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Unit 5: Implementing Ellipse Algorithm



                                                                                                  Notes
                                      2
                                    2
                                 =  b(x+ 2x +1)+ a(y -  y +  1 4 ) -  ab 2
                                                               2
                                                    2
                                                 2
                                                       k
                                          k
                                      k
                                                    k
            If P k  < 0, select E:
                              E
                             P k+1   = f(x k  + 2, y k  – ½)
                                                       2 2
                                   2
                                          2
                                 = b (x k  + 2)  + a (y k  – ½)  – a b
                                                    2
                                             2
                                 =  b(x+ 4x +4)+ a(y -  y +  1 4 ) - ab 2
                                      2
                                    2
                                                               2
                                                 2
                                                    2
                                                    k
                                      k
                                          k
                                                       k
                             E
                                  E
            Change of  P  is :  DP  =  P k+1  -  P =b (2x+ 3)
                      E
                                          2
                      k
                                      k
                             k
                                             k
            If P k  > 0, select SE:
                             P k+1   = f(x k  + 2, y k  – 3/2)
                              SE
                                 = b (x k  + 2)  + a (y k  – 3/2)  – a b
                                                         2 2
                                             2
                                   2
                                          2
                                                     2
                                 =  b(x+ 4x +4)a (y -  3y +9/4) - a b 2
                                      2
                                                  2
                                    2
                                                2
                                                                2
                                          k
                                                       k
                                      k
                                                  k
                      SE
                                    SE
            Change of  P  = is   DP  =  P k+1   – P k  = b (2x k  + 3) – 2a (y k  – 1)
                               SE
                                                        2
                                             2
                      k
                               k
            Calculate the changes of ∆Pk:
            If E is selected,
                            DP k+1   = b (2x k  + 5)
                                   2
                              E
                            D P  =  DP k+1  -  DP =2b 2
                             2
                               E
                                           E
                                     E
                               k
                                          k
                            DP k+1   = b (2x k  + 5) – 2a (y k  – 1)
                                              2
                                   2
                              SE
                           D P  =  DP k+1  -  D SE  = 2 b 2
                                     SE
                              SE
                             2
                                          k
                              k
            If SE is selected,
                                   2
                              E
                            DP k+1   = b (2x k  + 5)
                               E
                             2
                                           E
                                     E
                            D P  =  DP k+1  -  DP =2b 2
                                          k
                               k
                            DP k+1   = b (2x k  + 5) – 2a (y k  – 2)
                              SE
                                   2
                                              2
                                                    2
                           D P SE  DP SE  -  DP =2(a +b )
                                                2
                                           SE
                            2
                              k   =   k+1  k
            Initial values:
                                               2
                               x 0  = 0, y 0  = b, P 0  = b  + ¼a  (1 – 4b)
                                                    2
                                     2
                               E
                             DP  =  3b ,P =3b -  2a(b -  1)
                                      D
                                                  2
                                              2
                                        SE
                               0        0
            In region II (dy/dx < –1), all calculations are similar to that in region we except that y is
            decremented in each step.
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