# The Mid Point Algorithm Essay

835 words - 4 pages

* Midpoint Algorithm for the line of 1<m<∞

| | |
M1N pixel | M2NE pixel | |
M | | |

The equation for the line
ax + by + c = 0 1

The equation for the M point:
d – Equation for the first midpoint

d = a(x+1/2) + b(y+1) + c
d = ax + by + c + a/2 + b
d = 0 + a/2 + b
d = a/2+b
×2
d = a + 2b 2
By substituting the value fora and b we will able to find the value of d.
If d > 0, then we will select the N (North) pixel
If d < 0, then we will select NE (Northeast) pixel

Then there are two possibilities to go the line through M1 or M2
Equation for the above two points; M1 and M2
M1:
d1 – Equation for the M1 midpoint
d1 = a(x+1/2) ...view middle of the document...

In the first scenario we can choose East pixel and SouthEast pixels
d = initial decision variable.
d = X2b2 + Y2a2 - a2b2
d= (x+1)2b2 +(y-1/2)2 - a2b2
d = x2b2 + y2a2 - a2b2 + 2b2x + b2 - a2y + a2/4

d = 0 + 2b2x + b2 - a2y + a2/4 ;Since at the beginning x =0 and y = b the equation finally would be like below.
d = b2 –a2b + a2/4 1
if d < 0 we choose E pixel
if d >= 0 we choose SE pixel.
If we consider d<0 ; next pixel is E,
de = X2b2 + Y2a2 - a2b2
de = (x+2)2b2 +(y-1/2)2 - a2b2
de = x2b2 + y2a2 - a2b2 + 4b2x + 4b2 - a2y + a2/4
de = 0 + 4b2x +4b2 - a2y + a2/4
de-d = (4b2x +4b2 - a2y + a2/4) –( 2b2x + b2 - a2y + a2/4)
de= 2b2x +3b2 + d 2

If we consider d >= 0 ; next pixel is SE,
dse = X2b2 + Y2a2 - a2b2
dse = (x+2)2b2 +(y-3/2)2 - a2b2
dse = x2b2 + y2a2 - a2b2 + 4b2x + 4b2 - 3a2y + 9a2/4
dse = 0 + 4b2x +4b2 - 3a2y + 9a2/4
dse-d = (4b2x +4b2 -3 a2y + 9a2/4) –( 2b2x + b2 - a2y + a2/4)
dse= 2b2x +3b2 -2a2y + 2 a2+ d
dse= b2 (2x+3) +a2(-2y + 2) + d 3

Scenario 2 -...

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