Clipping Algorithms; 8-5 (contd.) Computer Graphics. Spring CS4815

Transcription

1 Computer Graphics Spring

2 Outline Clipping Algorithms; 8-5 (contd.) 1 Clipping Algorithms; 8-5 (contd.)

3 Announcements Clipping Algorithms; 8-5 (contd.) Mid-term: Tues, 12.00, Week07 (???)

4 Liang-Barsky Line Clipper Liang-Barsky algorithm is a better approach by avoiding repeated shortening of segment s e; like C-S also works for 3-D clipping regions Idea is to return to parametric form of line (see prev. lecture) where x = x s + u x y = y s + u y, 0 u 1 x = x e x s, y = y e y s

5 Liang-Barsky Line Clipper Liang-Barsky algorithm is a better approach by avoiding repeated shortening of segment s e; like C-S also works for 3-D clipping regions Idea is to return to parametric form of line (see prev. lecture) where x = x s + u x y = y s + u y, 0 u 1 x = x e x s, y = y e y s

6 Liang-Barsky Line Clipper Applying window test x bl x s + u x x ur y bl y s + u y y ur We break this in to four inequalities of the form u k p k q k p 1 = x, p 2 = x, p 3 = y, p 4 = y, q 1 = x s x bl (left border) q 2 = x ur x s (right border) q 3 = y s y bl (bot border) q 4 = y ur y s (top border) We can now evaluate the four u k s with u k = q k /p k Liang-Barsky: it s all about finding the 4 u k s

7 Liang-Barsky Line Clipper We want to find the points where line segment crosses borders Extension of line segment will cross all four borders somewhere! We do this by working with the u k s

8 Liang-Barsky Line Clipper We want to find the points where line segment crosses borders Extension of line segment will cross all four borders somewhere! We do this by working with the u k s

9 Liang-Barsky Line Clipper We want to find the points where line segment crosses borders Extension of line segment will cross all four borders somewhere! We do this by working with the u k s

10 If line is parallel to window border then x = 0 or y = 0; so, two of p i s will be 0, also (either p 1, p 2 or p 3, p 4 ) For each p i that is 0, if q i < 0 then line is entirely outside that clipping boundary; o/w, q i 0 and line is inside Proceeding from u = to u = + the infinite extension of every non-parallel (to borders) line will cut 2 borders from outside to inside and 2 borders from inside to out If p k < 0, with respect to that border, the infinite extension of the line moves from outside to inside; p k > 0 means the line proceeds from inside to outside So two p i s will be positive and two negative The value of u k = q k /p k tells us critical information about the line

11 Instead of maintaining four u i s, we only maintain 2: t 1 for the outside to inside cases (p k < 0) t 2 for the inside to outside cases (p k > 0) t 1 is initialised to 0 (corresponds to s) and is taken to be the larger of the two cases where p k < 0 t 2 is initialised to 1 (corresponds to e) and is taken to be the smaller of the two cases where p k > 0 u i : out-in u j : in-out u 2 e e t 2 =1 u 4 t 1 u 3 s u 1

12 Instead of maintaining four u i s, we only maintain 2: t 1 for the outside to inside cases (p k < 0) t 2 for the inside to outside cases (p k > 0) t 1 is initialised to 0 (corresponds to s) and is taken to be the larger of the two cases where p k < 0 t 2 is initialised to 1 (corresponds to e) and is taken to be the smaller of the two cases where p k > 0 u i : out-in u j : in-out u 2 e e t 2 =1 u 4 t 1 u 3 s u 1

13 Instead of maintaining four u i s, we only maintain 2: t 1 for the outside to inside cases (p k < 0) t 2 for the inside to outside cases (p k > 0) t 1 is initialised to 0 (corresponds to s) and is taken to be the larger of the two cases where p k < 0 t 2 is initialised to 1 (corresponds to e) and is taken to be the smaller of the two cases where p k > 0 u i : out-in u j : in-out u 2 e e t 2 =1 u 4 t 1 u 3 s u 1

14 Instead of maintaining four u i s, we only maintain 2: t 1 for the outside to inside cases (p k < 0) t 2 for the inside to outside cases (p k > 0) t 1 is initialised to 0 (corresponds to s) and is taken to be the larger of the two cases where p k < 0 t 2 is initialised to 1 (corresponds to e) and is taken to be the smaller of the two cases where p k > 0 u i : out-in u j : in-out u 2 e e t 2 =1 u 4 t 1 u 3 s u 1

15 Instead of maintaining four u i s, we only maintain 2: t 1 for the outside to inside cases (p k < 0) t 2 for the inside to outside cases (p k > 0) t 1 is initialised to 0 (corresponds to s) and is taken to be the larger of the two cases where p k < 0 t 2 is initialised to 1 (corresponds to e) and is taken to be the smaller of the two cases where p k > 0 u i : out-in u j : in-out u 2 e e t 2 =1 u 4 t 1 u 3 s u 1

16 u 1 s t 2 =1 e u 2 t 1 >1 u 4 u i : out-in u j : in-out if t 1 > t 2 line is completely outside the clipping region and it can be discarded Over Cohen-Sutherland, Liang-Barsky has been found to run 36% faster for 2-D lines and 40% faster for 3-D lines Note: alg doesn t depend on any ordering of start and end points of line See tutorial here

17 u 1 s t 2 =1 e u 2 t 1 >1 u 4 u i : out-in u j : in-out if t 1 > t 2 line is completely outside the clipping region and it can be discarded Over Cohen-Sutherland, Liang-Barsky has been found to run 36% faster for 2-D lines and 40% faster for 3-D lines Note: alg doesn t depend on any ordering of start and end points of line See tutorial here

18 u 1 s t 2 =1 e u 2 t 1 >1 u 4 u i : out-in u j : in-out if t 1 > t 2 line is completely outside the clipping region and it can be discarded Over Cohen-Sutherland, Liang-Barsky has been found to run 36% faster for 2-D lines and 40% faster for 3-D lines Note: alg doesn t depend on any ordering of start and end points of line See tutorial here

19 u 1 s t 2 =1 e u 2 t 1 >1 u 4 u i : out-in u j : in-out if t 1 > t 2 line is completely outside the clipping region and it can be discarded Over Cohen-Sutherland, Liang-Barsky has been found to run 36% faster for 2-D lines and 40% faster for 3-D lines Note: alg doesn t depend on any ordering of start and end points of line See tutorial here