Raeda Naamnieh 1. Outline Subdivision of Bezier Curves Restricted proof for Bezier Subdivision Convergence of Refinement Strategies 2.

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Subdivision of Bezier curves

Raeda Naamnieh

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Outline

Subdivision of Bezier Curves

Restricted proof for Bezier Subdivision

Convergence of Refinement Strategies

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MOTIVATION

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Definitions

• Definition 5.7 For , the functions

for where n is any nonnegative integer, are called the generalized Bernstein blending functions.

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Definitions

• Definition 16.11We call

the Bezier curve with control points on the interval .

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• For defined as above then

where

THE BEZIER CURVE SUBDIVISION THEOREM

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Outline




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Restricted Proof for Bezier Subdivision

• Lemma 16.22

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• Proof:

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• Proof:

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• Proof for Bezier Subdivision: induction on n, and for arbitrary c, a<c<b.If n=1

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• Proof for Bezier Subdivision:Now, assume the theorem holds for all

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• Proof for Bezier Subdivision:Now using the results from

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• Proof for Bezier Subdivision:-The second part of the proof is almost identical, hence left as exercise

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Outline




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SUBDIVISION AT THE MIDPOINT

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o Bezier polygon defined on .o the piecewise linear function given by the

original polygon.o the piecewise linear function formed with

vertices defined by concatenating together the control polygons for the two subdivided curves

and at the midpoint.o It has 2n+1 distinct points.

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o is a pisewise linear function defined by the ordered vertices of the control polygons of the Bezier curves whose composite is the original curve.

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o The subdivided Bezier curve at level is over the interval:

and has vertices: for oWe shall write

has distinct points which define it.

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Theorem 16.17:

That is, the polyline consisting of the union of all the sub polygons converges to the Bezier curve.

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Lemma 16.18:If is a Bezier curve, define . If Are defined by the rule in Theorem 16.12, then for

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Proof:By induction on the superscript, for ,

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Proof:Now, suppose that the conclusion has been shown for superscripts up to .Then,

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Lemma 16.19:Any two consecutive vertices of are no farther apart than ,where is independent of .That is, if and are two consecutive vertices of Then .

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Proof:Induction on ,Let First consider and

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Proof:Let

where

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Proof:Now, suppose

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Proof:Assume for . Now we show it is true for .The vertices in are defined by subdividing the Bezier polygons in .We see that are formed by subdividing the Bezier curve with control polygon where respectively.

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Proof:We shall prove the results for

Let us fix And call By the subdivision Theorem 16.12

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Proof:

Since this is proved for all the conclusion of the lemma holds for all

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Proof for convergence theorem:The subdivision theorem showed that over each subinterval , the Bezier curve resulting from the appropriate sub collection of is identical to the original We denote this by .

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Proof for convergence theorem:Any arbitrary value in the original interval is then contained in an infinite sequence of intervals, for which

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Proof for convergence theorem:Hence, the curve value, lies within the convex hull of the vertices of which correspond to the Bezier polygon over , for each .

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Proof for convergence theorem:Since the spacial extent of the convex hull of each Bezier polygon over , all and , gets smaller and converges to zero.

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Proof for convergence theorem:Consider the subsequence of polygons corresponding to the intervals containing . is contained in all of them, for all Further, if any other curve point were contained in all of them, say , then would be in

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Proof for convergence theorem:Since is the only point in that intersection, is the only point in the intersection of the convex hull of the Bezier polygons of these selected subintervals. The polygonal approximation converges.

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Summary

• Subdivision of Bezier Curves

• Restricted proof for Bezier Subdivision

• Convergence of Refinement Strategies

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Appendix

Geometric Modeling with Splines

Chapter 16Elaine CohenRichard F. RiesenfeldGershon Elber

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Q&A

Raeda Naamnieh 1. Outline Subdivision of Bezier Curves Restricted proof for Bezier Subdivision Convergence of Refinement Strategies 2.

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