12112.pdf

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)

(ISO/IEC – 27001 – 2005 Certified)

WINTER – 2012 EXAMINATION

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Subject Title: Computer Graphics Subject Code: 12112

Computer Engineering Group Page 1 of 28

Q.1 Attempt any Six of the Following (6 x 2 = 12) a. What is polygon? What are types of Polygon?

*(1 mark for definition; 1 mark for type’s list) Ans: Definition: - A closed plane figure is having three or more sides is called as polygon.

Triangles, rectangles, and octagons are all examples of polygons. A regular polygon is a polygon all of whose sides are the same lengths and all of whose interior angles are the same measures. Types of Polygon The polygons are divided into two categories as, 1) Convex Polygon. 2) Concave Polygon. 1) Convex Polygon:-

A convex polygon is a polygon in which the line segment joining any two points the polygon lies completely inside the polygon. So it is very easy to perform inside test on a pixel to be filled with given color.

Fig (a) Convex Polygon 2) Concave Polygon: -

A concave polygon is a polygon in which the line segment joining any two points within the polygon may or may not lay completely inside the polygon.

Fig (b) Conclave Polygon

b Give general criteria applied for line drawing algorithm. *(All four criteria are expected ½ mark for each criteria)

Ans: General criteria of any line drawing algorithm are as follow. 1. Line must appear straight. 2. Line must start and end properly. 3. Line should be drawn at rapid pace. 4. Line must have constant intensity to their length.

c. Define the terms windowing and clipping

*(1 mark for windowing and 1 mark for clipping) Ans: i Windowing: -A rectangular area specified in world co-ordinates is defined as window. The

window defines what is to be viewed. The window is generally rectangle in size.

ii. Clipping:- The method which determines the portions of a picture that is either inside or outside of a specified region of space is known as clipping. The region against which an object is to be clipped is called a clip window or clipping window. Normally clipping window




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is in a rectangular shape. This algorithm identifies which points, lines or portions of lines lies within the clipping window. These points, lines or portions of lines are displayed and remaining points, portion is discarded. This visible and invisible portions of picture is depend on whether elements are inside or outside to given region.

d. List any four properties of Bezier curve

*(Any 4 properties are expected ½ mark for one property) Ans: Properties of Bezier Curve

1. It always passes through first and last control point. 2. Tangent vectors at ends of the curve have same direction as first and last polygon

spans respectively. 3. It follows the shape of defining polygon. 4. It is contained within the convex hull of defining polygon. 5. Degree of polynomial defining the curve segments is equal to the total number of

control points minus 1. 6. Order of polynomial defining the curve segments is equal to the total number of control

points 7. Basic functions of Bezier curve are real.

e. List some graphics standards

*(four standards are expected ½ mark for one standard) Ans: 1. CORE

2. Programmer‟s Hierarchical Interactive Graphical Standard (PHIGS) 3. Initial Graphics Exchange Standard (IGES) 4. Computer Graphics Metafile Standard (CGM) 5. Virtual Device Metafile (VDM) 6 Virtual Device Interface (VDI)

f. Draw labeled diagram of shadow mask CRT.

*(1 mark for diagram, 1 mark for labeling) Ans:

Fig Shadow Mask CRT.




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g. Describe any one graphics mode graphics function with syntax. *(Any 1 function is expected; 1 mark for explanation, 1 mark for syntax)

Ans: Initgraph( ) : - It is used to initialize graphic mode. Syntax:- initgraph(int driver, int mode, char path);

Driver:- this argument gives the graphics driver to be used and it interfaces with display adapter. Some of available graphics drivers are CGA, EGA, VGA etc.

Closegraph( ):- It is used to close graphics mode, when one exit from graphics mode one shall restore the system to the previous display mode, closegraph function restores the previous display mode.

Syntax:- Closegraph( );

h. State two merits and demerits of DDA line drawing algorithm *(1 mark for merit and 1 mark for Demerit)

Ans: Merits of DDA:- It is simple algorithm

It is faster Demerits of DDA: - Floating point arithmetic in DDA algorithm is time consuming

Accumulation of round-off error in successive additions of floating point increment can cause the calculated point to drift away from the actual line path.

B) Attempt any two of the following ( 2 x 4 = 8) a. Explain 2D-transformation matrix for rotation about arbitrary point.

*(2 mark for explanation; 1 mark for diagram; 1 mark for matrix) Ans: Arbitrary point.

An arbitrary point or pivot point is one which is any point other than the origin. Since rotation of an object is done with respect to origin only hence to perform rotation about arbitrary point it shall be performed by shifting an object to the origin and then perform rotation after that the object is again shifted back to its original position. To perform rotation of an object about any pivot point P following set of operation is done with given sequence. Translate: In this operation an object is translated to origin from arbitrary point i.e. the base of an object is shifted to origin. Rotate: in this operation, rotate an object at give given angle. Translate: In this operation an object is translated back to arbitrary point from origin i.e. the base of an object is shifted to arbitrary point.




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Fig (a) Before Rotating of Object Fig (b) After Translation of an object to origin.

Fig (c) after rotating Fig (d) Translation of an object an object to origin. to original point

Fig gives an idea about rotation operation about arbitrary point in graphics. Fig (a) shows object at its original position. When rotation starts, the object has to be translated at origin as shown in fig (b). After that the object is rotated at given angle which is shown in fig (c). In last step the object is again moved back to its original position by performing translation given in fig (d).

1 0 0 cos sin 0 1 0 0

[ T1.R.T2] = 0 1 0 -sin cos 0 0 1 0 -Xp -Yp 1 0 0 1 Xp Yp 1

cos sin 0 1 0 0

= -sin cos 0 0 1 0

-Xp cos + Yp sin -Xp sin - Yp Cos 1 Xp Yp 1

cos sin 0

= -sin cos 0

-Xp cos + Yp sin + Xp -Xp sin - Yp Cos + Yp 1 1

This is how rotation about arbitrary point can be performed.




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b. Explain 3D translation and scaling transformation matrix *(2 marks for 3D translation and 2 marks for Scaling)

Ans: Translation: - In Euclidean geometry, a translation refers to moving every point at a constant distance in a specified direction. It is one of the rigid motions. A translation can also be interpreted as an addition of a constant vector to every point or shifting the origin of the coordinate system. If v is a fixed vector, then the translation Tv will work as

Tv (P) = P + v Translation is an affine transformation but not a linear transformation. Homogeneous coordinates are normally used to represent the translation operator by a matrix. Thus we write a 3D vector w= (Wx Wy Wz) using four homogeneous coordinates co-ordinates as w= (Wx Wy Wz 1) To translate an object by a vector v = [Tx Ty Tz], each homogeneous vector P = [x y z 1] would need to be multiplied by this translation matrix.

1 0 0 0 [ T] = 0 1 0 0

0 0 1 0 Tx Ty Tz 1

1 0 0 0 [ T * P] = [X Y Z 1] * 0 1 0 0

0 0 1 0 Tx Ty Tz 1

= [x + Tx y + Ty z + Tz 1] = P + v.

For Example:- Fig (a) Before Translation Fig (b) After Translation




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Scaling Transformation:- A scaling can be represented by a scaling matrix. To scale an object by a vector V= [Sx, Sy, Sz], each point P = [x y z] would need to be multiplied with this scaling matrix.

Sx 0 0 Sv = 0 Sy 0 0 0 Sz

Sx 0 0 P * Sv = [ X Y Z] * 0 Sy 0 = [SxX SyY SzZ ]

0 0 Sz

Such a scaling changes the diameter of an object by a factor between te scale factors, the area by a factor between the smallest and the largest product of two scale factors, and the volume by the product of all three. In homogeneous coordinates, since translation cannot be accomplished with a 3 X 3 matrix. To scale an object by a vector V= [Sx Sy Sz ] each homogeneous vector P = [X Y Z 1] would need to be multiplied with the scaling matrix.

Sx 0 0 0

Sv = 0 Sy 0 0 0 0 Sz 0

0 0 0 1

Sx 0 0 0

P * Sv = [ X Y Z 1] * 0 Sy 0 0 = [SxX SyY SzZ 1] 0 0 Sz 0

0 0 0 1

Fig (a) Before Scaling Fig (b) After Scaling




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c. Write a C code for scaling a line *(A demo code is given any relevant code shall be consider and assessed)

Ans: #include<stdio.h> #include<conio.h>

#include<graphics.h> void main() {

int gd=DETECT,gm=DETECT,sx,sy; clrscr();

printf("\nEnter Scaling factors (SX and SY)"); scanf("%d%d",&sx,&sy); initgraph(&gd,&gm,"D:\\TC\\BGI"); line(100,100,200,200); getch(); line(100,100,200*sx,200*sy); getch(); }

Q.2. Attempt any four of the following ( 4 x 4 = 16)

a. Write a C-program to draw a circle filled with blue color. *(A demo code is given any relevant code shall be consider and assessed)

Ans: #include<stdio.h> #include<conio.h> #include<graphics.h> void main()

{ int gd=DETECT,gm=DETECT,sx,sy; clrscr(); initgraph(&gd,&gm,"D:\\TC\\BGI"); circle(320,240,100); floodfill(320,240,1); getch(); }

b. Describe working of Direct View Storage tube (3 marks for explanation of DVST 1 mark for diagram) Ans: In Raster display and all refresh display one may need to refresh the screen to maintain

consistent image on screen else flickering occurs. An alternative solution to these display devices with inherent image storage capability, which stores image in display device for displaying. DVST uses a storage grid which stores picture information as a charge distribution just behind phosphor coated screen. It consists of two electron guns as writing gun and flood gun. Writing gun stores picture pattern as a positive charge on storage grid. This picture pattern is transferred to phosphor by continuous flood of electron generated from flood electrons. Electrons pass through




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collector at low speed and are attracted by positively charged picture pattern on storage grid and are repelled by rest. The attracted electrons by positive picture pattern pass right through it and strike on phosphor making it visible on screen.

c. Explain any two graphics file formats

*(Any two graphics file formats are expected 2 marks for each) Ans: BMP

Bitmap. This was probably the first type of digital image format that I can remember. Every picture on a computer seemed to be a BMP. In Windows XP the Paint program saves its images automatically in BMP, however now in Windows Vista images are saved to JPEG. BMP is the basis platform for many other file types.

JPG, JPEG

(Joint Photographic Experts Group) Jpeg format is used for color photographs, or any pictures with many blends or gradients. It is not good with sharp edges and tends to blur them a bit. This format became popular with the invention of the digital camera. Most, if not all, digital cameras download photos to your computer as a Jpeg file. Obviously the digital camera manufacturers see the value in high quality images that ultimately take up less space.

GIF

(Graphics Interchange Format) Gif format is best used for text, line drawings, screen shots, cartoons, and animations. Gif is limited to a total number of 256 colors or less. It is commonly used for fast loading web pages. It also makes a great banner or logo for your webpage. Animated pictures are also saved in GIF format. For example, a flashing banner would be saved as a Gif file.

PNG

(Portable Networks Graphic) Apparently this is one of the best image formats; however it was not always compatible with all web browsers or image software. Now days it is the best image format to use for website. I use .png for logos and screen shots.




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TIFF (Tagged Image File Format) This file format has not been updated since 1992 and

is now owned by Adobe. It can store an image and data (tag) in the one file. TIFF also stores files with virtually no compression and therefore is good for storing images that need to be edited and re-saved without suffering a compression loss. This file is commonly used for scanning, faxing, word processing, and so on. It is no longer a common file format to use with your digital photos as a jpeg is great quality and takes up less space.

d. Explain the syntax and use of following methods:

*(2 marks for ellipse and 2 marks for rectangle) Ans:

i. ellipse( ) ellipse draws an elliptical arc fillellipse draws and fills an ellipse sector draws and fills an elliptical pie slice

Declaration Syntax:

void far ellipse(int x, int y, int stangle, int endangle, int xradius, int yradius); void far fillellipse(int x, int y, int xradius, int yradius); void far sector(int x, int y, int stangle, int endangle, int xradius, int yradius);

Remarks:

ellipse draws an elliptical arc in the current drawing color. fillellipse draws an ellipse, then fills the ellipse with the current fill color and fill pattern. sector draws and fills an elliptical pie slice in the current drawing color, then fills it using the pattern and color defined by setfillstyle or setfillpattern.

Argument

(x,y) Center of ellipse xradius Horizontal axis yradius Vertical axis stangle Starting angle endangle Ending angle

The ellipse or sector travels from stangle to endangle.

If stangle = 0 and endangle = 360, the call to ellipse draws a complete ellipse.The linestyle parameter does not affect arcs, circles, ellipses, or pie slices. Only the thickness parameter is used.




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ii. rectangle( ) rectangle Draws a rectangle (graphics mode)

Declaration Syntax:

void far rectangle(int left, int top, int right, int bottom); Remarks: rectangle draws a rectangle in the current line style, thickness, and drawing color.

(left,top) is the upper left corner of the rectangle, and (right,bottom) is its lower right corner.

e. Draw and explain rotating memory frame buffer used for raster display

*(2 marks for diagram and 2 marks for explanation) Ans: Frame buffer is a large part of computer memory used to store display image. Different kind

of memory can be used for frame buffers like drums, disk or IC – shift registers. To generate a pixel of desired intensity to read the disk or drum. The information stored in disk or drum is in digital for, hence it is necessary to convert it into analog from using DAC and then this analog signal is used to generate the pixel. If only one bit is used to generate the pixel then only black and white picture is possible. The fig show three tracks used in parallel to provide a frame buffer with eight different intensity of pixel. Here it is necessary to read the disk or drum again and again to refresh the display. Therefore the rotating speed of drum is made to coincide with the refresh rate of the screen. Latency problem puts a limitation on these frame buffers as they require more time.

Digital Information Electronic gun 3 Bit Analog Screen DAC Signal

DISK OR Drum

D A C




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f. Given a window whose lower left corner is at (-3 , 1) and upper right corner is at (2, 6). Find the region code for following end points. (Two marks for each region)

Ans: From given data for window Xmin = -3 Ymin = 1 Xmax = 2 Ymax = 6

Bit 1 = 1 if y > Ymax

Bit 2 = 1 if y < Ymin Bit 3 = 1 if x > Xmax Bit 4 = 1 if x < Xmin

i. A( – 4 , 2) B( – 1 , 7) For line segment having end points as (– 4, 2) Substituting values in above four conditions we get, Code for A = 0001 For line segment having end points as (– 4, 2)

Substituting values in above four conditions we get, Code for B = 1000

ii. A( – 5 , 7) B( – 2 , 10)

For line segment having end points as (– 5, 7) Substituting values in above four conditions we get, Code for A = 1001 For line segment having end points as (– 2, 10) Substituting values in above four conditions we get, Code for B = 1000




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Q.3. Attempt any FOUR of the following. (4 x 4 = 16)

a) Write C-code for DDA circle drawing algorithm.

(Relevant code – 4Marks) Ans: #include <stdio.h>

#include <conio.h>

#include <math.h>

#include <graphics.h>

Void main ()

{

float x1, y1, x2, y2, x_start, y_start, epsilon;

int gd, gm, i, val;

int rad;

clrscr ();

printf (“\t\t CIRCLE DRAWN BY DDA ALGORITHM \n”);

printf (“\n ENTER THE RADIOUS OF CIRCLE: \n”);

scanf (“%d”, &rad);

detectgraph (&gd, &gm);

initgraph (&gd, &gm, “D:\TC3”);

x1 = rad*cos (0);

y1 = rad*sin (0);

x_start=x1;

y_start=y1;

i=0;

do

{

val=pow(2, i-1);

i++;

} while (val<rad);

epsilon=1/pow (2,i-1);

do

{

x2=x1+y1*epsilon;

y2=y1-epsilon*x2;

putpixel (200+x2, 200+y2,10);

x1=x2;

y1=y2;

} while ((y1-y_start) <epsilon II (x1-x_start) >epsilon);

getch ();

}




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b) Explain floodfill algorithm.

(Complete algorithm – 4Marks)

Ans: Flood fill algorithms starts from any speed point inside polygon & set it to fill colour.

In Flood Fill method user starts from any point inside polygon & fills the colour to the

neighbouring pixel till polygon boundary is reached. This method has one limitation that is

the polygon has any point of different colour then this method stops colouring the

neighbouring pixel.

Flood fill algorithm

Step 1: Read any seed pixel position (x, y)

Step 2: Check to see if this pixel (x, y) has old interior colour. If old colour, then set it to

new fillcolor.

Step 3: Repeat step 2 for neighbouring pixels of (x, y)

Step 4: If pixel (x, y) does not has old interior colour then stop. (Getpixel, Putpixel and

Flood fill statement can be used to increment the neighbouring pixel).

c) Describe inside/outside test used in polygon filling with example.

(Diagram – 2 Marks, Explanation – 2 Marks)

Ans: Once the polygon is entered in the display file, we can draw the outline of the polygon. To

show polygon as a solid object we have to set pixel inside the polygon as well as pixels on

the boundary of it. To determine whether or not a point is inside of a polygon, the

inside/outside test is performed.

Simple method of doing this is to construct a line segment between the point in question

and point known to be outside the polygon, as shown in fig. Now count how many

intersections of the line segments with the polygon boundary occur. If there are an odd

number of intersections, then the point in question is inside; otherwise it is outside. This

method is called the even-odd method (inside/outside test) of determining polygon

inside points.

The inside/outside test fails, if the intersection point is vertex of the polygon. Then we

have to look at the other endpoints of the two segments which meet at this vertex. If

these points lie on the same side of the constructed line, then the point in question counts

as an even number of intersections. If they lie on opposite sides of the constructed line,

then the point is counted as a single intersection.




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d) Write Bresenham line drawing algorithm.

(Relevant algorithm – 4Marks)

Ans:

P next = P0 + 2dy – 2dx




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e) Write midpoint circle algorithm.

(Relevant algorithm – 4Marks) Ans: Step1: Read the radius „r‟ of the circle

Step2: x=0, y=r

Step3: Calculate initial decision parameter P=1-r

Step4: do

{ putpixel (x, y)

If (P<0)

{

x=x+1

y=y

P= P+2x+1

}

Else

{

x=x+1

y=y-1

P=P+2(x-y) +1

}

} while(x<y)

Step5: Determine symmetrical points in other octants also.

Step 6: STOP

f) What is difference between VGA and SVGA?

(Any 4 valid differences-1 Mark each)

Ans:

VGA SVGA

1. Video graphics array used for Analog signals

1. Super Video graphics array used for Analog signals

2. Introduced by IBM in 1987 2. Introduced by IBM (in 1989) to provide inter operability and different standards

3. It has limited colours (256) 3. It provides various colour range – 16 million colours.

4. It has resolution 640 x 480. 4. It supports resolution 800 x 600 and 1024 x 768

5. It is compatible with graphics and text 5. It has better image quality




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Q.4. Attempt any TWO of the following. (2 x 8 =16)

a) Write C-code for DDA line drawing algorithm.

(Relevant code – 8Marks)

Ans:




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b) Explain midpoint subdivision line clipping algorithm.

(Diagram – 4 Marks, Explanation – 2 Marks, Algorithm – 2Marks)

Ans: Midpoint subdivision line clipping algorithm is as below:

Step1: Accept two end points of line say P1 (x1, y1) and P2 (x2, y2)

Step2: Read lower left and upper right co-ordinates of clipping window.

Step3: Assign 4-bit region code for points P1 and P2.

Step4: Check the category of this line for its visibility.

If line is completely visible draw it and stop.

If line is completely out of clipping window then discard it.

If the line is partially visible then continue the next step.

Step5: Find midpoint of line and divide it into two equal line segments.

Step6: For both divided line segments check the region code for visibility.

Repeat step 5 for partially visible line. And then repeat step 4, 5, and 6 till complete

line is divided into totally visible and invisible line segment.

The process is explained by following figures.




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c) Explain Hibert’s curve and Koch curve with example

(Hibert’s curve - Explanation 2 Marks & Diagram 2 Marks, Koch curve - Explanation 2

Marks & Diagram 2 Marks)

Ans: Hibert’s curve:

The Hibert‟s curve is built by successive approximation technique.

If square is divided into four quadrants, we can draw the first approximation of Hibert‟s

curve by connecting the centre point of each quadrant.

For 2nd approximation Hibert‟s curve can be drawn by further subdividing each of the

quadrants and connecting their centres before moving to the next major quadrant.

Such more further approximations level can be increased to achieve the accuracy of the

Hibert‟s curve.

The special property of Hibert‟s curves it that it is infinite in length when further

approximations are mode. As well it passes through each quadrant but it never crosses

the existing lines.

It is used in polygon filling object or to draw mountain shape object.

Koch curve:

The Koch curve can be drawn by dividing a line into four equal segments with scaling

factor 1/3 and middle two segments are so adjusted that they form adjacent sides of an

equilateral triangle as shown below.

For second approximation we repeat the procedure four times to draw a Koch curve.

The curve can be further extended by forming more number of approximations.

The curve does not fill any area. As well the Koch curve does not much deviate from its

original shape.




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Q.5. Attempt any FOUR of the following. ( 4 x 4 = 16) a. Describe features of GUI (1 M for each feature, Consider any 4 ) Ans: The graphical user interface, or “GUI”, is a computer interface that uses graphic icons and

controls in addition to text. Features of GUI

1. Pointer: A symbol that appears on the display screen and that you move to select object and commands.

2. Pointing device: A device, such as a mouse or trackball, that enables you to select objects on the display screen.

3. Icons: Small pictures that represent commands, files or windows. By moving the pointer to the icon and pressing a mouse button, you can execute a command or convert the icon into a window.

4. Desktop: the area on the display screen where icons are grouped is often referred to as the desktop because the icons are intended to represent real objects on a real desktop.

5. Windows: You can divide the screen into different areas. In each window, you can run a different program or display a different file.

6. Menus: Most graphical user interfaces let you execute commands by selecting a choice from a menu.

b. Describe any one of graphics standards. (Consider any one for 4 M) Ans: 1. CORE

2. Programmer‟s Hierarchical Interactive Graphical Standard (PHIGS) 3. Initial Graphics Exchange Standard (IGES) 4. Computer Graphics Metafile Standard (CGM) 5. Virtual Device Metafile (VDM) 6. Virtual Device Interface (VDI)

PHIGS: - PHIGS (Programmer's Hierarchical Interactive Graphics System) is an API standard for rendering 3D computer graphics, at one time considered to be the 3D graphics standard for the 1990s. Instead a combination of features and power led to the rise of OpenGL, which became the most popular professional 3D API of the 1990s. PHIGS is still widely used in the computer games and film industries. PHIGS was available as a standalone implementation (examples: Digital Equipment Corporation's DEC PHIGS, IBM's graPHIGS, Sun's SunPHIGS) and also used with the X Window system, supported via PEX, the "PHIGS Extension to X". PEX consisted of an extension to X, adding commands that would be forwarded from the X server to the PEX system for rendering. Workstations were placed in windows typically, but could also be forwarded to take over the whole screen, or to various printer-output devices. PHIGS was designed in the 1980s, inheriting many of its ideas from the Graphical Kernel System of the late 1970s, and became an ANSI (ANSI X3.144-1988), FIPS (FIPS 153) and then ISO standard (ISO/IEC 9592 and ISO/IEC 9593) by 1989. Due to its early gestation, the standard supports only the most basic 3D graphics, including basic geometry and




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meshes, and only the basic Gouraud, "Dot", and Phong shading for rendering scenes. Features considered "standard" today, notably texture mapping, were not supported, nor were many machines of the era physically capable of it (at least in real-time).

IGES:- The Initial Graphics Exchange Specification (IGES) (pronounced eye-jess) is a file format which defines a vendor neutral data format that allows the digital exchange of information among Computer-aided design (CAD) systems. The official title of IGES is Digital Representation for Communication of Product Definition Data, first published in January, 1980 by the U.S. National Bureau of Standards as NBSIR 80-1978. Many documents (like early versions of the Defense Standards MIL-PRF-28000 and MIL-STD-1840) referred to it as ASME Y14.26M, the designation of the ANSI committee that approved IGES Version 1.0. Using IGES, a CAD user can exchange product data models in the form of diagrams, wire frame, freeform surface or solid modelling representations. Applications supported by IGES include traditional engineering drawings, models for analysis, and other manufacturing functions. An IGES file is composed of 80-character ASCII records, a record length derived from the punched card era. Text strings are represented in "Hollerith" format, the number of characters in the string, followed by the letter "H", followed by the text string, e.g., "4HSLOT" (this is the text string format used in early versions of the FORTRAN language). Early IGES translators had problems with IBM mainframe computers because the mainframes used EBCDIC encoding for text, and some EBCDIC-ASCII translators would either substitute the wrong character, or improperly set the Parity bit, causing a misread. Here is a very small IGES file from 1987, containing only two POINTS (Type 116), two CIRCULAR ARC (Type 100), and two LINE (Type 110) entities. It represents a slot, with the points at the centers of the two half-circles that form the ends of the slot, and the two lines that form the sides. CGM:- The Computer Graphics Metafile (CGM) is the International Standard for storage and exchange of 2D graphical data. Although initially a vector format, it has been extended in 2 upwardly compatible extensions to include raster capabilities and provides a very useful format for combined raster and vector images. A metafile is a collection of elements. These elements may be the geometric components of the picture, such as polyline or polygon. They may be details of the appearance of these components, such as line colour. They may be information to the interpreter about how to interpret a particular metafile or a particular picture. The CGM standard specifies which elements are allowed to occur in which positions in a metafile. CGM also has profile rules and a Model Profile to attempt to solve the problem of flavours of standards. 4 Internationally Standardised Profiles (ISPs) have been developed for CGM. CGM has been accepted as a MIME data type.




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Type Metafile

Colors Unlimited

Compression RLE, CCITT Group3 and Group4

Maximum Image Size Unlimited

Multiple Images Per File Yes

Numerical Format NA

Originator ANSI, ISO

Platform All

Supporting Applications Too many to list

CORE:- The 3D Core Graphics System (a.k.a. Core) was the very first graphical standard ever developed. A group of 25 experts of the ACM Special Interest Group SIGGRAPH developed this "conceptual framework". The specifications were published in 1977 and it became a foundation for many future developments in the field of computer graphics.

c. What is open GL? Give basic drawing primitives of open GL. Enlist Features of open

GL. (For definition 2 M for primitives 1 M and for features 1 M) (Question is out of syllabus and any relevant answer mark should be given) Ans: “OpenGL is a software interface to graphics hardware.” This interface consists of about 150

distinct commands that you use to specify the objects and operations needed to produce interactive three dimensional application Basic drawing primitives of open GL

Points

Lines

Polygon

Triangle

Quad

Quad strip

Triangle strip

Triangle fan Features of OpenGL

Geometric Primitives

Colour coding

Viewing and Modeling

Texture mapping

Materials lighting




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Double buffering

Anti-aliasing

Z-buffering

Atmospheric effects

d. Describe shearing transformation. (For Diagram 1 M and Description 3 M) Ans: It is a transformation which slants or bends an object to specified direction. There are two

types of shearing transformation available in computer graphics. One which slants x coordinate values is known as X shearing and one that slants y coordinate values is known as Y shearing. Irrespective of shearing only one co-ordinate is change its coordinate and other values are same.

X-Shearing:- It is a process by which x coordinates are change to specified direction. In X shearing the values of y coordinates are kept same and values of x coordinates are change as a result of which vertical lines are tilt to right or left which results into slanting of entire object to one direction i.e. horizontally only.

Y Y

X X Fig Before Shearing of Object Fig After Shearing of object

Fig gives an idea about Y-shearing operation in computer graphics. Fig. gives an object without shear operation on it object look like a rectangle. When X-shearing operation is used on it object get slanted towards horizontally only. Comparing objects in both figures the base point is remains same just object gets slanted. The matrix for X-shearing is given by

1 0

X_sh = Shx 1




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Y Shearing: - It is a process by which y coordinates are change to specified direction. In Y shearing the values of x coordinates are kept same and values of y coordinates are change as a result of which horizontal lines are slanted to upward or downward which results into slanting of entire object to vertically only. Fig Before Shearing of Object Fig After Shearing of object Fig. gives an idea about Y-shearing operation in computer graphics. Fig gives an object without shear operation on it object look like a rectangle. When Y-shearing operation is used on it object get slanted towards vertically only. Comparing objects in both figures the base point is remains same just object gets slanted similar X-shearing. 1 0

Y_sh = ShY 1 e. Write C-code for translation transformation of 2D object. (Step Marks to be given 4 M ) Ans: #include<stdio.h>

#include<conio.h> #include<graphics.h> void main() { int gd=DETECT,gm=DETECT,tx,ty; clrscr(); printf("\nEnter Translation factors (TX and TY)"); scanf("%d%d",&tx,&ty); initgraph(&gd,&gm,"D:\\TC\\BGI"); line(100,100,200,200); getch(); line(100+tx,100+ty,200+tx,200+ty); getch(); }




Model Answer



Q.6. Attempt any FOUR of the following. ( 4 x 4 = 16) (Fractal line 2 M, Fractal surfaces 2 M) a. Describe fractal lines and fractal surfaces. Ans: Generally, object which is having smooth surface and regular shape are described by using

equation but natural object such as mountain, tree, waves and clouds have irregular shapes and it is very difficult to draw these shapes by using normal equation. There are may method of modeling these object but one of the important is by using fractal. From certain distance we will see a line as a simple, quite smooth line but as we go near to that lines. It will appear more rough.

We can draw coastline or lightning bolt by fractal lines. But if we want to draw a 3D object like, say mountain, then we have to use fractal surfaces. The concept of fractal lines can be extended to generate fractal surfaces. There are many by which we can do this. Here we are explaining a method which is based on triangle.




Model Answer



The fractal surfaces can be generated by performing following steps:

We will use fractal line algorithm to each edge of the triangle.

Compute its halfway point by the same logic as that of fractal lines.

Connect these halfway points by line segments. By doing this we are now having four small triangles. Now we can call this procedure recursively to each of small triangle until the triangles becomes too small.

b. Write short note Bezier curve. (For Diagram 1 M, Description 3 M) Ans: Bezier curve is an approach of constricting curve. It is determined by defining a polygon.

Bezier curve have properties that makes it highly useable and ease for curve design. These properties are also easy to implement. Hence Bezier curve is widely available in various CAD systems. This curve offers good flexibility and also avoids complex calculation. In this, four points are used to give complete curve. There is no addition of intermediate points like B-spline curve and also smoothly extensions of Bezier curve by picking four more points and construct second curve which can be join to first one.

Subsection of Bezier curve D

CD

C

BCD

BC ABCD

ABC

B

AB

A Fig. Subsection of Bezier Curve




Model Answer



Algorithm: - Step 1: Get four control points A (XA, YA), B (XB, YB), C (XC, YC), D (XD, YD) Step 2: Divide the curve represented by points A, B, C, and D in two sections

XAB = (XA + XB) / 2; YAB = (YA + YB) / 2; XBC = (XB + XC) / 2; YBC = (YB + YC) / 2; XCD = (XC + XD) / 2; YCD = (YC + YD) / 2; XABC = (XAB + XBC) / 2; YABC = (YAB + YBC) / 2; XBCD = (XBC + XCD) / 2; YBCD = (YBC + YCD) / 2; XABCD = (XABC + XBCD) / 2; YABCD = (YABC + YBCD) / 2;

Step 3: Repeat step 2 for section A, AB, ABC, and ABCD and section BCD, CD, D. Step 4: Repeat step 3 until section so short that they can be replace by straight lines. Step 5: Replace small sections by straight lines. Step 6: Stop.

c. How category of line is find out far its visibility using region code in cohen

Sutherland line clipping algorithm? (For region code 2 M and Description 2 M) Ans: Identify category of line and identify those lines which intersect the clipping window

category 3 and hence need to be clipped. Assign a 4-bit code called as region code to each endpoint of the line. These codes identily the location of the point relative to the boundaries of the clopping window. Compute 4-bit code. Starting from leftmost bit, each bit of code is set to 1 according to following scheme,

Bit 1: 1, if point is above window, i.e y > WT. (Ymax) Bit 2: 1, if point is below window, i.e y < WB. (Ymin) Bit 3: 1, if point is right of window, i.e x < WR (Xmax) Bit 4: 1, if point is left of window, i.e x < WL. (Xmin) All bits 0 i.e. 0000 – if point is inside window. 1001 1000 1010 WT 0001 Window 0010 0000 WB 0101 0100 0110

WL WR

Four bit region codes




Model Answer



Since codes are assigned to both end points of line them. Category1. The line is visible if region codes of both end points of line are 0000. i.e.

completely inside window. Here accept whole line to display.

Category2. The line is not visible i.e. line completely outside of window, if bitwise logical AND of codes of both end points is not 0000 and thus reject the line and do not display. Category3. If bitwise logical AND of the region codes of both end points is 0000 then it is a line intersecting boundaries and needs clipping and pass this line to phase 2 for clipping.

d. Explain viewing transformation and normalisation transformation. (Viewing 2 M, Normalisation 2 M) Ans: Viewing transformation Pictures are defined any stored into memory using any

convenient Cartesian co-ordinate System is called as World Co-ordinate Systems(WCS). When picture is to be displayed on display device it is measured in physical Device Co-ordinate System (PDCS) corresponding to the display device. Therefore to display any image on screen, we need to map it from stored World Co-ordinate Systems to appropriate Physical Device Co-ordinate system (PDCS). Therefore in general the process of mapping of picture from World Co-ordinate systems to Physical Device Co-ordinate system is referred as “Viewing Transformation”.

Normalisation transformation- It maps world coordinate system to normalized device coordinate system (NDCS), AS different display devices have different screen sizes measured in pixels we need to make our program display device independent. This display device independent unit system is called as normalised device coordinate system.

e. Describe B-spline curve. (For Diagram 1 M and Description 3 M)

Ans: B-Spline Curve Generation:- A B-spline is a spline function that has minimal support with respect to a given degree, smoothness, and domain partition. B-splines. A fundamental theorem states that every spline function of a given degree, smoothness, and domain partition can be uniquely represented as a linear combination of B-splines of that same degree and smoothness, and over that same partition. In the computer science subfields of computer-aided design and computer graphics, the term B-spline frequently refers to a spline curve parameterized by spline functions that are expressed as linear combinations of B-splines (in the mathematical sense above). A B-spline is simply a generalization of a Bézier curve




Model Answer



Fig B-Spline Curve

Advantages of B-Spline:

1. The Degree of a B-Spline polynomial can be set independently of the number of control points (with certain limitation)

2. B-Spline allow local control over the shape of spline curve of surface the trade-off is that B-Spline are more complex then Bezier Spline

f. Write DDA arc generation algorithm. (All correct steps 4 M) Ans: Algorithm:- Step 1 Read the center of a curvature say (x0 , y0)

Step 2 Read the are angle say Ә Step 3 Read the starting point of the arc i.e. (x,y) Step 4 Calculate d Ә

d Ә = Min (0.01, 1/(3.2 x (|x-x0|+|y – y0| ))) Step 5 Initialise Angle = 0 Step 6 While (Angle < Ә ) Do { Plot (x, y) X = x – (y - y 0) x d Ә Y = y + (x -x 0) x d Ә Angle = Angle + d Ә } Step 7 Stop.

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