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Published byIrea Reap Modified over 2 years ago

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Many slides are borrowed from Foley & Van Dam

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Topics Viewing Transformation Pipeline in 2D Line and polygon clipping (Slow Fast) – Simultaneous Equations – Cohen-Sutherland Line Clipping Algorithm – Cyrus-Beck / Liang-Barsky Line Clipping Algorithm Sutherland-Hodgman Polygon Clipping

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? ?

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Line Clipping Endpoints Configuration – Both inside: ? – One in, one out: ? – Both outside: ?

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Line Clipping Endpoints Configuration – Both inside: No need for clipping – One in, one out: ? – Both outside: ?

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Line Clipping Endpoints Configuration – Both inside: No need for clipping – One in, one out: ? – Both outside: ?

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Line Clipping Endpoints Configuration – Both inside: No need for clipping – One in, one out: 1 intersection point – Both outside: ?

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Line Clipping Endpoints Configuration – Both inside: No need for clipping – One in, one out: 1 intersection point – Both outside: 2/0 intersection points Is more than 2 intersection points possible? No, intersection of 2 convexes is still a convex.

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Simultaneous Equations

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Any better algorithm? P1P2Observation Out ? Exterior half plane Possible interior half plane

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Any better algorithm? P1P2Observation Out No intersection with the line(reject) In ? Exterior half plane Possible interior half plane

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Any better algorithm? P1P2Observation Out No intersection with the line(reject) In No intersection with the line Out/InIn/Out? Exterior half plane Possible interior half plane

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Any better algorithm? P1P2Observation Out No intersection with the line(reject) In No intersection with the line Out/InIn/OutOne intersection point with the line Exterior half plane Possible interior half plane

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Any better algorithm? P1P2Observation Out No intersection with the line(reject) In No intersection with the line OOne intersection point with the line Exterior half plane Possible interior half plane 1 0 11 00 01 10 Same encoding Different encoding

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Cohen-Sutherland for Line Clipping For line i from 1 to 4 switch P0.bit[i] &P1.bit[i ] case 11: trivially reject; //outside of the window return; case 00: continue; //no intersection point case 10 or 01: compute intersection point; refresh one endpoint by the intersection point; continue 0 0 1

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Cohen-Sutherland for Line Clipping For line i from 1 to 4 switch P0.bit[i] &P1.bit[i ] case 11: trivially reject; //outside of the window return; case 00: continue; //no intersection point case 10 or 01: compute intersection point; refresh one endpoint by the intersection point; 2 Replace endpoint with the intersection Pt 0 1

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Cohen-Sutherland for Line Clipping For line i from 1 to 4 switch P0.bit[i] &P1.bit[i ] case 11: trivially reject; //outside of the window return; case 00: continue; //no intersection point case 10 or 01: compute intersection point; refresh one endpoint by the intersection point; 3 continue 0 0

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Cohen-Sutherland for Line Clipping For line i from 1 to 4 switch P0.bit[i] &P1.bit[i ] case 11: trivially reject; //outside of the window return; case 00: continue; //no intersection point case 10 or 01: compute intersection point; refresh one endpoint by the intersection point; 4 Replace endpoint with the intersection pt 1 0

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Any even better algorithm? A trivial quiz…… There are multiple layers of doors for entering a room, and multiple layers of doors for exiting a room. Assume we dont know the order of the doors. There are only 4 doors A, B, C, D. The doors testify when the person P entered and left. DoorEntering TimeLeaving Time A0.1 B0.5 C0.6 D0.8

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Any even better algorithm? A trivial quiz…… There are multiple layers of doors for entering a room, and multiple layers of doors for exiting a room. Assume we dont know the order of the doors. There are only 4 doors A, B, C, D. The doors testify when the person P entered and left. DoorEntering TimeLeaving Time A0.1 B0.5 C0.6 D0.8 Time range of the person in the room Max(0.1,0.5) – Min(0.6,0.8)

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Any even better algorithm? A trivial quiz…… There are multiple layers of doors for entering a room, and multiple layers of doors for exiting a room. Assume we dont know the order of the doors. There are only 4 doors A, B, C, D. The doors testify when the person P entered and left. DoorEntering TimeLeaving Time A0.1 B0.6 C0.3 D0.8

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Any even better algorithm? A trivial quiz…… There are multiple layers of doors for entering a room, and multiple layers of doors for exiting a room. Assume we dont know the order of the doors. There are only 4 doors A, B, C, D. The doors testify when the person P entered and left DoorEntering TimeLeaving Time A0.1 B0.6 C0.3 D0.8 Time range of the person in the room Max(0.1,0.6) – Min(0.3,0.8) Never in the room!

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Cyrus-Beck/Liang-Barsky Line Clipping P1 P0 E (t = 0.03) L (t = 0.85) L(t = 0.8) E (t = -0.03) Entering Time: t = 0.13 Leaving Time: t = 0.8 Ignore ts 1 Max(Entering) ~Min(Leaving) Max(E) > Min(L) ? Totally outside!

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Cyrus-Beck/Liang-Barsky Line Clipping Questions we need to answer: – How to decide entering or leaving – Knowing entering or leaving, how to compute t?

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