Part A: Multiple Choice Questions (10 × 1 =
10 Marks)
1. The graphics pipeline consists of the following stages:
• Answer: (a) Modeling, Transformation, Clipping, Rasterization
2. What is the primary advantage of Bresenham’s line-drawing algorithm
over DDA?
• Answer: (a) It uses only integer arithmetic
3. What is the main difference between scan-line and flood-fill algorithms?
• Answer: (a) Scan-line works line by line, flood-fill fills connected
pixels
4. A Bezier curve is defined by:
• Answer: (c) Control and anchor points
5. In spline curves, continuity is defined as:
• Answer: (d) All of the above
6. In window-to-viewport mapping, the transformation ensures:
• Answer: (b) A proper aspect ratio of the image
7. The Cohen-Sutherland algorithm identifies regions using:
• Answer: (a) Outcodes
8. In perspective projection, objects further away appear:
• Answer: (b) Smaller
9. The viewing transformation involves:
• Answer: (d) All of the above
10. Which of the following is true about perspective projection?
• Answer: (a) Parallel lines converge
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Part B: Short Answer Questions (5 × 3 = 15
Marks)
1. Explain the region-filling algorithm using seed-fill or flood-fill
methods.
• Seed-Fill Algorithm: Starts at a ”seed point” inside the region
and spreads outward to fill all connected pixels until it hits the
boundary.
Recursive Approach: Recursively checks and fills neighboring
pixels.
Iterative Approach: Uses a stack to avoid recursion.
• Flood-Fill Algorithm: Fills all connected pixels with the same
color, commonly used in paint programs.
• Comparison: Seed-fill works well for defined boundaries. Flood-
fill is efficient for contiguous color regions but can overflow memory
in recursive methods.
2. What are modeling transformations? Explain their role in the
HCS (Homogeneous Coordinate System).
• Modeling transformations position, orient, and scale 3D ob-
jects in the scene.
• Translation: Moves the object to a new position.
• Scaling: Changes the object’s size.
• Rotation: Rotates the object about an axis.
• Role in HCS: Using HCS, coordinates are represented as [x, y, z, 1]
and transformations are performed using matrix multiplication.
HCS simplifies combining transformations into a single composite
matrix.
3. Write the steps involved in the Cohen-Sutherland line clipping
algorithm.
• Compute Outcodes: Calculate the region codes for each end-
point based on the window boundaries.
• Trivial Accept/Reject:
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– If both endpoints have outcodes 0000 (logical AND), the line
is completely inside.
– If the logical AND of outcodes 0, the line is completely out-
side.
• Partial Clipping: If neither trivial accept nor reject applies, cal-
culate intersections with window boundaries using edge equations.
• Update endpoints and recompute outcodes.
• Repeat until the line segment lies fully inside or is rejected.
4. What are the advantages of using curved surface representa-
tion in 3D modeling?
• Smoothness: Curved surfaces ensure smooth transitions and re-
alistic rendering.
• Compact Representation: Fewer control points represent the
surface, reducing memory usage.
• Flexibility: Ideal for organic shapes such as human faces and
vehicles.
• Continuity: Provides various levels of continuity (e.g., positional,
tangential).
5. Explain the viewing pipeline.
• Modeling Transformation: Converts object coordinates into
world coordinates. Places and orients objects in the scene using
translation, rotation, and scaling.
• Viewing Transformation: Converts world coordinates to cam-
era (view) coordinates. Positions the camera at the origin and
reorients the scene using a ”look-at” matrix.
• Projection Transformation: Maps 3D view coordinates to 2D
projection coordinates.
– Types:
∗ Perspective projection: Simulates depth by making dis-
tant objects appear smaller.
∗ Orthographic projection: Maintains the same size regard-
less of distance.
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• Clipping: Removes objects or parts outside the viewing volume
(e.g., frustum or cuboid).
• Window-to-Viewport Mapping: Maps the 2D projection win-
dow to the display screen (viewport) while maintaining the aspect
ratio.
Part C: Long Answer Questions (3 × 5 = 15
Marks)
1. Midpoint Circle Algorithm (Numerical)
Given r = 6, (h, k) = (0, 0):
Initialpoint :(x0 , y0 ) = (0, r) = (0, 6)
Initialdecisionparameter :p0 = 1 − r = −5
For each octant, calculate:
If pk < 0 : pk+1 = pk + 2xk + 3
Else :pk+1 = pk + 2xk − 2yk + 5
Steps to Plot Points Using Symmetry:
A circle is symmetric in all quadrants. If you calculate one point (x, y)
in the first octant (x ≤ y), you can use symmetry to find the corresponding
points in the other seven octants:
(x, y), (y, x), (−x, y), (−y, x), (−x, −y), (−y, −x), (x, −y), (y, −x)
Initial Setup:
Start at (x, y) = (0, r), where r is the radius of the circle. The initial
decision parameter is:
p0 = 1 − r
Iteration for the First Octant:
Calculate points in the first octant using the following rules:
If pk < 0 : T henextpointis(x + 1, y), pk+1 = pk + 2xk + 3
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If pk ≥ 0 : T henextpointis(x + 1, y − 1), pk+1 = pk + 2xk − 2yk + 5
Example: Plot Points for a Circle of Radius r = 6:
Step − by − StepCalculation :
InitialP oint :(x0 , y0 ) = (0, 6)
InitialDecisionP arameter :p0 = 1 − 6 = −5
Iteration pk (x, y)
0 −5 (0, 6)
1 −2 (1, 6)
2 3 (2, 6)
3 0 (3, 5)
4 7 (4, 5)
5 6 (5, 4)
Final Plot: After calculating points in the first octant, we plot the points
for all octants using symmetry.
2. Cohen-Sutherland Line Clipping Algorithm (Numer-
ical)
Given endpoints (5, 8) and (15, 20), and window boundaries xmin = 4, xmax =
12, ymin = 6, ymax = 18:
Compute Outcodes:
P1 (5, 8) : outcode = 0000
P2 (15, 20) : outcode = 1010
Clip with window boundaries:
Clip P2 at xmax = 12, compute intersection:
20 − 8
y = y1 + m(xmax − x1 ), m= = 1.2
15 − 5
N ewP2 = (12, 14.4)
Final clipped line: (5, 8) to (12, 14.4).
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3. Window-to-Viewport Transformation
Given window (2, 2) to (10, 8), and viewport (1, 1) to (5, 5):
Mapping Formula:
x − xw y − yw
x ′ = xv · + xmin
v , y ′ = yv · + yvmin
xmax
w − xwmin ywmax
− ywmin
Apply for point (4, 6):
4−2 6−2
x′ = 1 + (5 − 1) · = 2, y ′ = 1 + (5 − 1) · = 3.67
10 − 2 8−2
Mapped point: (2, 3.67).
