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Nira Dyn • Michael Floater • Kai Hormann. Dual 2n-Point Schemes. A Quadrilateral Rendering Primitive. Introduction. 1. 6. 1. 4. 4. Primal schemes one new vertex for each old vertex one new vertex for each old edge “keep old points, add edge midpoints” mask with odd length. - PowerPoint PPT Presentation
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A QuadrilateralRendering Primitive
Nira Dyn • Michael Floater • Kai Hormann
Dual 2n-Point Schemes
Dual 2n-Point Schemes
Primal schemesone new vertex for each old vertexone new vertex for each old edge
“keep old points, add edge midpoints”mask with odd length
Introduction
1 6 1
4 4
Dual 2n-Point Schemes
Dual schemesone new edge for each old vertexone new edge for each old edge
“add two edge-points, forget old points”mask with even length
Introduction
1 3
3 1
Dual 2n-Point Schemes
Known Schemes
1 33 1
1 6 14 4
1 105 10 5 1
1 15 156 20 16
1 12
B-Splines
linear
cubic
quintic
quadratic
quartic
2n-Point
-1 9 90 16 -10
-25 150 1500 256 -250 303 0
4-point
6-point?
Primal Dual
Dual 2n-Point Schemes
quintic precisioninterpolation
Primal 2n-Point Schemes
-1 9 9 -10 16 0
-25 150 150 -25 33 0 256 0 00
interpolation
cubic sampling
1
-1/16
9/161 1 1
quintic sampling
cubic precision
Dual 2n-Point Schemes
Dual 4-Point Scheme
-5 35 105 -7
-7 105 35 cubic sampling
cubic sampling
1
-7/128
1 1 1
cubic precision
-5
-5/128
105/128 35/12835/128 105/128
-7/128-5/128
Dual 2n-Point Schemes
Dual 4-Point Scheme
-5 35 105 -7-7 105 35 cubic precision-5
⇒ scheme is O(h4) and symbol contains (1+z)4
-5 37 37 -5-2 68 -2
-5 34 33 34 -5
-5 26 -58 8
-5 1313 -5
-5 -518
a(z) =
= ·(1+z)
= ·(1+z)2
= ·(1+z)3
= ·(1+z)4
= ·(1+z)5
⇒ scheme could be C4 and 4 µ span {(x-j)}
Dual 2n-Point Schemes
⇒ C2
|■| = 42/64 < 1⇒ C1
Smoothness Analysis
-5 35 105 -7-7 105 35 -5
-5 37 37 -5-2 68 -2
-5 34 33 34 -5
-5 26 -58 8
a(z) =
|■| = 84/128 < 1|■| = 72/128 < 1
⇒ C0
|■| = 42/64 < 1
|■| = 36/32 > 1
25 -170 103-40 24 272 272 24-596 103 -170 25-40
-5 26 -58 8 -5 26 -58 8× 2
|■| = 336/1024 < 1|■| = 256/1024 < 1|■| = 936/1024 < 1|■| = 336/1024 < 1
scheme is not C3
Dual 2n-Point Schemes
right and left eigenvector for 0:
Subdivision Matrix
-5 35 105 -7
-7 105 35 -5
-5 35 105 -7
-7 105 35 -5
-5 35 105 -7
-7 105 35 -5
0 0
0 0
0
0
0
0
0
0
0
0S =
-5 35 105 -7-7 105 35 -5
/128 ⇒
0 = 11 = 1/22 = 1/43 = 1/84 = 1/165 = 9/64
x0 = [1, 1, 1, 1, 1, 1]
y0 = [1, -27, 218, 218, -27, 1]/384
Dual 2n-Point Schemes
Limit Function
support size 7quasi-interpolation Q = I+R = I+I-T
[5, 866, -3509, 54428, -3509, 866, 5] / 49152
-5 -5-866-866 3509 35094387649152491524915249152 49152 49152 49152
0 0218384
218384
-27384
-27384
1384
1384
Dual 2n-Point Schemes
Limit Function
Dual 2n-Point Schemes
′
″
Dual 2n-Point Schemes
Dual 4-Point Scheme
Summary reproduces cubic polynomialsapproximation order O(h4)C2 continuoussupport size 7contains quartic polynomials
Dual 2n-Point Schemes
2n-Point-Schemes
2n-Point
2-Pointlinear
4-Pointcubic
6-Pointquintic
-1 9 90 16 -10
-25 150 1500 256 -250 303 0
Primal Dual
1 12
-5 35 105 -7-7 105 35 -5
1 33 1
⋯ ⋯ ⋯ ⋯⋯ ⋯ ⋯ ⋯⋯ ⋯ ⋯ ⋯
∶ ∶
Dual 2n-Point Schemes
Dual 4-Point Scheme
Dual 2n-Point Schemes
Dual 6-Point Scheme
Dual 2n-Point Schemes
Dual 8-Point Scheme
Dual 2n-Point Schemes
Examples
Dual 2n-Point Schemes
Examples
A QuadrilateralRendering Primitive
Thank You for Your Attention
Dual 2n-Point Schemes