Bresenham's Line Drawing Algorithm

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 • Bresenham's line drawing algorithm came into existence because DDA line drawing algorithm use to give more of decimal (floating) point numbers. 

• Which were round off to integer data type (int) because decimal number cannot be plotted on the screen.

•Due to rounding off the coordinate the output of line would be more in zig-zag form.


Explanation:

1) Consider 

    •dA= Above the line point.

    •dB= Below the line point.


2) so point can be above or below the line.

3) In this algorithm we will make a precise decision regarding the point above or below 

4) let decision variable be e 

    • e= dB - dA 

5) Where  if:      e<0 

    when        :     dB<dA

    Because: It will be below the point of line.

         •  if:          e>0

 When   

:         dB>=dA.  

 Because: Above the point of line.


Algorithm:

1) Start

2) Find slope of line i.e m, 

      m= y2-y1/x2-x1.

3) Find decisions variable e,

           e0= 2dy - dx.

4) To Find dy and dx,

            dy= y2-y1 

            dx= x2-x1.

5) if m<1 :

           a) if e0<0:

                  X2=x1 +1 

                  Y2= y1

//(No change in Y2)

              e1= e0 +2dy


          b) else e0>=0:

                  X2=x1 +1 

                 Y2= y1 +1

   e1= e0 +2dy - 2dx.

Note: x1 and y1 are the initial coordinates of line and will be input by the user itself. (Will also solve numerical there you will understand it more properly).

6) if m>=1 :

                a) if e0<0:

                      X2=x1(No change)

             Y2= y1 +1

             e1= e0 +2dx

                b) else e0>=0:

                X2=x1 +1 

                 Y2= y1 +1

   e1= e0 +2dx - 2dy.

7) repeat from 5 to 6 till you get the end points, but remember one thing slope of line m we will once and check once only.

Note: slope of line is always same for any two co-ordinates of line.

8) Stop.

                

Numerical:

1) Given: (1,1) (5,3).

 2) x1= 1  and y1= 1.

     x2= 5  and y2= 3.

3)  First slope:

 m= y2-y1/x2-x1 

                     = 3-1/5-1

                     = 1/2

4) dy= y2-y1= 2

     dx= x2-x1= 4

5)   m<1

      e0= 2dy-dx

          = 2(2)-4

     e0 = 0

6)  e0>=0

       x2= x1+1

            = 1+1

       x2= 2

       y2= y1+1

            = 1+1

             = 2

Next decision variable

    e1= e0 + 2dy -2dx 

        = 0 + 2(2) - 2(4)

        = -4

x2= 2 and y2= 2.

7)  e1<0

        x3= x2 + 1

             = 2 + 1

             = 3

        y3= y2 

             = 2

   e2= e1 + 2dy

        = -4 + 2(2)

        = 0

x3= 3 and y3= 2.

8) e2>=0

      x4= x3 +1

            = 3 +1

            =4

       y4= y3 +1

            = 2 +1

           =  3

   e3= e2 +2dy - 2dx

       = 0 + 2(2) - 2(4)

       = -4

x4= 4 and y4= 3.

9) e3 <0

      x5= x4 +1

           = 5

      y5= y4 

           = 3

  x5= 5 and y5= 3

10) Hence x5= x2 and y5= y2 we have reached the end points of the line we have stop here

 Coordinate:   x   y

                          1   1

                          2    2

                          3    2

                          4     3

                          5     3

    • Link to program:

  https://github.com/Rinky6767/Computer-Graphics/blob/main/Bresenham's%20Line%20Drawing%20program


*You can leave a comment if you have doubt in it.



  

   









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