Graphics Devices and Display Systems Questions and Answers

Answer: D

In the RGB colour model, a colour is given by the amounts of red, green and blue, each from 0 to 255. Pure green is 0 red, 255 green and 0 blue, so the answer is 0 : 255 : 0. The first option, 255 : 0 : 0, is red, 255 : 255 : 255 is white and 0 : 0 : 0 is black.

Answer: B

In the formula \(R_{D} = 1 - \dfrac{1}{C_{R}}\), \(R_{D}\) is the relative data redundancy of the first data set. When the compression ratio \(C_{R} = n_{1}/n_{2}\) equals 1, \(R_{D}\) is 0, meaning the first set has no redundant data relative to the second; as \(C_{R}\) grows, \(R_{D}\) approaches 1, meaning highly redundant data.

It is not called relative data compression, relation or representation, since the quantity measures the redundant part of the data.

Answer: D

Only statement III is not true, so the answer is III only. A trackball is a two dimensional positioning device, while a spaceball provides six degrees of freedom, so I is true. A spaceball does not actually move, because it senses the pressure applied to a sphere, so II is true. Statement III calls the trackball a three dimensional device, which contradicts I.

The options naming I or II as untrue are wrong because those statements are correct.

Answer: C

These rules are applied in animation. Squash and stretch, slow-in and slow-out, and staging the action properly are three of the basic principles of traditional animation, used to give motion a natural and readable look.

Rendering turns a scene description into an image, and morphing is a smooth transformation between two images or shapes, so neither of those is defined by these three rules.

Answer: C

The conversion is R, G, B equal to the column (1, 1, 1) minus the column (C, M, Y). The CMY components are the complements of the RGB components, so \(C = 1 - R\), \(M = 1 - G\) and \(Y = 1 - B\), and therefore \(R = 1 - C\), \(G = 1 - M\), \(B = 1 - Y\). This is option C.

Options A, B and D subtract the constants from the CMY column instead of subtracting the CMY column from 1, so they reverse the operation.

Answer: A

An actor in an animation is a small program invoked once per frame to determine the characteristics of some object in the animation: on each frame it is run a single time to decide where the object is and how it looks. The answer is (A).

Running it twice, 30 times or 60 times per frame would repeat the computation with no gain, since the frame shows only one state of the object. The rates of 30 or 60 relate to the number of frames per second, not to the invocations per frame.

Answer: B

Bresenham's line drawing algorithm is attractive because it uses integer arithmetic only. It keeps a decision variable that is updated by adding integers at each step, so it avoids the multiplications and the rounding of the floating-point calculation used in the direct and the DDA methods. The answer is (B).

Options (A), (C) and (D) all involve real or floating-point arithmetic, which is exactly what the algorithm is designed to avoid.

Answer: C

The refresh rate above which a picture stops flickering and fuses into a steady image is called the critical fusion frequency. It depends on the brightness of the picture and on the viewer, and it is generally in the range of 50 to 60 times a second for normal displays. The answer is (C).

'Crucial fusion frequency', 'current frequency fusion' and 'critically diffused frequency' are not terms in computer graphics.

Answer: A

If 40 black lines interleaved with 40 white lines can be distinguished across one inch, the resolution is 40 line-pairs per inch. A line pair is one black line together with the adjacent white line, so 40 black and 40 white lines make 40 pairs. The answer is (A).

80 line-pairs would need 160 lines, 1600 lines per inch is the square of 40 and has no meaning here, and 40 lines per inch counts only the black lines, but the standard unit of resolution is the line pair.

Answer: B

A 1024 by 1024 image has \(1024 \times 1024\) pixels, each held in n bits, so

\[ \text{bits} = 1024 \times 1024 \times n \]

\[ \text{bytes} = \frac{1024 \times 1024 \times n}{8} \]

and, since \(1 \text{ MB} = 1024 \times 1024\) bytes, the memory required is \(n / 8\) MB. The answer is (B). Multiplying by 8 gives bits, not megabytes, and (C) and (D) contain no n at all, so they ignore the depth of colour.