Simone Gasparini Analytical algorithm DDA algorithm ...home.deib.polimi.it/caglioti/ImageRasterization.pdf · Simone Gasparini [email protected] Image Analysis and Synthesis

Image Analysis and Synthesis - A.Y. 2006/2007

Simone GaspariniSimone [email protected]


Outline

Graphics Pipeline

Clipping

Rasterization

Line rasterization

• Analytical algorithm

• DDA algorithm

• Bresenham’s algorithm


Graphics pipeline

Steps required for rendering a 3D scene

Given a 3D (geometric) model of the scene and a viewpoint, generate the 2D image of the scene

Usually implemented as a mixture of software (CPU based) and hardware (GPU based) operations

Pipeline: a set of data processing elements connected in series, so that the output of one element is the input of the next one.

Graphics Pipeline: transform the 3D representation into a image


Graphics pipeline

Geometric model of the sceneRepresentation by means of points/vertex

M odeling

Geom etry

Rasterization

Display


Graphics pipeline

M odeling

Geom etry

Rasterization

Display

Clipping (field of view)Projection (3D 2D)Hidden surface removal (only visiblesurfaces)Shading & Shadowing (according tolight source(s) )


Graphics pipeline

M odeling

Geom etry

Rasterization

Display

From projected points to pixel (discretization )


Graphics pipeline

M odeling

Geom etry

Rasterization

Display

Image visualization (or printing)


Clipping

Consider only scene objects that are in the field of view of the camera

Two main approaches

BEFORE rasterization• For each objects evaluates intersections with the field of view• Possible intersections are new vertices of the object• It speeds up rasterization• It can be used for simple geometric primitives (segments)

AFTER rasterization (Scissoring)• Perform rasterization on all geometric primitives• Keep only the primitives inside the image borders


Rasterization

Last step in the graphics pipeline

Discretization:

• Projected points in float coordinates

• Pixel in int coordinates

Geometric primitives (segments) must be drawn into a pixel matrix (raster grid)

• A point is represented by a pixel

• A line segments as a chain of adjacent pixels

• Polygon as a set of connected pixels

If matrix is dense enough any kind of geometric primitive can bedrawn

Also known as Scan Conversion


Raster Grid

A grid of square pixelThe pixel centre has integer coordinatesA projected point is represented by the nearest pixel centreSegment line as a sequence of pixels such that:• Pixels lies as close to the ideal

line as possible• Sequence is straight as

possible

Pixel


Requirements

Clipping and Rasterization functions are called each time an image is created or modified

Hence… TIME MATTERS!

If Real-Time is required (e.g. animation), functions must be optimized

Optimization means:

• Use incremental algorithms− Exploits previous calculations− Minimize the use of time-consuming operation (multiplies and divides)

• Employ integer arithmetic− Integer arithmetic is faster than the floating-point one


Segments Rasterization

Line equation• y = mx + B• m = (x1-x2)/(y1-y2) = Δx/Δy

|m|<1 |m|>1|m| = 1 easy casem = 0 easy casem inf easy case

We deal with |m| < 1|m| < 1 is straightforward

|m|<1 a pixel foreach column

|m|>1 a pixel foreach rowFrom now on:

Pixel is “ON” Pixel is “OFF”


Analytical method – dummy method

Suppose:• P1(x1,y1) and • P2(x2,y2) segment extrema in integer coordinates

P1

P2

Starting from extremum having lowest x-coordinate:• xi= xi-1+1 • yi = mxi+ B• Round(yi)

Always select the closest pixel

Highly inefficient method!• Floating-point multiply (mx)• Floating-point addition (mx+B)• Round


DDA Algorithm

Floating-point multiply can be avoid using incremental technique

• The new point is calculated exploiting previous point

Digital Differential Analyzer algorithm (DDA)

• Mechanical device used to solve differential equation by numerical methods

• It trace out successive (x, y) values by simultaneously incrementing x and y by small steps proportional to the first derivative of x and y

• Line is the solution of

dy mdx

=


DDA Algorithm

Suppose |m|<1Note that:

But since

( )1 1

1

1

i i

i i

i i

y mx By m x x By y m x

+ +

+

+

= +

= + Δ +

= + Δ

1i iy y m+ = +

1xΔ =

1. Floating point addition

2. Round


DDA Algorithm

Pseudo-code

Line( int x0, int y0, int x1, int y1, int value)

{int x;float dy, dx, y, m;

dy = y1-y0;dx = x1-x0;m = dy/dx;y = y0;

for (x = x0; x <= x1; x++) {

WritePixel(x, floor(0.5 + y), value);y = y + m;

}}

// assumes |m|<1, x0 < x1

// initialization

// step y by slope m

// value to place in pixels

// set pixel to value


DDA Algorithm

|m|<1 and |m|>1 must be managed separately


Can we do better than DDA?

DDA uses

• Floating point addition

• Round function

Floating-point arithmetic should be avoided

• Approximation error

• Time consuming

We need an incremental method that uses only integer arithmetic

Bresenham’s algorithm (1965)


Bresenham’s Algorithm

Incremental methodEmploys only integer arithmeticHighly efficientCan be also applied to the rasterization of circles

At step n (assumes |m|<1):• P (xp,yp) pixel selected during previous step• For xp+1 we can select either E or NE pixel

NE

E

P (xp,yp)



Let Q intersection point between ideal line and x=xp+1 line

Let M the mid-point of the segment E-NE

Select the pixel minimizing the distance from the ideal line

Select the pixel on the same side of Q

NE

EP (xp,yp)

M

Q



To compute Q we need floating point arithmetic

Since the algorithms reduces to a choice between two pixel we can

exploit a decision variable.

E.g. we can choose the pixel according to the sign of the variable

NE

EP (xp,yp)

M

Q



Implicit equation of the line

The main idea is:

( , ) 0F x y ax by c= + + =

( , ) 0F x y >

( , ) 0F x y <

( , ) 0F x y =Let M=(xp+1,yp+1/2)

I can evaluate the sign of F(M) using the decision variable d=F(M)

Hence:

0 choose NE( ) 0 choose E

0 choose Ed F M

<⎧⎪= =⎨⎪ >⎩


Bresenham’s algorithm

Starting from

If E is chosen, for the new pixel it yields that

but

( ) ( )( , ) 0

( ) ( 1, 1/ 2) 1 1/ 2p p p p

F x y ax by c

d F M F x y a x b y c

= + + =

= = + + = + + + +

( ) ( )( 2, 1/ 2) 2 1/ 2new p p p pd F x y a x b y c= + + = + + + +

( ) ( )( 1, 1/ 2) 1 1/ 2old p p p pd F x y a x b y c= + + = + + + +

new oldd d a= +



Hence if we chose E we can evaluate the new value of d simply adding a to the previous value

If we had chosen NE:

Enew oldd d a= +

( ) ( )( 2, 3 / 2) 2 3/ 2new p p p pd F x y a x b y c= + + = + + + +

NEnew oldd d a b= + +



It sounds good… but a and b are not integer in general…If we consider the explicit equation of the line in this form:

Now a and b are integer!

1 2

1 2

dx x xdxy x B wheredy y ydy

= −⎧= + ⎨ = −⎩

( ), 0F x y dy x dx y B dx

a dyb dxc B dx

= ⋅ − ⋅ + ⋅ =

=⎧⎪ = −⎨⎪ = ⋅⎩



Calculate the initial value in the starting extremum

To avoid the fraction

Hence to evaluate dnew for any step a simple addition is needed

( ) ( )

( )

1 1 1 1

1 1

1 1

( 1, 1/ 2) 1 1/ 2/ 2

, / 2/ 2

F x y a x b y cax by c a bF x y a ba b

+ + = + + + +

= + + + +

= + +

= +

( )( , ) 2F x y ax by c= + +



MidpointLine(int x0, int y0, int x1, int y1, int value){

int dx, dy, incrE, incrNE, d, x, y;

dy = y1-y0;dx = x1-x0;d = 2*dy-dx;incrE = 2*dy;incrNE = 2*(dy-dx);x = x0;y = y0;

while (x < x1 ) {if ( d <= 0 ) {

d = d+incrE;x++;

} else {d = d+incrNE;x++;y++;

}WritePixel(x, y, value);

}}


Bresenham’s algorithm – Additional issues

Endpoint order

• Ensure that a line from P1 to P0 contains the same set of the line from P0 to P1

• The appearance of the line has to be independent of endpoints order

• Pixel choice is dependent on the order only when d=0 (the line passes exactly through the midpoint)

• Going left to right we chose E

• Hence going right to left we chose SW (by symmetry we would have chosen W)



Vertical Clipping

• What happens if the line has been analitycally clipped?• Suppose line clipped at x=xmin

E

NE

M(xmin,Round(mxmin+B))

(xmin, (mxmin+B))

Start from (xmin,Round(mxmin+B)) as the line was unclipped

Do not clip the line at x=xmin and the scan convert the line from (xmin,Round(mxmin+B)) to (x1,y1) !!

y = ymin

x = xmin



Horizontal Clipping

• What happens if the line intersects the horizontal border?• Suppose line clipped at y=ymin

• There are multiple pixel lying on the scan line y=ymin

• If scan conversion is applied at intersection point with y=ymin the line starts from A discarding leftmost pixels

• Hence find intersection with y=ymin -1/2 and round up the x value obtaining pixel B (Round(xymin-1/2), ymin)

y = ymin

x = xmin

y = ymin-1/2

B A

y = ymin-1


References

Principles of Computer Graphics (2nd Edition)J.D. Foley, A. Van-Dam, S.K. Feiner and J.F. Hughes - Addison-Wesley, 1990.

Principles of Digital Image SynthesisAndrew S. Glassner - Morgan Kaufmann,1995



Documents

Simone Gasparini Analytical algorithm DDA algorithm ...home.deib.polimi.it/caglioti/ImageRasterization.pdf · Simone Gasparini [email protected] Image Analysis and Synthesis