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- 1. Output primitives are the basic geometric structures which facilitate or describe a scene/picture. Example of these include: points, lines, curves (circles, conics etc), surfaces, fill colour, character string etc.
- 2. In order to draw the primitive objects, one has to first scan convert the object.Scan convert: Refers to the operation of finding out the location of pixels and then setting the values of corresponding bits, in the graphic memory, to the desired intensity code.
- 3. y means that for a unit (1) change in x mx there is m-change in y.x 1 means that for a unit (1) change in yy m there is 1/m change in x. 5
- 4. Uses differential equation of the line : m If slope |m| 1 then increment x in steps of 1 pixel and find corresponding y-values. If slope |m| 1 then increment y in steps of 1 pixel and find corresponding x-values. step through in x step through in y 6
- 5. A line with positive slope (Left to Right) If m <= 1, sample at unit x intervals (Δx = 1)and compute successive y values as: yk+1 = yk + m. If m >1, sample at unit y intervals (Δy =1) andcompute successive x values as: xk+1 = xk + 1/m.
- 6. A line with positive slope (Right to Left) If m <= 1, sample at unit x intervals (Δx =-1)and compute successive y values as: yk+1 = yk – m If m >1, sample at unit y intervals (Δy = -1) and compute successive x values as: xk+1 = xk - 1/m
- 7. A line with negative slope (Left to Right) If |m| <= 1, sample at unit x intervals (Δx = 1)and compute successive y values as: yk+1 = yk + m If |m| >1, sample at unit y intervals (Δy =-1)and compute successive x values as: xk+1 = xk - 1/m
- 8. A line with negative slope (Right to Left)If |m| <=1, sample at unit x intervals (Δx =1)and compute successive y values as: yk+1 = yk - m If |m| >1, sample at unit y intervals (Δy =1)and compute successive x values as: xk+1 = xk + 1/m
- 9. 1) Accept the end point co-ordinates of theline segment AB ie A(x1,x2) and B(x2,y2).2) Calculate dx and dy.3) If abs(dx)>=abs(dy) then , Step=dx else Step=dy.
- 10. 4) Let x increment = dx/Step ; Let y increment = dy/Step.5) Display the pixels at standing portion putpixel (x,y,white).6) Compute the next co-ordinate position alongthe line path xk+1 = xk + x increment yk+1 = yk + y increment Put Pixel(xk+1 , yk+1 ,white).
- 11. 7) If xk+1=x2 OR / AND yk+1=y2 then STOP else go to Step (4).
- 12. 1) In Bresenham’s Line Drawing Algorithm thepixel positions along a line path are obtained bydetermining the pixel i.e nearer the line at eachstep. 2) It is an efficient raster line generationalgorithm.
- 13. 1) Accept the end point co-ordinates of theline segment AB ie A(x1,x2) and B(x2,y2).2) Calculate dx and dy.3) If abs(dy)<=abs(dx) ie slope |m|<=1 (a) Compute initial decision parameter Po= 2abs(dy)-abs(dx).
- 14. (b) At each xk position perform the followingif Pk <0 i.e d1< d2 then, x increment=dx/abs(dx) AND y increment=0 Pk+1= Pk + 2abs(dy)else,Pk >0 ie d1>d2 X increment =dx/abs(dx) Y increment =dy/abs(dy) Pk+1= Pk +2abs(dy)-2abs(dx).
- 15. 4) If abs(dy) > abs(dx) ie slope |m|>1(a) Compute initial decision parameter Po= 2abs(dx)-abs(dy)(b) At each xk position,perform the following ifPk < 0 i.e d1 < d2 then x increment=0 y increment=dy/abs(dy) Pk+1= Pk + 2abs(dx)
- 16. Else, Pk >0 ie d1>d2 X increment=dx/abs(dx) Y increment=dy/abs(dy) Pk+1= Pk +2abs(dx)-2abs(dy).5) Calculate : xnext = xk + x increment ynext = yk + y increment. Display (xk+1, yk+1, white).
- 17. 6) Repeat Step (3) to (5) until xk+1=x2 AND /OR yk+1 =y27)STOP.
- 18. Midpoint CircleAlgorithm
- 19. Mid-point Circle Algorithm A method for direct distance comparison isto test the halfway position between twopixels to determine if this midpoint is insideor outside the circle boundary. This method is more easily applied toother conics, and for an integer circleradius.
- 20. 1) Accept the radius r and center (xc, yc).Thepoint of the circumference of a circle withcenter as origin (x0, y0) =(0,r).2) Calculate the initial decision parameter as Po =5/4 –r .3) At each xk position starting at k=0.Performthe following test.
- 21. If Pk <0 then, Xk+1= xk +1 and yk+1= yk Pk+1 = Pk +2xk +3Otherwise, Pk >0Xk+1= xk +1 and Yk+1= yk -1 Pk+1= Pk +2( xk - yk ) +54) Determine the symmetric points in other 7octants.
- 22. 5) Translate each calculated pixel position byT ( xk , yk ) and display the pixelX= xk+1 + xc AND Y= yk+1 + yc Put pixel ( x,y, white).6) Repeat step (3) to step (5) until x>=y.7)STOP.
- 23. BRESENHAM’S CIRCLEALGORITHM
- 24. 1) Accept the radius r and center (xc, yc).Thepoint of the circumference of a circle with centeras origin (x0, y0) =(0,r).2) Calculate the initial decision parameter as PO =3 – 2r.3) At each xk position starting at k=0.Performthe following test.
- 25. If Pk <0 then, Xk+1= xk +1 and yk+1= yk Pk+1 = Pk +4xk +6.Otherwise, Pk > 0Xk+1= xk +1 AND Yk+1= yk -1 Pk+1= Pk +4( xk - yk ) +104) Determine the symmetric points in other 7octants.
- 26. 5) Translate each calculated pixel position byT ( xk , yk ) and display the pixelX= xk+1 + xc AND Y= yk+1 + yc Put pixel ( x, y, white)6) Repeat step (3) to step (5) until x>=y7) STOP.
- 27. The DDA algorithm is accomplished by taking unit step in one direction and calculating other. The Bresenham’s Algorithm finds the closest integer co-ordinate to the actual part. DDA algorithm uses floating point calculation. Bresenham’s uses integer point calculation. DDA is more accurate compared to Bresenhams.
- 28. Circles and ellipses can be efﬁciently and accurately scan converted using midpoint methods. Bresenham’s line algorithm and the midpoint line algorithm methods are the most efﬁcient.
- 29. THANK YOU

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