CG Manual 2011 Only

7/28/2019 CG Manual 2011 Only

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Computer Graphics: Lab Manual 2011-12

Savitribai phule Shikshan Prasarak Mandals

S.K.N.Sinhgad College of Engineering,

Pandharpur

LAB MANUAL

Computer Graphics

Class: SE (CSE)

(2011-12)

Prepared By

Prof. Prakash R. Gadekar

1Dept.of CSE SKN Sinhgad College of Engineering,Pandharpur.


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Savitribai Phule Shikshan Prasarak Mandals


Pandharpur

CERTIFICATEThis is to certify that Mr. / Miss _________________________

of Class Second Year - CSE, Roll No. _________, Exam Seat No.

_________, has completed all the Term Work / Practical work in

the subject Computer Graphics, satisfactorily in the Department

of ComputerScience and Engineering as prescribed by University

of Solapur in the academic year 2011-2012.

Subject In-charge Head of Department Principal



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Savitribai phule Shikshan Prasarak Mandals


Pandharpur

INDEXSr.

No.Title of Experiment Date

Page

No.Remarks

Sign of

Teacher

01. Study of graphics in built function.

02. Program for:DDA line drawing algorithm

03. Program for:Bresenhams line drawing algorithm

04. Program to implement different 2-Dtransformations

05.Program for:Bresenhams circle generation algorithm

06. Program to transform object for gettingOrthographic projection of the object

07.Program for polygon Filling :1.Seed Fill Algorithm2.Edge Fill Algorithm

08.Program for Cohen-Sutherland 2-D LineClipping Algorithm

09. Program for the creation of Bezier Curve



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Lab Innovation

Sr.

No.

Title DatePage

No.Remarks

Sign of

Teacher

Signature



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Experiment No: - 1

Title : - Study of Graphics FunctionName :-

Roll No :-

Batch :-

Date :-

Marks :- Signature :-



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Experiment No: - 1

Aim:-Develop a program using a graphics inbuilt function.

Objectives:-Learn and understand the functions in graphics Library.

Theory:-

1) Loading computer graphics

When we start with the turbo c editor. We first includes the graphics library to convert the texteditor into graphics editor.To include the graphics library the steps are

i) Select the option menu from the menu bar

ii) Select linker option

iii) Select libraries from linker option

iv) Select graphics library from the libraries window

v) Click on Ok.

2) Initialized the graphics Mode:-

Before writing the computer graphics program, first we initialize the graphics driver and

graphics mode, for that purpose we use detectgraph() function and initgraph() function.Detectgraph():- determines the graphics driver and chooses mode by checking hardware. Detectgraphdetects your system graphics adaptor and chooses the mode that provide the highest resolution for thatadaptor. If no graphics hardware is detected, detectgraph sets *graphdriver to grNotDetected, andgraphresults returns grNotDetected.The main reason to call detectgraph directly is to override thegraphics mode that detectgraph recommends to initgraph.

Syntax:- detectgraph(&gdriver, &gmode);

Initgraph():-

It initializes graphics system, to start the graphic system , you must first call initgraph(),initgraph initializes the graphic system by loading a graphics driver from disk, then putting into thegraphics mode.

Initgraph resets all graphic settings like color,palette, current position, viewport etc. to their defaults.

Syntax:- initgraph(&gdriver,&gmode,C:\\tc\\bgi);



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Bgi is a file that stores the graphics driver and for each bgi driver there are different graphics mode. Inshort, it provides graphics interface.

Inbuilt functions :-

Following are inbuilt functions of graphics library:-

1) Line ():- Line () is used to draw a line.

Line(x1,y1,x2,y2);

2)Circle(): is used to draw the circle.

Syntax: circle(x1,y1,radius);

3)ellipse():- draws an elliptical arc.

Syntax:- ellipse(x,y,stangle,endangle,xradius,yradius);

fillellipse():- draws and fills an ellipse.

Syntax:- fillellipse(x,y,xradius,yradius);

4)drawpoly() :- draws the outline of a polygon, draws a polygon using the current line style and color.

Drawpoly(numpoints,*polypoints);

Numpoints:- specifies number of points

*polypoints :- points to a sequence of (numpoints*2) integers.

fillpoly():-draws and fills a polygon. It draws the outline of a polygon using the current line styleand color.

5)Line style:-

Line styles for getlinesettings and setlinestyle.

SOLID_LINE 0 Solid line

DOTTED_LINE 1 Dotted line

CENTER_LINE 2 Centered line

DASHED_LINE 3 Dashed line

USERBIT_LINE 4 User-defined line style

6)bar:- Draws a bar, bar draws a filled-in, rectangular, two-dimensional bar.

Declaration: void far bar(int left, int top, int right, int bottom);



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The bar is filled using the current fill pattern and fill color. bar doesnot outline the bar.

To draw an outlined two-dimensional bar, use bar3d with depth = 0.

7) Fill patterns:- for getfillsettings and setfillstyle.

EMPTY_FILL 0 Background color

SOLID_FILL 1 Solid fill

LINE_FILL 2 ---

LTSLASH_FILL 3

SLASH_FILL 4 , thick lines

BKSLASH_FILL 5 , thick lines

LTBKSLASH_FILL 6

HATCH_FILL 7 Light hatch

XHATCH_FILL 8 Heavy crosshatch

INTERLEAVE_FILL 9 Interleaving lines

WIDE_DOT_FILL 10 Widely spaced dots

CLOSE_DOT_FILL 11 Closely spaced dots

USER_FILL 12 User-defined fill pattern

All but EMPTY_FILL fill with the current fill color. EMPTY_FILL uses the current background color.

(left, top) the rectangle's upper left corner

(right, bottom) the rectangle's lower right corner

The coordinates are in pixels.To draw an outlined two-dimensional bar, use bar3d with depth = 0.

(left, top) the rectangle's upper left corner

(right, bottom) the rectangle's lower right corner

The coordinates are in pixels.

8)bar3d:=Draws a 3-D bar.

Declaration: void far bar3d(int left, int top, int right, int bottom, int depth, int topflag)

bar3d draws a three-dimensional rectangular bar, then fills it using the current fill pattern and fill color.



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The three-dimensional outline of the bar is drawn in the current line styleand color.

depth Bar's depth in pixels

topflag Governs whether a three-dimensional top is put on the bar

(left, top) Rectangle's upper left corner

(right, bottom) Rectangle's lower right corner

If topflag is non-zero, a top is put on the bar. If topflag is 0, no top is put on the bar: This makes itpossible to stack several bars on top of oneanother.

9) Getpixel: -get the color of specified pixel

Syntax: -usinged far getpixel(int x ,int y);

Example: -pcolor = getpixel(100,100)

10) Outtext: -Display a string in the viewport

Void far outtext(char far textstring);

Example: - outtext(display text in graphics);

11) Putpixel: - plots a pixel at specified point.

Syntax: void far putpixel (int x, int y,int pixelcolor);Example:putpixel(100,100,15);

12) Closegraph: shuts down the graphics mode

Syntax: -void far close graph( void );

Example: - closegraph( );

Conclusion:-



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Frequently Asked Questions:-

1) What is mean by Computer Graphics?

2) What is the need of Computer graphics?

3) What is interactive Computer graphics?

4) What are the different applications of Computer graphics?

5) What is the declaration for line? Describe each argument?

6) Why is it needed to load graphics library in Computer graphics?



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7) What is the job of initgraph() and Detectgraph() functions?

8) What is the job of graphics drivers and modes in computer graphics?

9) Howcan you fill the color in the ellipse? Are there different methods for filling the color?

10) What are the different ways to fill any picture with pattern?

Design Problem:-

1) Draw a penguin with the graphics function.2) Write a program to display a flag having more than 3 colors.3) Write a program to moving ball screen saver.



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Experiment No : - 2

Title :- DDA Line drawing algorithm

Name :-

Roll No :-

Batch :-

Date :-




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Experiment No: - 2

Problem Statement: - Write a program to develop a DDA line drawing algorithm

Objective: - To draw a rasterized straight line by making use of DDA line drawing algorithm

Theory:-

The process of turning on the pixel for a line segment is called Vector generation .Before discussing line drawing algorithm , it is useful to note the general requirements for such

algorithm. This algorithm specify the desired characteristics of line:1. The line should appear as a straight line and it should start and end accurately.2. The line should be displayed with constant brightness along its length independent of its

length and orientation.3. The line should be drawn rapidly.

Let us see the different lines drawn in following fig:-Vertical line 45

line


Horizantal Line


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The above differential equation can be used to obtain a rasterized straight line. For any given Xinterval X along a line, we can compute the corresponding y interval Y from euation (1) as

Y2-Y1 ----------------------(2)

X2-X1

Similarly, we can obtain X interval X corresponding to a specified Y as,

X2-X1

Y2-Y1

Once the intervals are known the values for next X and next Y on the straight line can be obtained asfollows:

Xi+1 = Xi + X -------(4) and Yi+1 = Yi + Y ------(5)Put the values of X and Y in the equations (4) and (5) , then these equation represent a recursion

for successive values of X and Y along the required line. Such a way of rasterizing a line is called Digital Differential Analyzer (DDA).

For simple DDA eitherX or Y , whichever is larger, is chosen as one raster unit i.e,

If |X| |Y| then

X = 1

Else Y = 1.

Let us see the algorithm for rasterizing a line.

Algorithm:

1) Read the line end points (x1,y1) and (x2,y2) such hat they are not equal.[If equalthen plot that point and exit]

2) X = |x2-x1| and Y = |y2-y1|

3) If (X Y) thenLength = X

ElseLength = Y

End if

4) X = (x2-x1) / length


Y = X

X = Y


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3) What is one complete raster unit?

4) What is the use of Sign() function?

5) What is the use of floor() function?

6) What are the disadvantages of DDA Algorithm?

7) Define 1) Line 2) Line segment 3) Vector



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Experiment No :- 3

Title :- Bresenhams Line drawing algorithm

Name :-

Roll No :-

Batch :-

Date :-




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DA Distance AboveDB Distance Below

As shown in gig, the line does not pass though all the raster points( pixels). It passes throughraster point(0,0) and subsequently crosses three pixels. It is seen that the intercept of line withthe line x = 1 is closer to the line y=0 i.e pixel (1,0) than to line y =1 i.e pixel(1,1). Theintercept of the line with line x = 2 is close to the line y = 1 i.e pixel (2,1) than to the line y=0 i.epixel(2,0). Hence , the raster point at (2,1) better represents the path of the line shown in fig.

e = Db Da or Da - Db

Let us define e = Db-Da, now if e>0, then it implies that Db>Da i.e pixel above the lineis closet to the true line if Db=0 , error is initialized with e = e-2* x. this is continued till error is negative. In eachiteration y is incremented by 1 . when e


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x = x1y = y1

4. e = 2* y-x [initialize value of decision variable or error to compensate for nonzerointercepts]

5. i=1 [initialize counter]

6. plot(x,y)7. while(e>=0)

{y = y+1;e = e-2 * x

}x=x+1;e =e + 2*y

8. i =i+1

9. if(i


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4) justify why Bresenhams line Drawing algorithm is superior than DDA line Drawingalgorithm?

Experiment No :- 4

Title :- Program to implement different 2D Transformations.

Name :-

Roll No :-

Batch :-

Date :-




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Experiment No: - 4

Problem Statement: - Program to implement different 2D Transformations

Objective: - To study the basic need of 2Dtransformation with matrices by using the basicscaling, rotation , translation , x shear and y shear operations.

Theory:-

A graphics system must allow the user to change the way object appears. In the realworld , we frequently rearrange objects or look at them from different angles. The computershould do the same with images on the screen, or its orientation. It is much more sensible tomake changes later to the objects original basic description. Implementing such a change is calla Transformation.

Basically transformation means making changes or allowing modifications to thepicture. we know that once the objects are created, the different applications may require thevariations in these. In some cases, the object size may be large, which needs to be stored incompressed form. All these requirements suggest us that we need to transform these objects.This process is carried out by transformation. For performing the different 2Dtransformations we need some mathematical operations, out of that , we are consideringmatrices.

Basically, transformation means change in image. We can modify the image byperforming some basic transformations, such as:

Scaling

Rotation Translation

Reflection

Shear TransformationNow , we will see the details of all the transformations.



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1. Scaling Transformation:-

Scaling is the process of changing the size of the object or image.Suppose we are having point p1 whose co-ordinates are (x1,y1) , then we can

represent this point p1 in matrix form as,

P1 = (x1,y1) = [x1 y1]

Now suppose we want to multiply this point i.e one P1s matrix by some matix,say matrix T of dimension (2*2), then we will get resultant point P2 with dimension as(1*2) i.e

P1 * T = P2[ x1 y1] * T = [ x2 y2]

Thus we can say, the matrix T gives the relation between original point P1 and the newpoint P2. But if the matrix T is identity matrix then point p2 will be the same as P1.

i.eP1.* T = P2

1 0P1 * = P2

0 1

1 0[x1 y1] * = [ x1 y1]

0 1

P2 = P1

But if the matrix T is not identity matrix , but it is2 0

T = Then P2 will become0 1

P1 * T = P22 0

[x1 y1] * = | 2x1 y1|

0 1P2 = | 2x1 y1|

i.e all x-co-ordinates will get double as compared to original one. But the y-co-ordinatesremains same as original. See fig, it means the width of the object becomes double butheight is not getting changed.

If matrix T is


Before After

i.e if all x co-ordinates becomes halfof the original and y co-ordinatesremains same as original


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0.5 0T = Then P2 will become

0 1

P1 * T = P20.5 0

[x1 y1] * = | 0.5x1 y1|0 1

P2 = | 0.5x1 y1|Similarly, we can change the height of image also by keeping width constant. In thatcase our transformation matrix will be,

1 0T = 0 2

Like this we can get both width and height double also by using matrix T as

2 0T =

0 2In general term, we can say matrix S is a scaling matrix and Sx is scale factor for x-cordinate and Sy will be the scale factor for y coordinate.

Sx 0

S = 0 Sy

When we scale image, all points will get changed except origin.

2. Rotation Transformation:-

For rotation we need some trigonometry. Suppose we have a point P1 = ( x1,y1)and we rotate it about the origin by an angle to get a new position P2 =(x2,y2). We wish to find the transformation which will change (x1,y1) into

(x2,y2).


Before After

P1(x1,y1)

Y P2(x2,y2) P1 * T = P2

But before we can check any transformation to seeif it is correct, we have to first know what (x2,y2)should be in terms of (x1,y1) and angle . Todetermine this we need trigonometric functionsline sin and cos.


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Let us say we have drawn a line from origin to point(x,y) of length L at an angle anticlockwise from x-axis , as shown in fig.

,

Sin = Opposite side/hypotenuse = y/L= y/(x+y)

Similarly, cos = adjacent side / hypotenuse = x/L

= x/(x+y)

If we consider length as exactly 1 then,

Sin = y/1 = y

Cos= x/1 = x

Now we will determine the transformation matrix for rotation.


x

(0,0)

(X ,Y)Y

X

L

The line segment L will have (0,0) and (x,y) as its two endpoints.

We can say,

L = (x+y)

X1

P2(x2,y2)

P1(x1,y1) = (1,0)

(0,0)

Y Consider a point p1 as (1,0) ref to fig. after rotating this point p1 inanticlockwise by an angle , we will get point P2. Now co-ordinates of P2 will be

X2 = Cos

Y2 = Sin

P1(x1,y1) = (0,1)

Similarly, if current point P1 is (0,1). Ref to fig and if we rotate P1

anticlockwise by angle then we will get P2 as ( - Cos , Sin). Hedistance of P1 from origin is 1 unit.

L = 1

As Sin = opposite side / hypotenuse = -x/L as L = 1

= -x

Cos = adjacent side / hypotenuse = y/L as L = 1

= y

X = - Sin and y = cos


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Coming back to original conceptP1 * T = P2

Where, P1 is original pointP2 is new pointT is transformation matrix.

For case 1 i.e when p1 = (1,0)P1* T = P2

1 0 * T = cos SinNow we know that original point i.e P1 , we also know what will be new point i.e P2. Here wehave to find what will be the transformation matrix , which will change P1 to P2. For simplicitylet us assume that transformation matrix (T) as

a bT =

c d

a bNow , 1 0 * = a bc d

cos = a and Sin = b

Similarly for case 2 i.e when P1 = ( 0,1)P1 * T = P2

0 1 * T = -Sin cos

a b0 1 * = c d

c d

c = -Sin d = cosNow , place the values of a ,b ,c d in transformation matrix.


(0,0)


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a b cos Sinc d = -Sin cos

For rotation in clockwise direction, the matrix will becos Sin

R = -Sin cos

The sign of angle determines the direction of rotation. When angle is positive i.ewhen we are following the direction exactly reverse as that of the direction of rotatinghands of a normal clock, then it is anticlockwise or sometimes it is called CounterClockwise see in fig. And when angle is negative I.e when we are following the samepath as that of hands of clock then it is called Clockwise, as shown in fig.

Anticlockwise Clockwise

So rotate a point in clockwise direction our matrix will become,Cos(-) Sin(-)

R =-Sin(-) cos(-)

Here, as we are rotating in clockwise direction, so we are replaceing by . Butcos(-)=cos() and sin(-) = -sin()

The transformation matrix for rotation in clockwise direction will be

Cos -SinR = Sin cos

The transformation matrix for roatation in anticlockwise direction will be as.Cos Sin

R = -Sin cos

3. Translation :-

Moving the whole image is called Translation. We can easily shift the imageby adding same value to all the points of image by which we want to shift the image.

Before After



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Here the magnitude of both x and y remains same. Only the sign of x co-ordinates gets changed.Now let us derive what will be the transformation matrix which gives us the desired result.

Our basic transformation rule will beP1 * T = P2

Now , P1 = [x y] and we need P2 as [-x,y]

(x y) * T = (- x y)

Now what should be the transformation matrix T, to get the desired result.So the T will be -1 0

0 1Let us check it out

P1 * T = P2(x y) * -1 0 = (- x y)

0 1After multiplying the transformation matrix with the point P1 we will get the desired point P2.

b.Reflection at x-axis:-

Similarly, the same rule you have to apply for 1) Reflection at X-axis and you will get the newpoint p2 as

(x y) * 1 0 = [x -y]0 -1

c. Reflection in the origin:-

We have seen reflection at x-axis and reflection at y-axis, if we want both these

transformation simultaneously then, we have to apply reflection in the origin. It is a combinationof both reflection at x-axis and reflection at y-axis.1 0 -1 0 = -1 00 -1 * 0 1 0 -1

(x y) * -1 0 = (-x -y)0 -1

Thus , in this we can perform the reflection transformation.

5. Shear transformation:-

The shear transformation means slanting the image. This shear transformation is of twotypes: Y-shear

X-shear1. Y-shear:-

In y-shear we are keeping x-co-ordinate values as it is and shifting the y-cordinate values only, which result in tilting of horizontal lines see fig.



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Before After

If P is (x,y) , then X-co-ordinate of P2 must be same as x co-ordinate of P1 . But y co-ordinateof P2 should change.

Transformation matrix for y-shear will be 1 a where a is constant factor, which will0 1

Decide , how much to tilt.Now , P1 * T = P2

x y * 1 a = x x.a+y0 1

So, x-co-ordinate of P2 are not changed but y-cordinate of P2 will be combination of constant

value a and x and u values of P1.P2 = ( x x.a+y)

So here we say that value of y will depend on value of x.2. X-shear:-

X-shear preserves y co-ordinate values and shifts x co-ordinate values causing verticalline to tilt. See in fig. means vice versa we are doing here.

Before After

So , you will get,x y * 1 0 = x+yb y

b 1

here, we have to say, the value of x will depend on value of y.

Conclusion:-

Frequently Asked Question:-

1) Describe with respect to 2D transformation:- 1) scaling 2) Rotation 3) Translation?



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2)What is mean by Transformation?

3) Compare homogeneous co-ordinate system and normalized co-ordinate system?

4) What is the need of homogeneous co-ordinates system?

5) What is composite transformation?

6) Explain the inverse transformation?

7) How can you perform homogeneous co-ordinates for scaling?



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8) What are the different steps for rotation about an arbitrary axis?

9) What is mean by shear and reflection?

10) Explain reflection about line?

Design Problems:

1. Implement rotation about all co-ordinate axes transformation on square in 3D.2. Implement rotation about any arbitrary axis transformation on square in 3D.3. Implement scaling and translation transformation on square in 3D.



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Experiment No :- 5

Title :- Bresenhams Circle Generation Algorithm

Name :-

Roll No :-

Batch :-

Date :-




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Experiment No:-5

Problem Statement:- Write a program to develop a Bresenhams Circle Generation Algorithm

Objective:- To draw a rasterized straight line by making use of Bresenhams Circle GenerationAlgorithm

Theory:

Bresenhams Circle Generation Algorithm:-

The Bresenhams circle drawing algorithm considers the eight-way symmetry of thecircle to generate it. It plots th part of the circle, i.e from 90 to 45, as shown in fig.


90 45

0

-y

+ x

y

To achieve best approximation to the true circle we have toselect those pixels in the raster that fall the least distance from thetrue circle. Ref to the fig, let us observe the 90 to 45 portion ofthe circle. It can be noticed that if points are generated from 90 to45, each new point closet to the true circle can be found byapplying either of the two options:

Increment in positive x direction by one unit or

Increment in positive x direction and negative y directionboth by one unit.


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Let us assume that point P in fig as a last scan converted pixel. Now we have two options eitherto choose pixel A or Pixel B. The closer pixel amongst these two can be determined as follows:

The distances of pixels A and B from the origin are given as,

Da =(xi+1)+(yi) and

Db =(xi+1)+(yi-1)

Now the distances of pixels A and B from the true circle are given as,

a = Da r and b = Db r

However, to avoid square root term in derivation of decision variable, i.e to simplify thecomputation and to make algorithm more efficient the a and b are defined as,

a = Da - rb =Db-r.


P


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from fig, we can observe that ais always positive and b always negative . Therefore,we can define Decision variable di as,

di = a + b

and we can say that , if di


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The remaining part of circle can be drawn by reflecting point about y-axis , x-axis and aboutorigin as shown in fig.

Therefore, by adding seven more plot commands after the plot command in step 4 of thealgorithm, the circle can be plotted . the remaining seven plot commands are:

Plot(y,x)Plot(y,-x)Plot(x,-y)Plot(-x,-y)Plot(-y,-x)Plot(-y,x) andPlot(-x,y)

Conclusion:-

Frequently Asked Question:-

1) What is one complete raster unit?

2) What is the use of Sign() function?

3) What is the use of floor() function?



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4) What is the need of Bresenhams Circle drawing algorithm?

9) What is the difference between DDA Circle generation algorithm & Bresenhams circlegeneration algorithm?

Design Problems:-

1) Design Isometric Cube by using complete straight line.

2) Write a program for animated circles.

Experiment No :- 6

Title :- Program for filling a polygon using scan-fill method

Name :-

Roll No :-

Batch :-

Date :-




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Experiment No: - 6

Problem Statement:- Program for filling a polygon using scan-fill methodObjective:- To fill the polygoin using scan fill method of raster display.Theory:-

Recursive Algorithm for seed fill methods have got two difficulties:1) The first difficulty is that if some inside pixels are already displayed in fill color then

recursive branch terminates, leaving further internal pixels that are initially set to the fill colorbefore applying the seed fill procedures.

2) Another difficulty with recursive seed fill method is that it cannot be used for largepolygons .This is because recursive seed fill procedures require stacking of neighboring pointsand in case of large polygons stack space may be insufficient for stacking of neighboring points.

To avoid this problem more efficient method can be used. Such method fills horizontalpixel spans across scan lines , instead of proceeding to 4-connected or 8-connected neighboringpoints. This is achieved by identifying the rightmost and leftmost pixels of the seed pixel andthen drawing a horizontal line between these two boundary pixels. This procedure is repeatedwith changing the seed pixel above and below the line just drawn until complete polygon isfilled. With this efficient method we have to stack only a beginning position for each horizontal



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pixel span, instead of stacking all unprocessed neighboring positions around the currentposition.

The fig.1. illustrates the scan line algorithm for filling of polygon. For each scan linecrossing a polygon, this algorithm locates the intersection points of the scan line with the

polygon edges. These intersections points are then sorted from left to right and thecorresponding positions between each intersection pair are set to the specified fill color.

In fig.1, we can see that there are two stretches of interior pixels from x = 6 to x = 9 andx = 12 to x = 15. The scan line algorithm first finds the largest and smallest y values of thepolygon. It then starts with the largest y value and works its way down scanning from left toright, in the manner of a raster display.

The important task in scan line algorithm is to find the intersection points of scan linewith the polygon boundary. When intersection points are even, they are sorted from left to right,paired and pixels between paired points are set to the fill colour. But in some cases intersectionpoint is a vertex. When scan line intersects polygon vertex special handling is required to findthe exact intersection points. To handle such cases, we must look at the other endpoints of thetwo line segments of the polygon which meet at this vertex. If these points lie on the same sideof the scan line, then the point in question counts as an even number of intersections. If they lieon opposite sides of the scan line, then the point is counted as single intersection.

We have seen that it is necessary to calculate x intersection points for scan line withevery polygon side. We can simplify these calculations by using Coherence properties.

A coherence property of a scene by which we can relate one part of the scene with theother parts of a scene. Here, we can use a slope of an edge as a coherence property. By using

this property we can determine the x intersection value on the lower scan line if the xintersection value for current scan line is known. This is given as,

Xi+1 = Xi - 1/m where, m is the slope of the edge.

As we scan from top to bottom value of y-co-ordinates between the two scan linechanges by 1.


Y

Scan line

X

Fig .1


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Yi+1 = Yi 1

Many times it is not necessary to compute the x intersections for scan line with everypolygon side. We need to consider only the polygon sides with endpoints straddling the currentscan line. See in fig:

It will be easier to identify which polygon sides should be tested for x-interception, if wefirst sort the sides in order of their maximum y value. Once the sides are sorted we can processthe scan lines from the top of the polygon to its bottom producing an active edge list for eachscan line crossing the polygon boundaries. The active edge list for scan line contains all edgescrossed by that scan line. The fig show sorted edges of the polygon with active edges.


BC

BA

DC

DE

AJ

GF

GH

EF

HI

JI

Scan

Consider only theses sides

Fig. consiser only the sides which intersect the scan line

A

C

B D

E

F

G

H

IJ

Scan line

Active ed

To

Bottom

Sorted List

Fig. sorted edges of the polygon with active edges


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In summary, a scan line algorithm for filling a polygon begins by ordering the polygonsides on the largest y value. It beginsWith the largest y value and scans down the polygon. Foreach y , it determines which sides can be intersected and finds the x values of these intersectionpoints. It then sorts, pairs and passes these x values to a line drawing routine.

Algorithm for polygon filling using scan line conversion:-

1. Read n, the number of vertices of polygon2. Read x and y co-ordinates of all vertices in array X[n] and Y[n]

3. Find Ymin and Ymax4. Store the initial x value (x1) y values y1 and y2 for two endpoints and x increment xfrom scan line to scan line for each edge in the array edges [n][4].

While doing this check that y1>y2, if not interchange y1 and y2 andcorresponding x1 and x2 so that for each edge, y1 represents its maximum y co-ordinate and y2 represents its minimum y co-ordinate.

5. Sort the rows of array, edges[n][4] in descending order of y1 , descending order of y2and ascending orer of x2.

6. Set Y = Ymax7. Find the active edges and update active edge list:

If( y>y2 and y


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}

8. Compute the s intersects for all active edges for current y value [initially x-intersect is x1and x intersects for successive y values can be given asXi+1 = Xi +X

Where, X = -1/m and m = y2-y1 / x2-x1 i.e slope of a line segment.9. If x intersect is vertes i.e x-intersect = x1 and y = y1 then apply vertex test to check

whether to consider one intersect or two intersects. Store all x intersects in the x-intersect[] array.

10. Sort x-intersct[] array in the ascending order11. Extract pairs of intersects from the sorted x-intersect [] array12. Pass pairs of x values to line drawing routine to draw corresponding line segments13. Set y = y 114. Repeat steps 7 through 13 unitl y>= ymin15. Stop

In step 7, we have checked for y


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3) Represent the polygon using display file structure?

5) Write down the different methods for testing a pixel inside of polygon?

6) Which are the different methods used for filling polygon?

7) Explain boundary fill method for polygon?

8) What is mean by Seed pixel and what is mean by four connected region and eight connectedregion?

9) Compare Seed fill , edge fill method and flood fill ?

10) What is the need of scan line algorithm?



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Design Problems:-

1) Implement flood fill algorithm for polygon having 8 vertex.

2) Implement Boundary fill algorithm for polygon using seed pixel and 8-connected region.

3) Implement scan line algorithm for hexagon

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