Clipping - 1

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Chapter 9Clipping

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The Rendering Pipeline

GeometryDatabaseGeometryGeometryDatabaseDatabase

Model/ViewTransform.Model/ViewModel/ViewTransform.Transform. LightingLightingLighting Perspective

Transform.PerspectivePerspectiveTransform.Transform. ClippingClippingClipping

ScanConversion

ScanScanConversionConversion

DepthTest

DepthDepthTestTestTexturingTexturingTexturing BlendingBlendingBlending

Frame-buffer

FrameFrame--bufferbuffer

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Problem:Given a 2D line/polygon and a window, clip the line/polygon to their regions that are inside the window.

Problem:Given a 2D line/polygon and a window, clip the line/polygon to their regions that are inside the window.

Line/Polygon Clipping (2D)

Objectives• Efficiency• Memory access

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Analytic Solution

Intersection of convex regions is convex Why?

L & D are convex - intersection is convex

single connected segment of LQuestion: Can boundary of two convex shapes intersect more than twice?Clipping - compute intersection of Lwith four boundary segments of window D

D

L

D

L

D

L

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Line-Line Intersection

( ) ( )( ) ( )

( ) ( )( ) ( )

Gx t x x x t

y t y y y tt G

x r x x x r

y r y y y rr1

101

11

01

101

11

01 2

202

12

02

202

12

02

0 1 0 1== + −

= + −

⎧⎨⎪

⎩⎪∈ =

= + −

= + −

⎧⎨⎪

⎩⎪∈[ , ] [ , ]

( )x y11

11,

( )x y01

01,

( )x y02

02,

( )x y12

12,

Γ1 Γ2

Intersection: x & y values equal in both representations - two linear equations in two unknowns (r,t)

( ) ( )( ) ( )

x x x t x x x r

y y y t y y y r

01

11

01

02

12

02

01

11

01

02

12

02

+ − = + −

+ − = + −

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Intersection with vertical/horizontal lines

( ) ( )( ) ( )

( )( ) ( ) ]1,0[]1,0[

20

21

20

2

20

2

210

11

10

1

10

11

10

1

1 ∈⎪⎩

⎪⎨⎧

−+=

==∈

⎪⎩

⎪⎨⎧

−+=

−+== r

ryyyry

xrxGt

tyyyty

txxxtxG

( )x y11

11,

( )x y01

01,

( )x y02

02,

( )21

20 , yx

Γ1 Γ2

Intersection: x & y values equal in both representations - two linear equations in two unknowns (r,t)

( )

( ) ( )ryyytyyy

xxxxt

xtxxx

20

21

20

10

11

10

10

11

10

20

20

10

11

10

−+=−+

−−

=

=−+

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Purpose:Fast treatment of line segments that are trivially inside/outside window.

Purpose:Fast treatment of line segments that are trivially inside/outside window.

Cohen-Sutherland Algorithm

P = (x,y) - point to be classified against window D

Idea: Assign to P a binary code consisting of a bit for each edge of D, using lookup table:

0000

0101

0001

1001

0100

1000 1010

0010

0110

01

1234

bit

y y< min

y y> max

x x> maxx x< min

y y≥ miny y≤ max

x x≤ maxx x≥ min

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Cohen-Sutherland Algorithm (cont’d)

Why?

Given L from ),( to),( 1100 yxyx

If bitwise and of the codes of ( , ) ( , )x y x y0 0 1 1 and is not zero,

& rectangle D.

then L can be trivially handled (it is either totally outside or totally inside D).

or the bitwise or is zero,

0000

0101

0001

1001

0100

1000 1010

0010

0110

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end); , , , , , (

;,

; ofbit zero-nonleftmost ofboundary Windowelseend

); , , , , , (;,

; ofbit zero-nonleftmost ofboundary Windowbegin

then) ) ndow((OutsideWi if else;) , draw( then ) 0 ) or ( ( if

;return then ) 0 ! ) and ( ( if);( code );( code

) , , , ),,( ),,( (

maxminmaxmin20

102

1

maxminmaxmin12

102

0

0

1010

10

1100

maxminmaxmin111000

yyxxPPEdgePPP

CEdge

yyxxPPEdgePPP

CEdge

PPPCC

CCPCPC

yyxxyxPyxP

Clip-S-C

Clip-S-C

Clip-S-C

∩⇐

⇐

∩⇐

⇐

===

⇐⇐==

0000

0101

0001

1001

0100

1000 1010

0010

0110A

B

CD

E

AB CB DB DE

Cohen-Sutherland Algorithm (cont’d)

01

1234

bit

y y< min

y y> max

x x> maxx x< min

y y≥ miny y≤ max

x x≤ maxx x≥ min

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3D clipping

Determine portion of line inside axis-aligned parallelpiped (viewing frustum in NDC)Simple extension to 2D algorithmsAfter perspective transform

means that clipping volume always the samexmin=ymin= -1, xmax=ymax= 1 in OpenGL

boundary lines become boundary planesbut bit-codes still work the same wayadditional front and back clipping plane

zmin = -1, zmax = 1 in OpenGL

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Triangle Clipping

How does intersection of rectangle & triangle looks like?

How many sides?

How to expand clipping to triangles?Hint: it is convexWill develop on the board…

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end); , , , ,,...,,, ,.., (

;,

;,

; ofbit zero-nonleftmost ofboundary Window begin

then) ) ndow((OutsideWi ifn to1ifor else

;)poly draw( then ) 0 )or ...or or ( ( if;return then ) 0 ! ) and ... and and ( ( if

);( code n to1ifor ) , , , , ,..,poly (

maxminmaxmin11,,110

11,

1,1

10

10

maxminmaxmin0

yyxxPPPPPPEdgePPP

EdgePPP

CEdge

P

CCCCCC

PCyyxxPP

niiiiii

iiii

iiii

i

i

n

n

ii

n

++−−

++

−−

∩⇐

∩⇐

⇐

=

===

⇐==

Clip-S-C

Clip-S-C

Cohen-Sutherland Algorithm for convex polygons

01

1234

bit

y y< min

y y> max

x x> maxx x< min

y y≥ miny y≤ max

x x≤ maxx x≥ min