Barycentric coordinates and transfinite interpolation

Barycentric coordinates and transfinite
interpolation
Michael Floater
Centre of Mathematics for Applications,
Department of Informatics,
University of Oslo

In this talk:
1. Barycentric coordinates
2. Generalization to polygons: Wachspress, mean value, etc.
3. Transfinite interpolation

Barycentric coordinates on a triangle
v 3
T
x
v 1
v 2
Given x ∈ T , want λ 1 , λ 2 , λ 3 ≥ 0 such that
λ 1 + λ 2 + λ 3 = 1,
and
λ 1 v 1 + λ 2 v 2 + λ 3 v 3 = x.

Linear system of three equations,
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
1 1 1 λ 1 1
⎝v1 1 v2 1 v3
1 ⎠ ⎝λ 2
⎠ = ⎝x 1 ⎠ .
v1 2 v2 2 v3
2 λ 3 x 2
By Cramer’s rule, unique solution is
λ 1 = A 1
A , λ 2 = A 2
A , λ 3 = A 3
A .
v 3
A
A 2
x
A 3
A 1
v 1
v 2

Properties
◮ Lagrange property: λ i (v j ) = δ ij .
◮ Interpolation: if
g(x) =
3∑
λ i (x)f (v i ),
i=1
then g(v i ) = f (v i ).
◮ Linear precision: if f is linear then g = f .
◮ Linearity: λ i is linear.
◮ Bernstein-Bezier basis for polynomials, degree d:
p(x) =
∑
i+j+k=d
n!
i!j!k! λi 1(x)λ j 2 (x)λk 3(x)c ijk .

Barycentric coordinates on polygons
Let Ω be a convex polygon.
v 4
v 3
v 5
v 1 v 2
Functions λ i : Ω → R, i = 1, . . . , n, are barycentric coordinates if
for x ∈ Ω, λ i (x) ≥ 0, i = 1, . . . , n, and
n∑
λ i (x) = 1,
i=1
n∑
λ i (x)v i = x.
i=1

Properties
◮ Lagrange property: λ i (v j ) = δ ij .
◮ Interpolation: if
n∑
g(x) = λ i (x)f (v i ),
i=1
then g(v i ) = f (v i ).
◮ Linear precision: if f is linear then g = f .
◮ λ i lies between the two functions:
1
1
0
0
0
0
0
0
◮ Generalized Bernstein-Bezier:
p(x) = ∑ ( ) d
λ i 1
i 1
(x) · · · λ in n (x)c i .
|i|=d
0
0
0
0

Coordinates for quadrilaterals
v 01
v 11
x
v 00
v 10
Can find λ, µ ∈ [0, 1] s.t.
x = (1 − λ)(1 − µ)v 00 + λ(1 − µ)v 10 + (1 − λ)µv 01 + λµv 11 ,
giving an interpolant
g(x) = (1 − λ)(1 − µ)f 00 + λ(1 − µ)f 10 + (1 − λ)µf 01 + λµf 11 .

To find λ and µ: let
a = v 00 −x, b = v 10 −v 00 , c = v 01 −v 00 , d = v 00 −v 10 −v 01 +v 11 .
Then need to solve
a + bλ + cµ + dλµ = 0.
Eliminating λ gives a quadratic in µ:
(a + cµ) × (b + dµ) = 0.
Similar for λ. The special cases c × d = 0 and b × d = 0 need to
be treated separately.

Wachspress coordinates: Wachspress 1973, Warren 1996
v i+1
x
v i
v i−1
For a convex polygon, n vertices, let
λ i (x) =
w i (x)
∑ n
j=1 w j(x) , w i(x) =
Then λ 1 , . . . , λ n are barycentric coordinates.
A(v i−1 , v i , v i+1 )
A(x, v i−1 , v i )A(x, v i , v i+1 ) .

Proof: Meyer et al. 2002
Let
A i = A(x, v i , v i+1 ) and B i = A(v i−1 , v i , v i+1 ).
Then express x as
and rearrange:
x = A i
v i−1 + (B i − A i−1 − A i )
v i + A i−1
v i+1 ,
B i B i
B i
B i
(v i − x) = 1 (v i − v i−1 ) − 1 (v i+1 − v i ).
A i−1 A i A i−1 A i
Summing both sides over i gives
∑ B i
(v i − x) = 0
A i−1 A i
or
i
∑
w i (x)(v i − x) = 0.
i

Wachspress coordinates are rational
λ i (x) =
B i
∑
k B k
∏
j≠i−1,i A j(x)
∏
j≠k−1,k A j(x)
=
degree (n − 2)
degree (n − 3) .

Application: barycentric mappings
Recipe: given x ∈ Ω,
1. express x in Wachspress form x = ∑ i λ i(x)v i ,
2. set x ′ = ∑ i λ i(x)v ′ i .

Application: curve deformation

Theorem (F. & Kosinka). Wachspress mappings are injective
Idea of proof: express Jacobian as
∣
∑
λ i λ j λ k ∣∣∣∣∣
J = 2
∂ 1 λ i ∂ 1 λ j ∂ 1 λ k A(v i, ′ v j, ′ v k ′
1≤i

Mean value coordinates
v i+1
v i
x
α i
α i−1
v i−1
For a convex polygon, n vertices, let λ i (x) = w i (x)/ ∑ n
j=1 w j(x),
where
w i (x) =
(
1
tan
‖v i − x‖
(
αi−1 (x)
2
)
+ tan
Then λ 1 , . . . , λ n are barycentric coordinates.
(
αi (x)
2
))
.

Proof:
With e i = (v i − x)/‖v i − x‖, need to show that
or
n∑ (
tan(αi−1 /2) + tan(α i /2) ) e i = 0,
i=1
n∑
tan(α i /2)(e i + e i+1 ) = 0. (1)
i=1
To show this, let e i = (cos θ i , sin θ i ). Then α i = θ i+1 − θ i , and so
( αi
)
( αi
)
tan (e i + e i+1 ) = tan (cos θ i + cos θ i+1 , sin θ i + sin θ i+1 )
2
2
= (sin θ i+1 − sin θ i , cos θ i − cos θ i+1 ).
Summing this over i = 1, . . . , n gives (1).

Application: mesh parameterization

Transfinite interpolation, Warren et al 2004
c ( t i+1 )
c ( t i)
For convex curve c : [a, b] → R 2 sample points v i = c(t i ) and take
limit of Wachspress interpolants as max i (t i+1 − t i ) → 0. Gives a
‘transfinite interpolant’
∫ /
b
∫ b
g(x) = w(x, t)f (c(t)) dt w(x, t) dt,
a
where w(x, t) = (c ′ (t) × c ′′ (t))/((c(t) − x) × c ′ (t)) 2 .
a

MV transfinite interpolation
c ( t i+1 )
c ( t i)
Analogously, the transfinite MV interpolant is
∫ /
b
∫ b
g(x) = w(x, t)f (c(t)) dt w(x, t) dt,
a
a
where
w(x, t) = (c(t) − x) × c′ (t)
‖c(t) − x‖ 3 .

Examples of MV interpolants

Non-convex curves? Angle formulation
p ( x, θ)
x
θ
For a convex curve, the MV interpolant is
g(x) =
∫ 2π
0
/ ∫
f (p(x, θ))
2π
‖p(x, θ) − x‖ dθ 1
‖p(x, θ) − x‖ dθ.
0

On arbitrary curves
p ( x, θ)
2
p 3
( x, θ)
p 1
( x, θ)
x
θ
where
g(x) =
∫ 2π
0
n(x,θ)
∑
j=1
φ(x) =
Hormann: φ(x) > 0.
/
(−1) j−1
‖p j (x, θ) − x‖ f (p j(x, θ)) dθ φ(x),
∫ 2π
0
n(x,θ)
∑
j=1
(−1) j−1
‖p j (x, θ) − x‖ dθ.

Application: image warping, Hormann

Application: mesh deformation, Ju, Schaefer, Warren

Hermite interpolation
Two methods:
◮ Use MV weight function (with Chris Dyken / Solveig Bruvoll)
◮ Radial minimization (with Christian Schulz)

Method 1: weight function
The cubic interpolant p to the data f (0), f ′ (0), f (1), f ′ (1), can be
expressed as
p(x) = g 0 (x) + ψ(x)g 1 (x),
where
g 0 (x) = (1 − x)f (0) + xf (1),
ψ(x) = x(1 − x),
g 1 (x) = (1 − x)m 0 + xm 1 ,
and
m 0 = f ′ (0) − f (1) + f (0), m 1 = −f ′ (1) + f (1) − f (0).

In R n we interpolate the data f and ∂f
∂n
on ∂Ω by the function
p(x) = g 0 (x) + ψ(x)g 1 (x), x ∈ Ω,
where
and
m(y) =
∫
/
f (p(x, v))
g 0 (x) =
S ‖p(x, v) − x‖ dv φ(x),
/
ψ(x) = 1 φ(x),
∫
g 1 (x) =
S
/
m(p(x, v))
‖p(x, v) − x‖ dv φ(x),
∫
φ(x) =
1
‖p(x, v) − x‖ dv.
S
( ∂f
∂n (y) − ∂g 0
∂n (y) ) / ∂ψ
∂n (y),
y ∈ ∂Ω.

To use this construction we need to find ∂ψ
∂n (y) and ∂g 0
∂n (y).
Theorem
If d(M E , ∂Ω) > 0 and d(M I , ∂Ω) > 0 and y ∈ ∂Ω then
∂ψ
∂n (y) = 1 ,
V n−1
∂g 0
∂n (y) = 1 f (p(y, v)) − f (y)
V n−1
∫H(n) ‖p(y, v) − y‖
dv,
where V n−1 is the volume of the unit sphere in R n−1 : V 1 = 2,
V 2 = π, V 3 = 4π/3, V 4 = π 2 /2,...

Hermite examples

Method 2: radial minimization
MV interpolation revisited: the value g(x) is the unique minimizer
a = g(x) of the local ‘energy’ function
where
∫
E x (a) =
q x,v (r) =
and ρ(x, v) = ‖p(x, v) − x‖.
S
∫ ρ(x,v)
0
(
q
′
x,v (r) ) 2 dr dv,
ρ(x, v) − r
a +
r f (p(x, v)),
ρ(x, v) ρ(x, v)

Hermite interpolation
Find a ∈ R and b ∈ R n that minimize
∫ ∫ ρ(x,v) (
E x (a, b) =
q
′′
x,v(r) ) 2 dr dv,
where q x,v is the cubic polynomial such that
S
0
q x,v (0) = a,
q ′ x,v(0) = v · b,
q x,v (ρ(x, v)) = f (p(x, v)),
q ′ x,v(ρ(x, v)) = D v f (p(x, v)),
and set g(x) = a.
Has cubic precision!

Example of radial minimization

Comparison of the two Hermite methods
Radial cosine Method 1 Method 2