Page 55 - DCAP313_LAB_ON_COMPUTER_GRAPHICS

Unit 3: Implementing Line Algorithm

• Subscript k takes integer values starting from 1, for the first point and increment by 1 Notes
until the final end point is reached.
• m->any real numbers between 0 and 1
• Calculate y values must be rounded to the nearest integer

Step 2
If the slope is greater than 1, the roles of x any y at the unit y intervals Dy = 1 and compute
each successive y values.
Dy = 1
m = Dy / Dx
m = 1/ (x 2 – x 1 )
m = 1 / (x k + 1 – x k )

x k + 1 = x k + (1/m) (7)
• Equation 6 and Equation 7 that the lines are to be processed from left end point to the
right end point.

Step 3
If the processing is reversed, the starting point at the right
Dx = – 1

m = Dy / Dx
m = ( y 2 – y 1 ) / – 1
y k + 1 = y k – m (8)
Intervals Dy = 1 and compute each successive y values.
Step 4

Here, Dy = – 1
m = Dy / Dx
m = – 1 / (x 2 – x 1 )
m = –1 / (x k + 1 – x k )

x k + 1 = x k + ( 1 / m ) (9)
Equation 6 and Equation 9 used to calculate pixel position along a line with –ve slope.
Advantage
Faster method for calculating pixel position then the equation of a pixel position.

Y = mx + b
Disadvantage
The accumulation of round of error is successive addition of the floating point increments is
used to find the pixel position but it take lot of time to compute the pixel position.
Algorithm: A naïve line-drawing algorithm
d x = x 2 – x 1
d y = y 2 – y 1
for x from x 1 to x 2
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