*+@:C:<?;B=B AG[ZFXZFUZPKNX

10 8D:>MTQ[9GCOE MTQ[JYRV[NIWSEOHLAB3! 110'%&'$%($'&PSD, NPR Shading, Stereoscopic 3D
Application Examples

SIGGRAPH Asia 2010

Course: Scattered Data Interpolation for Computer Graphics

Learning Doodle by Example
Application Examples

SIGGRAPH Asia 2010

Course: Scattered Data Interpolation for Computer Graphics

/, )#*#9+'1',+) Given N similar input drawings (doodles)) Construct more doodles that resemble the N inputs‣ Example resultsProposed Solution) Construct a space of doodles defined by the inputs) Use results from machine learning and statistics:) Consider drawings as sample points in some space.
) Similar doodles should be located nearby in the space.
) We wish to fit (or learn) a continuous function over the space that “explains”
the examples as well.
) We call the result a “Latent Doodle Space”(LDS).
‣ See the paper in EUROGRAPHICS06 (by Baxter and Anjyo) for details.

Two Main Challenges
To build a latent doodle space (LDS):) Find correspondences between two line drawings.
) Hard problem. No perfect solution.
‣ Do best possible, but still must have a good UI.
) Generate the space of similar drawings.
) Use Bayesian techniques and statistical methods to improve results.

How to construct the LDS?
To generate the space of similar drawings (LDS):) Two main tasks :) Finding latent coordinate systemPCA) Interpolating within LDSRBF) Three options in the EG06 paper: PCA+RBF, PCA+GP, GPLVM➡ This talk focuses on the first strategy: PCA+ RBF

Dimension reduction by PCA) A line drawing (doodle) has the structure:p) line drawing lk - stroke sp - linear segments av) After establishing correspondences among line drawings:) Construct the data matrix X (with zero-mean):stroke s1stroke s2(x, y)
line drawing l1
aa

a

line drawing l
a
a2
10211



1s1

20

b102

s1b 0




b1s 2b 2 s2

 ) Applying PCA: Eigen decomposition of the covariance matrix X T ⋅ X= X

2-d LDS and RBF) Make the LDS 2-dim by taking the eigenvectors with the first two largesteigenvalues.

) Interpolate each of s, t (scalar) values of the drawing samples by thin platespline:) space dimension n = 2 and smoothness m = 2 (explain later!)) The spline function is:where r := (s, t) =) The interpolant is of the form:f (s, t) := ∑ w iφ (s − si , t − t i ) + as + bt + ci

Demos2 + t 2

Locally Controllable Stylized Shading
Application Examples

SIGGRAPH Asia 2010

Course: Scattered Data Interpolation for Computer Graphics

Background: Cartoon shading? Based on thresholded  shading model
0.01.0  Motivation: Fake but expressive shading

With conventional shader

Artists want neat shading

Our Goal

conventionaldesired

8 Make toon shader artist-friendly– Edit undesirable shaded area
– Add artistic light and shade to original 3D lightingVideoThe shade painter (our approach)? The main idea:

Modify intensity according to painted strokes
Paint brush
operationIntensity  Keyframing

Key-frame Animation

Offset DataLinearinterpolation

+1#+0'16*,"'9!1',+ 6-'+1'+%
Paintedshade
Fall off

Boundary constraints
ThresholdFall-off constraints 

Interpolation by RBF
Threshold,41013-2"136/35167
  



RBF interpolation
Boundary constraints
Threshold

Fall off constraints )The discretized constraints for the RBF are!* ""!

The unknown values) We employ the following interpolation function:lf (x) = ∑ w iφ (x − c i ) + P(x)i=1

where x = (x1, x2, x3), and φ (x) := x ≡ x1 + x 2 + x 3 ; P(x) is a linear polynomial of2

x1, x2, x3; ci "means the constraint points (22and)) l is the number of all the!" ""!) We want to determine the weights {wk}"# *"! P(x)→Totally l + 4 #%$#!

For the given values hj (1 ≤ j ≤ l) , we have:l∑ w φ (c − c ) + P(c ) = h ,i=1ijijjfor 1 ≤ j ≤ l.

The unknown values♦ We further add the vanishing moment condition:l∑ w Q(c ) = 0. for any linear polynomial Q.
ii=1i→ Taking Q = 1, x1, x2, and x3 , we know that the above condition means:l∑w = 0i=1il∑ w c = 0, ( j =1, 2, 3)i=1iijThe linear equation for RBF
φ i j := φ (c i − c j )) By putting P(x) = p0 + p1x1 + p2x2 + p3x3""-3.""""""""""""""""""""""#"
we have the following linear equation:# φ11 φ12 %φ 21 %% % φ l1%11 %
% c11%c
% 12
$ c13φ1l1 c11 c12φ ll 1 c l1 c l 2
1 0 0 0c l1  
cl 2 
cl 3 0  

c13 &# w1 & # h &(% ( % 1 ( (% ( %  ((% (c l 3 (% w l ( % hl (% (0 (% p0 ( = % 0 ((% ( (% p1 ( %  (% ( (% p2 ( %  ((% ( % (0 '$ p3 ' $ 0 '