Subdivision curves. University of Texas at Austin CS384G - Computer Graphics

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1 Subdivision curves University of Texas at Austin CS384G - Computer Graphics Fall 2010 Don Fussell

2 Reading Recommended: Stollnitz, DeRose, and Salesin. Wavelets for Computer Graphics: Theory and Applications, 1996, section , A.5. Note: there is an error in Stollnitz, et al., section A.5. Equation A.3 should read: MV = VΛ University of Texas at Austin CS384G - Computer Graphics Fall 2010 Don Fussell 2

3 Subdivision curves Idea: repeatedly refine the control polygon P 1 P 2 P 2! curve is the limit of an infinite process Q = lim P j j University of Texas at Austin CS384G - Computer Graphics Fall 2010 Don Fussell 3

4 Chaikin s algorithm Chaikin introduced the following corner-cutting scheme in 1974: Start with a piecewise linear curve Insert new vertices at the midpoints (the splitting step) Average each vertex with the next (clockwise) neighbor (the averaging step) Go to the splitting step University of Texas at Austin CS384G - Computer Graphics Fall 2010 Don Fussell 4

5 Averaging masks The limit curve is a quadratic B-spline! Instead of averaging with the nearest neighbor, we can generalize by applying an averaging mask during the averaging step: r = (,r 1,r 0,r 1, ) In the case of Chaikin s algorithm: " r = 1 2, 1 % $ ' # 2& University of Texas at Austin CS384G - Computer Graphics Fall 2010 Don Fussell 5

6 Can we generate other B-splines? Answer: Yes Lane-Riesenfeld algorithm (1980) Use averaging masks from Pascal s triangle: Gives B-splines of degree n+1. n=0: 1 n=1: n=2: & & n# & n# & n# # r = $ $!, $!,!, $!! n 2 %% 0" % 1" % n" " University of Texas at Austin CS384G - Computer Graphics Fall 2010 Don Fussell 6

7 Subdivide ad nauseum? After each split-average step, we are closer to the limit curve. How many steps until we reach the final (limit) position? Can we push a vertex to its limit position without infinite subdivision? Yes! University of Texas at Austin CS384G - Computer Graphics Fall 2010 Don Fussell 7

8 One subdivision step Consider the cubic B-spline subdivision mask: Now consider what happens during splitting and averaging: A B C split ( ) A B a c average C A B a b c C a b c repeat University of Texas at Austin CS384G - Computer Graphics Fall 2010 Don Fussell 8

9 Math for one subdivision step Subdivision mask: One subdivision step: ( ) B split B a c average B a b c A C A C A C Split: 1 a = ( A + B ) 2 1 c = ( B + C ) 2 Average: a and c do not change 1 1 b = ( a + 2 B + c ) = ( A + 6 B + C ) 4 8 University of Texas at Austin CS384G - Computer Graphics Fall 2010 Don Fussell 9

10 Consolidated math for one step Subdivision mask: One subdivision step: ( ) B b a c A C Consolidated math for one subdivision step: P j + 1! a"! 4 4 0"! A" # $ 1 # $ # $ # b $ = 8 # $ # B $ #% c $ & #% 0 4 4$ & #% C$ & repeat P j Local subdivision matrix S University of Texas at Austin CS384G - Computer Graphics Fall 2010 Don Fussell 10

11 Local subdivision matrix, cont d Tracking the point s value through subdivision: P = SP = S SP = S S SP =! = S P j j j j The limit position of the point is then: P = S P 0 or as we d say in calculus P = lim j OK, so how do we apply a matrix an infinite number of times?? j S P 0 j University of Texas at Austin CS384G - Computer Graphics Fall 2010 Don Fussell 11

12 Eigenvectors and eigenvalues To solve this problem, we need to look at the eigenvectors and eigenvalues of S. First, a review Let v be a vector such that Sv = λv We say that v is an eigenvector with eigenvalue λ. An n x n matrix can have n eigenvalues and eigenvectors: Sv 1 = λ 1 v 1! Sv n = λ n v n If the eigenvectors are linearly independent (which means that S is non-defective), then they form a basis, and we can re-write P in terms of the eigenvectors: P = n i=1 a i v i University of Texas at Austin CS384G - Computer Graphics Fall 2010 Don Fussell 12

13 To infinity, but not beyond So, applying S to P: Applying it j times: Let s assume the eigenvalues are non-negative and sorted so that: λ 1 > λ 2 λ 3! λ n 0 Now let j go to infinity: If λ 1 > 1, then: If λ 1 < 1, then: If λ 1 = 1, then: 1 0 n n n P = SP = S a v = a Sv = a λ v j i i i i i i i i i i n n n j j j j 0 i i i i iλi i P = S P = S a v = a S v = a v i i i = j lim = 0 lim j j i n j iλi P S P a v i University of Texas at Austin CS384G - Computer Graphics Fall 2010 Don Fussell 13

14 Evaluation masks What are the eigenvalues and eigenvectors of our cubic B-spline subdivision matrix? # 1& % ( λ 1 =1 V 1 = 1 % ( $ 1' We re OK! But what is the final position? j j j ( 1λ1 1 2λ2 2 3λ3 3) P = lim a v + a v + a v P = j λ 2 = 1 2 $ 1' V = & ) 2 0 & ) % 1 ( λ 3 = 1 4 $ 2 ' & ) V 3 = 1 & ) % 2 ( Almost done from earlier we know that we can find a i, we but didn t give specifics. University of Texas at Austin CS384G - Computer Graphics Fall 2010 Don Fussell 14

15 Evaluation masks To finish up, we need to compute a 1. Remember: Rewrite as: P = a v + a v +! + a v n n We need to solve for the vector A. (This is really just a change of basis 1 A = V P0 for representing the vector P). T The solution is: " a1 # "! u1!# $ T a % $ % $ 2 % $! u2! = % P0 $ " % $ " % $ % $ T % & an ' $ &! un!%' Now we can compute the limit position: We call u 1 the evaluation mask.! a1 "!!!! " # a $ # $ # $ # $ V #! $ #%!!! $ & # $ % a & 2 P0 = v1 v2 " vn = A T 1 1 P = a = u P n 0 University of Texas at Austin CS384G - Computer Graphics Fall 2010 Don Fussell 15

16 Evaluation masks Note that we need not start with the 0 th level control points and push them to the limit. If we subdivide and average the control polygon j times, we can push the vertices of the refined polygon to the limit as well: T P = S P = u P So far we ve been looking at math for a subdivision function f(x). For a 2D parametric subdivision curve, (x(u), y(u)), just apply these formulas separately for the x(u) and y(u) functions. j 1 j University of Texas at Austin CS384G - Computer Graphics Fall 2010 Don Fussell 16

17 Recipe for subdivision curves The evaluation mask for the cubic B-spline is: ( ) Now we can cook up a simple procedure for creating subdivision curves: Subdivide (split+average) the control polygon a few times. Use the averaging mask. Push the resulting points to the limit positions. Use the evaluation mask. University of Texas at Austin CS384G - Computer Graphics Fall 2010 Don Fussell 17

18 Derivative of subdiv. function What is the tangent to the cubic B-spline function? Consider the formula for P again: Where: P j Derivative is just: P = a λ v + a λ v + a λ v! left " = # center $ # $ #% right $ & j j j j " 1# " 1# " 2 # j $ % 1 j $ % 1 j $ % Pj = a1 (1) 1 + a2( ) 0 a3( ) $ 1% $ 1 % $ 2 % & ' & ' & ' " P = lim Δx 0 center left Δx = lim j center left 1 # $ % & P ' = lim % a a u P j % & = = 2 1 & % ' ( & j ' 2 ( j # 1 $ T University of Texas at Austin CS384G - Computer Graphics Fall 2010 Don Fussell 18 2 j

19 Tangent analysis for 2D curve What is the tangent to a parametric cubic B-spline 2D curve? Using a similar derivation to what we just did for a 1D function (but omitting details): P P u P T Center, j Left, j 2 0 t = lim = T j PCenter, j P u Left, j 2 P0 Thus, we can compute the tangent using the second left eigenvector! This analysis holds for general subdivision curves and gives us the tangent mask. University of Texas at Austin CS384G - Computer Graphics Fall 2010 Don Fussell 19

20 Control Point Approximation vs. Interpolation Previous subdivision scheme approximated control points. Can we interpolate them? Yes: DLG interpolating scheme (1987) Slight modification to subdivision algorithm: splitting step introduces midpoints averaging step only changes midpoints For DLG (Dyn-Levin-Gregory), use: 1 r = ( 2,5,10,5, 2) 16 Since we are only changing the midpoints, the points after the averaging step do not move. University of Texas at Austin CS384G - Computer Graphics Fall 2010 Don Fussell 20