Question

Bresenham's line algorithm to generate a rectangle defined by its diagonal end points coordinates

Answer 1

Explanation:

only because the line is defined to start and end on integer coordinates (though it is entirely reasonable to want to draw a line with non-integer end points).

Answer 2

Bresenham's line algorithm :

Explanation:

• This algorithm is employed for scan changing a line.
• it had been developed by Bresenham. it's Associate in Nursing economical methodology as a result of it involves solely number addition, subtractions, and multiplication operations. These operations will be performed terribly apace thus lines will be generated quickly.

• In this method, the next pixel selected is that the one that has the least distance from the true line.  
• The method works as follows:

• Assume a pixel P1'(x1',y1'), then select subsequent pixels as we work our way to the night, one-pixel position at a time in the horizontal direction toward P2'(x2',y2').

• Once a component is chosen at any step

• The next pixel is:

1. The one to its right sides (lower-bound for the line)
2. On top, it's right and up (upper-bound for the line)

• The line is best approximated by those pixels that fall the smallest amount distance from the path between P1', P2'.

• To chooses the successive one between the bottom pixel S and top pixel T.

1.  If S is chosen

           We have xi+1=xi+1       and       yi+1=yi

    2.  If T is chosen

           We have xi+1=xi+1       and       yi+1=yi+1

• The actual y coordinates of the line at x = xi+1 is given by

           y=mxi+1+b

• Bresenham's Line Algorithm, the distance from S to the actual line in the y-direction.

           s = y-yi

• The distance from T to the actual line in the y-direction is given by

           t = (yi+1)-y

• Now consider the distinction between these 2 distance values

           s - t

• When (s-t) <0 ⟹ s < t  

         The closest pixel is S

         When (s-t) ≥0 ⟹ s < t

         The closest pixel is T

• This difference is

           s-t = (y-yi)-[(yi+1)-y]

                = 2y - 2yi -1

•  Bresenham's Line Algorithm

        by putting m by Bresenham's Line Algorithm and introducing decision variable

               di=△x (s-t)

               di=△x (2 Bresenham's Line Algorithm (xi+1)+2b-2yi-1)

                       =2△xyi-2△y-1△x.2b-2yi△x-△x

               di=2△y.xi-2△x.yi+c

Where c= 2△y+△x (2b-1)

• We can write the decision variable di+1 for the next slip on

               di+1=2△y.xi+1-2△x.yi+1+c

               di+1-di=2△y.(xi+1-xi)- 2△x(yi+1-yi)

• Since x_(i+1)=xi+1,we have

               di+1+di=2△y.(xi+1-xi)- 2△x(yi+1-yi)

Special Cases:

• If the chosen pixel is at the top pixel T (i.e., di≥0)⟹ yi+1=yi+1

               di+1=di+2△y-2△x

• If the chosen pixel is at the bottom pixel T (i.e., di<0)⟹ yi+1=yi

               di+1=di+2△y

• Finally, we calculate d1

               d1=△x[2m(x1+1)+2b-2y1-1]

               d1=△x[2(mx1+b-y1)+2m-1]

• Since mx1+b-yi=0 and m = Bresenham's Line Algorithm, we have

               d1=2△y-△x

Refer to the following attachment: