Interpolation - NUS Computingcs5240/lecture/interpolation.pdf · Weakness: More complex than...

Interpolation

CS5240 Theoretical Foundations in Multimedia

Leow Wee Kheng

Department of Computer Science

School of Computing

National University of Singapore

Leow Wee Kheng (NUS) Interpolation 1 / 44

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Fit the regional economic data.

0 10000 20000 30000 40000 50000 60000

GDP per capita

0

10000

20000

30000

40000

50000

60000

GN

I per

capit

a

GNI per capita

line t

quad t

cubic t


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Fit 2-D quadratic surface to 3-D data points.

3D points fitted surfaces


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What if you want to model the curve/surface of this aircraft exactly?

Nonlinear least squares approximate data points.Need interpolating curve/surface that fits exactly!

p4

p3

p0

p2p1

x

y

p4

p3

p2

p0

p1

x

y

approximating curve interpolating curve


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Approximation Problem

Given data points pi, i = 1, . . . , n, and their corresponding

data points qi, find a function f that minimizes the error

E =n∑

i=1

‖f(pi)− qi‖2. (1)

Interpolation Problem

Given data points pi, i = 1, . . . , n, and their corresponding

data points qi, find a function f such that f(pi) = qi for all i.

That means, error E = 0 over the data points pi.


Global Interpolation Linear Interpolation

Linear Interpolation

Given control points (x1, y1) and (x2, y2), find a line that passes them.

x2x1

y2

y1

y

xx

y

y − y1x− x1

=y2 − y1x2 − x1

. (2)

So,

y = y1 +y2 − y1x2 − x1

(x− x1). (3)



Alternatively,

x2x1

y2

y1

d1 d2

y

xx

y

y − y1x− x1

=y2 − y

x2 − x. (4)

That is,y − y1d1

=y2 − y

d2. (5)

So,

y =d1y2 + d2y1d1 + d2

. (6)



Linear interpolation can be extended to 2D data:Interpolate the height h(x, y) given heights at 4 neighbouring points.

y1

y2

x2x1

x y( , )

x

y

?

Interpolate along two directions: x and y.This is called bilinear interpolation (Assignment).

Similarly, 3D version is trilinear interpolation:interpolate along three directions x, y, z.


Global Interpolation Polynomial Interpolation

Polynomial Interpolation

Consider n+ 1 points pi = (xi, yi), i = 0, 1, . . . , n that lie on a curve.

p4

p3

p2

p0

p1

x

y

How to produce an interpolating curve that passes through the points?



Consider a polynomial of degree (order) d:

yi = a0 + a1xi + a2x2i + · · ·+ adx

di , for i = 0, 1, . . . , n. (7)

In matrix form,

1 x0 x20 · · · xd01 x1 x21 · · · xd1...

......

. . ....

1 xn x2n · · · xdn

a0a1...ad

=

y0y1...yn

(8)

◮ If d > n, no unique solution.

◮ If d = n, matrix has inverse: interpolating solution.

◮ If d < n, matrix has pseudo-inverse: approximating solution.



Interpolation vs. Approximation

0 0.5 1 1.5 2 2.5 3 3.5

x0.3

0.4

0.5

0.6

0.7

0.8

0.9

1y

control points (n = 6)degree n interpolationdegree n-1 approximationdegree n-3 approximation



Polynomial Interpolation

◮ Strength: Easy to implement.

◮ Weakness: Data matrix is often ill-conditioned when n is large.(Refer to optimization-2.pdf for condition number of matrix.)


Global Interpolation Lagrange Interpolation

Lagrange Interpolation

Suppose we have n+ 1 control points pi = (xi, yi), i = 0, 1, . . . , n.

An intuitive way to interpolate is to define a function

f(x) = y0L0(x) + y1L1(x) + · · ·+ ynLn(x) =

n∑

i=0

yiLi(x), (9)

such that

Li(xj) = δij =

{

1 if j = i,0 if j 6= i.

(10)

Then, obviouslyf(xi) = yiLi(xi) = yi. (11)



An easy way to define Li(x) is

Li(x) =(x− x0)(x− x1) · · · (x− xi−1)(x− xi+1) · · · (x− xn)

(xi − x0)(xi − x1) · · · (xi − xi−1)(xi − xi+1) · · · (xi − xn),

=n∏

j=0, j 6=i

x− xjxi − xj

. (12)

This Li(x) is called Lagrange interpolation polynomial.

Eq. 9 is already the interpolation formula for any x.No need to solve any other equation!



Example Lagrange interpolation polynomials

1 1.2 1.4 1.6 1.8 2x−0.5

0

0.5

1

Li

L0

L1

1 1.2 1.4 1.6 1.8 2x−0.5

0

0.5

1

Li

L0

L1

L2

degree-1: linear degree-2: quadratic




0 0.5 1 1.5 2 2.5 3 3.5

x0.3

0.4

0.5

0.6

0.7

0.8

0.9

1y

control points (n = 6)degree-n polynomial interpolation




◮ Strength: Easy to implement, no need optimization.

◮ Weakness: Does not take into account gradient at control points.


Global Interpolation Hermite Interpolation

Hermite Interpolation

Hermite interpolation is extension of Lagrange interpolation.

Given n+ 1 control points pi = (xi, yi), i = 0, 1, . . . , n, andthe gradients y′i = dy/dx at x = xi.

Define the interpolation function as

f(x) =n∑

i=0

(

yiHi(x) + y′iHi(x))

(13)

with appropriate Hi(x) and Hi(x):

Hi(xj) = δij , H ′i(xj) = 0,

Hi(xj) = 0, H ′i(xj) = δij.

(14)

Then,

f(xi) = yiHi(xi) = yi,df

dx

∣

∣

∣

∣

x=xi

= y′iHi(xi) = y′i. (15)



These Hi and Hi are called Hermite interpolation polynomials, andthey have these forms:

Hi(x) = (1− 2(x− xi)L′i(xi))L

2i (x)

Hi(x) = (x− xi)L2i (x).

(16)

Hermite Interpolation

◮ Strength: Takes into account gradient. Still no need optimization.

◮ Weakness: More complex than Lagrange interpolation.



Global Interpolation◮ Strength: Simple, single interpolating curve.◮ Weakness: Can have undesirable oscillations.

0 5 10 15 20 25 30

x−4

−2

0

2

4

6y

◮ Lagrange-1: control points at x = 0, 1, 8, 27.◮ Lagrange-2: control points at x = 0, 8, 16, 27.◮ Lagrange-3: control points at x = 0, 1, 8, 16, 27.


Piecewise Interpolation

Piecewise Interpolation

Interpolate using a combination of low-degree polynomials.

Piecewise Polynomial Interpolation

0 0.5 1 1.5 2 2.5 3 3.5

x0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

y

control points (n = 6)

p��

p��


Piecewise Interpolation Piecewise Lagrange Interpolation

Piecewise Lagrange Interpolation

Write Lagrange interpolation polynomials as functions of parameteru = [0, 1].

Example: Interpolate between 2 control points pi and pi+1:

x = x(u), x(0) = xi, x(1) = xi+1

y = y(u), y(0) = yi, y(1) = yi+1

(17)

xi i +1x

pi +1

pi

= 1u

yi

i +1y

= 0u

y

x



Lagrange Linear Polynomials

For interpolation with 2 control points, u = 0 at pi, u = 1 at pi+1.

0 0.2 0.4 0.6 0.8 1u0

0.2

0.4

0.6

0.8

1

Li

L0

L1

L0(u) =u− u1u0 − u1

=u− 1

0− 1= 1− u,

L1(u) =u− u0u1 − u0

=u− 0

1− 0= u.

(18)



p4

p3

p2

p0

p1

pi

pi +1

xi i +1x

= 0u

= 1u

x

y

Then, the linear interpolation between pi and pi+1 is

x = xiL0(u) + xi+1L1(u) = (1− u)xi + uxi+1,

y = yiL0(u) + yi+1L1(u) = (1− u)yi + uyi+1,(19)

which is standard linear interpolation.

Eq. 19 is applied to each segment between pi and pi+1 fori = 0, 1, . . . , n− 1.



Lagrange Quadratic Polynomials

For interpolation with 3 control points that are equally spaced in termsof xi, u = 0, 0.5, 1 at pi−1, pi, pi+1. Then, (Exercise)

L0(u) =(u− 0.5)(u− 1)

(0− 0.5)(0− 1)= 2u2 − 3u+ 1,

L1(u) =(u− 0)(u− 1)

(0.5− 0)(0.5− 1)= −4u2 + 4u,

L2(u) =(u− 0)(u− 0.5)

(1− 0)(1− 0.5)= 2u2 − u.

(20)



0 0.2 0.4 0.6 0.8 1u−0.2

0

0.2

0.4

0.6

0.8

1

Li

L0

L1

L2

Lagrange quadratic interpolation between pi, pi+1, and pi+2 is

x = xiL0(u) + xi+1L1(u) + xi+2L2(u),

y = yiL0(u) + yi+1L1(u) + yi+2L2(u),(21)

Eq. 21 is applied to each segment between pi, pi+1, pi+2, fori = 0, 2, 4, ..., n.



Global vs. Piecewise Interpolation

0 5 10 15 20 25 30x−4

−3

−2

−1

0

1

2

3

4y

◮ Lagrange-1: Global, control points at x = 0, 1, 8, 27.

◮ Lagrange-2: Global, control points at x = 0, 8, 16, 27.

◮ piecewise Lagrange: quadratic, 2 segments,control points at x = 0, 4, 8 and x = 8, 17.5, 27.


Piecewise Interpolation Piecewise Hermite Interpolation

Piecewise Hermite Interpolation

Hermite Cubic Polynomials

For interpolation with 2 control points, u = 0 at pi, u = 1 at pi+1.

From Eq. 18,L0(u) = 1− u, L1(u) = u.

So,L′0(u) = −1, L′

1(u) = 1.

Therefore, (Exercise)

H0(u) = (1− 2L′0(0)(u− 0))L2

o(u) = 2u3 − 3u2 + 1

H1(u) = (1− 2L′1(1)(u− 1))L2

1(u) = −2u3 + 3u2

H0(u) = (u− 0)L20(u) = u3 − 2u2 + u

H1(u) = (u− 1)L21(u) = u3 − u2.

(22)


Piecewise Interpolation Piecewise Hermite Interpolation

0 0.2 0.4 0.6 0.8 1u−0.2

0

0.2

0.4

0.6

0.8

1

Hermite cubic interpolation between pi and pi+1 is

y = yiH0(u) + yi+1H1(u) +dy

dx

dx

du

∣

∣

∣

∣

u=0

H0(u) +dy

dx

dx

du

∣

∣

∣

∣

u=1

H1(u)

= yiH0(u) + yi+1H1(u) + (xi+1 − xi)y′iH0(u)

+(xi+2 − xi+1)y′i+1H1(u). (23)


Piecewise Interpolation Splines

Splines

draftsman’s spline curve ruler


Piecewise Interpolation Cubic Spline

Cubic Spline

The point p(u) on a cubic spline between control points pi and pi+1 is:

p(u) = a0 + a1u+ a2u2 + a3u

3, for 0 ≤ u ≤ 1 (24)

where a0, . . . ,a3 are to be determined such that p(0) = pi,p(1) = pi+1.

Derivation of cubic spline for x and y are the same.

Suppose thatx = φ(u) = a0 + a1u+ a2u

2 + a3u3 (25)

and its 2nd derivative at pi is given by Di.

Since φ(u) is cubic, φ′′(u) is linear:

φ′′(u) = (1− u)Di + uDi+1 = Di + (Di+1 −Di)u. (26)



Integrating φ′′(u) twice gives

φ(u) = A+Bu+1

2Diu

2 +1

6(Di+1 −Di)u

3 (27)

for some constants A and B.

For u = 0, xi = φ(0) = A.For u = 1, xi+1 = φ(1) = A+B + 1

3Di +

16Di+1.

That is, B = xi+1 − xi −13Di −

16Di+1.

So, Eq. 27 becomes

φ(u) = xi +

(

xi+1 − xi −1

3Di −

1

6Di+1

)

u+1

2Diu

2 +1

6(Di+1 −Di)u

3.

(28)This is the interpolation function for x between pi and pi+1: x = φ(u).



Now, consider the two splines that meet at pi.

pi

pi −1

pi +1

= 0u

= 1u= 1u = 0u

The blue spline and its gradient are:

φ(u) = xi−1+

(

xi − xi−1 −1

3Di−1 −

1

6Di

)

u+1

2Di−1u

2+1

6(Di−Di−1)u

3

φ′(u) = xi − xi−1 −1

3Di−1 −

1

6Di +Di−1u+

1

2(Di −Di−1)u

2. (29)

The red spline is given by Eq. 28 and its gradient is:

φ′(u) = xi+1 − xi −1

3Di −

1

6Di+1 +Diu+

1

2(Di+1 −Di)u

2. (30)



At control point pi, u = 1 for Eq. 29 and u = 0 for Eq. 30:

φ′(1) = xi − xi−1 +1

6Di−1 +

1

3Di, (31)

φ′(0) = xi+1 − xi −1

3Di −

1

6Di+1. (32)

For continuity of gradient, equate Eq. 31 and 32, giving:

Di−1 + 4Di +Di+1 = 6(xi−1 − 2xi + xi+1), for i = 1, . . . , n− 1. (33)

So, have n− 1 equations, but have n+ 1 unknowns D0, D1, . . . , Dn.Need two more conditions, for example:

(a) natural spline: D0 = 0 or Dn = 0;

(b) specified gradient: φ′(u) = g0 at p0 or φ′(u) = gn at pn;

(c) quadratic end spans: D0 = D1 or Dn−1 = Dn.



Overall equation, in matrix form, is

a00 a01 0 0 · · · 0 0 01 4 1 0 · · · 0 0 00 1 4 1 · · · 0 0 0...

......

.... . .

0 0 0 0 · · · 1 4 10 0 0 0 · · · 0 an,n−1 ann

D0

D1

D2

...Dn−1

Dn

= 6

b0x0 − 2x1 + x2x1 − 2x2 + x3

...xn−2 − 2xn−1 + xn

bn

(34)



where (Exercise)

(a) a00 = 1 and a01 = b0 = 0 or ann = 1 and an,n−1 = bn = 0;

(b) a00 = 2, a01 = 1, and b0 = x1 − x0 − g0 oran,n−1 = 1, ann = 2, and bn = gn − xn + xn−1;

(c) a00 = 1, a01 = −1, and b0 = 0 oran,n−1 = −1, ann = 1, and bn = 0.

After solving D0, . . . , Dn, interpolation of x is computed using Eq. 28.Interpolation of y is computed in the same way as x using y0, . . . , yn.



Example:

0 5 10 15 20 25 30x0

1

2

3

y

piecewise Lagrangecubic spline 1cubic spline 3

◮ cubic spline 1: case 1, natural splines

◮ cubic spline 3: case 3, quadratic end spans



Cubic Spline Interpolation:

◮ Strength: 1st derivative is continuous.

◮ Weakness: oscillation can occur at points where 2nd derivative isnot continuous.

◮ Local modification of curve requires re-computing whole curve.

Other Splines

cubic B-spline, Bezier curve, NURBS (non-uniform rational B-splines).



Comparisons

◮ Piecewise Lagrange interpolation:1st and 2nd derivatives are discontinuous at control points.

◮ Piecewise Hermite interpolation:1st derivative is continuous but must be known at control points.2nd derivative is discontinuous at control points.

◮ Cubic spline interpolation:1st and 2nd derivatives are continuous at control points.


Surface Interpolation

Surface Interpolation

v

u

Surface can be represented using two spline curves on a grid.A point on the surface is parameterized by two parameters

x(u, v), y(u, v), z(u, v).

Analogous to bilinear interpolation.

But, this method cannot be applied to general non-rectangular meshes.Leow Wee Kheng (NUS) Interpolation 40 / 44

Summary

Summary

method property

polynomial unstable for large nLagrange stable for large nHermite stable for large nglobal have oscillations

piecewise reduce oscillationspiecewise linear discontinuous

piecewise polynomial discontinuouspiecewise Lagrange discontinuous 1st derivativespiecewise Hermite discontinuous 2nd derivatives

cubic spline continuous 1st & 2nd derivatives


Probing Questions

Probing Questions

◮ We have studied interpolation of planar curves, i.e., curves in 2D.Which of the methods studied can be easily applied to spacecurves, i.e., curves in 3D?

◮ Is it possible to adapt curve interpolation methods to curveapproximation? If yes, how? If no, why?

◮ We have studied the interpolation of values. Is it possible tointerpolate functions such as transformation functions? If yes,how? If no, why?


Homework

Homework

1. Describe the essence of global interpolation and piecewise (local)interpolation, each in one sentence.

2. Describe the essence of polynomial interpolation, Lagrangeinterpolation, Hermite interpolation and cubic B-splineinterpolation, each in one sentence.

3. Derive the Lagrange quadratic polynomials given in Eq. 20.

4. Derive the Hermite cubic polynomials given in Eq. 22.

5. For cubic spline interpolation, derive the values of the matrixelements a00, a01, an,n−1, ann, b0, and bn for the three cases asillustrated in page 36.


References

References

1. A. Davis and P. Samuels, An Introduction to Computational Geometry

for Curves and Surfaces, Clarendon Press, 1996.

2. B. I. Kvasov Methods of Shape-Preserving Spline Approximation, WorldScientific, 2000.


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