lecture 03 - lines

8/4/2019 Lecture 03 - Lines

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CS 430: Computer Graphics

1


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Assignment 1

Framework for World View Read XML Render Points in World

Window Save to an image

Details:Due April 12th, 2pmMakefile required Submit via email Attach zip/tar.gz


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Assignment 1: XML


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XML

Nodes define the XML docTags are nodes Even white space is a node!Nodes have children


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Parsing XML in JavaDocument doc = db.parse(file);

NodeList nodeLst =doc.getElementsByTagName("employee");

Tom

Cruise

George

Shanks

Gets all of the nodes with a specific name!


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Parsing XML in JavaNodeList nodeLst =

doc.getElementsByTagName("employee");

Node n = nodeLst.item(0);

if (n.getNodeType() ==Node.ELEMENT_NODE) {

Element el = (Element) n;

System.out.println(

el.getAttribute("type") );

Tom

Cruise

George

Shanks

Nodes can be anything, an element is a node corresponding to a TAG

Elements may have attributes (flags inside the tag)


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Parsing XML in JavaElement el = (Element) n;

NodeList children =el.getElementsByTagName("firstname");

Element child = (Element)children.item(1);

System.out.println("ChildNode: "+child.getNodeName());

Tom

Cruise

George

Shanks

Nodes also can have child nodes, here we access the firstname node.


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Writing Images in Java

BufferedImage code example


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Assignment Objectives

Read an xml file containing the world-viewcorrdinates, a set of colored points, and a listof transformation matricies.

Create a global transformation matrix from the

list of transformation matricies. Apply this transformation matrix to the points. Clip the points against the world window. Draw the points to the world window buffer. Save the results to an image


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Assignment 1

Skeleton Java Code on Assignment Page useit!

Start EARLY to avoid problems! Dont get caught up in XML, contact me if you

have issues!


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Line Drawing


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Scan-Conversion AlgorithmsScan-Conversion:

Computing pixelcoordinates forideal line on2D raster grid

Pixels bestvisualized as

circles/dotsWhy? Monitor

hardware

1994 Foley/VanDam/Finer/Huges/Phillips ICG


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In-Class Group Assignment


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Drawing a Line

y= mx+ B m = y /x Start at leftmostxand increment by 1

x =

1yi = Round(mxi + B)

This is expensive and inefficient Since x = 1,yi+1 =yi + m

No more multiplication!This leads to an incremental algorithm


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Digital Differential Analyzer (DDA) If |slope| is less then 1

x = 1 elsey = 1

Check for vertical line m = Compute corresponding

y(x) = m (1/m) xk+1 = xk +x

yk+1 = yk +y

Round (x,y) for pixellocation


xk+1

x

xk

yk+1

yky


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Generalizing DDA If |slope| is less than or equal to 1

Ending point should be right of starting point

If |slope| is greater than 1 Ending point should be above starting point

Vertical line is a special case x= 0


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Bresenhams Algorithm1965 @ IBMBasic Idea:

Only integerarithmetic Incremental

Consider theimplicitequationfor a line:



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Bresenhams AlgorithmGiven:implicitline equation:Let:

where rand q are points on the line and dx,dyare positiveThen:Observe that all of these are integersand: for points above the line

for points below the line

Pics/Math courtesy of Dave Mount @ UMD-CP


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Bresenhams AlgorithmSuppose we just

finished(assume 0 slope

1) other casessymmetric

Which pixel next?

E or NE


MQ


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Assume: Q = exacty value atymidway between E and NE:Observe:If , then pick EElse pick NE

If ,it doesnt matter

Bresenhams Algorithm


M

Q < M

Q = M MQ


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Create modified implicit function (2x)

Create a decision variable D to select, where D

is the value offat the midpoint:



M


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If then M is below the lineNE is the closest pixel

If then M is above the line E is the closest pixel


M

Note: because we multiplied by 2x, D is now aninteger which is very good news

How do we make this incremental??


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What increment for computing a new D? Next midpoint is:

Hence, increment by:

Case I: When E is next


M


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Case II: When NE is next

What increment for computing a new D? Next midpoint is:

Hence, increment by:


M


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How to get an initial value for D? Suppose we start at: Initial midpoint is:Then:



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The Algorithm

Assumptions:

0 slope 1

Pre-computed:


void bresenham( point q, point r ){int dx,dy,D,x,y, incrE, incrNE;if( r.x


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Problem!

Only works for In class exercise: extend so it works for void bresenham( point q, point r ){

int dx,dy,D,x,y, incrE, incrNE;if( r.x


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Problem!

In class exercise: change so it works for void bresenham( point q, point r ){

int dx,dy,D,x,y, incrE, incrNE;if( r.x


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Generalize Algorithm

Ifqx > rx, swap points

If slope > 1, always increment y, conditionallyincrement x

If -1 < slope < 0, always increment x,conditionally decrement y

If slope > 1, always increment y, conditionallyincrement x (Rework D increments)

If slope < -1, always decrement y,conditionally increment x (Rework Dincrements)


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Some issues with Bresenhams

AlgorithmsPixel density varies

based on slopestraight lines look

darker, morepixels per unitlength

Endpoint orderLine from P1 to P2

should match P2to P1

Always choose E

when hitting M, 1994 Foley/VanDam/Finer/Huges/Phillips ICG


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Some issues with Bresenhams

AlgorithmsHow to handle the line when it hits the clip

window?

Vertical intersectionsCould change line slopeNeed to change init cond.

Horizontal intersections

Again, changes in theboundary conditions

Cant just intersectthe line w/ the box



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Line clipping

Lecture Credits: Most pictures are fromFoley/VanDam; Additional and extensive thanks

also goes to those credited on individual slides


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Scissoring Clipping

Pics/Math courtesy of Dave Mount @ UMD CP

Performed during scan conversion of

the line (circle, polygon)

Compute the next point (x,y)If xmin x x max and ymin y y max Then output the point.Else do nothing

Issues with scissoring:

Too slowDoes more work then necessary

Better to clip lines to window, thandraw lines that are outside of

window


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The Cohen-Sutherland Line Clipping

AlgorithmHow to clip lines to fit

in windows?easy to tell if whole

line falls w/inwindow

harder to tell whatpart falls inside



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Cohen-SutherlandBasic Idea:First, do easy test

completelyinsideor outside thebox?

If no, we need a

more complextestNote: we will also

need to figure out

how line Pics/Math courtesy of Dave Mount @ UMD CP


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Cohen-Sutherland

Perform trivial accept and reject Assign each end point a location code

Perform bitwise logical operations on a lineslocation codes Accept or reject based on result Frequently provides no information

Then perform more complex line intersection


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Cohen-SutherlandThe Easy Test:Compute 4-bit code

based onendpointsP1 and

P2



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Cohen-Sutherland

Line is completelyvisible iff both codevalues of endpointsare 0, i.e.

If line segments arecompletely outsidethe window, then



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Cohen-SutherlandOtherwise,

we clip the lines:

We know that thereis a bit flip, w.o.l.g.assume its (x0,x1)

Which bit? Try `em all!

suppose its bit 4Thenx0


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Cohen-SutherlandClearly:Using similartriangles

Solving foryc gives



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Cohen-SutherlandReplace

(x0,y0) with (xc,yc)Re-compute codesContinue until all

bit flips (clip lines)are processed,i.e. all points areinside the clipwindow



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Cohen Sutherland

0101

1010


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Cohen Sutherland

0101

0010


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Cohen Sutherland

0001

0010


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Cohen Sutherland

0001

0000


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Cohen Sutherland

0000

0000


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Parametric Line Clipping

Developed by Cyrus and Beck in 1978Used to clip 2D/3D lines against convex

polygon/polyhedron

Liang and Barsky (1984) algorithmefficient in clipping upright 2D/3Dclipping regions

Cyrus-Beck may be reduced to more

efficient Liang-Barsky caseBased on parametric form of a line

Line: P(t) = P0 +t(P1-P0)


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Parametric Line Equation

Line: P(t) = P0 +t(P1-P0)

t value defines a point on the line going throughP0 and P1

0


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The Cyrus-Beck Technique

Cohen-Sutherland algorithm computes (x,y)intersections of the line and clipping edge

Cyrus-Beck finds a value of parameter t for

intersections of the line and clipping edges Simple comparisons used to find actual

intersection points Liang-Barsky optimizes it by examining tvalues

as they are generated to reject some linesegments immediately


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Finding the Intersection PointsLine P(t) = P

0

+t(P1

-P0

)Point on the edge PeiNi Normal to edge i

Make sure1.D 0, or P 1 P02.Ni D 0, lines are not

parallel


DN

]P[PNt

)-P(PLet D

]-Pt[PN]-P[PN])-P-Pt(P[PN

][P(t)-PN

i

Eii

iEii

Eii

Eii

=

=

=+

=+

=

0

01

010

010

0

0

0


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Calculating Ni

Ni for window edges WT: (0,1) WB: (0, -1) WL: (-1,0) WR: (1,0)

Ni

for arbitrary edges Calculate edge direction

E = (V1 - V0) / |V1 - V0|

Be sure to process edges in CCW order

Rotate direction vector -90 Nx = Ey Ny = -Ex


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Finding the Line Segment

Calculate intersection points between line and every windowline

Classify points as potentially entering (PE) or leaving (PL)

PE if crosses edge into inside half plane => angle P0P1 and Ni

greater 90 => Ni

D < 0

PL otherwise.

Find Te = max(te)

Find Tl = min(tl)

Discard ifTe

> Tl

IfTe < 0 Te = 0

IfTl > 1 Tl = 1

Use Te, Tlto compute intersection coordinates (xe,ye), (xl,yl)

lecture 03 - lines

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