Computer Graphics Lec_1

8/10/2019 Computer Graphics Lec_1

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Computer Graphics & its ApplicationsSubject Code: CSE -710



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Course Content• Introduction• Scan Conversion – Lines, Circles, Ellipses:

Filling polygons and clipping algorithms• Two Dimensional Transformations

• Three dimensional Transformations• Projections• Plane Curves and surfaces

• Solid Modeling• Visible Surface Determination



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Books• Principles of Interactive Computer Graphics

W. M. Newman, Robert E. Sproull• Mathematical Elements for Computer Graphics

D.F. Rogers and J. A. Adams• Computer Graphics

Steven Harrington

• Computer GraphicsDonald Hearn, M. Pauline Baker

• Computer Graphics: Principles and PracticeJ. D. Foley, A Van Dam, S. K. Feiner, and J. F. Hughes



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• What is Computer Graphics ?• What is image processing ?• Some Applications of Computer Graphics

– CAD/CAM

– Entertainment – Education and Training – Animation/Video Games – 3D Surface Modeling

– Visualization



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Computer graphics

The core technological issues in computer generated imagery.

Concept Model Image



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Modeling

Concept Model Image

MODELS (in computer graphics): A Digital, Descriptive orMathematical representation of a scene or object that can bestored (efficiently) in computer memory.



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Rendering

The rendered IMAGE is a visual representation of themodel on digital output media.





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Devices

• Keyboard• Mouse• Joystick • Trackball and Trace ball

• Digitizer• Scanners• Touch Panels• Light Pen

• Voice System



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Introduction

• Pixels• Straight Lines• Line Drawing algorithms, straight

– Digital Differential Analyzer DDA – Bresenham

• Second order Curves – Mid point Algorithm





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Pixel Coordinate Systems

X

Y



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Straight Lines

• A straight line , say ‘s, has two end points, sayp1, p 2 and a collection of points connecting thetwo end points.

• Geometrically a line is defined using the

parametric representation –S = p 1 + t*(p 2-p 1), t [0,1]Or an equation with constants a,b,c and

end points:

A*x + b*y + c = 0 and p1,p2



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Line drawing Algorithm

• Using the line equations• DDA Algorithm

• Bresenham’s Line drawing Algorithm



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Line Equation

The equation of a Line is given byy = mx + b ………………. (1)

Rearranging we getb = y – mx ---------------------(2)

where m = (y 2-y1) / (x 2- x1) = dy/dx ≈ ∆y/∆x

For any given small x interval ∆x along the line, compute thecorresponding y interval i.e.

∆y = m ∆x --------------------- (3)Similarly for any given small y interval ∆y along the line, compute thecorresponding x interval i.e.

∆x = ∆y / m---------------------( 4)



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DDA Algorithm

(Xk, Yk)

X

Y

(Xk+1 ,Yk+1 )



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DDA

• This is an incremental Scan conversion linedrawing algorithm based on calculating either ∆y or ∆x .

• Sample the line at unit intervals in one coordinate and determine the corresponding

integer values nearest line path for all the otherco ordinates

• Case :1 If |m| ≤ 1X k+1 = X k + 1 Y k+1 = Y K + m

• Case :2 If |m| > 1 Y k+1 = Y k + 1Xk+1 = X K + 1 / m



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DDA

• Case :3 If |m| < 1 and start point is on the righthand side

X k+1 = X k - 1 Y k+1 = Y K - m

• Case :4: If |m| > 1 and start point is on theright hand side

Y k+1

= Y k

- 1Xk+1 = X K – 1/ m



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Assignment Problems1) Using DDA line drawing algorithm find the pixel co-

ordinate to draw a line from (2,3) to (9,7).

2) Find the Pixel co-ordinates using DDA to draw la linewhose end points are (2,3) , (5,9).

3) Consider a line from (0,0) to (5,5), use DDA to rasterizethe line.

4) Write a pseudo code to describe the DDA for scan

converting a line whose slope is betweena) -45 deg and +45 degb) 45 deg and -45 deg

3) Consider a line froma) (0,0) to (-8,-4)b) (1,1) to (8,5)c) (0,0) to (5,5)

use any line drawing algorithm.



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Bresenham’s Line drawing Algorithm

Initialpixel (12,12)

(13,12) (13,13)

Either or

Next

(Xk,Yk)

(Xk+1 ,Yk+1 )Either

Next Pixel

Xk+1

Either or Either or Either or Either or Either

(Xk,Yk)

Next Pixel

(Xk,Yk)

(Xk+1 ,Yk+1 )

Next Pixel

(Xk+1 ,Yk+1 )

Next PixelNext PixelNext Pixel

(Yk) (Yk+1 )Either or

Xk Xk+1

Yk

Yk+1

Y

d1d1

d2

Xk Xk+1Xk+2 Xk+A

Yk

Yk+1

Yk+2

Yk+3



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• Testing the sign of an integer parameter whose value isproportional to the difference between the two pixelposition from the actual line.

• Consider a line with |m| < 1, pixels positions along a lineare determined by sampling at unit x intervals.



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The y co ordinate on this sampling line w.r.t pixel column

positions x k + 1 is calculated asy = m.(x k + 1) + b

Then d 1 = y – y k = m.(x k +1) + b – y k

d2 = (y k + 1) - y

= ( y k + 1) – [ m.(x k +1)+b]If d 1 < d 2 , next point is (x k +1, y k )If d1 ≥ d2 , next point is (x k +1, y k +1)

The differenced1 – d2 = m(x k +1) + b – y k – [(y k +1) – m.(x k +1) - b]











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P (x,y)6=6 (11,16)

6+(-4)=2 (12,17)2+(-4)=-2 (13,17)-2+(16)=14 (14,18)14+(-4)=10 (15,19)10+(-4)=6 (16,20)6+(-4)=2 (17,21)2+(-4)=-2 (18,21)-2+(16)=14 (19,22)14+(-4)=10 (20,23)

Computer Graphics Lec_1

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