CPCS 391 Computer Graphics 1

transcript

Instructor: Dr. Sahar ShabanahLecture 3

+Scan conversion Algorithms

Primitives and Attributes Why Scan Conversion? Algorithms for Scan Conversion: Lines Circles Ellipses Filling Polygons

+Scan Conversion Problem

To represent a perfect image as a bitmapped image.

+Line Drawing Algorithms Lines are used a lot - want to get them right. Lines should appear straight, not jagged. Horizontal, vertical and diagonal easy, others difficult Lines should terminate accurately. Lines should have constant density. Line density should be independent of line length or

angle. Lines should be drawn rapidly. Efficient algorithms.

+DDA: Digital Differential Analyzer

y i1 mx i1 Bm(x i x) Bmx i mx B(mx i B) mxy i mx

y i1 y i m x 1

(xi,yi)

(xi,Round(yi))

(xi,Round(yi+m))

(xi,yi +m)

Desired Line

Line: Left to Right:1- Slope m>0: sample at unit x intervals ( Δx= 1), calculate each succeeding y value as

x i1 x i 1m

2- Slope m<0: sample at unit y

intervals ( Δy= 1), calculate each

succeeding x value as

Line: from Right to Left3- Slope m> 0:

y i1 y i m x 1

4- Slope m< 0:

x i1 x i 1m

Faster than brute force.

Based on Calculating either ∆x or ∆y.

Mathematically well defined

Floating point Round off error. Time consuming

arithmetic

Advantages Disadvantages

+Bresenhams Line Algorithm Accurate Efficient Integer Calculations Uses Symmetry for other lines Adapted to display circles, ellipses and curves It has been proven that the algorithm gives an

optimal fit for lines

+Bresenhams Line Algorithm

y m(x i 1) bd1 y ym(xk1) b yk

d2 (yk1) yyk1 m(xk1) bd1 d2 2m(xk1) 2y 2b 1 (3 11)

decision parameter, use m yx

pk x(d1 d2)2yxk 2xy c (3 12)

+Bresenhams Line Algorithm

The sign of pk is the same as the sign of d1 – d2,

since Δx> 0 for our example. Parameter c is independent and will be eliminated in the recursive calculations for pk.

If the pixel at yk is closer to the line path than the pixel at yk+l (that is, d1 < d2), then decision parameter pk is negative. In that case, we plot the lower pixel; otherwise, we plot the upper pixel.

pk1 2yxk1 2xyk c

+ Bresenhams Line Algorithm

This recursive calculation of decision parameters is performed at each integer x position, starting at the left coordinate endpoint of the line. The first parameter, po is evaluated from Eq. 3-12 at the starting pixel position (xo, yo) and with m evaluated as Δy/Δx:

pk1 pk 2y(xk1 xk ) 2x(yk1 yk )but xk1 xk 1pk1 pk 2y 2x(yk1 yk )

p0 2y 2x

yk1 yk 0or1

+Bresenhams Line Drawing Algorithm1. Input the two line endpoints, store the left endpoint

(x0,y0).

2. Plot the first point (x0,y0).

3. Calculate constants ∆x, ∆y, and 2∆y - 2∆x and 2∆y, get starting values for decision parameter pk, p0=2∆y-∆x

4. At each xk along the line, starting at k = 0, do the following test: if pk < 0, the next point to plot is(xk+1, yk)

pk+1 = pk + 2∆yelse, the next point to plot is(xk+1, yk+1) pk+1=pk +2∆y-2∆x

5. Repeat step 4. ∆x times.

+Bresenhams Line Algorithm14

+Midpoint Line Algorithm

If (BlueLine < Midpoint) Plot_East_Pixel();

Else Plot_Northeast_Pixel();

Find an equation, given a line and a point, that will tell us if the point is above or below that line?

Midpoint Line Algorithm

xy yx Bxyx xy BxF(x,y) ax by c 0F(x,y) yx xy Bxnow,d F(M) M midpoint

F(x p 1,y p 12)

y(x p 1) x(y p 12) Bx

y d y d

If F(x,y) ==0 (x,y) on the line <0 for points below the

line >0 for points above the

line d=F(M)

+Midpoint Line Algorithm P=(xp, yp) is pixel chosen by the algorithm in previous step

To calculate d incrementally we require dnew

If d > 0 then choose NE

d y(x p 1) x(y p 12) Bx

dnew F(x p 2,y p 32)

y(x p 2) x(y p 32) Bx

dnew d y xNE

1 2 4 3 4 dnew d NE

NE y x

P=(xp, yp)

xp+1xp xp+2

),1( 21 yxp

),2( 23 yxp

+Midpoint Line Algorithm If d < 0 then choose E

d y(x p 1) x(y p 12) Bx

dnew F(x p 2, y p 12)

y(x p 2) x(y p 12) Bx

dnew d yE{

dnew d EE y

P=(xp, yp)

xp+1xp xp+2

),1( 21 pp yx

),2( 21 pp yx

+Midpoint Line Algorithm To find Initial value of d

d0 F(x0 1,y0 12)

y(x p 1) x(y p 12) Bx

yx p xy p Bx y 12x

F(x0, y0) y 12x

d0 y 12x

[as (x0, y0) is on the line]

P=(x0, y0)

x0+1x0

),1( 21

Only fractional value

d0 2y xNE 2(y x)E 2y

Multiply by 2 to avoid fractions. Redefine d0, E, NE

+Midpoint Line Algorithm

Midpoint: Looks at which side of the line the mid point falls on.

Bresenham: Looks at sign of scaled difference in errors.

It has been proven that Midpoint is equivalent to Bresenhams for lines.

+void MidpointLine(int x0, int y0, int x1, int y1, int color){

int dx = x1 – x0, dy = y1 – y0;

int d = 2*dy – dx;

int dE = 2*dy, dNE = 2*(dy – dx);

int x = x0, y = y0;

WritePixels(x, y, color);

while (x < x1) {

if (d <= 0) { // Current d

d += dE; // Next d at E

} else {

d += dNE; // Next d at NE

y++} Write8Pixels(x, y, color);}}

+Midpoint Circle Algorithm Implicit of equation of circle is: x2 + y2 - R2

= 0, at origin

Eight way symmetry require to calculate one octant

For each pixel (x,y), there are 8 symmetric pixels

In each iteration only calculate one pixel, but plot 8 pixels

23+Midpoint Circle Algorithm Define decision variable d as:

SE Choose Circle outside s

E Choose Circle inside s

yxFMFd

),1()(22

P=(xp, yp)

xp+1xp xp+2

),1( 21 pp yx ),2( 2

1 pp yx

yp – 1

yp – 2

SE MSE

E choose weeither Choose

(x p + 2,y p − 32)

24+Midpoint Circle Algorithm If d <= 0 then midpoint m is inside circle

we choose E Increment x y remains unchanged

22212 P=(xp, yp)

xp+1xp xp+2

),1( 21 pp yx ),2( 2

1 pp yx

yp – 1

yp – 2

25+Midpoint Circle Algorithm If d > 0 then midpoint m is outside circle

we choose E Increment x Decrement y

d = x p +1( )2

+ y p − 12( )

2− R2

dnew = F(x p + 2, y p − 32)

= x p + 2( )2

+ y p − 32( )

2− R2

dnew − d = 2x p − 2y p + 5ΔSE

1 2 4 4 3 4 4

dnew = d + ΔSE

P=(xp, yp)

xp xp+2

),1( 21 pp yx

(x p + 2,y p − 32)

yp – 1

yp – 2

26+Midpoint Circle AlgorithmInitial condition Starting pixel (0, R) Next Midpoint lies at (1, R – ½) d0 = F(1, R – ½) = 1 + (R2 – R + ¼) – R2 = 5/4 – R To remove the fractional value 5/4 :

Consider a new decision variable h as, h = d – ¼ Substituting d for h + ¼,

d0=5/4 – R h = 1 – R d < 0 h < – ¼ h < 0 Since h starts out with an integer value and is incremented by

integer value (E or SE), e can change the comparison to just h < 0

27+Midpoint Circle Algorithmvoid MidpointCircle(int radius, int value) {

int x = 0;int y = radius ;int d = 1 – radius ;CirclePoints(x, y, value);while (y > x) {

if (d < 0) { /* Select E */d += 2 * x + 3;

} else { /* Select SE */

d += 2 * ( x – y ) + 5;y – –;

}x++;CirclePoints(x, y, value);

28+Midpoint Circle Algorithm

Void CirclePoints(int x, int y, float value){

SetPixel(x,y);SetPixel(x,-y); SetPixel(-x,y); SetPixel(-x,-y); SetPixel(y,x);

SetPixel(y,-x); SetPixel(-y,x);

SetPixel(-y,-x);}

29+ Midpoint Circle Algorithm Second-order differences can be used to enhance

performance.

ppp yxSE

52)1(23)1(2

SESEEE

yxSExE

pnewpp

5)1(2)1(23)1(2

SESEEE

yxSExE

pnewpp

E is chosen

SE is chosen 523

:value Initial

30+Midpoint Circle Algorithmvoid MidpointCircle(int radius, int value) {

int x = 0;int y = radius ;int d = 1 – radius ;int dE = 3;int dSE = -2*radius +5;CirclePoints(x, y, value);while (y > x) {

if (d < 0) { /* Select E */d += dE;dE += 2;dSE += 2;

} else { /* Select SE */d += dSE;dE += 2;dSE += 4;y – –;}

x++;CirclePoints(x, y, value);}

31+Midpoint Ellipse Algorithm Implicit equation is: F(x,y) = b2x2 + a2y2 – a2b2 = 0 We have only 4-way symmetry There exists two regions

In Region 1 dx > dy Increase x at each step y may decrease

In Region 2 dx < dy Decrease y at each step x may increase

(x1,y1)(-x1,y1)

(x1,-y1)(-x1,-y1)

(-x2,y2)

(-x2,-y2)

(x2,y2)

(x2,-y2)

32+Midpoint Ellipse Algorithm

Region 1

Region 2S SE

Gradient Vector

TangentSlope = -1

dxdyaybx

22 022

1 1 Region In

33+Midpoint Ellipse Algorithm In region 1

yaxbdd

bayaxb

SEto move then 0 d if

bayaxb

E to move then 0 d if

bayaxb

)22()32(

222322

222122

P=(xp, yp)

s Curre

),1( 21 pp yx ),2( 2

1 pp yx

yp – 1

yp – 2

),2( 23 pp yx

34+Midpoint Ellipse Algorithm In region 2

yaxbdd

bayaxb

SEto move then 0 d if

bayaxb

Sto move then 0 d if

bayaxb

)32()22(

222232

222212

P=(xp, yp)

s Curre

nt Nex

)1,( 21 pp yx

)2,( 21 pp yx

yp – 1

yp – 2

)2,( 23 pp yx

35+Midpoint Ellipse AlgorithmDPx=2*ry*ry; Dpy =2*rx*rx; x=0; y=ry; Px=0; Py =2*rx*rx*ry; f =ry*ry +rx*rx(0.25-ry ); ry2=ry *ry; Set4Pixel(x,y); while (px<py ) //Region I{ x=x+1; Px=Px+DPx; if (f>0) // Bottom case{y=y -1; Py =Py -Dpy ; f=f+ry2+Px-Py;} else // Top casef=f+ry2+Px; Set4Pixel(x,y);

CPCS 391 Computer Graphics 1

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