CS335
Fall, 2008
Homework 6 Solution Set (20 points)
Due: November 6, 2008 (Thursday)

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1. (Polygon Filling)
   Let V0, V1, ... V9 be the vertices of a polygon (in counter-clock-
   wise order) defined as follows: 

        V0 = (1, 2)
        V1 = (6, 11)
        V2 = (11, 2)
        V3 = (11, 12)
        V4 = (9, 21)
        V5 = (9, 9)
        V6 = (6, 21)
        V7 = (3, 9)
        V8 = (3, 21)
        V9 = (1, 12)

   Build a Bucket-Sorted Edge Table for this polygon. Assuming the
   resolution of the screen is  20x22.   (10 points)

   Solution:
   First note that two edges: e9=V8V9 and e4=V3V4, have to be
   shortened by one unit in y-direction (see the following figure).

   The corresponding Bucket-Sorted Edge Table is shown below:

2. (Bezier Curves)
   Bezier curves satisfy "convex hull" property. That is, the curve
   is always contained in the convex hull of its control points.
   Why?  Justify your answer. (5 points)

   Solution:
   Because each point of the curve is a linear combination of its
   control points.

   The four weights of a cubic Bezier curve are:

      (1-t)**3 ,  3t(1-t)**2 ,  3t**2(1-t) ,   t**3
        
   
   It is easy to see that these weights are non-negative for 
    0 <= t <= 1 

   The sum of these weights is equal to 1.  This follows from the
   following observation

      (1-t)**3 + 3t(1-t)**2 + 3t**2(1-t) + t**3
   
     = [ (1-t) + t ]***3

     = 1***3

     = 1

   Hence, each point of a cubic Bezier curve is indeed a linear (convex)
   combination of its control points.



3. There are several methods to compute points of a cubic Bezier curve.
   These methods include: Horner's rule, Forward differencing, and
   recurrence formula. Horner's rule and recurrence formula have been
   discussed in class. Which method is more efficient to compute a
   point of a cubic Bezier curve and why?  (5 points)

   Solution:
   Horner's rule requires 3 multiplications and 3 additions for the
   evaluation of a point.

   Recurrence formula needs 6 multiplications and 12
   additions/subtractiona for the eavluation of a point.

   Hence Horner's rule is more efficient for the evaluation of a
   single point.