1. Clipping
Jian Huang, CS594
This set of slides reference slides devised at Ohio State and MIT.

2. Viewing Pipeline Revisited
Object Space
World Space
Eye Space
Clipping Space
Canonical view volume
Screen Space
Object space: coordinate where each component is defined
World space: all components put together via affine transformation. (camera, lighting defined in this space)
Eye space: camera at the origin, view direction coincides with the z axis. Hither and Yon perpendicular to the z axis
Clipping space: All point is in homogeneous coordinate. Perspective division gets everything into 3D image space.
3D image space (Canonical view volume): a parallelpipied shape defined by (-1:1,-1:1,0,1). Objects distorted.
Screen space: x and y mapped to screen pixel coordinates
January 17, 2019

3. Why do clipping
Clipping is a visibility preprocess. In real-world scenes clipping can remove a substantial percentage of the environment from consideration.
Clipping offers an important optimization
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4. What is clipping, two views
Clipping is to spatially partition geometric primitives, according to their containment within some region. Clipping can be used to:
Distinguish whether geometric primitives are inside or outside of a viewing frustum or picking frustum
Detecting intersections between primitives
Clipping is to subdivide geometric primitives. Several other potential applications.
Binning geometric primitives into spatial data structures
computing analytical shadows.
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5. Point Clipping
(x, y)
is inside iff
AND
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6.  = Line Clipping - Half Plane Tests
Modify endpoints to lie in rectangle
“Interior” of rectangle?
Answer: intersection of 4 half-planes
3D ? (intersection of 6 half-planes)
y < ymax
y > ymin
x > xmin
x < xmax = 
interior
xmin
xmax
ymin
ymax
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7. Line Clipping Is end-point inside a clip region? - half-plane test
If outside, calculate intersection betwee line and the clipping rectangle and make this the new end point
Both endpoints inside: trivial accept
One inside: find intersection and clip
Both outside: either clip or reject (tricky case)
January 17, 2019

8. Cohen-Sutherland Algorithm (Outcode clipping)
Classifies each vertex of a primitive, by generating an outcode. An outcode identifies the appropriate half space location of each vertex relative to all of the clipping planes. Outcodes are usually stored as bit vectors.
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9. Cohen-Sutherland Algorithm (Outcode clipping)
if (outcode1 == '0000' and outcode2 == ‘0000’) then
line segment is inside
else
if ((outcode1 AND outcode2) == 0000) then
line segment potentially crosses clip region
line is entirely outside of clip
region
endif
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10. The Maybe cases? If neither trivial accept nor reject:
Pick an outside endpoint (with nonzero outcode)
Pick an edge that is crossed (nonzero bit of outcode)
Find line's intersection with that edge
Replace outside endpoint with intersection point
Repeat outcode test until trivial accept or reject
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11. The Maybe case
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12. The Maybe Case
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13. One Plane At a Time Clipping (a.k.a. Sutherland-Hodgeman Clipping)
The Sutherland-Hodgeman triangle clipping algorithm uses a divide-and-conquer strategy.
Clip a triangle against a single plane. Each of the clipping planes are applied in succession to every triangle.
There is minimal storage requirements for this algorithm, and it is well suited to pipelining.
It is often used in hardware implementations.
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14. Sutherland-Hodgman Polygon Clipping Algorithm
Subproblem:
clip a polygon (input: vertex list) against a single clip edges
output the vertex list(s) for the resulting clipped polygon(s)
Clip against all four planes
generalizes to 3D (6 planes)
generalizes to any convex clip polygon/polyhedron
Used in viewing transforms
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15. Polygon Clipping At Work
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16. Sutherland-Hodgman
SHclippedge(var: ilist, olist: list; ilen, olen, edge : integer)
s = ilist[ilen];     olen = 0;
for i = 1 to ilen do
d := ilist[i];
if (inside(d, edge) then
if (inside(s, edge) then           case 1
addlist(d, olist);     olen := olen + 1;
else                                          case 4
n := intersect(s, d, edge);
addlist(n, olist); addlist(d, olist);    olen = olen + 2;
else if (inside(s, edge) then            -- case 2
n := intersect(s, d, edge); addlist(n, olist);  olen ++; s = d;
end_for;
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17. Sutherland-Hodgman
SHclip(var: ilist, olist: list; ilen, olen : integer)
SHclippedge(ilist, tmplist1, ilen, tlen1, RIGHT);
SHclippedge(tmplist1, tmplist2, tlen1, tlen2, BOTTOM);
SHclippedge(tmplist2, tmplist1, tlen2, tlen1, LEFT);
SHclippedge(tmplist1, olist, tlen1, olen, TOP);
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18. With Pictures
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19. Sutherland-Hodgman Advantages: Disadvantages:
Elegant (few special cases)
Robust (handles boundary and edge conditions well)
Well suited to hardware
Canonical clipping makes fixed-point
implementations manageable
Disadvantages:
Only works for convex clipping volumes
Often generates more than the minimum number of triangles needed
Requires a divide per edge
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20. Interpolating Parameters
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21. 3D Clipping (Planes) Red Polygon – Clipx y z
image plane
near
far
Red Polygon – Clip
Transform into 4D Clipping space (canonical viewing volume) Homogenous co-ordinates
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22. Naïve 3D Euclidean Space Clipping
After perspective projection, Euclidean space is not linear!!
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23. Difficulty with Euclidean Space Clipping
Clipping will handle most cases. However, there is one case in general that cannot be handled this way.
Parts of a primitive lie both in front of and behind the viewpoint. This complication is caused by our projection stage.
It has the nasty habit of mapping objects in behind the viewpoint to positions in front of it.
Solution: clip in homogeneous coordinate
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24. 4DPolygonClip Use Sutherland Hodgman algorithm
Use arrays for input and output lists
There are six planes of course !
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25. 4D Clipping OpenGL uses -1<=x<=1, -1<=y<=1, -1<=z<=1
We use: -1<=x<=1, -1<=y<=1, -1<=z <=0
Must clip in homogeneous coordinates:
w>0: -w<=x<=w, -w<=y<=w, -w<=z<=0
w<0: -w>=x>=w, -w>=y>=w, -w>=z>=0
Consider each case separately
What issues arise ?
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26. 4D Clipping Clip boundary: x/w = 1 i.e. (x–w=0);
Point A is inside, Point B is outside. Clip edge AB
x = Ax + t(Bx – Ax)
y = Ay + t(By – Ay)
z = Az + t(Bz – Az)
w = Aw + t(Bw – Aw)
Clip boundary: x/w = 1 i.e. (x–w=0);
w-x = Aw – Ax + t(Bw – Aw – Bx + Ax) = 0
Solve for t.
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27. Still have issues with 4D Clipping
W=-X
W=X
P1 and P2 map to same physical point !
Solution:
Clip against both regions
Negate points with negative W
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28. Still have issues with 4D Clipping
Inf
-Inf P2
Line straddles both regions
After projection one gets two line segments
How to do this? Only before the perspective division
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29. More on Perspective Transform
There are a number of perspective matrices depending on the field of view desired, and the near and far plane. But same essential idea: a perspective matrix moves the depth value z into the fourth column, where it will used to divide through the x and y values
when the final homogeneous coordinate is translated back into a 3D point (3D image space), z is usually referred to as ‘depth’ of the point
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30. More on perspective transforms
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31. More on Perspective Transform
Perspective projections categorized by the number of axis the view plane cuts (ie 1-point, 2-point or 3-point perspective)
the plane cuts the z axis, lines parallel to the z meets at infinity; lines parallel to the x or y axis will not meet at infinity. 1-point perspective.
the plane cuts the x and z axis, lines parallel to the x/z axis meet at infinity; lines parallel to the y axis will not meet at infinity. 2-point perspective.
if the plane cuts the x, y, and z axis then lines parallel to the x, y, or z axis will meet at infinity. This is 3-point perspective.
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32. More on Homogeneous Coordinates
To 4D: (x,y,z) -> (x,y,z,1)
Back to 3D: (x,y,z,w) -> (x/w, y/w, z/w)
A point is on a plane if the point satisfies 0 == A*x + B*y + C*z + D
Point P: (x,y,z,1).
Representing a plane N = (A,B,C,D). Point P is on the plane, if P dot N == 0
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33. Transforming Normals
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34. Transforming Normals Transform P to P’ -> P’ = M * P (M is known)
and transform N to N’ -> N’ = Q * N
Let Q be our transformation matrix for N.
We want to make sure that after transformation, N’ is the normal of the transformed plane. That is, N’T * P’ = 0
We get:
N’T * P’ = (Q * N)T * (M * P) = NT * QT * M * P = 0
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35. Transforming Normals So, need QT *M = Identity Then, QT = M –1
Still, we want N’ = Q * N.
Q = (M –1)T
January 17, 2019

36. Viewing Pipeline Revisited
Object Space
World Space
Eye Space
Clipping Space
Canonical view volume
Screen Space
Object space: coordinate where each component is defined
World space: all components put together via affine transformation. (camera, lighting defined in this space)
Eye space: camera at the origin, view direction coincides with the z axis. Hither and Yon perpendicular to the z axis
Clipping space: All point is in homogeneous coordinate. Perspective division gets everything into 3D image space.
3D image space (Canonical view volume): a parallelpipied shape defined by (-1:1,-1:1,0,1). Objects distorted.
Screen space: x and y mapped to screen pixel coordinates
January 17, 2019

37. Right-Handed Or Left-Handed
Usually use right-handed coordinate (convention in math)
Left-handed good for screen
To convert, just flip x or y or z. (any one of the three)
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38. How about the viewing pipeline?
The range of z for the canonical view volume is [0,1]. x and y still remain the same.
Is converting back and forth (flipping) a major issue?
January 17, 2019