CG08 - Bresenham’s Line Algorithm Data structure.pdf

Computer Graphics
CS-443
Lecture 8
“Bresenham’s Line Algorithm”
By:
Faiz.Ur.Rehman
Lecturer in Computer Science
faiz.rehmanpk@yahoo.com

Index
o The Bresenham algorithm
o Review
o The Big Idea
o Deriving The Bresenham Line Algorithm
o The Bresenham Line Algorithm
o Examples
o Tasks
o Summary
Faiz.Ur.Rehman, Lecturer CS-GPGC Kohat
2

Objective of the today’s topic
• To Understand the basic concept of line drawing
algorithm.
• To Study that how line is draw on raster display.
• To understand a mathematical background of
Bresenham’s line algorithm.
3

The Bresenham Line Algorithm
• The Bresenham algorithm is
another incremental scan
conversion algorithm
• The big advantage of this algorithm
is that it uses only integer
calculations
4
Jack Bresenham worked for 27
years at IBM before entering
academia. Bresenham developed
his famous algorithms at IBM in
t h e e a r l y 1 9 6 0 s

Review
• Point Slope Line Equation:
• y = m.x + b
• Slope:
• m = y2 − y1 / x2 − x1 = Δy / Δx
• y-intercept
• b = y1 − m.x1
• x-interval
• Δx = Δy / m
• y-interval
• Δy = m Δx
5

The Big Idea
• Move across the x axis in unit intervals and at each step
choose between two different y coordinates
6
2 3 4 5
2
4
3
5
(xk, yk)
(xk+1, yk)
(xk+1, yk+1)
(Previous)
NE (next)
E (next)
• For example, from
position (2, 3) we have to
choose between (3, 3)
and (3, 4)
• We would like the pixel
that is closer to the
original line

Deriving the Bresenham line algorithm
• At sample position xk+1 the vertical separations from
the mathematical line are labelled dupper and dlower
• The y coordinate on the mathematical line at xk+1 is:
7
b
x
m
y k +
+
= )
1
(
y
yk
yk+1
xk+1
dlower
dupper

Deriving the Bresenham line algorithm (cont…)
• So, dupper and dlower are given as follows:
• and:
• We can use these equations to make a simple decision
about which pixel is closer to the mathematical line.
8
k
lower y
y
d −
=
k
k
Lower y
b
x
m
d −
+
+
= )
1
(
y
y
d k
upper −
+
= )
1
(
b
x
m
y k
k −
+
−
+
= )
1
(
1

Deriving the Bresenham line algorithm (cont…)
• This simple decision is based on the difference between
the two pixel positions:
• Let’s substitute m with ∆y/∆x where ∆x and ∆y are the
differences between the end-points:
9
1
2
2
)
1
(
2 −
+
−
+
=
− b
y
x
m
d
d k
k
upper
lower
)
)
1
(
1
(
)
)
1
(
( b
x
m
y
y
b
x
m
d
d k
k
k
k
Uper
Lower −
+
−
+
−
−
+
+
=
−
)
1
2
2
)
1
(
2
(
)
( −
+
−
+



=
−
 b
y
x
x
y
x
d
d
x k
k
upper
lower
)
1
2
(
2
2
2 −

+

+


−


= b
x
y
y
x
x
y k
k
c
y
x
x
y k
k +


−


= 2
2

The Bresenham line algorithm
• BRESENHAM’S LINE DRAWING ALGORITHM
(for |m| < 1.0)
1. Input the two line end-points, storing the left end-
point in (x0, y0)
2. Plot the point (x0, y0)
3. Calculate the constants Δx, Δy, 2Δy, and (2Δy - 2Δx)
and get the first value for the decision parameter as:
4. At each xk along the line, starting at k = 0, perform the
following test. If pk < 0, the next point to plot is
(xk+1, yk) and:
10
x
y
p 
−

= 2
0
y
p
p k
k 
+
=
+ 2
1

The Bresenham line algorithm
Otherwise, the next point to plot is (xk+1, yk+1) and:
5. Repeat step 4 (Δx – 1) times
• The algorithm and derivation above assumes slopes are
less than 1. for other slopes we need to adjust the
algorithm slightly.
11
x
y
p
p k
k 
−

+
=
+ 2
2
1

Example 1
• Let the given end points for the line be (30,20) and (40, 28)
• M = dy = y2 – y1 = 28 – 20 = 8
• dx x2 – x1 40 – 30 = 10
• m = 0.8
• dy = 8 and dx = 10
• The initial decision parameter P0 is
• P0 = 2dy – dx = 2(8) – 10 = 16 – 10 = 6
• P0 = 6>0, there x and y will be increase with interval one.
• The constants 2dy and 2dy-2dx are
• 2dy = 2(8) = 16 2dy-2dx = 2(8)- 2(10) =16 – 20
• 2dy = 16 2dy – 2dx = - 4
12

Example 1 (cont…)
• End points (30,20) (40,28)
• dx=x2-x1=40-30=10 dy= y2-y1=28-20=8
• P0=2dy-dx=2X8-10 = 6 >0 pt(31,21)
• Pk+1= pk+2dy-2dx= 6+16-20 =2 >0 (32,22)
• Pk+1=2+16-20 =-2 <0 (33,22)
• Pk+1= pk+2dy =-2+16 = 14 >0 (34,23)
• Pk+1= pk+2dy-2dx=14+16-20=10>0 (35,24)
• Pk+1=10+16-20=6 >0 (36,25)
• Pk+1= 6+16-20=2>0 (37,26)
• Pk+1=2+16-20=-2 <0 (38,26)
• Pk+1= pk+2dy=-2+16=14>0 (39,27)
• Pk+1=pk+2dy-2dx= 14+16-20=10>0 (40,28)
13

Example 1 (cont…)
• The starting point (x0, y0)=(30,20) and the successive
pixel positions are given in the following table
14
K Pk (Xk+1, Yk+1)
0 6 (31,21)
1 2 32,22
2 -2 33,22
3 14 34,23
4 10 35,24
5 6 36,25
6 2 37,26
7 -2 38,26
8 14 39,27
9 10 40,28

Example 2
• We plot the initial point (x0 , y0)=(20,10) and (x1 , y1)=
(30,18)
• determine successive pixel positions along the line path
from the decision parameter as.
15
K pk (xk +1, yk +1) K pk (xk +1, yk +1)
0 6 (21,11) 5 6 (26,15)
1 2 (22,12) 6 2 (27,16)
2 -2 (23,12) 7 -2 (28,16)
3 14 (24,13) 8 14 (29,17)
4 10 (25,14) 9 10 (30,18)

16

Example 2
17

Task 1
• Consider the line from (5, 8) to (9,11),use general
Bresenham’s line algorithm to rasterize this line.
Evaluate and tabulate all the steps involved.
18

Task 2
• Consider the line from (0, 0) to (-8,-4), use general
Bresenham’s line algorithm to rasterize this line.
Evaluate and tabulate all the steps involved.
19

Bresenham Line Algorithm Summary
• The Bresenham line algorithm has the following
advantages:
• A fast incremental algorithm
• Uses only integer calculations
• Comparing this to the DDA algorithm, DDA has the
following problems:
• Accumulation of round-off errors can make the pixelated line
drift away from what was intended
• The rounding operations and floating-point arithmetic
involved are time consuming
20

Assignment No.1
• Explore the Bresenham’s line algorithm procedure
(for |m| >1.0)
21

CG08 - Bresenham’s Line Algorithm Data structure.pdf

CG08 - Bresenham’s Line Algorithm Data structure.pdf

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