Interpolation and polynomial approximation Interpolation

Transcription

1 Outline Interpolation and polynomial approximation Interpolation Lagrange Cubic Approximation Bézier curves B- 1

2 Some vocabulary (again ;) Control point : Geometric point that serves as support to the curve Knot : a specific value of the parameter u corresponding to a joint between pieces of a curve Knot sequence : the set of knots values (in an increasing order). Wolfram research 2

3 Interpolation The curve passes through the control points Approximation The curve doesn't necessary passes through the control points But these have an influence... Statistic approaches? Least squares «Kriging» Not very adapted to geometric modelling 3

4 Interpolation 4

5 Interpolation Interpolation We want draw a regular and smooth parametric curve through a certain number n of points Pi Several families of base functions are available Most obvious are polynomials There are others Trigonometric functions (by mean of a Fourier decomposition for instance) Power functions etc... 5

6 Interpolation We must choose the parametrization (nodal sequence) Uniform u2=2 u1=1 un-1=n-1 u0=0 6

7 Interpolation Or non-uniform u2=7 u1=2.5 un-1=15 u0=0 7

8 Interpolation Can we choose u as a curvilinear abscissa? In principle no since we don't know the final shape of the curve beforehand (with the exception of interpolation points) We will see later that it is often impossible that u corresponds with s exactly along the whole curve using analytical functions. But nothing forbids to get close to that numerically... 8

9 Interpolation Parametrization as an approximate arc length u2=d1+d2 d2 u1=d1 d1 dn-1 u0=0 un-1=d1+...+dn-1 9

10 Interpolation In all cases, we want to have : x u i = xi P u i =P i y u i = y i { We are going to interpolate the functions x(u) and y(u) with ONE polynomial with n parameters This one must be of order p=n-1 : 2 x u =a 0 a 1 u a 2 u a n 1 u n 1 n 1 = a j u j j=0 We set the linear system and solve... { x u 1 =x 1 x u n 1 =x n 1 10

11 Interpolation Vandermonde matrix { 1 u0 u u n 1 0 n 1 1 a0 x0 x u 1 =x 1 1 u1 u u a1 = x 1 x u n 1 =x n 1 2 n 1 x n 1 1 u n 1 u n 1 u n 1 a n 1 Can be solved by classical numerical methods, but... ( )( ) ( ) The condition number of this matrix is VERY bad It must be solved for each RHS member ( ( xi ) or ( yi ) ), or have to take the inverse of this matrix, or perform an LU decomposition. 11

12 Interpolation Instead of setting the polynomial in x and in y and solving, we can put it under the following n 1 form : p { x u = x i l i u 0 n 1 p i i y u = y l u 0 n 1 P u = P i l ip u 0, where the lip(u) are a polynomial basis of order p=n-1. These polynomials verify, for an interpolation : p i l u j = ij We have only one computation to do for any position of the interpolation points, knowing the ui (the parametrization) 12

13 Interpolation Lets determine the li(u) : they are of order n-1 and such that : l p u =1 ; i= j l ip u j = ij { i p i j l u j =0 ; i j lip may be factored by (u-uj) for j i p i l u =k u u 0 u u 1 u u i 1 u u i 1 u u n 1 k is such that l ip ui =1 k = ui u 0 ui u 1 ui u i 1 u i u i 1 ui un 1 1 p i l (u)= n 1 j=0, j i (u u j ) (ui u j ) The lip are the Lagrange polynomials of order p 13

14 Interpolation Lagrange polynomials p n 1 u u j l u l u = p ' = l u i u ui j=0,i j u i u j p i p n 1 l u = u u j p' j=0 n 1 l (u i )= (u i u j ) j=0, j i (first derivative w/r to u at u=u i ) Properties Interpolation l ip u j = ij Partition of unity n 1 n 1 p l i (u)=1 0 x(u)= x i l ip (u) with x i =1 i 0 14

15 Lagrange polynomials Order 4 The ui are evenly distributed between u=0 and u=1 The sum is equal to 1 (partition of unity) Presence of negative values 15

16 Lagrange polynomials Order 10 Presence of huge overshoots near the boundaries 16

17 Lagrange polynomials The interpolation is represented by the following form : n 1 p i i p i P (u)= P l (u) with l (u)= i=0 n 1 j=0, i j (u u i ) (u j u i ) Two things worth noting: The curve depends linearly on the position of the points It is formed by a weighted sum of basis functions that express the influence of each point on the curve 17

18 Lagrange polynomials An experiment We approximate a circle by an increasing number of points Simultaneously, approximation order increases In every case, the curve is C 18

19 Lagrange polynomials 3 points, order 2 (a parabola!) 19

20 Lagrange polynomials 5 points, order 4 20

21 Lagrange polynomials 11 points, order 10 21

22 Lagrange polynomials 21 points, order 20 22

23 Lagrange polynomials Does it work? The points were set exactly on the circle What occurs if their position is inaccurate? Or if the approximated shape is not so simple? We are going to see two cases The coordinates of the points are perturbed randomly A deterministic increase and decrease of the radius 23

24 Lagrange perturbation Random perturbation Each point is moved radially by a value between -0.5 and +0.5 % of the circle's radius 24

25 Lagrange polynomials 3 points, random perturbation 1% 25

26 Lagrange polynomials 5 points, random perturbation 1% 26

27 Lagrange polynomials 11 points, random perturbation 1% 27

28 Lagrange polynomials 21 points, random perturbation 1% 28

29 Lagrange polynomials Runge phenomenon Similar to Gibbs phenomenon of the decompositions in harmonic functions 29

30 Lagrange polynomials Deterministic perturbation Each point is moved radially by a value of -5 or +5 % of the circle's radius 30

31 Lagrange polynomials 3 points, deterministic perturbation 10% 31

32 Lagrange polynomials 5 points, deterministic perturbation 10% 32

33 Lagrange polynomials 11 points, deterministic perturbation 10% 33

34 Lagrange polynomials 21 points, deterministic perturbation 10% 34

35 Lagrange polynomials How to minimize Runge's phenomenon? The problem here is the use of a unique polynomial and regular intervals between knots. Instead, if we concentrate knots at the extremities, the interpolation is less prone to Runge's phenomenon. Make use of Chebyshev knots: 2i 1 u =cos π 2n ( n i ) in the interval [-1,1] or uin = cos 2i 1 π 2 2 ( 2n ) in the interval [0,1] 35

36 Lagrange polynomials 21 points, random perturbation 1% Uniform nodal intervals lagrange interpolation 36

37 Lagrange polynomials 21 points, random perturbation 1% Using Chebychev knots 37

38 Lagrange polynomials Perturbation of a point We shift one point by a big displacement 38

39 Lagrange polynomials 11 points, using Chebychev knots 39

40 Lagrange polynomials 11 points, using Chebychev knots 40

41 Lagrange polynomials 99 points, using Chebychev knots 41

42 Lagrange polynomials 499 points, using Chebychev knots 42

43 Lagrange polynomials Even without perturbations, we still do numerical approximations Up to how many points can we pass through? 43

44 Lagrange polynomials 51 points (lagrange polynomial of order 50) 44

45 Lagrange polynomials 61 points (lagrange polynomial of order 60) 45

46 Lagrange polynomials Here it has nothing to do with Runge's phenomenon It's rather an accumulation of errors due to arithmetic calculations done in finite precision In our case, near the boundaries, we do an inexact sum of very big numbers in absolute value, but whose sum (the exact sum) is close to the unit (partition of unity). n 1 n 1 p i i p i P (u)= P l (u) with l (u)= i=0 j=0, i j (u u i ) (u j u i ) Big numbers in absolute value, computed with a good relative accuracy Sum increasing the relative error 46

47 Lagrange polynomials How to evaluate a polynomial? Under its Lagrange form (here) : robust except at the boundaries Alternative : newton's formula and divided differences Horner's scheme n 1 P u = i u i P u = a 0 u a 1 u a 2 u a n 1 i=0 Others techniques adapted to particular forms of polynomials later in the course e.g. Optimal method with respect to the number of arithmetic operations for a polynomial expressed as a sum of monomials may be numerically unstable Algorithm of De Casteljau for Bernstein polynomials Algorithm of De Boor for the evaluation of Bsplines In any case, one should be cautious with numerical errors!47

48 Numerical errors Numerical errors Floating point calculations are (most of the time) inaccurate Analysis of the rounding done during operations in floating point : x y= x y 1 1, 1 2 x y= x y 1 2, 2 2 x y= xy 1 3, 3 Mathematically equivalent calculations but expressed differently give distinct results An example with x 2 y 2 = x y x y 48

49 Numerical errors Error made with the expression x y x y x y x y = x y 1 1 x y = x y x y No catastrophic increase of the relative error Error made with the expression x 2 y 2 = [ x 2 (1+ δ1 ) y 2 (1+ δ 2 )](1+ δ3 ) = (( x 2 y 2 )(1+ δ 1)+ (δ1 δ 2 ) y 2 )(1+ δ 3 ) = (( x y )(1+ δ1+ δ3+ (δ1 δ2 ) y + δ1 δ3 + (δ1 δ 2 ) y δ 3 )) ( x x) ( y y) When x is close to y, the error can be of the order of magnitude of the theoretical result... 49

50 Numerical errors Some useful rules (not a comprehensive list!) Prefer x y x y to 2 x y 2 Lagrange form more accurate than horner's scheme N E. g. sum of many terms S = X [ j ] Naive algorithm : j=1 S=0; for (j=1;j<=n;j++){ S=S+X[j] ; } return S ; involves an error N Kahan's summation algorithm S=X[1];C=0 for (j=2;j<=n;j++) { Y=X[j]-C; T=S+Y; C=(T-S)-Y; S=T } return S ; involves an error 2 50

51 Numerical errors Example of catastrophic rounding nx*ny samples Computation of an integral : ymax f ( x, y)dxdy with f (x, y)=x + y 2 S= 2 Ω nx 1 ny 1 S i=0 f x i, y j det J ymin j=0 dx= x max x min / nx dy= y max y min / ny x i = x min i dx dx / 2 y j = y min j dy dy /2 det J =dx dy xmin y xmax x 51

52 Numerical errors Computations made with the following parameters: xmin= ymin= 0.0 ; xmax= ymax= 1.0 ; nx = ny = 10, Sexact=2/3 1) Single precision floating point numbers 2) Double precision floating point numbers 3) Quad precision floating point numbers 4) Single precision floating points numbers with Kahan's summation algorithm bechet@yakusa:floating_error$./test 10 1) sum (float )= ) sum (double )= ) sum (ldouble)= ) sum (kahan )= Note : Program should be compiled without optimization! 52

53 Numerical errors 100 1) sum (float )= ) sum (double )= ) sum (ldouble)= ) sum (kahan )= ) sum (float )= ) sum (double )= ) sum (ldouble)= ) sum (kahan )=

54 Numerical errors ) sum (float )= ) sum (double )= ) sum (ldouble)= ) sum (kahan )= ) sum (float )= ) sum (double )= ) sum (ldouble)= ) sum (kahan )= Compensated summation algorithm coming from : William Kahan. Further remarks on reducing truncation errors. Comm. ACM, 8(1):40,

55 Numerical errors y=p(x)=(1-x)n for x=1.333 and 2<n<41 S. Graillat, Ph. Langlois, N. Louvet Compensated Horner Scheme Research Report RR , LP2A, University of Perpignan, France, july

56 Numerical errors Definition of the condition number of a numerical expression Ratio of the direct error to the inverse error y K P, x =lim sup 0 x D x For a polynomial K P, x = i=0 ai x p p i n inverse error δ x x +δ x ( monomial form ) n p x y= p( x ) δy direct error y = p ( x )= y +δ y = p( x +δ x ) i a x i=0 i 56

57 Numerical errors There are various compensated algorithms to carry out calculations on floating point numbers. Ex. Kahan summation, Compensated Horner scheme... In general, they allow to have similar results as when using internal floating point with a precision twice that of the input data, following by a final rounding. See references available on the course's website for the compensated Horner scheme and other documents. 57

58 Lagrange polynomials Morality : Lagrange interpolation is not suited beyond 10's of control points because of the Runge phenomenon A modification of the position of a control point leads to global changes of the curve. The evaluation of high order polynomials expressed as monomials leads to numerical problems. No control of the slopes at the boundary of the curve (start and finish). 65

59 66

60 Motivation Let us imagine that we have many (100's of) control points But we don't want a Lagrange interpolation! We should stay with a low order scheme but conserve enough freedom to pass through every point Curve defined by pieces... and of low order (1) 67

61 We are going to build a low order interpolation for each knot interval, such that we can impose slopes at the knots. P'=P'i+1 u=ui+1 P'=P'i P=Pi+1 P=Pi u=ui 68

62 In each range [i,i+1], we want to have an independent polynomial We have 4 parameters : position at each knot and associated tangents. The basis must have 4 degrees of freedom, thus be of order 3 in the case of polynomials. x (u i )=x i P (ui )=P i y (u i )= y i... { x [i ] u = A[i ]0 A[i ]1 u A[i ]2 u2 A[i ]3 u3, u [ui, u i 1 ] 69

63 First, every interval has a unit length i.e. ui 1 ui =1 Then we ensure identical intervals [0...1] between each interpolation point : u u i u = =u u i u i 1 u i d u =1 du On each interval i, we thus have the following relation: 2 3 x[i] ( u )=a + a u +a u + a u [i] 0 [i]1 [i ]2 [i] 3,u [0,1] 70

64 P'=P'i+1 u =1 P'=P'i P=Pi+1 P=Pi u =0 71

65 We pass through both control points: 2 3 P ( u =0)=P i a [i ]0 +a [i ]1 u + a[i ]2 u + a[i ]3 u =x i 2 3 P ( u =1)=P i+1 a [i]0 +a [i]1 u +a [i] 2 u +a [i ]3 u = x i+1 We impose both slopes : P ' ( u 0 =0)=P 'i a[i ]1 +2 a [i ]2 u +3 a [i ]3 u 2= xi' ' 2 ' P ( u =1)=P 'i+1 a[i ]1 +2 a [i] 2 u +3 a[i ]3 u = x i +1 At the end : { a[i]0 = xi a[i]1 =x 'i a[i] 2=3 xi 1 x i 2 x'i x 'i 1 a[i]3 =2 xi x i 1 xi' x 'i 1 72

66 We have continuity We have continuity of the derivatives But how to choose the slopes? Let the user choose ( artistic freedom) Automatically... 73

67 By finite differences with three points : xi 1 xi xi xi 1 x= 2 u i 1 u i 2 u i u i 1 ' i At the boundaries, we use finite differences (asymmetric) x1 x 0 x= u 1 u0 ' 0 x ' n 1 x n 1 x n 2 = u n 1 u n 2 The result depends on the parametrization! Cardinal spline x i 1 xi 1 x = 1 c, 0 c 1 2 ' i x '0=(1 c)( x 1 x 0 ) ' x n 1 =(1 c)( x n 1 x n 2 ) c is a «tension» parameter. c=0 gives yields the so called Catmull-Rom spline, c=1 a zigzagging line. 74

68 Continuity of the curve/parameter but loss of regularity (and of geometric continuity in many cases) 5 points, finite differences by varying the parametrization [0..1], [0..2], [0..5], [0..10] 75

69 5 points, Cardinal Spline (Catmull-Rom) c=0 76

70 Catmull-Rom are widely used in computer graphics Simple to compute, effective Local control (price to pay : discontinuous secd derivative) Animations with keyframing Ensures a fluid motion because of the continuity of the slope 77

71 5 points, Cardinal Spline c=

72 5 points, Cardinal Spline c=0.5 79

73 5 points, Cardinal Spline c=

74 5 points, Cardinal Spline c=1.0 81

75 We can impose the continuity of second derivatives... On a curve with n points, we have n extra relations to impose We may impose the continuity of the second derivative only on the n-2 interior knots What about the 2 points on the boundary? Impose a vanishing second derivative. We obtain what is called «natural spline» We could also impose the slopes (i.e. only n-2 relations remaining) Or, impose that the third derivative is zero on the points 1 and n-2 That means a single polynomial expression for the first two knot intervals, and the last two. 82

76 Natural Spline : mathematical approximation of the spline historically used in naval construction. C. DeBoor 83

77 Boeing 84

78 We impose the continuity of the second derivatives '' x[i 1] 1 =x ''[i ] 0 2a[i 1] 2 6a [i 1]3 =2a [i ]2 We substitute in the internal equations 2[3 x i x i 1 2 x 'i 1 x 'i ] 6[2 x i 1 x i x 'i 1 x 'i ] ' ' =2[3 x i 1 x i 2 x i x i 1 ] Finally we obtain : x ' i 1 ' i 4 x x ' i 1 =3 xi 1 x i 1 85

79 At the boundaries we want '' x [0] 0 =0 2a[0] 2=0 '' x [n 2] 1 =0 2a[n 2] 2 6a[ n 2] 3=0 ' ' 2 x0 x1 =3 x 1 x0 ' ' x n 2 2 x n 1=3 x n 1 x n 2 We have then a linear system with n unknowns : ' x0 3 x 1 x0 x '1 3 x 2 x0 1 x '2 = 3 x 3 x x' 3 x n 1 x n 3 n ' 3 x n 1 x n 2 x n 1 86

80 By solving the system, we have : x '0 ' 1 ' 2 x x ' x n 2 x 'n 1, which is substituted in { a [i]0 = xi ' a [i]1 =x i ' ' a [i] 2=3 xi 1 x i 2 xi x i 1 ' ' a [i]3 =2 xi x i 1 xi x i 1,to get the polynomial in each portion : x [i ] u =a [i] 0 a[i ]1 u a [i ]2 u 2 a[i]3 u 3, 0 u 1 From the global parameter u, we have to find in which portion we are ( the value of i ), then compute right polynomial... 87

81 Catmull-Rom spline 5 control points, natural spline 88

82 Another experiment We approximate a circle by a number of increasing points Simultaneously, the order of the approximation the number of pieces increases. In all the cases, the curve is C C2 89

83 3 points, order 3! 90

84 5 points 91

85 11 points 92

86 21 points 93

87 Random perturbation Each point is moved radially by a value between -0.5 and +0.5 % of the circle's radius 94

88 3 points, random perturbation 1% 95

89 5 points, random perturbation 1% 96

90 11 points, random perturbation 1% 97

91 21 points, random perturbation 1% 98

92 Deterministic perturbation Each point is shifted radially depending on its position by -5 or +5 % of the circle's radius 99

93 3 points, deterministic perturbation 5% 100

94 5 points, deterministic perturbation 5% 101

95 11 points, deterministic perturbation 5% 102

96 21 points, deterministic perturbation 5% 103

97 non local control 11 points 104

98 105

99 Perturbation of a point We shift one point by a significant amount 106

100 21 points 107

101 99 points 108

102 999 points 109

103 Stable interpolation scheme Weak Runge phenomenon The displacement of a point yet affects all the curve Nevertheless, the perturbation fades very quickly further away from the shifted point «Overshoots» are limited. 110

104 Closed curve? The curve can be closed, just impose everywhere that the second derivative is continuous. 111

105 Instead of '' x [0] 0 =0 2a[0] 2=0 x ''[n 2] 1 =0 2a[n 2] 2 6a[ n 2] 3=0... we have x '' [n 2] '' [0 ] (1)= x (0) 2a[ n 2 ]2 +6a[ n 2 ]3=2a [0] 2 ' 0 ' 1 ' 2 3( x1 x n 2 ) x x 3( x 2 x 0 ) Circulant x = 3( x 3 x 1) x n 1 = x 0 and ' ' matrix x n 1 = x x 'n 3 3( x n 2 x n 4 ) x' 3( x 0 x n 3 ) n 2 ( )( ) ( ) 112

106 General case : arbitrary parametrization P (ui )= P i { x (ui )= xi y (ui )= y i x[i] (u)= A[i] 0 + A[i ]1 u+ A[i ] 2 u 2 + A[i ]3 u 3, u [u i, u i+1 ] We again change the parametrization... u ui u = u i +1 u i du = 1 = 1 du u i+1 ui hi 2 3 x[i] ( u )=a + a u +a u + a u [i] 0 [i]1 [i ]2 [i] 3,u [0,1] 113

107 We pass through both points : P u0 =0 = P i a [i ]0 a [i]1 u 0 a [i ] 2 u a [i ]3 u = xi 2 3 P u1=1 = P i 1 a [i ]0 a [i ]1 u1 a [i] 2 u 1 a [i]3 u1= x i 1 We impose both slopes : dp dp 1 2 ' P u 0 =0 = 0 = 0 = P 'i a[i ]1 2a [i] 2 u 0 3a [i] 3 u 0 = x i hi du d u hi dp dp 1 P ' u 1=1 = 1 = 1 = P ' i 1 a [i ]1 2a[i ]2 u 1 3a[i ]3 u 21 = x 'i 1 h i du d u hi a[i ]0 = x i Finally : a = x ' h [i ]1 i i a[i ]2 =3 x i 1 xi 2 x 'i h i x 'i 1 h i ' { ' i a[i ]3=2 x i x i 1 x h i x ' i 1 hi 114

108 We impose the second derivative for a natural spline d P [i] d P [i] d 2 u d P [i] 1 2 u = 2 2 = du d u d u d u 2a[i 1] 2 6a [i 1] 3 2a [i] 2 '' '' x[i 1] 1 = x[i ] 0 = 2 2 hi 1 hi 2 hi 2 We substitute in the internal equations 2[3 xi x i 1 2 x 'i 1 hi 1 x 'i hi 1 ] 6[ 2 x i 1 x i x 'i 1 hi 1 x 'i hi 1 ] 2 i 1 = h 2[3 x i 1 x i 2 x 'i hi x 'i 1 hi ] 2 i h 115

109 We obtain finally : x 'i 1 2 x 'i 2 x 'i x 'i 1 x i x i 1 x i 1 x i =3 3 2 hi 1 hi hi 1 hi2 hi x ' i 1 ' i ' i 2 x hi 1 2 x x ' i 1 hi hi 1 =3 x i x i 1 3 x i 1 x i h i 1 hi 116

110 At the boundaries we want a vanishing second derivative '' [0] x (0)=0 x '' [n 2] (1)=0 2a [0] =0 (h ) 2a[ n 2 ]2 +6a[ n 2 ]3 (h 2 n 2 ) h0 (2 x '0 + x '1)=3( x 1 x 0 ) =0 ' ' h n 2 ( x n 2 +2 x n 1 )=3( x n 1 x n 2 ) We have then a linear system of n unknowns: (next page) 117

111 ( ' 0 ' 1 2 h0 h0 h1 2 h1 + 2 h 0 h0 x x = x 'n 2 )( ) h n 2 2 hn h n 3 h n 3 h n 2 2 h n 2 x 'n 1 ( 3( x 1 x 0 ) h1 h0 3 ( x 1 x 0 )+3 ( x 2 x 1 ) h0 h1 h n 2 h n 3 3 ( x n 2 x n 3 )+3 ( x n 1 x n 2 ) h n 3 h n 2 3( x n 1 x n 2 ) ) 118

112 u=3 Natural Spline (uniform parametrization) u=4 u=1 u=4 u=5 u=2 u=0 Natural Spline (non uniform parametrization) 119

113 Compact notation We would like a compact representation as for the Lagrange interpolation. a [i]0 = xi 2 x [i ] ( u )=a [i ]0 +a [i]1 u +a [i ]2 u +a [i ]3 u 0 u <1 { n p i0 n ' i p i1 x[i] ( u )= xi h ( u )+ x h ( u ) 0 n p i0 0 n 0 { ' a [i]1 =x i ' ' a [i] 2=3 xi 1 x i 2 xi x i 1 a [i]3 =2 xi x i 1 xi' x 'i 1 n ' i p i1 y[i ] ( u )= yi h ( u )+ y h ( u ) 0 3 p i0 n ' i p i1 P [i ] ( u )= P i h ( u )+ P h ( u ) This on each interval [i,i+1]. 120

114 We have on each interval : { a [i]0 = xi a [i]1 =x 'i ' ' a [i] 2=3 xi 1 x i 2 xi x i 1 a [i]3 =2 xi x i 1 xi' x 'i 1 x [i ] u = xi x 'i u 3 x i 1 x i 2 x 'i x'i 1 u 2 2 xi x i 1 x'i x 'i 1 u 3 By rearranging equations x [i ] u = xi 1 3 u 2 2 u 3 x 'i u 2 u 2 u ' 2 3 x i 1 3 u 2 u xi 1 u u 121

115 Hermite polynomials (for two points) { p h 00 =1 3 u 2 2 u 3 p 2 3 h10 =3 u 2 u p 2 3 h 01= u 2 u u p 2 3 h11= u u 122

116 The more general Hermite basis of degree 2n-1 allows to do an interpolation of n points on an interval, as the Lagrange basis, additionally imposing slopes at each point. [ hi0n (u)= 1 n i1 n '' l (u i ) n' l (u i ) n i ] (u ui ) [l in (u)]2 h u = u ui [l u ] 2 n 1 n u ui l u n l i u = n ' = l ui u ui j=0,i j u j u i n n 1 l u = u u j j=0 123

117 Properties of the Hermite basis Interpolation of the positions hi0n u j = ij h i0n ' u j =0 n hi1 u j =0 h i1n ' u j = ij Interpolation of the slopes n i0 h i u =1 124

118 Property of affine invariance It is a useful property such that the curves we define for a set of control points can undergo linear affine transformations without hassle. Let Pi* the affine transformation of the control points Pi Let P*(u,Pi) the affine transformation of the points of the curve P(u,Pi) defined from the original points Pi Let P(u,Pi*) the new curve based on the modified control points Pi*, with the same parametrization. The affine invariance is verified iff P*(u,Pi) = P(u,Pi*) for all u. 125

119 Affine transformations P A P u Translation a u= b ; A= c [] [ ] Scaling 3 rotations d 0 0 u=0 ; A= 0 e f Shear 1 g h u=0 ; A= 0 1 i [ ] 12 degrees of freedom [ ] [ ] cos sin 0 u=0 ; A= sin cos

120 Let P a parametric curve built this way : n 1 P (u, P i )= P i K ni (u) 0 Let's verify the invariance by a translation t: * i n 1 n 1 n i n 1 n i n i P (u, P )= ( P i +t) K (u)= P i K (u)+ t K (u) 0 n 1 0 n i = P (u, P i )+ t K (u) 0 = P (u, P i )+t=p * (u, P i ) 0 Partition of unity n 1 iff n K i (u)=

121 For the other multiplicative transformations * i n 1 n i n 1 n i P (u, P )= (A P i ) K (u)=a P i K (u) 0 0 =A P (u, P i )=P * (u, P i ) (no particular conditions except linearity with respect to the coordinates of the control points) Consequently, iff the basis functions form a partition of unity, and the dependence with respect to the control points is linear, then the representation is invariant by any affine transformation. 128

122 Do Hermite's basis form a partition of unity? n 1 { n h?? u =1 0 F u =1 F 0 =1 F 1 =1 ' p h 00 =1 3 u 2 2 u 3 p 2 3 h10 =3 u 2 u p 2 3 h 01= u 2 u u p 2 3 h11= u u F 0 =0 ' F 1 =0 129

123 Let's check the invariance If we apply a translation to the control points Pi, the derivatives Pi' should not change... * i '* i P = P i +t P =P ' i P ( P *i )= ( P i +t ) h ni0 (u)+ P i' hi1n (u) 1 n i0 1 n i0 1 ' i n i1 = th (u)+ P i h (u)+ P h (u) * =t + P (P i )= P ( P i ) 130

124 We must also check the invariance for the other multiplicative transformations : those affect both the coordinates and the derivatives * '* P i = A P i * ' P i = A P i 1 1 n ' n P P i = A P i h i0 u A P i hi1 u n i0 ' i n i1 = A P i h u P h u 0 0 * = A P ( P i )= P (P i ) QED 131

125 Beware of the computation of the slopes x'i... Natural : x ' 0 ' 1 ' 2 3 x 1 x0 x 3 x 2 x0 1 x = 3 x 3 x x' 3 x n 1 x n 3 n ' 3 x n 1 x n 2 x n 1 Linear operator P 'i = L( P i P j ) P 'i * = L (P *i P *j )= L(( A P i +t) ( A P j +t)) ' = A L( P i P j )= A P i It is OK in this case 132

126 Rotation of 45 Scaling x direction (times 0.5) Followed by a translation 133

127 3D Curves Minimal order so that a curve can have a torsion (non planar curve) n 1 p Let's consider a Lagrange interpolation P u = P i l i u i=0 2 points on a straight line (no curvature) 3 points in a plane (no torsion) 4 points torsion becomes possible Minimal order to join smoothly two arbitrarily oriented curves = 3 134