hw3 - CS4620 Fall 2013 HW3 Due in class Show your work for...

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CS4620 Fall 2013 HW3 Due in class, Nov 25 2013 Show your work for every question. 1 B-Spline and Bezier Curves As we discussed in lecture, the B-spline curve is a series of segments, each defined by a cubic polynomial in the form f Bsp ( u ) = u 3 u 2 u 1 M Bsp P Bsp where P Bsp is a 4-by-2 matrix whose rows are the four control points that influence the segment, and M Bsp is the 4-by-4 spline matrix for the B-spline: M Bsp = 1 6 2 6 6 4 - 1 3 - 3 1 3 - 6 3 0 - 3 0 3 0 1 4 1 0 3 7 7 5 . On the other hand, a B´ezier spline segment has the form f Bez ( u ) = u 3 u 2 u 1 M Bez P Bez where M Bez = 2 6 6 4 - 1 3 - 3 1 3 - 6 3 0 - 3 3 0 0 1 0 0 0 3 7 7 5 . Suppose we have the four control points p 0 , . . . , p 3 for a segment of a B- spline curve (these are the rows of P Bsp ) and we would like to compute the B´ezier control points p 0 0 , . . . , p 0 3 that describe exactly the same curve as a B´ezier segment. (This amounts to solving for the value of P Bez that makes f Bez equal to f Bsp .) 1. Find the matrix M Bsp-to-Bez so that if P Bez = M Bsp-to-Bez P Bsp then f Bez ( u ) = f Bsp ( u ) for all u . You will find that this matrix has nice numbers. 2. Write expressions for p 0 0 , . . . , p 0 3 as functions of p 0 , . . . , p 3 . 1
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3. Draw a diagram that illustrates both sets of control points and the spline segment for the case p 0 = (0 , 0) p 1 = (0 , 1) p 2 = (1 , 0) p 3 = (1 , 1) Hint: While it is possible to solve this through extensive algebraic ma- nipulation of polynomials, using matrices is much more effective and less error-prone. The second part of this problem concerns designing a B-spline segment that interpolates one of its control points. Consider a B-spline segment f ( u ) that has four control points p 0 , . . . , p 3 . Suppose p 1 , p 2 , and p 3 are given, and we want to position p 0 so that the curve starts at p 1 . In other words, we want f (0) = p 1 .
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