Chapter 3 Scan Conversion Lecture-02

1. Chapter 3Scan ConversionLecture-02

2. 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.

3. 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. p2 p1

4. Bresenham’s Line Algorithm • 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)

5. 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 . p2 p1

6. Bresenham’s Line Algorithm • 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

7. Bresenham’s Line Algorithm • 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); • } • 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++;

8. Scan-Converting a Circle

9. P=(xp, yp) E yp ME M yp – 1 SE MSE yp – 2 xp xp+1 xp+2 Next Current Previous 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:

10. P=(xp, yp) E yp ME d < 0 M yp – 1 yp – 2 xp xp+1 xp+2 Next Current Previous Midpoint Circle Algorithm • If d <= 0 then midpoint m is inside circle • we choose E • Increment x • y remains unchanged

11. P=(xp, yp) yp M d > 0 yp – 1 SE MSE yp – 2 xp xp+1 xp+2 Next Current Midpoint Circle Algorithm • If d > 0 then midpoint m is outside circle • we choose S • Increment x • Decrement y Previous

12. 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

13. Midpoint Circle Algorithm • 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 : Else Line 8 : { Line 9 : P=p+2(x-y)+5; Line 10 : y--; Line 11 : } Line 12 : x++; Line 1 3: }