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Lines and curves algorithms
Lines and curves algorithms
Lines and curves algorithms
Lines and curves algorithms
Lines and curves algorithms
Lines and curves algorithms
Lines and curves algorithms
Lines and curves algorithms
Lines and curves algorithms
Lines and curves algorithms
Lines and curves algorithms
Lines and curves algorithms
Lines and curves algorithms
Lines and curves algorithms
Lines and curves algorithms
Lines and curves algorithms
Lines and curves algorithms
Lines and curves algorithms
Lines and curves algorithms
Lines and curves algorithms
Lines and curves algorithms
Lines and curves algorithms
Lines and curves algorithms
Lines and curves algorithms
Lines and curves algorithms
Lines and curves algorithms
Lines and curves algorithms
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Lines and curves algorithms

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  • 1. WHERE TO DRAW A LINE?? Line drawing is accomplished by calculating intermediate positions along the line path between specified end points. Precise definition of line drawing Given two points P and Q in the plane, both with integer coordinates, determine which pixels on a raster screen should be on in order to make a picture of a unit-width line segment starting from P and ending at Q.
  • 2. 6 (3,5 3)43210 0 1 2 3 4 5 6
  • 3. Line drawing (contd.) The thinnest line is of one-pixel wide. We willconcentrate on drawing a line of 1 pixel resolution.The Cartesian slope-intercept equation for a straightline is y= m. ∆x + bm is the slope of the line and b is the y intercept.Given the endpoints of a line segment. m = (y2-y1) / (x2-x1) b= y1-m.x1
  • 4. Line Drawing (cont) Also for any given interval ∆x along a line, we can compute the corresponding y interval ∆y from ∆y= m. xSimilarly we can obtain the x interval ∆x corresponding to a specified ∆y as ∆x= ∆y / mThese equations form the basis for determining deflection voltages in analog devices.
  • 5. Line Drawing (cont) On Raster systems, lines are plotted with pixels, and step sizes in the horizontal and vertical directions are constrained by pixel separations. Hence we ought to “sample” a line at discrete positions and determine the nearest pixel to the line at each sampled position.
  • 6. Bresenham’s Line AlgorithmAn accurate, efficient raster line drawing algorithmdeveloped by Bresenham, scan converts lines usingonly incremental integer calculations that can beadapted to display circles and other curves.Keeping in mind the symmetry property of lines, letsderive a more efficient way of drawing a line. Starting from the left end point (x0,y0) of a given line , we step to each successive column (x position) and plot the pixel whose scan-line y value closest to the line path Assuming we have determined that the pixel at (xk,yk) is to be displayed, we next need to decide which pixel to plot in column xk+1.
  • 7. Bresenham Line Algorithm (cont)Choices are(xk +1, yk) and (xk+1, yK+1) d1 = y – yk = m(xk + 1) + b – yk d2 = (yk + 1) – y = yk + 1- m(xk + 1) – bThe difference between these 2 separations is d1-d2 = 2m(xk + 1) – 2 yK + 2b – 1 A decision parameter pk for the kth step in the line algorithmcan be obtained by rearranging above equation so that itinvolves only integer calculations
  • 8. Bresenham’s Line AlgorithmDefine Pk = Δx ( d1-d2) = 2Δyxk-2 Δxyk + cThe sign of Pk is the same as the sign of d1-d2, since Δx > 0. Parameter c is a constant and has the value 2Δy + Δx(2b-1) (independent of pixel position)If pixel at yk is closer to line-path than pixel at yk +1 (i.e, if d1 < d2) then pk is negative. We plot lower pixel in such acase. Otherwise , upper pixel will be plotted.
  • 9. Bresenham’s algorithm (cont)At step k + 1, the decision parameter can be evaluated as, pk+1 = 2Δyxk+1- 2Δxyk+1+ cTaking the difference of pk+ 1 and pk we get the following. pk+1 – pk = 2Δy(xk+1- xk)-2Δx(yk+1 – yk)But, xk+1 = xk +1, so that pk+1 = pk + 2Δy - 2 Δx(yk+1 – yk)Where the term yk+1-yk is either 0 or 1, depending on the sign ofparameter pk
  • 10. Bresenham’s Line AlgorithmThe first parameter p0 is directly computed p0 = 2 Δyx0 - 2 Δxy0 + c = 2 Δyx0 + 2 Δxy0 +2 Δy + Δx (2b-1)Since (x0,y0) satisfies the line equation , we also have y0 = Δy/ Δx * x0 + bCombining the above 2 equations , we will have p0 = 2Δy – Δx The constants 2Δy and 2Δy-2Δx are calculated once for eachtime to be scan converted
  • 11. Bresenham’s Line Algorithm So, the arithmetic involves only integer addition and subtraction of 2 constantsSTEP 1 Input the two end points and store the left end point in (x0,y0)STEP 2 Load (x0,y0) into the frame buffer (plot the first point)STEP 3 Calculate the constants Δx, Δy, 2Δy and 2Δy-2Δx and obtain the starting value of the decision parameter as p0 = 2Δy- Δx
  • 12. Bresenham’s Line AlgorithmSTEP 4 At each xk along the line, starting at k=0,perform the following test:If pk < 0 , the next point is (xk+1, yk) and pk+1 = pk + 2ΔyOtherwisePoint to plot is (xk+1, yk+1) pk+1 = pk + 2Δy - 2ΔxRepeat step 4 (above step) Δx times
  • 13. Loading the frame buffer•Calculate frame-buffer addresses by incrementalmethods•For a bilevel system(1 bit per pixel), address forpixel(x,y) Addr(x,y) = addr(0,0)+y(xmax +1)+x Addr(x+1,y) = addr(x,y)+1 Addr(x+1,y+1) = addr(x,y)+ x max +2
  • 14. Where do we draw a circle??? Properties of a circle:A circle is defined as a set of points that are all the given distance (x ,y ). c cThis distance relationship is expressed by the pythagorean theorem inCartesian coordinates as (x – xc)2 + (y – yc) 2 = r2We could use this equation to calculate the points on the circlecircumference by stepping along x-axis in unit steps from xc-r to xc+r andcalculate the corresponding y values at each position as y = yc +(- ) (r2 – (xc –x )2)1/2This is not the best method: Considerable amount of computation Spacing between plotted pixels is not uniform
  • 15. Polar co-ordinates for a circleWe could use polar coordinates r and θ, x = xc + r cosθ y = yc + r sinθ A fixed angular step size can be used to plot equally spaced points along the circumferenceA step size of 1/r can be used to set pixel positions to approximately 1 unit apart for a continuous boundaryBut, note that circle sections in adjacent octants within one quadrant are symmetric with respect to the 45 deg line dividing the two octantsThus we can generate all pixel positions around a circle by calculating just the points within the sector from x=0 to x=yThis method is still computationally expensive
  • 16. Bresenham to Midpoint Bresenham requires explicit equation  Not always convenient (many equations are implicit)  Based on implicit equations: Midpoint Algorithm (circle, ellipse, etc.)  Implicit equations have the form F(x,y)=0.
  • 17. Midpoint Circle AlgorithmWe will first calculate pixel positions for a circle centered around theorigin (0,0). Then, each calculated position (x,y) is moved to its properscreen position by adding xc to x and yc to yNote that along the circle section from x=0 to x=y in the first octant,the slope of the curve varies from 0 to -1Circle function around the origin is given by fcircle(x,y) = x2 + y2 – r2Any point (x,y) on the boundary of the circle satisfies the equationand circle function is zero
  • 18. Midpoint Circle Algorithm For a point in the interior of the circle, the circle function is negative and for a point outside the circle, the function is positive Thus,  f (x,y) < 0 if (x,y) is inside the circle boundary circle  fcircle(x,y) = 0 if (x,y) is on the circle boundary  fcircle(x,y) > 0 if (x,y) is outside the circle boundary
  • 19. Midpoint Circle AlgorithmAssuming we have just plotted the pixel at (xk,yk) , we next need todetermine whether the pixel at position (xk + 1, yk-1) is closer to the circleOur decision parameter is the circle function evaluated at the midpointbetween these two pixels pk = fcircle (xk +1, yk-1/2) = (xk +1)2 + (yk -1/2)2 – r2 If pk < 0 , this midpoint is inside the circle and the pixel on the scan line ykis closer to the circle boundary. Otherwise, the mid position is outside or on the circle boundary, and we select the pixelon the scan line yk-1
  • 20. Midpoint Circle AlgorithmSuccessive decision parameters are obtained using incrementalcalculations Pk+1 = fcircle(xk+1+1, yk+1-1/2) = [(xk+1)+1]2 + (yk+1 -1/2)2 –r2OR Pk+1 = Pk+2(xK+1) + (yK+12 – yk2) – (yK+1- yk)+1Where yk+1 is either yk or yk-1, depending on the sign of pkIncrements for obtaining Pk+1: 2xk+1+1 if pk is negative 2xk+1+1-2yk+1 otherwise
  • 21. Midpoint circle algorithmNote that following can also be done incrementally: 2xk+1 = 2xk +2 2 yk+1 = 2yk – 2At the start position (0,r) , these two terms have the values 2 and 2r-2respectivelyInitial decision parameter is obtained by evaluating the circle function atthe start position (x0,y0) = (0,r) p0 = fcircle(1, r-1/2) = 1+ (r-1/2)2-r2OR P0 = 5/4 -rIf radius r is specified as an integer, we can round p0 to p0 = 1-r
  • 22. The actual algorithm 1: Input radius r and circle center (xc,yc) and obtain the first point on the circumference of the circle centered on the origin as (x0,y0) = (0,r) 2: Calculate the initial value of the decision parameter as P0 = 5/4 - r 3: At each xk position starting at k = 0 , perform the followingtest: If pk < 0 , the next point along the circle centered on (0,0) is(xk+1, yk) and pk+1 = pk + 2xk+1 + 1
  • 23. The algorithmOtherwise the next point along the circle is (xk+1, yk-1) and pk+1 = pk + 2xk+1 +1 -2yk+1Where 2xk+1 = 2xk+2 and 2yk+1 = 2yk-24: Determine symmetry points in the other seven octants5: Move each calculated pixel position (x,y) onto thecircular path centered on (x,yc) and plot the coordinatevalues x = x+ xc , y= y+ yc 6: Repeat steps 3 through 5 until x >= y

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