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The monotonicity of a family of barycentric coordinates for quadrilaterals
Computer Aided Geometric Design 51 (2017) 1–3
Contents lists available at ScienceDirect
Computer Aided Geometric Design www.elsevier.com/locate/cagd
Short Communication
The monotonicity of a family of barycentric coordinates for quadrilaterals ✩ Chongyang Deng ∗ , Feifan Shi School of Science, Hangzhou Dianzi University, Hangzhou 310018, China
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 31 March 2016 Received in revised form 25 January 2017 Accepted 29 January 2017 Available online 14 February 2017
Recently, Floater (2016) proved that four well-known kinds of generalized barycentric coordinates in convex polygons share a simple monotonicity property. In this note we proved that a family of barycentric coordinates for quadrilaterals (Floater, 2015a) are monotonic, too. © 2017 Elsevier B.V. All rights reserved.
Keywords: Generalized barycentric coordinates Monotonicity
1. Introduction For a convex polygon P ⊂ R2 with vertices { v i }ni=1 , in some anticlockwise ordering, any functions φi : P → R, i = 1, 2, · · · , n, are generalized barycentric coordinates (GBCs) if, for all x ∈ P , {φi (x) ≥ 0}ni=1 , and n i =1
φ i ( x ) = 1,
n
φi (x) v i = x.
i =1
It is easy to see that φi is increasing along the edges e i −1 = v i −1 v i and e i = v i v i +1 in the direction towards v i . Recently, Floater (2016) extended the monotonicity along edges of polygon to lines from the polygon boundary to polygon vertex and proved that four well-known kinds of generalized barycentric coordinates share the monotonicity property. In this note we showed that a family of barycentric coordinates for quadrilaterals (Floater, 2015a) also have the monotonicity property. 1.1. A family of barycentric coordinates for quadrilaterals Given a convex quadrilateral P ⊂ R2 with vertices { v i ∈ R2 }4i =1 , four weights { w i > 0}4i =1 assigned for each vertex and a point x ∈ P , let { A i , B i , C i }4i =1 be the signed triangle area,
A i = A i (x) = A (x, v i , v i +1 ), B i = B i (x) = A (x, v i −1 , v i +1 ), C i = A ( v i −1 , v i , v i +1 ), ✩
*
This paper has been recommended for acceptance by Konrad Polthier. Corresponding author. E-mail address: [email protected] (C. Deng).
http://dx.doi.org/10.1016/j.cagd.2017.01.003 0167-8396/© 2017 Elsevier B.V. All rights reserved.
2
C. Deng, F. Shi / Computer Aided Geometric Design 51 (2017) 1–3
where vertices are indexed cyclically, v i +4 = v i , i ∈ Z, and further let
G = B 21 + ρ 2 B 22 + 2ρ ( A 1 A 3 + A 2 A 4 ), F2 = −B1 + ρ B2 +
√
G, F3 = −B1 − ρ B2 +
√
(1) G,
λ1 = F 2 F 3 , λ2 = 2F 3 A 4 , λ3 = 4ρ A 4 A 1 , λ4 = 2 A 1 F 2 , λi φi = 4 , j =1 λ j
ρ=
where
x=
w1 w3 w2 w4
4
(2) (3) (4)
> 0 is a parameter. Floater (2015a) defined a one-parameter family of barycentric coordinates for x ∈ P as
φi v i .
(5)
i =1
1.2. Monotonicity of generalized barycentric coordinates Floater (2016) proposed the definition of monotonicity of generalized barycentric coordinates as follows. Definition 1. Let the boundary of the polygon be ∂ P . For each i and y ∈ ∂ P , y = v i , if the coordinate φi is increasing along the line segment from y to v i , then generalized barycentric coordinates {φi }ni=1 are monotonic. Obviously, this definition can also be given as Definition 2. Let the boundary of the polygon be ∂ P . For each i, and y ∈ ∂ P , y = v i , x = (1 − t ) y + t v i , if ψi (t ) := φi (x) is an increasing function of t ∈ [0, 1], then generalized barycentric coordinates {φi }ni=1 are monotonic. 2. The monotonicity of a family of barycentric coordinates for quadrilaterals Theorem 3. For convex quadrilateral, the coordinates defined by (5) are monotonic. Proof. We prove that φ1 is monotonic, i.e., φ1 > 0. The proofs of φi , i = 2, 3, 4 are similar. Letting x = (1 − t ) y + t v 1 , then the functions of x can be expressed as functions of t. For convenience, we omit variable t in functions of t. For example, B 1 , B 1 denote B 1 (t ), B 1 (t ) , respectively. From the definition of φ1 (see Eq. (4)) we have
φ1 =
4
i =1 (λ1 λi − λ1 λi ) ( 4j =1 λ j )2
4 =
i =2 (λ1 λi − λ1 λi ) . ( 4j =1 λ j )2
Based on the definitions of {λi }4i =1 (see Eq. (3)) we express λ 1 λi − λ1 λ i (i = 2, 3, 4) as follows,
λ 1 λ2 − λ1 λ 2 = 2F 32 ( F 2 A 4 − F 2 A 4 ), λ 1 λ3 − λ1 λ 3 = 4ρ [ A 1 F 3 ( F 2 A 4 − F 2 A 4 ) + A 4 F 2 ( F 3 A 1 − F 3 A 1 )], λ 1 λ4 − λ1 λ 4 = 2F 22 ( F 3 A 1 − F 3 A 1 ). So we only need to prove
A 1 F 3 > 0, A 4 F 2 > 0;
(6)
F 2 A 4 − F 2 A 4 > 0, F 3 A 1 − F 3 A 1 > 0. Without loss of generality, we assume that the anticlockwise direction is positive direction. Then
A i > 0, C i > 0, i = 1, 2, 3, 4 . Combining with (Floater, 2015b)
B 1 B 2 = A2 A4 − A1 A3 we get
|B 1 B 2| < | A2 A4| + | A1 A3| = A2 A4 + A1 A3.
(7)
C. Deng, F. Shi / Computer Aided Geometric Design 51 (2017) 1–3
3
With the definition of G (see Eq. (1)) we have
√
G>
B 21 + ρ 2 B 22 + 2ρ | B 1 B 2 | = | B 1 | + ρ | B 2 | ≥ | B 1 |,
and it follows that F 2 > 0, F 3 > 0. Thus (6) holds. To prove (7), we let B i , y = A ( y , v i −1 , v i +1 ), A i , y = A ( y , v i , v i +1 ), i = 1, 2, 3, 4. According to the linearity of area and x = (1 − t ) y + t v 1 , we have
B 1 = A ( x, v 4 , v 2 )
= (1 − t ) A ( y , v 4 , v 2 ) + t A ( v 1 , v 4 , v 2 ) = (1 − t ) B 1, y − tC 1 .
(1 − t ) B 1 + B 1 = (1 − t )[− B 1, y − C 1 ] + (1 − t ) B 1, y − tC 1 = −C 1 . Similarly, we have
B 2 = (1 − t ) B 2, y , (1 − t ) B 2 + B 2 = 0, A 1 = (1 − t ) A 1, y , (1 − t ) A 1 + A 1 = 0, A 4 = (1 − t ) A 4, y , (1 − t ) A 4 + A 4 = 0. Based on these equations, we have
√
√
√
√
F 2 A 4 − F 2 A 4 = B 1 A 4 − B 1 A 4 + ρ ( B 2 A 4 − B 2 A 4 ) + ( G ) A 4 −
√ A 4, y = √ [(1 − t )G + 2G + 2C 1 G ], 2 G
F 3 A 1 − F 3 A 1 = B 1 A 1 − B 1 A 1 − ρ ( B 2 A 1 − B 2 A 1 ) + ( G ) A 1 −
√ A 1, y = √ [(1 − t )G + 2G + 2C 1 G ].
G A 4
G A 1
2 G
Then Eq. (7) holds because
(1 − t )G + 2G = 2B 1 [(1 − t ) B 1 + B 1 ] + 2ρ 2 B 2 [(1 − t ) B 2 + B 2 ] + 2ρ A 1 [(1 − t ) A 3 + A 3 ] +2ρ A 2 [(1 − t ) A 4 + A 4 ] + 2ρ A 3 [(1 − t ) A 1 + A 1 ] + 2ρ A 4 [(1 − t ) A 2 + A 2 ] √ √ = −2B 1 C 1 + 2ρ A 1 C 4 + 2ρ A 4 C 2 + (2C 1 G − 2C 1 G ) √ √ > 2C 1 ( G − B 1 ) − 2C 1 G √ > −2C 1 G . 2 Acknowledgements This work is supported by NSFC (61370166, 61379072, 61272300, 61502128). References Floater, M.S., 2015a. The inverse of a rational bilinear mapping. Comput. Aided Geom. Des. 33, 46–50. Floater, M.S., 2015b. Generalized barycentric coordinates and applications. Acta Numer. 24, 161–214. Floater, M.S., 2016. On the monotonicity of generalized barycentric coordinates on convex polygons. Comput. Aided Geom. Des. 42, 34–39.
Contents lists available at ScienceDirect
Computer Aided Geometric Design www.elsevier.com/locate/cagd
Short Communication
The monotonicity of a family of barycentric coordinates for quadrilaterals ✩ Chongyang Deng ∗ , Feifan Shi School of Science, Hangzhou Dianzi University, Hangzhou 310018, China
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 31 March 2016 Received in revised form 25 January 2017 Accepted 29 January 2017 Available online 14 February 2017
Recently, Floater (2016) proved that four well-known kinds of generalized barycentric coordinates in convex polygons share a simple monotonicity property. In this note we proved that a family of barycentric coordinates for quadrilaterals (Floater, 2015a) are monotonic, too. © 2017 Elsevier B.V. All rights reserved.
Keywords: Generalized barycentric coordinates Monotonicity
1. Introduction For a convex polygon P ⊂ R2 with vertices { v i }ni=1 , in some anticlockwise ordering, any functions φi : P → R, i = 1, 2, · · · , n, are generalized barycentric coordinates (GBCs) if, for all x ∈ P , {φi (x) ≥ 0}ni=1 , and n i =1
φ i ( x ) = 1,
n
φi (x) v i = x.
i =1
It is easy to see that φi is increasing along the edges e i −1 = v i −1 v i and e i = v i v i +1 in the direction towards v i . Recently, Floater (2016) extended the monotonicity along edges of polygon to lines from the polygon boundary to polygon vertex and proved that four well-known kinds of generalized barycentric coordinates share the monotonicity property. In this note we showed that a family of barycentric coordinates for quadrilaterals (Floater, 2015a) also have the monotonicity property. 1.1. A family of barycentric coordinates for quadrilaterals Given a convex quadrilateral P ⊂ R2 with vertices { v i ∈ R2 }4i =1 , four weights { w i > 0}4i =1 assigned for each vertex and a point x ∈ P , let { A i , B i , C i }4i =1 be the signed triangle area,
A i = A i (x) = A (x, v i , v i +1 ), B i = B i (x) = A (x, v i −1 , v i +1 ), C i = A ( v i −1 , v i , v i +1 ), ✩
*
This paper has been recommended for acceptance by Konrad Polthier. Corresponding author. E-mail address: [email protected] (C. Deng).
http://dx.doi.org/10.1016/j.cagd.2017.01.003 0167-8396/© 2017 Elsevier B.V. All rights reserved.
2
C. Deng, F. Shi / Computer Aided Geometric Design 51 (2017) 1–3
where vertices are indexed cyclically, v i +4 = v i , i ∈ Z, and further let
G = B 21 + ρ 2 B 22 + 2ρ ( A 1 A 3 + A 2 A 4 ), F2 = −B1 + ρ B2 +
√
G, F3 = −B1 − ρ B2 +
√
(1) G,
λ1 = F 2 F 3 , λ2 = 2F 3 A 4 , λ3 = 4ρ A 4 A 1 , λ4 = 2 A 1 F 2 , λi φi = 4 , j =1 λ j
ρ=
where
x=
w1 w3 w2 w4
4
(2) (3) (4)
> 0 is a parameter. Floater (2015a) defined a one-parameter family of barycentric coordinates for x ∈ P as
φi v i .
(5)
i =1
1.2. Monotonicity of generalized barycentric coordinates Floater (2016) proposed the definition of monotonicity of generalized barycentric coordinates as follows. Definition 1. Let the boundary of the polygon be ∂ P . For each i and y ∈ ∂ P , y = v i , if the coordinate φi is increasing along the line segment from y to v i , then generalized barycentric coordinates {φi }ni=1 are monotonic. Obviously, this definition can also be given as Definition 2. Let the boundary of the polygon be ∂ P . For each i, and y ∈ ∂ P , y = v i , x = (1 − t ) y + t v i , if ψi (t ) := φi (x) is an increasing function of t ∈ [0, 1], then generalized barycentric coordinates {φi }ni=1 are monotonic. 2. The monotonicity of a family of barycentric coordinates for quadrilaterals Theorem 3. For convex quadrilateral, the coordinates defined by (5) are monotonic. Proof. We prove that φ1 is monotonic, i.e., φ1 > 0. The proofs of φi , i = 2, 3, 4 are similar. Letting x = (1 − t ) y + t v 1 , then the functions of x can be expressed as functions of t. For convenience, we omit variable t in functions of t. For example, B 1 , B 1 denote B 1 (t ), B 1 (t ) , respectively. From the definition of φ1 (see Eq. (4)) we have
φ1 =
4
i =1 (λ1 λi − λ1 λi ) ( 4j =1 λ j )2
4 =
i =2 (λ1 λi − λ1 λi ) . ( 4j =1 λ j )2
Based on the definitions of {λi }4i =1 (see Eq. (3)) we express λ 1 λi − λ1 λ i (i = 2, 3, 4) as follows,
λ 1 λ2 − λ1 λ 2 = 2F 32 ( F 2 A 4 − F 2 A 4 ), λ 1 λ3 − λ1 λ 3 = 4ρ [ A 1 F 3 ( F 2 A 4 − F 2 A 4 ) + A 4 F 2 ( F 3 A 1 − F 3 A 1 )], λ 1 λ4 − λ1 λ 4 = 2F 22 ( F 3 A 1 − F 3 A 1 ). So we only need to prove
A 1 F 3 > 0, A 4 F 2 > 0;
(6)
F 2 A 4 − F 2 A 4 > 0, F 3 A 1 − F 3 A 1 > 0. Without loss of generality, we assume that the anticlockwise direction is positive direction. Then
A i > 0, C i > 0, i = 1, 2, 3, 4 . Combining with (Floater, 2015b)
B 1 B 2 = A2 A4 − A1 A3 we get
|B 1 B 2| < | A2 A4| + | A1 A3| = A2 A4 + A1 A3.
(7)
C. Deng, F. Shi / Computer Aided Geometric Design 51 (2017) 1–3
3
With the definition of G (see Eq. (1)) we have
√
G>
B 21 + ρ 2 B 22 + 2ρ | B 1 B 2 | = | B 1 | + ρ | B 2 | ≥ | B 1 |,
and it follows that F 2 > 0, F 3 > 0. Thus (6) holds. To prove (7), we let B i , y = A ( y , v i −1 , v i +1 ), A i , y = A ( y , v i , v i +1 ), i = 1, 2, 3, 4. According to the linearity of area and x = (1 − t ) y + t v 1 , we have
B 1 = A ( x, v 4 , v 2 )
= (1 − t ) A ( y , v 4 , v 2 ) + t A ( v 1 , v 4 , v 2 ) = (1 − t ) B 1, y − tC 1 .
(1 − t ) B 1 + B 1 = (1 − t )[− B 1, y − C 1 ] + (1 − t ) B 1, y − tC 1 = −C 1 . Similarly, we have
B 2 = (1 − t ) B 2, y , (1 − t ) B 2 + B 2 = 0, A 1 = (1 − t ) A 1, y , (1 − t ) A 1 + A 1 = 0, A 4 = (1 − t ) A 4, y , (1 − t ) A 4 + A 4 = 0. Based on these equations, we have
√
√
√
√
F 2 A 4 − F 2 A 4 = B 1 A 4 − B 1 A 4 + ρ ( B 2 A 4 − B 2 A 4 ) + ( G ) A 4 −
√ A 4, y = √ [(1 − t )G + 2G + 2C 1 G ], 2 G
F 3 A 1 − F 3 A 1 = B 1 A 1 − B 1 A 1 − ρ ( B 2 A 1 − B 2 A 1 ) + ( G ) A 1 −
√ A 1, y = √ [(1 − t )G + 2G + 2C 1 G ].
G A 4
G A 1
2 G
Then Eq. (7) holds because
(1 − t )G + 2G = 2B 1 [(1 − t ) B 1 + B 1 ] + 2ρ 2 B 2 [(1 − t ) B 2 + B 2 ] + 2ρ A 1 [(1 − t ) A 3 + A 3 ] +2ρ A 2 [(1 − t ) A 4 + A 4 ] + 2ρ A 3 [(1 − t ) A 1 + A 1 ] + 2ρ A 4 [(1 − t ) A 2 + A 2 ] √ √ = −2B 1 C 1 + 2ρ A 1 C 4 + 2ρ A 4 C 2 + (2C 1 G − 2C 1 G ) √ √ > 2C 1 ( G − B 1 ) − 2C 1 G √ > −2C 1 G . 2 Acknowledgements This work is supported by NSFC (61370166, 61379072, 61272300, 61502128). References Floater, M.S., 2015a. The inverse of a rational bilinear mapping. Comput. Aided Geom. Des. 33, 46–50. Floater, M.S., 2015b. Generalized barycentric coordinates and applications. Acta Numer. 24, 161–214. Floater, M.S., 2016. On the monotonicity of generalized barycentric coordinates on convex polygons. Comput. Aided Geom. Des. 42, 34–39.