1 CSCE 441 Computer Graphics: Clipping Lines Jinxiang Chai
1 CSCE 441 Computer Graphics: Clipping Lines Jinxiang Chai.
Dec 20, 2015
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1
CSCE 441 Computer Graphics:Clipping Lines
Jinxiang Chai
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OpenGL Geometric Primitives
All geometric primitives are specified by vertices
GL_QUAD_STRIPGL_QUAD_STRIP
GL_POLYGONGL_POLYGON
GL_TRIANGLE_STRIPGL_TRIANGLE_STRIP
GL_TRIANGLE_FANGL_TRIANGLE_FAN
GL_POINTSGL_POINTSGL_LINESGL_LINES
GL_LINE_LOOPGL_LINE_LOOPGL_LINE_STRIPGL_LINE_STRIP
GL_TRIANGLESGL_TRIANGLES
GL_QUADSGL_QUADS
3/135
Review: Drawing General Polygons
How to draw every interior pixels?
?
Scanline conversion of polygons!
4/135
Review: Curved Boundaries
How to deal with curved boundaries?
Boundary fill algorithm!
5/135
Review: How to deal with this?
Multiple color boundaries?
Flood fill algorithm!
6/135
Why Clip?
We do not want to waste time drawing objects that are outside of viewing window (or clipping window)
7/135
Clipping Points
Given a point (x, y) and clipping window (xmin, ymin), (xmax, ymax), determine if the point should be drawn
(xmin,ymin)
(xmax,ymax)
(x,y)
8/135
Clipping Points
Given a point (x, y) and clipping window (xmin, ymin), (xmax, ymax), determine if the point should be drawn
(xmin,ymin)
(xmax,ymax)
(x,y)
xmin=<x<=xmax?ymin=<y<=ymax?
9/135
Clipping Points
Given a point (x, y) and clipping window (xmin, ymin), (xmax, ymax), determine if the point should be drawn
(x1,y1)
(x2,y2)
(xmin,ymin)
(xmax,ymax)
xmin=<x<=xmax?ymin=<y<=ymax?
10/135
Clipping Points
Given a point (x, y) and clipping window (xmin, ymin), (xmax, ymax), determine if the point should be drawn
(x1,y1)
(x2,y2)
(xmin,ymin)
(xmax,ymax)
xmin=<x1<=xmax Yes
ymin=<y1<=ymax Yes
11/135
Clipping Points
Given a point (x, y) and clipping window (xmin, ymin), (xmax, ymax), determine if the point should be drawn
(x1,y1)
(x2,y2)
(xmin,ymin)
(xmax,ymax)
xmin=<x2<=xmax Noymin=<y2<=ymax No
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Clipping Lines
1P
2P
3P4P
5P
6P
7P
8P 9P
10P
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Clipping Lines
5P
6̂P
7P
8P
9̂P
10P̂
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Clipping Lines
Given a line with end-points (x0, y0), (x1, y1)
and clipping window (xmin, ymin), (xmax, ymax),
determine if line should be drawn and clipped end-points of line to draw.
(x0, y0)
(x1, y1)
(xmin,ymin)
(xmax,ymax)
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Clipping Lines
1P
2P
3P4P
5P
6P
7P
8P 9P
10P
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Outline
Simple line clipping algorithm Cohen-Sutherland Liang-Barsky
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Clipping Lines
1P
2P
3P4P
5P
6P
7P
8P 9P
10P
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Clipping Lines – Simple Algorithm
If both end-points inside rectangle, draw line If one end-point outside,
intersect line with all edges of rectangle
clip that point and repeat test
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Clipping Lines – Simple Algorithm
),( 00 yx
),( minmin yx
),( maxmax yx
),( 11 yx
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Clipping Lines – Simple Algorithm
),( 00 yx
),( minmin yx
),( maxmax yx
),( 11 yx
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Intersecting Two Lines
),( 00 yx
),( 22 yx
),( 11 yx
),( 33 yx
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Intersecting Two Lines
),( 00 yx
),( 22 yx
),( 11 yxtxxxtx )()( 010 ),( 33 yx
tyyyty )()( 010
10 t
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Intersecting Two Lines
),( 00 yx
),( 22 yx
),( 11 yxtxxxtx )()( 010 ),( 33 yx
tyyyty )()( 010
0)0( xx 0)0( yy
1)1( xx 1)1( yy
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Intersecting Two Lines
),( 00 yx
),( 22 yx
),( 11 yx
sxxxtxxx )()( 232010
),( 33 yx
syyytyyy )()( 232010
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Intersecting Two Lines
),( 00 yx
),( 22 yx
),( 11 yx
02
02
3201
3201
yy
xx
s
t
yyyy
xxxx
),( 33 yx
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Intersecting Two Lines
),( 00 yx
),( 22 yx
),( 11 yx
02
02
3201
3201
yy
xx
s
t
yyyy
xxxx
),( 33 yx
Substitute t or s back into equation to find intersection
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Clipping Lines – Simple Algorithm
),( 00 yx
),( minmin yx
),( maxmax yx
),( 11 yx
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Clipping Lines – Simple Algorithm
),( 00 yx
),( minmin yx
),( maxmax yx
),( 11 yx
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Clipping Lines – Simple Algorithm
),( 00 yx
),( minmin yx
),( maxmax yx
),( 11 yx
30/135
Clipping Lines – Simple Algorithm
),( 00 yx
),( minmin yx
),( maxmax yx
),( 11 yx
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Clipping Lines – Simple Algorithm
)ˆ,ˆ( 00 yx
),( minmin yx
),( maxmax yx
),( 11 yx
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Clipping Lines – Simple Algorithm
)ˆ,ˆ( 00 yx
),( minmin yx
),( maxmax yx
),( 11 yx
33/135
Clipping Lines – Simple Algorithm
)ˆ,ˆ( 00 yx
),( minmin yx
),( maxmax yx
),( 11 yx
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Clipping Lines – Simple Algorithm
)ˆ,ˆ( 00 yx
),( minmin yx
),( maxmax yx
),( 11 yx
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Clipping Lines – Simple Algorithm
),( 00 yx
),( minmin yx
),( maxmax yx
),( 11 yx
36/135
Clipping Lines – Simple Algorithm
),( 00 yx
),( minmin yx
),( maxmax yx
),( 11 yx
37/135
Clipping Lines – Simple Algorithm
)ˆ,ˆ( 00 yx
),( minmin yx
),( maxmax yx
),( 11 yx
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Clipping Lines – Simple Algorithm
)ˆ,ˆ( 00 yx
),( minmin yx
),( maxmax yx
),( 11 yx
39/135
Clipping Lines – Simple Algorithm
)ˆ,ˆ( 00 yx
),( minmin yx
),( maxmax yx
),( 11 yx
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Clipping Lines – Simple Algorithm
)ˆ,ˆ( 00 yx
),( minmin yx
),( maxmax yx
)ˆ,ˆ( 11 yx
41/135
Clipping Lines – Simple Algorithm
)ˆ,ˆ( 00 yx
),( minmin yx
),( maxmax yx
)ˆ,ˆ( 11 yx
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Clipping Lines – Simple Algorithm
Lots of intersection tests makes algorithm expensive
Complicated tests to determine if intersecting rectangle
Is there a better way?
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Trivial Accepts
Big Optimization: trivial accepts/rejects How can we quickly decide whether line
segment is entirely inside window Answer: test both endpoints
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Trivial Accepts
Big Optimization: trivial accepts/rejects How can we quickly decide whether line
segment is entirely inside window Answer: test both endpoints
45/135
Trivial Rejects
How can we know a line is outside of the window
Answer:
46/135
Trivial Rejects
How can we know a line is outside of the window
Answer:
Wrong side
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Trivial Rejects
How can we know a line is outside of the window
Answer: both endpoints on wrong side of same edge, can trivially reject the line
48/135
Cohen-Sutherland Algorithm
Classify p0, p1 using region codes c0, c1
If , trivially reject If , trivially accept Otherwise reduce to trivial cases by splitting
into two segments
010 cc
010 cc
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Cohen-Sutherland Algorithm
Every end point is assigned to a four-digit binary value, i.e. Region code
Each bit position indicates whether the point is inside or outside of a specific window edges
bit 4 bit 3 bit 2 bit 1
leftrightbottomtop
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Cohen-Sutherland Algorithm
bit 4 bit 3 bit 2 bit 1
leftrightbottomtop
top
rightleft
bottom
Region code?
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Cohen-Sutherland Algorithm
bit 4 bit 3 bit 2 bit 1
leftrightbottomtop
top
rightleft
bottom
0010
?
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Cohen-Sutherland Algorithm
bit 4 bit 3 bit 2 bit 1
leftrightbottomtop
top
rightleft
bottom
0010
0100
53/135
Cohen-Sutherland Algorithm
00000001
0101
1001 1000 1010
0010
01100100
54/135
Cohen-Sutherland Algorithm
00000001
0101
1001 1000 1010
0010
01100100
),( 00 yx
),( 11 yx
55/135
Cohen-Sutherland Algorithm
Classify p0, p1 using region codes c0, c1
If , trivially reject If , trivially accept Otherwise reduce to trivial cases by splitting
into two segments
010 cc
010 cc
56/135
Cohen-Sutherland Algorithm
Classify p0, p1 using region codes c0, c1
If , trivially reject If , trivially accept Otherwise reduce to trivial cases by splitting
into two segments
010 cc
010 cc
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Cohen-Sutherland Algorithm
Classify p0, p1 using region codes c0, c1
If , trivially reject If , trivially accept Otherwise reduce to trivial cases by splitting
into two segments
010 cc
010 cc
00000001
0101
1001 1000 1010
0010
01100100
Line is outside the window! reject
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Cohen-Sutherland Algorithm
Classify p0, p1 using region codes c0, c1
If , trivially reject If , trivially accept Otherwise reduce to trivial cases by splitting
into two segments
010 cc
010 cc
00000001
0101
1001 1000 1010
0010
01100100
Line is outside the window! reject
59/135
Cohen-Sutherland Algorithm
Classify p0, p1 using region codes c0, c1
If , trivially reject If , trivially accept Otherwise reduce to trivial cases by splitting
into two segments
010 cc
010 cc
00000001
0101
1001 1000 1010
0010
01100100
Line is inside the window! draw
60/135
Cohen-Sutherland Algorithm
Classify p0, p1 using region codes c0, c1
If , trivially reject If , trivially accept Otherwise reduce to trivial cases by splitting
into two segments
010 cc
010 cc
00000001
0101
1001 1000 1010
0010
01100100
61/135
Window Intersection
(x1, y1), (x2, y2) intersect with vertical edge at xright
yintersect = y1 + m(xright – x1)
• where m=(y2-y1)/(x2-x1)
(x1, y1), (x2, y2) intersect with horizontal edge at ybottom
xintersect = x1 + (ybottom – y1)/m
• where m=(y2-y1)/(x2-x1)
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Example 1
00000001
0101
1001 1000 1010
0010
01100100
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Example 1
00000001
0101
1001 1000 1010
0010
01100100
64/135
Example 1
00000001
0101
1001 1000 1010
0010
01100100
65/135
Example 1
00000001
0101
1001 1000 1010
0010
01100100
66/135
Example 2
00000001
0101
1001 1000 1010
0010
01100100
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Example 2
00000001
0101
1001 1000 1010
0010
01100100
68/135
Example 2
00000001
0101
1001 1000 1010
0010
01100100
69/135
Example 2
00000001
0101
1001 1000 1010
0010
01100100
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Example 2
00000001
0101
1001 1000 1010
0010
01100100
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Example 2
00000001
0101
1001 1000 1010
0010
01100100
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Example 2
00000001
0101
1001 1000 1010
0010
01100100
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Example 3
00000001
0101
1001 1000 1010
0010
01100100
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Example 3
00000001
0101
1001 1000 1010
0010
01100100
75/135
Example 3
00000001
0101
1001 1000 1010
0010
01100100
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Example 3
00000001
0101
1001 1000 1010
0010
01100100
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Cohen Sutherland
Java applet: click here
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Cohen-Sutherland Algorithm
Extends easily to 3D line clipping27 regions6 bits
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Cohen-Sutherland Algorithm
Use region codes to quickly eliminate/include lines
Best algorithm when trivial accepts/rejects are common
Must compute viewing window clipping of remaining lines
Non-trivial clipping cost More efficient algorithms exist
80/135
Liang-Barsky Algorithm
Parametric definition of a line: x = x1 + uΔx
y = y1 + uΔy
Δx = (x2-x1), Δy = (y2-y1), 0<=u<=1
Lines are oriented: classify lines as moving inside to out or outside to in
Goal: find range of u for which x and y both inside the viewing window
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Liang-Barsky Algorithm
Current clipped line (u1=0,u2=1)
(-.5,-.5)
(2,1)
82/135
Liang-Barsky Algorithm
Current clipped line (u1=0.333,u2=0.6)
(-.5,-.5)
(2,1)
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Liang-Barsky Algorithm
For lines starting outside of boundary, update its starting point (u1)
For lines starting inside of boundary, update end point (u2)
For lines paralleling the boundaries and outside window,
reject it.
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Liang-Barsky Algorithm
For lines starting outside of boundary, update its starting point (u1)
For lines starting inside of boundary, update end point (u2)
For lines paralleling the boundaries and outside window,
reject it.
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Liang-Barsky Algorithm
Mathematically: xmin <= x1 + uΔx <= xmax
ymin <= y1 + uΔy <= ymax
Rearranged 1: u*(-Δx) <= (x1 – xmin) 2: u*(Δx) <= (xmax – x1) 3: u*(-Δy) <= (y1 – ymin) 4: u*(Δy) <= (ymax – y1) gen: u*(pk) <= (qk), k=1,2,3,4
86/135
Liang-Barsky Algorithm
Rules:
1) pk = 0: the line is parallel to boundaries If for that same k, qk < 0, it’s outside Otherwise it’s inside
2) pk < 0: the line starts outside this boundary rk = qk/pk
u1 = max(0, rk, u1) /**update starting point**/ pk > 0: the line starts inside the boundary
rk = qk/pk
u2 = min(1, rk, u2) /** update end point**/ If u1 > u2, the line is completely outside
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Liang-Barsky Algorithm
Current clipped line (u1=0,u2=1)
(-.5,-.5)
(2,1)
88/135
Liang-Barsky Algorithm
Current clipped line (u1=0,u2=1)
(-.5,-.5)
(2,1)
Δx = (x2-x1)=2.5
Δy = (y2-y1)=1.5
p1 = -Δx =-2.5<0
q1 = (x1-xmin)=-.5
r1 = q1/p1=0.2
89/135
Liang-Barsky Algorithm
Current clipped line (u1=0,u2=1)
(-.5,-.5)
(2,1)
Δx = (x2-x1)=2.5
Δy = (y2-y1)=1.5
p1 = -Δx =-2.5<0
q1 = (x1-xmin)=-.5
r1 = q1/p1=0.2line starts outside this boundary
90/135
Liang-Barsky Algorithm
Current clipped line (u1=0,u2=1)
(-.5,-.5)
(2,1)
Δx = (x2-x1)=2.5
Δy = (y2-y1)=1.5
p1 = -Δx =-2.5<0
q1 = (x1-xmin)=-.5
r1 = q1/p1=0.2
91/135
Liang-Barsky Algorithm
Current clipped line (u1=0,u2=1)
(-.5,-.5)
(2,1)
Δx = (x2-x1)=2.5
Δy = (y2-y1)=1.5
p1 = -Δx =-2.5<0
q1 = (x1-xmin)=-.5
r1 = q1/p1=0.2
92/135
Liang-Barsky Algorithm
Current clipped line (u1=0,u2=1)
(-.5,-.5)
(2,1)
Δx = (x2-x1)=2.5
Δy = (y2-y1)=1.5
p1 = -Δx =-2.5<0
q1 = (x1-xmin)=-.5
r1 = q1/p1=0.2u1 = max(0, r1, u1)
93/135
Liang-Barsky Algorithm
Current clipped line (u1=0.2,u2=1)
(-.5,-.5)
(2,1)
u1<u2?
- yes: continue
- no: the line is completely outside window
94/135
Liang-Barsky Algorithm
Current clipped line (u1=0.2,u2=1)
(-.5,-.5)
(2,1)
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Liang-Barsky Algorithm
Current clipped line (u1=0.2,u2=1)
(-.5,-.5)
(2,1)
Δx = (x2-x1)=2.5
Δy = (y2-y1)=1.5
p2 = Δx =2.5>0
q2 = (xmax-x1)=1.5
r2 = q2/p2=0.6
96/135
Liang-Barsky Algorithm
Current clipped line (u1=0.2,u2=1)
(-.5,-.5)
(2,1)
Δx = (x2-x1)=2.5
Δy = (y2-y1)=1.5
p2 = Δx =2.5>0
q2 = (xmax-x1)=1.5
r2 = q2/p2=0.6u2 = min(1, r2, u2)
97/135
Liang-Barsky Algorithm
Current clipped line (u1=0.2,u2=0.6)
(-.5,-.5)
(2,1)
Δx = (x2-x1)=2.5
Δy = (y2-y1)=1.5
p2 = Δx =2.5>0
q2 = (xmax-x1)=1.5
r2 = q2/p2=0.6
98/135
Liang-Barsky Algorithm
Current clipped line (u1=0.2,u2=0.6)
(-.5,-.5)
(2,1)
u1<u2?
- yes: continue
- no: the line is completely outside window
99/135
Liang-Barsky Algorithm
Current clipped line (u1=0.2,u2=0.6)
(-.5,-.5)
(2,1)
Δx = (x2-x1)=2.5
Δy = (y2-y1)=1.5
p3 = -Δy =-1.5<0
q3 = (y1-ymin)=0.5
r3 = q3/p3=0.333
100/135
Liang-Barsky Algorithm
Current clipped line (u1=0.2,u2=0.6)
(-.5,-.5)
(2,1)
Δx = (x2-x1)=2.5
Δy = (y2-y1)=1.5
p3 = -Δy =-1.5<0
q3 = (y1-ymin)=0.5
r3 = q3/p3=0.333u1 = max(0, r3, u1)
101/135
Liang-Barsky Algorithm
Current clipped line (u1=0.333,u2=0.6)
(-.5,-.5)
(2,1)
Δx = (x2-x1)=2.5
Δy = (y2-y1)=1.5
p3 = -Δy =-1.5<0
q3 = (y1-ymin)=0.5
r3 = q3/p3=0.333
102/135
Liang-Barsky Algorithm
Current clipped line (u1=0.333,u2=0.6)
(-.5,-.5)
(2,1)
103/135
Liang-Barsky Algorithm
Current clipped line (u1=0.333,u2=0.6)
(-.5,-.5)
(2,1)
Δx = (x2-x1)=2.5
Δy = (y2-y1)=1.5
p3 = -Δy =-1.5<0
q3 = (y1-ymin)=0.5
r3 = q3/p3=0.333
104/135
Liang-Barsky Algorithm
Current clipped line (u1=0.333,u2=0.6)
(-.5,-.5)
(2,1)
Δx = (x2-x1)=2.5
Δy = (y2-y1)=1.5
p4 = Δy =1.5>0
q4 = (ymax-y1)=1.5
r4 = q4/p4=1
105/135
Liang-Barsky Algorithm
Current clipped line (u1=0.333,u2=0.6)
(-.5,-.5)
(2,1)
Δx = (x2-x1)=2.5
Δy = (y2-y1)=1.5
p4 = Δy =1.5>0
q4 = (ymax-y1)=1.5
r4 = q4/p4=1u2 = min(1, r4, u2)
106/135
Liang-Barsky Algorithm
Check the left edge (u1=0.333,u2=0.6)
(-.5,-.5)
(2,1)
u1<u2
xnew1=x1+Δx*u1
ynew1=y1+Δy*u1
xnew2=x2+Δx*u2
ynew2=y2+Δy*u2
107/135
Liang-Barsky Algorithm
Faster than Cohen-Sutherland
Extension to 3D is easy
- Parametric representation for 3D lines
- Compute u1,u2 based on the intersection between line and plane
108/135
Comparison
Cohen-Sutherland Repeated clipping is expensive Best used when trivial acceptance and rejection is
possible for most lines Liang-Barsky
Computation of t-intersections is cheap (only one division)
Computation of (x,y) clip points is only done once
Algorithm doesn’t consider trivial accepts/rejects Best when many lines must be clipped
109/135
Curve Clipping
110/135
Curve Clipping
111/135
Curve Clipping
Approximate a curve using a set of straight-line segments
Apply line clipping for curve clipping
112/135
Next Lecture
Polygon fill-area clipping
113/135
Next Lecture
Polygon fill-area clipping
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