Working principle of dda and bresenham line drawing explaination with example

Computer Graphics
PRESENTATION
Presenter :
• Aashish Adhikari
AASHISH ADHIKARI [mail@aashishadhikari.info.np]

• Digital Differential Analyzer (DDA) Algorithm
a. Working Principle
b. Examples
• Bresenham’s Line Drawing Algorithm (BSA)
a. Working Principle
b. Examples
Contents

Working Principle of DDA Algorithm
Algorithm
Step 1: Input the line endpoints and store the left endpoint in (x0, y0) and right endpoint in (x1, y1).
Step 2: Calculate the values of Δx and Δy,
Δx = x1 – x0, Δy = y1 – y0
Step 3: Based on the calculated difference in step 2, now we need to identify the number of steps
to put pixel as follows:
If Δx > Δy, then we need more steps in x co-ordinate; otherwise in y
coordinate.
I.e. if (absolute(Δx) > absolute (Δy) ), then steps = absolute(Δx) else steps =
absolute(Δy)
Step 4: We calculate the increment in x coordinate and y coordinate as follows:
xincreament =
Δx
𝑆𝑡𝑒𝑝𝑠 𝑓𝑙𝑜𝑎𝑡
Yincreament =
Δy
Step 5: Perform round off, and plot each calculated (x, y) i.e. Plot(round(x), round(y)).
Step 6: END
Definition: The Digital Differential Analyzer(DDA) is a scan conversion line drawing algorithm based on calculating
either Δx or Δy from the equation m = Δy/Δx.

Example 1 : ( For m<1 OR Δx>Δy)
* Using DDA Plot (5,4) to (12,7)
Solution:
Given,
(x0, y0) = ( 5,4)
(x1, y1) = ( 12,7)
Δx = x1 – x0 = 12-5 = 7
Δy = y1 – y0 = 7-4 = 3
Since, Δx > Δy or, Slope m=
Δy
Δx
=
3
7
i.e. <1
So, no. of Steps= absolute (Δx)
= |7| = 7
Now,
xincreament =
Δx
=
7
7
= 1
Yincreament =
Δy
=
3
7
= 0.4
Steps X Y Y(Rounded off)
0 5 4 4
1 6 4.4 4
2 7 4.8 5
3 8 5.2 5
4 9 5.6 6
5 10 6 6
6 11 6.4 6
7 12 6.8 7

Example 2 : ( For m>1 OR Δx<Δy)
* Using DDA Plot (5,7) to (10,15)
Solution:
Given,
(x0, y0) = ( 5,7)
(x1, y1) = ( 10,15)
Δx = x1 – x0 = 10-5 = 5
Δy = y1 – y0 =15-7 = 8
Since, Δx < Δy or, Slope m=
Δy
Δx
=
8
5
i.e. >1
So, no. of Steps= absolute (Δy)
= |8| = 8
Now,
xincreament =
Δx
=
5
8
= 0.6
Yincreament =
Δy
=
8
8
=1
Steps X Y Y(Rounded off)
0 5 7 5
1 5.6 8 6
2 6.2 9 6
3 6.8 10 7
4 7.4 11 7
5 8 12 8
6 8.6 13 9
7 9.2 14 9
8 9.8 15 10

Working Principle of Bresenham’s Line Drawing
Algorithm (BSA) [for positive slope questions]
Definition: An accurate and efficient raster line-drawing algorithm, developed by Bresenham‘s. It only uses integer
calculations so can be adapted to display circles and other curves.
AlgorithmAlgorithm
Step 1: Input the line endpoints and store the left
endpoint in (x0, y0) and right endpoint in (x1, y1).
Δx = abs(x1 – x0), Δy = abs(y1 – y0)
Step 3: Calculate initial decision parameter as,
p = 2 Δy- Δx
Step 4: if x1 > x0 then
Set x= x0
Set y=y0
Set xend=x1
Otherwise
Set x= x1
Set y=y1
Set xend=x0
Step 5: Draw pixel at (x,y)
Step 6: while(x<xend)
x++;
if p<0 then
p=p+2Δy
Otherwise
y++
p=p+2Δy-2Δx
Draw pixel at (x,y)
Step 7: END

Example 1 : ( For m<=1 OR Δx>Δy)
* Digitize a line with end points (10,15) to (15,18) using BSA
Solution:
Given,
(x0, y0) = ( 10,15)
(x1, y1) = ( 15,18)
Δx = x1 – x0 = |15-10| = 5
Δy = y1 – y0 = |18-15| = 3
Since, Slope m=
Δy
Δx
<=1
So, we sample at x direction i.e., at each step we simply increment x-coordinate by 1 and find appropriate y-coordinate.
First pixel to plot is (10,15), which is the starting endpoint.
Now, initial decision parameter, p0 = 2 Δy- Δx = 2*3-5 = 1
We know that,
If pk < 0, then we need to set pk+1 = pk+2△y and plot pixel(xk+1, yk)
And if pk ≥ 0, then we need to set pk+1 = pk + 2 △y – 2 △x then plot pixel(xk+1, yk+1)
Here p0 = 1, we have i.e., pk ≥0
So,
p1 = p0 + 2 △y – 2 △x = 1 + 2 * 3 – 2 x 5 = -3
p2 = p1+ 2 △y = -3+2 * 3 = 3.
p3 = p2 + 2 △y – 2 △x = 3+6-10 = -1.
p4 = p3 + 2 △y = -1+ 6 = 5
Successive pixel positions and decision parameter calculation is shown in the table.

Working Principle of Bresenham’s Line Drawing
Algorithm (BSA) [for negative slope questions]
Algorithm
Step 1: Input the line endpoints and store the left endpoint in (x0, y0) and right endpoint in (x1, y1).
Δx = abs(x1 – x0),
Δy = abs(y1 – y0)
Step 3: Calculate initial decision parameter as,
pk = 2 Δy- Δx
Step 4: for (i=1 to Δx)
{
set pixel (xs, ys)
while (pk>0)
y0=y0-1;
pk= pk-2 Δx
end while, (pk>0)
x0=x0+1
pk= pk+2 Δy
i++;
}
set pixel (x0, y0)

Example 2 : ( For m>=1 OR Δx>Δy)
Solution:
Given,
(x0, y0) = ( 6,12), (x1, y1) = ( 10,05)
Now, We calculate the Δx and Δy as follows:
Δx = x1 – x0 = |10-6| = 4, Δy = y1 – y0 = |5-12| = 7
Since, Slope m=
Δy
Δx
>=1
So, we sample at x direction i.e., at each step we simply increment x-coordinate by 1 and find appropriate y-coordinate.
First pixel to plot is (10,15), which is the starting endpoint.
Now, initial decision parameter, p0 = 2 Δy- Δx = 2*7-4 = 10
For (i=1 to 4) using the condition we proceed as follows:
For i=1
set pixel (6,12) while (10>0)
y0= 12-1 = 11
pk= 10-2*4 = 2
while (2>0)
y0= 11-1 =10
pk= 2-2*4 = -6
end while
x0= x0+1 = 6+1 = 7
pk = pk+ 2 Δy
= -6+2*7 = 8
Successive pixel positions and decision parameter calculation is shown in the table.
I X0 Y0 Pk Points
1 6 12 10 (7,10)
2 7 10 8 (8,9)
3 8 9 14 (9,7)
4 9 7 12 (10,5)

Working Principle of Bresenham’s Line Drawing Algorithm (BSA) [for negative
slope questions]
( For m>=1 OR Δx>Δy)
Alternative Method

Thanks
2020.04.19

Working principle of dda and bresenham line drawing explaination with example

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