Student: joycelin
Assignment: 2
Grader: Charlie

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Bresenham:

program runs, buggy (55):                   55
correct implementation of Bresenham (20):   15
no floats and divides (20):                 20
extra credit (5):                            0
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total (100):                                90

Cohen-Sutherland:

program runs, buggy (55):                   50
inside case correct (10):                   10
outside case correct (10):                  10
one clip correct (10):                      10
two clips correct (10):                     10
extra credit (5):                            0
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total (100):                                90


Comments:
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Bresenham:
Oops... you did the case for (dy<dx) and (dy>dx)
and left out the case where (dy==dx)
So, you are not able to draw lines wiht slope
equal to 1.

You really did not need to separate the algorithm
into separate cases. 1 loop is enough.

Cohen-Sutherland:
You're implementation of Cohen-Sutherland
is actually a little off. To calculate the outcodes,
you should actually bitwise or together the constants.
This would lead to a more efficient algorithm.

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