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Mid Point Circle Drawing Derivation

Mid Point Circle Drawing Derivation (Algorithm)
The mid point circle algorithm is used to determine the pixels needed for rasterizing a circle while drawing a circle on a pixel screen.
In this technique algorithm determines the mid point between the next 2 possible consecutive pixels and then checks whether the mid point in inside or outside the circle and illuminates the pixel accordingly.



This is how a pixel screen is represented:
pixel representation on computer screen
A circle is highly symmetrical and can be divided into 8 Octets on graph. Lets take center of circle at Origin i.e (0,0) :
Mid Point Circle Drawing Derivation Algorithm




We need only to conclude the pixels of any one of the octet rest we can conclude because of symmetrical properties of circle.


Let us Take Quadrant 1: 
quadrant of mid point circle algorithm

Radius = OR = r
Radius = x intercept = y intercept
At point R
     coordinate of x = coordinate of y   or we can say  x=y
let us take Octet 2 of quadrant 1
here first pixel would be (0,y)
            here value of y intercept = radius (r)
 as circle’s center is at origin

         DERIVATION
Derivation of mid point circle drawing algorithm
let us assume we have plotted Pixel P whose coordinates are  
Now we need to determine the next pixel. 

We have chosen octet 2 where circle is moving forward and downwards so y can never be increased, either it can be same or decremented. Similarly x will always be increasing as circle is moving forward too.

So y is needed to be decided.
Now we need to decide whether we should go with point N or S.

For that decision Mid Point circle drawing technique will us decide our next pixel whether it will be N or S.

As   is the next most pixel of   therefore we can write,

                    

And similarly   in this case.

Let M is the midpoint between   and 
And coordinates of point (M) are
                        Mid point Circle formula

                                  Formula of mid point circle


                                 final formula of mid point circle algorithm

Equation of Circle with Radius r

(x– h)2 + (y – k)2 = r2

When coordinates of centre are at Origin i.e., (h=0, k=0)

     x2 + y2 = r2    (Pythagoras theorem)
Function of Circle Equation

F(C)  x2 + y2 - r2
Function of Midpoint M (xk+1 ,  yk -1/2) in circle equation

F(M)= xk+12 + (yk -1/2)2 - r
          = (xk+1)2 + (yk -1/2)2 - r

The above equation is too our decision parameter pk

Pk = (xk+1)2 + (yk -1/2)2 - r2           …….(i)

To find out the next decition parameter we need to get Pk+1

Pk+1 = (xk+1+1)2 + (yk+1 -1/2)2 - r2

Now,
Pk+1- Pk =  (xk+1+1)2 + (yk+1 -1/2)2 - r2
                          -[(xk+1)2 + (yk -1/2)2 - r2]

                     =  ((xk+1)+1)2 + (yk+1 -1/2)2
                            - (xk+1)2   - (yk -1/2)2 

                 = (xk+1)2 + 1 +2(xk+1) + yk+12 +(1/4) - yk+1
                      - (xk+1)2   - yk2 – (1/4) + yk+1

                = 2(xk+1) + yk+12 - yk2 - yk+1 + yk +1

                =  2(xk+1) +( yk+12 - yk2 ) - (yk+1 - yk) +1


Pk+1 = Pk + 2(xk+1)+( yk+12 - yk2 ) - (yk+1 - yk) +1     …..(ii)


Now let us conclude the initial decision parameter
For that we have to choose coordinates of starting point i.e. (0,r)
Put this in (i) i.e. Pk
Pk = (xk+1)2 + (yk -1/2)2 - r2
P0 = (0+1)2 + (r -1/2)2 - r2
        = 1 + r2 + ¼ - r – r2
        = 1 + ¼ - r

               …..(initial decision parameter)

Now If Pk ≥ 0 that means midpoint is outside the circle and S is closest pixel so we will choose S (xk+1,yk-1)
That means yk+1 = yk-1
Putting coordinates of S in (ii) then,

Pk+1 = Pk + 2(xk+1)+( yk-12 - yk2 ) - (yk-1 - yk) +1
        = Pk + 2(xk+1)+( yk -1)2 - yk2 ) - ((yk-1) - yk) +1
        = Pk + 2(xk+1)+yk 2+1-2yk - yk2 - yk+1 + yk +1
        = Pk + 2(xk+1)-2yk +2 + 1
        = Pk + 2(xk+1)-2(yk -1) + 1

As we know (xk+1 = xk+1) and (yk-1 = yk-1)
Therefore,
                Pk+1 = Pk + 2xk+1 - 2yk-1 + 1

And if Pk < 0 that means midpoint is inside the circle and N is closest pixel so we will choose N (xk+1 , yk)
i.e. yk+1 = yk
        Now put coordinates of N in (ii)
Pk+1 = Pk + 2(xk+1)+( yk2 - yk2 ) - (yk - yk) +1
        = Pk + 2(xk+1)+( yk2 - yk2 ) - (yk - yk) +1
        = Pk + 2(xk+1) +1

as xk+1 = xk+1 , therefore,
                        Pk+1 = Pk + 2xk+1 +1


      Hence we have derived the mid point circle drawing algorithm.





                               








Comments

  1. Thank you so much

    yk-1 = y(k-1) ===> this should be yk-1 = y(k+1) right?

    ReplyDelete

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