Computer Graphics
Q.1 The process of selecting and viewing the picture with
different views is called _________
A.
Windowing B. Clip window C. Clipping D. Rasterization
Q.2 A process which
divides each element of the picture into its visible and invisible portions,
allowing
the invisible portion to be
discarded is called _________
A.
Windowing B. Clip window C. Clipping D. Rasterization
Q.3 The region against which an object is to be clipped is
called _________
A. .
Windowing B. Clip window C. Clipping D. Rasterization
Q.4 The four bit codes that are used to indicate the end
point of line is called ________
A. Region
codes B. outcodes C. Either a or b D. None
Q.5 The _______ can be clipped against rectangle by first
performing a trival accept/ reject test .
A. Circles B. Rectangles C. Squares D. Polygons
Q.6 A _______ is nothing but the collection of lines.
A. Circles B. Rectangles C. Squares D. Polygons
Q.7 the memory area which is dedicated to graphics only is
called _____
A. Frame
buffer B. RAM C. ROM D. None
Q.8 Shift register is operated as _______
A. LIFO B. FILO C. FIFO D. LILO
Q.9 What is the polynomial equation for calculating circle.
A. X2+y2 = r2 B. X2 = y2 -
r2 C. X2 + r2 = y2
D. None
Q.10
In the display controller architecture –
A. the memory mapping allows the CPU
to access the frame buffer directly.
B. the display controller to access
the CPUs memory.
C. there is a separate provision to
avoid memory conflicts.
D. All the above
Q.11
The desired characteristics of line are –
1. The line should appear as a
straight line.
2. It should start and end at the
same point.
3. The line should be drawn rapidly.
4. The line should be displayed with
fading brightness.
A. 1,2,3,4 B. 2,3,4 C.
1,3,4 D. None of the
above
Q.12
The slope of a straight line is given as –
A. m = Δx / Δy = y2-y1 / x2-x1
B. m = Δx / Δy = x2-x1 / y2-y1
C. m = Δy / Δx = x1 – x2 / y1 – y2 D.
m = Δx / Δy = y2-y1
/ x2-x1
Q.13
Consider the line from (0,0) to (4,6). Using simple DDA algorithm rasterize
this line up to third level.
What would be the output.
A. (0,0) = x = 0.5
y = 0.5 B. (0,0) = x
= 0.5 y= 0.5
(0,1) = x = 0.5 y = 1.0 (1,1) = x = 1.167 y = 1.5
(1,1) = x = 1.0 y = 1.0 (1,2) = x= 1.833 y = 2.5
Q.14
Sutherland Cohen algorithm was developed by which of the following scientist
A. Dan Cohen and Ivan Sutherland B. Ivan Cohen and Dan Sutherland
Q.15
Advantages of DDA algorithm are –
1. It is the simplest algorithm and
does not require special skills for implementation.
2. It is a faster method for
calculating pixel positions than the direct use of equation y = mx + b
3. Floating point arithmetic in DDA
is still time consuming.
4. The algorithm is orientation
dependent.
A. 1,2,3,4 B. 1,3,4 C.
1,2 D. All are
disadvantages
Q.16 If the midpoint lies above the line __A____,
if the midpoint lies below the line ____B____
1. pixel B is closer to the line 2. pixel A is closer
to the line.
A. A – 1 B – 2 B. A – 2 B
– 1
Q.17
Three different circle drawing algorithms are –
1. DDA algo. 2. Bresenham algo. 3.
Midpoint circle drawing algo. 4.
Polynomial algo
A. 1,2,3 B. 2,3,4 C.
1,3,4 D. 1,2,4
Q.18
A _a_ polygon is a polygon in which the line segment joining any 2
points within the polygon lies
completely inside the polygon. A _b_
polygon is a polygon in which the line segment joining any
2 points within the polygon may not
lie completely inside the polygon.
A. a – concave b – convex B. a –
convex b – concave
Q.19
Match the following
SET A SET
B
1. Coherence a. it exists when all pixels on a span
are set to the same value.
2. Primitive’s spatical coherence b. it exists when consecutive scan lines
that intersect the rectangle are identical.
3. Span coherence c. It is the fact
that primitives often do not change from pixel to pixel with in a span or from
scan line to scan line.
4. Scan –line coherence d. it is a property of a
scene by which we can relate one part of a scene with the other parts of a
scene.
A. 1-a, 2-b, 3-c, 4-d B.
1-b, 2-c, 3-a, 4-d C. 1-c, 2-d, 3-b, 4-a D.
1-a, 2-d, 3-c, 4-b
Q.20
Match the following (Draw the lines in between)
SET A SET
B
1. SET BIT 1 a. if the end point is above
the window.
2. SET BIT 2 b. if the end point is to the
left of the window.
3. SET BIT 3 c. if the end point is below
the window.
4. SET BIT 4 d. if the end point is to the right
of the window.
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