### Bresenham's Line Drawing Derivation

Bresenham's Line Drawing Algorithm Derivation
Bresenham Line drawing algorithm is used to determine closest points to be illuminated on the screen to form a line.

As we know a line is made by joining 2 points, but in a computer screen, a line is drawn by illuminating the pixels on the screen.

(Here pixel (1,2), (3,1) and (5,5) are illuminated and others are non-illuminated)
A line from pixel (2,2) to (7,5) will be shown like this on the screen.

The slope of a line plays a major role in the line equation that's why Bresenham line drawing algorithm calculates the equation according to the slope of the line.

The slope of the line can be greater than 1 (m>1) or less than or equal to 1 (m<=1).
Now enough talking let's derive the equations.

Derivation:

Let's say we want to draw a line on the screen.
so, to draw a line we have to calculate the points or pixels to be illuminated on the screen.

Now while drawing a line a sometimes it passes through 2 pixels at the same time then we have to choose 1 pixel from the 2 to illuminate it.

so, the bresenham algorithm calculates the distance from the intersection point y to both the pixels and whichever is smaller we choose that pixel.

The equation of Line is:

Y=mx+c           ....(1)
Here m is the slope of Line and
C is y-Intercept of line

The slope can also be written as

First we calculate the slope of the line if slope is less than 1, then X will always be incremented.
so at (m<=1)
we calculate the d1 which is distance between intersection point y to the pixel yk

So, d1 = y-yk
d1 = mxk+1 +c -yk         By using ....(1)
Similarly d2 is the distance between pixel yk+1 and intersection point y.
d2 = yk+1-y
d2 = yk+1-(mxk+1+c)       By using ....(1)
Note: here x is always incrementing so we can write xk+1 as xk +1 and here yk+1 is next pixel so we can write it as yk+1.
subtracting d2 from d1
d1-d2 = m(xk+1) +c -yk – [yk+1-(mxk+1+c)]
= m(xk+1)+c -yk yk-1+m(xk+1)+c
d1-d2 = 2m(xk+1)-2yk+2c-1           .....(3)
Multiplying both side by (dx)

dx(d1-d2) = 2dy(xk+1)-2dx(yk)+2dx(c)-dx

Now ­we need to find decision parameter PK
PK= dx(d1-d2) and,
C = 2dy+2dx(c)-dx which is constant
So new equation is.
PK =2dy(xk)-2dx(yk) +C      .......(4)
Now our next parameter will be
PK+1 =2dy(xk+1)-2dx(yk+1) +C .....(5)
Subtracting Pk from PK+1
Pk+1-Pk= 2dy(xk+1­-xk)-2dx(yk+1­-yk) +C-C
Note: here x is always incrementing so we can write xk+1 as xk +1
Pk+1-Pk= 2dy(x-xk+1)-2dx(yk+1­-yk)
Pk+1 = Pk+2dy-2dx(yk+1­-yk)........(6)
when Pk<0 then (d1-d2)<0
So d1<d2 then we will write yk+1 as yk­ because current pixel’s distance from intersection point y is smaller.so, we will have to choose current pixel.

Then our formula will be:
Pk+1 = Pk+2dy-2dx(y-yk)
Pk+1 = Pk+2dy
And when Pk>0 then (d1-d2)>0
So d1>d2 then we will write yk+1 as yk+1­ because current pixel’s distance from intersection point y is larger.so, we will have to choose upper pixel.
At that time our formula will be:
Pk+1 = Pk+2dy-2dx(yk+1­-yk)
Pk+1 = Pk+2dy-2dx
We can say that (yk+1­-yk) value can either be 0 or 1.

For Initial decision parameter
From 4th equation
PK = 2dy(xk)-2dx(yk) +C
PK = 2dy(xk)-2dx(yk)+2dx(c)-dx+2dy
By using 1st equation
yk=m(xk)+c
c=yk-m(xk)
P0 = 2dy(xk)-2dx(yk)+2dx(yk-m(xk)-dx (By using 2)
= 2dy(xk)-2dx(yk)+2dxyk-2dyxk+2dy-dx

P0 =2dy-dx

But if the slope of line is greater then 1 (m>1).
then our Y coordinate will always be incremented and we have to choose between xk or xk+1.
So, our Line equation will be:
Yk+1 = m(x)+c

Now, In this case our d1 will be the distance between intersection point x and pixel xk
d1= x-xk                By using ....(7)
d1= 1/m(yk+1-c)-xk
And similarly our d2 will be the distance between intersection point x and pixel xk+1
d2 = xk+1-x             By using ....(7)
d2 = xk+1-1/m(yk+1-c)

Note: here y is always incrementing so we can write yk+1 as yk+1 and here xk+1 is next pixel so we can write it as xk+1.
subtracting d2 from d1
d1-d2 =1/m(yk+1-c)-xk – [xk+1-1/m(yk+1-c)]
= 1/m(yk+1-c)-xk xk-1+1/m(yk+1-c)
d1-d2 = 2/m(yk+1-c)-2xk-1
d1-d2 = 2dx(yk+1-c)-2dyxk-dy

Multiplying both side by (dy)

dy(d1-d2) = 2dxyk+2dx-2dxc-2dyxk-dy

Now ­we need to find decision parameter PK
PK = dy(d1-d2)
C = 2dx-2dxc-dy
which is constant
So new equation is
PK = 2dxyk-2dyxk +C   ......(8)

Now our next parameter will be
Pk+1 = 2dx(yk+1)-2dy(xk+1)+C
Substracting Pk from Pk+1
Pk+1-Pk = 2dx(yk+1-yk)-2dy(xk+1-xk)+C-C
Note: here x is always incrementing so we can write yk+1 as yk +1
Pk+1-Pk = 2dx(yk +1-yk)-2dy(xk+1-xk)
Pk+1=Pk+2dx-2dy(xk+1-xk)
when Pk<0 then (d1-d2)<0
So d1<d2 then we will write xk+1 as xk­ because current pixel’s distance from intersection point x is smaller.so, we will have to choose current pixel.

Then our formula will be:

Pk+1 = Pk+2dx-2dy(xk-xk)
Pk+1 = Pk+2dx
And when Pk>0 then (d1-d2)>0
So d1>d2 then we will write xk+1 as xk­+1 because current pixel’s distance from intersection point x is Larger.so, we will have to choose next pixel.

At that time our formula will be:
Pk+1= Pk+2dx-2dy(xk+1-xk)
Pk+1= Pk+2dx-2dy

For Initial decision parameter
From 8th equation
PK = 2dxyk-2dyxk +C
P0 = 2dxy0+2dx-2dxc-2dyx0-dy
By using 1st equation
yk=m(xk)+c
c=yk-m(xk)P0 = 2dxy0-2dyx0+2dx-2dx(y0-m(x0))-dy
P0 = 2dxy0-2dyx0+2dx-2dxy0+2dyx0-dy (By using 2)
P0 = 2dx-dy

hence formulas for bresenham is derived.

1. derivation part of bresenhams algorithm is high

2. Very nice and easily understandable .Thank uh

3. Best derivation for bresenham algo on internet AFAIK

4. best derivation ever

### Computer Graphics and its Applications

We all use Computers and Mobile phones for communication all these devices have displays that show content to you but, Have you ever thought that how these displays show images and videos on screen? These images are displayed on the screens using some programming algorithms. (which we will discuss later) What is Computer Graphics?It is an art of drawing images, lines, charts, and other visuals on the computer screens with the help ofprogramming. The visuals/images on the computer screens are made up of numerous pixels. A pixel is a dot or square on the computer screen, it is the smallest unit of graphic that is displayed on the computer screen. The number of individual non-overlapping pixels contained on a computer screen is called Resolution. Now we know what is computer graphics lets know about its applications Computer Graphics ApplicationsThe Computer Graphics is used in numerous areas some of them are as follows: Graphical User Interface (GUI): The interaction of users interacts with e…

### Bresenham's Circle Drawing Algorithm

Bresenham’s Circle Drawing Algorithm
A circle is made up of 8 Equal Octets so we need to find only coordinates of any one octet rest we can conclude using that coordinates.
We took octet-2. Where X and Y will represent the pixel Let us make a function Circle() with parameters coordinates of Centre (Xc,Yc) and pixel point (X,Y) that will plot the pixel on screen.

We will find pixels assuming that Centre is at Origin (0,0) then we will add the coordinates of centre to corresponding X and Y while drawing circle on screen.
Circle (Xc,Yc,X,Y){
Plot (Y+Xc , X+Yc)……Octet-1 Plot (X+Xc , Y+Yc)……Octet-2 Plot (-X+Xc , Y+Yc)……Octet-3 Plot (-Y+Xc , X+Yc)…..Octet-4 Plot (-Y+Xc , -X+Yc)……Octet-5

### Bresenham's Circle Drawing Derivation

Bresenham's Circle Drawing Algorithm Derivation
Bresenham circle drawing algorithm is used to determine the next pixel of screen to be illuminated while drawing a circle by determining the closest nearby pixel.
Let us first take a look how a circle is drawn on a pixel screen
(this is how pixel graph is represented)
As Circles are symmetrical so the values of y-intercept and x-intercept are are same if circle's Center coordinates are at Origin (0,0).

Here,  Radius = OA = r Due to symmetrical property of Circle we don't need to calculate all the pixels of all the octets and quadrants We need to find the pixels of only one octet, rest we can conclude through this.

Lets take the Octet 2 which is in quadrant 1 here both x and y are positive here the initial pixel would be (0,y) coordinate

At point R both the value of both x and y coordinates would be same as R is at same distance of Both X and Y axis.

Derivation of Bresenham circle algorithm: