2. Module – 1(Basics of Computer graphics and
Algorithms)
• Basics of Computer Graphics and its applications.
Video Display devices- Refresh Cathode Ray Tubes,
Random Scan Displays and systems, Raster scan
displays and systems.
• Line drawing algorithms- DDA, Bresenham’s algorithm.
Circle drawing algorithms- Midpoint Circle generation
algorithm, Bresenham’s algorithm.
60. • 1. Beam Penetration Method :
Beam Penetration Method is quite similar to the normal CRT and it uses only one electron
gun. In this, different colors of multi-layered are coated on inner surface of screen,
normally the two layers of phosphorus i.e., red and green are coated. It is a method used
for displaying color pictures that has been used with random scan monitors.
• 2. Shadow Mask Method :
Shadow Mask Method is the method which is used in raster scan system which includes
color TVs. In this the pixel is made up of three -colors. So due to three colors it uses three
electron guns one for producing each color. The colors are red, green and blue. In this the
important consideration for a color monitor is the setting of electron guns and the
phosphor dots forming a pixel.
61.
62. Line drawing algorithms
• The Line drawing algorithm is a graphical algorithm which is used to
represent the line segment on discrete graphical media, i.e., printer and
pixel-based media.
• A line contains two points. The point is an important element of a line
Equation of the straight line
We can define a straight line with the help of the following equation.
y= mx + a
• Where,
64. DDA (Digital Differential Analyzer)
• Simplest line drawing algorithm.
• Digital Differential Analyzer algorithm is also known as an incremental
method of scan conversion. Previous step results are used in the next
step.
• Digital Differential Analyzer algorithm is used to perform rasterization
on polygons, lines, and triangles etc
65.
66.
67.
68.
69.
70.
71. Algorithm of Digital Differential Analyzer (DDA) Line Drawing
Step 1: Start.
Step 2: We consider Starting point as (x1, y1), and ending point (x2, y2).
Step 3: Now, we have to calculate Δx and Δy.
Δx = x2-x1
Δy = y2-y1
m = Δy/Δx
72. Step 4: Now, we calculate three cases.
If m < 1
Then x change in Unit Interval
y moves with deviation
(xk+1, yk+1) = (xk+1, yk+m)
If m > 1
Then x moves with deviation
y change in Unit Interval
(xk+1, yk+1) = (xk+1/m, yk+1)
If m = 1
Then x moves in Unit Interval
y moves in Unit Interval
(xk+1, yk+1) = (xk+1, yk+1)
Step 5: We will repeat step 4 until we find the ending point of the line.
Step 6: Stop.
73.
74. Problems
• There is a system with resolution 640 X 480. Calculate the size of
the frame buffer to store 12 bits per pixel?
Solution
Resolution = 640 X 480
Number of bits/pixel = N = 12
Required Frame Buffer Memory = N X Resolution
= 12 X 640 X 480
Memory in bytes = (12 x 640 X 480) / 8
= 460800 Bytes
Memory in KB = 460800/1024
= 450 KB
75. There is a system with resolution 1280 X 1024. Find out the size of frame buffer in
bytes and kilobytes, if 9 bits per pixel are stored. Also find out how many colors can
be displayed?
Solution
Resolution = 1280 X 1024
Number of bits/pixel = N = 9
Required Frame Buffer Memory = N X Resolution
= 9 X 1280 X 1024
Memory in bytes = (9 X 1280 X 1024) / 8
= 1474560 Bytes
Memory in KB = 1474560/1024 = 1440 KB
• If there are n bits per pixel then we can have a total of 2n colors.
Here we have 9 bits per pixel, so,
Total number of colors = 2n = 29= 512 colors
76. There are two raster systems with resolutions of 1280×1024, and 2560×2048.
a) Tell the size of the frame buffer (in bytes) for each of these systems to store 12
bits/pixel? b) How much storage is required for each system if we store 24 bits per
pixel?
1280 x 1024 x 12 bits / 8 = 1920KB
2560 x 2048 x 12 bits / 8 = 7680KB
77. Consider two raster systems with the resolutions of 640 x 480 and 1280 x 1024.
a) How many pixels could be accessed per second in each of these systems by a
display controller that refreshes the screen at a rate of 60 frames per second?
Solution
Since 60 frames are refreshed per second and each frame consists of 640 x 480 pixels,
the access rate of such a system is (640 x 480) * 60 = 1.8432 x 107 pixels/second.
Likewise, for the 1280 x 1024 system, the access rate is (1280 x 1024) * 60 = 7.86432 x 107 pixels/second.
79. Line equation
The slope-intercept form of a line is written as
• To illustrate Bresenham’s approach, we first consider the scan conversion
process for lines with positive slope less than 1
• Pixel positions along a line path are the determined by sampling at unit x
intervals.
• Starting from the left end point( x0,y0) of a given line, we step to each
successive column(x position) and plot the pixel whose scan line y value is
closest to the line path
89. Calculate the points between the starting coordinates (9, 18) and
ending coordinates (14, 22).
Calculate ΔX and ΔY from the given input.
ΔX = Xn – X0 = 14 – 9 = 5
ΔY =Yn – Y0 = 22 – 18 = 4
Calculate the decision parameter.
Pk= 2ΔY – ΔX = 2 x 4 – 5 = 3
So, decision parameter Pk = 3
90. As Pk >= 0, so case-02 is satisfied.
Thus,
k+1 k
• P = P + 2ΔY – 2ΔX = 3 + (2 x 4) – (2 x 5) = 1
• Xk+1 = Xk + 1 = 9 + 1 = 10
• Yk+1 = Yk + 1 = 18 + 1 = 19
Similarly, Step-03 is executed until the end point is reached or number of iterations
equals to 5 times.
(Number of iterations = ΔX =5 times)
x y p
9 18 3
10 19 1
11 20 -1
12 20 7
13 21 5
14 22 3
91. Calculate the points between the starting coordinates (20, 10) and ending coordinates (30, 18).
dx=10 dy=8 2dy=16
Pk=2dy-dx= 16-10=6
2dy-2dx= 16-20 = -4
x y p
20 10 6
21 11 2
22 12 -2
23 12 14
24 13 10
25 14 6
26 15 2
27 16 -2
28 16 14
29 17 10
30 18 6
109. Píoblem-
02:
Given the centre point coordinates (4, 4) and radius as 10, generate all
the points to form a circle.
Given-
Centre Coordinates of Circle (X0, Y0) = (4, 4)
Radius of Circle = 10
The following table shows the generation of points
for Quadrant-1-
Xplot = Xc + X0 = 4 + X0
Yplot = Yc + Y0 = 4 + Y0
110.
111.
112. Bresenham’s Circle Drawing Algorithm
Bresenham’s algorithm is also used for circle drawing.
It is known as Bresenhams’s circle drawing algorithm.
Let us assume we have a point p (x, y) on the boundary of the circle and with r radius
satisfying the equation fc (x, y) = 0
113. We assume,
The distance between point P3 and circle boundary = d1
The distance between point P2 and circle boundary = d2
Now, if we select point P3 then circle equation will be-
d1 = (xk +1)2 + (yk)2 – r2 {Value is +ve, because the point is
outside the circle}
if we select point P2 then circle equation will be-
d2 = (xk +1)2 + (yk –1)2 – r2 {Value is -ve, because the point is
inside the circle}
Now, we will calculate the decision parameter
(dk) = d1 + d2
dk =(xk +1)2 + (yk)2 – r2 + (xk +1)2 + (yk –1)2 – r2
= 2(xk +1)2 + (yk)2+ (yk -1)2 – 2r2 ……… (1)
114. If
dk < 0
then
Point P3 is closer to circle boundary, and the final coordinates are-
(xk +1, yk) = (xk +1, yk)
If
dk >= 0
then
Point P2 is closer to circle boundary, and the final coordinates are-
(xk +1, yk) = (xk +1, yk -1)
Now, we will find the next decision parameter (dk+1)
(dk+1) = 2(xk+1 +1)2 + (yk+1)2+ (yk+1 -1)2 – 2r2 …………… (2)
Now, we find the difference between decision parameter equation (2) – equation
(1)
(dk+1) – (dk) = 2(xk+1+1)2 + (yk+1)2+ (yk+1 –1)2 – 2r2 – 2(xk +1)2 + (yk)2+ (yk– 1)2 –
2r2
(dk+1) = dk + 4xk + 2(yk+1
2– yk
2) –2 (yk+1 – yk) + 6
117. Algorithm of Bresenham’s Circle Drawing
Step 1: Start.
Step 2: First, we allot the starting coordinates (x1, y1) as follows-
x1 = 0
y1 =r
Step 3: Now, we calculate the initial decision parameter d0 –
d0 = 3 – 2 x r
Step 4: Assume,the initial coordinates are (xk, yk)
The next coordinates will be (xk+1, yk+1)
Now, we will find the next point of the first octant according to the value of the decision parameter
(dk).
Step 5: Now, we follow two cases-
Case 1: If
dk < 0
then
xk+1 =xk + 1
yk+1 =yk
dk+1 = dk + 4xk+1 + 6
Case 2: If
dk >= 0
then
xk+1 =xk + 1
yk+1 =yk –1
dk+1 = dk + 4(xk+1 – yk+1)+ 10
Step 6: If the center coordinates (x1, y1) is not at the origin (0, 0), then we will draw the points as follow-
X coordinate = xc + x1
y coordinate = yc + y1 {xc andyc representsthe current value of x and y coordinate}
Step 7: We repeat step 5 and 6 until we get x>=y.
Step 8: Stop.
118. Example
The radius of a circle is 8, and center point coordinates are (0, 0). Apply
bresenham’s circle drawing algorithm to plot all points of the circle.
Solution:
Step 1: The given stating points of the circle (x1, y1) = (0, 0)
Radius of the circle (r) = 8
Step 2: Now, we will assign the starting point (x1, y1) as follows-
x1 = 0
y1 = r (radius) = 8
Step 3: Now, we will calculate the initial decision parameter (d0)
d0 = 3 – 2 x r
d0 = 3 – 2 x 8
d0 = -13
Step 4: The value of initial parameter d0 < 0. So, case 1 is satisfied.
Thus,
xk+1 =xk + 1 = 0 + 1 = 1
yk+1 =yk = 8
dk+1 = dk + 4xk+1 + 6 = –13 + (4 x 1) + 6 = –3
Step 5: The center coordinates are already (0, 0) so we will move to next step.
Step 6: Follow step 4 until we get x >= y.
