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March 1, 2009 Dr. Muhammed Al-Mulhem 1
ICS 415Computer Graphics
Clipping
Dr. Muhammed Al-MulhemDr. Muhammed Al-Mulhem
March 1, 2009March 1, 2009
Dr. Muhammed Al-MulhemDr. Muhammed Al-Mulhem
March 1, 2009March 1, 2009
March 1, 2009 Dr. Muhammed Al-Mulhem 2
The Rendering Pipeline
Transform
Illuminate
Transform
Clip
Project
Rasterize
Model & CameraModel & CameraParametersParameters Rendering PipelineRendering Pipeline FramebufferFramebuffer DisplayDisplay
ModelWorld
WorldCamera
March 1, 2009 Dr. Muhammed Al-Mulhem 3
Why clip?
We don’t want to waste time rendering objects that are We don’t want to waste time rendering objects that are outside the outside the viewing window (or clipping window)viewing window (or clipping window)
We don’t want to waste time rendering objects that are We don’t want to waste time rendering objects that are outside the outside the viewing window (or clipping window)viewing window (or clipping window)
March 1, 2009 Dr. Muhammed Al-Mulhem 4
What is clipping?
Analytically calculating the portions of primitives within Analytically calculating the portions of primitives within the view windowthe view window
Analytically calculating the portions of primitives within Analytically calculating the portions of primitives within the view windowthe view window
March 1, 2009 Dr. Muhammed Al-Mulhem 5
Clip to what?
View Window
Eye position(focal point)
Viewing Frustum
March 1, 2009 Dr. Muhammed Al-Mulhem 6
Why illuminate before clipping?
Transform
Illuminate
Transform
Clip
Project
Rasterize
Model & CameraModel & CameraParametersParameters Rendering PipelineRendering Pipeline FramebufferFramebuffer DisplayDisplay
ModelWorld
WorldCamera
March 1, 2009 Dr. Muhammed Al-Mulhem 7
Why WorldCamera before clipping?
Transform
Illuminate
Transform
Clip
Project
Rasterize
Model & CameraModel & CameraParametersParameters Rendering PipelineRendering Pipeline FramebufferFramebuffer DisplayDisplay
ModelWorld
WorldCamera
March 1, 2009 Dr. Muhammed Al-Mulhem 8
Why?
Why illuminate before clipping?Why illuminate before clipping?
Why WorldWhy WorldCamera before clipping?Camera before clipping?
EfficiencyEfficiency
Why illuminate before clipping?Why illuminate before clipping?
Why WorldWhy WorldCamera before clipping?Camera before clipping?
EfficiencyEfficiency
March 1, 2009 Dr. Muhammed Al-Mulhem 10
Why clip before project and rasterize?
Transform
Illuminate
Transform
Clip
Project
Rasterize
Model & CameraModel & CameraParametersParameters Rendering PipelineRendering Pipeline FramebufferFramebuffer DisplayDisplay
ModelWorld
WorldCamera
March 1, 2009 Dr. Muhammed Al-Mulhem 11
Why Clip?
• Bad idea to rasterize outside of framebuffer bounds Bad idea to rasterize outside of framebuffer bounds
• Also, don’t waste time scan converting pixels Also, don’t waste time scan converting pixels outside windowoutside window
• Bad idea to rasterize outside of framebuffer bounds Bad idea to rasterize outside of framebuffer bounds
• Also, don’t waste time scan converting pixels Also, don’t waste time scan converting pixels outside windowoutside window
March 1, 2009 Dr. Muhammed Al-Mulhem 12
Clipping
The naïve approach to clipping lines:The naïve approach to clipping lines:
for each line segmentfor each line segment
for each edge of view_windowfor each edge of view_window
find intersection pointfind intersection point
pick “nearest” pointpick “nearest” point
if anything is left, draw itif anything is left, draw it
What do we mean by “nearest”?What do we mean by “nearest”?
How can we optimize this?How can we optimize this?
The naïve approach to clipping lines:The naïve approach to clipping lines:
for each line segmentfor each line segment
for each edge of view_windowfor each edge of view_window
find intersection pointfind intersection point
pick “nearest” pointpick “nearest” point
if anything is left, draw itif anything is left, draw it
What do we mean by “nearest”?What do we mean by “nearest”?
How can we optimize this?How can we optimize this?A
B
CD
March 1, 2009 Dr. Muhammed Al-Mulhem 13
Trivial Accepts
Big optimization: trivial accept/rejectsBig optimization: trivial accept/rejects
How can we quickly determine whether a line segment is How can we quickly determine whether a line segment is entirely inside the view window?entirely inside the view window?
A: test both endpoints.A: test both endpoints.
Big optimization: trivial accept/rejectsBig optimization: trivial accept/rejects
How can we quickly determine whether a line segment is How can we quickly determine whether a line segment is entirely inside the view window?entirely inside the view window?
A: test both endpoints.A: test both endpoints.
March 1, 2009 Dr. Muhammed Al-Mulhem 14
Trivial Rejects
How can we know a line is outside view window?How can we know a line is outside view window?
A: if both endpoints on wrong side of A: if both endpoints on wrong side of samesame edge, can edge, can trivially reject linetrivially reject line
How can we know a line is outside view window?How can we know a line is outside view window?
A: if both endpoints on wrong side of A: if both endpoints on wrong side of samesame edge, can edge, can trivially reject linetrivially reject line
March 1, 2009 Dr. Muhammed Al-Mulhem 15
Clipping Lines To View Window
Combining trivial accepts/rejectsCombining trivial accepts/rejects
• Trivially Trivially acceptaccept lines with both endpoints lines with both endpoints inside all edges of the view windowinside all edges of the view window
• Trivially Trivially reject reject lines with both endpoints lines with both endpoints outside the same edge of the view outside the same edge of the view windowwindow
• Otherwise, reduce to trivial casesOtherwise, reduce to trivial cases by splitting into two segmentsby splitting into two segments
Combining trivial accepts/rejectsCombining trivial accepts/rejects
• Trivially Trivially acceptaccept lines with both endpoints lines with both endpoints inside all edges of the view windowinside all edges of the view window
• Trivially Trivially reject reject lines with both endpoints lines with both endpoints outside the same edge of the view outside the same edge of the view windowwindow
• Otherwise, reduce to trivial casesOtherwise, reduce to trivial cases by splitting into two segmentsby splitting into two segments
March 1, 2009 Dr. Muhammed Al-Mulhem 16
Cohen-Sutherland Line Clipping• Divide Divide view windowview window into regions defined by into regions defined by windowwindow edges edges
• Assign each region a 4-bit Assign each region a 4-bit outcodeoutcode: Bit 1 indicates y-value of points are : Bit 1 indicates y-value of points are above yabove ymaxmax
• Divide Divide view windowview window into regions defined by into regions defined by windowwindow edges edges
• Assign each region a 4-bit Assign each region a 4-bit outcodeoutcode: Bit 1 indicates y-value of points are : Bit 1 indicates y-value of points are above yabove ymaxmax
0000 00100001
1001
0101 0100
1000 1010
0110
ymax
xmax
March 1, 2009 Dr. Muhammed Al-Mulhem 17
Cohen-Sutherland Line Clipping
For each line segmentFor each line segment
• Assign an outcode to each vertex Assign an outcode to each vertex
• If both outcodes = 0, trivial acceptIf both outcodes = 0, trivial accept
– Same as performing Same as performing if (bitwise OR = 0)if (bitwise OR = 0)
• ElseElse
– bitwise AND vertex outcodes togetherbitwise AND vertex outcodes together
– if result if result 0, trivial reject 0, trivial reject
For each line segmentFor each line segment
• Assign an outcode to each vertex Assign an outcode to each vertex
• If both outcodes = 0, trivial acceptIf both outcodes = 0, trivial accept
– Same as performing Same as performing if (bitwise OR = 0)if (bitwise OR = 0)
• ElseElse
– bitwise AND vertex outcodes togetherbitwise AND vertex outcodes together
– if result if result 0, trivial reject 0, trivial reject
March 1, 2009 Dr. Muhammed Al-Mulhem 18
Cohen-Sutherland Line Clipping
• If line cannot be trivially accepted or rejected, subdivide so If line cannot be trivially accepted or rejected, subdivide so that one or both segments can be discardedthat one or both segments can be discarded
– Pick an edge of view window that the line crosses (Pick an edge of view window that the line crosses (how?how?))
– Intersect line with edge (Intersect line with edge (how?how?))
– Discard portion on wrong side of edge and assign new Discard portion on wrong side of edge and assign new outcode to new vertexoutcode to new vertex
– Apply trivial accept/reject tests; repeat if necessary Apply trivial accept/reject tests; repeat if necessary
• If line cannot be trivially accepted or rejected, subdivide so If line cannot be trivially accepted or rejected, subdivide so that one or both segments can be discardedthat one or both segments can be discarded
– Pick an edge of view window that the line crosses (Pick an edge of view window that the line crosses (how?how?))
– Intersect line with edge (Intersect line with edge (how?how?))
– Discard portion on wrong side of edge and assign new Discard portion on wrong side of edge and assign new outcode to new vertexoutcode to new vertex
– Apply trivial accept/reject tests; repeat if necessary Apply trivial accept/reject tests; repeat if necessary
March 1, 2009 Dr. Muhammed Al-Mulhem 19
Cohen-Sutherland Line Clipping
If line cannot be trivially accepted or rejected, subdivide so If line cannot be trivially accepted or rejected, subdivide so that one or both segments can be discardedthat one or both segments can be discarded
Pick an edge that the line crossesPick an edge that the line crosses• Check against edges in same order each time Check against edges in same order each time
• ((For example: top, bottom, right, left)For example: top, bottom, right, left)
If line cannot be trivially accepted or rejected, subdivide so If line cannot be trivially accepted or rejected, subdivide so that one or both segments can be discardedthat one or both segments can be discarded
Pick an edge that the line crossesPick an edge that the line crosses• Check against edges in same order each time Check against edges in same order each time
• ((For example: top, bottom, right, left)For example: top, bottom, right, left)
A
B
D E
C
March 1, 2009 Dr. Muhammed Al-Mulhem 20
Cohen-Sutherland Line Clipping
Intersect line with edge (Intersect line with edge (how?how?) – Chapter 6.) – Chapter 6.Intersect line with edge (Intersect line with edge (how?how?) – Chapter 6.) – Chapter 6.
A
B
D E
C
March 1, 2009 Dr. Muhammed Al-Mulhem 21
Discard portion on wrong side of edge and assign outcode to new Discard portion on wrong side of edge and assign outcode to new vertexvertex
Apply trivial accept/reject tests and repeat if necessary Apply trivial accept/reject tests and repeat if necessary
Discard portion on wrong side of edge and assign outcode to new Discard portion on wrong side of edge and assign outcode to new vertexvertex
Apply trivial accept/reject tests and repeat if necessary Apply trivial accept/reject tests and repeat if necessary
Cohen-Sutherland Line Clipping
A
B
DC
March 1, 2009 Dr. Muhammed Al-Mulhem 22
Discard portion on wrong side of edge and assign outcode to new Discard portion on wrong side of edge and assign outcode to new vertexvertex
Apply trivial accept/reject tests and repeat if necessary Apply trivial accept/reject tests and repeat if necessary
Discard portion on wrong side of edge and assign outcode to new Discard portion on wrong side of edge and assign outcode to new vertexvertex
Apply trivial accept/reject tests and repeat if necessary Apply trivial accept/reject tests and repeat if necessary
Cohen-Sutherland Line Clipping
A
B
C
March 1, 2009 Dr. Muhammed Al-Mulhem 23
View Window Intersection Code
• (x(x11, y, y11), (x), (x22, y, y22) intersect with vertical edge at x) intersect with vertical edge at xrightright
– yyintersectintersect = y = y11 + m(x + m(xrightright – x1) – x1)
where m=(ywhere m=(y22-y-y11)/(x)/(x22-x-x11))
• (x(x11, y, y11), (x), (x22, y, y22) intersect with horizontal edge at y) intersect with horizontal edge at ybottombottom
– xxintersectintersect = x = x11 + (y + (ybottombottom – y1)/m – y1)/m
where m=(ywhere m=(y22-y-y11)/(x)/(x22-x-x11))
• (x(x11, y, y11), (x), (x22, y, y22) intersect with vertical edge at x) intersect with vertical edge at xrightright
– yyintersectintersect = y = y11 + m(x + m(xrightright – x1) – x1)
where m=(ywhere m=(y22-y-y11)/(x)/(x22-x-x11))
• (x(x11, y, y11), (x), (x22, y, y22) intersect with horizontal edge at y) intersect with horizontal edge at ybottombottom
– xxintersectintersect = x = x11 + (y + (ybottombottom – y1)/m – y1)/m
where m=(ywhere m=(y22-y-y11)/(x)/(x22-x-x11))
March 1, 2009 Dr. Muhammed Al-Mulhem 24
Cohen-Sutherland Review
• Use Outodes to quickly eliminate/include linesUse Outodes to quickly eliminate/include lines
– Best algorithm when trivial accepts/rejects are commonBest algorithm when trivial accepts/rejects are common
• Must compute viewing window clipping of remaining linesMust compute viewing window clipping of remaining lines
– Non-trivial clipping costNon-trivial clipping cost
– Redundant clipping of some linesRedundant clipping of some lines
More efficient algorithms existMore efficient algorithms exist
• Use Outodes to quickly eliminate/include linesUse Outodes to quickly eliminate/include lines
– Best algorithm when trivial accepts/rejects are commonBest algorithm when trivial accepts/rejects are common
• Must compute viewing window clipping of remaining linesMust compute viewing window clipping of remaining lines
– Non-trivial clipping costNon-trivial clipping cost
– Redundant clipping of some linesRedundant clipping of some lines
More efficient algorithms existMore efficient algorithms exist
March 1, 2009 Dr. Muhammed Al-Mulhem 25
Solving Simultaneous Equations
Equation of a lineEquation of a line
• Slope-intercept (explicit equation): y = mx + bSlope-intercept (explicit equation): y = mx + b
• Implicit Equation: Ax + By + C = 0Implicit Equation: Ax + By + C = 0
• Parametric Equation: Line defined by two points, PParametric Equation: Line defined by two points, P00 and P and P11
– PP(t) = (t) = PP00 + ( + (PP11 - - PP00) t, where ) t, where PP is a vector [x, y] is a vector [x, y]TT
– x(t) = xx(t) = x00 + (x + (x1 1 - x- x00) t) t
– y(t) = yy(t) = y00 + (y + (y1 1 - y- y00) t) t
Equation of a lineEquation of a line
• Slope-intercept (explicit equation): y = mx + bSlope-intercept (explicit equation): y = mx + b
• Implicit Equation: Ax + By + C = 0Implicit Equation: Ax + By + C = 0
• Parametric Equation: Line defined by two points, PParametric Equation: Line defined by two points, P00 and P and P11
– PP(t) = (t) = PP00 + ( + (PP11 - - PP00) t, where ) t, where PP is a vector [x, y] is a vector [x, y]TT
– x(t) = xx(t) = x00 + (x + (x1 1 - x- x00) t) t
– y(t) = yy(t) = y00 + (y + (y1 1 - y- y00) t) t
March 1, 2009 Dr. Muhammed Al-Mulhem 26
Parametric Line EquationDescribes a finite lineDescribes a finite line
Works with vertical lines (like the view window edge)Works with vertical lines (like the view window edge)
• 0 <=t <= 10 <=t <= 1
– Defines line between PDefines line between P00 and P and P11
• t < 0t < 0
– Defines line before PDefines line before P00
• t > 1t > 1
– Defines line after PDefines line after P11
Describes a finite lineDescribes a finite line
Works with vertical lines (like the view window edge)Works with vertical lines (like the view window edge)
• 0 <=t <= 10 <=t <= 1
– Defines line between PDefines line between P00 and P and P11
• t < 0t < 0
– Defines line before PDefines line before P00
• t > 1t > 1
– Defines line after PDefines line after P11
March 1, 2009 Dr. Muhammed Al-Mulhem 27
Parametric Lines and Clipping
Define each line in parametric form: Define each line in parametric form:
• PP00(t)…P(t)…Pn-1n-1(t)(t)
Define each edge of view window in parametric form:Define each edge of view window in parametric form:
• PPLL(t), P(t), PRR(t), P(t), PTT(t), P(t), PBB(t)(t)
Perform Cohen-Sutherland intersection tests using Perform Cohen-Sutherland intersection tests using appropriate view window edge and lineappropriate view window edge and line
Define each line in parametric form: Define each line in parametric form:
• PP00(t)…P(t)…Pn-1n-1(t)(t)
Define each edge of view window in parametric form:Define each edge of view window in parametric form:
• PPLL(t), P(t), PRR(t), P(t), PTT(t), P(t), PBB(t)(t)
Perform Cohen-Sutherland intersection tests using Perform Cohen-Sutherland intersection tests using appropriate view window edge and lineappropriate view window edge and line
March 1, 2009 Dr. Muhammed Al-Mulhem 28
Line / Edge Clipping EquationsFaster line clippers use parametric equationsFaster line clippers use parametric equations
Line 0:Line 0:• xx00 = x = x00
00 + (x + (x0011 - x - x00
00) t) t00
• yy00 = y = y0000 + (y + (y00
11 - y - y0000) t) t00
xx0000 + (x + (x00
11 - x - x0000) t) t0 = 0 = xxLL
00 + (x + (xLL11 - x - xLL
00) t) tLL
yy0000 + (y + (y00
11 - y - y0000) t) t0 = 0 = yyLL
00 + (y + (yLL11 - y - yLL
00) t) tLL
• Solve for tSolve for t00 and/or t and/or tLL
Faster line clippers use parametric equationsFaster line clippers use parametric equations
Line 0:Line 0:• xx00 = x = x00
00 + (x + (x0011 - x - x00
00) t) t00
• yy00 = y = y0000 + (y + (y00
11 - y - y0000) t) t00
xx0000 + (x + (x00
11 - x - x0000) t) t0 = 0 = xxLL
00 + (x + (xLL11 - x - xLL
00) t) tLL
yy0000 + (y + (y00
11 - y - y0000) t) t0 = 0 = yyLL
00 + (y + (yLL11 - y - yLL
00) t) tLL
• Solve for tSolve for t00 and/or t and/or tLL
View Window Edge L:View Window Edge L:• xxLL = x = xLL
00 + (x + (xLL11 - x - xLL
00) t) tLL
• yyLL = y = yLL00 + (y + (yLL
11 - y - yLL00) t) tLL
View Window Edge L:View Window Edge L:• xxLL = x = xLL
00 + (x + (xLL11 - x - xLL
00) t) tLL
• yyLL = y = yLL00 + (y + (yLL
11 - y - yLL00) t) tLL
March 1, 2009 Dr. Muhammed Al-Mulhem 29
Cyrus-Beck Algorithm
We wish to optimize line/line intersectionWe wish to optimize line/line intersection
• Start with parametric equation of line:Start with parametric equation of line:
– P(t) = PP(t) = P00 + (P + (P1 1 - P- P00) t) t
• And a point and normal for each edgeAnd a point and normal for each edge
– PPLL, N, NLL
We wish to optimize line/line intersectionWe wish to optimize line/line intersection
• Start with parametric equation of line:Start with parametric equation of line:
– P(t) = PP(t) = P00 + (P + (P1 1 - P- P00) t) t
• And a point and normal for each edgeAnd a point and normal for each edge
– PPLL, N, NLL
March 1, 2009 Dr. Muhammed Al-Mulhem 30
Cyrus-Beck AlgorithmFind t such thatFind t such that
NNLL [P(t) - P [P(t) - PLL] = 0] = 0
Substitute line equation for P(t): Substitute line equation for P(t):
• NNLL [ [PP00 + (P + (P1 1 - P- P00) t) t - P - PLL] = 0] = 0
Solve for tSolve for t
• t = Nt = NLL [P [PLL – P – P00] / -N] / -NLL [P [P11 - P - P00]]
Find t such thatFind t such that
NNLL [P(t) - P [P(t) - PLL] = 0] = 0
Substitute line equation for P(t): Substitute line equation for P(t):
• NNLL [ [PP00 + (P + (P1 1 - P- P00) t) t - P - PLL] = 0] = 0
Solve for tSolve for t
• t = Nt = NLL [P [PLL – P – P00] / -N] / -NLL [P [P11 - P - P00]]
PL
NL
P(t)Inside
P0
P1
March 1, 2009 Dr. Muhammed Al-Mulhem 31
Cyrus-Beck Algorithm
Compute t for line intersection with all four edgesCompute t for line intersection with all four edges
Discard all (t < 0) and (t > 1)Discard all (t < 0) and (t > 1)
Classify each remaining intersection asClassify each remaining intersection as
• Potentially Entering (PE)Potentially Entering (PE)
• Potentially Leaving (PL)Potentially Leaving (PL)
NNLL [P [P11 - P - P00] > 0 implies PL] > 0 implies PL
NNLL [P [P11 - P - P00] < 0 implies PE] < 0 implies PE
Compute t for line intersection with all four edgesCompute t for line intersection with all four edges
Discard all (t < 0) and (t > 1)Discard all (t < 0) and (t > 1)
Classify each remaining intersection asClassify each remaining intersection as
• Potentially Entering (PE)Potentially Entering (PE)
• Potentially Leaving (PL)Potentially Leaving (PL)
NNLL [P [P11 - P - P00] > 0 implies PL] > 0 implies PL
NNLL [P [P11 - P - P00] < 0 implies PE] < 0 implies PE
March 1, 2009 Dr. Muhammed Al-Mulhem 32
Compute PE with largest tCompute PE with largest t
Compute PL with smallest tCompute PL with smallest t
Clip to these two pointsClip to these two points
Cyrus-Beck Algorithm
PE
PLP1
PL
PE
P0
March 1, 2009 Dr. Muhammed Al-Mulhem 33
Cyrus-Beck AlgorithmBecause of horizontal and vertical clip lines:Because of horizontal and vertical clip lines:
• Many computations reduceMany computations reduce
Normals: (-1, 0), (1, 0), (0, -1), (0, 1)Normals: (-1, 0), (1, 0), (0, -1), (0, 1)
Pick constant points on edgesPick constant points on edges
solution for t:solution for t:
• -(x-(x00 - x - xleftleft) / (x) / (x11 - x - x00))
• (x(x00 - x - xrightright) / -(x) / -(x11 - x - x00))
• -(y-(y00 - y - ybottombottom) / (y) / (y11 - y - y00))
• (y(y00 - y - ytoptop) / -(y) / -(y11 - y - y00))
Because of horizontal and vertical clip lines:Because of horizontal and vertical clip lines:• Many computations reduceMany computations reduce
Normals: (-1, 0), (1, 0), (0, -1), (0, 1)Normals: (-1, 0), (1, 0), (0, -1), (0, 1)
Pick constant points on edgesPick constant points on edges
solution for t:solution for t:
• -(x-(x00 - x - xleftleft) / (x) / (x11 - x - x00))
• (x(x00 - x - xrightright) / -(x) / -(x11 - x - x00))
• -(y-(y00 - y - ybottombottom) / (y) / (y11 - y - y00))
• (y(y00 - y - ytoptop) / -(y) / -(y11 - y - y00))
March 1, 2009 Dr. Muhammed Al-Mulhem 34
ComparisonCohen-SutherlandCohen-Sutherland
• Repeated clipping is expensiveRepeated clipping is expensive
• Best used when trivial acceptance and rejection is possible for most linesBest used when trivial acceptance and rejection is possible for most lines
Cyrus-BeckCyrus-Beck
• Computation of t-intersections is cheapComputation of t-intersections is cheap
• Computation of (x,y) clip points is only done onceComputation of (x,y) clip points is only done once
• Algorithm doesn’t consider trivial accepts/rejectsAlgorithm doesn’t consider trivial accepts/rejects
• Best when many lines must be clippedBest when many lines must be clipped
Liang-Barsky: Optimized Cyrus-BeckLiang-Barsky: Optimized Cyrus-Beck
Nicholl et al.: Fastest, but doesn’t do 3DNicholl et al.: Fastest, but doesn’t do 3D
Cohen-SutherlandCohen-Sutherland
• Repeated clipping is expensiveRepeated clipping is expensive
• Best used when trivial acceptance and rejection is possible for most linesBest used when trivial acceptance and rejection is possible for most lines
Cyrus-BeckCyrus-Beck
• Computation of t-intersections is cheapComputation of t-intersections is cheap
• Computation of (x,y) clip points is only done onceComputation of (x,y) clip points is only done once
• Algorithm doesn’t consider trivial accepts/rejectsAlgorithm doesn’t consider trivial accepts/rejects
• Best when many lines must be clippedBest when many lines must be clipped
Liang-Barsky: Optimized Cyrus-BeckLiang-Barsky: Optimized Cyrus-Beck
Nicholl et al.: Fastest, but doesn’t do 3DNicholl et al.: Fastest, but doesn’t do 3D
March 1, 2009 Dr. Muhammed Al-Mulhem 35
Clipping Polygons
• Clipping polygons is more complex than clipping the Clipping polygons is more complex than clipping the individual linesindividual lines
• Input: polygonInput: polygon
• Output: Original polygon, new polygon, or nothingOutput: Original polygon, new polygon, or nothing
• OptimizationOptimization• The biggest line optimizer we had was trivial accept or reject. The biggest line optimizer we had was trivial accept or reject.
• When can we trivially accept/reject a polygon as opposed to the When can we trivially accept/reject a polygon as opposed to the line segments that make up the polygon?line segments that make up the polygon?
• Clipping polygons is more complex than clipping the Clipping polygons is more complex than clipping the individual linesindividual lines
• Input: polygonInput: polygon
• Output: Original polygon, new polygon, or nothingOutput: Original polygon, new polygon, or nothing
• OptimizationOptimization• The biggest line optimizer we had was trivial accept or reject. The biggest line optimizer we had was trivial accept or reject.
• When can we trivially accept/reject a polygon as opposed to the When can we trivially accept/reject a polygon as opposed to the line segments that make up the polygon?line segments that make up the polygon?
March 1, 2009 Dr. Muhammed Al-Mulhem 36
What happens to a triangle during clipping?What happens to a triangle during clipping?
Possible outcomesPossible outcomes::
What happens to a triangle during clipping?What happens to a triangle during clipping?
Possible outcomesPossible outcomes::
triangle triangle
Why Is Clipping Hard?
triangle quad triangle 5-gon
How many sides can a clipped triangle have?How many sides can a clipped triangle have?
March 1, 2009 Dr. Muhammed Al-Mulhem 38
A really tough case: A really tough case: A really tough case: A really tough case:
Why Is Clipping Hard?
March 1, 2009 Dr. Muhammed Al-Mulhem 39
A really tough case: A really tough case: A really tough case: A really tough case:
Why Is Clipping Hard?
concave polygon multiple polygons
March 1, 2009 Dr. Muhammed Al-Mulhem 40
Sutherland-Hodgman Clipping
Basic idea:Basic idea:
• Consider each edge of the view window individuallyConsider each edge of the view window individually
• Clip the polygon against the view window edge’s equationClip the polygon against the view window edge’s equation
Basic idea:Basic idea:
• Consider each edge of the view window individuallyConsider each edge of the view window individually
• Clip the polygon against the view window edge’s equationClip the polygon against the view window edge’s equation
March 1, 2009 Dr. Muhammed Al-Mulhem 41
Sutherland-Hodgman Clipping
Basic idea:Basic idea:
• Consider each edge of the viewport individuallyConsider each edge of the viewport individually
• Clip the polygon against the edge equationClip the polygon against the edge equation
Basic idea:Basic idea:
• Consider each edge of the viewport individuallyConsider each edge of the viewport individually
• Clip the polygon against the edge equationClip the polygon against the edge equation
March 1, 2009 Dr. Muhammed Al-Mulhem 42
Sutherland-Hodgman Clipping
Basic idea:Basic idea:
• Consider each edge of the viewport individuallyConsider each edge of the viewport individually
• Clip the polygon against the edge equationClip the polygon against the edge equation
Basic idea:Basic idea:
• Consider each edge of the viewport individuallyConsider each edge of the viewport individually
• Clip the polygon against the edge equationClip the polygon against the edge equation
March 1, 2009 Dr. Muhammed Al-Mulhem 43
Sutherland-Hodgman Clipping
Basic idea:Basic idea:
• Consider each edge of the viewport individuallyConsider each edge of the viewport individually
• Clip the polygon against the edge equationClip the polygon against the edge equation
Basic idea:Basic idea:
• Consider each edge of the viewport individuallyConsider each edge of the viewport individually
• Clip the polygon against the edge equationClip the polygon against the edge equation
March 1, 2009 Dr. Muhammed Al-Mulhem 44
Sutherland-Hodgman Clipping
Basic idea:Basic idea:
• Consider each edge of the viewport individuallyConsider each edge of the viewport individually
• Clip the polygon against the edge equationClip the polygon against the edge equation
Basic idea:Basic idea:
• Consider each edge of the viewport individuallyConsider each edge of the viewport individually
• Clip the polygon against the edge equationClip the polygon against the edge equation
March 1, 2009 Dr. Muhammed Al-Mulhem 45
Sutherland-Hodgman Clipping
Basic idea:Basic idea:
• Consider each edge of the viewport individuallyConsider each edge of the viewport individually
• Clip the polygon against the edge equationClip the polygon against the edge equation
Basic idea:Basic idea:
• Consider each edge of the viewport individuallyConsider each edge of the viewport individually
• Clip the polygon against the edge equationClip the polygon against the edge equation
March 1, 2009 Dr. Muhammed Al-Mulhem 46
Sutherland-Hodgman Clipping
Basic idea:Basic idea:
• Consider each edge of the viewport individuallyConsider each edge of the viewport individually
• Clip the polygon against the edge equationClip the polygon against the edge equation
Basic idea:Basic idea:
• Consider each edge of the viewport individuallyConsider each edge of the viewport individually
• Clip the polygon against the edge equationClip the polygon against the edge equation
March 1, 2009 Dr. Muhammed Al-Mulhem 47
Sutherland-Hodgman Clipping
Basic idea:Basic idea:
• Consider each edge of the viewport individuallyConsider each edge of the viewport individually
• Clip the polygon against the edge equationClip the polygon against the edge equation
Basic idea:Basic idea:
• Consider each edge of the viewport individuallyConsider each edge of the viewport individually
• Clip the polygon against the edge equationClip the polygon against the edge equation
March 1, 2009 Dr. Muhammed Al-Mulhem 48
Sutherland-Hodgman Clipping
Basic idea:Basic idea:
• Consider each edge of the viewport individuallyConsider each edge of the viewport individually
• Clip the polygon against the edge equationClip the polygon against the edge equation
Basic idea:Basic idea:
• Consider each edge of the viewport individuallyConsider each edge of the viewport individually
• Clip the polygon against the edge equationClip the polygon against the edge equation
March 1, 2009 Dr. Muhammed Al-Mulhem 49
Sutherland-Hodgman Clipping
Basic idea:Basic idea:
• Consider each edge of the viewport individuallyConsider each edge of the viewport individually
• Clip the polygon against the edge equationClip the polygon against the edge equation
• After doing all edges, the polygon is fully clippedAfter doing all edges, the polygon is fully clipped
Basic idea:Basic idea:
• Consider each edge of the viewport individuallyConsider each edge of the viewport individually
• Clip the polygon against the edge equationClip the polygon against the edge equation
• After doing all edges, the polygon is fully clippedAfter doing all edges, the polygon is fully clipped
March 1, 2009 Dr. Muhammed Al-Mulhem 50
Sutherland-Hodgman Clipping
• Input/output for algorithm:Input/output for algorithm:
• Input: List of polygon vertices in order. Input: List of polygon vertices in order.
• Output: List of clipped polygon vertices consisting of Output: List of clipped polygon vertices consisting of old vertices (maybe) and new vertices (maybe).old vertices (maybe) and new vertices (maybe).
• Note: this is exactly what we expect from Note: this is exactly what we expect from the clipping operation against each edgethe clipping operation against each edge
• Input/output for algorithm:Input/output for algorithm:
• Input: List of polygon vertices in order. Input: List of polygon vertices in order.
• Output: List of clipped polygon vertices consisting of Output: List of clipped polygon vertices consisting of old vertices (maybe) and new vertices (maybe).old vertices (maybe) and new vertices (maybe).
• Note: this is exactly what we expect from Note: this is exactly what we expect from the clipping operation against each edgethe clipping operation against each edge
March 1, 2009 Dr. Muhammed Al-Mulhem 51
Sutherland-Hodgman Clipping
Sutherland-Hodgman basic routine:Sutherland-Hodgman basic routine:
• Go around polygon one vertex at a timeGo around polygon one vertex at a time
• Current vertex has position Current vertex has position pp
• Previous vertex had position Previous vertex had position ss, and it has been added to , and it has been added to the output if appropriate.the output if appropriate.
Sutherland-Hodgman basic routine:Sutherland-Hodgman basic routine:
• Go around polygon one vertex at a timeGo around polygon one vertex at a time
• Current vertex has position Current vertex has position pp
• Previous vertex had position Previous vertex had position ss, and it has been added to , and it has been added to the output if appropriate.the output if appropriate.
March 1, 2009 Dr. Muhammed Al-Mulhem 52
Sutherland-Hodgman Clipping
• Edge from Edge from ss to to pp takes one of four cases: takes one of four cases:
(Orange line can be a line or a plane)(Orange line can be a line or a plane)
• Edge from Edge from ss to to pp takes one of four cases: takes one of four cases:
(Orange line can be a line or a plane)(Orange line can be a line or a plane)
inside outside
s
p
p output
inside outside
s
p
no output
inside outside
sp
i output
inside outside
sp
i outputp output
March 1, 2009 Dr. Muhammed Al-Mulhem 53
Sutherland-Hodgman ClippingFour cases:Four cases:
• ss inside plane and inside plane and pp inside plane inside plane
– Add Add pp to output to output
– Note: Note: ss has already been added has already been added
• ss inside plane and inside plane and pp outside plane outside plane
– Find intersection point Find intersection point ii
– Add Add ii to output to output
• ss outside plane and outside plane and pp outside plane outside plane
– Add nothingAdd nothing
• ss outside plane and outside plane and pp inside plane inside plane
– Find intersection point Find intersection point ii
– Add Add ii to output, followed by to output, followed by pp
Four cases:Four cases:
• ss inside plane and inside plane and pp inside plane inside plane
– Add Add pp to output to output
– Note: Note: ss has already been added has already been added
• ss inside plane and inside plane and pp outside plane outside plane
– Find intersection point Find intersection point ii
– Add Add ii to output to output
• ss outside plane and outside plane and pp outside plane outside plane
– Add nothingAdd nothing
• ss outside plane and outside plane and pp inside plane inside plane
– Find intersection point Find intersection point ii
– Add Add ii to output, followed by to output, followed by pp