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Bresenham’s Midpoint Algorithm

Bresenham’s Midpoint Algorithm. CS5600 Computer Graphics Rich Riesenfeld Spring 2006. Lecture Set 1. Line Characterizations. Explicit: Implicit: Constant slope: Constant derivative:. Discrete Lines. Lines vs. Line Segments What is a discrete line segment?

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Bresenham’s Midpoint Algorithm

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  1. Bresenham’s Midpoint Algorithm CS5600Computer Graphics Rich Riesenfeld Spring 2006 Lecture Set 1

  2. Line Characterizations • Explicit: • Implicit: • Constant slope: • Constant derivative: CS 5600

  3. Discrete Lines • Lines vs. Line Segments • What is a discrete line segment? • This is a relatively recent problem • How to generate a discrete line? CS 5600

  4. “Good” Discrete Line - 1 • No gaps in adjacent pixels • Pixels close to ideal line • Consistent choices; same pixels in same situations CS 5600

  5. “Good” Discrete Line - 2 • Smooth looking • Even brightness in all orientations • Same line for as for • Double pixels stacked up? CS 5600

  6. Restricted Form • Line segment in first octant with 0 < m < 1 • Let us proceed CS 5600

  7. Two Line Equations • Explicit: • Implicit: Define: Hence, CS 5600

  8. From previous We have, Hence, CS 5600

  9. Relating Explicit to Implicit Eq’s Recall, Or, where, CS 5600

  10. Verify that Look at extreme values of y Investigate Sign ofF below line on line above line CS 5600

  11. The Picture above line below line CS 5600

  12. “Reasonable assumptions” have reduced the problem to making a binary choice at each pixel: Key to Bresenham Algorithm NE (next) E (next) (Previous) CS 5600

  13. Define a logical decision variable d linear in form incrementally updated (with addition) tells us whether to go E or NE Decision Variable d(logical) CS 5600

  14. The Picture NE Q midpoint M idealline previous E CS 5600

  15. The Picture (again) NE Q midpoint M idealline E previous CS 5600

  16. Observe the relationships • Suppose Q is above M, as before. • Then , M is below the line • So, means line is above M, • Need to move NE, increasey value CS 5600

  17. The Picture (again) NE midpoint Q idealline E previous CS 5600

  18. Observe the relationships • Suppose Q is below M, as before. • Then F(M) < 0 , implies M is above the line • So, F(M) < 0 , means line is below M, • Need to move to E; don’t increasey CS 5600

  19. M= Midpoint = : • Want to evaluate at M • Will use an incr decision var d • Let, CS 5600

  20. How will d be used? Recall, Therefore, CS 5600

  21. Case 1: Suppose E is chosen • Recall • ... CS 5600

  22. ... Case 1: Suppose E is chosen CS 5600

  23. Recall, Or, where, Review of Explicit to Implicit CS 5600

  24. Case 1: . CS 5600

  25. Case 2: Suppose NE chosen Recall ... CS 5600

  26. Case 2: Suppose NE ... CS 5600

  27. Case 2: . CS 5600

  28. Case 2: . CS 5600

  29. Summary • At each step of the procedure, we must choose between moving E or NE based on the sign of the decision variable d • Then update according to CS 5600

  30. What is initial value of d ? • First point is • First midpoint is • What is initial midpoint value? CS 5600

  31. What is initial value of d ? CS 5600

  32. What is initial value of d ? Hence, CS 5600

  33. What is initial value of d ? Hence, CS 5600

  34. What Does “2 x ” Do ? • Has the same 0-set • Changes the slope of the plane • Rotates plane about the 0-set line CS 5600

  35. What is initial value of d ? CS 5600

  36. What is initial value of d ? CS 5600

  37. More Summary • Initial value • Case 1: • Case 2: CS 5600

  38. More Summary Choose CS 5600

  39. Example • Line end points: • Deltas: CS 5600

  40. Graph 13 12 11 10 9 8 7 6 4 5 6 7 8 9 10 11 CS 5600

  41. Example ( dx = 4; dy = 3 ) • Initial value of CS 5600

  42. Graph 13 12 11 10 9 8 7 6 4 5 6 7 8 9 10 11 CS 5600

  43. Example (dx=4; dy=3) • Update value of d CS 5600

  44. Example (dx=4; dy=3) -2 • Update value of d • Last move was NE, so 2d(6,9) = 2d(y - dy) = 2(4 - 3) = - 2 d = 2 – 2 = 0  E CS 5600

  45. Graph 13 12 11 10 9 8 7 6 4 5 6 7 8 9 10 11 CS 5600

  46. Example (dx=4; dy=3 ) • Previous move was E CS 5600

  47. Graph 13 12 11 10 9 8 7 6 4 5 6 7 8 9 10 11 CS 5600

  48. Example (dx=4; dy=3 ) • Previous move was CS 5600

  49. Graph 13 12 11 10 9 8 7 6 4 5 6 7 8 9 10 11 CS 5600

  50. Graph 13 12 11 10 9 8 7 6 4 5 6 7 8 9 10 11 CS 5600

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