B. SC CSIT Computer Graphics Unit1.2 By Tekendra Nath Yogi

1.2: Raster Graphics
Presented By : Tekendra Nath Yogi
Tekendranath@gmail.com
College Of Applied Business And Technology

Basic Raster Algorithms for 2D Primitives
• In Raster graphic system a picture can be describe in
two ways:
– A Set of intensities for the pixel positions in the display or
– A set of complex objects consisting of set of basic
geometric objects.
– These basic geometric objects are called output primitives;
– Includes:
• Point ,Line , Circle, Ellipse, Polygon, etc.
2By: Tekendra Nath Yogi2/9/2019

Contd….
• If the picture(scene) is described as a set of intensities for the
pixel positions in the display then the scene is displayed by
loading the pixel arrays into the frame buffer.
• If the picture(scene) is described as a set of complex objects
consisting of set of basic geometric objects then the scene is
displayed by scan converting the basic geometric-structure
specifications into pixel patterns.
• Normally picture is described in terms of output primitives.

Points
• Point plotting is accomplished by converting a single
coordinate position furnished by an application program into
appropriate operations for the output device in use.
• For example:
– For a black and- white raster system a point is plotted by setting the bit
value corresponding to a specified screen position within the frame
buffer to 1.
– Then, as the electron beam sweeps across each horizontal scan line, it
emits a burst of electrons (plots a point) whenever a value of 1 is
encountered in the frame buffer.

Contd….
Fig: point on a display device

Lines
• A line segment is completely defined in terms of its two
endpoints.
• Line segment is thus defined as:
– Line_seg ={(x1,y1), (x2, y2)}

Contd….
• Digital devices display a straight line by plotting (illuminating)
discrete coordinate points along the line path which are
calculated from the equation of the line.

Contd….
• Screen locations are referenced with integer values, so plotted
positions may only approximate actual line positions between
two specific endpoints.
• E.g.,: A computed line position of (10.48, 20.51) will be
converted to pixel position (10, 21). This rounding of
coordinate values to integers causes lines to be displayed with
a stairstep appearance (the “jaggies”).

Line Drawing Algorithms
• If the coordinates of end points of the line segment are known,
there are several methods for selecting the pixels between the
end pixels(line drawing).
• Two main algorithms :
– Digital Difference Analyzer (DDA algorithm)
– Bresenham Line Drawing Algorithm
• Based on the slope-intercept equation of a straight line.

Contd…
10
• The slope-intercept equation of a straight line is:
– Where, m = slope of line and b = y-intercept.
– For any two given points (x1, y1) and (x2, y2)
– Therefore, equation (1) becomes,
– At any point, (xk, yk)
)1(.............  bmxy
12
12
xx
yy
m



x
xx
yy
yb
12
12



)2(.............yk  bmxk

Contd..
11
• At (xk+1, yk+1),
• Subtracting (2) from (3) we get,
• Therefore,
– ∆y = m.∆x
– Or, ∆x = ∆y/ m
– Or, m = ∆y /∆x
These equations form the basis for sampling point between give two end point of a
line.
)3(............11   bmxy kk
)( 11 kkkk xxmyy  

DDAAlgorithm
• The digital differential analyzer(DDA) is a scan conversion
line algorithm based on calculating either:
– ∆y = m.∆x or
– ∆x = ∆y/m
• Sample the line at unit intervals in one coordinate
• Determine the corresponding integer values nearest the line
path in another co-ordinate

Case1: A line with positive slope
• a) assumption 1: line are processed left end to right end point:
– i) if (m<=1 ) then sampling in x-direction
• ∆x =1, xk+1 = xk+1 and
• yk+1 =yk+ m
– ii)if (m>1) then sampling in y-direction(i.e.,
reverse role of x and y)
• ∆y = 1, yk+1 =yk+1 and
• xk+1 = xk+(1/m)

Contd….
• b) assumption 2: line are processed right end to left end point:
– i) if (m<=1 ) then sampling in x-direction
• ∆x =-1,
• xk+1 = xk -1 and
• yk+1 =yk - m
– ii)if (m>1) then sampling in y-direction(i.e.,
• ∆y = - 1, yk+1 =yk-1 and
• xk+1 = xk -(1/m)

Case1: A line with negative slope
• a) assumption 1: line are processed left end to right end point:
– i) if (|m|<=1 ) then sampling in x-direction
• ∆x =1, xk+1 = xk+1 and
• yk+1 =yk+ m
– ii)if (|m|>1) then sampling in y-direction(i.e.,
• ∆y = -1, yk+1 =yk-1 and
• xk+1 = xk-(1/m)

Contd….
• b) assumption 2: line are processed right end to left end point:
– i) if (|m|<=1 ) then sampling in x-direction
• ∆x =-1,
• xk+1 = xk -1 and
• yk+1 =yk – m
– ii)if (|m|>1) then sampling in y-direction(i.e., reverse
role of x and y)
• ∆y = 1, yk+1 =yk+1 and
• xk+1 = xk +(1/m)

Contd….
DDAAlgorithm:
1. Input the two line endpoints (x1, y1) and (x2, y2).
2. Plot first point (x1, y1).
3. Calculate constants Δx = (x2- x1), and Δy =(y2-y1).
4. If |Δx| > |Δy|
then steps = |Δx|
else steps = |Δy|
5. Calculate XInc = Δx / steps and YInc = Δy / steps
6. for (k=0; k<steps; k++)
{
x =x+ xInc;
y =y+ yInc;
plot(ROUND(x), ROUND(y));
}

Contd….
• Advantages of DDA algorithm:
– It is the simplest algorithm and it does not require special skills for
implementation.
– It is a faster method for calculating pixel positions than the direct use of
equation y= mx+b.
• Disadvantages of DDA algorithm:
– Floating point arithmetic in DDA algorithm is still time consuming.
– The algorithm is orientation dependent. So the end point accuracy is
poor.

Contd….
• Example: Rasterize/digitize the line with end point(10, 20) and
(20, 30) using DDA algorithm.
• solution:
– Given two line end points are: (x1,y1) = (10, 20) and (x2, y2) = (20,30)
– Plot (10,20)
– Now calculating Δx = (x2- x1) =(20-10) = 10 and
– Δy =(y2-y1)= (30-20)= 10.
– Therefore, step =10
– Calculating XInc = Δx / steps = 10/10 = 1and
– YInc = Δy / steps = 10/10 =1

Contd….
Iteration no.(K) X = X+ xinc Y = Y+yinc Pixel(x,y)
0 11 21 (11,21)
1 12 22 (12,22)
2 13 23 (13,23)
3 14 24 (14,24)
4 15 25 (15,25)
5 16 26 (16,26)
6 17 27 (17,27)
7 18 28 (18,28)
8 19 29 (19,29)
9 20 30 (20,30)

Class work
• Example1: Rasterize the line with end point (10,20) and
(14,30).
• Example2: Rasterize the line with end point (20,30) and
(15,34).
• Example3: Rasterize the line with end point (0,0) and (-6,-6).

Bresenham's Line algorithm(BLA)
• An accurate and efficient line generating algorithm, developed
by Bresenham.
• scan conversion algorithm.
• scan converts lines only using integer calculation to find the
next (x, y) position to plot.
• Hence, It avoids incremental error accumulation.

Contd….
• Bresenham's scan-conversion process for lines with positive slope less than 1:
– Pixel positions along a line path are determined by sampling at unit x intervals.
– Starting from the left end point (x0, y0) of a given line
– step to each successive x position and
– y plot the pixel whose scan-line value is closest to the line path.

Contd….
• i.e.,:
– Assuming we have determined that the pixel at (xk, yk) is to be
displayed.
– we next need to decide which pixel to plot in column xk+1.
– Our choices are the pixels at positions (xk+1, yk)and (xk+1, yk+1).
– In this case choose the pixel positions that is closest to the
mathematical line

Contd….
• For example, as shown in the following illustration, from
position (2, 3) you need to choose between (3, 3) and (3, 4).
You would like the point that is closer to the original line.

Contd….
• At sampling position xk+1, vertical pixel separation from
mathematical line path is labeled as d1& d2 as shown in figure
below .
• The y-coordinate on the mathematical line path at pixel column xk+1
is calculated as:
y=m(xk+1)+b

Contd….
• Then d1 = y - yk
= m(xk + 1)+b – yk
And d2 =( yk + 1) - y
= (yk + 1)-m ( xk + 1 )-b
• Now,
d1-d2 = 2m ( xk + 1 ) - ( yk + 1 )-yk+2b
= 2m (xk + 1) - 2yk + 2b – 1

Contd….
•
• u

Contd….
• J
• Note: For a line with positive slope greater than 1, we simply interchange
the role of x & y.

Contd….
• Algorithm:
contd….

Contd….
• Contd…
contd….

Contd….
• Contd…

Contd….
• Example: Digitize the line with end point (10,20) and (20,26)
using BLA.
• Solution:
1. Given, (x1,y1) = (10, 20) and (x2,y2) = (20,26)
2. dx= |20-10| =10 and dy =|26-20| =6
3. Here x2>x1 is true i.e., 20>10 is true so set a = 1
4. y2>y1 is true i.e., 26>20 is true so set b =1
5. Here (dx>dy) is true i.e., 10>6 is true so, initial decision parameter
p0= 2dy-dx =2*6-10= 2

Contd….
• For k= 0; k <10; k++
K xk yk pk Pk+1 Xk+1 Yk+1 Pixel(x,y)
0 10 20 2 -6 11 21 (11,21)
1 11 21 -6 6 12 21 (12,21)
2 12 21 6 -2 13 22 (13,22)
3 13 22 -2 10 14 22 (14,22)
4 14 22 10 2 15 23 (15,23)
5 15 23 2 -6 16 24 (16,24)
6 16 24 -6 6 17 24 (17,24)
7 17 24 6 -2 18 25 (18,25)
8 18 25 -2 10 19 25 (19,25)
9 19 25 10 2 20 26 (20,26)

Contd….
• For k = 0
– (Pk=2<0) is false so
– Pk+1 =pk + 2dy-2dx =2+ 2*6-2*10 =-6
– Xk+1 = xk+a =10+1 =11
– Yk+1 = yk+b =20+1 =21

Contd….
• For k = 1
– (Pk=-6<0) is true so
– Pk+1 =pk + 2dy =-6+ 2*6 =6
– Xk+1 = xk+a =11+1 =12
– Yk+1 = yk=21

Contd….
• For k = 2
– Pk+1 =pk + 2dy-2dx =6+ 2*6-2*10 =-2
– Xk+1 = xk+a =12+1 =13
– Yk+1 = yk+b =21+1 =22

Contd….
• For k = 3
– Pk+1 =pk + 2dy =-2+ 2*6 =10
– Xk+1 = xk+a =13+1 =14
– Yk+1 = yk=22

Contd….
• For k = 4
– Pk+1 =pk + 2dy-2dx =10+ 2*6-2*10 =2
– Xk+1 = xk+a =14+1 =15
– Yk+1 = yk+b =22+1 =23

Contd….
• For k = 5
– Pk+1 =pk + 2dy-2dx =2+ 2*6-2*10 =-6
– Xk+1 = xk+a =15+1 =16
– Yk+1 = yk+b =23+1 =24

Contd….
• For k = 6
– Pk+1 =pk + 2dy =-6+ 2*6 =6
– Xk+1 = xk+a =16+1 =17
– Yk+1 = yk=24

Contd….
• For k = 7
– Pk+1 =pk + 2dy-2dx =6+ 2*6-2*10 =-2
– Xk+1 = xk+a =17+1 =18
– Yk+1 = yk+b =24+1 =25

Contd….
• For k = 8
– Pk+1 =pk + 2dy-2dx =-2+ 2*6-2*10 =10
– Xk+1 = xk+a =18+1 =19
– Yk+1 = yk+b =25

Contd….
• For k = 9
– Pk+1 =pk + 2dy-2dx =10+ 2*6-2*10 =2
– Xk+1 = xk+a =19+1 =20
– Yk+1 = yk+b =25+1 =26

Contd….
• Example1: Digitize the line with endpoints (3,10) and (6,2)
using Bresenham line algorithm.

Circle Generating algorithms
• Circle is a frequently used component in pictures and graphs.
• A circle is defined as the set of points that are all at a given
distance r from the center position (xc, yc) as shown in figure
below:
Fig: Circle with center coordinates (xc, yc) and radius r

Contd….
• The equation of circle is:
• Where (x, y) = point on circle circumference can be calculated
by stepping along x-axis in unit steps from xc-r to xc+r and
corresponding y-value can be calculated as:
222
)()( ryyxx cc 
22
)( xxryy cc 

Contd….
• Problems in above method:
– Square root calculation at each step increase computational complexity.
– Spacing between plotted pixel positions is not uniform.

Computation can be reduced by considering the symmetry properties of circle!
• 4-way symmetry: The shape of circle is similar in each quadrant.
• i.e., If (x, y) is pixel position on circumference of circle in 1st quadrant
then corresponding symmetric positions are:
• 2nd quadrant = (-x, y)
• 3rd quadrant = (-x, -y )
• 4th quadrant = (x, -y )

Contd….
• 8-way symmetry: circle sections in adjacent octants within one
quadrant are symmetric with respect to 450 line dividing the two
octants:
•A point at position (x,y)
on a one-eighth circle
sector is mapped into the
seven circle points in the
other octants of the xy-
plane.
•So, generate all pixel
positions around a circle
by calculating only the
points within the sector
from x= 0 to x=y
Problem of computation still persists !

Contd….
• Midpoint circle algorithm:
– In mid point circle algorithm, we sample at unit intervals and determine
the closest pixel position to the specified circle path at each step.
– For given radius r and screen center position (xc, yc), first calculate
pixel positions around a circle path centered at the coordinate origin
(0,0) by using 8-way symmetry.
– Then, move each calculated position (x, y) to its proper screen position
by adding xc to x and yc to y.

Contd….
• To apply the mid point method, circle function is defined as:
• For any point (x, y) :
222
),( ryxyxfcircle 








boundarycircletheoutsideisyxif
boundarycircletheonisyxif
boundarycircletheinsideisyxif
yxfcircle
),(,0
),(,0
),(,0
),(

Contd….
• In midpoint algorithm decision parameter is a circle function.
• Test are performed for midpoint between pixels near the circle
path at each sampling step.
222
)
2
1
()1(
)
2
1
,1(
ryx
yxfp
kk
kkcirclek


Fig: midpoint between the two
candidate pixels at sampling
position xk+1
Decision:
yk+1 = yk if pk<0
yk+1 = yk-1 otherwise

Contd….
• Successive decision parameter at sampling position xk+1+1
=xk+2 is:
1)()()1(2
)
2
1
(]1)1[(
)
2
1
,1(
1
22
11
22
1
2
111






kkkkkkk
kk
kkcirclek
yyyyxpp
ryx
yxfp

Contd….
• If pk<0 then yK+1 = yk Thus
Pk+1 = Pk + 2xk+1+1
• Otherwise , yK+1 = yk -1 Thus
Pk+1 = Pk + 2xk+1+1-2yk+1
• Also incremental evaluation of 2xk+1 and 2yk+1
2xk+1 = 2xk + 2
2yk+1 = 2yk – 2 if pk >0
• At start position (x0,y0) = (0,r)
2x0 = 0 and 2y0 = 2r

Contd….
• Initial decision parameter:
• For r specified as an integer, round p0 to
P0 = 1-r
(because all increments are integers)
r
rr
rfp circle



4
5
)
2
1
(1
)
2
1
,1(
22
0

Midpoint circle algorithm
1. Input radius r and circle center (xc, yc) and obtain the first point on the circumference of a
circle centered on the origin as
(x0,y0) = (0,r)
2. Calculate the initial value of the decision parameter as
P0 = 5/4 – r = 1-r
3. At each xk position, starting at k = 0, perform the following test:
If pk < 0, the next point along the circle centered on (0,0) is (xk+1,yk) and
Pk+1 = pk + 2xk+1 + 1
Otherwise, the next point along the circle is (xk+1,yK-1) and
Pk+1 = pk + 2xk+1 + 1 -2yk+1
Where 2xk+1 = 2xk + 2 and 2yk+1 = 2yk-2
4. Determine the symmetry points in the other seven octants.
5. Move each calculated pixel position (x,y) onto the circular path
centered on (xc,yc) and plot the co-ordinate values:
x = x + xc, y = y+yc
6. Repeat steps 3 through 5 until x ≥ y

Contd….
• Example1: Draw a circle with center at (2,3) and radius 4
using midpoint circle drawing algorithm.
– Solution: Step 1 and 2:
• Given, radius(r) = 4
• Circle center (xc, yc) = (2, 3)
• Initial point on boundary of circle (x0, y0) =(0,r)= (0, 4)
• Initial decision parameter = 1-r = 1-4= -3
• Successive decision parameter and positions along the circular path
are calculated as until x>= y i.e. while(x<y) is true.

Contd….
K Xk Yk Pk Xk+1 Yk+1 2yk+1 2xk+1 Pk+1 Pixel Symmetric pixels
0 0 4 -3 1 4 8 2 0 (1, 4) (4,1), (-1,4), (-4,1), (-1,
-4), (-4, -1), (1,-4), (4,-
1)
1 1 4 0 2 3 6 4 -1 (2,3) (3,2,), (-2,3), (-3,2), (-
2, -3), (-3, -2), (2,-3),
(3, -2)
2 2 3 -1 3 3 6 6 6 (3,3) (3,3,), (-3,3), (-3, 3), (-
3,-3), (-3,-3), (3,-3),
(3,-3)

Contd….
• Iteration 0:
– Step 3:
• Xk = x0= 0
• Yk= y0= 4
• Pk= p0= 1-r = 1-4= -3
• Here pk< 0 is true
• So, next point is (xk+1, yk) = (0+1, 4) = (1, 4)
• and pk+1= pk + 2xk+1+1= -3+2*1 +1 =0

Contd….
• Iteration 0 Contd….:
– Step4: Now calculating symmetric pixels:
• For pixel (1,4) : (4,1), (-1,4), (-4,1), (-1, -4), (-4, -1), (1,-4), (4,-1)
– Step 5: Moving each calculated pixel position (x,y) onto the circular
path centered on (2,3) as: x = x + xc, y = y+yc
– Move (1,4) To(1+2, 4+3)= (3,7)
– Move (4,1)to (4+2, 1+3) =(6, 4),
– Move (-1,4) To (-1+2, 4+3)= (1,7)
– Move (-4,1) To (-4+2, 1+3)= (-2,4)
– Move (-1, -4) To (-1+2, -4+3)= (1,-1)
– Move (-4, -1) To (-4+2, -1+3)= (-2,2)
– Move (1,-4) To (1+2, -4+3)= (3, -1)
– Move (4,-1) To (4+2, -1+3)= (6,2)
– Step 6: x< y is true so continue

Contd….
• Iteration 1:
– Step3:
• Xk = x1= 1
• Yk= y1= 4
• Pk= p1= 0
• Here pk< 0 is false
• So, next point is (xk+1, yk -1) = (1+1, 4-1) = (2, 3)
• and pk+1= pk + 2xk+1+1 – 2yk+1= 0+ 2*2 +1 – 2*3 = -1

Contd….
• Iteration 1 Contd…:
– Step 4: Now calculating symmetric pixels:
• For pixel (2,3): (3,2,), (-2,3), (-3,2), (-2, -3), (-3, -2), (2,-3), (3, -2)
– Step 5: Moving each calculated pixel position (x,y) onto the circular path
centered on (2,3) as: x = x + xc, y = y+yc
• Move (2,3) To (2+2, 3+3)= (4, 6)
• Move (3,2) To ( 3+2, 2+3)= (5,5)
• Move (-2,3) To (-2+2, 3+3)= (0,6)
• Move (-3,2) To (-3+2, 2+3)= (-1,5)
• Move (-2, -3) To (-2+2, -3+3)= (0,0)
• Move (-3, -2) To (-3+2, 3+3)= (-1,6)
• Move(2,-3) To (2+2, -3+3)= (4,0)
• Move (3, -2) To (3+2, -2+3)= (5,1)
– Step6: x< y is true so continue

Contd….
• Iteration 2:
– Step 3:
• Xk = x2= 2
• Yk= y2= 3
• Pk= p2= -1
• Here pk< 0 is true
• So, next point is (xk+1, yk ) = (2+1, 3) = (3,3)
• And pk+1= pk + 2xk+1+1 = -1+ 2*3 +1 = 6

Contd….
• Iteration 2 Contd…..:
– Step 4:Now calculating symmetric pixels:
• For pixel (3,3): (3,3,), (-3,3), (-3, 3), (-3,-3), (-3,-3), (3,-3), (3,-3)
– Step 5: Moving each calculated pixel position (x,y) onto the circular
path centered on (2,3) as: x = x + xc, y = y+yc
• Move (3,3) To (3+2, 3+3) = (5,6)
• Move (3,3) To (3+2, 3+3) = (5,6)
• Move (-3,3) To (-3+2, 3+3)= (-1,6)
• Move (-3,3) To (-3+2, 3+3)= (-1,6)
• Move (-3,-3) To (-3+2, -3+3)= (-1, 0)
• Move (-3,-3) To (-3+2, -3+3)= (-1, 0)
• Move (3,-3) To (3+2, -3+3)= (5, 0)
• Move (3,-3) To (3+2, -3+3)= (5, 0)
– Step 6: x< y is false so terminate!

Contd….
• Example2: calculate pixel position of circle with radius 10
and center at origin i.e. (xc, yc)= (0,0).
– Solution:
• Given, radius(r) = 10
• Circle center (xc, yc) = (0, 0)
• Initial point on boundary of circle (x0, y0) =(0,r)= (0, 10)
• Initial decision parameter = 1-r = 1-10 = -9
• Successive decision parameter and positions along the circular path
are calculated as until x>= y i.e. while(x<y) is true.

Contd….
K Xk Yk Pk Xk+1 Yk+1 Pk+1 pixel 2xk+1 2yk+1 Symmetric pixels
0 0 10 -9 1 10 -6 1,10 2 10 (10,1), ( 1,- 10), (-1, 10), (-1, -10), (-10,-
1), (-10,1), (10,-1)
1 1 10 -6 2 10 -1 2,10 4 20 (10, 2),(-2,10) ,(-10,2), (-2,-10), (-10,-
2),(2, -10),(10, -2)
2 2 10 -1 3 10 6 3,10 6 20 (10, 3), (-10, 3), (-3, 10), (-3, -10), (-10,
-3), (3, -10), (10, -3)
3 3 10 6 4 9 -3 4,9 8 18 (9,4), (-4, 9), (-9,4), (-4, -9), (-9, -4), (4,
-9), (9, 4)
4 4 9 -3 5 9 8 5,9 10 18 (9,5), (-5,9), (-9,5), (-5, -9), (-9, -5), (5, -
9), (9,-5)
5 5 9 8 6 8 5 6,8 12 16 (8,6), (-6,8), (-8,6) , (-6,-8), (-8, -6), (6, -
8), (8, -6)
6 6 8 5 7 7 6 7,7 14 14 (7,7), (-7,7) (-7, 7), (-7, -7), (-7, -7),(7, -
7) (7, -7)

Contd….
y
x1 2 3 4 5 6 7 8 90 10
1
2
3
8
9
0
10
5
6
7
4

Contd….
• Iteration 0:
– Xk = x0= 0
– Yk= y0= 10
– Pk= p0= 1-r = 1-10= -9
– Here pk< 0 is true
– So, next point is (xk+1, yk) = (0+1, 10) = (1, 10)
– Now pk+1= pk + 2xk+1+1= -9+2*1 +1 =-6
– Symmetric pixels: (10,1), ( 1,- 10), (-1, 10), (-1, -10), (-10,-1), (-10,1), (10,-1)
– Here circle center is at origin so directly plot the symmetric pixels
– x< y is true so continue

Contd….
• Iteration 1:
– Xk = x1= 1
– Yk= y1= 10
– Pk= p1= -6
– Now pk+1= pk + 2xk+1+1= -6+2*2 +1 =-1
– Symmetric pixels:(10, 2),(-2,10) ,(-10,2), (-2,-10), (-10,-2),(2, -10),(10, -2)

Contd….
• Iteration 2:
– Xk = x2= 2
– Yk= y2= 10
– Pk= p2= -1
– Now pk+1= pk + 2xk+1+1= -1+2*3+1 =6
– Symmetric pixels:(10, 3), (-10, 3), (-3, 10), (-3, -10), (-10, -3), (3, -10), (10, -
3)

Contd….
• Iteration 3:
– Xk = x3= 3
– Yk= y3= 10
– Pk= p3= 6
– Here pk< 0 is false
– So, next point is (xk+1, yk -1) = (3+1, 10-1) = (4, 9)
– Now pk+1= pk + 2xk+1+1 – 2yk+1= 6+ 2*4 +1 – 2*9 = -3
– Symmetric pixels: (9,4), (-4, 9), (-9,4), (-4, -9), (-9, -4), (4, -9), (9, 4)

Contd….
• Iteration 4:
– Xk = x4= 4
– Yk= y4= 9
– Pk= p4= -3
– So, next point is (xk+1, yk) = (4+1, 9) = (5,9)
– Now pk+1= pk + 2xk+1+1= -3+2*5+1 =8
– Symmetric pixels: (9,5), (-5,9), (-9,5), (-5, -9), (-9, -5), (5, -9), (9,-5)

Contd….
• Iteration 5:
– Xk = x5= 5
– Yk= y5= 9
– Pk= p5= 8
– Now pk+1= pk + 2xk+1+1 – 2yk+1= 8+ 2*6 +1 – 2*8 = 5
– Symmetric pixels:(8,6), (-6,8), (-8,6) , (-6,-8), (-8, -6), (6, -8), (8, -6)

Contd….
• Iteration 6:
– Xk = x6= 6
– Yk= y6= 8
– Pk= p6= 5
– Now pk+1= pk + 2xk+1+1 – 2yk+1= 5+ 2*7 +1 – 2*7 = 6
– Symmetric pixels: (7,7), (-7,7) (-7, 7), (-7, -7), (-7, -7),(7, -7) (7, -7)
– x< y is false so terminate !

Homework
using midpoint circle drawing algorithm. (Note: pixel = (x+xc,
y+yc)
y+yc)
y+yc)

Implementation

Contd….

Contd….
• Output:

Ellipse Generating Algorithm
• Loosely stated:
– Ellipse is an elongated circle. So, ellipse can be generated by modifying
circle –drawing procedures.
• A precise definition:
– An ellipse is defined as the set of points such that the sum of the
distances from two fixed positions(foci) is the same for all points.
– i.e., d1+d2= constant for all point on the boundary of the ellipse

Contd..
• let F1= (x1, y1) and F2= (x2, y2) and boundary point P(x, y)
• Then d1 = and d2 =
• So,
        constant
2
2
2
2
2
1
2
1  yyxxyyxx

Contd….
• Assumption for simplification: major and minor axes are
oriented to align with the coordinate axes as shown in figure
below.
• For above ellipse, equation of the ellipse can be written as:
• Problem: computational complexity!
1
22








 





 
y
c
x
c
r
yy
r
xx

Contd….
• To reduce computational complexity symmetric property can
be used.
– An ellipse in standard position is symmetric between quadrants as
shown in figure below:
(x,y)
(x,-y)
(-x,y)
(-x,-y)
rx
ry

Contd….
• Midpoint ellipse algorithm:
– Given rx , ry and (xc, yc)
– Midpoint ellipse algorithm first determine the boundary points (x, y) of
the ellipse in standard position centered on the origin, and
– Then shift the points so the ellipse is centered at (xc, yc)

Contd….
• The midpoint ellipse algorithm started either at(0, ry) or at (rx,
0), and is applied throughout the first quadrant by taking unit
steps in the x-direction where the slope of the curve has a
magnitude less than 1(region1), and taking unit steps in the y-
direction where the slope has a magnitude greater than
1(region2).

Contd….
• Consider an ellipse centered at the origin, (xc,yc)=(0,0) then the ellipse
equation becomes:
• To apply the midpoint method Ellipse function is defined as:
• For any point (x,y)
– fe(x,y) < 0 if (x,y) is inside the ellipse
– fe(x,y) = 0 if (x,y) is on the ellipse
– fe(x,y) > 0 if (x,y) is outside the ellipse
• Thus, the ellipse function serves as a decision
Parameter in midpoint ellipse algorithm.
1
22















yx r
y
r
x
  222222
, yxxye rryrxryxf 

Contd….
• Decision parameter at sampling
position (xk+1) in region1:
Midpoint between candidate
pixels at sampling position
xK+ 1 along an elliptical
path
 
22
2
222
2
1
2
1
)1(
,11
yxkxky
kkek
rryrxr
yxfp








Decision:
yk+1 = yk if p1k<0 , i.e., select pixel( xK+1,yk) to plot
yk+1 = yk-1 otherwise, i.e., select pixel( xK+1,yk -1 ) to plot

Contd….
• At the next sampling position (xk+1 + 1 = xk + 2), the decision
parameter for region 1 is evaluated as
where yk+1 is either yk or yk -1 depending on the sign of p1k.
 
































22
1
222
1
22
2
1
222
2
1
111
2
1
2
1
)1(211
2
1
]1)1[(
,11
kkxykykk
yxkxky
kkek
yyrrxrpp
or
rryrxr
yxfp

Contd….
• where yk+1 is either yk or yk -1 depending on the sign of p1k.
• So, successive decision parameter:
• At the initial position (x0, y0)= (0, ry), the two terms evaluate to
2ry
2x = 0 ……. (i) & 2rx
2y = 2rx
2ry ……. (ii)
• As x and y are incremented, updated values are obtained by adding 2ry
2 to
(i) and subtracting 2rx
2 from (ii).






011
01
1
kk
kk
k
pify
pify
y










01221
0121
1
22
1
2
2
1
2
1
kkxykyk
kykyk
k
pifyrrxrp
pifrxrp
P

Contd….
• Updated values are compared at each step, and we move from
region 1 to region 2 when 2ry
2x >= 2rx
2y
• In region 1, initial value of decision parameter is obtained by
evaluating the ellipse function at the start position (x0, y0) = (0,
ry):
222
0
22
2
22
0
0
4
1
1
2
1
1
2
1
,11
xyxy
yxyxy
ye
rrrrp
or
rrrrrp
rfp
















Contd….
• Decision parameter at sampling position (yk-1) in region2:
– Over region 2, we sample at unit steps in the negative y direction, and
the midpoint is now taken between horizontal pixels at each step.

Contd….
• For region 2 the decision parameter is evaluated as
 
  2222
2
2
2
1
1
2
1
1,2
yxkxky
kkek
rryrxr
yxfp








Decision:
xk+1 = xk if p2k > 0 , i.e., select pixel( xK,yk-1) to plot
xk+1 = xk+1 otherwise, i.e., select pixel( xK+1,yk -1 ) to plot

Contd….
• At the next sampling position (yk+1 - 1 = yk - 2), the decision
parameter for region 2 is evaluated as:
• where xk+1 is set either to xk or to xk+1depending on the sign of
p2k.
 
































22
1
222
1
2222
2
1
2
12
1
11
2
1
2
1
)1(222
]1)1[(
2
1
1,2
kkyxkxkk
yxkxky
kkek
xxrryrpp
or
rryrxr
yxfp

Contd….
• where xk+1 is set either to xk or to xk+1depending on the sign of
p2k.
– If p2k > 0, next point is (xk, yk-1) and
– else, next point is (xk+1, yk-1) and
2
1
2
1 222 xkxkk ryrpp  
2
1
2
1
2
1 2222 xkxkykk ryrxrpp  

Contd….
• In region 2, initial value of decision parameter is obtained by
evaluating the ellipse function at the initial position (x0, y0) that
is the last position selected in region 1
  222
0
2
2
0
2
0
000
1
2
1
2
1,
2
1
2
yxxy
e
rryrxrp
yxfp















Midpoint Ellipse Algorithm
1. Input rx, ry and ellipse center (xc, yc). Obtain the first point on an ellipse
centered on the origin (0, 0),
(x0, y0) = (0, ry).
1. Calculate initial value for decision parameter in region 1 as:
2
4
122
01 xyxy rrrrp 

3. At each xk in region 1, starting from k = 0, test p1k :
If p1k < 0, next point (xk+1, yk) and
else, next point (xk+1, yk-1) and
with 2ry
2xk+1 = 2ry
2xk + 2ry
2, 2rx
2yk+1 = 2rx
2yk – 2rx
2
and repeat step 3 until 2ry
2x  2rx
2y
2
1
2
1 211 ykykk rxrpp  
2
1
2
1
2
1 2211 ykxkykk ryrxrpp  

4. Calculate the initial value for decision parameter in region 2 using the last point
calculated in region 1 as:
where (x0, y0) is the last position calculate in region 1
5. At each yk in region 2, starting from k = 0, test p2k:
If p2k > 0, next point is (xk, yk-1) and
else, next point is (xk+1, yk-1) and
continue until y=0
2
1
2
1 222 xkxkk ryrpp  
    222
0
22
2
1
0
2
0 12 yxxy rryrxrp 
2
1
2
1
2
1 2222 xkxkykk ryrxrpp  

6. For both region determine symmetry points in the other 3 quadrants
7. Move each calculated pixel position (x,y) onto the elliptical path centered
on (xc,yc) and plot the coordinate values
x=x + xc, y =y + yc

Example
•Digitize an ellipse with center (0,0) and x-radius (rx )=8 and y-
radius (ry)=6, using the midpoint ellipse algorithm.
•For region 1:
–Initial point for ellipse centered at origin is (x0, y0) =(0, ry)= (0, 6)
2ry
2x = 0 and
2rx
2ry = 768
p10 = ?
–The initial value of decision parameter is:
3321
2
4
122
0  xyxy rrrrp

Contd….
• Successive decision parameter values and positions along the
ellipse path are calculated using the midpoint method as:
k p1k (xk+1,yk+1) 2ry
2 xk+1 2 rx
2yk+1
0 -332 (1,6) 72 768
1
2
3
4
5
6

Contd….
k p1k
(xk+1,yk+1) 2ry
2 xk+1 2 rx
2yk+1
0 -332 (1,6) 72 768
1 -224 (2,6) 144 768
2 -44 (3,6) 216 768
3 208 (4,5) 288 640
4 -108 (5,5) 360 640
5 288 (6,4) 432 512
6 244 (7,3) 504 384

Contd….
• For region 2, initial point is (x0, y0) = (7, 3), and initial
decision parameter is :
• The remaining pixels along the elliptical path in the first
quadrant are then calculated as:
• Now stop calculation, since y= 0 .
  23)2,7(1,2 2
1
2
1  ekkek fyxfp
k p2k (xk+1,yk+1) 2rv
2 xk+1 2 rx
2yk+1
0 -23 (8, 2) 576 256
1 361 (8, 1) 576 128
2 297 (8, 0) ---- ----

Plot pixel positions
6
5
4
3
2
1
0
0 1 2 3 4 5 6 7 8

Contd….
• Symmetric point calculation:
– For (0, 6) : (0, 6), (0, -6), (0, -6)
– For (1,6) : (-1, 6), (-1,-6), (1, -6)
– For (2,6): (-2,6), (-2, -6), (2,-6)
– For (3,6): (-3,6), (-3, -6), (3, -6)
– For(4,5): (-4, 5), (-4,-5) (4,-5)
– For (5, 5): (-5,5), (-5,-5) (5,-5)
– For (6,4):(-6,4) (-6,-4) (6,-4)
– For (7,3): (-7,3), (-7,-3), (7,-3)
– For (8,2):(-8,2), (-8,-2), (8,-2)
– For (8,1): (-8,1), (-8,-1), (8,-1)
– For (8,0): (-8,0), (-8,0), (8,0)

Homework
• Example2: Digitize an ellipse with rx=8 and ry=6, and
centered at (2,3) using the midpoint ellipse algorithm.
• Example 3: Digitize an ellipse with center (20, 20) and x-
radius= 8 and y-radius= 6.

Midpoint Ellipse algorithm implementation

Contd….

Contd….
• Output:

Thank You !

B. SC CSIT Computer Graphics Unit1.2 By Tekendra Nath Yogi

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B. SC CSIT Computer Graphics Unit1.2 By Tekendra Nath Yogi