Line Polygon Clipping

8/13/2019 Line Polygon Clipping

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Topics

Viewing Transformation Pipeline in 2D

Line and polygon clipping

Brute force analytic solution Cohen-Sutherland Line Clipping Algorithm

Cyrus-Beck Line Clipping Algorithm Sutherland-Hodgman Polygon Clipping

Sampling Theorem (Nyquist Frequency)


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Viewing Transformation in 2D

x world

y world

xview

yview

World and Viewing

Coordinates

Normalized DeviceCoordinates

1

1

x view

y view

Clipping

Device Coordinates


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Viewing Transformation in 2D

A scene in World-Coordinates

World-Coordinates to ViewingCoordinates

Viewing Coordinates toNormalized Device Coordinates

Normalized Device Coordinates

to Device Coordinates

WC

VC

NDCDC

Clipping

Objects are given in world coordinates The world is viewed through a window

The window is mapped onto a device viewport


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Line and Polygon Clipping

The problem:

Given a set of 2D lines or polygons and a

window, clip the lines or polygons to theirregions that are inside the window


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Motivations

Efficiency

Display in portion of a screen

Occlusions

clip rectangle


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Line Clipping

We will deal only with lines (segments)

Our window can be described by twoextreme points:

(xmin,ymin) and (xmax,ymax)

A point (x,y) is in the window iff:

xmin x xmax and ymin y ymax


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Brute Force Analytic Solution

The intersection of convex regions is always convex

Since both Wand S are convex, their intersection is

convex, i.e a single connected segment of S

Question: Can the boundary of two convex shapes

intersect more than twice?

0, 1, or 2 intersections between a line and a window

W W W

S S S


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Pseudo Code for Midpoint Line Drawing

Assume x1>x0 and 0 < slope 1

Line(x0,y0,x1,y1)begin

int dx, dy, x, y, d, E, E ;x:= x0; y=y0;

dx := x1-x0; dy := y1-y0;

d := 2*dy-dx;E := 2*dy; := 2*(dy-dx);

PlotPixel(x,y);

while(x < x1) do

if (d < 0) thend:=d+ E;x:=x+1;

end;

else

d:=d+ E;x:=x+1;

y:=y+1;

end;

PlotPixel(x,y);

end;end;


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Line Clipping

Q E

NEM

y = ymin

x = xmin

(xmin,Round(mxmin+B))

(xmin, mxmin+B)

Midpoint Algorithm: Intersection with a vertical edge


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Line Clipping

Midpoint Algorithm: Intersection with a horizontal edge

AB y = ymin

x = xmin

y = ymin-1/2


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Cohen-Sutherland for Line Clipping

Clipping is performed by computing intersections with four

boundary segments of the window:

Li, i=1,2,3,4

Purpose: Fast treatment of lines that are trivially

inside/outside the window

Let P=(x,y) be a point to be classified against window W

Idea: Assign Pa binary code consisting of a bit for eachedge of W. The bit is 1 if the pixel is in the half-plane that

does not contain W


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bit 1 0

123

y < yminy > ymax

x > xmax

y yminy ymax

x xmax4 x < xmin x xmin

0101

0001

01100100

1001

0010

1010

0000

1000ymin

ymax

xmin xmax

1

234


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Given a line segment S fromp0=(x0,y0) top1=(x1,y1)to be clipped against a window W

If code(p0) AND code(p1) is not zero, then S istrivially rejected

If code(p0) OR code(p1) is zero, then S is triviallyaccepted

0101

0001

01100100

1001

0010

1010

0000

1000


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Otherwise: let assume w.l.o.g. thatp0 is outside W

Find the intersection of S with the edge

corresponding to the MSB in code(p0) that is equal to1. Call the intersection pointp2.

Run the procedure for the new segment (p1 ,p2).

A

BC

D

EF

GH

I

1

2

34


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Cyrus-Beck Line Clipping

V0

V1

inside

N

U

Inside/Outside Test:

Assume WLOG that V=(V1-V0) is the border vector where

"inside" is to its right

If V=(Vx

,Vy

), N is the normal to V, pointing outside,

defined by N=(-Vy,Vx)

Vector U points "outside" if NU > 0

Otherwise U points "inside"


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P0

inside

N QP1V(t)

L

The parametric line P(t)=P0+(P1-P0)t

The parametric vector V(t)=P(t)-Q

The segment P0P1 intersects the line L at t0 satisfying V(t0)N=0

The intersection point is P(t0)

=P1-P0 points inside if (P1-P0)N


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p0p1

Ni Qi

Denote p(t)=p0+(p1-p0)t t[0..1]

Let Qi

be a point on the edge Li

with outside

pointing normal Ni

V(t) = p(t)-Qi is a parameterized vector from Qi to

the segment P(t)

Ni V(t) = 0 iff V(t) Ni

We are looking for tsatisfying Ni V(t) = 0


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0 = Ni V(t)

= Ni (p(t)-Qi)

= Ni (p0+(p1-p0)t-Qi)= Ni (p0-Qi) + Ni (p1-p0)t

Solving for t we get:

where =(p1-p0)

Comment: If Ni=0, t has no solution (V(t) Ni)

Ni(p0-Qi)

-Ni(p1-p0)t =

Ni(p0-Qi)

-Ni=


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p1

p0

PL

PE

p0

p1

PE

PE

PL

PL

p1

PL

PEp0


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The intersection of p(t) with all four edges Li is

computed, resulting in up to four ti values

If ti1 , ti can be discarded

Based on the sign of Ni, each intersection

point is classified as PE (potentially entering) orPL (potentially leaving)

PE

with the largest tandPL

with the smallest tprovide the domain of p(t) inside W

The domain, if inverted, signals that p(t) is

totally outside


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Sutherland-Hodgman

Polygon-Clipping Algorithm


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Sutherland-Hodgman


right clip boundary bottom clip boundary

left clip boundary top clip boundary

Idea: Clip a polygon by successively clipping against each

(infinite) clip edge

After each clipping a new set of vertices is produced.


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Sutherland-Hodgman


clipboundary

inside outsides

clipboundary

inside outside

s

p

clipboundary

inside outsidep

si

sclipboundary

inside outside

pi

p added tooutput list

i added tooutput list

no output i and p added to

output list

p

For each clip edge - scan the polygon and consider the

relation between successive vertices of the polygon

Each iteration adds 0, 1 or 2 new vertices

Assume vertex s has been dealt with, vertex p follows:


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Sutherland-Hodgman


Window

V3

V'2

V2

V1

V3

V2

V1

LeftClipping

RightClipping

BottomClipping

TopClipping

V1V

2V3

V1V

2V2V3

V1V

2V2V3

V1V

2V2V2


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Sampling Theorem

Question: How dense should be the pixel grid in

order to draw properly a drawn object?

Given a sampling at intervals equal to d then one

may recover frequencies of wavelength > 2d

Aliasing: If the sampling interval is more than 1/2

the wavelength, erroneous frequencies may be

produced


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Sampling Theorem

2D Example: Moire Effect

Documents

Line Polygon Clipping