Implicitizing rational surfaces without base points by moving planes and moving quadrics

Computer Aided Geometric Design 70 (2019) 1–15

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

Implicitizing rational surfaces without base points by moving planes and moving quadrics ✩ Yisheng Lai a , Falai Chen b,∗ , Xiaoran Shi c a b c

Department of Mathematics, Zhejiang Gongshang University, Hangzhou, Zhejiang 310018, China Department of Mathematics, University of Science and Technology of China, Hefei, Anhui 230026, China Department of Mathematics, Harbin Institute of Technology, Harbin 150001, China

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 5 August 2018 Received in revised form 1 March 2019 Accepted 2 March 2019 Available online 19 March 2019

It was proven by Cox, Goldman and Zhang that a tensor product rational surface without base points can be implicitized by moving quadrics whenever the rational surface doesn’t contain low degree moving planes following it. However, when the rational surface does have low degree moving planes, Cox, Goldman and Zhang’s method fails. In this paper, we show that a rational surface without base points can always be implicitized by moving quadrics together with moving planes whether the rational surface has low degree moving planes or not. A specific method is also provided to construct the moving planes and moving quadrics that comprise a compact determinantal representation of the implicit equation of the rational surface. © 2019 Elsevier B.V. All rights reserved.

Keywords: Rational surface Implicitization Moving plane Moving quadric Syzygy module

1. Introduction Given a rational parametric surface as defined by

x=

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

, y=

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

, z=

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

where a, b, c , d are bivariate polynomials in s and t, the implicitization problem consists in finding a trivariate polynomial f (x, y , z) such that the equation f (x, y , z) = 0 defines the closed image of the given parameterization. f (x, y , z) = 0 is called the implicit equation of the parametric surface. Implicitization is a fundamental problem in Computer Aided Geometric Design and Geometric Modeling and there are numerous applications related to it. Typical methods for solving the implicitization problem include resultants, Gröbner bases, Wu’s characteristic sets, etc. The resultant-based methods have been widely used in geometric modeling (Sederberg et al., 1984; Chionh and Goldman, 1992; Chionh et al., 2000; Zhang, 2000), but they are generally invalid when the rational surface has base points. Methods based on Gröbner bases and Wu’s characteristic sets can find the implicit equation of the rational surface even when the rational surface contains base points, yet they are too time-consuming to be practical (Buchberger, 1989; Cox et al., 1992; Gao and Chou, 1992; Li, 1989; Wu, 1994). One innovation for efficiently computing the implicit equation is the method of moving surfaces proposed by Sederberg and Chen (Sederberg and Chen, 1995). The method of moving surfaces can express the implicit equation of a rational surface

✩

*

This paper has been recommended for acceptance by Rida Farouki. Corresponding author. E-mail addresses: [email protected] (Y. Lai), chenfl@ustc.edu.cn (F. Chen), [email protected] (X. Shi).

https://doi.org/10.1016/j.cagd.2019.03.001 0167-8396/© 2019 Elsevier B.V. All rights reserved.

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Y. Lai et al. / Computer Aided Geometric Design 70 (2019) 1–15

as a determinant of smaller size, and is even simpler when base points exist. Unfortunately, while extensive experiments show that the method of moving surfaces never fails, there lack explicit constructions of moving surfaces and rigorous proofs of the corresponding determinant being always non-vanishing. Thereafter a lot of work has emerged on the validity, compactness and efficiency of implicitization method by moving surfaces (Adkins et al., 2005; Botbol and Dickenstein, 2016; Busé et al., 2003; Busé and Dohm, 2007; Chen et al., 2001, 2003, 2005, 2007 and 2008; Cox et al., 2000, and 2003; Hong et al., 2017; D’Andrea, 2001; Deng et al., 2005; Dohm, 2009; Jia, 2014; Khetan and D’Andrea, 2006; Lai and Chen, 2016, 2017; Shen and Goldman, 2017a, 2017b; Shi and Goldman, 2012; Wang and Chen, 2012; Zheng et al., 2009; Zhang, 2000). For a more complete review of the most recent developments on surface implicitization, the reader is referred to Chen (2014) and Jia et al. (2018). For a rational surface without base points, Cox et al. (2000) proved that the method of moving quadrics is valid provided that there is no moving plane of low degree that follows the surface (here ’low degree’ will be specified in later context). In this case, the authors of the present paper recently developed efficient algorithms to compute the moving quadrics from moving planes (Lai and Chen, 2016 and 2017). However, when the rational surface contains low degree moving planes, the method of moving quadrics fails to implicitize the rational surface. In this paper, we extend the result in Cox et al. (2000) to a more general case–the parametric surface without base points may contains low degree moving planes, and show that it can be always implicitized by moving quadrics together with moving planes. Specific method is also provided to construct the moving planes and the moving quadrics that form a determinant representation of the implicit equation of the rational surface. We organize the manuscript as follows. Section 2 recalls some preliminary knowledge about moving surface method for implicitization, and provides an example to demonstrate that the method of moving quadrics (Cox et al., 2000) is invalid when the rational surface contains low degree moving planes. Section 3, 4 and 5 present theoretic results to validate our method of implicitizing rational surfaces without base points by a combination of moving planes and moving quadrics. Some examples are also provided to demonstrate the correctness of our method. We conclude the paper in Section 6 with some future research problems. 2. Preliminary knowledge This section briefly reviews some preliminary knowledge about the implicitization method of moving planes and moving surfaces (Sederberg and Chen, 1995). An example is also provided to demonstrate that the method of moving quadrics (Cox et al., 2000) fails when there is a low degree moving plane following the parametric surface. Let C[s], C[s, t ] and C[X] be the polynomial rings over the complex number filed C , where X = (x, y , z, w ) are homogeneous variables in a three dimensional space. For a polynomial ring R, R k refers to the set of k-tuples whose entries are in R, and R k×l denotes the set of k × l matrices whose elements are in R. We further use R l to denote the set of polynomials in R whose degree is less than or equal to l, and R lk to denote the set of k-tuples whose elements are in R l . A rational tensor product surface of bi-degree (m, n) in homogeneous form is defined by

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

(1)

where a, b, c , d ∈ C[s, t ] are bi-degree (m, n) polynomials and gcd(a, b, c , d) = 1. We always assume that m ≥ n ≥ 1 and the rational surface (1) is properly parameterized. We can write P(s, t ) with homogeneous parameters P(s, u ; t , v ). A base point of the surface P(s, t ) is a parameter pair (s0 , u 0 ; t 0 , v 0 ) in P 1 × P 1 for which P(s0 , u 0 ; t 0 , v 0 ) = (0, 0, 0, 0). A moving surface of degree k is a family of algebraic surfaces

S (X; s, t ) :=

σ 

f i (X)b i (s, t ) = 0

(2)

i =1

with one algebraic surface corresponding to each parameter pair (s, t ), where f i (X) ∈ R[X]k , b i (s, t ) ∈ C[s, t ], i = 1, . . . , σ . When f i (X) is linear or quadric, the moving surface ia called a moving plane or a moving quadric respectively. b i (s, t ) ∈ C[s, t ], i = 1, . . . , σ , are called blending functions which are linearly independent. σ ,σ A typical choice for the blending functions is {bk (s, t )}kσ=1 = {si t j }i =1 0, 2j =0 . In this case, a moving plane of bi-degree (σ1 , σ2 ) takes the form

L (X; s, t ) := A (s, t )x + B (s, t ) y + C (s, t ) z + D (s, t ) w

=

σ2 σ1   ( A i j x + B i j y + C i j z + D i j w ) s i t j = 0,

(3)

i =0 j =0

where A (s, t ), B (s, t ), C (s, t ), D (s, t ) ∈ C[s, t ] and A i j , B i j , C i j , D i j ∈ C . Sometimes we also write a moving plane in a vector form L(s, t ) = ( A (s, t ), B (s, t ), C (s, t ), D (s, t )) ∈ C[s, t ]4 .

Y. Lai et al. / Computer Aided Geometric Design 70 (2019) 1–15

3

Similarly, a moving quadric of bi-degree (σ1 , σ2 ) can be written as

Q (X; s, t ) :=

σ2 σ1  

( A i j x2 + B i j y 2 + C i j z2 + D i j xy + E i j xz + F i j yz

i =0 j =0

+G i j xw + H i j y w + I i j zw + J i j w 2 )si t j = 0.

(4)

A moving surface S (X; s, t ) is said to follow the rational surface P(s, t ) defined by (1) if

S (P(s, t ); s, t ) ≡ 0.

(5)

In the following, when we say a parametric surface has a moving surface, it means that the moving surface follows the parametric surface. In order to compute the bi-degree (σ1 , σ2 ) moving planes of the parametric surface P(s, t ), we substitute X = P(s, t ) into the equation (3) to get L (P(s, t ); s, t ) ≡ 0, and set the coefficients of all the monomials si t j in L (P(s, t ); s, t ) to be zero. Then a linear system of equations is obtained with A i j , B i j , C i j , D i j , i = 0, 1, . . . , σ1 , j = 0, 1, . . . , σ2 being unknowns. Solving the linear system of equations arrives at the required moving planes. Moving quadrics of bi-degree (σ1 , σ2 ) can be similarly found from Q (P(s, t ); s, t ) ≡ 0. The core stone of the moving surface method is the following proposition. Proposition 2.1. (Sederberg and Chen, 1995) Let S i (X; s, t ) :=

σ 

h i j (X)b j (s, t ) = 0,

i = 1, 2, . . . , σ ,

j =1

be a set of moving surfaces following P(s, t ). Define

  h11 (X) · · ·   .. .. f (X) =  . .   hσ 1 (X) · · ·



h1σ (X) 

 .  hσ σ (X)  .. .

(6)

If the degree of f (X) equals to the degree of the implicit equation of P(s, t ), then f (X) = 0 is the implicit equation of P(s, t ). In the absence of base points and if there is no moving plane of bi-degree (m − 1, n − 1) following P(s, t ), Cox et al. (2000) proved that P(s, t ) can be implicitized by mn linearly independent moving quadrics of bi-degree (m − 1, n − 1), that is, the implicit equation of P(s, t ) can be written as an mn × mn determinant whose entries are quadratic polynomials in X. However, the following example demonstrates that this conclusion does not extend to the case where the parametric surface P(s, t ) contains bi-degree (m − 1, n − 1) moving planes. Example 2.2. Given a parametric surface P(s, t ) of bi-degree (2, 2):

a = (s + 2t )(1 − s + 2t ), b = (s + t )(−2 + s + t ), c = (s + t )(1 − s + t ), d = (1 + t 2 )(s2 + 1). It is easy to find one moving plane of bi-degree (1, 1):

P 1 = (−1 + s − t ) y + (−2 + s + t ) z, and 6 moving quadrics Q 1 , Q 2 , Q 3 , Q 4 , Q 5 , Q 6 of bi-degree (1, 1) (which are linearly independent over C ), where

Q 1 = xP 1 , Q 2 = y P 1 , Q 3 = z P 1 , Q 4 = w P 1 . One can check that any four of the six moving quadrics are linearly dependent over C[X] (since four of them are multiples of the moving plane P 1 ), and this indicates that the maximum number of C[X]-linearly independent moving quadrics of bi-degree (1, 1) is less than 4. Therefore the surface P(s, t ) can’t be implcitized only by moving quadrics of bi-degree (1, 1). The purpose of the current paper is to prove that a rational surface without base points can always be implcitized by moving quadrics together with moving planes even if the rational surface contains low degree moving planes (that is, degree less than or equal to (m − 1, n − 1)). Thus the result in the current paper generalizes that of Cox et al. (2000). In the following, we shall present the main theoretic results in the subsequent sections. We assume that the parametric surface P(s, t ) doesn’t contain any base point throughout the paper. We further assume the resultant Res(a, b, c ) of a(s, t ), b(s, t ), c (s, t ) with respect to s, t is nonzero, and the leading terms of a(s, t ), b(s, t ), c (s, t ) are all nonzero, since after a suitable projective transformation of coordinates these conditions will be satisfied.

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3. Theoretic results on moving planes Before presenting the main results on moving planes, we provide several lemmas. Lemma 3.1. Let ( f 1 , f 2 , f 3 , f 4 ) ∈ C[s]4 and gcd( f 1 , f 2 , f 3 , f 4 ) = 1. Assume ( g 1 , g 2 , g 3 , g 4 ) ∈ C[s]4 such that f 1 g 1 + f 2 g 2 + f 3 g 3 + f 4 g 4 ≡ 0. Then there exist polynomials h i (s) ∈ C[s], i = 1, 2, . . . , 6 such that

= h 1 ( f 2 , − f 1 , 0, 0) + h 2 ( f 3 , 0, − f 1 , 0) + h 3 ( f 4 , 0, 0, − f 1 )

( g1 , g2 , g3 , g4 )

+h4 (0, f 3 , − f 2 , 0) + h5 (0, f 4 , 0, − f 2 ) + h6 (0, 0, f 4 , − f 3 ). Proof. The proof is similar to Lemma 1 in Cox et al. (1998a, 1998b). The details are omitted.

2

Based on Lemma 3.1, we can prove the following lemma which is crucial to prove Theorem 3.9. Lemma 3.2. Let a, b, c , d ∈ C[s, t ], and assume a, b, c , d do not have any common solution in P 1 × P 1 . Let f (s, t ) ∈ C[s, t ] be a polynomial whose degree in t is at most 2n − 1. Then there exist polynomials A , B , C , D ∈ C[s, t ] whose degree in t is at most n − 1 such that

f = a A + bB + cC + dD .

(7)

Proof. Since a, b, c , d do not have any common solution in P 1 × P 1 , the variety V(a, b, c , d) = ∅. By Theorem 1 in Chapter 4 of Cox et al. (1992), the ideal generated by a, b, c , d is a, b, c , d = C[s, t ]. Hence for any f (s, t ) ∈ C[s, t ], there exist polynomials A , B , C , D ∈ C[s, t ] such that

f = a A + bB + cC + dD . Now we prove that we can find A , B , C , D ∈ C[s, t ] such that degt ( A , B , C , D ) ≤ n − 1. Assume degt ( A , B , C , D ) = k. Write

A=

k 

A i (s)t i ,

B=

i =0

k 

B i (s)t i ,

C=

i =0

k 

C i (s)t i ,

D=

k 

i =0

D i (s)t i ,

i =0

where A i , B i , C i , D i ∈ C[s], i = 0, 1, . . . , k, and

a=

n 

ai (s)t i ,

b=

n 

i =0

b i (s)t i ,

i =0

c=

n  i =0

c i (s)t i ,

d=

n 

di (s)t i ,

i =0

where ai , b i , c i , di ∈ C[s], i = 0, 1, . . . , n. We claim that gcd(an , bn , cn , dn ) = 1. Otherwise there exists s0 ∈ C such that an (s0 ) = bn (s0 ) = cn (s0 ) = dn (s0 ) = 0, which implies that (s0 , 1; 1, 0) is a common solution of a, b, c , d in P 1 × P 1 . This contradicts with the assumption that a, b, c , d do not have any common solution in P 1 × P 1 . If k > n − 1, then the leading term in t of a A + bB + cC + dD should be zero, that is,

an A k + bn B k + cn C k + dn D k ≡ 0. By Lemma 3.1, there exist polynomials h i ∈ C[s], i = 1, 2, . . . , 6 such that

( A k , B k , C k , D k ) = h1 (bn , −an , 0, 0) + h2 (cn , 0, −an , 0) + h3 (dn , 0, 0, −an ) + h4 (0, cn , −bn , 0) + h5 (0, dn , 0, −bn ) + h6 (0, 0, dn , −cn ). Let

( A , B , C , D ) = ( A , B , C , D ) − h1 (b, −a, 0, 0) − h2 (c , 0, −a, 0) − h3 (d, 0, 0, −a) − h4 (0, c , −b, 0) − h5 (0, d, 0, −b) − h6 (0, 0, d, −c ). Then it is easy to check that

f = a A + bB + cC + dD

and degt ( A , B , C , D ) ≤ k − 1. This process can be continued until degt ( A , B , C , D ) ≤ n − 1.

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Y. Lai et al. / Computer Aided Geometric Design 70 (2019) 1–15

5

In the following, we use M P k to denote the set of moving planes of bi-degree (k, n − 1) following P(s, t ). Obviously, M P k can be regarded as a vector space over C . If dim( M P k ) = l, then we usually say P(s, t ) has l moving planes of bi-degree (k, n − 1). The following two lemmas are fundamental to prove Theorem 3.13–the main result on moving planes. Lemma 3.3. dim( M P 2m−1 ) = 2mn. That is, there are 2mn moving planes of bi-degree (2m − 1, n − 1) following P(s, t ). Proof. A moving plane of bi-degree (2m − 1, n − 1) takes the form 2m −1 n −1  

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

i =0 j =0

where A i j , B i j , C i j , D i j ∈ C , i = 0, 1, . . . , 2m − 1, j = 0, 1, . . . , n − 1, are undetermined coefficients. The moving plane follows the rational surface P(s, t ) if and only if 2m −1 n −1  

( A i j a(s, t ) + B i j b(s, t ) + C i j c (s, t ) + D i j d(s, t ))si t j ≡ 0.

i =0 j =0 2m−1,n−1

The above equation derives a linear system of equations with { A i j , B i j , C i j , D i j }i , j =1

being unknowns:

Gx = 0, where x = ( A 00 , B 00 , C 00 , D 00 , . . . , A 2m−1,n−1 , B 2m−1,n−1 , C 2m−1,n−1 , D 2m−1,n−1 ) T , and G ∈ C 6n×8n is the coefficient matrix indexed by polynomials si t j a, si t j b, si t j c , si t j d, 0 ≤ i ≤ 2m − 1, 0 ≤ j ≤ n − 1, that is,

(a, b, c , d, · · · , s2m−1t n−1 a, s2m−1t n−1 b, s2m−1t n−1 c , s2m−1t n−1 d) = (1, t , · · · t 2n−1 , · · · , · · · , s3m−1 , s3m−1t , · · · , s3m−1t 2n−1 )G . Thus to prove the conclusion, we only need to show that rank(G ) = 6mn. By assumption, Resultant (a, b, c ) = 0. It follows that the Sylvester matrix S yl(a, b, c ) of a, b, c with respect to s, t has a full rank which is 6mn (see Dixon (1908)). Since S yl(a, b, c ) is a submatrix of G, 6mn = rank( S yl(a, b, c )) ≤ rank(G ) ≤ 6mn. Thus rank(G ) = 6mn. This completes the proof. 2 Lemma 3.4. P(s, t ) doesn’t have any moving plane of bi-degree at most (2m − 1, n − 1) with the following form:

L (X; s, t ) := A (s, t )x + B (s, t ) y + C (s, t ) z = 0.

(8)

Proof. The proof is similar to that of Lemma 3.3. The moving plane L (X; s, t ) can be rewritten as

L (X; s, t ) :=

2m −1 n −1  

( A i j x + B i j y + C i j z ) s i t j = 0,

i =0 j =0 2m−1,n−1

where A i j , B i j , C i j ∈ C are unknown coefficients. From L (P(s, t ); s, t ) ≡ 0, a linear system with { A i j , B i j , C i j }i , j =1 unknowns is obtained:

being

 G x˜ = 0, where x˜ = ( A 00 , B 00 , C 00 , . . . , A 2m−1,n−1 , B 2m−1,n−1 , C 2m−1,n−1 ) T , and  G is a 6mn × 6mn matrix which is indexed by polynomials si t j a, si t j b, si t j c, 0 ≤ i ≤ 2m − 1, 0 ≤ j ≤ n − 1. It is easy to see that  G is exactly the Sylvester matrix S yl(a, b, c ). By assumption, Res(a, b, c ) = 0. Hence det( G ) = 0, which implies that the above linear system has only the trivial solution. Thus there doesn’t exist any moving plane of the form A (s, t )x + B (s, t ) y + C (s, t ) z = 0. 2 Now we are ready to present the main results on moving planes. To begin with, we introduce the notion of syzygy module. Definition 3.5. Let R be a polynomial ring, and M be a submodule of R l . Let F := ( f 1 , f 2 , . . . , f k ) be an ordered k-tuple of elements of M. A syzygy of F is an ordered k-tuple of polynomials g 1 , g 2 , . . . , gk ∈ R such that

g 1 f 1 + g 2 f 2 + . . . + g k f k = 0. The set of all the syzygies of F is a submodule of R k , called the syzygy module of F , and denoted S yz( f 1 , f 2 , . . . , f k ).

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Y. Lai et al. / Computer Aided Geometric Design 70 (2019) 1–15

A core idea of our method is to consider the set of moving planes of P(s, t ) whose degree in t is less than or equal to n − 1:

M n−1 := {( A , B , C , D ) ∈ C[s, t ]4 | a A + bB + cC + dD ≡ 0, degt ( A , B , C , D ) ≤ n − 1}.

(9)

It turns out that M n−1 is isomorphic to a syzygy module over C[s]. In fact, write P(s, t ) = (a, b, c , d) in the following form

a=

n 

ai (s)t i , b =

i =0

n 

b i (s)t i , c =

i =0

n 

c i (s)t i , d =

i =0

n 

di (s)t i ,

i =0

where ai (s), b i (s), c i (s), di (s) ∈ C[s]m , i = 0, 1, . . . , n, and for any L(s, t ) ∈ M n−1 , write L(s, t ) in the form

L(s, t ) =

n −1 

Li (s)t i ,

i =0

where Li (s) ∈ C[s]4 , i = 0, 1, . . . , n − 1. From P(s, t ) · L(s, t ) ≡ 0, we obtain

M (s) · L(s) = 0,

(10)

where M (s) ∈ C[s]2n×4n is a polynomial matrix:

⎛

a0 ⎜ a1

b0 b1

c0 c1

d0 d1

⎞ a

0 ⎜ ⎜ ⎜ a2 b 2 c 2 d2 a1 ⎜ .. .. .. ⎜ .. ⎜ . . . . a2 M (s) = ⎜ ⎜ .. ⎜a b c d ⎜ n n n n . ⎜ an ⎜ ⎜ ⎝

b0 b1

c0 c1

d0 d1

b2

c2

d2

.. .

.. .

.. .

bn

cn

dn

..

..

.

.. ..

..

.

..

.

..

. . a0 b 0 c 0 . a1 b 1 c 1 .

a2

b2

c2

.. .

.. .

.. .

an

bn

cn

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ d0 ⎟ ⎟ ⎟ d1 ⎟ ⎟ d2 ⎟ ⎟ .. ⎟ . ⎠ dn

(11)

2n×4n

and L(s) = (L0 (s), L1 (s), . . . , Ln−1 (s)) T ∈ C[s]4n . It is easy to see that M n−1 is isomorphic to the solution of the equation (10) which is a syzygy module over C[s], denoted by S yz( M (s)). Proposition 3.6. S yz( M (s)) is a free module of dimension 2n. Proof. Since S yz( M (s)) is a module over a univariate polynomial ring C[s], by Theorem 2.1 in Chapter 6 of Cox et al. (1998a), S yz( M (s)) is a free module and its dimension is 4n − rank( M (s)). In the following we will show that rank( M (s)) = 2n and hence the dimension of S yz( M (s)) is 4n − 2n = 2n. Since the rational surface P(s, t ) = (a, b, c , d) doesn’t contain any base point, for any fixed s ∈ C , gcdt (a, b, c , d) = 1. Here gcdt means the GCD with respect to the variable t. Then there exist nonzero constants αi , βi , i = 1, 2, 3, 4 (which may depend on s) such that f := α1 a + α2 b + α3 c + α4 d and g := β1 a + β2 b + β3 c + β4 d have no common factor in t. Thus it is easy to see that

rank( M (s)) = rank( S yl( f , g )) = 2n, where S yl( f , g ) refers to the Sylvester resultant matrix of the polynomials f and g with respect to t. This completes the proof. 2 Definition 3.7. Let p1 (s), p2 (s), . . . , p2n (s) be a basis of the syzygy module S yz( M (s)) with minimal degree in s. Then p1 (s), p2 (s), . . . , p2n (s) is called a minimal basis of S yz( M (s)). Furthermore, denote μi = deg(pi (s)), i = 1, 2, . . . , 2n, and without loss of generality, we assume μ1 ≤ μ2 ≤ . . . ≤ μ2n . Then the rational surface P(s, t ) (or S yz( M (s))) is called to have a minimal basis of type (μ1 , μ2 , . . . , μ2n ).

l

Definition 3.8. For a polynomial vector v(s) = i =0 vi si ∈ C[s]k , where vi ∈ Ck , we call vl = 0 the leading vector of v(s), and denote it by L V (v). The degree of v(s) is the maximum degree of all its entries, and denote it by deg(v). The minimal basis of S yz( M (s)) has the following important properties.

Y. Lai et al. / Computer Aided Geometric Design 70 (2019) 1–15

7

Theorem 3.9. Let p1 (s), p2 (s), . . . , p2n (s) be a minimal basis of the syzygy module S yz( M (s)) with type (μ1 , μ2 , . . . , μ2n ). Then 1. L V (p1 ), L V (p2 ), . . . , L V (p2n ) are C -linearly independent. 2. For any L(s) ∈ S yz( M (s)) with deg L(s) = k, L(s) can be written uniquely as

L(s) =

2n 

h i (s)pi (s),

(12)

i =1

3.

where 2n hi (s) ∈ C[s] and deg(hi pi ) ≤ k, i = 1, 2, . . . , n. i =1 μi = 2mn.

Proof.

(p2n ) are linearly dependent, then there exist constants λi , i = 1, 2, . . . , 2n, at least 1. Suppose that L V (p1 ), L V (p2 ), . . . , L V 2n

one of them is nonzero, such that i =1 λi L V (pi ) = 0. Let k be the largest index such that λk = 0, and set pk (s) = k

i =1 λi s

μk −μi p (s). Since L V (p ) = k λ L V (p ) = 0, one has deg(p ) < μ . We replace p by p , and it is easy to i i k k i =1 i k k k

see that p1 , . . . , pk−1 , pk , pk+1 , . . . , p2n also form a basis of S yz( M (s)), but one of its elements has a lower degree. This contradicts with the degree minimality of the minimal basis. 2. Since p1 (s), p2 (s), . . . , p2n (s) are a minimal basis of the syzygy module S yz( M (s)), one immediately arrives at the equation (12). Thus we only have to prove the degree bounds on the polynomials h i (s), i = 1, 2, . . . , 2n. Suppose on the contrary, max{deg(h i pi )}2n = l > k. Let I := {i | deg(hi pi ) = l}. Comparing the leading term on the both sides of the i =1 equation (12), we obtain



λi L V (pi ) = 0,

i∈I

where λi is the leading coefficient of h i , i ∈ I . This contradicts with the linear independency of L V (p1 ), L V (p2 ), . . . , L V (p2n ). 3. We homogenize ai (s), b i (s), c i (s), di (s) to ai (s, u ), b i (s, u ), c i (s, u ), di (s, u ) respectively, i = 0, 1, . . . , n. Accordingly, the matrix M (s) defined in (11) is homogenized to M (s, u ). Now consider the map of graded R-modules (here R = C[s, u ]) induced by the matrix M (s, u ):

φ : R (−m)4n −→ R 2n ,

φ(v(s, u )) = M (s, u )v(s, u ),

where v(s, u ) ∈ R (−m) and R (−m) indicates the graded R-module such that its part in degree k is R (−m)k = R k−m . The null space of the map φ is exactly the syzygy module S yz( M (s, u )), and by Proposition 3.6, it can be written as a direct sum 4n

S yz( M (s, u )) = R (−m − μ1 ) ⊕ R (−m − μ2 ) ⊕ · · · ⊕ R (−m − μ2n ).

φ induces a map

φk : R (−m)k4n −→ R k2n . By the statement just proved in item 2, the null space of φk is isomorphic to

R k−m−μ1 ⊕ R k−m−μ2 ⊕ · · · ⊕ R k−m−μ2n for sufficiently large k (specifically, k ≥ max (m + μi )). 1≤i ≤2n

By Lemma 3.2, for any g(s, u ) ∈ R k2n , the polynomial f (s, t ) = (1, t , . . . , t 2n−1 ) · g(s, 1) can be written as

f (s, t ) = a A + bB + cC + dD , where A , B , C , D ∈ C[s, t ] and degt ( A , B , C , D ) ≤ n − 1. Expanding the above equation and comparing the terms in t on both sides of the equation, one obtains

M (s) · L(s) = g(s, 1) where L(s) ∈ C[s]4n . Homogenize the above equation, we have

M (s, u ) · L(s, u ) = g(s, u ) which implies that the map φk is onto when k is sufficiently large. Therefore the dimension formula from linear algebra indicates that

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dim( R k−m−μ1 ⊕ R k−m−μ2 ⊕ · · · ⊕ R k−m−μ2n ) = dim R k4n−m − dim R k2n for sufficiently large k. Since dim R i = i + 1 for i ≥ 0, the above equation reduces to 2n 

(k − m − μi + 1) = 4n(k + 1 − m) − 2n(k + 1)

i =1

Simplifying the equation we arrive at

2n

i =1

μi = 2mn. 2

Corollary 3.10. The number of linearly independent moving planes of bi-degree (k, n − 1) following P(s, t ) is Specially, the number of linearly independent moving planes of bi-degree (μ2n − 1, n − 1) is 2n(μ2n − m).

2n

i =1 (k

+ 1 − μi )+ .

Proof. The number of linearly independent moving planes of bi-degree (k, n − 1) is exactly the dimension of the vector space S yz( M (s, u ))k ,  i.e., the set of elements of S yz( M (s, u )) whose degree is less than 2n or equal to k. By item 2 of Theorem 3.9, 2n the number is i =1 (k + 1 − μi )+ . When k = μ2n − 1, the number reduces to i =1 (μ2n − μi ) = 2n(μ2n − m). 2 Corollary 3.11. m ≤ μ2n ≤ 2m. Proof. By Corollary 3.10 and item 3 of Theorem 3.9, the number of linearly independent moving planes of bi-degree (2m − 1, n − 1) is

dim( M P 2m−1 ) =

2n 2n   (2m − μi )+ = 2mn + (μi − 2m)+ . i =1

i =1

On the other hand, dim( M P 2m−1 ) = 2mn by Lemma 3.3. Therefore 2nμ2n ≥

2n

i =1

μi = 2mn, we immediate have μ2n ≥ m. 2

(13)

2n

μi − 2m)+ = 0, and hence μ2n ≤ 2m. Next from

i =1 (

Corollary 3.12. There doesn’t exist any moving plane of bi-degree (μ2n − 1, n − 1) with the following form:

L (X; s, t ) := A (s, t )x + B (s, t ) y + C (s, t ) z = 0 which follows the surface P(s, t ). Proof. The result follows direct from Lemma 3.4 since

μ2n − 1 ≤ 2m − 1 by Corollary 3.11. 2

Hereafter, we use  to denote the set of indices  = {(i , j ) | 0 ≤ i ≤ μ2n − 1, 0 ≤ j ≤ n − 1}, and set γ = 2n(μ2n − m). Now we are able to present the main result in this section–there exist γ moving planes which will be used to construct the implicitization matrix. Theorem 3.13. There exists a subset B = {(i 1 , j 1 ), (i 2 , j 2 ), . . . , (i γ , j γ )} of  such that 1. P(s, t ) doesn’t have moving planes of bi-degree (μ2n − 1, n − 1) in the form



di j wsi t j +

(i , j )∈\B



(ai j x + bi j y + c i j z)si t j = 0.

(14)

(i , j )∈

2. There exists a basis P = { P 1 , P 2 , . . . , P γ } of the vector space M P μ2n −1 , where each moving plane P k takes the form



P k := wsik t jk +

(i , j )∈\B

dkij wsi t j +



(akij x + bkij y + ckij z)si t j = 0,

(15)

(i , j )∈

k = 1, 2, . . . , γ . Proof. Consider a moving plane of bi-degree (μ2n − 1, n − 1) following P(s, t ):

L (X; s, t ) :=

μ n −1 2n −1  i =0

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

j =0

where A i j , B i j , C i j , D i j are unknown coefficients. From L (P(s, t ); s, t ) ≡ 0, one obtains a linear system with 2n(μ2n + m) equations and 4nμ2n unknowns:

Y. Lai et al. / Computer Aided Geometric Design 70 (2019) 1–15

9

Gx T = 0,

(16)

here x = ( A 00 , B 00 , C 00 , D 00 , . . . , . . . , A μ2n −1,n−1 , B μ2n −1,n−1 , C μ2n −1,n−1 , D μ2n −1,n−1 ), and G is a 2n(μ2n + m) × 4nμ2n coefficient matrix indexed by the polynomials {si t j a, si t j b, si t j c , si t j d | (i , j ) ∈ }:

(a, b, c , d, · · · , sμ2n −1t n−1 a, sμ2n −1t n−1 b, sμ2n −1 t n−1 c , sμ2n −1t n−1 d) = (1, t , · · · , t 2n−1 , · · · · · · , sμ2n +m−1 , sμ2n +m−1t , · · · , sμ2n +m−1t 2n−1 )G . By Corollary 3.10, the linear system (16) has

γ = 2n(μ2n + m), that is, G has a full rank.

γ = 2n(μ2n − m) linearly independent solutions. Hence rank(G ) = 4nμ2n −

Let  G 1 be the coefficient matrix indexed by the polynomials {si t j a, si t j b, si t j c | (i , j ) ∈ }.  G 1 is a submatrix of G with size 2n(μ2n + m) × 3nμ2n . Now consider moving planes of bi-degree (μ2n − 1, n − 1) with the form A (s, t )x + B (s, t ) y + C (s, t ) z = 0. Using a similar argument above and Corollary 3.12, we can show that  G 1 also has a full rank, that is, rank( G1) = 3nμ2n . Thus the column vectors of G indexed by {si t j a, si t j b, si t j c | (i , j ) ∈ } are linearly independent. Now since rank(G ) = 2n(μ2n + m) ≥ 3nμ2n , there must exist an index set

B = {(i 1 , j 1 ), . . . , (i γ , j γ )} ⊂ , such that the square matrix  G 2 obtained from G by deleting columns indexed by {si t j d : (i , j ) ∈ B } is nonsingular. In other  words, G 2 is an order 2n(μ2n + m) submatrix of G which is indexed by {si t j a, si t j b, si t j c : (i , j ) ∈ } {si t j d : (i , j ) ∈  \ B }. This implies that P(s, t ) doesn’t have moving planes of bi-degree (μ2n − 1, n − 1) with the form





di j si t j w +

(i , j )∈\B

(ai j x + bi j y + c i j z)si t j = 0.

(i , j )∈

Item 1 is thus proved. Next we come to prove item 2. Let x˜ be the 2n(μ2n + m)-dimensional vector by deleting elements { D i j : (i , j ) ∈ B } from x. By moving all the terms involving { D i j : (i , j ) ∈ B } on the left-hand side of the equation (16) to the right-hand side, (16) can be reformulated as

 G 2 x˜ T = − De T ,

(17)

here D represents the coefficient matrix indexed by the polynomials si 1 t j 1 d, . . . , si γ t j γ d, and e denotes the vector ( D i 1 , j 1 , D i 2 , j 2 , . . . , D i γ , j γ ), where (ik , jk ) ∈ B , k = 1, 2, . . . , γ . G 2 is nonsingular, for any specific choice of e, the equation (17) has a unique solution. Choose e to be the basis Since  vectors εk = (0, . . . , 1, . . . , 0), 1 ≤ k ≤ γ , respectively, we obtain γ linearly independent solutions which correspond to γ moving planes { P 1 , P 2 , . . . , P γ } of bi-degree (μ2n − 1, n − 1) following the surface P(s, t ), and P k takes the form

P k := sik t jk w +



dkij wsi t j +

(i , j )∈\B



(akij x + bkij y + ckij z)si t j = 0,

(i , j )∈

k = 1, 2, . . . , γ . Obviously, P = { P 1 , P 2 , . . . , P γ } forms a basis of M P μ2n −1 . This completes the proof.

2

4. Theoretical results on moving quadrics In this section, we shall present theoretic results on moving quadrics. We will show that independent moving quadrics can be found to compose the implicitization matrix.

η = (2m − μ2n )n linearly

Theorem 4.1. There are no moving quadrics of bi-degree (μ2n − 1, n − 1) of the form

Q :=



( A i j x2 + B i j y 2 + C i j z2 + D i j xy + E i j xz + F i j yz)si t j

(i , j )∈

+



(G i j xw + H i j y w + I i j zw )si t j = 0,

(i , j )∈\B

following P(s, t ), where B is defined in Theorem 3.13. Proof. The moving quadric Q can be rewritten as

Q := ( g 1 x + g 4 y + g 5 z + g 7 w )x + ( g 2 y + g 6 z + g 8 w ) y + ( g 3 z + g 9 w ) z = 0,

(18)

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Y. Lai et al. / Computer Aided Geometric Design 70 (2019) 1–15

where g i (s, t ) ∈ C[s, t ] is a bi-degree (μ2n − 1, n − 1) polynomial, i = 1, 2, . . . , 9 and g 7 (s, t ), g 8 (s, t ), g 9 (s, t ) do not contain terms si t j , (i , j ) ∈ B . Since Q follows P(s, t ), we have

( g 1 a + g 4 b + g 5 c + g 7 d)a + ( g 2 b + g 6 c + g 8 d)b + ( g 3 c + g 9 d)c ≡ 0. Thus

( g 1 a + g 4 b + g 5 c + g 7 d, g 2 b + g 6 c + g 8 d, g 3 c + g 9 d) ∈ S yz(a, b, c ) is a syzygy of bi-degree (μ2n + m − 1, 2n − 1). Since Res(a, b, c ) = 0, by Theorem 3.10 in Adkins et al. (2005), there are polynomials h1 , h2 , h3 ∈ C[s, t ] of bi-degree at most (μ2n − 1, n − 1) such that

g1a + g4 b + g5 c + g7 d = h1 c + h2 b,

g 2 b + g 6 c + g 8 d = −h2 a + h3 c ,

g 3 c + g 9 d = −h1 a − h3 b.

That is, there exist three moving planes of bi-degree (μ2n − 1, n − 1) following P(s, t ):

g 1 x + ( g 4 − h 2 ) y + ( g 5 − h 1 ) z + g 7 w = 0,

h 2 x + g 2 y + ( g 6 − h 3 ) z + g 8 w = 0,

h 1 x + h 3 y + g 3 z + g g w = 0, and each of them takes the form



di j wsi t j +

(i , j )∈\B



(ai j x + bi j y + c i j z)si t j = 0

(i , j )∈

since g 7 (s, t ), g 8 (s, t ), g 9 (s, t ) do not contain terms si t j , (i , j ) ∈ B . It is easy to verify that at least one of the three moving planes is not identically zero, which contradicts Theorem 3.13. 2 The next theorem provides the required number of moving quadrics for constructing the implicitization matrix. Theorem 4.2. There are (2m − μ2n )n linearly independent moving quadrics { Q α β : (α , β) ∈  \ B } of bi-degree (μ2n − 1, n − 1) following the surface P(s, t ), where each moving quadric Q α β takes the form

Q α β : = sα t β w 2 +

+





(G i j xw + H i j y w + I i j zw )si t j

(i , j )∈\B

( A i j x2 + B i j y 2 + C i j z2 + D i j xy + E i j xz + F i j yz)si t j = 0.

(19)

(i , j )∈

Proof. Consider a moving quadric of bi-degree (μ2n − 1, n − 1) following P(s, t ):

Q (X; s, t ) :=

μ n −1 2n −1  i =0

( A i j x2 + B i j y 2 + C i j z2 + D i j xy + E i j xz + F i j yz

j =0

+G i j xw + H i j y w + I i j zw + J i j w 2 )si t j = 0. From Q (P(s, t ); s, t ) ≡ 0, a linear system with 3n(2m + μ2n ) equations and 10nμ2n unknowns is obtained:

Hx = 0,

(20)

where x = ( A 00 , B 00 , . . . , J 00 , . . . , . . . , A μ2n −1,n−1 , B μ2n −1,n−1 , . . . , J μ2n −1,n−1 ) , and H is the 3n(2m + μ2n ) × 10nμ2n coefficient matrix indexed by the polynomials si t j a2 , si t j b2 , si t j c 2 , si t j ab, si t j ac, si t j ad, si t j bc , si t j bd, si t j cd, si t j d2 , (i , j ) ∈ B : T

(a2 , b2 , c 2 , ab, ac , bc , ad, bd, cd, d2 , · · · , · · · , sμ2n −1t n−1 a2 , · · · , sμ2n −1t n−1 d2 ) = (1, t , · · · , t 3n−1 , · · · , · · · , sμ2n +2m−1 , sμ2n +2m−1t , · · · , sμ2n +2m−1t 3n−1 ) H .

Let  H be the matrix obtained from H by deleting the columns indexed by {si t j d2 : (i , j ) ∈ } {si t j ad, si t j bd, si t j cd : (i , j ) ∈ B }. Then  H is a square matrix of order 3n(μ2n + 2m). Now we claim  H is nonsingular. Otherwise, there exists a bi-degree (μ2n − 1, n − 1) moving quadric Q with the form (18) that follows the rational surface P(s, t ). This contradicts Theorem 4.1. Next we let x˜ be the subvector of x by deleting elements {G i j , H i j , I i j : (i , j ) ∈ B } and { J i j | (i , j ) ∈ }. By moving all the terms involving {G i j , H i j , I i j : (i , j ) ∈ B } and { J i j | (i , j ) ∈ } on the left-hand side of equation (20) to its right-hand side, the equation (20) is written as

Y. Lai et al. / Computer Aided Geometric Design 70 (2019) 1–15

11

 H x˜ = − De

(21)

where D is the coefficient matrix indexed by the polynomials {s t d | (i , j ) ∈ } and {s t ad, s t bd, s t cd | (i , j ) ∈ B }, and e is the subvector of x whose elements consist of { J i j | (i , j ) ∈ } and {G i j , H i j , I i j : (i , j ) ∈ B }. Since  H is nonsingular, we can solve for x˜ for any choice of e. Thus we can obtain |e| = (7μ2n − 6m)n linearly independent moving quadrics of bi-degree (μ2n − 1, n − 1) following the surface P(s, t ). Among these moving quadrics, 4γ moving quadrics {xP i , y P i , z P i , w P i | i = 1, . . . , γ } come from the moving planes { P 1 , P 2 , . . . , P γ } defined in Theorem 3.13 by multiplying each moving plane P i with x, y , z, w respectively. In the following, we are interested in moving quadrics which are not linear multiple of the moving planes P i , i = 1, . . . , γ . For each (α , β) ∈  \ B , set i j 2

i j

i j

i j

J α β = 1, J i j = 0 for all (i , j ) = (α , β) and

G i j = H i j = I i j = 0 for all (i , j ) ∈ B . Solving the linear system (21), we obtain (2m − μ2n )n linearly independent moving quadrics { Q α β : (α , β) ∈  \ B } of bi-degree (μ2n − 1, n − 1) of the form (19). It is easy to see that neither of these moving quadrics is a linear multiple of the moving planes P i , i = 1, . . . , γ . The theorem is thus proved. 2 In the remainder of the paper, we set

η = (2m − μ2n )n and  \ B = {(α1 , β1 ), (α2 , β2 ), . . . , (αη , βη )}.

5. Validity of the method and algorithm In Section 3 and 4, we have obtained γ moving planes and η moving quadrics. In this section, we will show that these moving planes and moving quadrics compose a square matrix whose determinant gives the implicit equation of the rational parametric surface P(s, t ). A specific algorithm is also proposed to compute the implicitization matrix. Theorem 5.1. A tensor product parametric surface P(s, t ) can be implicitized by γ moving planes and η moving quadrics of bi-degree

(μ2n − 1, n − 1) following P(s, t ). Proof. According to Theorems 3.13 and 4.2, P(s, t ) has



P k := wsik t jk +

dkij wsi t j +

(i , j )∈\B



γ moving planes of bi-degree (μ2n − 1, n − 1):

(akij x + bkij y + ckij z)si t j = 0, (ik , jk ) ∈ B,

(i , j )∈

k = 1, 2, . . . , γ , and

η moving quadrics of bi-degree (μ2n − 1, n − 1):  Q k : = sαk t βk w 2 + (G kij xw + H kij y w + I kij zw )si t j +



(i , j )∈\B

( A kij x2

+ B kij y 2 + C ikj z2 + D kij xy + E kij xz + F ikj yz)si t j = 0,

(i , j )∈

k = 1, 2, . . . , η, (αk , βk ) ∈  \ B . These moving planes and moving quadrics form a square matrix M (X):

( P 1 , . . . , P γ , Q 1 , . . . , Q η )T = M (X)ξ T , where ξ is the vector comprised of blending functions {si t j | (i , j ) ∈ }. M (X) is an order γ + η = nμ2n matrix whose entries are linear and quadratic polynomials in X. By Proposition 2.1, the determinant det( M (X)) gives the implicit equation of P(s, t ), provided that the degree of det( M (X)) is 2mn. It is easy to check that the degree of det( M (X)) equals to γ + 2η = 2mn if det( M (X)) ≡ 0. Thus we only need to show det( M (X)) doesn’t vanish indeed. By ordering the index set  into (i 1 , j 1 ), . . . , (i γ , j γ ), (α1 , β1 ), . . . , (αη , βη ), the matrix M (X) takes the form

⎛ ⎜ ⎜ ⎜ ⎜ M (X) = ⎜ ⎜ ⎜ ⎜ ⎝

⎞

w + ···

..

⎟ ⎟ ⎟ ⎟ ⎟, ⎟ ⎟ ⎟ ⎠

. w + ···

w2 + · · ·

..

. w2 + · · ·

12

Y. Lai et al. / Computer Aided Geometric Design 70 (2019) 1–15

where in the upper left γ × γ block, only diagonal entries contain terms involving with w, and w 2 term appears only in the last η diagonal entries. Then the determinant of M (X) contains the term w 2mn , which means det( M (X)) ≡ 0. Therefore det( M (X)) = 0 is the implicit equation of the parametric surface P(s, t ), and we call M (X) the implicitization matrix of P(s, t ) hereafter. The theorem is proved. 2 The above theorem proves the validity of the implicitization method for some specific choice of moving planes and moving quadrics. In the following, we show that the method is also valid for generic choice of moving planes and moving quadrics. Theorem 5.2. Let P := { P 1 , P 2 , . . . , P γ } be an arbitrary basis of M P μ2n −1 , and Q (P ) be the set of moving quadrics generated by P , that is,

Q (P ) :=

γ 



li (X) P i | li (X) ∈ C[X]1 ,

i =1

where li (X) is a linear function in X, i = 1, 2, . . . , γ . Let M Q μ2n −1 be the set of moving quadrics of bi-degree at most (μ2n − 1, n − 1) following P(s, t ), and C Q μ2n −1 be a subspace of M Q μ2n −1 complement to Q (P ), that is,

M Q μ2n −1 = Q (P ) ⊕ C Q μ2n −1 . We further assume that Q := { Q 1 , Q 2 , . . . , Q η } is a basis of C Q μ2n −1 . Then P(s, t ) can be implicitized by the

γ moving planes

{ P 1 , P 2 , . . . , P γ } and the η moving quadrics { Q 1 , Q 2 , . . . , Q η }.

Proof. It is easy to see that Q (P ) is a subspace of M Q μ2n −1 with dimension 4γ , and

X (P ) := {xP 1 , y P 1 , z P 1 , w P 1 , . . . , xP γ , y P γ , z P γ , w P γ } is a basis of Q (P ). Since dim( M Q μ2n −1 ) = 4γ + η as proved in Theorem 4.2, dim(C Q μ2n −1 ) = η . Furthermore, η X (P ) { Q i }i =1 forms a basis of M Q μ2n −1 . Let P 1 , P 2 , . . . , P γ be the γ moving planes defined in Theorem 3.13, and Q 1 , Q 2 , . . . , Q η be the η moving quadrics as defined in Theorem 4.2. Furthermore, let M (X) be the implicitization matrix constructed from moving planes P 1 , P 2 , . . . , P γ and moving quadrics Q 1 , Q 2 , . . . , Q η . By Theorem 5.1, det( M (X)) = 0 defines the implicit equation of P(s, t ). Next we denote M (X) the implicitization matrix constructed from moving planes P 1 , P 2 , . . . , P γ and moving quadrics Q 1 , Q 2 , . . . , Q η . Our goal is to show that det( M (X)) = C det( M (X)) for some nonzero constant C . Thus det( M (X)) also implcitizes P(s, t ). We approach the goal as follows. Since { P 1 , P 2 , . . . , P γ } and { P 1 , P 2 , . . . , P γ } are two bases of M P μ2n −1 , there is an order γ invertible matrix H such that

( P 1, P 2, . . . , P γ ) = ( P 1, P 2, . . . , P γ )H . γ

as

On the other hand, since {xP i , y P i , z P i , w P i }i =1

Qk =

γ 

lki (X) P i +

i =1

η 

λki Q i ,

(22)



η

{ Q i }i =1 forms a basis of M Q μ2n −1 , each Q k can be expressed uniquely

k = 1, . . . , η ,

i =1

where lki (X) is a linear function in X, and λki is a constant. Or in matrix form

( Q 1 , . . . , Q η )T = (U , V )( P 1 , . . . , P γ , Q 1 , . . . , Q η )T ,

(23)

where U is an η × γ matrix whose elements are linear functions in X, and V is an order η constant matrix. We claim that the square matrix V is nonsingular. If on the contrary, V is singular, and let vi , i = 1, . . . , η be η rows of V . Then there exist constants νi , i = 1, . . . , η , at η least one of them is nonzero, such that ν v = 0. From (23), one has i =1 i i η 

νk Q k = (ν1 , . . . , νη )( Q 1 , . . . , Q η )T =

γ 

li (X) P i ,

i =1

k =1

where li (X) is a linear function in X. Since Q (P ) = Q (P ), here Q (P ) is similarly defined as Q (P ), η  k =1

νk Q k ∈ Q (P ).

Y. Lai et al. / Computer Aided Geometric Design 70 (2019) 1–15

13

η

On the other hand, 0 = k=1 νk Q k ∈ C Q μ2n−1 . This is impossible since Q (P ) and C Q μ2n−1 are complement to each other. Thus V is nonsingular. Combining equations (22) and (23) gives



M (X) =

H U

0 V



M (X).

It follows that

det( M (X)) = C det( M (X)), where C = det( H ) det( V ) is a nonzero constant since H and V are invertible matrices. The theorem is thus proved.

2

The following corollary confirms that our method generalizes the result in Cox et al. (2000). Corollary 5.3. Suppose the rational surface P(s, t ) doesn’t have moving planes of bi-degree (m − 1, n − 1). Then P(s, t ) can be implicitized by mn linearly independent moving quadrics of bi-degree (m − 1, n − 1) following P(s, t ). Proof. When P(s, t ) doesn’t have moving planes of bi-degree (m − 1, n − 1), it is easy to see that Theorem 3.9, and in this case, γ = 0 and η = mn. The statement follows from Theorem 5.2. 2

μi = m, i = 1, 2, . . . , 2n by

Based on Theorem 5.2, we can derive an algorithm to compute the implicit equation of a rational surface without base points. The key step is to compute a minimal basis of the syzygy module S yz( M (s)), where M (s) is defined in (11). This can be done by first computing a generating set for S yz( M (s)) using the Groebner basis computation as described in Section 3, Chapter 5 of Cox et al. (1998a), and then applying the Gaussian elimination technique adopted in Chen and Wang (2003) to compute a minimal basis from the generating set. After a minimal basis pi , i = 1, 2, . . . , 2n of S yz( M (s)) is obtained, we can compute γ moving planes of bi-degree (μ2n − 1, n − 1) by multiplying pi with s j , j = 0, 1, . . . , μ2n − μi − 1, i = 1, 2 . . . , 2n. The η moving quadrics can be found by solving a linear system of equations from Q (P(s, t ); s, t ) ≡ 0. We outline the algorithm in Algorithm 1. In the following, we provide several examples to demonstrate our algorithm. Algorithm 1 Computing the implicit equation of a rational surface without base points. Input: A rational parametric surface P(s, t ) without base points. Output: A square matrix whose determinant gives the implicit equation of P(s, t ). 1: Compute a minimal basis p1 , p2 , . . . , p2n of the syzygy module S yz( M (s)), where M (s) is defined in (11). Denote μi = deg (pi ), i = 1, 2, . . . , 2n, and without loss of generality, assume μ1 ≤ μ2 ≤ . . . , μ2n . 2: Construct γ = 2n(μ2n − m) moving planes of bi-degree (μ2n − 1, n − 1). This can be done by multiplying each pi by s j , j = 0, 1, . . . , μ2n − μi − 1, i = 1, 2, . . . , 2n. Denote the moving planes by P := { P 1 , P 2 , . . . , P γ }. 3: Compute a basis of the set of moving quadrics M Q μ2n −1 of bi-degree (μ2n − 1, n − 1). 4: Compute a basis Q := { Q 1 , Q 2 , . . . , Q η } of the subspace of M Q μ2n −1 complement to Q (P ). Here η = n(2m − μ2n ), and Q (P ) is defined in Theorem 5.2. 5: Construct the implcitization matrix M (X) from the γ moving planes P and η moving quadrics Q. Then det( M (X)) = 0 is the required implicit equation.

Example 5.4. Consider a tensor product surface P(s, t ) of bi-degree (2, 1):

a = ( s + 1) + t ,

b = 1 + t (s2 + 1),

c = (s2 − s + 1) + ts,

d = s + t (s + 1).

P (s, t ) has a minimal basis respect to the variable s with two elements:

P 1 = (x − w )s − x + y ,

P 2 = 7s3 w − 7( y + z)s2 + (5x − 7z + 9w )s + 2x + 5 y − 7z − 7w .

Thus P(s, t ) has a type (1, 3), and P( s , t ) :

γ = 2, η = 1. By computation, we find one moving quadric of bi-degree (2, 0) following

Q = 7( zw − xz − xy + xw )s2 + (5x − 7z + 9w )(x − w )s + (2x + 5 y − 7z − 7w )(x − w ) which is not a linear multiple of P 1 , P 2 . Then the two moving planes P 1 , s P 1 plus the moving quadric Q form an implicitization matrix:

⎛

x− w 0 M (X) = ⎝ 7( zw − xz − xy + xw )

⎞ −x + y 0 ⎠ x− w −x + y (5x − 7z + 9w )(x − w ) (2x + 5 y − 7z − 7w )(x − w )

and the determinant of M (X) gives the implicit equation of P(s, t ):

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det( M (X)) := −x4 + x3 y + 3x3 z − 2x2 y 2 − 3x2 yz + xy 3 + xy 2 z

+ x3 w + 4x2 y w − 6x2 zw − xy 2 w + 4xyzw − y 2 zw − 2x2 w 2 − 4xy w 2 + 4xzw 2 − yzw 2 + 2xw 3 + 2 y w 3 − zw 3 − w 4 = 0. Example 5.5. Given a biquadratic rational surface P(s, t ):

a = (s + 2t )(1 − s + t ), c = (s + t )(1 − s + t ),

b = (s + t )(−2 + s + t ), d = (1 + t 2 )(s2 + 1),

it has a minimal basis with respect to s of type (1, 1, 3, 3):

P 1 = (−s − t )x + (s + 2t ) z, 2

P 2 = (−1 + s − t ) y + (−2 + s + t ) z,

3

P 3 = (−3s − 2 + 2s t + st )x + (5s2 + 4 − 2s3 t ) z + 2s(−1 + s − t ) w , P 4 = (3 − 2s)(s + t )x − 2(2s2 − 2s + s3 t − s2 t ) y

−(2 − 2s2 − 3s + 2s3 t − 4s2 t ) z − 2(3 − 2s)(s + t ) w . and γ = 4 and η = 2. P 1 , P 2 produces 4 moving planes of bi-degree (2, 1): P 1 , s P 1 , P 2 , s P 2 . We then compute two moving quadrics Q 1 , Q 2 of bi-degree (2, 1) which are complement to Q ( P 1 , P 2 ). The four moving planes P 1 , s P 1 , P 2 , s P 2 and two moving quadrics Q 1 , Q 2 form an order 6 implicitization matrix M (X) which gives the implicit equation of P(s, t ). We omit the detailed expressions for Q 1 and Q 2 since they are too large. Example 5.6. We reconsider Example 2.2. It is easy to compute that P(s, t ) has a minimal basis with respect to s of type (1, 2, 2, 3). The minimal basis produces 4 moving planes P 1 , s P 1 , P 2 , P 3 of bi-degree (2, 1), where P 1 = (−1 + s − t ) y + (−2 + s + t )z. Compute two moving quadrics Q 1 , Q 2 of bi-degree (2, 1) which are complement to Q ( P 1 , P 2 , P 3 ). Then the four moving planes and two moving quadrics form a 6 × 6 determinant, and setting the determinant to zero gives the implicit equation of P(s, t ). Again, the detailed expressions for P 2 , P 3 and Q 1 , Q 2 are omitted. Example 5.7. Consider a bi-cubic Bézier patch of the well known Utah teapot model (Crow, 1987). The handle of the teapot consists of four bicubic Bézier surface patches which do not have base points. One can compute that the minimal basis of each surface patch with respect to s has the type (2, 2, 3, 3, 4, 4). Thus γ = 6 and η = 6, and the surface patch can be implicitized by 6 moving planes and 6 moving quadrics, that is, the implicit equation can be expressed as a determinant of 12 × 12 with 6 linear rows and 6 quadratic rows. 6. Conclusion In this paper, we propose a method to implicitize tensor product rational surfaces without base points using a combination of moving planes and moving surfaces. The implicit equation is expressed as a compact determinant which has a smaller size than the classic Dixon resultant. The method works whether the rational surface has lower degree moving planes or not, and thus generalizes a previous result by Cox et al. Some examples are provided to demonstrate the validity of our method. It is interesting to generalize the current method to the case where the rational surface contains base points. Acknowledgements The authors are grateful to professor David Cox for helpful discussions with him. This work is supported by The National Natural Science Foundation of China (Grant Nos. 11571338, 11271328). References Adkins, W.A., Hoffman, J.W., Wang, H., 2005. Equations of parametric surfaces with base points via syzygies. J. Symb. Comput. 39 (1), 73–101. Botbol, N., Dickenstein, A., 2016. Implicitization of rational hypersurfaces via linear syzygies: a practical overview. J. Symb. Comput. 74, 493–512. Buchberger, B., 1989. Applications of Groebner bases in nonlinear computational geometry. In: Kapur, D., Mundy, J. (Eds.), Geometric Reasoning. Elsevier Science Publisher, MIT Press, pp. 413–446. Busé, L., Cox, D.A., D’andrea, C., 2003. Implicitization of surfaces in P 3 in the presense of base points. J. Algebra Appl. 2, 189–214. Busé, L., Dohm, M., 2007. Implicitization of bihomegenous parametrizations of algebraic surfaces via linear syzygies. In: Proceedings of the 2007 International Symposium on Symbolic and Algebraic Computation. ACM, pp. 69–76. Chen, F., 2014. Recent advances on surface implicitization. J. Univ. Sci. Technol. China 44 (5), 345–361. Chen, F., Cox, D.A., Liu, Y., 2005. The μ-bases and implicitization of a rational parametric surface. J. Symb. Comput. 39, 689–706. Chen, F., Shen, L., Deng, J., 2007. Implicitization and parametrization of quadratic and cubic surfaces by μ-bases. Computing 79, 131–142. Chen, F., Wang, W., 2003. Revisiting the μ-basis of a rational ruled surface. J. Symb. Comput. 36 (5), 699–716.

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