# Computer Graphics 4 Bresenham Line Drawing Algorithm, Circle

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```					Computer Graphics 4:
Bresenham Line
Drawing Algorithm,
Circle Drawing &
Polygon Filling

Course Website: http://www.comp.dit.ie/bmacnamee
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Contents
In today’s lecture we’ll have a look at:
– Bresenham’s line drawing algorithm
– Line drawing algorithm comparisons
– Circle drawing algorithms
• A simple technique
• The mid-point circle algorithm
– Polygon fill algorithms
– Summary of raster drawing algorithms
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The Bresenham Line Algorithm
The Bresenham algorithm is
another incremental scan
conversion algorithm
algorithm is that it uses only
integer calculations
Jack Bresenham
worked for 27 years at
IBM before entering
developed his famous
algorithms at IBM in
the early 1960s
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The Big Idea
Move across the x axis in unit intervals and
at each step choose between two different y
coordinates
For example, from
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(xk+1, yk+1)
position (2, 3) we
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have to choose
(xk, yk)                      between (3, 3) and
3                            (3, 4)
(xk+1, yk)
2
We would like the
point that is closer to
2     3      4     5
the original line
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Deriving The Bresenham Line Algorithm

At sample position
yk+1
xk+1 the vertical               dupper
y
separations from the                     dlower
mathematical line are yk
labelled dupper and dlower           xk+1
The y coordinate on the mathematical line at
xk+1 is:
y  m( xk  1)  b
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Deriving The Bresenham Line Algorithm
39                                  (cont…)
So, dupper and dlower are given as follows:
d lower  y  yk
 m( xk  1)  b  yk
and:
d upper  ( yk  1)  y
 yk  1  m( xk  1)  b
We can use these to make a simple decision
about which pixel is closer to the mathematical
line
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Deriving The Bresenham Line Algorithm
39                                  (cont…)
This simple decision is based on the difference
between the two pixel positions:
d lower  d upper  2m( xk  1)  2 yk  2b  1
Let’s substitute m with ∆y/∆x where ∆x and
∆y are the differences between the end-points:
y
x(dlower  dupper )  x(2 ( xk  1)  2 yk  2b  1)
x
 2y  xk  2x  yk  2y  x(2b  1)

 2y  xk  2x  yk  c
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Deriving The Bresenham Line Algorithm
39                                  (cont…)
So, a decision parameter pk for the kth step
along a line is given by:
pk  x(dlower  dupper )
 2y  xk  2x  yk  c
The sign of the decision parameter pk is the
same as that of dlower – dupper
If pk is negative, then we choose the lower
pixel, otherwise we choose the upper pixel
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Deriving The Bresenham Line Algorithm
39                                  (cont…)
Remember coordinate changes occur along
the x axis in unit steps so we can do
everything with integer calculations
At step k+1 the decision parameter is given
as:
pk 1  2y  xk 1  2x  yk 1  c
Subtracting pk from this we get:
pk 1  pk  2y ( xk 1  xk )  2x( yk 1  yk )
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Deriving The Bresenham Line Algorithm
39                                  (cont…)
But, xk+1 is the same as xk+1 so:
pk 1  pk  2y  2x( yk 1  yk )
where yk+1 - yk is either 0 or 1 depending on
the sign of pk
The first decision parameter p0 is evaluated
at (x0, y0) is given as:
p0  2y  x
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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:
p0  2y  x
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:
pk 1  pk  2y
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The Bresenham Line Algorithm (cont…)

Otherwise, the next point to plot is (xk+1, yk+1) and:
pk 1  pk  2y  2x
5.   Repeat step 4 (Δx – 1) times

ACHTUNG! The algorithm and derivation
above assumes slopes are less than 1. for
other slopes we need to adjust the algorithm
slightly
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Bresenham Example
Let’s have a go at this
Let’s plot the line from (20, 10) to (30, 18)
First off calculate all of the constants:
– Δx: 10
– Δy: 8
– 2Δy: 16
– 2Δy - 2Δx: -4
Calculate the initial decision parameter p0:
– p0 = 2Δy – Δx = 6
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Bresenham Example (cont…)

18                                                        k   pk   (xk+1,yk+1)

17                                                        0
16                                                        1

15                                                        2

14                                                        3
4
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5
12
6
11
7
10
8
20   21   22   23 24   25   26   27   28   29   30   9
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Bresenham Exercise
Go through the steps of the Bresenham line
drawing algorithm for a line going from
(21,12) to (29,16)
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Bresenham Exercise (cont…)

18                                                        k   pk   (xk+1,yk+1)

17                                                        0
16                                                        1

15                                                        2

14                                                        3
4
13
5
12
6
11
7
10
8
20   21   22   23 24   25   26   27   28   29   30
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Bresenham Line Algorithm Summary

The Bresenham line algorithm has the
– An 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
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A Simple Circle Drawing Algorithm
The equation for a circle is:
x y r
2    2     2

where r is the radius of the circle
So, we can write a simple circle drawing
algorithm by solving the equation for y at
unit x intervals using:
y   r 2  x2
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A Simple Circle Drawing Algorithm
39                            (cont…)

y0  202  02  20

y1  202 12  20
y2  202  22  20

y19  202  192  6

y20  202  202  0
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A Simple Circle Drawing Algorithm
39                                     (cont…)
However, unsurprisingly this is not a brilliant
solution!
Firstly, the resulting circle has large gaps
where the slope approaches the vertical
Secondly, the calculations are not very
efficient
– The square (multiply) operations
– The square root operation – try really hard to
avoid these!
We need a more efficient, more accurate
solution
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Eight-Way Symmetry
The first thing we can notice to make our circle
drawing algorithm more efficient is that circles
centred at (0, 0) have eight-way symmetry
(-x, y)      (x, y)

(-y, x)                          (y, x)

R
2
(-y, -x)                          (y, -x)

(-x, -y)     (x, -y)
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Mid-Point Circle Algorithm
Similarly to the case with lines,
there is an incremental
algorithm for drawing circles –
the mid-point circle algorithm
In the mid-point circle algorithm
we use eight-way symmetry so
The mid-point circle
only ever calculate the points      algorithm was
for the top right eighth of a       developed by Jack
Bresenham, who we
circle, and then use symmetry       heard about earlier.
Bresenham’s patent
to get the rest of the points       for the algorithm can
be viewed here.
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Mid-Point Circle Algorithm (cont…)
Assume that we have
just plotted point (xk, yk) (xk, yk) (xk+1, yk)

The next point is a
choice between (xk+1, yk)            (xk+1, yk-1)
and (xk+1, yk-1)
We would like to choose
the point that is nearest to
the actual circle
So how do we make this choice?
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Mid-Point Circle Algorithm (cont…)
Let’s re-jig the equation of the circle slightly
to give us:
f circ ( x, y )  x 2  y 2  r 2
The equation evaluates as follows:
 0, if ( x, y) is inside the circle boundary

f circ ( x, y )  0, if ( x, y) is on the circle boundary
 0, if ( x, y) is outside the circle boundary

By evaluating this function at the midpoint
between the candidate pixels we can make
our decision
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Mid-Point Circle Algorithm (cont…)
Assuming we have just plotted the pixel at
(xk,yk) so we need to choose between
(xk+1,yk) and (xk+1,yk-1)
Our decision variable can be defined as:
pk  f circ ( xk  1, yk  1 )
2
 ( xk  1) 2  ( yk  1 ) 2  r 2
2
If pk < 0 the midpoint is inside the circle and
and the pixel at yk is closer to the circle
Otherwise the midpoint is outside and yk-1 is
closer
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Mid-Point Circle Algorithm (cont…)
To ensure things are as efficient as possible
we can do all of our calculations
incrementally
First consider:

pk 1  f circ xk 1  1, yk 1  1
2


 [( xk  1)  1]  yk 1  1
2
2
2
 r   2

or:
pk 1  pk  2( xk  1)  ( yk 1  yk )  ( yk 1  yk )  1
2       2

where yk+1 is either yk or yk-1 depending on
the sign of pk
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Mid-Point Circle Algorithm (cont…)
The first decision variable is given as:
p0  f circ (1, r  1 )
2
 1  (r     1 )2  r 2
2
 5 r
4
Then if pk < 0 then the next decision variable
is given as:
pk 1  pk  2 xk 1  1
If pk > 0 then the decision variable is:
pk 1  pk  2 xk 1  1  2 yk  1
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The Mid-Point Circle Algorithm
MID-POINT CIRCLE ALGORITHM
•   Input radius r and circle centre (xc, yc), then set the
coordinates for the first point on the circumference of a
circle centred on the origin as:
( x0 , y0 )  (0, r )
•   Calculate the initial value of the decision parameter as:
p0  5  r
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•   Starting with k = 0 at each position xk, perform the
following test. If pk < 0, the next point along the circle
centred on (0, 0) is (xk+1, yk) and:
pk 1  pk  2 xk 1  1
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The Mid-Point Circle Algorithm (cont…)

Otherwise the next point along the circle is (xk+1, yk-1)
and:
pk 1  pk  2 xk 1  1  2 yk 1
4.   Determine symmetry points in the other seven octants
5.   Move each calculated pixel position (x, y) onto the
circular path centred at (xc, yc) to plot the coordinate
values:
x  x  xc       y  y  yc
6.   Repeat steps 3 to 5 until x >= y
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Mid-Point Circle Algorithm Example
To see the mid-point circle algorithm in
action lets use it to draw a circle centred at
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Mid-Point Circle Algorithm Example
39                                           (cont…)

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k   pk   (xk+1,yk+1)   2xk+1   2yk+1
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8                                             0
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1
6
5                                             2
4
3
3
2                                             4
1                                             5
0
6
0   1   2   3 4   5   6   7   8   9 10
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Mid-Point Circle Algorithm Exercise
Use the mid-point circle algorithm to draw
the circle centred at (0,0) with radius 15
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Mid-Point Circle Algorithm Example
39                                          (cont…)
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k    pk   (xk+1,yk+1)   2xk+1   2yk+1
15
14
0
13
12                                               1
11                                               2
10                                               3
9                                               4
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5
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6                                               6
5                                               7
4                                               8
3
9
2
1                                              10
0                                              11
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16   12
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Mid-Point Circle Algorithm Summary
The key insights in the mid-point circle
algorithm are:
– Eight-way symmetry can hugely reduce the
work in drawing a circle
– Moving in unit steps along the x axis at each
point along the circle’s edge we need to
choose between two possible y coordinates
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Filling Polygons
So we can figure out how to draw lines and
circles
How do we go about drawing polygons?
We use an incremental algorithm known as
the scan-line algorithm
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Scan-Line Polygon Fill Algorithm
10                      Scan Line

8

6

4

2

0
2   4   6   8       10      12   14   16
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Scan-Line Polygon Fill Algorithm
The basic scan-line algorithm is as follows:
– Find the intersections of the scan line with all
edges of the polygon
– Sort the intersections by increasing x
coordinate
– Fill in all pixels between pairs of intersections
that lie interior to the polygon
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Scan-Line Polygon Fill Algorithm
39                           (cont…)
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Line Drawing Summary
Over the last couple of lectures we have
looked at the idea of scan converting lines
The key thing to remember is this has to be
FAST
For lines we have either DDA or Bresenham
For circles the mid-point algorithm
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Anti-Aliasing
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Summary Of Drawing Algorithms
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Mid-Point Circle Algorithm (cont…)

6

5

4

3

1    2    3    4
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Mid-Point Circle Algorithm (cont…)

6

M
5

4

3

1       2   3   4
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Mid-Point Circle Algorithm (cont…)

6
M
5

4

3

1    2    3       4
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0   1   2   3   4   5   6   7   8   9 10
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