©zachary wartell, larry f. hodges scan conversion i revision 1.6 8/9/2007 copyright 2006, dr....

©Zachary Wartell, Larry F. Hodges

Scan Conversion I

Revision 1.68/9/2007

Copyright 2006, Dr. Zachary Wartell, UNCC, All Rights Reserved


Scan Conversion

• recall slide 23 and slide 24 from ITCS 4120-Introduction.ppt• scan conversion – algorithm with parameters:

CoordinateGeometryDescription D [IN], // !2D!

Raster R [IN/OUT] // typically the framebuffer

alter R by setting the appropriate pixel’s data values (RGB, etc) to represent pixels of D

• scan conversion of– 1D geometry

• line segments• curves (circle, ellipse, etc.)

– 2D geometry• axis-aligned rectangle• polygons

b

a

D=Line(a,b)

a

b

Pixels: P0,0, P1,1, P2,2, P3,2, …. P7,7

Scan-Conversion


(0,0) 1 2 3 4

1

2

3

4

0.5 1.5 2.5 3.5

0.5

1.5

2.5

3.5

0 1 2 3

1

2

3

0.5 1.5 2.5 3.5

0.5

1.5

2.5

3.5

0

VS

Center Lower-Left Corner

(2.5,3.5)

(2,3)

LLC: Area of pixel (x,y) is squarewith lower-left corner at (x,y)

(OpenGL)

Scan Convert a Point

(2,3)(2.5,3.5)

- location of pixel address

- location of “diagonal” point

Address-Coordinate Mapping Conventions


4

4(0,0)

(Lower-Left Corner)

4

40

(Center)

0

Scan Convert a Vertical/Horizontal Line

4

4

(Lower Left Corner)

Pixel Length = Coordinate Length 4 = 4

(0,0)

Line 1: ((0,0),(4,0))Line 2: ((4,0),(4,4))

Overlap No Overlap

Pixel Length ≠ Coordinate Length 5 ≠ 4

Gap

(OpenGL)


3

30

(Center)

0

Scan Convert an Axis-Aligned Rectangle

Pixel Width & Area ≠ Coord. Width & Area 4 ≠ 3, 16 ≠ 9

Rect 1: ((0,0),(3,3))Rect 2: ((3,0),(6,3))

(Lower Left Corner)

(OpenGL)

(Lower Left Corner)

3

3(0,0) 3(0,0)

3

Overlap

No Overlap

Pixel Width = Coord. Width Pixel Area = Coord. Area (3 = 3, 9 = 9)


Scan Convert Example (LLC/OpenGL)Rect 1: ((0,0),(4,4))

4

4(0,0)

Line 1: ((4,0),(4,4))

(0,0)

4

4

1) 2)

Line 2: ((4,4),(0,4))

(0,0)

4

4

3) Line 3: ((0,4),(0,0))

(0,0)

4

4

4) Line 4: ((0,0),(4,0))

(0,0)

4

4

5)

Overlap

Gap


So do we go with Center or Lower-Left Corner?

• It depends …– actual API’s specification (such as OpenGL) are

generally equivalent to the lower-left corner convention

• easy to mathematically specify• adjacent, filled primitives’ pixels don’t overlap

– but when first learning scan conversion algorithms its easier to use the center convention


Explicit (Slope-Intercept) Equation

Slope-Intercept Formula For A Line

Given a third point on the line: P = (X,Y)

Slope: (Y - Y1)/(X - X1)

= (Y2 - Y1)/(X2 - X1)

Solving For Y

Y = [(Y2-Y1)/(X2-X1)]X

+ [-(Y2-Y1)/(X2-X1)]X1 + Y1

orY = mx + b

Cartesian Coordinate System

2

4

1 2 3 4 5 6

3

5

6

1 P1 = (X1,Y1)

P2 = (X2,Y2)

P = (X,Y)

SLOPE =

RISE

RUN=

Y2-Y1X2-X1

L


Implicit Equation

Cartesian Coordinate System

2

4

1 2 3 4 5 6

3

5

6

1 P1 = (X1,Y1)

P2 = (X2,Y2)

F(X,Y) = 0

F(x,y): mx + b – y = 0

For arbitrary (x,y):

F(x,y) < 0 ↔ (x,y) above LF(x,y) = 0 ↔ (x,y) on LF(x,y) > 0 ↔ (x,y) below L

F(X,Y) < 0

F(X,Y) > 0

L


Parametric Form of a 2D Line Equation

Given points P1 = (X1, Y1) and P2 = (X2, Y2)

X = X1 + t(X2-X1)

Y = Y1 + t(Y2-Y1)

t is called the parameter. When

t = 0 we get (X1,Y1)

t = 1 we get (X2,Y2)

As 0 < t < 1 we get all the other points on the line segment between (X1,Y1) and (X2,Y2).


Other Helpful Formulas

Length of line segment between P1 and P2:

L = [ (X2-X1)2 + (Y2-Y1)2 ]1/2

Midpoint of a line segment between P1 and P3:

P2 = ( (X1+X3)/2 , (Y1+Y3)/2 )

Two lines are perpendicular iff

1) M1 = -1/M2

2) Cosine of the angle between them is 0.


Basic Line and Circle Algorithms

1. Must compute integer coordinates of pixels which lie on or near a line or circle.

2. Pixel level algorithms are invoked millions or billions of times when an image is created or modified.

3. Lines must create visually satisfactory images.• Lines should appear straight• Lines should terminate accurately• Lines should have constant density• Line density should be independent of line length and angle.• Lines sharing vertex should avoid/minimize overlap• Lines sharing vertex should avoid/minimize gaps!

4. Line algorithm should always be defined.


Simple DDA Line Algorithm

-Based on the parametric equation of a line: P = P1 + t(P2-P1)

Procedure DDA(X1,Y1,X2,Y2 :Integer);

Var Length, I :Integer;

X,Y,Xinc,Yinc :Real;

Begin

Length := ABS(X2 - X1);

If ABS(Y2 - Y1) > Length Then

Length := ABS(Y2-Y1);

Xinc := (X2 - X1)/Length;

Yinc := (Y2 - Y1)/Length;

X := X1;

Y := Y1;

For I := 0 To Length Do

Begin

Plot(Round(X), Round(Y));

X := X + Xinc;

Y := Y + Yinc

End {For}

End; {DDA}

avoidgaps


DDA Example

Compute which pixels should be turned on to represent the line from (6,9) to (11,12).

Length := Max of (ABS(11-6), ABS(12-9)) = 5

Xinc := 1

Yinc := 0.6

Values computed are:

(6,9),

(7,9.6),

(8,10.2),

(9,10.8),

(10,11.4),

(11,12) 6 7 8 9 10 11 12 13

9

10

11

12

13

DDA creates good lines but it is too time consuming due to the round function and long operations on floating point values.


The Midpoint Method

Assumptions: line with slope m between 0 and 1 (i.e. line lies in first octant).

The general formula for a line is y = mx + b where m is the slope of the line and b is the y-intercept.

The other octants algorithms are similar based on symmetry. Also we distinguish between x-major (|Δx|>|Δy|) and y-major (|Δy|>|Δx| lines.

+x-x

-y

+y


Bresenham Line (Midpoint formulation)

xk

yk

yk+1

xk+1 =xk+1

Pk+1

L: y=mx+b

Pk

?

Pk – current pixel at (xk,yk)Pk+1 – next pixel at (xk+1,yk+1)


From Pk:(xk,yk) do we step to Pα::(xk+1, yk) or Pβ:(xk+1, yk+1) ? -Is Pk+1(xk, yk+1) = Pα or Pβ

-Is Pά or Pβ closer to line L? dl > du ↔ Δd > 0 -Is mid-point Mk below or above line L? f(xk+1, yk + ½) > 0

Bresenham Line (Midpoint formulation)

xk

yk

yk+1

xk+1 =xk+1

Pβ

Pα

Mk

L: y=mx+b

Pk

du

dl

y=m(xk+1)+bΔd=dl-du

Distance=1

yk+1 = (?)


Goal: Is mid-point Mk below or above line L? f(xk+1, yk + ½) > 0

f(xk+1, yk + ½) = m (xk+1) + b – (yk + ½) = Δy / Δx (xk+1) + b – (yk + ½)

Insight #1: avoid fractions, so use: g(…) > 0, K > 0 → K * g(…) > 0

Let K = 2 Δ x and define “decision parameter” pk pk = 2 Δ x * f(…) = 2 Δy (xk+1) + 2 Δx b – Δx (2 yk + 1)

= 2 Δy xk - 2 Δx yk + [2 Δy + Δx (2 b -1)] (1)

Avoid Fractions

pk

yk

yk+1Pβ

Pα

MkPk

yk

yk+1Pβ

Pα

MkPk


Avoid Multiplications – Base Case

Insight #2: Formulate as only iterative additions instead of multiplications

#2.1 Base Case: p0 = 2 Δy x0 - 2 Δx y0 + [2 Δy + Δx (2 b -1)] = 2 Δy - Δx

p0= 2 Δy - Δx

xk=0

yk=0

yk+1

xk+1 =xk+1

Pβ

Pα

Mk=0

L: y=mx+b

P0

xk=0

yk=0

yk+1

xk+1 =xk+1

Pβ

Pα

Mk=0

L: y=mx+b

P0


Insight #2 : Formulate as iterative addition instead of multiplication

#2.2 Induction Case: Compute pk+1 from pk

pk = 2 Δy xk - 2 Δx yk + [2 Δy + Δx (2 b -1)] (2) from (1) [18]pk+1 = 2 Δy xk+1 - 2 Δx yk+1 + [2 Δy + Δx (2 b -1)] (3)

so pk+1 = pk + 2 Δy - 2 Δx (yk+1 – yk) - from (2) and (3)

yields 2 situations:

i) yk+1 – yk = 0 iff we step to Pα

ii) yk+1 – yk = 1 iff we step to Pβ

yk,yk+1

xk

Pk

L

Pα=Pk+1=(xk+1,yk+1)

Pβ

Mk

xk+1 =xk+1

Mk+1

pkpk+1

xk+2

yk+1

yk,yk+1

xk

Pk

L

Pα=Pk+1=(xk+1,yk+1)

Pβ

Mk

xk+1 =xk+1

Mk+1

pkpk+1

xk+2

yk+1

Avoid Multiplications – Induction Case

Figure 20: “yk+1 – yk =0” situation


yk

xk

Pk

L'

Pα

xk+1 =xk+1

pk

xk+2

yk+1=yk+1

Mk

Mk+1

pk+1

Pβ=Pk+1=(xk+1,yk+1)

yk

xk

Pk

L'

Pα

xk+1 =xk+1

pk

xk+2

yk+1=yk+1

Mk

Mk+1

pk+1

Pβ=Pk+1=(xk+1,yk+1)

Avoid Multiplications – Induction Case (Situation ii)

Figure 21: “yk+1 – yk =1” situation

Insight #2 : Formulate as iterative addition instead of multiplication

#2.2 Induction Case: Compute pk+1 from pk

pk = 2 Δy xk - 2 Δx yk + [2 Δy + Δx (2 b -1)] (2) from (1) [18]pk+1 = 2 Δy xk+1 - 2 Δx yk+1 + [2 Δy + Δx (2 b -1)] (3)

so pk+1 = pk + 2 Δy - 2 Δx (yk+1 – yk) - from (2) and (3)

yields 2 situations:

i) yk+1 – yk = 0 iff we step to Pα

ii) yk+1 – yk = 1 iff we step to Pβ


Bresenham’s Line Algorithm

Begin {Bresenham for lines with slope between 0 and 1}

Δx := ABS(xend - xstart);

Δy := ABS(yend - ystart);

p := 2*Δy - Δx;

If xstart > xend Then Begin

x := xend;

y := yend

End

Else Begin

x := xstart;

y := ystart

End;

For I := 0 to Δx Do Begin

Plot(x,y);

x := x + 1;

If p 0 Then Begin

y := y + 1;

p := p + 2*(Δy- Δx);

Else

p := p + 2*Δy End

End {For Loop}

End; {Bresenham}


In Class Exercise

3,5

9,9

Line: ((3,5),(9,9))

I p step x y

0

1

2

3

4

5

6

xstart = , ystart =xend = , yend =

Δx = , 2*(Δy- Δx) =Δy = , 2*Δy =

p0 = 2*Δy- Δx =

{major, major_minor}


In Class Exercise (Answer)

3,5

9,9

Line: ((3,5),(9,9))

xstart = 3 , ystart = 5xend = 9 , yend = 9

Δx = 6 , 2*(Δy- Δx) = -4Δy = 4 , 2*Δy = 8

p0 = 2*Δy- Δx = 2

I p step x y

0 2 major_minor 3 5

1 -2 major 4 6

2 6 major_minor 5 6

3 2 major_minor 6 7

4 -2 major 7 8

5 6 major_minor 8 8

6 2 major_minor 9 9


Generalizing to other Octant’s

Octant 1: (x-major)

Octant 2: (y-major)Octant 3: (y-major)

Octant 4: (x-major)

Octant 5: (x-major)

Octant 6: (y-major)

Octant 7: (y-major)

Octant 8: (x-major)


Generalizing to other Octant’s

octant major minor midpoint offset

major-minor step cond.

p0 Δpmajor Δpmajor_minor

1 x +=1 y +=1 (+1,+1/2) p > 0 2*Δy - Δx 2*Δy 2*(Δy- Δx)

2

… … … … … … … …

8

O1

O2O3

O4

O5

O6 O7

O8

O1

O2O3

O4

O5

O6 O7

O8


Simple Circle Algorithms

Since the equation for a circle on radius r centered at (0,0) is

x2 + y2 = r2,

an obvious choice is to plot

y = ±(r2 - x2)1/2

as -r ≤ x ≤ r.

This works, but is inefficient because of the

multiplications and square root

operations. It also creates large gaps in the

circle for values of x close to R.

A better approach, which is still inefficient but avoids the gaps is to plot

x = r cosø

y = r sinø

as ø takes on values between 0 and 360 degrees.


Circle Drawing Algorithm

We only need to calculate the values on the border of the circle in the first octant. The other values may be determined by symmetry. Assume a circle of radius r with center at (0,0).

Procedure Circle_Points(x,y :Integer);

Begin

Plot(x,y);

Plot(y,x);

Plot(y,-x);

Plot(x,-y);

Plot(-x,-y);

Plot(-y,-x);

Plot(-y,x);

Plot(-x,y)

End;

(a,b)

(b,a)

(a,-b)

(b,-a)

(-a,-b)

(-a,b)

(-b,-a)

(-b,a)


Fast Circles

Consider only the second octant of a circle of radius r centered on the origin. We begin by plotting point (0,r) and end when x > y.

The decision at each step is whether to choose the pixel directly right the current pixel or the pixel which is below and to the right (8-way stepping).

Assume Pk = (xk, yk) is the current pixel.Pα = (xk +1, yk) is the pixel directly rightPβ = (xk +1, yk-1) is the pixel right and below

y=x

x2 + y2 – r2 = 0


Fast Circles - The Decision VariableImplicit Equation: f(x,y) = x2 + y2 - r2

f(x,y) = 0 ↔ (x,y) on C f(x,y) < 0 ↔ (x,y) in C (“below”) f(x,y) > 0 ↔ (x,y) outside C (“above”)

Mk: f(xk + 1, yk - 1/2) = (xk + 1)2 + (yk - 1/2)2 - r2

pk = (xk + 1)2 + (yk - 1/2)2 - r2

What about pk+1? pk+1 = (xk+1 + 1)2 + (yk+1 – 1/2)2 - r2

= [(xk + 1) + 1]2 + (yk+1 – 1/2)2 - r2

= pk + 2(xk + 1) + (y2k+1 - y2

k) – (yk+1 - yk) + 1

yk+1 = yk iff we step to Pα

yk+1 = yk - 1 iff we step to Pβ

C: x2 + y2 - r2 =0

PαPk

Pβ

Mk

yk-1

yk

xk+1

(=xk+1)

xk

PαPk

Pβ

Mk

yk-1

yk

xk+1

(=xk+1)

xk


Fast Circles - Decision Variable (cont.)

The initial value of pk is

p0 = f(1,r - 1/2) = 1+ (r - 1/2)2 - r2

= 5/4 - r {1-r can be used if r is an integer}

When point Pα = (xk + 1, yk) is chosen then

pk+1 = pk + 2xk+1 + 1

When point Pβ = (xk+ 1, yk - 1) is chosen then

pk+1 = pk + 2xk+1 + 1 – 2yk+1


Fast Circle Algorithm

Begin {Circle}

x := 0;

y := r;

p := 1 - r;

Repeat

Circle_Points(x,y);

x := x + 1;

If p < 0 Then

p := p + 2*x + 1

Else Begin

y := y - 1;

p := p + 2*(x-y) + 1

End

Until x > y

End; {Circle}

y=x

x2 + y2 – r2 = 0


Revisions• 1.5

– typos– fixing math variable fonts– added answer slide to In Class Exercise– Dumped #34, #35, #36 Antialiasing – Solutions

• 1.6– Dumped slide 33 on ellipsoid rasterization, want to spend class time on other things

©zachary wartell, larry f. hodges scan conversion i revision 1.6 8/9/2007 copyright 2006, dr....

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