Scan Conversion 2

7/30/2019 Scan Conversion 2

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I N T R O D U C T I O N T O C O M P U T E R G R A P H I CS

Andries van Dam September 23, 2008 ScanConversion2 #/26

Scan Conversion 2

(pages 81-91 in the textbook)


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Version 1: really bad

Forx= R to R

y = sqrt(R * Rx*x);

Pixel (round(x), round(y));

Pixel (round(x), round(-y));

Version 2: slightly less bad

Forx= 0 to 360

Pixel (round (R cos(x)), round(R sin(x)));

(17, 0)

(0, 17)

(17, 0)

(0, 17)

Scan Converting Circles


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Symmetry: If (x0 + a, y0 + b) is on circle

also (x0 a, y0 b) and (x0 b, y0 a);hence 8-way symmetry.

Reduce the problem to finding the pixelsfor 1/8 of the circle

R

(x0 + a,y0 + b)

(x-x0)2 + (y-y0)

2 =R2

(x0,y0)

Version 3 Use Symmetry


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Scan top right 1/8 of circle of radius R

Circle starts at (x0, y0 + R)

Lets use another incremental algorithmwith decision variable evaluated atmidpoint

(x0,y0)

Using the Symmetry


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E

SE

x = x0; y = y0 + R; Pixel(x, y);

for (x = x0+1; (x

x0) > (yy0); x++) {

if (decision_var < 0) {

/* move east */

update decision_var;

}

else {

/* move south east */

update decision_var;

y--;

}

Pixel(x, y);

}

Note: can replace all occurrences ofx0, y0 with 0, 0and Pixel (x0 + x, y0 + y) with Pixel (x, y)

Shift coordinates by (-x0,-yo)

Sketch of Incremental Algorithm


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Decision variable

negative if we move E, positive if wemove SE (or vice versa).

Follow line strategy: Use implicit equationof circle

f(x,y) = x2 + y2 R2 = 0

f(x,y) is zero on circle, negative inside,

positive outside

If we are at pixel (x, y)

examine (x + 1, y) and (x + 1, y1)

Compute fat the midpoint

What we need for Incremental

Algorithm


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P = (xp, yp)M ME

MSE

SE

Decision Variable

E

Evaluate f(x,y) =x2 + y2 R2

at the point

We are asking: Is

positive or negative? (it is zero on circle)

Ifnegative, midpoint inside circle, choose E

verticaldistance to the circle is less at

(x + 1, y) than at (x + 1, y1).

Ifpositive, opposite is true, choose SE

2

2

2

2

1)1(

2

1,1 Ryxyxf -

-++=

-+

-+

2

1,1 yx


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Decision based on vertical distance Ok for lines, since dand dvertare

proportional

For circles, not true:

Which dis closer to zero? (i.e. which of thetwo values below is closer to R):

RyxCircyxd

RyxCircyxd

--++=-+

-++=+

22

22

)1()1()),1,1((

)1()),,1((

2222)1()1()1( -++++ yxoryx

The right decision variable?


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We could ask instead:

Is (x+ 1)2 + y2 or (x+ 1)2 + (y 1)2closer to R2?

The two values in equation above differ by

12])1()1[(])1[(2222 -=-++-++ yyxyx

fE fSE = 290 257 = 33

(0, 17) (1, 17)

(1, 16)

fE = 12 + 172 = 290

E

SE

fSE = 12 + 162 = 257

2y 1 = 2(17) 1 = 33

Alternate Phrasing (1/3)


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The second value, which is always less,is closerif its difference from R2 is less than

i.e., if

then

so

so

so


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How to compute the value of

at successive points?

Answer: Note that

is just

and that

is just

(1/2)

2

2

2

2

1)1(),( Ryxyxf -

-++=

2232),(

),()1,1(

32),(

),(),1(

SE

E

+-+=D

--+

+=D

-+

yxyx

yxfyxf

xyx

yxfyxf

Incremental Computation, Again


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FF

F

Incremental Computation(2/2)

If we move E, update by adding 2x+ 3

If we move SE, update by adding2x+ 3 2y+ 2.

Forward differences of a 1st degree polynomialare constants and those of a 2nd degreepolynomial are 1st degree polynomials

this first order forward difference, like apartial derivative, is one degree lower


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The function is linear, henceamenable to incremental computation, viz:

Similarly

Second Differences(1/2)32),(E +=D xyx

2),()1,1( EE =D--+D yxyx2),(),1( EE

=D-+Dyxyx

4),()1,1(SESE

=D--+D yxyx

2),(),1( SESE =D-+D yxyx


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For any step, can compute new E(x, y) from oldE(x, y) by adding appropriate second constantincrement update delta terms as we move.

This is also true ofSE(x, y)

Having drawn pixel (a,b), decide location of newpixel at (a + 1, b) or (a + 1, b 1), usingpreviously computed d(a, b).

Having drawn new pixel, must update d(a, b) fornext iteration; need to find either d(a + 1, b) or

d(a + 1, b 1) depending on pixel choice

Must add E(a, b)or SE(a, b) to d(a, b)

So we

Look at d(i) to decide which to draw next,updatexand y

Update dusing E(a,b) or SE(a,b)

Update each ofE(a,b) and SE(a,b) for futureuse

Draw pixel

Second Differences(2/2)


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Midpoint Eighth Circle Algorithm

MEC (R) /* 1/8th of a circle w/ radius R */

{

int x = 0, y = R;

int delta_E, delta_SE;

float decision;

delta_E = 2*x + 3;

delta_SE = 2(x-y) + 5;

decision = (x+1)*(x+1) + (y + 0.5)*(y + 0.5) R*R;

Pixel(x, y);

while( y > x ) {

if (decision > 0) {/* Move east */

decision += delta_E;

delta_E += 2; delta_SE += 2; /*Update delta*/}

else {/* Move SE */

y--;

decision += delta_SE;

delta_E += 2; delta_SE += 4; /*Update delta*/

}x++;

Pixel(x, y);

}

}


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Uses floats! 1 test, 3 or 4 additions per pixel

Initialization can be improved

Multiply everything by 4 No Floats!

Makes the components even, but sign of

decision variable remains same

Questions

Are we getting all pixels whose distancefrom the circle is less than ?

Why is y > xthe right stopping criterion?

What if it were an ellipse?

Analysis


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Patterned primitives

Aligned Ellipses

Non-integerprimitives

General conics

Other Scan Conversion Problems


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Patterned line from Pto Q is not sameas patterned line from Q to P.

Patterns can be geometric or cosmetic

Cosmetic: Texture applied aftertransformations

Geometric: Pattern subject totransformations

Cosmetic patterned line

Geometric patterned line

Patterned Lines

P Q

P Q


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Geometric Pattern vs. Cosmetic Pattern

+

Geometric(Perspectivized/Filtered)

Cosmetic(Contact Paper)


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Equation is

i.e,

Computation of and is similar

Only 4-fold symmetry

When do we stop stepping horizontally andswitch to vertical?

Aligned Ellipses

12

2

2

2

=+b

y

a

x

222222bayaxb =+

ED

SED


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When absolute value of slope of ellipse ismore than 1, viz:

How do you check this? At a point (x,y) forwhich f(x,y) = 0, a vector perpendicular tothe level set is f(x,y) which is

This vector points more right than up when

Direction Changing Criterion (1/2)

),(),,( yx

y

fyx

x

f

0),(),( >

-

yx

y

fyx

x

f


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In our case,

and

so we check for

i.e.

This, too, can be computed incrementally

Direction Changing Criterion (2/2)

xayxx

f 22),( =

ybyxy

f 22),( =

02222

>- ybxa

022 >- ybxa


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Now in ENE octant, not ESE octant

This problem is artifact ofaliasing muchmore on this later

Problems with Aligned Ellipses


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Non-Integer Primitives

Initialization is harder

Endpoints are hard, too

making Line (P,Q) and Line (Q,R)join properly is a good test

Symmetry is lost

Non Integer Primitives and

General Conics

General Conics

Very hard--the octant-changing test istougher, the difference computationsare tougher, etc.

do it only if you have to.

Note that drawing gray-scale conics is

easier than drawing B/W conics


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Generic Polygons

(More information and these pictures on page92-93 of textbook)

Scan Conversion 2

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