Page 48 - DCAP313_LAB_ON_COMPUTER_GRAPHICS

Lab on Computer Graphics

Notes and
m = −1024 (∆x (1 − 0.125) − ∆y (1 − 0.75))
= −1024 (3.55(1 − 0.125) − 2.55(1 − 0.75)
= −1024 (2.6625 − .6375)

= −2073.6
And so we initialize the integer quantity Œ = −2073 and the algorithm proceeds as follows:
Thus the integer version for Bresenham’s algorithm with arbitrary endpoints is constructed as
follows:
Integer Bresenham’s Algorithm

The points (x 1 , y 1 ) and (x 2 , y 2 ) are assumed not equal and have arbitrary real coordinates Πis
(x, y) Πdescription
(0, 0) –2073 illuminate pixel (0, 0)
538 increment ε by 2611
(1, 0) increment x by 1
(1, 0) 538 illuminate pixel (1, 0)
since Π> 0
(1, 1) increment y by 1
–3097 decrement Œ by 3635
–486 increment ε by 2611
(2, 1) increment x by 1
(2, 1) –486 illuminate pixel (2, 1)
2125 increment ε by 2611
(3, 1) increment x by 1
(3, 1) 2125 illuminate pixel (3, 1)
since Π> 0
(3, 2) increment y by 1
–1510 decrement Œ by 3635
1101 increment ε by 2611
(4, 2) increment x by 1
(4, 2) 1101 illuminate pixel (4, 2)

assumed to be integer valued.
Let ∆x = m (x 2 – x 1 )
Let ∆y = m (y 2 – y 1 )
Let i 1 = x Í Î 1 ˙ ˚
x Í Î
Let i 2 = 2 ˙ ˚
Let j = y Í Î 1 ˙ ˚
Let Œ = Dy (1 - x ( 1 - i) - Dx(1 - y - j)˙ ˚
Í
Î
1
i
for i = 1 to i 2
illuminate (x, y)
if ( Œ ≥ 0)
y + = 1

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