has a

⎞T −s2 α 2 + α 2 t 2 + s2 α β + t 2 α β − α β ⎟ ⎜ −2 sα 2 t 2 − sα 2 + 2 sα β ⎟ , P(s, t ) = ⎜ 2 2 2 ⎠ ⎝ −2 ts α − t α − 2 t α β 2 2 −α s − α t − α ⎛

where α , β are two constants. The corresponding implicit equation is F (x, y , z, w ) ≡ α 2 β w 3 − β 3 w 3 − α 2 w 2 x − β 2 w 2 x + w y 2 α − z2 w α + β wx2 + w y 2 β + z2 w β + x3 + xy 2 + xz2 = 0. A μ-basis for the parabolic dupin cyclide P(s, t ) is p = (−t , 0, 1, − (α + β) t ), q = (s, 1, 0, − (α − β) s), r = (1, −s, t , −β). This μ-basis is parametrically strong since bideg(p) + bideg(q) + bideg(r) = (0, 1) + (1, 0) + (1, 1) = (2, 2) = bideg(P). We can also check that Res(s,t ) (p · X, q · X, r · X) is equal to F (x, y , z, w ) up to a nonzero scalar. Hence this μ-basis is also algebraically strong. A surface that is algebraically strong need not be parametrically strong – see Example 3 in Section 4.2. Conversely, a surface that is parametrically strong need not be algebraically strong – see Example 3 in Shen and Goldman (2017). We shall need the following theorem about the degree of the resultant from Shen and Goldman (2017). Theorem 1. Let p, q, r be three moving planes with bidegrees (σ11 , σ12 ), (σ21 , σ22 ), (σ31 , σ32 ) each consisting of a dense set of monomials. Then

DegX (Res(s,t ) (p · X, q · X, r · X)) = (σ11 σ22 + σ12 σ21 ) + (σ11 σ32 + σ12 σ31 ) + (σ31 σ22 + σ32 σ21 ).

(14)

Corollary 2. A μ-basis p, q, r with bidegrees (σ11 , σ12 ), (σ21 , σ22 ), (σ31 , σ32 ) each consisting of a dense set of monomials is algebraically strong if and only if

Deg(P) = (σ11 σ22 + σ12 σ21 ) + (σ11 σ32 + σ12 σ31 ) + (σ31 σ22 + σ32 σ21 ).

(15)

3. Existence of moving planes Consider a rational parametrization

P(s, t ) = (a(s, t ), b(s, t ), c (s, t ), d(s, t )) with bidegree (m, n). A moving plane of bidegree (σ1 , σ2 ) (7) with blending functions si t j , i = 0, . . . , σ1 , j = 0, . . . , σ2 can be written in the form

( x y z w · · · s σ1 t σ2 x s σ1 t σ2 y s σ1 t σ2 z s σ1 t σ2 w ) · ( A 0,0

B 0,0

C 0,0

D 0,0 · · · A σ1 ,σ2

B σ1 ,σ2

C σ1 ,σ2

D σ1 ,σ2 ) = 0.

(16)

Since (x, y , z, w ) = (a, b, c , d) is bidgree (m, n), there is a (m + σ1 + 1)(n + σ2 + 1) × 4(σ1 + 1)(σ2 + 1) matrix M P such that

(x y z w · · · sσ1 t σ2 x sσ1 t σ2 y sσ1 t σ2 z sσ1 t σ2 w ) = (1 · · · sm+σ1 · · · t n+σ2 · · · sm+σ1 t n+σ2 ) · M P .

(17)

The columns of the matrix M P are indexed by the polynomials

x, y , z, w , . . . , sσ1 t σ2 x, sσ1 t σ2 y , sσ1 t σ2 z, sσ1 t σ2 w ,

0 ≤ i ≤ σ1 , 0 ≤ j ≤ σ2 ,

and the rows of M P are indexed by the monomials

1, · · · , sm+σ1 , · · · , t n+σ2 , · · · , sm+σ1 t n+σ2 . Note that M P is the coefficient matrix of the linear system generated by the moving planes of bidgree (σ1 , σ2 ) that follow P(s, t ) and the coefficient vectors of these moving planes are solutions of

M P · ( A 0,0

B 0,0

C 0,0

D 0,0 · · · A σ1 ,σ2

B σ1 ,σ2

C σ1 ,σ2

D σ1 ,σ2 ) T = 0.

(18)

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Table 1 Deficiency values of triangular base points with multiplicity ki and rectangular base points with multiplicity ki × li (Zheng et al., 2003). ki -ple

ki × li -ple

2 2k i 2

2(ki − 1)(li − 1) + ki li

ki

bi pi

(ki − 1)(li − 1)

If the blending functions are γi (s, t ), i = 0, . . . , σ instead of si t j , we can also form the coefficient matrix M P by a similar construction. The blending functions γi (s, t ), i = 0, . . . , σ are determined by the types of the base points, and the blending functions are si t j if all the base points are simple (Zheng et al., 2003). We shall adopt the following notation:

N P (σ1 , σ2 ) = the number of linearly independent moving planes of bidegree (σ1 , σ2 ) that follow the surface P(s, t ) Proposition 3. N P (σ1 , σ2 ) = max{4(σ1 + 1)(σ2 + 1) − Rank( M P ), 0}. Proof. This proposition follows from Equation (18).

2

The rank of M P is determined by the surface P(s, t ). Each base point of P(s, t ) reduces the rank of M P . For instance, a base point induces at least one nonzero vector v such v · M P = 0. Let (s0 , t 0 ) be a base point of P(s, t ). Then by (17)

(0

···

0) = (1

s0

t0

s20

s0 t 0

m +σ 1 n +σ 2 t0 )·

t 02 · · · s0

MP = v · MP,

i.e., v is a correlation coefficient vector for the row vectors of M P . Even if the base points are given, there are different surfaces with these given base points since we have freedom in the choices of many of the coefficients of the surface P(s, t ) (see the discussion in Section 2.1). Therefore, the rank of M P can be further reduced by appropriately choosing the coefficients of the surface P(s, t ) (see Example 3 below). 3.1. With the assumption of regularity It is not easy to describe the mathematical relationship between the rank of M P and the coefficients of P(s, t ). However, with the assumption that the ideals of certain vector spaces are regular (see Adkins et al., 2005; Zheng et al., 2003 for details on regularity), there are some results concerning dimension that do not involve these coefficients. Here we review and summarize some useful results from Zheng et al. (2003). Let

I = a(s, t ), b(s, t ), c (s, t ), d(s, t ) be the ideal generated by P(s, t ). The moving planes will use blending functions belonging to the derivative deal

I = a, b, c , d, as , b s , c s , d s , at , bt , ct , dt , where the subscripts s and t denote partial derivatives with respect to s and t. For a bidegree (m, n) parametrization P(s, t ) with ηt triangular ki -ple base points i = 1, . . . , ηt and ηr rectangular ki × li -ple base points i = ηt + 1, . . . , η, η = ηt + ηr in general position, consider the moving planes with bidegree (σ1 , σ2 ) that follow the surface P (s, t ). We say that I is (σ1 , σ2 )-regular if the dimension of the vector space of blending functions is

η (σ1 + 1)(σ2 + 1) − i =1 bi ; I I is (m + σ1 , n + σ2 )-regular if the dimension of the vector space of constraint equations for the moving planes that follow the surface is (m + σ1 + 1)(n + σ2 + 1) − in Table 1.

η

i =1

p i , where b i , p i are the deficiency values listed

Theorem 2. (Zheng et al., 2003) For a bidegree (m, n) parametrization P(s, t ) with η base points in general position, if I is (σ1 , σ2 )-regular and I I is (m + σ1 , n + σ2 )-regular, then the number of linearly independent moving planes with bidegree (σ1 , σ2 ) that follow the surface is

N P (σ1 , σ2 ) = max{4(σ1 + 1)(σ2 + 1) − (m + σ1 + 1)(n + σ2 + 1) +

η  ( p i − 4bi ), 0}, i =1

where b i , p i are the deficiency values listed in Table 1.

(19)

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L.-Y. Shen, R. Goldman / Computer Aided Geometric Design ••• (••••) •••–•••

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Without the assumptions of regularity, the actual number of moving planes that follow the surface may be greater than the number of moving planes that follow the surface given by formula (19) (see Example 4 in subsection 4.2). The existence of the moving planes that follow the surface can be checked by formula (19) (or Proposition 3). In particular, the existence conditions for three such moving planes p, q, r with bidegrees (σ11 , σ12 ), (σ21 , σ22 ), (σ31 , σ32 ) are equivalent to

⎧ ⎨ N P (σ11 , σ12 ) ≥ 1, N P (σ21 , σ22 ) ≥ 1, N P (σ31 , σ32 ) ≥ 1 if (σi1 , σi2 )i =1,2,3 are different; N P (σ11 , σ12 ) ≥ 2, N P (σ31 , σ32 ) ≥ 1 if (σ11 , σ12 ) = (σ21 , σ22 ) = (σ31 , σ32 ); ⎩ N P (σ11 , σ12 ) ≥ 3 if (σ11 , σ12 ) = (σ21 , σ22 ) = (σ31 , σ32 ).

(20)

Proposition 4. For a bidegree (m, n) parametrization P(s, t ) with base points in general position, if I is (σ1 , σ2 )-regular and I I is (m + σ1 , n + σ2 )-regular, then the number of moving planes with bidegree (σ1 , σ2 ) that follow the surface given by formula (19) is less than or equal to the number of moving planes with bidegree (σ1 , σ2 ) that follow the surface given in Proposition 3. Proof. This proposition holds because the moving planes can be computed by solving (2). The following corollaries are helpful for reducing the computations in finding strong

2

μ-bases.

Corollary 3. For a bidegree (m, n) parametrization P(s, t ) with base points in general position, consider the moving planes with bidegree (σ1 , σ2 ) that follow the surface P(s, t ). If I is (σ1 , σ2 )-regular and I I is (m + σ1 , n + σ2 )-regular for any bidegree (σ1 , σ2 ), then 1. N P (σ1 , σ2 ) ≥ N P (σ1 , σ2 ) if σ1 ≥ σ1 and σ2 ≥ (n − 3)/3; 2. N P (σ1 , σ2 ) ≥ N P (σ2 , σ1 ) if m ≥ n and σ1 ≥ σ2 . Proof. By formula (19),

N P (σ1 , σ2 ) − N P (σ1 , σ2 ) = (σ1 − σ1 )(3σ2 − n + 3). Therefore we have statement 1 of this corollary. Similarly,

N P (σ1 , σ2 ) − N P (σ2 , σ1 ) = (m − n)(σ1 − σ2 ). Therefore we have we have statement 2 of this corollary.

2

Corollary 4. For a bidegree (m, n) parametrization P(s, t ) with ηt triangular ki -ple base points i = 1, . . . , ηt and ηr rectangular ki × li -ple base points i = ηt + 1, . . . , η, η = ηt + ηr in general position, consider the moving planes with bidegree (σ1 , σ2 ) that follow the surface P(s, t ). Suppose that I is (σ1 , σ2 )-regular and I I is (m + σ1 , n + σ2 )-regular.

ηt 2

η k are fixed, then maxki ,ηt i =t 1 ki = ηt occurs at ki = 1, i = 1, . . . , ηt ;

ηi =t 1 ki i +1

η 2. If i =1 2 are fixed, then maxki ,ηt i =t 1 ki = ηt occurs at ki = 1, i = 1, . . . , ηt ;

η

η 3. If i =η +1 ki li are fixed, then maxki ,ηr i =η +1 (ki li − 2(ki − 1)(li − 1)) = ηr occurs at ki = li = 1, i = ηt + 1, . . . , η ; t t 4. If all the base points are simple, then the number of the linearly independent moving planes that follow P(s, t ) with bidegree (σ1 , σ2 ) is maximal. In particular N P (σ1 , σ2 ) = max{4(σ1 + 1)(σ2 + 1) − (m + σ1 + 1)(n + σ2 + 1) + η, 0}. 1. If

Proof. For statement 1, if one base point is not simple, without loss of generality let k1 = 2. Then ηt decreases by three ηt ηt since k2 is fixed. Hence k decreases by two. Similar remarks hold for other multiple base points so we get i =1 i i =1 i statement 1. A similar discussion leads to statement 2. Statement 3 is straightforward since minli ≥1,ki ≥1 (2(ki − 1)(li − 1)) = 0 occurs when li ≥ 1, ki ≥ 1.

η For statement 4, reviewing N P (σ1 , σ2 ) = max{4(σ1 + 1)(σ2 + 1) − (m + σ1 + 1)(n + σ2 + 1) + i =1 ( p i − 4b i ), 0}, we need to get a maximum of

ηt

i =1 k i +

for each

η

η

i =1

(4bi − p i ) =

ηt

i =1

 

η ( 2k2 i − 4 k2i ) + i =ηt +1 (2(ki − 1)(li − 1) + ki li − 4(ki − 1)(li − 1)) =

i =ηt +1 (k i l i − 2(k i − 1)(l i − 1)). From statements 1 and 3, the maximum

ηt 2

η fixed pair of i =1 k i and i =ηt +1 k i l i . We get the formula in statement

occurs when all the base points are simple 4 if all the base points are simple.

2

We close this section with a theorem constraining when strong μ-bases can exist. We show that if a rational surface has a parametrically strong μ-basis, then the bidegrees (m, n) cannot both be very high. Theorem 3. For a bidegree (m, n) parametrization P(s, t ) with base points in general position, suppose I is (σ1 , σ2 )-regular and I I

is (m + σ1 , n + σ2 )-regular for any bidegree (σ1 , σ2 ). If P(s, t ) has a strong μ-basis then (m − 3)(n − 3) < 3.

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Proof. First of all, formula (19) holds since the assumptions are given same as those of Theorem 2. Suppose three moving planes p, q, r form a parametrically strong μ-basis with bidegrees (σ11 , σ12 ), (σ21 , σ22 ), (σ31 , σ32 ). Then m = σ11 + σ21 + σ31 , n = σ12 + σ22 + σ32 . Moreover by formula (19) the number of moving planes with these bidegrees that follow this surface should be

4(σ11 + 1)(σ12 + 1) − (m + σ11 + 1)(n + σ12 + 1) + 4(σ21 + 1)(σ22 + 1) − (m + σ21 + 1)(n + σ22 + 1) + 4(σ31 + 1)(σ32 + 1) − (m + σ31 + 1)(n + σ32 + 1) +

η

ηi =1

ηi =1 i =1

( p i − 4bi ) ≥ 1, ( p i − 4bi ) ≥ 1, ( p i − 4bi ) ≥ 1.

By Proposition 2 and statement 4 of Corollary

η 4, the number of these moving planes is maximal when the number of simple base points reaches the maximum value ( p i − 4bi ) = η = (m + 1)(n + 1) − 4. Therefore i =1

4(σ11 + 1)(σ12 + 1) − (m + σ11 + 1)(n + σ12 + 1) + (m + 1)(n + 1) − 4 ≥ 1, 4(σ21 + 1)(σ22 + 1) − (m + σ21 + 1)(n + σ22 + 1) + (m + 1)(n + 1) − 4 ≥ 1, 4(σ31 + 1)(σ32 + 1) − (m + σ31 + 1)(n + σ32 + 1) + (m + 1)(n + 1) − 4 ≥ 1.

(21)

Summing over these three inequalities we get

3(σ11 σ12 + σ21 σ22 + σ31 σ32 ) + 3m + 3n − 2mn ≥ 3

(22)

Moreover by (15) the implicit degree of the surface P(s, t ) is

Deg(P) = σ11 σ22 + σ12 σ21 + σ11 σ32 + σ12 σ31 + σ31 σ22 + σ32 σ21 . Therefore by Corollary 1 and formula (4), when the number moving planes is maximal

2mn − (m + 1)(n + 1) + 4 = σ11 σ22 + σ12 σ21 + σ11 σ32 + σ12 σ31 + σ31 σ22 + σ32 σ21 . Since p, q, r from a parametrically strong

(23)

μ-basis,

mn = (σ11 + σ21 + σ31 )(σ12 + σ22 + σ32 ). Therefore

mn − (σ11 σ22 + σ12 σ21 + σ11 σ32 + σ12 σ31 + σ31 σ22 + σ32 σ21 ) = σ11 σ12 + σ21 σ22 + σ31 σ32 . Substituting this result into Equation (23), we get

m + n − 3 = σ11 σ12 + σ21 σ22 + σ31 σ32

(24)

Combining Equations (22) and (24) yields

(m − 3)(n − 3) ≤ 3. Note that the conditions in (21) are relaxed when we deduce (22) from (21). Hence, we can check the condition (m − 3)(n − 3) ≤ 3 especially in boundary situations. We find that P(s, t ) has no strong μ-basis if (m, n) = (6, 4) (resp. (4, 6)). We omit the verification, which is similar to the tests in Algorithm 1. Finally, we conclude that if P(s, t ) has a strong μ-basis, then (m − 3)(n − 3) < 3. 2 Remark 1. Note that Theorem 3 is based on formula (19) in Theorem 2 for the number of moving planes. For a numerically specified surface of bidegree (m, n) but with (m − 3)(n − 3) ≥ 3, there still possibly exists a strong μ-basis (see Example 5). 4. Computing strong μ-bases We can easily implicitize a rational parametric surface if we can find a strong μ-basis for this surface. We are now ready to find a strong μ-basis if one exists for a given rational tensor product surface P(s, t ) with bidegree (m, n) that has base points in general position. 4.1. With the assumption of regularity The base points and their multiplicities are not easy to determine. Here we present algorithms to find surfaces that have strong μ-bases without computing their base points. We could achieve this goal by analyzing condition (13) with all the possible types of base points. Here, however, we prefer to find the strong μ-bases using Theorem 2 because there is much less computation for the number of moving planes in Theorem 2 than in Proposition 3. By Proposition 4, we may miss some cases but we can find surfaces having strong μ-bases quickly. Throughout Section 4.1 we shall assume that I is (σ1 , σ2 )-regular and I I is (m + σ1 , n + σ2 )-regular for any bidegree (σ1 , σ2 ).

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L.-Y. Shen, R. Goldman / Computer Aided Geometric Design ••• (••••) •••–•••

Algorithm 1. Find the surfaces P(s, t ) with bidegree (m, n) that have a strong

μ-basis.

1. Find the set Expected Moving Planes (E M P ) of possible triples [(σ11 , σ12 ), (σ21 , σ22 ), (σ31 , σ32 )] that satisfy Equation (13); 2. Compute the expected degrees of the resultants of each triple bidegree in E M P by formula (15) and for each expected degree find the set Expected Base Points (E B P ) of possible base point types by the degree formula (4) and Proposition 2; 3. For each base point type in E B P , check the number of linearly independent moving planes with associated bidegree in E M P by formula (19); 4. If by (20) the number of linearly independent moving planes is sufficient, compute the moving planes by solving (18) and check formula (12). 5. Output the pairs of base point type and associated triple of bidegrees for the μ-basis if any such pairs exist. Remark 2. We can reduce some computations by using Corollary 3. If we fail to find moving planes with bidegree (σ1 , σ2 ), then moving planes do not exist with lower bidegree by statement 2 of Corollary 3. Moreover we can first check the situation with simple base points for the given degree in step 3. By statement 4 of Corollary 4, if there are not enough moving planes to construct a strong μ-basis for simple base points, then we need not check other situations with multiple base points for this given degree. Example 2 (Bicubic surfaces). To illustrate Algorithm 1, we check surfaces with bidegree (3, 3). From Equation (13), we find that

E M P = {[(0, 1), (1, 0), (2, 2)], [(1, 1), (1, 1), (1, 1)], [(0, 1), (0, 1), (3, 1)], [(0, 1), (1, 1), (2, 1)],

[(0, 2), (1, 0), (2, 1)], [(0, 1), (0, 2), (3, 0)]} (symmetrically equivalent situations are omitted). By Equation (15), we need to consider only the surfaces with implicit degrees {5, 6, 6, 6, 7, 9} associated to EMP. By the degree formulas in Theorem 2 and the upper bound on the number of base points in Proposition 2 for a rational surface P(s, t ) of bidegree (3, 3), the possible types of base points are listed in

B P = {1-ple, 2-ple, 3-ple, 1 × 2-ple, 1 × 3-ple, 2 × 1-ple, 2 × 2-ple, 2 × 3-ple, 3 × 1-ple, 3 × 2-ple, 3 × 3-ple}. Note that by Zheng et al. (2003) the maximum multiplicity of a base point of a surface with bidegree (3, 3) is three. We illustrate one case with implicit degree 6. By the implicit degree formula (4), the possible type set EBP of base points is

{(η1 , · · · , η11 )|η1 + 4η2 + 9η3 + 2η4 + 3η5 + 2η6 + 4η7 + 6η8 + 3η9 + 6η10 + 9η11 = 12} where ηi , i = 1, . . . , 11 are the numbers of base points in B P . We find 268 elements in EBP by solving for non-negative values of (η1 , · · · , η11 ) in this linear Diophantine equation. Now we check the number of moving planes with expected degree in E M P with implicit degree 6, i.e.,

{[(1, 1), (1, 1), (1, 1)], [(0, 1), (0, 1), (3, 1)], [(0, 1), (1, 1), (2, 1)]}. By formula (19) and equations (20), the number of moving planes with bidegree [(1, 1), (1, 1), (1, 1)] associated to some types of base points E B P is sufficient. There are precisely 105 base point types among E P B having strong μ-bases formed by three moving planes with bidegrees (1, 1), (1, 1), (1, 1). For the other cases with implicit degree {5, 7, 9}, we can apply a similar analysis. We find that for bidegree (3, 3) only the surfaces with implicit degree 6 can have strong μ-bases. We now summarize the results in Example 2 as a proposition. Proposition 5. A rational tensor product surface with bidegree (3, 3) having base points in general positions of the types B P = {1-ple; 1 × 2-ple, 1 × 3-ple, 2 × 1-ple, 3 × 1-ple}, where the numbers of base points in B P are {(η1 , η2 , η3 , η4 , η5 )|η1 + 2η2 + 3η3 + 2η4 + 3η5 = 12}, has a strong μ-basis with bidegrees (1, 1), (1, 1), (1, 1). There are similar propositions for the commonly used surfaces with bidegrees (3, 2) and (2, 2). Proposition 6. A rational tensor product surface with bidegree (3, 2) having base points in general positions of the types B P = {1-ple; 1 × 2-ple, 2 × 1-ple, 3 × 1-ple}, where the numbers of base points in B P are {(η1 , η2 , η3 , η4 )|η1 + 2η2 + 2η3 + 3η4 = 8}, has a strong μ-basis with bidegrees (1, 0), (1, 1), (1, 1). Proposition 7. A rational tensor product surface with bidegree (2, 2) having base points in general positions of the types B P = {1-ple; 1 × 2-ple, 2 × 1-ple}, where the numbers of base points in B P are {(η1 , η2 , η3 )|η1 + 2η2 + 2η3 = 5}, has a strong μ-basis with bidegrees (0, 1), (1, 0), (1, 1).

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Table 2 The strong μ-bases for bidegree (m, n), m − n ≤ 1 for parametrizations with simple base points.

μ-basis

Surface

#Base points

Implict degree

Strong

(2, 2) (3, 2) (3, 3) (4, 3) (4, 4) (5, 4)

5 8 12 16 21 26

3 4 6 8 11 14

Bidegrees Bidegrees Bidegrees Bidegrees Bidegrees Bidegrees

Table 3 The strong points.

μ-bases for bidegree (m, 3) for parametrizations with simple base μ-basis

Surface

#Base points

Implict degree

Strong

(5, 3) (6, 3) (7, 3) (8, 3) . . .

20 24 28 32

10 12 14 16

Bidegrees Bidegrees Bidegrees Bidegrees

. . .

. . .

. . .

Table 4 The strong points.

(1, 1), (1, 0), (0, 1) (1, 1), (1, 1), (1, 0) (1, 1), (1, 1), (1, 1) (2, 1), (1, 1), (1, 1) (2, 1), (1, 2), (1, 1) (2, 1), (2, 1), (1, 2)

(2, 1), (2, 1), (1, 1) (2, 1), (2, 1), (2, 1) (3, 1), (2, 1), (2, 1) (3, 1), (3, 1), (2, 1)

μ-bases for bidegree (m, 2) for parametrizations with simple base μ-basis

Surface

#Base points

Implict degree

Strong

(4, 2)

11 10

5 6

Bidegrees (2, 1), (1, 1), (1, 0) Bidegrees (2, 0), (1, 1), (1, 1)

(5, 2)

14 13

6 7

Bidegrees (2, 1), (2, 1), (1, 0) Bidegrees (2, 1), (2, 0), (1, 1)

(6, 2)

17 16

7 8

Bidegrees (3, 1), (2, 1), (1, 0) Bidegrees (2, 1), (2, 1), (2, 0)

(7, 2)

20 19 18

8 9 10

Bidegrees (3, 1), (3, 1), (1, 0) Bidegrees (3, 1), (2, 1), (2, 0) Bidegrees (3, 0), (2, 1), (2, 1)

. . .

. . .

. . .

. . .

Theorem 3 indicates that we can find a strong μ-basis by assuming that the surface has a sufficient number of simple base points. In Tables 2–5, we list the surfaces having strong μ-bases computed by Algorithm 1. The base points in these tables are all simple. Thus if we know the number of simple base points of a rational tensor product surface or the implicit degree of a rational tensor product surface (which is easy to compute – see Section 4.2), we can use these tables to predict whether or not the surface has a strong μ-basis. Moreover if a strong μ-basis exists, then we can simply solve the linear system of equations in (18) to compute the elements of this μ-basis. If we relax certain conditions we can modify Algorithm 1 to compute algebraically strong μ-bases. In this case, we would start with the possible implicit degree Deg(P) of a surface P(s, t ). Algorithm 2. Find the surfaces P(s, t ) with bidegree (m, n) that have an algebraically strong

μ-basis.

1. For each possible implicit degree in [2, 2mn], find the set E B P of possible base point types by the degree formula (4) and Proposition 2; 2. For each possible implicit degree in [2, 2mn], find the set E M P of possible triples [(σ11 , σ12 ), (σ21 , σ22 ), (σ31 , σ32 )] satisfying formula (15) with (i , j ) ≥ (0, 0) in (11); 3. For each pair of base point type and associated triple of bidegrees from E B P and E M P corresponding to the same implicit degree, check the number of linearly independent moving planes by formula (19); 4. If by (20) the number of linearly independent moving planes is sufficient, compute the moving planes by solving (18) and check formula (12); 5. Output the pairs of base point type and associated triple of bidegrees for the μ-basis if any such pairs exist. There are simplifications for Algorithm 2 but we omit this discussion which is similar to Remark 2.

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L.-Y. Shen, R. Goldman / Computer Aided Geometric Design ••• (••••) •••–•••

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Table 5 The strong points.

μ-bases for bidegree (m, 1) for parametrizations with simple base μ-basis

Surface

#Base points

Implict degree

Strong

(2, 1) (3, 1) (4, 1)

2 4 6 5 8 7 6

2 2 2 3 2 3 4

Bidegrees Bidegrees Bidegrees Bidegrees Bidegrees Bidegrees Bidegrees

. . .

. . .

. . .

(5, 1)

. . .

(1, 0), (1, 0), (0, 1) (1, 1), (1, 0), (1, 0) (2, 1), (1, 0), (1, 0) (2, 0), (1, 1), (1, 0) (3, 1), (1, 0), (1, 0) (2, 1), (2, 0), (1, 0) (2, 0), (2, 0), (1, 1)

Remark 3. The μ-bases are computed in both Algorithm 1 and Algorithm 2. We can output these μ-bases whenever they are needed. Alternatively, if we know the number of base points, or if we know the implicit degree of the surface, then we can use Tables 2–5 to predict the bidegrees of the elements of the μ-basis and then solve the linear system of equations in (18) to compute a strong μ-basis. Remark 4. The μ-bases for rational ruled surfaces are well studied in Chen and Wang (2003), Chen et al. (2001), where the computation of μ-bases is based on Gaussian elimination on vectors. A rational ruled surface always has an algebraically strong μ-basis (Shen and Goldman, 2017). The μ-bases computed in Chen and Wang (2003), Chen et al. (2001) are algebraically strong but not necessary parametrically strong. In Table 5, we list the cases where a rational ruled surface has a strong μ-basis. 4.2. For a numerically specified surface In the previous subsection, we give algorithms to find surfaces that have parametrically or algebraically strong μ-bases. To apply formula (19) and reduce the amount of computation, we assume that the base points are of two possible types, I is (σ1 , σ2 )-regular and I I is (m + σ1 , n + σ2 )-regular. Using Algorithm 1 and Algorithm 2, we can find some surface types having parametrically or algebraically strong μ-bases. In practical computations, however, we need to deal with surfaces with numerically specified coefficients. In fact, there are specific surfaces having parametrically or algebraically strong μ-bases that are not included in Tables 2–5. Given a rational surface P(s, t ), we can compute its implicit degree Deg(P) instead of checking all possible degrees in [2, 2mn]. Deg(P) is the number of intersection points of a random line generated by the intersection of two random planes

α1 x + β1 y + γ1 z + δ1 w = α2 x + β2 y + γ2 z + δ2 w = 0 with the surface P(s, t ) (Chionh and Goldman, 1992). The s values corresponding to intersection points are roots of the univariate resultant

H (s) = Rest (α1 a + β1 b + γ1 c + δ1 d, α2 a + β2 b + γ2 c + δ2 d). However, H (s) has extraneous roots coming from the base points. To remove these extraneous roots, make a different ˜ 1 , . . . , δ˜2 . Using these constants in the above resultant formula, we get a polynomial H˜ (s) with the same random choice α extraneous roots as H (s). Then it follows easily that

˜ (s))) Deg(P) = Deg( H (s)) − Deg(gcd( H (s), H since, by assumption, the surface is properly parametrized by P(s, t ). Algorithm 3. Find the algebraically strong

μ-bases for a surface P(s, t ) with bidegree (m, n) if any exist.

1. Compute the implicit degree Deg(P) of the surface; 2. For the implicit degree Deg(P), find the set E B P of possible base point types by the degree formula (4) and Proposition 2; 3. For the implicit degree Deg(P), find the set E M P of possible triples [(σ11 , σ12 ), (σ21 , σ22 ), (σ31 , σ32 )] satisfying formula (15) with (i , j ) ≥ (0, 0) in (11); 4. For each pair of base point type and associated triple of bidegrees from E B P and E M P corresponding to the same implicit degree, compute the linearly independent moving planes by solving (18); 5. If the number of linearly independent moving planes is sufficient for (20), then check formula (12); 6. Output the pairs of base point type and associated triple of bidegrees for the μ-basis if any such pairs exist.

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Remark 5. For a given rational surface P(s, t ) with bidegree (m, n), the computations are much simpler if we know the base points and their multiplicities. The number of elements of E M P will be much reduced since the implicit degree is fixed. One significant difference between Algorithm 3 and Algorithm 2 is that we compute the number of moving planes in Step 3 of Algorithm 3 by formula (19) and in Step 4 of Algorithm 2 by counting the moving planes found by solving the linear system (18). Without the regularity assumptions, formula (19) may fail for a specific numerical parametrization. We give two examples to show that formula (19) may fail. This situation is not covered by Algorithms 1 and 2, so in this case we must use Algorithm 3. Example 3. Consider the following rational surface with bidegree (2, 2)

⎛ ⎜ ⎜ ⎝

⎞T

50st 2 + 118st − s2 − t 2 − 2s + 2t 3 3 ⎟ 70st 2 167st 2 2 −2s t − 1 + 3 − 3 − t 2 + s − 2t ⎟ ⎟ 2 2 2 2 ⎠

2s2 t 2 + 3 −

P(s, t ) = ⎜

s t + 2st − 8st − 2t − 2t

2s2 t 2 + 2 −

28st 2 3

59st 3

+

.

− s2 − 2t 2 − s

P(s, t ) has four simple base points (1, 0), (2, 3), (−1, 2), (0, 1) and they are in general position since using Equation (5) we can check that Rank( M B ) = 4. The implicit degree of this surface is four. There is a triple of moving planes that follow this surface with bidegrees (1, 1), (1, 1), (1, 0) whose resultant degree is also four. By formula (19), there should be N P (1, 0) = 8 − 12 + 4 = 0 moving planes that follow this surface with bidegree (1, 0). However, we can actually find such a moving plane by solving (18)



p=

215 s 59

+

330 59

,s +

18 59

,2s +

312 59

,−

215 s 59

−

486



.

59

There are four moving planes with bidegree (1, 1) that follow this surface. We can select two of these moving planes that follow this surface

q = (−st +

295 36

+

2365 t 295 , 12 3

+

295 s 36

r = (−10/3 − 215 t , −8 − 7/3 s +

−

3991 t , − 9263 54 54

215 t 472 , 9 9

+

295 s 18

− 14/3 s −

+

18313 t , st 18

−

148457 t ), 108

860 t t , 1 + 3440 ), 3 9

such that p, q, r are R[s, t ]-linearly independent. The sum of the bidegrees of these three moving planes is (3, 2) which is greater than (2, 2). One can check that these three moving planes satisfy equation (9) and their resultant is exactly the implicit equation of P(s, t ); hence they form an algebraically strong μ-basis for P(s, t ). We omit the long expressions here. Thus here we have an example of a μ-basis that is algebraically strong, but not parametrically strong. See Shen and Goldman (2017) for a theoretical explanation of this anomaly. Example 4. Consider the following rational surface with bidegree (2, 2)

⎞T

⎛

s2 t − 2 st 2 − s2 + t 2 ⎜ 2 s2 t + 3 st 2 + 12 s2 + 2 t 2 ⎟ ⎟ P(s, t ) = ⎜ ⎝ −2 s2 t + 2 st 2 − 2 s2 + 3 t 2 ⎠ . 2 s2 t + st 2 + s2 + 2 t 2 P(s, t ) has a double base point at (0, 0) and a simple base point at (∞, ∞) and these base points are in general position since using Equation (5) we can check that Rank( M B ) = 4. The implicit degree of this surface is three. There is a triple of moving planes that follow this surface with bidegrees (1, 1), (1, 0), (0, 1) whose resultant degree is also three. By formula (19), there should be N P (1, 0) = max(8 − 12 + 3, 0) = 0 and N P (0, 1) = 0 moving planes with bidegrees (1, 0) and (0, 1) that follow this surface. However, we can actually find such moving planes that follow this surface by solving (18)

p = (70 s + 110, 14 s + 15, 49 s, −70) , q = (20 t + 25, 25 t + 19, 49, −35 t − 105). By Theorem 2 there should be only three moving planes that follow this surface with bidegree (1, 1) but actually by Proposition 3 there are five such moving planes. We select one of them



r=

3390 st 133

+

490 s 19

+ 40 t ,

48 st 133

+ t , 21 st −

210 s 19

− 14 t ,

150 st 19

+

70 s



19

such that p, q, r are R[s, t ]-linearly independent. One can check that these three moving planes form a strong μ-basis of P(s, t ). This situation with a double base point is not included in Algorithms 1 and 2, but we can find this strong μ-basis using Algorithm 3.

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Example 5. Consider the following rational surface with bidegree (6, 4) P(s, t ) = (4 s6 t 4 − 10 s6 t 3 − 2 s5 t 4 + 8 s6 t 2 − 4 s4 t 4 − 2 s6 t + 12 s5 t 2 + 16 s4 t 3 − 18 s3 t 4 − 6 s5 t − 10 s4 t 2 + 2 s3 t 3 + 16 s2 t 4 − 3 s4 t + 7 s3 t 2 − 12 s2 t 3 − 8 st 4 + s3 t − 2 t 2 s2 − 2 st 3 + 4 t 4 + 2 s3 + 20 s2 t − 18 t 3 + 11 st + 10 t 2 − 5 s − 9 t − 2, −4 s6 t 3 + 6 s6 t 2 − 2 s5 t 3 + 4 s4 t 4 − 2 s6 t + 4 s5 t 2 + 4 s4 t 3 − 10 s3 t 4 − 2 s5 t − 20 s4 t 2 + 38 s3 t 3 − 4 s2 t 4 + 4 s4 t + 4 s3 t 2 − 20 s2 t 3 + 10 st 4 − 2 s3 t − 2 t 2 s2 − 16 st 3 + 14 t 4 + 2 s3 + 13 s2 t − 28 st 2 + 8 t 3 + 2 st + 21 t 2 − 2 s − 4 t − 4, −4 s6 t 3 + 8 s5 t 4 + 10 s6 t 2 − 22 s5 t 3 − 4 s6 t + 22 s5 t 2 − 2 s4 t 3 − 12 s3 t 4 − 8 s5 t + 6 s4 t 2 + 24 s3 t 3 − 18 s2 t 4 − 6 s4 t + 10 s3 t 2 − 12 s2 t 3 − 10 st 4 − 16 s3 t + 28 t 2 s2 + 20 st 3 − 8 t 4 + 4 s3 + 16 s2 t − 11 st 2 − 2 t 3 + 4 s2 + 8 st + 4 t 2 − 2 s − 15 t − 4, 4 s5 t 4 − 8 s5 t 3 − 4 s4 t 4 + 4 s5 t 2 + 2 s4 t 3 + 2 s3 t 4 − 4 s5 t + 14 s4 t 2 − 4 s3 t 3 − 22 s2 t 4 − 6 s4 t + 14 s3 t 2 + 20 s2 t 3 + 10 st 4 − 2 s3 t + 6 t 2 s2 − 20 st 3 + 4 t 4 − 18 s2 t − 5 st 2 + 16 t 3 + 4 s2 + 6 st − 16 t 2 + 2 s − 3 t + 4). For a general rational parametrization with bidegree (m, n) = (6, 4), by formula (19) there should be at most N P (2, 1) = max(24 − 54 + 31, 0) = 1 moving planes with bidegree (2, 1) that follow this surface occurring with a maximum of (m + 1)(n + 1) − 4 = 31 simple base points. However, for the numerically specified surface P(s, t ) given in this example, we can actually find two moving planes with bidegree (2, 1) that follow P(s, t ) by solving (18)





p = −2 st − 2 t − 2, −st − 1, st + s − t + 2, 2 s2 t − s2 + st − s , q = (2 st − 2 s − t + 2, s2 t − 2 s2 + 2 t − 1, −s2 t + s2 + 2 t − 2, st + s − 2 t − 2). Hence, formula (19) does not hold for this example. We can also find a moving plane with bidegree (2, 2) that follows P( s , t )



r = 2 s2 t + 2 t 2 + 2 t , 2 t 2 s2 − 2 s2 t + 2, −2 st 2 − 2 st + 2 t 2 − 1, 2 t 2 + 1



such that p, q, r are R[s, t ]-linearly independent. One can check that these three moving planes form a strong μ-basis of P(s, t ). This example is not included in Theorem 3 since (m − 3)(n − 3) = 3 < 3. Certainly too, the situation in this example is not included in Algorithm 1. Using Algorithms 1 and 2 in subsection 4.1 we can find surfaces having strong μ-bases or algebraically strong μ-bases. However, as we have just seen, there may be more surfaces having strong μ-bases or algebraically strong μ-bases when the coefficients are specified numerically. For a surface with numerical coefficients we can use Algorithm 3 to find an algebraically strong μ-basis. Clearly, we can find strong μ-bases for surfaces with numerical coefficients by applying similar approaches. 5. Conjecture The degrees of the elements of a μ-basis for a fixed rational curve are unique, but the degrees of the elements of a μ-basis for a fixed rational surface need not be unique. Indeed there are examples of rational surfaces that have two very different μ-bases: one parametrically and algebraically strong and another neither parametrically nor algebraically strong (Shen and Goldman, 2017). As we observed in the Introduction, arbitrary μ-bases for rational surfaces fail to share many of the characteristic properties of μ-bases for rational curves. Strong μ-bases for rational surfaces seem more like the natural analogues of μ-bases for rational curves. If this insight is valid, then just as the degrees of the elements of a μ-basis for rational curves are unique, we would also expect the following conjecture to hold for strong μ-bases for rational tensor product surfaces. Conjecture. For rational tensor product surfaces, the bidegrees of the elements of a strong μ-basis are unique. We have verified this conjecture numerically with many examples – see Tables 2–5. We can also check the validity of this conjecture with randomly chosen (m, n) if the computing time is sustainable. However, in the future we would, of course, prefer to find a rigorous proof. One possible way to prove this conjecture is to first check for the existence of the moving planes that can form a μ-basis and second check the independence of these moving planes. To check for the existence of the moving planes that can form a μ-basis is equivalent to determining the existence of solutions of a semi-algebraic system consisting of equalities and inequalities. To check for the independence of these moving planes requires additional algebraic analysis. In the following example, we show that checking for the independence of moving planes is indeed necessary. Example 6. For an example of the Conjecture, consider a surface P(s, t ) with bidegree (m, n) = (5, 2). Suppose P(s, t ) has 14 simple base points in general position. If we care only about the number of moving planes that follow this surface, then by formula (19) there are potentially three possible strong μ-bases

p, q, r with bidegrees (2, 1), (2, 1), (1, 0); p, q, r with bidegrees (2, 2), (2, 0), (1, 0); p, q, r with bidegrees (3, 1), (1, 1), (1, 0) since bideg(p) + bideg(q) + bideg(r) = bideg(P) and DegX (Res(s,t ) (p · X, q · X, r · X)) = Deg(P) = 6.

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However, for the last two triples of these moving planes, the three moving planes are R[s, t ]-linearly dependent, i.e., [p, q, r] = 0. Hence, we do finally obtain a strong μ-basis formed by p, q, r with unique bidegrees (2, 1), (2, 1), (1, 0) (see Table 4). 6. Conclusion We presented fast algorithms for finding strong μ-bases when they exist for rational tensor product surfaces. Using these algorithms, we generated tables of rational tensor product surfaces with strong μ-bases based on the bidegree of the rational surface and the number of simple base points or the degree of the implicit equation of the parametrization. We have implemented these algorithms using Maplesoft 2015 on an Acer Aspire S7 Ultrabook with RAM 4G and Intel(R) Core(TM)@1.8GHz. We can quickly find a strong μ-basis for a given rational tensor product surface if one exists, since we need only solve a simple system of linear equations. The algorithms avoid computing the base points explicitly by enumerating all the possible types of the base points. For each of these types, we need only solve linear systems to establish the existence of a strong μ-basis. If we know the types of the base points of a rational tensor product surface or the implicit degree of a rational tensor product surface, we can reduce the amount of computation by avoiding checking some redundant types of base points. Acknowledgements The first author would thank Dr. Chun-ming Yuan for helpful discussions concerning the proof of Theorem 3. The authors wish to thank the anonymous reviewers for their comments and suggestions. References Adkins, W.A., Hoffman, J.W., Wang, H.H., 2005. Equations of parametric surfaces with base points via syzygies. J. Symb. Comput. 39 (1), 73–101. Busé, L., Cox, D.A., D’andrea, C., 2003. Implicitization of surfaces in P 3 in the presence of base points. J. Algebra Appl. 2 (2), 189–214. 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