Chapter 3Scan Conversion

Lecture-02

Bresenham’s Line Algorithm

• Bresenham’s line algorithm– is a highly efficient incremental method for scan-

converting lines.– It produces mathematically accurate results

using only integer addition, subtraction and multiplication by 2, which can be accomplished by a simple arithmetic shift operation.

Bresenham’s Line Algorithm• Scan convert the line in the Figure, • 0<m<1. • Start with pixel P1(x1,y1).

• Choose either the pixel on the right or the pixel right and up.

• The coordinates of the last chosen pixel upon entering step i are (xi,yi).

• Choose the next between the bottom pixel S and the top pixel T.

p1

p2

• The difference is –s – t = (y-yi)-[(yi+1)-y]

= y-yi-yi-1+y

= 2y-2yi-1

= 2(y-yi)-1

= 2 [ { m(xi + 1) + b } - yi) ] - 1

= 2 m(xi + 1) + 2b - 2yi – 1

∆x( s – t)=∆x [2m(xi + 1) + 2b - 2yi – 1]

So di = ∆x [2m(xi + 1) + 2b - 2yi – 1] ……………. (i)

• If the chosen pixel is the top pixel T (di ≥ 0) then xi+1 = xi+1 and yi+1 = yi + 1 and so

di+1 = di + 2 (Δy - Δx)

Bresenham’s Line Algorithm

Bresenham’s Line Algorithm• If the chosen pixel is pixel S (di < 0) then

xi+1= xi+1 and yi+1 = yi and so • di+1 = di + 2Δy • where di = 2Δy * xi - 2Δx * yi +C • and C = 2Δy + Δx (2b – 1) • We use here a decision variable di . For

the value of each di we calculate the corresponding value of di+1 .

p1

p2

• Hence, we have

• Finally, we calculate d1 ,the base case value for this recursive formula, from the original definition of the decision variable di

• di = ∆x [ 2m (x1 + 1) + 2b - 2y1 – 1]

= ∆x [ 2( mx1 + b - y1 ) + 2m – 1]

= ∆x [ 2*0 + 2m – 1] = 2(∆y / ∆x)* ∆x - ∆x

=2∆y - ∆x

Bresenham’s Line Algorithm

Bresenham’s Line Algorithm• Void Bresenham( ) • { • Line 1: dx=x2-x1; • Line 2: dy=y2-y1; • Line 3: dT=2*(dy-dx); • Line 4: dS=2*dy; • Line 5: d=(2*dy)-dx; • Line 6: putpixel(x1,y1); • Line 7: while(x1<x2) • Line 8: { • Line 9: x1++;

• Line 10: if(d<0) • Line 11: { • Line 12: d=d+dS; • Line 13: putpixel(x1,y1); • Line 14: } • Line 15: else • Line 16: { • Line 17: y1++; • Line 18: d=d+dT; • Line 19: putpixel(x1,y1); • Line 20: } • Line 21: } • Line 22: putpixel(x2,y2); • }

Scan-Converting a Circle

Midpoint Circle Algorithm• Implicit of equation of circle is:

x2 + y2 - R2 = 0

• Eight way symmetry require to calculate one octant

• Define decision variable d as:

SE Choose

Circle outside s

E Choose

Circle inside s

iM

d

iM

d

Ryx

yxFMFd

pp

pp

0

0

1

),1()(

22

212

21

P=(xp, yp)

ME

xp+1xp xp+2

Pre

vio

us

Curr

ent

Next

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

1 pp yx

ME

yp

yp – 1

yp – 2

SE

MSE

E choose we

either Choose

0d

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

– we choose E– Increment x– y remains unchanged

Edd

xE

xdd

Ryx

yxFd

Ryxd

new

p

pnew

pp

ppnew

pp

32

32

2

),2(

1

22

212

21

22

212

P=(xp, yp)

ME

xp+1xp xp+2

Pre

vio

us

Curr

ent

Next

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

1 pp yx

ME

yp

yp – 1

yp – 2

d < 0

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

– we choose S– Increment x– Decrement y

SEdd

yxSE

yxdd

Ryx

yxFd

Ryxd

new

pp

ppnew

pp

ppnew

pp

522

522

2

),2(

1

22

232

23

22

212

Pre

vio

us

P=(xp, yp)

MSE

xp+1xp xp+2

Curr

ent

Next

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

3 pp yx

MSE

yp

yp – 1 yp – 2

d > 0

Midpoint Circle Algorithm

Initial 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 :

• d0= 1– R

• We can now summarize the algorithm for generating all the pixel coordinates in the 900 to 450 octant :

Line 1 : Int x=0,y=r,p=1-r;Line 2 : While(x<=y)Line 3 : {Line 4 : setPixel(x,y);Line 5 : If(p<0)Line 6 : p=p+2x+3;Line 7 : ElseLine 8 : {Line 9 : P=p+2(x-y)+5;Line 10 : y--;Line 11 : }Line 12 : x++;Line 1 3: }

Midpoint Circle Algorithm