1. AASHISH ADHIKARI [mail@aashishadhikari.info.np]
PRESENTATION
Computer Graphics
Lumbini City College
BSc. CSIT Third Semester
Presenter :
Aashish Adhikari
AASHISH ADHIKARI

2. AASHISH ADHIKARI [mail@aashishadhikari.info.np]
Contents
Digital Differential Analyzer (DDA) Algorithm
Working Principle
Examples
Bresenham’s Line Drawing Algorithm (BSA)
AASHISH ADHIKARI

3. Working Principle of DDA Algorithm
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.
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 𝑆𝑡𝑒𝑝𝑠 𝑓𝑙𝑜𝑎𝑡
Step 5: Perform round off, and plot each calculated (x, y) i.e. Plot(round(x), round(y)).
Step 6: END
AASHISH ADHIKARI

4. Example 1 : ( For m<1 OR Δx>Δy)
* Using DDA Plot (5,4) to (12,7)
Steps X Y
Y(Rounded off) 5 4 1 6
4.4 2 7
4.8 3 8
5.2 9
5.6 10 11
6.4 12
6.8
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
= i.e. <1
So, no. of Steps= absolute (Δx)
= |7| = 7
Now,
xincreament = Δx 𝑆𝑡𝑒𝑝𝑠 𝑓𝑙𝑜𝑎𝑡 = = 1
Yincreament = Δy 𝑆𝑡𝑒𝑝𝑠 𝑓𝑙𝑜𝑎𝑡 = = 0.4
AASHISH ADHIKARI

5. Example 2 : ( For m>1 OR Δx<Δy)
* Using DDA Plot (5,7) to (10,15)
Steps X Y
Y(Rounded off) 5 7 1
5.6 8 6 2
6.2 9 3
6.8 10 4
7.4 11 12
8.6 13
9.2 14
9.8 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
= i.e. >1
So, no. of Steps= absolute (Δy)
= |8| = 8
Now,
xincreament = Δx 𝑆𝑡𝑒𝑝𝑠 𝑓𝑙𝑜𝑎𝑡 = = 0.6
Yincreament = Δy 𝑆𝑡𝑒𝑝𝑠 𝑓𝑙𝑜𝑎𝑡 = =1
AASHISH ADHIKARI

6. AASHISH ADHIKARI [mail@aashishadhikari.info.np]
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.
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 = 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)
Algorithm
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
AASHISH ADHIKARI

7. 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 = * 3 – 2 x 5 = -3
p2 = p1+ 2 △y = -3+2 * 3 = 3.
p3 = p2 + 2 △y – 2 △x = = -1.
p4 = p3 + 2 △y = = 5
Successive pixel positions and decision parameter calculation is shown in the table.
AASHISH ADHIKARI

8. AASHISH ADHIKARI [mail@aashishadhikari.info.np]
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).
Step 2: Calculate the values of Δx and Δy,
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)
AASHISH ADHIKARI

9. Example 2 : ( For m>=1 OR Δx>Δy)
* Digitize a line with end points (6,12) to (10,5) using BSA
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 8
(8,9) 3 9 14
(9,7) 4
(10,5)
AASHISH ADHIKARI

10. Working Principle of Bresenham’s Line Drawing Algorithm (BSA) [for negative slope questions]
* Digitize a line with end points (6,12) to (10,5) using BSA
( For m>=1 OR Δx>Δy)
Alternative Method
AASHISH ADHIKARI

11. AASHISH ADHIKARI [mail@aashishadhikari.info.np]
Thanks
AASHISH ADHIKARI