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Computer Graphics
- 2. Output primitivesare 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.
- 3. In order todraw 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.
- 5. y means that for a unit (1) change in x m x there is m-change in y. x 1 means that for a unit (1) change in y y m there is 1/m change in x. 5
- 6. Uses differential equationof 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
- 7. A linewith 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) and compute successive x values as: xk+1 = xk + 1/m.
- 8. A linewith 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
- 9. A linewith 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
- 10. A linewith 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
- 11. 1) Accept theend point co-ordinates of the line 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.
- 12. 4) Let xincrement = dx/Step ; Let y increment = dy/Step. 5) Display the pixels at standing portion put pixel (x,y,white). 6) Compute the next co-ordinate position along the line path xk+1 = xk + x increment yk+1 = yk + y increment Put Pixel(xk+1 , yk+1 ,white).
- 13. 7) If xk+1=x2OR / AND yk+1=y2 then STOP else go to Step (4).
- 15. 1) In Bresenham’sLine Drawing Algorithm the pixel positions along a line path are obtained by determining the pixel i.e nearer the line at each step. 2) It is an efficient raster line generation algorithm.
- 16. 1) Accept theend point co-ordinates of the line 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).
- 17. (b) At eachxk position perform the following if 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).
- 18. 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 if Pk < 0 i.e d1 < d2 then x increment=0 y increment=dy/abs(dy) Pk+1= Pk + 2abs(dx)
- 19. 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).
- 20. 6) Repeat Step(3) to (5) until xk+1=x2 AND /OR yk+1 =y2 7)STOP.
- 21.
- 22. Mid-point Circle Algorithm A method for direct distance comparison is to test the halfway position between two pixels to determine if this midpoint is inside or outside the circle boundary. This method is more easily applied to other conics, and for an integer circle radius.
- 23. 1) Accept theradius r and center (xc, yc).The point of the circumference of a circle with center 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.Perform the following test.
- 24. If Pk <0then, Xk+1= xk +1 and yk+1= yk Pk+1 = Pk +2xk +3 Otherwise, Pk >0 Xk+1= xk +1 and Yk+1= yk -1 Pk+1= Pk +2( xk - yk ) +5 4) Determine the symmetric points in other 7 octants.
- 25. 5) Translate eachcalculated pixel position by T ( xk , yk ) and display the pixel X= 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.
- 26.
- 27. 1) Accept theradius r and center (xc, yc).The point of the circumference of a circle with center as origin (x0, y0) =(0,r). 2) Calculate the initial decision parameter as PO =3 – 2r. 3) At each xk position starting at k=0.Perform the following test.
- 28. If Pk <0then, Xk+1= xk +1 and yk+1= yk Pk+1 = Pk +4xk +6. Otherwise, Pk > 0 Xk+1= xk +1 AND Yk+1= yk -1 Pk+1= Pk +4( xk - yk ) +10 4) Determine the symmetric points in other 7 octants.
- 29. 5) Translate eachcalculated pixel position by T ( xk , yk ) and display the pixel X= 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.
- 30. The DDAalgorithm 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.
- 31. Circles andellipses can be efficiently and accurately scan converted using midpoint methods. Bresenham’s line algorithm and the midpoint line algorithm methods are the most efficient.
- 33.