Transcript Slide 1

Rasterization
From Geometry to Fragments
Chapter 7 (kinda)
1
University of North Carolina – Chapel Hill
COMP 136
7/17/2015
Announcements
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These Slides
• Chapter 7 of Angel
• Chapters 3, 6, 13 of
Computer Graphics (C
Edition), 2nd ed. by
Hearn and Baker
– ISBN: 0-13-530924-7
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Pipelines I
• Pipeline – set of operations which
occurring in a set order
– Why use pipelines?
• Separate problem into manageable pieces
• Each piece can be handled separate from
others  parallelism
• Important measures
– Throughput
– Latency
– Bottleneck
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Pipelines II
•
•
1)
2)
3)
4)
5
Hardware has long been pipelined
Graphics hardware is no exception:
Modeling
Geometry Processing
Rasterization
Fragment Processing
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Pipelines III
• Modeling (chapters 10 and 11)
– The conversion of analog (real world)
objects into discrete data
• i.e. creating vertices and connectivity via
range scanning
– The design of a complex structure from
simpler primitives
• i.e. architecture and engineering designs
– Done Offline
• We will ignore this step for now
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Pipelines IV
• Application programmer pipes
modeling output into…
• Geometry Processing
–
–
–
–
7
Animate objects
Move objects into camera space
Project objects into device coordinates
Clip objects external to viewing window
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Pipelines V
• Rasterization
– Conversion of geometry in device
coordinates into fragments (or pixels) in
screen coordinates
– After this step there is no notion of a
“polygon”
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Pipelines VI
• Fragment Processing
–
–
–
–
9
Texture lookups
Special ops (like XOR)
Programmable GPU steps
Chapters 8 and 9
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Aside
• These last 3 steps need to be FAST
• Developed 20-40 years ago… but little
has changed
• Efficient memory use speeds things up
– Cache, cache, cache
• Integers and bit ops over floating point
• Fewer bits usually faster
– float over double, half over float
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Geometry Processing I
• First step is to transform the objects
to device coordinates
– View volume becomes cube from -1..1
– Modelview and projection matrices
• IMPORTANT NOTE:
– Earlier projection matrix mapped to
plane z = d, not a range
– Range necessary for z-buffer technique
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Geometry Processing II
[
[
[
[
1
0
0
0
0
1
0
0
0
0
-1
0
0
β
0
]
]
]
]
– = (zfar + znear) / (zfar - znear)
– β = (2zfarznear) / (zfar - znear)
– z values now map to a range
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Geometry Processing III
• Not all primitives map to inside window
– Cull those that are completely outside
– Clip those that are partially inside
• 2D v. 3D
– Projection plane v. projection cube
– Clipping can occur in either space
– Choice of visible surface algorithm used
forces one or the other
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Clipping
• Rasterization is very expensive
– More or less linear w/ number of
fragments created
– Consists of adds, rounding and logic
branches per pixel
– Only rasterize that which is in viewable
region
• A few operations now saves many
later
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Point Clipping
• Okay, this one’s easy
• Who wants to define the algorithm?
• Given (xp, yp, zp), check the following:
– -1 <= xp <= 1
– -1 <= yp <= 1
– -1 <= zp <= 1
• Only if in 3D, otherwise, z is unnecessary
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Line Clipping I
• Line exits the viewable volume/plane
– Part is visible
– Find the real world location of exit point
• Shortened line passed to rasterization
• There must be no difference to the
final image
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Line Clipping II
• We’ll do 2D first
• Clip a line against 1
edge of a square
• Similar Triangles
–
–
–
–
–
–
–
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B
A/B = C/D
Which do we know?
B = (y1 – y2)
D = (x1 – x2)
A = (y1 – ymax)
C = AD/B
(x’, y’) = (x1-C, ymax)
(x1, y1)
A
C
(x’, y’)
D
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(x2, y2)
7/17/2015
Line Clipping III
• Similarly handled for the other cases
• Extends easily to 3D
• EXPENSIVE! (below for 2D)
–
–
–
–
3
2
1
4
floating point subs
floating point mults
floating point div
times (for each edge!)
• We need to save ourselves some ops
• Following algs. in Hearn/Baker pp 224-235
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Cohen-Sutherland Line Clipping I
•
•
Split plane into 9 regions
Assign each a 4 bit value
1001
1000
1010
0001
0000
Window
0010
0101
0100
0110
– (above,below,right,left)
•
1.
2.
3.
4.
5.
Assign the vertices of line 4
bit value
If both 0000, trivial accept
If (v1&v2 != 0), trivial reject
Clip against one side
(where one is non-zero)
Assign new vertex 4 bit
Go to 1
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Cohen-Sutherland Line Clipping IV
• What are the vertex codes for these
lines?
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Cohen-Sutherland Line Clipping III
• Extends easily to 3D line clipping
– 27 regions
– 6 bits
– Similar triangles still works
• but must be done for 2 sets of similar
triangles
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Liang-Barsky Line Clipping I
• Parametric definition of a line:
– x = x1 + uΔx
– y = y1 + uΔy
– Δx = (x2-x1), Δy = (y2-y1), 0<=u<=1
• Goal: find range of u for which x and y
both inside the viewing window
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Liang-Barsky Line Clipping II
• Mathematically:
– xmin <= x1 + uΔx <= xmax
– ymin <= y1 + uΔy <= ymax
• Rearranged
–
–
–
–
–
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1:
2:
3:
4:
gen:
u*(-Δx) <= (x1 – xmin)
u*(Δx) <= (xmax – x1)
u*(-Δy) <= (y1 – ymin)
u*(Δy) <= (ymax – y1)
u*(pk) <= (qk)
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Liang-Barsky Line Clipping III
•
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
•
•
–
pk > 0: the line starts inside the boundary
•
•
–
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rk = qk/pk
u1 = max(0, rk, u1)
rk = qk/pk
u2 = min(1, rk, u2)
If u1 > u2, the line is completely outside
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Liang-Barsky Line Clipping IV
• Also extends to 3D
– Add z = z1 + uΔz
– Add 2 more p’s and q’s
– Still only 2 u’s
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Liang-Barsky v. Cohen-Sutherland
• According to Hearn-Baker:
– Generally, Liang-Barsky is more efficient
• Requires only one division
• Find intersection values for (x,y) only at end
• My view
– Understand what you implement
• Debugging is the bigger cost!
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Nicholl-Lee-Nicholl Line Clipping I
•
•
•
•
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This test is most complicated
Also the fastest
Only works well for 2D
Quick overview here
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Nicholl-Lee-Nicholl Line Clipping II
• Divide the region
based on p1
– Case 1: p1 inside
– Case 2: p1 across
edge
– Case 3: p1 across
corner
T
R
L
B
LT
L
LR
L
T
L T
LB
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TR
L
LB
TB
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Nicholl-Lee-Nicholl Line Clipping IV
• Symmetry handles other cases
• Find slopes of the line and 4 region
bounding lines
• Find which region P2 is in
– If not in any labeled, the line is discarded
• Subtractions, multiplies and divisions
can be carefully used to minimum
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A note on redundancy
• Why present multiple forms of clipping?
– Why do you learn multiple sorts?
– Not always easy to do the fastest
– The fastest for the general case isn’t always
the fastest for every specific case
• Mostly sorted list  bubble sort
– History repeats itself
• You may need something similar in a different
area. Grab the one that maps best.
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Polygon Inside/Outside
• Polygons have a distinct inside and
outside…
• Decided by
– Even/Odd
– Winding Number
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Polygon Clipping I
• Note the difference between clipping
lines and polygons:
NOTE!
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Polygon Clipping II
• Some difficulties:
– Maintaining correct inside/outside
– Variable number of vertices
– Handle screen corners
correctly
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Sutherland-Hodgeman Polygon
Clipping I
• Simplify via separation
• Clip whole polygon against one edge
– Repeat with output for other 3 edges
– Similar for 3D
• You can create intermediate vertices
that get thrown out
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Sutherland-Hodgeman Polygon
Clipping II
• Example I
out  in
in  in
save new clip vert save ending vert
and ending vert
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in  out
out  out
save new clip vert save nothing
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Sutherland-Hodgeman Polygon
Clipping II
• Example 2
Start
Left
Right
Bottom
Top
Note that the point one of the points added when clipping
on the right gets removed when we clip with bottom
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Weiler-Atherton Polygon
Clipping I
• Problem with Sutherland-Hodgeman:
– Concavities can end up linked
Remember
me?
• Weiler-Atherton creates separate
polygons in such cases
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Weiler-Atherton Polygon
Clipping II
• Example
add clip pt.
and end pt.
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add end pt.
add clip pt.
cache old dir.
follow clip edge until
a) new crossing
found
b) reach pt. already
added
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Weiler-Atherton Polygon
Clipping III
• Example (cont)
continue from add clip pt.
cached location and end pt.
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add clip pt.
cache dir.
follow clip edge until
a) new crossing
found
b) reach pt. already
added
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Weiler-Atherton Polygon
Clipping IV
• Example (concluded)
continue from nothing added
cached location finished
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Final result:
Two unconnected
polygons
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Weiler-Atherton Polygon
Clipping V
• Difficulties:
– What if the polygon recrosses edge?
– How many “cached” crosses?
– Your geometry step must be able to
create new polygons
• Instead of 1-in-1-out
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Geometry section ended
• Any questions?
– Affine transforms
• Animation
• Move to camera space
– Projective transform
• Now in [-1..1] device coordinates
– Clipping
• Save rasterization time later
• Must be exactly the same shape still
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Rasterization I
• All geometry received is inside normalized
viewing region
• We only care about the (x, y) portion now
– z doesn’t matter
• Direct map from (x, y) location to (s, t)
pixel
– s = ((x+1)/2)*numPixelsWide
• Don’t round to int here… keep floating point
• Convert continuous geometry to discrete
pixels
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Rasterization II
• Continuous  Discrete not simple:
• Just something to think about
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Line Algorithms
• Length of a line != number of pixels
– Ex) line from (0, 0) to (10, 10)
– Length = 10*sqrt(2)
– NumPixels = 10
• Longest Extent of line
– (Δx > Δy) ? Δx : Δy;
• After finding start pixel, we loop:
– For each pixel in longest extent, do we go
up/down or not in secondary direction?
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DDA Lines I
• Digital Differential Analyzer
–
–
–
–
Find longest extent
Find amount to alter secondary (amt)
primary, secondary = startPt(x, y)
For each in longest extent
• secondary += amt;
• setPixel(round(primary), round(secondary))
– if primary is y, then setPixel would change order 
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DDA Lines II
• Finding the amount to increase
secondary:
– Δprimary = (primaryEnd-primaryStart)
– Δsecondary = (secondaryEnd-secondaryStart)
– amt = Δsecondary/Δprimary
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DDA Lines III
•
•
•
•
•
•
•
startPt = (1, 1) endPt = (10, 7)
Δx = 9, Δy = 6
x is primary, y secondary
amt = 2/3
int prim = round(1), float sec = 1
setPixel(prim, round(sec))
for (prim = 2..10)
– sec += amt
– setPixel(prim, round(sec))
• Simulate the loop
–
–
–
–
–
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sec
sec
sec
sec
sec
=
=
=
=
=
5/3 ~ 2
7/3 ~ 2
9/3 ~ 3
11/3 ~ 4
13/3 ~ 4
–
–
–
–
–
sec =
sec =
sec =
sec =
END
15/3
17/3
19/3
21/3
~
~
~
~
5
6
6
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DDA Lines IV
• Cost:
– Init: 2 fl.pt. subs 1 fl.div.
– Loop: 1 fl.pt. add 1 round
• Can it be done better?
– Of Course! What kind of question is
that?
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Brensenham’s Lines I
• Remove all fl.pt. arithmetic from loop
• DDA stores fl.pt. loc. of secondary and
fl.pt. increment amt
• Bresenham stores an integer decision
parameter
– When below 0, don’t go up/down
– When above 0, move up/down and
subtract from the decision parameter
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Bresenham’s Lines II
• y = mx + b
– m = Δy/Δx
• Δx*y = Δy*x + Δx*b
– Δx*b a constant independent of slope
• Throw it away
– We’re left with integers only
– Some math with (xk, yk) and (xk+1, yk+1)
finds the decision parameter
• Not to be covered here
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Brensenham’s Lines III
•
•
•
•
•
•
Decision parameter  pk
int prim = round(primStart)
int sec = round(secStart)
p0 = 2*Δsec – Δprim
setPixel(prim, sec)
for (prim = primStart+1..primEnd)
– if (pk < 0)
//write these on the board
• pk+1 = pk + 2Δsec
– else
• ++sec;
//or --sec; depending on slope up or down
• pk+1 = pk + 2Δsec - 2Δprim
– setPixel(prim, sec)
52
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Brensenham’s Lines IV
•
•
•
•
•
startPt = (1, 1) endPt = (10, 7)
Δx = 9, Δy = 6
x is primary, y secondary
p0 = 2*6-9 = 3
Simulate:
–
–
–
–
–
–
–
–
–
53
p0
p1
p2
p3
p4
p5
p6
p7
p8
>
<
>
>
<
>
>
<
>
0
0
0
0
0
0
0
0
0









p1=3+2*6-2*9=-3
p2=-3+2*6=9
p3=9+2*6-2*9=3
p4=3+2*6-2*9=-3
p5=-3+2*6=9
p6=9+2*6-2*9=3
p7=3+2*6-2*9=-3
p8=-3+2*6=9
p9=9+2*6-2*9=3
University of North Carolina – Chapel Hill
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7/17/2015
Line Algorithms Closing Up
• Horizontal and vertical lines
• Note that you can start with either
vertex as the first one
– Choose consistently and it simplifies the
problem
• Test slopes < 1, > 1, negative and
positive
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Circles and Curves
• No longer have one principle direction
which describes the number of pixels
• Do have redundancy so that we only
need to solve part of the problem
– Draw 1/8 of circle and mirror
– Draw 1/4 of ellipse and mirror
• Bresenham made integer only
versions for some of them too
• See pp 98-110 of Hearn/Baker
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Polygons
• All of polygon is inside the viewing
region now
• We must find edges and fill them
• Edges - a connected series of lines
• How do we fill?
– Suggestions?
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Scan Line Polygons I
• Mark all minima/maxima
• Create all distinct y values for edges
• For each scan line (horizontal) of polygon
– Pair up edge pixels in order
• For each pair of edge pixels
– Fill the region in-between
• Problems:
– Local maxima/minima – double entry
– Coincident pixels – be careful
– Horizontal lines – make sure it starts and stops
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7/17/2015
Scan Line Polygons II
• “Create all distinct y values for edges”
– Difficult to handle properly
– Make sure endpoints are only in the
correct number of times
• minimum/maximum – 2 times
• others – 1 time
– Take care NOT to add those on the same
level as the endpoint
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7/17/2015
Scan-Line Polygons III
• Example:
Input Vertices
Rasterized Edges
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Scan-Line Fill
Local Maximum/minimum
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7/17/2015
Scan-Line Polygons III
• Data Structure
– struct slpPixel{ int x, int y, float z }
– create an ordered list first sorted on descending
y then increasing x
• Test this a LOT
– Saw structures as previous
– Horizontal edges
• And just off horizontal
– Verticle edges
– Degenerate cases (two edges coincide)
60
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7/17/2015
Flood Fill I
• Draw the entirety of all edges to some
buffer
• Start from some “seed” position inside
• “Flood out” from that position
– 4-Fill, 8-Fill, Boundary Fill
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7/17/2015
Flood Fill II
void boundaryFill4(int x, int y, int fill, int
boundary)
{
int curr;
curr = getPixel(x, y);
if ((current != boundary) && (current != fill)) {
setColor(fill);
setPixel(x, y);
boundaryFill4(x+1, y, fill, boundary);
boundaryFill4(x-1, y, fill, boundary);
boundaryFill4(x, y+1, fill, boundary);
boundaryFill4(x, y-1, fill, boundary);
}
}
62
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Flood Fill III
• Problems with the above (4 Fill):
– Stack overflow
– How choose the
start point?
– Buffer used?
– See example:
• 8 Fill solves the example’s problem
– But you have to handle the lines
differently
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7/17/2015
Flood Fill IV
5
3
4
6
2
1
6
5
2
4
3
1
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University of North Carolina – Chapel Hill
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Polygon Algorithms
• Scan-line is generally used for
rasterization
• Flood-Fills are used in things like MS
Paint
– Where user is inputting values and
interacting in the creation of the polygons
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7/17/2015
Rasterization Done
• Any questions?
– DDA Lines v. Bresenham lines
– Curve algorithms
– Polygon filling
66
University of North Carolina – Chapel Hill
COMP 136
7/17/2015
Fragment Processing
• As mentioned long ago…
–
–
–
–
Texture lookups here
Special ops (XOR)
Stencil buffer can be checked here
Depth buffer check
• We will only discuss depth alg, though
• For convenience, we discuss others, but they
would be executed at other times
67
University of North Carolina – Chapel Hill
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7/17/2015
Hidden Surface Removal
• Visible Surface Detection
– Depends on your half-full/half-empty
preference
• Object Space
– Some algorithms are executed on the
polygons at previous stages
• Fragment Space
– Some are executed on the fragments in
this stage
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Back Face Culling
• Where? Object Space
• When? Before clipping in geometry
step
• What? If normal dot toViewer < 0
throw away polygon
• Does it solve all problems? NO:
– Which of these two front
faces goes in front?
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Z-Buffering I
•
•
•
•
Where? Fragment Space
When? Immediately after rasterization
What? See next slide
Does it solve all problems?
– Yes
– Expensive with memory
• used to be a problem
– Easy to code
– Does not need universal knowledge
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Z-Buffering II
• So far, it’s been unspecified, but we must
maintain the z for all fragments
– Maintain floating point here!
• Store completely separate buffer with same
resolution as color buffer
• Initialize to -1.1 (z only went from [-1..1])
• As each fragment comes down the pipe,
check depth[s][t] < z
– If yes, depth[s][t] = z, color[s][t] = fragmentColor
– Else, throw away fragment
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7/17/2015
Painter’s Algorithm I
•
•
•
•
Where? Object Space
When? Before rasterization
What? See following slides
Does it solve all problems?
– Yes
– Somewhat laborious
– Requires all geometry at once
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7/17/2015
Painter’s Algorithm II
• General idea:
– Sort all objects so that you are rendering
those farthest away first
– Nearer objects will overwrite farther
objects as they are rasterized/drawn
• Problem:
– Lines/Planes have an extent in z, not just
a single value
73
University of North Carolina – Chapel Hill
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7/17/2015
Painters Algorithm III
• Sort all object’s zmin and zmax
• If one object’s zmin and zmax are uninterrupted,
it is fine
• For all pairs of objects with z-overlap
– Check to see if they overlap in x
• If No, it is fine
• If Yes, check to see if they overlap in y
– If No, it is fine
– If Yes, you must find a dividing plane and split one
• Last step is the beast
– They may intersect, so you must split one plane by
the other
– After split, you can resort and should be fine
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University of North Carolina – Chapel Hill
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7/17/2015
Other Fragment Ops
• If you are doing other fragment
operations, you may need to maintain
other values during rasterization
– Texture coordinates must be interpolated
– Normals may need to be interpolated
• Phong Shading
– Etc.
75
University of North Carolina – Chapel Hill
COMP 136
7/17/2015
Are we confident?
• We’ve just completed the process of
geometry to fragments
• Any questions?
76
University of North Carolina – Chapel Hill
COMP 136
7/17/2015