Solving problems on encoding graphic information

Solving problems on coding graphic information.

Raster graphics.

Vector graphics.

Introduction

This electronic manual contains a group of tasks on the topic “Coding graphic information.” The collection of problems is divided into types of problems based on the specified topic. Each type of task is considered taking into account a differentiated approach, i.e. tasks of a minimum level (score “3”), general level (score “4”), and advanced level (score “5”) are considered. The given problems are taken from various textbooks (list attached). Solutions to all problems are examined in detail, methodological recommendations are given for each type of problem, and brief theoretical material is given. For ease of use, the manual contains links to bookmarks.

Raster graphics.

Types of tasks:

1. Finding the amount of video memory.

2. Determining screen resolution and setting graphics mode.

1. Finding the amount of video memory

In tasks of this type the following concepts are used:

· video memory volume,

· graphics mode,

· color depth,

· screen resolution,

· palette.

In all such problems, you need to find one or another quantity.

Video memory - this is special RAM, in which a graphic image is formed. In other words, in order to receive a picture on the monitor screen, it must be stored somewhere. That's what video memory is for. Most often, its value is from 512 KB to 4 MB for the best PCs with the implementation of 16.7 million colors.

Video memory capacity calculated by the formula: V=I*X*Y, whereI– color depth of an individual point, X,Y – screen dimensions horizontally and vertically (the product of x and y is the screen resolution).

The display screen can operate in two main modes: text And graphic.

IN graphic mode the screen is divided into separate luminous points, the number of which depends on the type of display, for example 640 horizontally and 480 vertically. Glowing dots on the screen are usually called pixels, their color and brightness may vary. It is in the graphic mode that all complex graphic images created by the computer appear on the computer screen. special programs, which control the parameters of each pixel of the screen. Graphic modes are characterized by such indicators as:

- resolution(the number of dots with which the image is reproduced on the screen) - currently typical resolution levels are 800 * 600 dots or 1024 * 768 dots. However, for monitors with a large diagonal, a resolution of 1152 * 864 pixels can be used.

- color depth(number of bits used to encode the color of a dot), e.g. 8, 16, 24, 32 bits. Each color can be considered as a possible state of a point. Then the number of colors displayed on the monitor screen can be calculated using the formula K=2 I, Where K– number of flowers, I– color depth or bit depth.

In addition to the knowledge listed above, the student should have an idea of the palette:

- palette(number of colors used to reproduce an image) e.g. 4 colors, 16 colors, 256 colors, 256 shades gray, 216 colors in a mode called High color or 224, 232 colors in True color mode.

The student must also know the connections between units of measurement of information, be able to convert from small units to larger ones, Kbytes and Mbytes, use a regular calculator and Wise Calculator.

Level "3"

1. Determine the required amount of video memory for various graphic modes of the monitor screen, if the color depth per dot is known. (2.76)

Screen mode	Color depth (bits per dot)

Solution:

1. Total dots on the screen (resolution): 640 * 480 = 307200
2. Required amount of video memory V= 4 bits * 307200 = 1228800 bits = 153600 bytes = 150 KB.
3. The required amount of video memory for other graphics modes is calculated in the same way. When making calculations, the student uses a calculator to save time.

Answer:

Screen mode	Color depth (bits per dot)

	150 KB	300 KB	600 KB	900 KB	1.2 MB
	234 KB	469 KB	938 KB	1.4 MB	1.8 MB
	384 KB	768 KB	1.5 MB	2.25 MB
	640 KB	1.25 MB	2.5 MB	3.75 MB

2. Black and white (no grayscale) raster graphic image has a size of 10 ´10 points. How much memory will this image take up?(2.6 8 )

Solution:

1. Number of points -100

2. Since there are only 2 colors: black and white. then the color depth is =2)

3. The amount of video memory is 100*1=100 bits

Problem 2.69 is solved in a similar way

3. To store a bitmap of size 128 x 128 pixels took up 4 KB of memory. What is the maximum possible number of colors in the image palette. (USE_2005, demo, level A). (See also Problem 2.73 )

Solution:

1. Determine the number of image points. 128*128=16384 points or pixels.

2. The amount of memory for a 4 KB image can be expressed in bits, since V=I*X*Y is calculated in bits. 4 KB=4*1024=4096 bytes = 4096*8 bits =32768 bits

3. Find the color depth I =V/(X*Y)=32768:16384=2

4. N=2I, where N is the number of colors in the palette. N=4

Answer: 4

4. How many bits of video memory does information about one pixel on a b/w screen (without halftones) take?(, P. 143, example 1)

Solution:

If the image is B/W without halftones, then only two colors are used - black and white, i.e. K = 2, 2i = 2, I = 1 bits per pixel.

Answer: 1 pixel

5. How much video memory is needed to store four pages of images if the bit depth is 24 and the display resolution is 800 x 600 pixels? (, No. 63)

Solution:

1. Find the amount of video memory for one page: 800*600*24= bits = 1440000 bytes = 1406.25 KB ≈1.37 MB

2. 1.37*4 =5.48 MB ≈5.5 MB for storing 4 pages.

Answer: 5.5 MB

Level "4"

6. Determine the amount of computer video memory that is required to implement the monitor’s graphic mode High Color with a resolution of 1024 x 768 pixels and a color palette of 65536 colors. (2.48)

If the student remembers that the High Color mode is 16 bits per dot, then the amount of memory can be found by determining the number of dots on the screen and multiplying by the color depth, i.e. 16. Otherwise, the student can reason like this:

Solution:

1. Using the formula K=2I, where K is the number of colors, I is the color depth, we determine the color depth. 2I =65536

The color depth is: I = log= 16 bits (calculated using programsWiseCalculator)

2.. The number of image pixels is: 1024´768 =

3. The required amount of video memory is: 16 bits ´ = 12 bits = 1572864 bytes = 1536 KB = 1.5 MB (»1.2 MB. The answer was given in the workshop by Ugrinovich). We teach students, when converting to other units, to divide by 1024, not 1000.

Answer: 1.5 MB

7. During the raster conversion process graphic image the number of colors has decreased from 65536 to 16. How many times will the amount of memory it takes up decrease? (2.70, )

Solution:

To encode 65536 different colors for each point, 16 bits are needed. To encode 16 colors, only 4 bits are needed. Consequently, the amount of memory occupied has decreased by 16:4=4 times.

Answer: 4 times

8. Is 256 KB of video memory enough to operate the monitor in 640 mode? ´ 480 and a palette of 16 colors? (2.77)

Solution:

1. Find out the amount of video memory that will be required to operate the monitor in 640x480 mode and a palette of 16 colors. V=I*X*Y=640*480*4 (24 =16, color depth is 4),

V= 1228800 bits = 153600 bytes = 150 KB.

2. 150 < 256, значит памяти достаточно.

Answer: enough

9. Specify the minimum amount of memory (in kilobytes) required to store any 256 x 256 pixel bitmap image if you know the image uses a palette of 216 colors. There is no need to store the palette itself.

1) 128

2) 512

3) 1024

4) 2048

(USE_2005, level A)

Solution:

Let's find the minimum amount of memory required to store one pixel. The image uses a palette of 216 colors, therefore, one pixel can be associated with any of 216 possible color numbers in the palette. Therefore, the minimum amount of memory for one pixel will be equal to log2 216 = 16 bits. The minimum amount of memory sufficient to store the entire image will be 16 * 256 * 256 = 24 * 28 * 28 = 220 bits = 220: 23 = 217 bytes = 217: 210 = 27 KB = 128 KB, which corresponds to point number 1.

Answer: 1

10. Graphic modes with color depths of 8, 16, 24, 32 bits are used. Calculate the amount of video memory required to implement these color depths at different screen resolutions.

Note: the task ultimately comes down to solving problem No. 1 (level “3”, but the student himself needs to remember the standard screen modes.

11. How many seconds will it take for a modem transmitting messages at a speed of 28800 bps to transmit color raster image 640 x 480 pixels in size, assuming that the color of each pixel is encoded in three bytes? (USE_2005, level B)

Solution:

1. Determine the image volume in bits:

3 bytes = 3*8 = 24 bits,

V=I*X*Y=640*480*24 bits =7372800 bits

2. Find the number of seconds to transmit an image: 7372800: 28800=256 seconds

Answer: 256.

12. How many seconds will it take for a modem transmitting messages at a speed of 14400 bps to transmit a color bitmap image measuring 800 x 600 pixels, assuming that there are 16 million colors in the palette? (USE_2005, level B)

Solution:

To encode 16 million colors, 3 bytes or 24 bits are required (True Color graphics mode). The total number of pixels in the image is 800 x 600 = 480000. Since there are 3 bytes per 1 pixel, then for 480,000 pixels there are 480,000 * 3 = 1,440,000 bytes or bits. : 14400 = 800 seconds.

Answer: 800 seconds.

13. Modern monitor allows you to get different colors on the screen. How many bits of memory does 1 pixel take? ( , p.143, example 2)

Solution:

One pixel is encoded by a combination of two characters “0” and “1”. We need to find out the pixel code length.

2x =, log2 =24 bits

Answer: 24.

14. What is the minimum amount of memory (in bytes) sufficient to store a black-and-white raster image measuring 32 x 32 pixels, if it is known that the image uses no more than 16 shades of gray. (USE_2005, level A)

Solution:

1. The color depth is 4, because 16 color gradations are used.

2. 32*32*4=4096 bits of memory for storing black and white images

3. 4096: 8 = 512 bytes.

Answer: 512 bytes

Level "5"

15. The monitor works with a 16 color palette in 640*400 pixels mode. Encoding an image requires 1250 KB. How many pages of video memory does it take up? (Task 2, Test I-6)

Solution:

1. Because page – a section of video memory that contains information about one screen image of one “picture” on the screen, i.e. several pages can be placed in video memory at the same time, then to find out the number of pages you need to divide the amount of video memory for the entire image by the amount of memory per 1 page. TO-number of pages, K=Vimage/V1 page

Vimage = 1250 KB according to condition

1. To do this, let's calculate the amount of video memory for one image page with a 16 color palette and a resolution of 640*400.

V1 page = 640*400*4, where 4 is the color depth (24 =16)

V1 page = 1024000 bits = 128000 bytes = 125 KB

3. K=1250: 125 =10 pages

Answer: 10 pages

16. A video memory page is 16,000 bytes. The display operates in 320*400 pixel mode. How many colors are in the palette? (Task 3, Test I-6)

Solution:

1. V=I*X*Y – volume of one page, V=16000 bytes = 128000 bits according to the condition. Let's find the color depth I.

I= 128000 / (320*400)=1.

2. Let us now determine how many colors are in the palette. K =2 I, Where K– number of flowers, I– color depth . K=2

Answer: 2 colors.

17. A color image of size 10 is scanned ´10 cm. Scanner resolution 600 dpi and color depth 32 bits. What information volume will the resulting graphic file have? (2.44, , problem 2.81 is solved similarly )

Solution:

1. A scanner resolution of 600 dpi (dots per inch) means that in a 1-inch segment the scanner is able to distinguish 600 dots. Let's convert the scanner resolution from dots per inch to dots per centimeter:

600 dpi: 2.54 » 236 dots/cm (1 inch = 2.54 cm)

2. Therefore, the image size in pixels will be 2360´2360 pixels. (multiplied by 10 cm.)

3. The total number of image pixels is:

4. The information volume of the file is:

32 bits ´ 5569600 = bits » 21 MB

Answer: 21 MB

18. The amount of video memory is 256 KB. The number of colors used is 16. Calculate display resolution options. Provided that the number of image pages can be 1, 2 or 4. (, No. 64, p. 146)

Solution:

1. If the number of pages is 1, then the formula V=I*X*Y can be expressed as

256 *1024*8 bits = X*Y*4 bits, (since 16 colors are used, the color depth is 4 bits.)

i.e. 512*1024 = X*Y; 524288 = X*Y.

The ratio between the height and width of the screen for standard modes does not differ from each other and is equal to 0.75. This means that to find X and Y, you need to solve the system of equations:

Let's express X=524288/Y, substitute it into the second equation, we get Y2 =524288*3/4=393216. Let's find Y≈630; X=524288/630≈830

630 x 830.

2. If the number of pages is 2, then one page with a volume of 256:2 = 128 KB, i.e.

128*1024*8 bits = X*Y*4 bits, i.e. 256*1024 = X*Y; 262144 = X*Y.

We solve the system of equations:

X=262144/Y; Y2 =262144*3/4=196608; Y=440, X=600

The resolution option could be 600 x 440.

4. If the number of pages is 4, then 256:4 =64; 64*1024*2=X*Y; 131072=X*Y; We solve the system and the screen point size is 0.28 mm. (2.49)

Solution:

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1. The problem comes down to finding the number of dots across the width of the screen. Let's express diagonal size in centimeters. Considering that 1 inch = 2.54 cm, we have: 2.54 cm 15 = 38.1 cm.

2. Let's define ratio between screen height and width ana for the frequently used screen mode 1024x768 pixels: 768: 1024 = 0.75.

3. Let's define screen width. Let the screen width be L, and the height h,

h:L =0.75, then h= 0.75L.

According to the Pythagorean theorem we have:

L2 + (0.75L)2 = 38.12

1.5625 L2 = 1451.61

L ≈ 30.5 cm.

4. The number of dots across the screen width is:

305 mm: 0.28 mm = 1089.

Therefore, the maximum possible monitor screen resolution is 1024x768.

Answer: 1024x768.

26. Determine the relationship between the height and width of the monitor screen for various graphics modes. Does this ratio differ for different modes? a) 640x480; b)800x600; c) 1024x768; a) 1152x864; a) 1280x1024. Determine the maximum possible screen resolution for a 17" diagonal monitor with a screen dot size of 0.25 mm. (2.74 )

Solution:

1. Let's determine the relationship between the height and width of the screen for the listed modes; they hardly differ from each other:

2. Let's express the diagonal size in centimeters:

2.54 cm 17 = 43.18 cm.

3. Let's determine the width of the screen. Let the screen width be L, then the height is 0.75L (for the first four cases) and 0.8L for the last case.

According to the Pythagorean theorem we have:

Therefore, the maximum possible resolution of the monitor screen is. 1280x1024

Answer: 1280x1024

3. Color and image coding.

Students use the knowledge acquired previously Number systems, converting numbers from one system to another.

Theoretical material of the topic is also used:

A color raster image is formed in accordance with the RGB color model, in which the three basic colors are Red, Green and Blue. The intensity of each color is specified by an 8-bit binary code, which is often expressed in terms of hexadecimal system Reckoning. In this case, the following record format is RRGGBB.

Level "3"

27. Write down the red color code in binary, hexadecimal and decimal notation. (2.51)

Solution:

Red color corresponds to the maximum value of red color intensity and minimum values intensities of green and blue basic colors , which corresponds to the following data:

Codes/Colors	Red	Green	Blue
binary
hexadecimal
decimal

28. How many colors will be used if 2 levels of brightness gradation are taken for each pixel color? 64 brightness levels for each color?

Solution:

1. In total, for each pixel a set of three colors is used (red, green, blue) with their own brightness levels (0-on, 1-off). So K=23 =8 colors.

Answer: 8; 262,144 colors.

Level "4"

29. Fill out the color table at 24-bit color depth in hexadecimal notation.

Solution:

With a color depth of 24 bits, 8 bits are allocated for each color, i.e. for each color there are 256 intensity levels possible (28 = 256). These levels are specified in binary codes (minimum intensity, maximum intensity). IN binary representation The following color formation is obtained:

Color name	Intensity
Red	Green	Blue
Black
Red
Green
Blue
White

Converting to hexadecimal number system we have:

Color name	Intensity
Red	Green	Blue
Black
Red
Green
Blue
White

30. On a “small monitor” with a 10 x 10 raster grid there is a black and white image of the letter “K”. Represent the contents of video memory as a bit matrix in which the rows and columns correspond to the rows and columns of the raster grid. ( , p.143, example 4)

9 10

Solution:

Encoding the image on such a screen requires 100 bits (1 bit per pixel) of video memory. Let "1" mean a filled pixel, and "0" mean an unfilled pixel. The matrix will look like this:

0001 0001 00

0001 001 000

0001 01 0000

00011 00000

0001 01 0000

0001 001 000

0001 0001 00

Experiments:

1. Search for pixels on the monitor.

Arm yourself with a magnifying glass and try to see the triads of red, green and blue (RGB – from English. "Red –Green –Blue" dots on the monitor screen. (, .)

As the primary source warns us, the results of experiments will not always be successful. The reason is this. What exist different technologies manufacturing cathode ray tubes. If the tube is made using technology "shadow mask" then you can see a real mosaic of dots. In other cases, when instead of a mask with holes, a system of phosphor threads of three primary colors is used (aperture grille), the picture will be completely different. The newspaper provides very visual photographs of three typical paintings that “curious students” can see.

It would be useful for the guys to be informed that it is advisable to distinguish between the concepts of “screen points” and pixels. The concept of "screen points"- physically really existing objects. Pixels- logical elements of the image. How can this be explained? Let's remember. That there are several typical image configurations on a monitor screen: 640 x 480, 600 x 800 pixels and others. But you can install any of them on the same monitor. This means that pixels are not monitor points. And each of them can be formed by several neighboring luminous points (within one). Upon command to color one or another pixel blue, the computer, taking into account the set display mode, will color one or more adjacent points on the monitor. Pixel density is measured as the number of pixels per unit length. The most common units are called briefly as (dots per inch - the number of dots per inch, 1 inch = 2.54 cm). The dpi unit is generally accepted in the field computer graphics and publishing. Typically, the pixel density for a screen image is 72 dpi or 96 dpi.

2. Conduct an experiment in graphic editor what if for each pixel color there are 2 levels of brightness gradation? What colors will you receive? Present it in the form of a table.

Solution:

Red	Green	Blue	Color





			Turquoise

			Crimson

Vector graphics:

1. Vector image coding tasks.

2. Getting a vector image using vector commands

With the vector approach, the image is considered as a description of graphic primitives, lines, arcs, ellipses, rectangles, circles, shades, etc. The position and shape of these primitives in the graphic coordinate system are described.

Thus, a vector image is encoded with vector commands, that is, described using an algorithm. A straight line segment is determined by the coordinates of its ends, circle – center coordinates and radius, polygon– coordinates of its corners, shaded area- border line and shading color. It is advisable for students to have a vector graphics command system table (, p.150):

Team	Action
Line to X1, Y1	Draw a line from the current position to position (X1, Y1).
Line X1, Y1, X2,Y2	Draw a line with start coordinates X1, Y1 and end coordinates X2, Y2. The current position is not set.
Circle X, Y, R	Draw a circle; X, Y are the coordinates of the center, and R is the length of the radius.
Ellipse X1, Y1, X2,Y2	Draw an ellipse bounded by a rectangle; (X1, Y1) are the coordinates of the upper left, and (X2, Y2) are the coordinates of the lower right corner of the rectangle.
Rectangle X1, Y1, X2,Y2	Draw a rectangle; (X1, Y1) - coordinates of the left top corner, (X2,Y2) - coordinates of the lower right corner of the rectangle.
Drawing Color Color	Set the current drawing color.
Fill color Color	Set current fill color
Fill X, Y, BORDER COLOR	Shade arbitrary closed figure; X, Y – coordinates of any point inside a closed figure, BORDER COLOR – color of the boundary line.

1. Vector image coding tasks.

Level "3"

1. Describe the letter “K” with a sequence of vector commands.

Literature:

1., Computer Science for Lawyers and Economists, p. 35-36 (theoretical material)

2. , Computer Science and IT, pp. 112-116.

3. N. Ugrinovich, L. Bosova, N. Mikhailova, Workshop on Informatics and IT, pp. 69-73. (problems 2.67-2.81)

4. Popular lectures on computer design. – St. Petersburg, 2003, pp. 177-178.

5. In search of a pixel or types of cathode ray tubes. // Computer science. 2002, 347, pp. 16-17.

6. I. Semakin, E Henner, Computer Science. Problem book-workshop, vol. 1, Moscow, LBZ, 1999, pp. 142-155.

Electronic textbooks:

1. , Information in the school computer science course.

2. , A workbook on the topic “Information Theory”

Tests:

1. Test I-6 (Coding and Measuring Graphic Information)

Encoding graphic information

Example 1. The computer's video memory has a capacity of 512Kb, the graphics grid size is 640´200, and the palette has 16 colors. How many screen pages can be simultaneously stored in the computer's video memory?

Given :

K=640 ´ 200=128000 pixels;

N=16 colors;

Solution

Calculation of the information volume of a raster graphic image (the amount of information contained in a graphic image) is based on counting the number of pixels in this image and determining the color depth (the information weight of one pixel).

So, to calculate the information volume of a raster graphic image, the formula V=K*i is used, where V is the information volume of a raster graphic image, measured in bytes, kilobytes, megabytes; K – the number of pixels (dots) in the image, determined by the resolution of the information carrier (monitor screen, scanner, printer); i – color depth, which is measured in bits per pixel.

Color depth is specified by the number of bits used to encode the color of a dot.

Color depth is related to the number of colors displayed by the formula

N=2 i, where N is the number of colors in the palette, i is the color depth in bits per pixel.

We use the formulas:

V=K*i; N=2 i ; m= VVP/V, where m is the number of screen pages

16=2 4 ® i=4 bits/pixel;

K=640 ´ 200=128000pixels

V=128000*4=512000bit=64000byte=62.5Kb per screen

M=512/62.5=8 pages

Answer: 8 full screen pages can be stored simultaneously in the computer's video memory.

Example 2. During the raster conversion process graphic file the number of colors decreased from 1024 to 32. How many times did the information volume of the file decrease?

a) 5 b) 2 c)3 d)4

Solution

We use the formula N=2 i, where N is the number of colors in the image, i– number of bits allocated for each pixel. Then 1024=2 i, hence, i= 10 bits. After converting the file 32=2 i, i=5, the information volume of the file has decreased by 2 times.

Answer: the information volume of the file has decreased by 2 times.

Example 3. Monitor screen resolution is 1024x768 pixels, color depth is 16 bits. What is the required amount of video memory for this graphics mode?

a) 256 bytes b) 4 KB c) 1.5 MB d) 6 MB

Solution

Total dots on screen: 1024×768=786432

Required amount of video memory 16 bits × 786432 = 12,582,912 bits = 1,572,864 bytes = 1536 KB = 1.5 MB

Answer: 1.5 MB required amount of video memory for this graphics mode.

Coding audio information

Solution

Audio adapter (sound card) – a device that converts electrical vibrations of sound frequency into a numerical binary code and vice versa.

Sampling frequency– is the number of measurements of the input signal in 1 second.

Typical sampling rates of audio adapters:

11 kHz (8 bits) - conversation,

Formula for calculating the size of a digital audio file:

V = f * I * t,

Where N is the number of volume levels, i is the sound depth (bits).

Duration of a musical piece: 4 min = 240 sec;

Sound quality = Quantization level ⋅ Sampling frequency.

Quantization level: 16 bits (depth of each piece);

Sampling frequency: 44.1 kHz = 44100 Hz = 44100 s -1 ;

Sound quality = 16 bits ⋅ 44100 s -1 = 705600 bps;

The volume of a musical work = 240 seconds ⋅ 705600 bits/sec = 169344000 bits = 19.6 MB.

Answer: 19.6 MB volume of a piece of music.

Example 2. It is necessary to estimate the information volume of a stereo audio file with a sound duration of 1 second at high quality audio (16 bits, 48 kHz). To do this, the number of bits per sample must be multiplied by the number of samples per second and multiplied by 2 (stereo):

Solution

V = f * I * t,

where: f – sampling frequency (Hz); I – sound depth (bits); t – playing time (sec).

V=16 bits × 48,000 × 2 = 1,536,000 bits = 192,000 bytes = 187.5 KB.

Answer: 187.5 K bytes is the information volume of a stereo audio file.

Example 3. Estimate the information volume of a digital stereo sound file with a sound duration of 1 minute with average sound quality (16 bits, 24 kHz).

Solution

To solve, we use the formula for calculating the size of a digital audio file:

V = f * I * t,

where: f – sampling frequency (Hz); I – sound depth (bits); t – playing time (sec).

V =16 bits × 24,000 × 2 × 60 = 46,080,000 bits = 5,760,000 bytes =

5625 KB ≈ 5.5 MB.

Monitor screen resolution is 1024 * 768 pixels, color depth is 16 bits. What is the required amount of video memory for this graphics mode?

1) 256 bytes 2) 4 KB 3) 1.5 MB 4) 6 MB

The RGB color model uses 3 bytes to encode one pixel. A photograph measuring 2048 * 1536 pixels was saved as an uncompressed file using RGB encoding. Determine the size of the resulting file in megabytes.

What is the minimum amount of memory (in bytes) sufficient to store any 32*32 pixel black and white bitmap image if the image is known to use no more than 16 shades of gray?

Specify the minimum amount of memory (in kilobytes) sufficient to store any 64 * 64 pixel bitmap image if the image is known to have a palette of 256 colors. There is no need to store the palette itself.

1) 128 2) 256 3) 2 4) 4.

Specify the minimum amount of memory (in bytes) required to store any 8 * 32 pixel bitmap image if the image is known to have a palette of 256 colors. There is no need to store the palette itself.

1) 128 2) 256 3) 512 4) 1024.

Specify the minimum amount of memory that is sufficient to store a 64-color, 32-by-128-pixel bitmap graphic. There is no need to store the palette itself.

1) 32 KB 2) 64 bytes 3) 4096 bytes 4) 3 KB.

1) 2 2) 3 3) 4 4) 5.

In the process of converting a raster graphics file, the number of colors decreased from 1024 to 32. How many times did the information volume of the file decrease?

1) 2 2) 3 3) 4 4) 5.

In the process of converting a raster graphics file, the number of colors decreased from 512 to 8. How many times did the information volume of the file decrease?

1) 5 2) 2 3) 3 4) 4.

After converting a 256-color raster graphic file to black and white (2 colors), its size was reduced by 70 bytes. What was the size of the original file?

1) 70 bytes 2) 640 bits 3) 80 bits 4) 560 bits.

The book, consisting of 1360 pages, occupies a volume of 40 MB. Some of the book's pages are color images in 320 * 640 pixels format. One page of a book with text contains 1024 characters. Each character is encoded with 1 byte. Number of pages with text per 560 more quantity pages with color images. How many colors are used to represent the book's images? Write your answer as an integer.

Color depth/ Palette

The color of a pixel produced by a printer is determined by three components: cyan, magenta and yellow. For each component of one pixel, 4 bits were allocated. How many colors can a pixel be painted in?

The color of a monitor pixel is determined by three components: green, blue and red. 5 bits were allocated for the red and blue components of one pixel. How many bits are allocated for the green component of one pixel if a raster image of 8 * 8 pixels takes up 128 bytes of memory?

1) 5 2) 6 3) 8 4) 16.

In the process of converting a raster graphic file, its volume decreased by 1.5 times. How many colors were in the palette initially if, after conversion, a raster image of the same resolution was obtained in a 256-color palette?

To store a raster image measuring 64 by 64 pixels, 512 bytes of memory were allocated. What is the maximum possible number of colors in the image palette?

1) 16 2) 2 3) 256 4) 1024.

To store a raster image measuring 128 * 128 pixels, 4 kilobytes of memory were allocated. What is the maximum possible number of colors in the image palette?

1) 8 2) 2 3) 16 4) 4.

The monitor allows you to receive 16,777,216 colors on the screen. How much memory in bytes does 1 pixel take up?

1) 2 2) 3 3) 4 4) 5.

Definition of color

1) white; 2) red; 3) green; 4) blue.

What color will the page specified by the tag have?

1) white; 2) green; 3) purple; 4) black.

What color will the page specified by the tag have?

1) white; 2) blue; 3) gray; 4) black.

1) black; 2) yellow; 3) purple; 4) white.