g.h. raisoni college of engineering, nagpur. department of ... · g.h. raisoni college of...

G.H. Raisoni College of Engineering, Nagpur.

Department of Information Technology

Fifth Semester Computer Graphics




Practical List

1) WAP to implement line generation using DDA algorithm

2) WAP to implement line using Bresenham’s line generation algorithm.

3) WAP to generate circle using circle generation algorithm

4) WAP to draw a polygon using DDA

5) Write a Program to fill polygon using Edge flag algorithm.

6) WAP to implement ellipse using ellipse generation algorithm

7) WAP to perform translation, rotation scaling on 2-D transformation

8) WAP to fill polygon using seed fill algorithm




EXPREMENT NO.1

Aim : - Write a program to implement line generation using DDA

Theory :- The simplest way to rasterize a line is to solve the governing differential

equation:

Equation of the straight line is

Y = mx + c

Differential with respect to x

∆dy/ ∆dy = m = constant --------------(1)

Assume that the ∆y and ∆x are the small increment in x and y direction along the path

respectively along the path.

Equation (1) can be written as ∆y/ ∆x =y2 - y1/ x2 – x1 = constant --------- (2)

Equation (2) has one solution

Yi +1 = yi + ∆y

= yi + (y2 - y1/ x2 – x1) ---------------- (3)

equation (3) represents a recursive relation for successive values of y along the required

line x1, y1, x2, y2 are the end points of the required line .

when equation (3) is used to rasterise the line then it is called as digital

differential analyzer (DDA).

ALGORITHM

Step 1). Initialize(x1,y1), (x2,y2)

if abs(x2 – x1) ≥ abs(y2 – y1) then

len = abs(x2 – x1)

else

len = abs(y2 – y1)

∆x = (x2 – x1)/len

∆y = (y2 – y1)/len

x = x1 + 0.5 * sign(∆x)

y = y1 + 0.5 * sign(∆y)

Step 2). begin main loop

i=1

while( i ≤ len)

plot (integer(x),integer(y))

x = (x + ∆x)

y = (y + ∆y)

i = i + 1

end while

finish




VIVA VOICE

Q 1. What will happen if the line endpoints are assumed equal?

Q 2. What is the value of Integer( -8.5 )?

Q 3. What is the value of Integer( 8.5 )?

Q 4. Which modifications are necessary to generate dashed line from DDA algorithm.

Q 5. Which modifications are necessary to generate dotted line from DDA algorithm.

Q 6. Write the algorithm for DDA.

Q 7. What is rasterization?

Q 8. Give the drawbacks of DDA algorithm.

Q9. What is DDA?




Experiment No.2

Aim : - To implement Bresenham’s line generation algorithm

Theory : - This algorithm seeks to select the optimum raster location to represents a

straight line. To achieve this, the algorithm always implemented by one unit in either x or

y directions depending on the slope of the line.

The increment in the other variables either 0 or 1, is determined by examining the

distance between the actual line location and the nearest grid location. This distance is

called the error and only error term is needed to be examined.

If the slope of the line through (0,0) is greater than ½ i.e. m>1/2 then its

intercept with the x=1 which is closer to the line y=1 then y=0 hence pixel(1,1) is

selected. Similarly , if the slope of the line is less than ½ i.e. m<1/2 than its intercept with

the line is x=1 is closer to the line y=0 hence select pixel (1,0).

ALGORITHM

The line end point are (x1, y1) and (x2,y2) assumed not equal

All variables are assumed integer

The sign function returns -1, 0, 1 as its argument is <0, =0, or >0

Step 1). Initialize variables

x = x1

y = y1

∆x = abs(x2 – x1)

∆y = abs(y2 – y1)

S1 = sign(x2 – x1)

S2 = sign(y2 – y1)

0≤ ∆y/ ∆x<1/2(error<0)

1/2≤ ∆y/ ∆x≤1 (error≥0)




Step 2). Interchange ∆x and ∆y depending on slope of the line

Temp = ∆x

∆x = ∆y

∆y = Temp

Interchange = 1

else Interchange = 2

end if

Step 3). Initialize the error term to compensate for a nonzero intercept

ē = 2 * ∆y – ∆x

Step 4). Main loop

for I = 1 to ∆x

plot(x,y)

while ( ē ≥ 0)

if interchange = 1 then

x = x + s1

else

y = y + s2

end if

ē = ē – 2 * ∆x

end while

if interchange = 1 then

y = y + s2

else

x = x +s2

end if

ē = ē – 2 * ∆y

next i

finish

VIVA VOICE

1) What is an Error term in Bresenham’s algo?

2) What will happen if we will not use error term?

3) Give the advantages of Bresenham’s algo over DDA algo.

4) Give the significance of error term.

5) Give the Advantages of Bresenham’s algorithm.




EXPERIMENT NO 3

AIM :- To generate a circle using Bresenham’s circle algorithm

THEORY : - To draw a circle, only first octant of the circle is required to be generated

and other parts can be obtained by successive reflections. The circle generation is done

by using Bresenham’s algorithm.

If the first octant is generated, the second octant is obtained by reflection through

the line y = x to yield the first quadrant . The results in the first quadrant are reflected

through the line x = 0 to obtain those in second quadrant. The combined results in the

upper semicircle are reflected through the line y = 0 to complete the circle.

ALGORITHM

Initialize the variables

xi = 0

yi = R

∆i = 2(1- R)

Limit = 0

Step 1). Plot(xi , yi)

if yi ≤ Limit then 4

Determine the case 1 or 2, 4 or 5, or 3

if ∆i < 0 then 2

if ∆i > 0 then 3

if ∆i = 0 then 20

Determine whether case 1 or 2

1) δ = 2 ∆i + 2 yi -1

if δ ≤ 0 then 10

if δ > 0 then 20

determine whether case 4 or 5

3) δ’ = 2 ∆i - 2 xi -1

if δ’ ≤ 0 then 20

if δ’ >0 then 30

perform the moves

move mH

10) xi = 2 xi + 1

∆i = ∆i + 2 xi +

go to 1

move mD

20) xi = 2 xi + 1

yi = 2 yi - 1

∆i = ∆i + 2 xi - 2 yi + 2

go to 1

move mv

30) yi = yi – 1

∆i = ∆i - 2 yi + 1

go to 1

finish




VIVA VOICE

1. How to generate the circle in 2nd

quadrant from 1st quadrant clockwise ?

2. How to generate the circle in 2nd

quadrant from 1st quadrant anticlockwise?

3. How to generate the circle in 3rd

quadrant from 1st quadrant clockwise ?

4. How to generate the circle in 3rd

quadrant from 1st quadrant anticlockwise?

5. Explain the Bresenham’s circle algorithm.

6. Explain the different moves in this algorithm.

7. When to choose vertical move?

8. When to choose horizontal move?

9. When to choose diagonal move?

10. Generate the circle in 1st quadrant having radius 8 and center is origin.




EXPERIMENT NO 4

AIM :- To generate a polygon using any line generation algorithm

THEORY :- A polygon can be represented as a number of line segments connected end

to end to form a close figure. Alternatively, it may be represented as the points where the

sides of the polygon are connected. The line segments which make up the the polygon

boundary are called sides or edges. The end points of the sides are called the polygon

vertices.

ALGORITHM:

Step 1. Eenter number of vertices in n

Step2. Enter the co-ordinates of each vertex

Step 3. Repeat the following steps for n number of times

Step 4. Initialize(x1,y1), (x2,y2)

if abs(x2 – x1) ≥ abs(y2 – y1) then

len = abs(x2 – x1)

else

len = abs(y2 – y1)

∆x = (x2 – x1)/len

∆y = (y2 – y1)/len

x = x1 + 0.5 * sign(∆x)

y = y1 + 0.5 * sign(∆y)

Step 5. begin main loop

i=1

while( i ≤ len)

plot (integer(x),integer(y))

x = (x + ∆x)

y = (y + ∆y)

i = i + 1

end while

finish

VIVA QUESTION

1) What is scan conversion?

2) How will you generate a polygon?

3) Which algorithms can be used to generate a polygon?




EXPERIMENT NO 5

Aim :- Write a Program to fill polygon using Edge flag algorithm.

Theory :- It’s a technique to fill the polygon by two steps.The first step is to outline the

contour.This establishes pairs of span bounding pixels on each scanline.The second step

is to fill between these bounding pixels.

Algorithm :- 1> For contour outline :

Using the half scan line convention for each edge intersecting the polygon, set the

leftmost pixel whose midpoint lies to the right of the intersection, i.e. for x+1/2 >

Xintersection ,to the boundary value.

2> For Filling :

For each scanline intersecting the polygon

Inside = FALSE

For X = 0 (left) to X = X max(right)

If the pixel at X is set to the boundary value then negate inside

If Inside = TRUE then

Set the pixel at X to the Polygon value

Else

Reset the pixel at X to the Background value

Endif

Next x

Viva-voce :-

Qu 1 Can this algo visits each pixel more than once?

Qu 2 Explain this algo by giving example.

Qu 3 Give the advantages of this algorithm

.

Qu 4 Give the Disadvantages of this algorithm.

Qu 5 How can we overcome the disadvantage of this algorithm.

Qu 6 Do you feel that this algo is best suited amongst all ?

Qu 7. Why this algo is best suited?




EXPERIMENT NO. 6

AIM : - Write program to generate an ellipse

THEORY : - Equation of an origin centered axis aligned ellipse is given by

x2/a

2 + y

2/b

2 = 1

Where, a is the semi major axis and b is semi minor axis

Rewrite the equation as

1. b2x

2 + a

2y

2 – a

2b

2 = 0

Consider the generation of ellipse in first quadrant in clockwise direction.

∆i is a distance of diagonal pixel from the centre of ellipse.

∆i = b2 (xi + 1)

2 +a

2 (y1 -1)

2 – a

2b

2

If ∆i < 0, means diagonal pixel lies inside the ellipse, so consider case 1

If ∆i > 0 , means diagonal pixel lies outside the ellipse

If ∆i = 0 , means ellipse passes through the diagonal pixel.

ALGORITHM

Step 1). Initialize the variables

xi = 0

yi = b

∆i = a2 + b

2 – 2ab

2

Limit = 0

Step 2). if yi ≥ limit

then plot (xi, yi)

else

exit

Step 3). if ∆i < 0

Then go to step 4

else

if ∆i > 0

then go to step 5

else

if ∆i = 0

then xi = xi + 1

yi = yi + 1

∆i = ∆i + b2(2(xi + 1) + 1) – a

2 (2(y-1) – 1)

else

then go to step 6




Step 4). d1 = 2 ∆i + (2yi – 1) a2

if d1 ≤ 0 then

xi = xi +1

∆i = ∆i + b2 *(2*(x+1)+1)

Step 5) d2 = 2*∆i – (2 * x + 1) * b2

If d2 <=0 then

x = x + 1

y = y + 1

∆i = ∆i + b2 [2(x + 1) + 1] – a

2 (2(y – 1)- 1)

Else

y = y – 1

Step 6) x = x + 1

y = y + 1

∆i = ∆i + b2[2(x + 1) + 1] – a2(2(y – 1)- 1)

VIVA VOICE

1) What is the equation of origin centered axis.

2) How will you generate an ellipe.

3) What is the difference between circle generation and ellipse generation

algorithm

4) Explain the different moves in this algorithm.

5) When to choose vertical move?

6) When to choose horizontal move?

7) When to choose diagonal move?




EXPERIMENT NO. 7

Aim :-Write a program 2-D transformation on a polygon namely Translation,

Rotation,Scaling and Reflection

Theory :-Following are the different Transformations

1>Translation

1 0 0

0 1 0

Tx Ty 1

Where Tx and Ty are the translation factors.

2>Rotation

i> in clockwise direction

Cosθ -Sinθ 0

Sinθ Cosθ 0

0 0 1

ii> in anticlockwise direction

Cosθ -Sinθ 0

Sinθ Cosθ 0

0 0 1

3>Scaling

Sx 0 0

0 Sy 0

0 0 1

Where Sx and Sy are the Scaling factors.




4>.Reflection

i> about X-axis

1 0 0

0 -1 0

0 0 1

ii> about Y-axis

-1 0 0

0 1 0

0 0 1

iii> about Origin

-1 0 0

0 -1 0

0 0 1

iv> about Y = X axis

0 1 0

1 0 0

0 0 1

v> about Y = -X axis

0 -1 0

-1 0 0

0 0 1




Viva-voce

Q1 Which are the real life examples of transformations?

Q2 Write the transformation matrix for Translation.

Q3 Write the transformation matrix for Rotation in clockwise direction.

Q4 Write the transformation matrix for Rotation in anticlockwise direction.

Q5 Write the transformation matrix for Scaling.

Q6 Write the transformation matrix for reflection about X-axis.

Q 7 Write the transformation matrix for reflection about Y-axis.

Q8 Write the transformation matrix for reflection about Origin.

Q9 Write the transformation matrix for reflection about X=Y axis

Q10 Write the transformation matrix for reflection about Y=-X axis.




EXPERIMENT NO. 8

AIM : - Write a program to implement seed fill algorithm for polygon filling

THEORY : - Generation of solid areas from the simple edge of the vertex description is

referred as solid scan conversion or polygon filling or contour filling.

There are two techniques used to fill a polygon namely scan conversion and seed

fill technique.

The seed fill algorithm required one of the pixel position inside the polygon or

region to be filled. This pixel is inside the region and set to polygon value. This pixel is

referred as seed pixel. The algorithm continues to search the pixels which are adjacent to

the current seed pixel to see whether they are boundary pixels or the pixels inside the

region. All the pixels interior to the region and are adjacent to the seed pixel are set to the

polygon value.

ALGORITHM : -

Seed(x,y) is the seed pixel

Push is a function for placing a pixel on the stack

Pop is a function for removing a pixel from the stack

Pixel(x,y) = seed(x,y)

Initialize stack

Push pixel(x,y)

While (stack not empty)

get a pixel from the stack

Pop Pixel (x,y)

If Pixel(x,y) <> New value then

Pixel(x,y) = New value

Else

If (Pixel(x+1,y) <> New value and

Pixel(x+1,y) <> Boundary value then

Push Pixel(x+1,y)

if Pixel(x,y+1) <> New value and

Pixel(x,y+1) <> Boundary value then

Push Pixel(x,y+1)

If Pixel(x-1,y) <> New value and

Pixel(x+1,y) <> Boundary value then

Push Pixel(x-1,y)

If Pixel(x,y-1) <> New value and

Pixel(x,y-1) <> Boundary value then

Push pixel(x,y-1)

End if

End while




VIVA VOICE

Q 1 Explain the Seed fill algorithm.

Q 2 Give the advantages of this algorithm.

Q 3 Give the disadvantages of this algorithm.

Q 4 Can we choose any pixel as seed pixel?

Q 5 Which are the two conditions are required before setting the pixel?

Q 6 If the pixel is Boundary pixel ,can we set it ?

Q 7 If the pixel is already setted , can we set it ?

Q 8 Is there any modifications are required in this algorithm?

Q 9 Which modifications are required?

Q 10 Is there any other algorithm available to avoid the drawbacks ? Explain.

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