EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 7, No. 2, 2014, 191-200 ISSN 1307-5543 – www.ejpam.com Intersections of Rational Parametrized Plane Curves Mohammed Tesemma 1, Haohao Wang 2,∗ 1 Department of Mathematics, Spelman College, Atlanta, GA 30314 2 Department of Mathematics, Southeast Missouri State University, Cape Girardeau, MO, USA Abstract. In this paper, we introduce and compare three different methods of computing the intersec- tions of rational parametrized plane curves. The common approach of these methods is to apply the µ-basis of the plane curves, and avoid of computing the implicit equations of the curves, which increase the computation efficiency. 2010 Mathematics Subject Classifications: 14Q05, 13D02 Key Words and Phrases: Syzygy, Smith Normal Form, Resultants, Curves 1. Introduction Through this paper, we shall consider two rational plane curves C1 and C2 in the complex projective two-space given as the image of generic one-to-one rational parametrizations: F(s, t) = ( f0(s, t), f1(s, t), f2(s, t)), (s, t) 6= (0, 0), (1) and G(u, v) = (g0(u, v), g1(u, v), g2(u, v)), (u, v) 6= (0, 0), (2) where f0, f1, f2 (respectively, g0, g1, g2) are linearly independent homogeneous polynomials of the same degree d ≥ 2 (respectively, d ′ ≥ 2), and gcd( f0, f1, f2) = 1 (respectively, gcd(g0, g1, g2) = 1). The implicit equation of a parametric curve is a polynomial f in the polynomial ring C[x0, x1, x2] such that f (a0, a1, a2) = 0 whenever [a0, a1, a2] is a point on the parametrized curve, and f is irreducible. Without loss of generality, we let F(x , y, z) = 0 and G(x , y, z) = 0 be the implicit equation of the parametrized curves F(s, t) and G(u, v) respectively. It is known that the intersection number of C1 and C2 is dd ′ counting the intersection multiplicity. In general, one can obtain this information by computing the implicit equation G(x , y, z) = 0 of the curveC2, then solve the equation G( f0(s, t), f1(s, t), f2(s, t)) = 0 for (s, t). ∗Corresponding author. Email addresses: mtesemma@spelman.edu (M. Tesemma), hwang@semo.edu (H. Wang) http://www.ejpam.com 191 c© 2014 EJPAM All rights reserved. M. Tesemma, H. Wang, / Eur. J. Pure Appl. Math, 7 (2014), 191-200 192 (Or computing the implicit equation F(x , y, z) = 0 of the curve C1, then solve the equation F(g0(u, v), g1(u, v), g2(u, v)) = 0 for (u, v).) The aim of this paper is to find the intersection of the curve C1 and the curve C2 only use the parametrizations of the curves without computing the implicit equation of the curves. We will accomplish this by using the µ-basis of the syzygies of the curve. Moving lines and moving planes (syzygies) were introduced into Computer Aided Geo- metric Design by Sederberg, Cox and their collaborators in order to develop robust, efficient algorithms for implicitizing rational curves and surfaces [5, 7–9]. Their success motivated people to develop fast algorithms for computing special bases, called µ-bases, for moving lines and moving planes [1–3, 10]. We begin in Section 2 with a brief review of moving lines and µ-basis of the moving lines or rational parametrized plane curves. In Sections 3, we describe three different different methods of computing the intersections of two parametric curves. For each method, we first provide the theoretical background behind the algorithm, then we give the specific algorithm, and then we illustrate the algorithm via an example. In Section 4, we provide a brief summary of the paper, and a open questions for future study. 2. Moving Lines A moving line is a family of lines with each pair of parameters (s, t) corresponding to a line: L(x0, x1, x2; s, t) = 2 ∑ i=0 Ai(s, t)x i = A0 x0 + A1 x1 + A2 x2, (3) where A0, A1, A2 are homogeneous polynomials in s, t of the same degree. For simplicity, sometimes we write a moving plane as L = (A0, A1, A2). The moving plane (3) follows the parametrization (1) if L · F= 2 ∑ i=0 Ai(s, t) fi(s, t) = A0 f0 + A1 f1 + A2 f2 ≡ 0, (4) that is, if the point on the curve at the parameter (s, t) lies on the plane at the parameter (s, t). Definition 1. Two moving lines p(s, t),q(s, t) are called a µ-basis of the rational plane curve F(s, t) if p,q are moving lines that follow F(s, t) and satisfy the following two conditions: 1. [p,q] = κF(s, t), 2. deg(p) + deg(q)) = deg(F), where κ is some nonzero constant and [p,q] is the outer product of p,q. The notation of a µ-basis for rational space curves can be generalized in an obvious way to rational curves of arbitrary dimension. The existence of a µ-basis for a rational curve in any dimension follows directly from the Hilbert-Burch Theorem [Theorem 20.15, 6]. In particular, M. Tesemma, H. Wang, / Eur. J. Pure Appl. Math, 7 (2014), 191-200 193 the proof of the existence of a µ-basis for a rational curve in an affine n-space is given in [Exercise 17, Page 286, 4]. An alternative existence proof as well as a simple algorithm to compute a µ-basis based solely on Gaussian elimination is presented in [10]. The µ-basis elements p,q for a rational plane curve are not unique. But the degrees of the µ-basis elements µ1 = deg(p), µ2 = deg(q) for a rational plane curve are unique. Song- Goldman [10] proved this result and the following proposition for rational space curves. For the convenience of the reader, we will focus our attention to plane curves, and below is the statement of their result for plane curves. Proposition 1. Suppose p(s, t), q(s, t) are a µ-basis of degrees µ1,µ2 for the rational plane curve F(s, t) and let L(s, t) be a moving curve of degree m that follows the curve. Then there exist polynomials α(s, t), β(s, t) such that L= αp+ βq, where deg(α) = m−µ1, deg(β) = m−µ2. 3. Comparison of Computational Methods In this section, we will introduce three different methods, GCD method, Resultant matrix method, and Smith normal form method, to compute the intersection of two parametrized curves C1 and C2, where C1 and C2 are given as the image of generic one-to-one rational parametrizations: F(s, t) = ( f0(s, t), f1(s, t), f2(s, t)), (s, t) 6= (0, 0), and G(u, v) = (g0(u, v), g1(u, v), g2(u, v)), (u, v) 6= (0, 0), where f0, f1, f2 (respectively, g0, g1, g2) are linearly independent homogeneous polynomials of the same degree d ≥ 2 (respectively, d ′ ≥ 2), and gcd( f0, f1, f2) = 1 (respectively, gcd(g0, g1, g2) = 1). These three methods only use the parametrizations of the curves without computing the implicit equation of the curves. We will accomplish this by using the µ-basis of the syzygies of the curve. For each of the method, we will first prove the validity of the method, and then provide the computational algorithm for each of the method. 3.1. GCD Method First, we will study the GCD method. Theorem 1. With about notation, if p(s, t) and p(s, t) are the µ-basis for the rational plan curve F(s, t), then {(u0, v0) 6= (0,0) | p(s, t) ·G(u0, v0) = q(s, t) ·G(u0, v0) = 0} = {(u0, v0) 6= (0,0) | gcd(Ress(p(s, t) ·G(u0, v0),q(s, t) ·G(u0, v0)), Rest(p(s, t) ·G(u0, v0),q(s, t) ·G(u0, v0))) = 0} ⇔G(u0, v0) ∈ C1 ∩C2. M. Tesemma, H. Wang, / Eur. J. Pure Appl. Math, 7 (2014), 191-200 194 Proof. First, we will show that {(u0, v0) 6= (0,0) | p(s, t) ·G(u0, v0) = q(s, t) ·G(u0, v0) = 0} ⇔ G(u0, v0) ∈ C1 ∩C2. To do so, we let (u0, v0) 6= (0, 0) be such that p(s, t) · G(u0, v0) = q(s, t) · G(u0, v0) = 0. This means that G(u0, v0) ‖ [p(s, t) q(s, t)]. Hence, G(u0, v0) = kF(s, t), and G(u0, v0) ∈ C1 ∩C2. On the other hand, if G(u0, v0) ∈ C1 ∩C2, then G(u0, v0) = F(s0, t0) for some (s0, t0) 6= (0,0). Thus, Definition 1 yields that p(s, t) ·G(u0, v0) = q(s, t) ·G(u0, v0) = 0. Therefore, the claim is proved. Now, we will show that {(u0, v0) 6= (0,0) | p(s, t) ·G(u0, v0) = q(s, t) ·G(u0, v0) = 0} = {(u0, v0) 6= (0,0) | gcd(Ress(p(s, t) ·G(u0, v0),q(s, t) ·G(u0, v0)), Rest(p(s, t) ·G(u0, v0),q(s, t) ·G(u0, v0))) = 0}. Let Gp = p(s, t) ·G(u, v) and Gq = q(s, t) ·G(u, v) be polynomials in variables s, t, u, v. Then f = Ress(Gp, Gq) is a polynomial in t, u, v and g = Rest(Gp, Gq) is a polynomial in s, u, v. Moreover, gcd( f , g) is a polynomial in u, v, and the solution set {(u0, v0) 6= (0,0) | gcd( f , g) = 0} is exactly the set of {(u0, v0) 6= (0,0) | p(s, t) ·G(u0, v0) = q(s, t) ·G(u0, v0) = 0. Hence, the natural algorithm to compute the intersection of two plane curves via GCD are stated below: GCD Algorithm Input: Parametrized curves F(s, t) and G(u, v). Output: The set of parameters and their corresponding points of the intersection of two curves given by parametrization F(s, t) and G(u, v). Procedure: 1. Compute p(s, t),q(s, t), the µ-basis for F(s, t). 2. Compute Gp = p(s, t) ·G(u, v), and Gq = q(s, t) ·G(u, v). 3. Compute f = Ress(Gp, Gq) and g = Rest(Gp, Gq), where Ress( f , g) stands for the resultant of polynomial f , g with respect to s. 4. The solution to gcd( f , g) = 0 corresponding the the parameters (u0, v0) such that G(u0, v0) are the intersection points of the two curves with correct multiplicity. Example 1. Below are two rational paramatrized plane curves given by Wang-Goldman [11]: Lemniscate of Bernoulli, a rational quartic curve F and a rational cubic curve G(u, v). F(s, t) =(s4 − t4, −2st(t2 − s2), t4 + 6s2 t2 + s4), G(u, v) =(−3v3 − 2uv2 + 4u2v + 2u3, −v3 + 3uv2 − u3 + 2u2v, 2uv2 + 2u2v + u3). M. Tesemma, H. Wang, / Eur. J. Pure Appl. Math, 7 (2014), 191-200 195 GCD Method: First, compute a µ-basis for F(s, t): p(s, t) = (−s2 − t2, −2st, s2 − t2), q(s, t) = (2st, −s2 − t2, 0). Compute Gp :=p(s, t) ·G(u, v) =− s2(u3 + 2u2v − 4uv2 − 3v3)− 3t2(u3 + 2u2v − v3) + 2st(u3 − 2u2v − 3uv2 + v3), Gq :=q(s, t) ·G(u, v) =2st(2u3 + 4u2v − 2uv2 − 3v3) + s2(u3 − 2u2v − 3uv2 + v3) + t2(u3 − 2u2v − 3uv2 + v3). And f :=Ress(Gp, Gq) =4t4(u2 + 2uv + 2v2)2(22u8 − 86u6v2 − 10u5v3 + 131u4v4 + 34u3v5 − 84u2v6 − 20uv7 + 25v8), g :=Rest(Gp, Gq) =4s4(u2 + 2uv + 2v2)2(22u8 − 86u6v2 − 10u5v3 + 131u4v4 + 34u3v5 − 84u2v6 − 20uv7 + 25v8). Hence, the solutions to gcd( f , g) = (u2+2uv+2v2)2(22u8−86u6v2−10u5v3+131u4v4+34u3v5−84u2v6−20uv7+25v8) = 0 are the parameters (u0, t0) such that G(u0, v0) are the intersections of the two curves. The inter- sections are listed below: G(−1− 1i, 1) =(3+ 6i,−6+ 3i, 0) double points, G(−1+ 1i, 1) =(3− 6i,−6− 3i, 0) double points, G(−1.163− 0.135i, 1) =(1.645+ 0.436i,−0.311+ 0.771i,−1.166− 0.187i), G(−1.163+ 0.135i, 1) =(1.645− 0.436i,−0.311− 0.771i,−1.166+ 0.187i), G(−0.840− 0.483i, 1) =(0.566+ 2.396i,−2.572+ 1.089i,−0.739− 0.253i), G(−0.840+ 0.483i, 1) =(0.566− 2.396i,−2.572− 1.089i,−0.739+ 0.253i), G(0.661− 0.145i, 1) =(−2.166− 0.855i, 1.567− 0.635i, 2.399− 0.865i), G(0.661+ 0.145i, 1) =(−2.166+ 0.855i, 1.567+ 0.635i, 2.399+ 0.865i), G(1.343− 0.343i, 1) =(4.954− 6.630i, 4.452− 1.056i, 8.006− 4.344i), G(1.343+ 0.3430i, 1) =(4.954+ 6.630i, 4.452+ 1.056i, 8.006+ 4.344i). We may check our solution by a conventional computation via implicit equation of F(s, t). We note that the implicit equation of F(s, t) is: F(x , y, z) = 4(x2 + y2)2 + 4w2(y2 − x2) = 0, and F(G(u, v)) = 4(u2+2uv+2v2)2(22u8−86u6v2−10u5v3+131u4v4+34u3v5−84u2v6−20uv7+25v8). Obviously, gcd( f , g) = F(G(u, v)), hence, the solutions to gcd( f , g) = 0 are indeed the parameters of G(u, v) corresponding to the intersections of the two curves. M. Tesemma, H. Wang, / Eur. J. Pure Appl. Math, 7 (2014), 191-200 196 3.2. Resultant Matrix Method Now, we will use resultant matrix to find the intersection of two parametrized curves. First, we let p(x , y, z; s, t) = p(s, t) · (x , y, z) and q(x , y, z; s, t) = q(s, t) · (x , y, z) be a µ-basis of a rational parametric plane curve C1 of degree d given by F(s, t) = ( f0(s, t), f1(s, t), f2(s, t)), (s, t) 6= (0, 0). It is know that the implicit equation of the curve is F(x , y, z) = Ress(p(x , y, z; s, 1), q(x , y, z; s, 1)). Let the resultant matrix of the curve C1 be denoted by M(x , y, z). We note that M(x , y, z) is a square matrix of size d × d, and the rank of the matrix rankM(x , y, z) = d. Let the rational parametric plane curve C2 of degree d ′ be given by G(u, v) = (g0(u, v), g1(u, v), g2(u, v)), (u, v) 6= (0, 0). By substituting the variables x , y, z in matrix M(x , y, z) with (g0(u, v), g1(u, v), g2(u, v)), we obtain a matrix M(u, v) = M(G(u, v)) = M(g0(u, v), g1(u, v), g2(u, v)). We can derive the following relationship between the rank of the matrix M(u, v) and the intersection of the two curves: Theorem 2. {(u0, v0) 6= (0, 0) | rankM(u0, v0)< d} ⇔ G(u0, v0) ∈ C1 ∩C2. Proof. G(u0, v0) ∈ C1 ∩C2 if and only if G(u0, v0) is such that det M(G(u0, v0)) = 0, which is equivalent to the condition that rankM(u0, v0)< d. Hence, the computational algorithm follows directly as below: Resultant Matrix Algorithm Input: Parametrized curves F(s, t) and G(u, v). Output: The set of parameters and their corresponding points of the intersection of two curves given by parametrization F(s, t) and G(u, v). Procedure: 1. Compute p(s, t),q(s, t), the µ-basis for F(s, t). 2. Construct the resultant matrix of the curve F(s, t) denoted by M(x , y, z) using information of the µ-basis. 3. Substituting the variables x , y, z in matrix M(x , y, z) with (g0(u, v), g1(u, v), g2(u, v)). 4. The solution to det(M(g0(u, v), g1(u, v), g2(u, v)) = 0 corresponding the the parameters (u0, v0) such that G(u0, v0) are the intersection points of the two curves with correct multi- plicity. M. Tesemma, H. Wang, / Eur. J. Pure Appl. Math, 7 (2014), 191-200 197 Example 2. To compare the algorithm, we will use the same example as before, two rational paramatrized plane curves given by Wang-Goldman [11]: Lemniscate of Bernoulli, a rational quartic curve F and a rational cubic curve G(u, v). F(s, t) =(s4 − t4, −2st(t2 − s2), t4 + 6s2 t2 + s4), bG(u, v) =(−3v3 − 2uv2 + 4u2v + 2u3, −v3 + 3uv2 − u3 + 2u2v, 2uv2 + 2u2v + u3). Resultant Matrix Method: Compute the µ-basis of the parametrized curve F(s, t) in terms of the moving lines as below: µ-basis: p(s, t) = (−s2 − t2, −2st, s2 − t2), q(s, t) = (2st, −s2 − t2, 0) moving line form: p(x , y, z; s, 1) = s2(−x + z)− 2s y − (x + z), q(x , y, z; s, 1) = −s2 y + 2sx − y. Then, construct matrix M(x , y, z) as M(x , y, z) =     −x + z 0 −y 0 −2y −x + z 2x −y −(x + z) −2y −y 2x 0 −(x + z) 0 −y     . And M(u, v) =M(G(u, v)) =4(u2 + 2uv + 2v2)2(22u8 − 86u6v2 − 10u5v3 + 131u4v4 + 34u3v5 − 84u2v6 − 20uv7 + 25v8). Since M(u, v) is exactly the same as F(G(u, v)), the solutions to M(u, v) = 0 are indeed the parameters of G(u, v) corresponding to the intersections of the two curves. 3.3. Smith Normal Form Method Recall that for every nonzero square univariate polynomial matrix A (that is the entries of matrix A are polynomials in one variable) with rank(A) = r, there exist invertible polynomial matrices P, Q (P is invertible if det(P) is a non-zero constant) such that PAQ = Diag( f1, f2, . . . , fr , 0, 0, . . . , 0) where f1, . . . , fr are non-zero polynomials with fk | fk+1 for 1 ≤ k ≤ r, and Diag(a1, . . . , an) means the diagonal matrix with diagonal entries a1, . . . , an. The diagonal matrix is called the Smith normal form of the polynomial matrix A is denoted by S(A). The polynomials fk for k = 1, . . . , r are called the k-th invariant factors of A, and Dk = ∏k i=1 fi are called k-th determinant factors of the matrix A. Since the implicit equation of the curve is F(x , y, z) = Ress(p(x , y, z; s, 1), q(x , y, z; s, 1)) is obtained by M(x , y, z), the determinant of the resultant matrix of the curveC1. In the previous subsection, substituting the variables x , y, z in matrix M(x , y, z)with (g0(u, v), g1(u, v), g2(u, v)), we obtain a matrix M(u, v) = M(G(u, v)) = M(g0(u, v), g1(u, v), g2(u, v)). M. Tesemma, H. Wang, / Eur. J. Pure Appl. Math, 7 (2014), 191-200 198 Theorem 3. S(M(u, 1)) = Diag( f1(u), . . . , fd(u)) and S(M(1, v)) = Diag(g1(v), . . . , gd(v)). And {(u0, 1) or (1, v0) | fd(u0) = 0 or gd(v0) = 0}⇔ G(u0, 1) or G(1, v0) ∈ C1 ∩C2. Proof. Since det(M(u, v)) is not identically zero, rankM(u, v) = d, therefore, the Smith normal forms S(M(u, 1)) = Diag( f1(u), . . . , fd(u)) and S(M(1, v)) = Diag(g1(v), . . . , gd(v)). By the property of Smith normal form, det S(M) = det(PMQ) = c detM = 0, and the solutions to the d-th invariant factor ofM is exactly the solutions to det S(M) = 0. Therefore, G(u0, v0) ∈ C1 ∩C2 ⇔ {(u0, v0) 6= (0,0) | det S(M(u0, v0)) = 0} ⇔ {(u0, 1) or (1, v0) | fd(u0) = 0 or gd(v0) = 0}. Hence, the algorithm via Smith normal form follows directly as below: Smith Normal Form Algorithm Input: Parametrized curves F(s, t) and G(u, v). Output: The set of parameters and their corresponding points of the intersection of two curves given by parametrization F(s, t) and G(u, v). Procedure: i) Compute p(s, t),q(s, t), the µ-basis for F(s, t). ii) Construct the resultant matrix of the curve F denoted by M(x , y, z) using information of the µ-basis. iii) Substituting the variables x , y, z in matrix M(x , y, z) with (g0(u, v), g1(u, v), g2(u, v)). iv) Compute the Smith normal form S(M(u, 1)) and S(M(1, v)). The solution for fd(u) = 0 and gd(v) = 0 corresponding the the parameters (u0, 1) and (1, v0) such that G(u0, 1) and G(1, v0) are the intersection points of the two curves with correct multiplicity. Example 3. To compare the algorithm, we will use the same example as before, two rational paramatrized plane curves given by Wang-Goldman [11]: Lemniscate of Bernoulli, a rational quartic curve F and a rational cubic curve G(u, v). F(s, t) =(s4 − t4, −2st(t2 − s2), t4 + 6s2 t2 + s4), G(u, v) =(−3v3 − 2uv2 + 4u2v + 2u3, −v3 + 3uv2 − u3 + 2u2v, 2uv2 + 2u2v + u3). Smith Normal Form Method: Compute the µ-basis of the parametrized curve F(s, t) in terms of the moving lines as below: µ-basis: p(s, t) = (−s2 − t2, −2st, s2 − t2), q(s, t) = (2st, −s2 − t2, 0) M. Tesemma, H. Wang, / Eur. J. Pure Appl. Math, 7 (2014), 191-200 199 moving line form: p(x , y, z; s, 1) = s2(−x + z)− 2s y − (x + z), q(x , y, z; s, 1) = −s2 y + 2sx − y. Then, construct matrix M(x , y, z) as M(x , y, z) =     −x + z 0 −y 0 −2y −x + z 2x −y −(x + z) −2y −y 2x 0 −(x + z) 0 −y     . And M(u, v) =M(G(u, v)) =4(u2 + 2uv + 2v2)2(22u8 − 86u6v2 − 10u5v3 + 131u4v4 + 34u3v5 − 84u2v6 − 20uv7 + 25v8). Then, we compute the Smith normal form ofM(u, 1) S(M(u, 1)) =     1 0 0 0 0 1 0 0 0 0 2+ 2u+ u2 0 0 0 0 1 22(2+ 2u+ u2)(25− 20u− 84u2 + 34u3 + 131u4 − 10u5 − 86u6 + 22u8)     . The last diagonal entry is the polynomial 1 22 (2+ 2u+ u2)(25− 20u− 84u2 + 34u3 + 131u4 − 10u5 − 86u6 + 22u8), which is exactlyM(u, 1) and F(G(u, 1)). Hence the solution to the equation 1 22 (2+ 2u+ u2)(25− 20u− 84u2 + 34u3 + 131u4 − 10u5 − 86u6 + 22u8) = 0 are the parameters of G(u, v) corresponding to the intersection of the two curves. 4. Conclusion In this paper, we compared three different methods of finding the intersections of two rational parametrized plane curves. These methods only use the µ-basis of one curve without finding the implicit equations of the curves. Since finding implicit equation is not an easy task, these three algorithms increase the computation efficiency. 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