Midpoint circle algo

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Midpoint circle algo

  1. 1. Midpoint Circle AlgorithmMid point algorithm is very similar toBresenham’s approach. It is based on thefollowing function for testing the spatialrelationship between an arbitrary point(x,y)and a circle of radius centered at the origin: < 0 (x,y) inside circlef(x,y) = x2 + y2 – r2 = 0 (x,y) on the circle > 0 (x,y) outside circle`
  2. 2. Now consider the coordinates of the pointhalfway between pixel T & S (xi + 1, yi – ½)This is called the midpoint and we use it todefine a decision parameter: pi = f( xi +1, yi – ½) = (xi + 1)2 + (yi – ½)2 – r2If pi is –ve , the midpoint is inside the circle,then we choose pixel T. If pi is +ve, themidpoint is outside the circle, & we choose S.
  3. 3. Similarly, the decision parameter for the nextstep is:pi+1 = (xi+1 + 1)2 + (yi+1 – ½)2 – r2Since x i+1 = xi + 1 pi+1 - pi = [(xi + 1)+ 1]2 – (xi + 1)2 + (y i+1 – ½)2 – (yi – ½)2Hence pi+1 = pi + 2(xi + 1) + 1 + (y i+12 – yi2) – (y i+1 – yi)
  4. 4. If pixel T is chosen (meaning pi < 0), we have yi+1 = yi.If pixel S is chosen (meaning pi > 0), we have yi+1 = yi – 1. Thus, pi+1= pi + 2(xi + 1) +1 if pi<0 pi + 2(xi + 1) +1 – 2(yi – 1) if pi>0In terms of (xi, yi), we have pi+1= pi + 2xi + 3 if pi<0 pi + 2(xi - yi ) + 5 if pi>0
  5. 5. Finally, we compute the initial value for thedecision parameter using the originaldefinition of pi and (0,r): pi = (0 + 1)2 + (r – ½)2 – r2 = 5/4 – rOne can see that this is not really integercomputation. However, when r is an integerwe can simply set p1= 1 – r. The error ofbeing ¼ less than the precise value does notprevent p1 from getting the appropriate sign.
  6. 6. It does not affect the rest of the scanconversion process, because the decisionvariable is only updated with integer incrementin subsequent steps.
  7. 7. Algorithm:The following is a description of this midpointcircle algorithm that generates the pixelcoordinates in the 90˚ to 45 ˚ octant: int x = 0, y = r, p = 1-r while (x <=y) { setpixel(x,y); if (p < 0) p = p + 2x + 3
  8. 8. else { p = p + 2(x – y) + 5 y- - } x++ }

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