interpolation and polynomial approximation interpolation - … · 2019-07-15 · 1 computer aided...

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Computer Aided Design

Outline

Interpolation and polynomial approximation Interpolation

Lagrange Cubic Splines

Approximation Bézier curves B-Splines

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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

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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

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Interpolation

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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...

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Interpolation

We must choose the parametrization (nodal sequence)

Uniform

u0=0

u1=1

u2=2

un-1

=n-1

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Interpolation

Or non-uniform

u0=0

u1=2.5

u2=7

un-1

=15

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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...

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Interpolation

Parametrization as an approximate arc length

u0=0

u1=d

1

u2=d

1+d

2

d1

dn-1

d2

un-1

=d1+...+d

n-1

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Interpolation

In all cases, we want to have :

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 :

We set the linear system and solve...

P ui=P i ≡ {xui=xi

y ui= y i

x u=a0a1 ua2 u2⋯an−1 un−1

=∑j=0

n−1

a j uj

{x u1=x1

⋮

x un−1=xn−1

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Interpolation

Vandermonde matrix

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 ( y

i ) )

, or have to take the inverse of this matrix, or perform an LU decomposition.

{x u1=x1

⋮

x un−1=xn−1 (1 u0 u0

2⋯ u0

n−1

1 u1 u12

⋯ u1n−1

⋮ ⋮ ⋮ ⋮ ⋮

1 un−1 un−12

⋯ un−1n−1)(

a0

a1

⋮an−1

)=(x0

x1

⋮xn−1

)

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Interpolation

Instead of setting the polynomial in x and in y and solving, we can put it under the following form :

, where the lip(u) are a polynomial basis of order p=n-1.

These polynomials verify, for an interpolation :

We have only one computation to do for any position of the interpolation points, knowing the u

i

(the parametrization)

{x u=∑

0

n−1

x i l ipu

y u=∑0

n−1

yi l ipu

⇔ P u=∑0

n−1

Pi l ipu

l ipu j= ij

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Interpolation

Lets determine the li(u) : they are of order n-1

and such that :l i

pu j= ij⇔{l i

pu j=1 ; i= j

l ipu j=0 ; i≠ j

lip may be factored by (u-u

j) for j≠i

l ipu =k u−u0u−u1⋯u−ui−1u−ui1⋯u−un−1

k is such that l ipui=1

k=ui−u0ui−u1⋯ui−ui−1ui−ui1⋯ui−un−1−1

l ip(u)= ∏

j=0, j≠i

n−1 (u−u j)

(ui−u j)

The lip are the Lagrange polynomials of order p

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Interpolation

Lagrange polynomials

Properties

Interpolation Partition of unity

l ipu=

l pu

l p 'uiu−ui

= ∏j=0, i≠ j

n−1 u−u j

ui−u j

l pu=∏

j=0

n−1

u−u j

l ipu j=ij

l p '(ui)= ∏

j=0 , j≠i

n−1

(ui−u j) (first derivative w/r to u at u=ui)

∑0

n−1

l ip(u)=1

x(u)=∑0

n−1

x i l ip(u) with x i=1∀i

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Order 4 The u

i are evenly

distributed between u=0 and u=1

The sum is equal to 1 (partition of unity)

Presence of negative values

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Order 10 Presence of huge

overshoots near the boundaries

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The interpolation is represented by the following form :

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

P (u)=∑i=0

n−1

P i l ip(u) with l i

p(u)= ∏

j=0, i≠ j

n−1 (u−ui)

(u j−ui)

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An experiment We approximate a circle by an increasing number

of points Simultaneously , approximation order increases

In every case, the curve is C

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3 points, order 2 (a parabola !)


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5 points, order 4


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11 points, order 10


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21 points, order 20


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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

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Lagrange perturbation

Random perturbation Each point is moved radially by a value between

-0.5 and +0.5 % of the circle's radius

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3 points, random perturbation 1%


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28




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Runge phenomenon Similar to Gibbs phenomenon of the

decompositions in harmonic functions

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Deterministic perturbation Each point is moved radially by a value of -5 or +5

% of the circle's radius

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3 points, deterministic perturbation 10%


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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:

in the interval [-1,1]

or in the interval [0,1]

uin=cos ( 2 i−1

2 nπ)

uin=

12

+12

cos( 2 i−12 n

π)

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Uniform nodal intervals – lagrange interpolation


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Using Chebychev knots


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Perturbation of a point We shift one point by a big displacement

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11 points, using Chebychev knots


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Even without perturbations, we still do numerical approximations

Up to how many points can we pass through ?

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51 points (lagrange polynomial of order 50)


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61 points (lagrange polynomial of order 60)


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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).

Big numbers in absolute value, computed with a good relative accuracy

Sum increasing the relative error

P (u)=∑i=0

n−1

P i l ip(u) with l i

p(u)= ∏

j=0, i≠ j

n−1 (u−ui)

(u j−ui)

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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

Optimal method with respect to the number of arithmetic operations for a polynomial expressed as a sum of monomials – may be numerically unstable

Others techniques adapted to particular forms of polynomials later in the course e.g.

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 !

P u=∑i=0

n−1

i ui⇔ P u =a0u a1ua2u an−1

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Numerical errors

Numerical errors Floating point calculations are (most of the time)

inaccurate Analysis of the rounding done during operations in

floating point :

Mathematically equivalent calculations but expressed differently give distinct results

An example with

x⊖ y=x− y 11 , ∣1∣≤2

x⊕ y= x y 12 , ∣2∣≤2

x2− y2

=x yx− y

x⊗ y= xy13 , ∣3∣≤

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Numerical errors

Error made with the expression

No catastrophic increase of the relative error Error made with the expression

When x is close to y, the error can be of the order of magnitude of the theoretical result...

x2− y2

(x⊗ x)⊖( y⊗ y) = [ x2(1+ δ1)− y2

(1+ δ2)](1+ δ3)

= ((x2− y2

)(1+ δ1)+ (δ1−δ2) y2)(1+ δ3)

x yx− y

x⊕ y⊗x⊖ y = x− y 11x y1213

= x y x− y 1123122 31312 3

≈ 5

= ((x2− y2

)(1+ δ1+ δ3+ (δ1−δ2) y2+ δ1 δ3+ (δ1−δ2) y2

δ3))

50


Numerical errors

Some useful rules (not a comprehensive list !) Prefer to

Lagrange form more accurate than horner's scheme … E. g. sum of many terms

Naive algorithm :

involves an error Kahan's summation algorithm

involves an error

x2− y2x yx− y

S=0;for (j=1;j<=N;j++){ S=S+X[j] ; }return S ;≈ N

S=X[1];C=0for (j=2;j<=N;j++) { Y=X[j]-C; T=S+Y; C=(T-S)-Y; S=T }return S ;

≈2

S=∑j=1

N

X [ j ]

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Numerical errors

Example of catastrophic rounding Computation of an integral :

S=∫Ω

f (x , y)dxdy with f (x , y)=x2+ y2

x

y

S≈ ∑i=0

nx−1

∑j=0

ny−1

f x i , y j det J

dx=xmax−xmin/nxdy= ymax− ymin/ny

x i =xmini dxdx /2y j = ymin j dydy /2

det J =dx dy

nx*ny samples

xmin

xmax

ymin

ymax

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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 numbers2) Double precision floating point numbers3) Quad precision floating point numbers4) Single precision floating points numbers with Kahan's summation algorithm

bechet@yakusa:floating_error$ ./test 101) sum (float )=0.665000021457672119142) sum (double )=0.665000000000000257573) sum (ldouble)=0.664999999999999999974) sum (kahan )=0.66499999999999999997Note : Program should be compiled without optimization !

53


Numerical errors

bechet@yakusa:floating_error$ ./test 1001) sum (float )=0.666649818420410156252) sum (double )=0.666650000000000519943) sum (ldouble)=0.666650000000000000934) sum (kahan )=0.66664993762969970703


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Numerical errors



William Kahan. Further remarks on reducing truncation errors. Comm. ACM, 8(1):40,1965.

Compensated summation algorithm coming from :

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Numerical errors

y=p(x)=(1-x)n for x=1.333 and 2<n<41

S. Graillat, Ph. Langlois, N. LouvetCompensated Horner SchemeResearch Report RR2005-04, LP2A, University of Perpignan, France, july 2005

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Numerical errors

Definition of the condition number of a numerical expression

Ratio of the direct error to the inverse error

For a polynomial

( monomial form )

K P , x=lim 0

sup x∈D

∣ y∣∣ x∣

K P , x=∑i=0

n

∣ai∣∣x∣i

∣∑i=0

nai x i∣

x y= p(x)

y= p(x)= y+δ y= p(x+δ x)

x+δ x

p

p

p

δ x

δ ydirecterror

inverseerror

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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.

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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).

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Splines

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Splines

MotivationLet 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)

68


Splines

We are going to build a low order interpolation for each knot interval, such that we can impose slopes at the knots.

u=ui+1

u=ui

P=Pi+1

P=Pi

P'=P'i

P'=P'i+1

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Splines

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 [i ]u=A[i ]0A[i ]1 uA[ i ]2 u2A[i ]3 u3 , u∈[ui , ui1]

P (ui)=P i ≡ {x (ui)=x i

y (ui)= y i

...

70


Splines

First, every interval has a unit length i.e.

Then we ensure identical intervals [0...1] between each interpolation point :

On each interval i , we thus have the following relation:

ui1−ui=1

u=u−ui

u i1−ui

=u−ui

x[ i ](u)=a[ i ] 0+a[ i]1 u+a[ i ]2 u2+a[i ] 3 u3 , u∈[0,1]

d udu

=1

71


Splines

P=Pi+1

P=Pi

P'=P'i

P'=P'i+1

u=0

u=1

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Splines

We pass through both control points:

We impose both slopes :

At the end :

P (u=0)=P i ⇔a[ i ]0+a[ i ]1 u+a[ i ]2 u2+a[ i ]3 u3

=x i

P (u=1)=P i+1 ⇔a[ i ]0+a[ i ]1 u+a[ i ] 2 u2+a[ i ]3 u3

=x i+1

P '(u0=0)=P 'i ⇔ a[ i ]1+2a[ i ]2 u+3 a[i ]3 u2

=xi'

P '(u=1)=P 'i+1⇔a[ i ]1+2 a[ i ] 2 u+3 a[ i ]3 u2

=xi +1'

{a[ i ]0=xi

a[ i ]1=x i'

a[ i ] 2=3 xi1−x i−2 xi'−x i1

'

a[ i ]3=2xi−x i1xi'x i1

'

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Splines

We have continuity We have continuity of the derivatives But how to choose the slopes ?

Let the user choose ( “artistic” freedom) Automatically ...

74


Splines

By finite differences with three points :

At the boundaries, we use finite differences (asymmetric)

The result depends on the parametrization ! Cardinal spline

c is a « tension » parameter. c=0 gives yields the so called “Catmull-Rom” spline, c=1 a zigzagging line.

x i'=

xi1−xi

2 ui1−ui

xi−xi−1

2 ui−ui−1

x0'=

x1−x0

u1−u0

xn−1'

=xn−1−xn−2

un−1−un−2

x i'=1−c

x i1−xi−1

2 , 0≤c≤1

x0'=(1−c)(x1− x0)

xn−1'

=(1−c)(xn−1−xn−2)

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Splines

5 points, finite differences by varying the parametrization [0..1] , [0..2] , [0..5] , [0..10]

Continuity of the curve/parameter but loss of regularity (and of geometric continuityin many cases)

76


Splines

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

77


Splines

Catmull-Rom Splines 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

78


Splines

5 points, Cardinal Spline c=0.25

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Splines


80


Splines


81


Splines


82


Splines

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.

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Splines

Natural Spline : mathematical approximation of the spline historically used in naval construction.

C. D

eBoor

84


Splines

Boeing

85


Splines

We impose the continuity of the second derivatives

We substitute in the “internal” equations

Finally we obtain :

x[ i−1]

''1=x [i ]

''0⇔ 2a[ i−1] 26a[ i−1]3=2a[i ]2

2 [3 x i−x i−1−2 x i−1'

−x i']6 [2 x i−1−x ix i−1

'x i

']

=2 [3 x i1−x i−2 x i'−x i1

']

x i−1'

4 x i'x i1

'=3xi1−x i−1

86


Splines

At the boundaries we want

We have then a linear system with n unknowns :

x [0]

''0=0⇔ 2a[0] 2=0

x [n−2]

''1=0⇔ 2a[n−2] 26a[ n−2] 3=0

2 x0'x1

'=3x1−x0

xn−2'

2 xn−1'

=3xn−1−xn−2

2 11 4 1

1 4 1⋱

1 4 11 2

x0

'

x1'

x2'

⋮

xn−2'

xn−1'

=3x1−x0

3 x2−x0

3x3−x1

⋮

3xn−1−xn−3

3 xn−1−xn−2

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Splines

By solving the system, we have :

, which is substituted in

,to get the polynomial in each portion :

From the global parameter u, we have to find in which portion we are ( the value of i ) , then compute right polynomial...

x0

'

x1'

x2'

⋮

xn−2'

xn−1'

{a[ i ]0=xi

a[ i ]1=x i'

a[ i ] 2=3 xi1−x i−2 xi'−x i1

'


'

x [i ]u =a[i ] 0a[ i ]1ua[ i ]2 u2a[ i ]3 u

3 , 0≤u1

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Splines

5 control points, natural spline

Catmull-Rom spline

89


Splines

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 C

2

90


Splines

3 points, order 3!

91


Splines

5 points

92


Splines

11 points

93


Splines

21 points

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Splines

Random perturbation Each point is moved radially by a value between

-0.5 and +0.5 % of the circle's radius

95


Splines


96


Splines


97


Splines


98


Splines


99


Splines

Deterministic perturbation Each point is shifted radially depending on its

position by -5 or +5 % of the circle's radius

100


Splines


101


Splines


102


Splines


103


Splines


104


Splines

11 points

non local control

105


Splines

106


Splines

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

107


Splines

21 points

108


99 points

Splines

109


Splines

999 points

110


Splines

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.

111


Splines

Closed curve ? The curve can be closed, just impose everywhere

that the second derivative is continuous.

112


Splines

Instead of

... we have

(4 1 11 4 1

1 4 1⋱

1 4 11 1 4

)(x0

'

x1'

x2'

⋮

xn−3'

x n−2'

)=(3(x1−xn−2)

3(x2−x0)

3(x3−x1)

⋮3( xn−2− xn−4)

3(x0−x n−3))

x[n−2]

''(1)=x[0 ]

''(0)⇔ 2a[ n−2 ]2+6a[ n−2 ]3=2a [0] 2

x [0]

''0=0⇔ 2a[0] 2=0

x [n−2]

''1=0⇔ 2a[n−2] 26a[ n−2] 3=0

Circulantmatrix

xn−1=x0

xn−1'

=x0'and

113


Splines

General case : arbitrary parametrization

We again change the 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 u2+ A[ i ]3 u3 , u∈[ui , ui+1]

u=u−ui

ui +1−ui

x[i ](u)=a[ i ]0+a[ i ]1 u+a[ i ]2 u2+a[ i ] 3 u3 , u∈[0,1]

d udu

=1

ui+1−ui

=1hi

114


Splines

We pass through both points :

We impose both slopes :

Finally :

P u0=0=P i ⇔a[ i ]0a[ i ]1 u0a[i ] 2u02a[ i ]3 u0

3=xi

P u1=1=Pi1⇔a[ i ]0a[i ]1 u1a[i ] 2 u12a[i ]3 u1

3=x i1

P 'u0=0=

d Pd u

0=d Pd u

01hi

=P 'i ⇔ a[i ]12a [ i ] 2 u03a [ i] 3u02=x i

' hi

P 'u1=1=

d Pd u

1=d Pd u

11hi

=P 'i1⇔a[i ]12a[ i ]2 u13a[ i ]3 u12=x i1

' hi

{a[i ]0=x i

a[i ]1=x i' hi

a[i ]2=3 x i1− xi−2 x i' hi−x i1

' hi

a[i ]3=2 x i−x i1x i' hix i1

' hi

115


Splines

We impose the second derivative for a natural spline

We substitute in the internal equations

d 2 P [i ]

du2 u=d 2 P[i ]

d u2

d 2u

d u2 =d 2 P [ i]

d u2

1

hi2

2 [3 xi−x i−1−2 x i−1' hi−1−x i

' hi−1]6 [2 x i−1−x ix i−1' hi−1x i

' hi−1]

hi−12

=2 [3 x i1−x i−2 x i

' hi−x i1' hi ]

hi2

x[ i−1]

''1=x[i ]

''0⇔

2a[ i−1] 26a [ i−1] 3

hi−12

=2a [i] 2

hi2

116


Splines

We obtain finally :

x i−1'

2 x i'

hi−1

2 x i

'x i1

'

hi

=3 x i−x i−1

hi−12

3 x i1−x i

hi2

hi x i−1'

2 x i'hi−12 xi

' xi1

'=3

hi

hi−1

x i−x i−13hi−1

hi

x i1−x i

117


Splines

At the boundaries we want a vanishing second derivative …

We have then a linear system of n unknowns:

(next page)

x[0]

''(0)=0⇔

2a [0] 2

(h02)

=0

x[n−2]

''(1)=0⇔

2a[ n−2 ]2+6a[ n−2 ]3

(hn−22

)=0

h0(2 x0'+ x1

')=3(x1−x0)

hn−2(xn−2'

+2 xn−1'

)=3(xn−1−xn−2)

118


Splines

(2h0 h0

h1 2 h1+2h0 h0

⋱

hn−2 2 hn−2+2 hn−3 hn−3

hn−2 2hn−2

)(x0

'

x1'

⋮

xn−2'

xn−1'

)=

(3(x1−x0)

3h1

h0

(x1− x0)+3h0

h1

(x 2−x1)

⋮

3hn−2

hn−3

(xn−2−xn−3)+3hn−3

hn−2

(xn−1−xn−2)

3(xn−1−xn−2)

)

119


Splines

Natural Spline(uniform parametrization)

Natural Spline(non uniform parametrization)

u=0

u=1

u=2

u=4

u=5

u=3

u=4

120


Splines

Compact notation We would like a compact representation as for the

Lagrange interpolation.

This on each interval [i,i+1].

x[i ](u)=a[ i ]0+a[ i ]1 u+a[i ]2 u2+a[i ]3 u3

0≤u<1

{x[ i](u)=∑

0

n

xi hi 0p(u)+∑

0

n

x i' hi 1

p(u)

y[ i ](u)=∑0

n

yi hi 0p(u)+∑

0

n

y i' hi 1

p(u)

...

⇔ P[i ](u)=∑0

n

P i hi 0p(u)+∑

0

n

Pi' hi 1

p(u)

{a[ i ]0=xi

a[ i ]1=x i'

a[ i ] 2=3 xi1−x i−2 xi'−x i1

'


'

121


Splines

We have on each interval :

By rearranging equations

{a[i ]0=xi

a[i ]1=x i'

a[i ] 2=3 xi1−x i−2 xi'−x i1

'

a[i ]3=2xi−x i1xi'x i1

'

x [i ]u =xix i'u3x i1−x i−2 x i

'−xi1

' u

22 xi−x i1xi

'x i1

'u

3

x [i ]u=xi 1−3u22u

3x i

'u−2u

2u

3

x i13u2−2 u

3xi1

'−u

2u

3

122


Splines

Hermite polynomials (for two points)

{h00

p=1−3u

22u

3

h10p=3u

2−2u

3

h01p=u−2u

2u

3

h11p=−u

2u

3

123


Splines

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−

l n ''(ui)

l n '(ui)

(u−ui)][ l in(u)]

2

hi1nu =u−ui[l i

nu]

2

l inu=

l nu

l n 'uiu−ui

= ∏j=0, i≠ j

n−1 u−ui

u j−ui

l nu =∏

j=0

n−1

u−u j

124


Splines

Properties of the Hermite basis

hi0n

u j= ij

hi0n

'u j=0

hi1n'u j=ij

hi1nu j=0

∑i

hi0nu=1

Interpolation of the positions

Interpolation of the slopes

125


Splines

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 P

i

Let P*(u,Pi) the affine transformation of the points of the

curve P(u,Pi) defined from the original points P

i

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,P

i*)

for all u.

126


Splines

Affine transformations

Translation

Scaling

3 rotations

Shear

12 degrees of freedom

P ≡A⋅Pu

u=[abc ] ; A=[

1 0 00 1 00 0 1 ]

u=0 ; A=[d 0 00 e 00 0 f ]

u=0 ; A=[cos −sin 0sin cos 0

0 0 1]⋯u=0 ; A=[

1 g h0 1 i0 0 1]

127


Splines

Let P a parametric curve built this way :

Let's verify the invariance by a translation t:

P (u , Pi)=∑0

n−1

P i K in(u)

P (u , Pi*)=∑

0

n−1

(P i+t) K in(u)=∑

0

n−1

P i K in(u)+∑

0

n−1

t K in(u)

=P (u , P i)+∑0

n−1

t K in(u)

=P (u , P i)+t=P*(u , Pi) iff ∑

0

n−1

K in(u)=1

Partition of unity

128


Splines

For the other multiplicative transformations

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.

P (u , Pi*)=∑

0

n−1

(A⋅P i) K in(u)=A⋅∑

0

n−1

P i K in(u)

=A⋅P (u , P i)=P*(u , P i)

(no particular conditions except linearity with respect to the coordinates of the control points)

129


Splines

Do Hermite's basis form a partition of unity ?

∑0

n−1

h??n

u=1

{h00

p=1−3u

22u

3

h10p=3u

2−2u

3

h01p=u−2u

2u

3

h11p=−u

2u

3F u=1

F 0=1F 1=1 F '

0=0F '

1=0

130


Let's check the invariance If we apply a translation to the control points P

i, the

derivatives Pi' should not change ...

P (Pi*)=∑

0

1

(P i+t )hi0n(u)+∑

0

1

Pi' hi1

n(u)

=∑0

1

thi0n(u)+∑

0

1

P i hi0n(u)+∑

0

1

P i' hi1

n(u)

=t+ P (P i)=P*(P i)

P i*=P i+t P i

' *=P i

'

Splines

131


Splines

We must also check the invariance for the other multiplicative transformations : those affect both the coordinates and the derivatives

P P i*=∑

0

1

A⋅P ihi0nu∑

0

1

A⋅Pi'hi1

nu

P i*= A⋅P i P i

' *=A⋅P i

'

=A⋅∑0

1

P i hi0nu∑

0

1

P i' hi1

nu

=A⋅P (P i)=P*(P i)

QED

132


Splines

Beware of the computation of the slopes x'i ...

Natural Splines :

2 11 4 1

1 4 1⋱

1 4 11 2

x0

'

x1'

x2'

⋮

xn−2'

xn−1'

=3x1−x0

3 x2−x0

3x3−x1

⋮

3xn−1−xn−3

3 xn−1−xn−2

P i'=L(P i−P j)⇒ P i

' *=L (P i

*−P j

*)=L((A⋅Pi+t)−( A⋅P j+t))

=A⋅L(P i−P j)=A⋅Pi'

Linear operator

It is OK in this case

133


Splines

Rotation of 45° Scaling x direction

(times 0.5) Followed by a

translation

134


Splines

3D Curves Minimal order so that a curve can have a torsion

(non planar curve) Let's consider a Lagrange interpolation 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

P u=∑i=0

n−1

P i l ipu

interpolation and polynomial approximation interpolation - … · 2019-07-15 · 1 computer aided...

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