Review of Interpolation

1. Review of Interpolation A method of constructing a function that crosses through a discrete set of known data points. .

2. Spline (general case) • Given n+1 distinct knotsxi such that: with n+1 knot values yi find a spline function with each Si(x) a polynomial of degree at most n.

3. Spline Interpolation • Linear, Quadratic, Cubic • Preferred over other polynomial interpolation • More efficient • High-degree polynomials are very computationally expensive • Smaller error • Interpolant is smoother

4. Natural Cubic Spline Interpolation • Si(x) = aix3 + bix2 + cix + di • 4 Coefficients with n subintervals = 4n equations • There are 4n-2 conditions • Interpolation conditions • Continuity conditions • Natural Conditions • S’’(x0) = 0 • S’’(xn) = 0

5. Natural Conditions

6. Bezier Curves • A similar but different problem: Controlling the shape of curves. Problem: given some (control) points, produce and modify the shape of a curve passing through the first and last point. http://www.ibiblio.org/e-notes/Splines/Bezier.htm

7. Bezier Curves Idea: Build functions that are combinations of some basic and simpler functions. Basic functions: B-splines Bernstein polynomials

8. Bernstein Polynomials • Bernstein polynomials of degree n are defined by: • For v = 0, 1, 2, …, n, • In general there are N+1 Bernstein Polynomials of degree N. For example, the Bernstein Polynomials of degrees 1, 2, and 3 are: • 1. B0,1(t) = 1-t, B1,1(t) = t; • 2. B0,2(t) = (1-t)2, B1,2(t) = 2t(1-t), B2,2(t) = t2; • 3. B0,3(t) = (1-t)3, B1,3(t) = 3t(1-t)2, B2,3(t)=3t2(1-t), B3,3(t) = t3;

9. Deriving the Basis Polynomials • Write 1 = t + (1-t) and raise 1 to the third power for cubic splines (2 for quadratic, 4 for quartic, etc…) • 1^3 = (t + (1-t))^3 = t^3 + 3t^2(1-t) + 3t(1-t)^2 + (1-t)^3 • The basis functions are the four terms: • B0(t) = (1-t)^3 • B1(t) = 3t(1-t)^2 • B2(t) = 3t^2(1-t) • B3(t) = t^3

10. Bernstein Polynomials Bernstein polynomials of degree 3

11. Using the Basis Functions • To use the Bernstein basis functions to make a spline we need 4 “control” points for a cubic. (3 for quadratic, 5 for quartic, etc…) • Let (x0,y0), (x1,y1), (x2,y2), (x3,y3) be the “control”points • The x and y coordinates of the cubic Bezier spline are given by: • x(t) = x0B0(t) + x1B1(t) +x2B2(t) + x3B3(t) • y(t) = y0B0(t) + y1B1(t) + y2B2(t) + y3B3(t) • Where 0<= t <= 1.

12. Bernstein Polynomials • Given a set of control points {Pi}Ni=0, where Pi = (xi, yi). A Bezier curve of degree N is: P(t) = Ni=0 PiBi,N(t), • Where Bi,N(t), for I = 0, 1, …, N, are the Bernstein polynomials of degree N. • P(t) is the Bezier curve Since Pi = (xi, yi) • x(t) = Ni=0xiBi,N(t) and y(t) = Ni=0yiBi,N(t) • The spline goes through (x0,y0) and (x3,y3) • The tangent at those points is the line connecting (x0,y0) to (x1,y1) and (x3,y3) to (x2,y2). • The positions of (x1,y1) and (x2,y2) determine the shape of the curve.

13. Bernstein Polynomials Example • Find the Bezier curve which has the control points (2,2), (1,1.5), (3.5,0), (4,1). Substituting the x- and y-coordinates of the control points and N=3 into the x(t) and y(t) formulas on the previous slide yields x(t) = 2B0,3(t) + 1B1,3(t) + 3.5B2,3(t) + 4B3,3(t) y(t) = 2B0,3(t) + 1.5B1,3(t) + 0B2,3(t) + 1B3,3(t)

14. Bernstein Polynomials Example • Substitute the Bernstein polynomials of degree three into these formulas to get x(t) = 2(1 – t)3 + 3t(1 – t)2 + 10.5t2(1 – t) + 4t3 y(t) = 2(1-t)3 + 4.5t(1 – t)2 + t3 • Simplify these formulas to yield the parametric equation: P(t) = (2 – 3t + 10.5t2 – 5.5t3, 2 – 1.5t – 3t2 + 3.5t3) where 0 < t < 1 P(t) is the Bezier curve http://www.cs.brown.edu/exploratories/freeSoftware/repository/edu/brown/cs/exploratories/applets/bezierSplines/bezier_splines_java_browser.html

15. Composite Bezier Curves • In practice, shapes are often constructed of multiple Bezier curves combined together. • Allows for flexibility in constructing shapes. (Consecutive Bezier curves need not join smoothly.)

16. Composite Bezier Curves • Linking curves: • Example: Find the composite Bezier curve for the four sets of control points: {(-9,0), (-8,1), (-8,2.5), (-4,2.5)}, {(-4,2.5), (-3,3.5), (-1,4), (0,4)} {(0,4), (2,4), (3,4), (5,2)}, {(5,2), (6,2), (20,3), (18,0)}

17. Composite Bezier Curves Use the same process as earlier to yield these four parametric equations: • P1(t) = (-9 + 3t - 3t2 + 5t3, 3t + 1.5t2 - 2t3) • P2(t) = (-4 + 3t + 3t2 - 2t3, 2.5 + 3t - 1.5t2) • P3(t) = (6t - 3t2 + 2t3, 4 - 2t3) • P4(t) = (5 + 3t + 39t2 - 29t3, 2 + 3t2 - 5t3)

18. Composite Bezier Curves Linking Curves Smoothly • To have two Bezier curves P(t) and Q(t) meet smoothly would require: • PN = Q0 • P’(PN) = Q’(Q0) • It is sufficient to require that the control points PN-1, PN = Q0, and Q1 be collinear.

19. Composite Bezier Curves • Example of collinear control points: {(0,3), (1,5), (2,1), (3,3)} and {(3,3), (4,5), (5,1), (6,3)} P(t) = (3t, 3 + 6t - 18t2 + 12t3) Q(t) = (3 + 3t, 3 + 6t – 18t2 + 12t3) and P’(t) = (3, 6 – 36t + 36t2) Q’(t) = (3, 6 – 36t + 36t2) Substituting t = 1 and t = 0 into P’(t) and Q’(t), yields P’(1) = (3,6) = Q’(0)

20. Composite Bezier Curves It’s collinear!

21. Practical Applications

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23. Practical Applications

24. Practical Applications • 4.35 0.00 4.35 0.00 4.35 0.19 4.35 0.19 • 4.35 0.19 3.53 0.23 3.39 0.36 3.39 1.09 • 3.39 1.09 3.39 1.08 3.39 6.20 3.39 6.20 • 3.39 6.20 3.39 6.20 3.93 6.20 3.93 6.20 • 3.93 6.20 5.07 6.20 5.29 6.02 5.52 4.92 • 5.52 4.92 5.52 4.92 5.76 4.92 5.76 4.92 • 5.76 4.92 5.76 4.92 5.70 6.62 5.70 6.62 • 5.70 6.62 5.70 6.62 0.30 6.62 0.30 6.62 • 0.24 4.92 0.24 4.92 0.30 6.62 0.30 6.62 • 0.48 4.92 0.48 4.92 0.24 4.92 0.24 4.92 • 2.07 6.20 0.93 6.20 0.71 6.02 0.48 4.92 • 2.61 6.20 2.61 6.20 2.07 6.20 2.07 6.20 • 2.61 1.09 2.61 1.08 2.61 6.20 2.61 6.20 • 1.65 0.19 2.47 0.23 2.61 0.36 2.61 1.09 • 1.65 0.00 1.65 0.00 1.65 0.19 1.65 0.19 • 1.65 0.00 1.65 0.00 4.35 0.00 4.35 0.00

25. Bézier surfaces in computer graphics