PageDate/Bresenham's Line Drawing AlgorithmThis is an accurate and efficient rasterIt scan converts line using incrementalline generating algorithminteger calculation that can be adaptedto display circles and other curve.In this algorithm we have to decide whichpixel position is closest to the scan line.By using a decision parameter we candetermine the losest one to the scan lineyktzy K4TfigYKxk n xK+1 K+2when dieda,then Picof negativeykthOR d2>d1d2yfig.2diYKnknxk+1 Camlin PageDateThe Figure 1 shows a positive slope that isless than 1.Here (xk yk is the starting position Nextpixel positionsarexandYk+1In Figure 2, y is the scan lineie y = mn+bdi and d2 are the vertical pixel separationfrom the scan line y.From figure 2 , theequationoftheslopewillbeymexdi =y-yk =di = m / n +1) +b'-Ykd2=lyk +1) yd2 = yk+1 - [m (nk +1) + b]d2 = YK+ mv(x) +1) Tbdi - d2 = m (n K +1) tb - Yk - ykt1-m(x+1)-b)m/xi 1)+b-y y, K - 1+m(xx+1) + bdi--dz=2m(nK1)-Qyk+2b-1Substitute m = DyAx =2 Dy (n+1) K - 2yk 2b -Dn= An (2Dy (nk +1)] - 20 xyx - +2Dnb - AnDn= 20y (nk 11)- - 2Dnyk + 2Dnb - An=2Dyx k + 2Dy -2Dxyk +2 Anb. - An= 2Dynk + 2 Dy -2D ny +(2b-1) AnPk 20yx K - 2Dxy 20y + (2b-1) DnP Kfl = 2Dy n - 2Dny +2 Dy + (2b - 1) 02k+1K+1An(di-d2) is called as the decision parameterPK.Here An will always be greater than o,because it is a positive slopeyie The sign will of Pk will depend on sign of (di-diie . Pk always depend on sign (di-diPk is negative when t d2 > di) of or (diddsSo di 0+weie. YK+1 yk+1then PR71 PK + 2 Dy - 2D ly+1-yr) KPktiPk +2 Dy -whenPKweplotthelowerpixelYk40,thenPk+1 PK + 2Dy - 2An lyk -yk)Pk+1 Pk + 2DyInitialDecisionParameterPo 2Dy - AxBresenham'sLinedrawingalgorithm[Im/cl]IInputthe2lineendpointsandstoretheleft endpoint in (no,yo)2plot the first point.Load (no, yo) into the frame buffer / ie,3CalculateAnandobtainthetstartingDy2Dy, 20y - 20xvalueforthe Camlin PageDate / /decision parameter as Po = 20y - on4At each n along the line startingperform the following test(a) If PK 20, the next point to plot is(XR+1) yk) and Pkti = Pk + 2Dyb)therwise, the next point is (xkti )when PR 30and PR+1 = Pk + 20y - 2 An5Repeat step 4 On times or till the reachingof last endpoint.Exampleeg:-Drawa line using Bresenham's algorithmgiven the starting and ending point as (20,10)and (30,18) The line has a slope of 0.8(xo, yo) = (20590)slope = 0.8 M <1Dn = x2 - NIS= 30-20102 Dy -IDn= 16 - 20By = Y2-y1- -418-10$ 8initial decision parameter , PoPo z 2 Dy - An2 x8-10Po 16-10 =6 K.O , po - PKpk = 6Pk>oP1 = 2 Dy - 2 Dn64-42(P1,70)KPk(nkth ykti"I506(21,11)1 2(22,12)2-2(23, 12)314(24,13)4 10(25,14)5 6(26,15)6 2(27, 16)7-2(28,16)8 14(29,17)9 10(30,18) P2 Pk + 20y - Risix= 2+-A-2CP2LO)P3 = Pk + 2Dy-2-4-16p14 (P370)-PqPk 20y- 25x14+-4610 (P470)Ps = PK +2 A y - 202= 10 +16 20 = 10-4 -=6 ( Ps 70)=P6 = Pk 4 2 A y - - 2An=6+-42(P6>0)=P7 = Pk+2Dy - 20x22-4= -2('P7<0)P8 PK + 2 Dy-2 +16+= 14(P8 >0)Pq = Pk+2Dy- 2 AXAS=14 + - 4= 10( Pa >0)-2. eg:- (9,18) (14, 22)no,yo)(9,18)An = N2-X1Dy = y2-y1= 14-922-18=542Dy = 2x4= 82Dn= 2x5 =10=aPo=20y -An20x-2Ay = +8-10= =2= S-53 (Po20)ikPk(nkm)Pr = Po + 2Dy-20o03(9,18)3 + 8 - 10I1(10,19)3 - 3 - 221-1(11,19)P121 (PI>O)37(12,20)45(13,21)P2 = Pit2Dy-2Dn53(14,22)=1+8-10 NAS - NAS+= 1t-2P2 =-1 (P2<0)P3 P2+2DyP5:P4+2Dy-24n-1-827 (P3>0)1-5-+-2=$ + 3-P4 =27+8-10P5 70)=7-2 Cac or=.5 (P4>0) DateAdvantages of Bresenham1It is easy to implement2It is fast and incremental3'It executes fast but less faster than DDAalgorithm4The points generated by this algorithm aremore accurate than DDA algorithm5.It uses fixed points only.Disadvantages of Bresenham1'The result line is not smooth2.This algorithm is for the basic line drawing3 It cannot handle diminishing jaggies.Circle Drawing Algorithmcircle is defined as the set of points that areall at a given distance 'r' from a centreposition ( N.C, yc)The general equation of a circle is:-n2 ty = rYcrXcX from figure, the distance relationship isexpressed as(x - x1)2 (+(y-yc) = 92To calculate the position of points on acircle circumference by stepping alongthe x -axis - in unit steps from mc r toxctr and the corresponding Y values ateach position as y = y t J 92 - (nc-x)Problem with this approach is that it involvesconsiderable computation at each step.Also the spacing between plotted pixel positionis not uniformTo eliminate the unequal spacing, calculatepolar coordinates r and O.points along the circular boundary usingExpressing the circle equation in polar formx = xc + rcasoy= yc + r sin Oby using equally spaced points along theBy using with these equations, a circle is plotteddepends circumference The step size chosen for odevice on the application and display

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