The analysis of the task is 5 EGE on computer science.

The unified state exam in computer science consists of 27 tasks. In task 5, the skills of encoding and decoding information are checked. The schoolboy must be able to encode and decode information in various number systems, as well as decrypt messages and choose the optimal code. Here you can learn how to solve the task of 5 EGE on computer science, as well as study examples and ways to solve on the basis of detailed disassembled tasks.

All tasks EGE All tasks (107) Ege Quest 1 (19) Ege Quest 3 (2) EE Assignment 4 (11) EE Assignment 5 (10) EE Assignment 6 (7) EE Assignment 7 (3) EE Assignment 9 (5) EEG TASK 10 (7) EGE QUESTION 11 (1) EGE TASK 12 (3) EGE TASK 13 (7) EGE TASK 16 (19) EGE OPTION 17 (4) EGE without number (9)

For encoding letters decided to use binary performance

For encoding letters, the binary representation of numbers 0, 1, 2, 3 and 4 was decided to use, respectively (with the preservation of one inconcent zero in the case of a single-digit presentation). If you encode the sequence of letters in this way and the result is to record the octal code, it will turn out ...

For transmission over the communication channel, the message consisting only of characters

For transmission over the communication channel, a message consisting only of symbols A, B, B, and G is used by seductive encoding. A message is transmitted via the communication channel. Code the message to this code. The resulting binary number is transferred to a hexadecimal view.

The task enters the exam in computer science for grade 11 at number 5.

For encoding letters a, b, in, g decided to use two-digit

For encoding letters A, B, B, G decided to use two-digit sequential binary numbers (from 00 to 11, respectively). If in this way to encode the sequence of characters and record the resulting binary number in a hexadecimal number system, then it turns out ...

The task enters the exam in computer science for grade 11 at number 5.

Messages containing only 5 letters are transmitted via communication channel.

On the communication channel, messages containing only 5 letters are transmitted. For encoding letters used uneven binary code. Among the words below, specify this that can be decoded only in one way. If there are several such words, specify the first alphabet.

The task enters the exam in computer science for grade 11 at number 5.

You need to use uneven binary code to send messages

On the communication channel, messages containing only 4 letters are transmitted. To send messages, you need to use an unequomerous binary code that allows unambiguous decoding; At the same time, messages should be as short as possible. Encryptor can use one of the following codes. What code should he choose?

The task enters the exam in computer science for grade 11 at number 5.

For encoding a message consisting of only letters A, B, B and G

For encoding a message consisting of only letters A, B, B and G, an uneven binary code is used. If this way encodes the sequence of characters and record the result in hexadecimal code, then it will turn out ...

The task enters the exam in computer science for grade 11 at number 5.

For 5 letters of the Latin alphabet are given their binary codes.

For 5 letters of the Latin alphabet, their binary codes are set (for some letters - from two bits, for some of three). These codes are presented in the table. Determine what set of letters is encoded binary string?

The task enters the exam in computer science for grade 11 at number 5.

To transfer numbers by channel with interference, the parity check code is used

To transmit numbers through a channel with interference, the parity check code is used. Each digit is written in a binary representation, with the addition of leading zeros to length 4, and the sum of its elements modulo is added to the resulting sequence 2. Determine what number was transmitted via channel?

The task enters the exam in computer science for grade 11 at number 5.

5-bit code is used to transmit data via communication channel

A 5-bit code is used to transfer data via communication channel. The message contains only letters a, b and c, which are encoded by code words. When transmitted, interference is possible. However, some errors can try to fix. Any two of these three code words differ from each other at least in three positions. Therefore, if an error occurred in no more than one position when transferring the word, then you can make a reasonable assumption about what letter was transmitted. If the accepted code word differs from code words for letters a, b, in more than one position, it is believed that an error occurred (it is denoted by "x"). Received a message. Decod this message - select the correct option.

The task enters the exam in computer science for grade 11 at number 5.

For encoding some sequence consisting of letters

For encoding some sequence consisting of letters, an uneven binary prefix code is used. Is it possible to cut for one of the letters of the length of the code word so that the code remains to remain prefix? The codes of the remaining letters should not change. Select the correct answer option. Note. The prefix code is a code in which no codeword is the beginning of another; Such codes can uniquely decode the resulting binary sequence.

The task enters the exam in computer science for grade 11 at number 5.

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To encode some sequence consisting of letters to, l, m, n, decided to use an uneven binary code that satisfies the Fano condition. For the letter N used code word 0, for the letter to - code word 10. What is the smallest possible total length of all four code words?

Note.

Decision.

Find the shortest views for all letters. The code words 01 and 00 cannot be used, since the condition of Fano is disturbed. We use, for example, for the letter L code word 11. Then for the fourth letter it is impossible to choose a code word without disturbing the condition of Fano. Consequently, for the remaining two letters you need to use three-digit code words. Correct the letters L and M code words 110 and 111. Then the total length of all four code words is 1 + 2 + 3 + 3 \u003d 9.

Answer: 9.

Answer: 9.

For encoding some sequence consisting of letters A, B, B, G and D, an uneven binary code is used, which allows you to unambiguously decode the resulting binary sequence. This code: a - 1; B - 0100; In - 000; G - 011; D - 0101. It is required to reduce for one of the letters of the length of the code word so that the code can still be decoded unambiguously. The codes of the remaining letters should not change. What of the specified ways can this be done?

1) for the letter G - 11

2) for the letter in - 00

3) for the letter G - 01

4) it is impossible

Decision.

For one-to-member decoding, the code word resulting as a result should not be the beginning of any other. The first answer is not suitable, because the code of the letter A is the beginning of the code of the letter G. The second answer is suitable. The third version of the answer is not suitable, since, in this case, the code of the letter G is the beginning of the letter D.

The correct answer is subject to number: 2.

Answer: 2.

To encode some sequence consisting of letters and, k, l, m, n, decided to use an uneven binary code that satisfies the Fano condition. For the letter H used code word 0, for the letter K - code word 10. What is the smallest possible total length of all five code words?

Note. The Fano Condition means that no code word is the beginning of another code word. This provides the ability to unambiguously decrypt encoded messages.

Decision.

You can not use code words that start with 0 or from 10. 11 We also can't use, because then we will no longer be able to take any other codeword, and we need five. Therefore, we take three-digit 110. 111. Again, we cannot use it, because you need another code word, and at the same time there will be no more free. Now it remains to take only two words and it will be 1110 and 1111. Total we have 0, 10, 110, 1110 and 1111 - 14 characters.

Answer: 14.

Answer: 14.

To encode some sequence consisting of letters and, k, l, m, n, decided to use an uneven binary code that satisfies the Fano condition. For the letter l used code word 1, for the letter M - code word 01. What is the smallest possible total length of all five code words?

Note. The Fano Condition means that no code word is the beginning of another code word. This provides the ability to unambiguously decrypt encoded messages.

Decision.

Fano's condition - no code word can be the beginning of another code word. Since there is already a code word 1, no other can start with 1. only with 0. Also, it can not start with 01, since we already have 01. That is, any new code word will start from 00. But it can not To be 00, since otherwise we will not be able to take a single code word, because all the longer words start either from 1, or from 00, or from 01. We can take either 000 or 001. But not both immediately, since again In this case, we will no longer be able to take a single new code. Then Take 001. And since we have only two codes left, we can take 0000 and 0001. Total we have: 1, 01, 001, 0000, 0001. Total 14 characters.

Disassembly 5 of the tasks of the 2016 year on computer science from demoralism. This task is to encode and decode information (able to interpret the results obtained during the simulation of real processes). This is the task of the basic level of complexity. An approximate time of execution of the task is 2 minutes.

Task 5:

On the communication channel, messages containing only four letters are transmitted: P, O, C, T; For transmission, a binary code that allows unambiguous decoding is used. For the letters T, O, P code words are used: T: 111, A: 0, P: 100.
Specify the shortest code word for the letter C, in which the code will allow unambiguous decoding. If there are several such codes, specify the code with the smallest numeric value.

Answer: ________

HAPPENING 5 TASKS EGE 2016:

To solve this task, you need to know the condition of Fano.

Fano condition:
The encoded message can be unambiguously decoded if no code word is the end of another code word.

Reverse condition Fano:
The encoded message can be unambiguously decoded from the end if no code word is the end of another code word.

Let's start checking in order:

0 - can not be, since the O-0 (also the code word cannot begin with 0, since the condition of Fano is not completed),

1 - can not be, because from the unit begins T-111 and P-100,

10 - can not be, because with 10 begins P-100,

11 - can not be, since the T-111 begins with 11,

100 - can not be because P-100,

101 — suitable, since the condition of Fano is performed,

110 — suitableSince the Fano condition is performed.

By the condition of the task, if the words are somewhat, you need to select the code with the smallest numerical value - so we choose 101 .

The lesson is dedicated to how to solve 5 task of the EGE on computer science

The 5th theme is characterized as the tasks of the basic level of complexity, the execution time is about 2 minutes, the maximum score - 1

• Coding - This is the presentation of information in the form, convenient for its storage, transmission and processing. The rule of transformation of information to this representation is called code.
• Coding happens uniformand uneven:
• with uniform coding, all characters correspond to the codes of the same length;
• with uneven encoding, different symbols correspond to the codes of different lengths, it makes it difficult to decoding.

Example: We encrypt the letters A, B, B, G using binary encoding uniform code and consider the number of possible messages:

So we got uniform codebecause the length of each codeword is the same for all codes. (2).

Coding and message decoding

Decoding (decoding) - This is the recovery of the message from the sequence of codes.

To solve problems with decoding, you need to know the condition of Fano:

Fano condition: No codeword should be the beginning of another code word (which provides unambiguous decoding of messages from the beginning)

Prefix code - This is a code in which no codeword coincides with the beginning of another code word. Messages when using such a code are decoded uniquely.

Unambiguous decoding is provided:

Solution 5 assignments of the exam

EGE 5.1: For encoding letters O, B, D, P, and they decided to use the binary representation of numbers 0, 1, 2, 3 and 4, respectively (with the preservation of one inconcent zero in the case of a single-digit presentation).

Encoding the sequence of the waterfall letters in this way and the result record the octal code.

✍ Solution:

• We translate the number into binary codes and put them in compliance with our letters:

O -\u003e 0 -\u003e 00 in -\u003e 1 -\u003e 01 d -\u003e 2 -\u003e 10 P -\u003e 3 -\u003e 11 A -\u003e 4 -\u003e 100

Now we encode the sequence of letters from the word Waterfall:

010010001110010

We break the result on groups of three characters to right to left to translate them into an octal number system:

010 010 001 110 010 ↓ ↓ ↓ ↓ ↓ 2 2 1 6 2

Result: 22162

The decision of the exam of this task on computer science, video:

Consider another analysis of 5 assignments of the exam:

EGE 5.2: For 5 letters of the Latin alphabet, their binary codes are set (for some letters - from two bits, for some of three). These codes are presented in Table:

a.	b.	c.	d.	e.
000	110	01	001	10

What set of letters is encoded by a binary string 1100000100110?

✍ Solution:

• First, check the Fano condition: no code word is the beginning of another code word. The condition is true.

✎ 1 Solution Option:

• The code is divided from left to right according to the data presented in the table. Then we translate it into the letters:

110 000 01 001 10 ↓ ↓ ↓ ↓ ↓ b a c d e

Result: b a c d e.

✎ 2 Solution Option:

110 000 01 001 10

Result: b a c d e.

In addition, you can watch video solutions to this task EGE on computer science:

Resist the following 5 Task:

Ege 5.3:
To transmit numbers through a channel with interference, the parity check code is used. Each digit is recorded in a binary representation, with the addition of leading zeros to length 4, and the sum of its elements 2 is added to the resulting sequence (for example, if we transmit 23, then we obtain the sequence 0010100110).

Determine what number was transmitted via the channel in the form 01100010100100100110.

✍ Solution:

• Consider example From the terms of the problem:

It was 23 10 became 0010100110 2

Where the numbers themselves themselves (highlight them in red):

0010 10011 0 (0010 - 2, 0011 - 3)

First added digit 1 After binary two - this is a parity check (1 unit in 0010 - it means odd), 0 After binary triple - it is also an odd check (2 units in 0011 So - even).

Based on the analysis of the example, we solve our task as follows: Since the "necessary" figures are formed from groups of 4 numbers each plus a single number to check parity, then we will break the encoded message to groups 5, and throw from each group the last symbol:

we smash at 5:

01100 01010 01001 00110

return the last symbol from each group:

0110 0101 0100 0011

Resulttransfer to the decimal system:

0110 0101 0100 0011 ↓ ↓ ↓ ↓ 6 5 4 3

Answer: 6 5 4 3

You can watch video solutions to this task EGE on computer science:

Ege 5.4:

For encoding some sequence consisting of letters to, l, m, N decided to use an uneven binary code that satisfies the Fano condition. For the letter H used code word 0, for the letter K - code word 10.

What is the smallest possible total length of all four code words?

✍ Solution:

✎ 1 solution option Based on logical conclusions:

• Find the shortest possible code words for all letters.
• Code words. 01 and 00 It is impossible to use, since then the condition of Fano is disturbed (begin with 0, and 0 - this is N.).
• Let's start with two-sided code words. Take for the letter L. codeword 11 . Then for the fourth letters you can not pick up the code word, without disturbing the condition of the Fano (if you take 110 or 111, then they begin with 11).
• So, you need to use three-digit code words. Clean the letters L. and M. code words 110 and 111 . Fano condition is respected.

(H) 1 + (K) 2 + (L) 3 + (M) 3 \u003d 9

✎ 2 Solution Option:

(N) -\u003e 0 -\u003e 1 symbol (K) -\u003e 10 -\u003e 2 symbols (L) -\u003e 110 -\u003e 3 characters (m) -\u003e 111 -\u003e 3 characters

The total length of all four code words is equal to:

(H) 1 + (K) 2 + (L) 3 + (M) 3 \u003d 9

Answer: 9

5.5: EGE on computer science 5 Task 2017 FIPI option 2 (edited by Krylova S.S., Churkina i.e.):

Messages containing only 4 letters are transmitted over the communication channel: A, B, B, G; For transmission, a binary code that allows unambiguous decoding is used. For letters A B C These code words are used:

A: 101010, B: 011011, Q: 01000

Specify the shortest code word for the letter G, in which the code will allow unambiguous decoding. the smallestnumerical value.

✍ Solution:

• The smallest codes could look like 0 and 1 (disposable). But it would not satisfy the condition of Fano ( BUT begins with a unit - 101010 , B. starts from scratch - 011011 ).
• The next smallest code would be a two-letter word 00 . Since it is not a prefix of any of the codewords presented, R \u003d 00..

Result:00

5.6: EGE on computer science 5 Task 2017 FIPI version 16 (edited by Krylova S.S., Churkina i.e.):

To encode some sequence consisting of letters A, B, B, G and D, decided to use an uneven binary code, which allows you to unambiguously decode the binary sequence appearing on the receiving side of the communication channel. Used code:

A - 01 b - 00 in - 11 g - 100

Specify what code should be encoded by the letter D. Lengththis code word should be the smallestof all possible. The code must satisfy the property of unambiguous decoding. If there are several such codes, specify the code with the smallest numeric value.

✍ Solution:

Result: 101

More detailed lesson analysis can be viewed on the video of the exam in computer science 2017:

5.7: 5 task. DEVEROVESIA EGE 2018 Informatics (FIPI):

The encrypted messages containing only ten letters are transmitted over the communication channel: A, B, E, and, K, L, P, C, T, W. The uneven binary code is used to transmit. For nine letters, code words are used.

The encrypted messages containing only four letters are transmitted over the communication channel: A, B, B, G; For transmission, a binary code that allows unambiguous decoding is used. For letters BUT, B., IN Code words are used:

A: 00011 B: 111 V: 1010

Specify the shortest code word for the letter G.In which the code will allow unambiguous decoding. If there are several such codes, specify the code with the smallestnumerical value.

✍ Solution:

Result: 00

5.9: Training option number 3 of 01.10.2018 (FIPI):

On the communication channel, messages containing only letters are transmitted: A, e, d, k, m, r; For transmission, use binary code that satisfies the Fano condition. It is known that the following codes are used:

E - 000 d - 10 k - 111

Specify the smallest possible length of the encoded message. Dedmakar.
In response write the number - the number of bits.

✍ Solution:

D E D M A K A R 10 000 10 001 01 111 01 110

Consider the number of numbers in the final code and get 20 .

Result: 20

See the following task solution:

To see a presentation with pictures, design and slides, download its file and open in PowerPoint on your computer.
Text Content Slides Presentation: Preparations for the Egressant Informatiki Society No. 1 G. Azov Balamutova Irina Aleksandrovna2015. Encoding and decoding information. (Tasks 5) Data encoding, combinatorics, number system (task 10) Content of the topic "Coding and decoding." Theory 1 Outcoming 2 Outcoming 3 Outcomes for training Code: Coding Data, Combinatorics, Systems Systems 1 Out 2 Options 3 Options 4 Options 5 Options For Training Site Literature Sites EEH2 decoded from the beginning if the Fano condition is satisfied: no code word is the beginning of another code word; The encoded message can be unambiguously decoded from the end if the reverse Fano condition is performed: no code word is the end of the other code word; the Fano condition is sufficient, But not the necessary condition of the unequivocal decodingTeoria3 encoding is the transfer of information from one language to another. Coding can be uniform and uneven. After uniform coding, all characters are encoded by codes of equal length. In uneven coding, different characters can be encoded by codes of different lengths. This connection channels are transmitted to the communication channel, each of which contains 16 letters A, 8 letters b, 4 letters in and 4 letters g (there are no other letters in messages). Every letter is encoded by a binary sequence. When choosing the code, two requirements were taken into account: a) no codeword is the beginning of another (it is necessary that the code allows unambiguous decoding); b) The total length of the encoded message must be as small as possible. How the code from the above should be selected for encoding letters a, b, in and g? 555551) A: 0, b: 10, in: 110, g: 1112) A: 0, b: 10, in: 01, g: 113) a: 1, b: 01, in: 011, g: 0014) A: 00, B: 01, C: 10, G: 11 Beading 15 We first choose codes, In which no codeword coincides with the beginning of another (such codes call the prefix) for code 2, the condition "A" is not performed, since the code word of the letter in (01) begins with the code word of the letter A (0) for code 3 condition " A "is not performed, since the code word of the letter B (011) begins with the code word Letters B (01) for codes 1 and 4, the condition is performed, they are considered having fun about the total number of bits in the message for code 1: 16 ∙ 1 + 8 · 2 + 4 ∙ 3 + 4 ∙ 3 \u003d 56 Bitching the total number of bits in a message for code 4: 16 ∙ 2 + 8 · 2 + 4 ∙ 2 + 4 ∙ 2 \u003d 64 Tyakod 1 gives the smallest length of the message, so we choose its response: 1.6 Tasks 1 for encoding some sequence consisting of letters A, B, B, G, decided to use an uneven binary code that satisfies the Fano condition. For the letter A, I used code word 0, for the letter B - code word 110.Kova The smallest possible total length of all four code words? 1) 7 2) 8 3) 9 4) 107 Output 2 solution (method 1, exception options): Fano condition This means that no code word coincides with the beginning of another code list, there is already a code word 0, no other code word begins with 0 since there is code 110, code words 1, 11 are prohibited; In addition, no other code word can begin with 110tically, you need to select two more code words for which these limitations are performed. There is one permissible code word from two characters: 10 If you select a code word 10 for the letter in, then one remains The permissible threeximwall code word - 111, which can be selected for the letter of the G8 value of Task 2 by selecting code words A - 0, b - 110, in - 10, g - 111, we obtain the total length of the code words 9 characters. If you do not choose in - 10, That is, there are three permissible threeximilical code words: 100, 101 and 110; When choosing any two of them for letters in and g, we obtain the total length of the code words 10, which is more than 9; Therefore, select Option 3 (9 characters) Answer: 3. Solution Tape 2 (continued) 9 ab10100Recution (Method 2, Tree construction): Fano Correction means that no code word coincides with the beginning of another code word; At the same time in the code tree, all code words should be located in the leaves of the tree that do not have descendants; we construct a tree for the specified code words A - 0 and b - 110: 10 Feed 2 Stroke lines 2 "Empty" branches are marked for which you can "attach" leaves For code words of letters in (10) and (111) ab10100vgvybrav code words A - 0, b - 110, B - 10, G - 111, we obtain the total length of the code word code 9, the symptoms: 3. Task 2 method 2, construction of a tree continuation11 On the communication channel, messages containing only 4 letters P, O, C, T; For transmission, a binary code that allows unambiguous decoding is used. For the letters T, O, P code words are used: T: 111, ABOUT: 0, P: 100. For the shortest code word for the letter C, in which the code will allow unambiguous decoding. If there are several such codes, specify the code with the smallest numeric value. 12 Output 3 OT101000P1Recution (Method 2, Tree Building): Fano Correction means that no codeword coincides with the beginning of another code word; At the same time, in the code tree, all code words should be located in the leaves of the tree, that is, in nodes that do not have descendants; we construct a tree for the specified code words about - 0, T - 111 and P - 100: 13 Operations of the task 3 were marked by two " empty "branches that can be" attached "a sheet for code word letter C: 101 or 110; Of these, the minimum value of the code 101rection of the problem 3 (continued) 14 15 masters are marked two "empty" branches, for which you can "attach" a sheet for the code word letter C: 101 or 110; Of these, the minimum value has code 101. From 101000P1Sebving code words A - 0, b - 110, B - 10, g - 111, we obtain the total length of the code word code words 9. The symptoms: 101. Solution of the problem 3 (continued) 15 The black and white raster image is encoded line, starting from the left upper angle and Finishing in the lower right corner. When encoding 1 denotes black, and 0 - white. BD9AA5 2) BDA9B5 3) BDA9D5 4) DB9DAB 16 Following 4 "Extend" Raster Image into a chain: First first (top) line, then - second, etc.: In this strip 24 cells, black fill in units, and white - zeros: Since each figure in the hexadecimal system is unfolded exactly in 4 binary figures, we break the strip on the notebooks - groups of four cells (in this case, it is still to start the breakdown, because a whole number of Tetrad - 6): Translating the tetrad to a hexadecimal system , We receive successively numbers B (11), D (13), A (10), 9, D (13) and 5, that is, the value of the BDA9D5 invesuette the correct answer - 3.17. Tasks 4 1 line2 line3 line4110101010101010101010101014 (continued) Task 5 No. 7746. For encoding some sequence consisting of letters A, B, B, G, and D, an uneven binary code is used, allowing uniquely decoders The resulting binary sequence. This code: a - 1; B - 0100; In - 000; G - 011; D - 0101. It is required to reduce for one of the letters of the length of the code word so that the code can still be decoded unambiguously. The codes of the remaining letters should not change. What of the specified ways can this be done? 1) For the letter G - 112) for the letter in - 003) for the letter G - 014) it is not possible: 19 tasks for self-decisions2
Task 5 No. 1104. For encoding the letters X, E, L, O, D, decided to use the binary representation of numbers 0, 1, 2, 3 and 4, respectively (with the preservation of one inconcent zero in the case of a single-digit presentation). If you encode the sequence of Icewright letters in this way and the result will be recorded by hexadecimal code, it will turn out 1) 999С2) 32541453) 123F 4) 2143034 Answer: 20 Answers Task 5 No. 1104Helode0123400011011100Snamed You should submit data to the number of the number in binary code: Code the sequence of letters: Ice - 100110011111100 . Now we will break this view on the fourths to the right left and transfer the resulting set of numbers in the decimal code, then in hexadecimal. 1001 1001 1001 1100 - 9 9 9 12 - 999С. The correct answer is specified at number 1.21 Task 5 No. 7193 for transmission over the communication channel of the message consisting only of characters A, B, B, and G, is used uneven (in length) Code: a - 0; B - 100; In - 101. What kind of code word you need to encode the symbol g, so that it is minimal, and the code allows you to unambiguously split the encoded message to the characters? 1) 12) 113) 01 Decision4) 010 http://inf.reshuege.ru/test?Theme\u003d232 Answer: 222
Task 5 No. 9293.23 for encoding some sequence consisting of letters and, k, l, m, n, decided to use an uneven binary code that satisfies the Fano condition. For the letter l used code word 1, for the letter M codeword 01. What is the smallest possible total length of all five code words? Note. The Fano Condition means that no code word is the beginning of another code word. This provides the ability to unambiguous decryption of encoded messages. RESULT: 4Entens http://inf.reshuege.ru/test?Theme\u003d23123
24 Promotive training Video Tutorial LinksLinkHttps: //www.youtube.com/watch? V \u003d bobnzjwlsnu Topic: data encoding, combinatorics, number systems (tasks 10) 25 What you need to know: Russian alphabet Principles of work with numbers recorded in positional specificity systems The word consists from l letters, and there are N1 options for selecting the first letter, N2 options for choosing a second letter, etc., the number of possible words is calculated as a product N \u003d N1 · N2 · ... · NLAI word consists of l letters, and each letter may be Native n methods, the number of possible words is calculated as N \u003d NL26Teorya Vasya is 5-letter words in which there are only letters C, L, O, N, and the letter C is used in each word exactly 1 time. Each of the other allowable letters can occur in the Word any number of times or not to meet at all. A word is considered to be any permissible sequence of letters, not necessarily meaningful. How many words are there, which can write Vasya? 27 Feed 1 letter C can stand at one of five places: from ****, * with ***, ** s **, *** s * and **** With, where * denotes any of the remaining three characters in each case, in each other four positions, any of the three letters l, o, H, therefore, at a given location, the letter C we have 34 \u003d 81 variants of the entire variants 5 · 81 \u003d 405.Wrant: 405.28Recuration How many different symbolic sequences of length 5 in a four-letter alphabet (A, C, G, T), which contain exactly two letters a? 29 Candle 2 solution (option 1, bust): Consider various options for words of 5 letters that contain two Letters A and begin with A: AA *** A * A ** A ** A * A *** The stars refers to any character from the set (C, G, T), that is, one of three characters. So, in each template there are 3 positions, each of which can be fill in three ways, therefore the total number of combinations (for each template!) Is 33 \u003d 27 Total 4 templates, they give 4 · 27 \u003d 108 combinations. Positions are now considering templates where the first The score of the letter A is on the second position, there are only three of them: * AA ** * A * A * * A ** AONI gives 3 · 27 \u003d 81 combination of template, where the first in the score of the letter A is on the third position: ** AA * ** A * and they give 2 · 27 \u003d 54 combinations and one template, where the combination of AA is at the end of *** AA, they give 27 combinations. Total we get (4 + 3 + 2 + 1) · 27 \u003d 270 combinations: 270 . Making (continued) 31 All 4-letter words made up of letters to, l, p, t, recorded in alphabetical order and are numbered. Here is the beginning of the list: kkkk2. Kkl3. Kkkr4. CKT ...... write down the word that stands at 67th place from the beginning of the list. 32Read 3 The easiest solution to this task is to use the number systems; Indeed, here the alphabetical alphabetical order here is equivalent to the arrangement in an increase in the numbers recorded in the chimeful number system (the base of the number system is equal to the number of letters used). Fill the replacement of K0, L1, P2, T3; Since the numbering of words begins with a unit, and the first number of KKKK0000 is 0, number 67 will stand the number 66, which must be translated into the fourth system: 66 \u003d 10024 After performing a reverse replacement (numbers per letter), we get the word LCKR. RESULT: LCKR .333Production 34 Options 4 Task 10 No. 6777. How many words of length 5 can be made up of the letters E, G, E? Each letter can enter the word several times. 35Recel in alphabet M symbols, the number of all possible "words" (messages) length n is q \u003d Mn. In our case, n \u003d 5, m \u003d 3. Consequently, Q \u003d 35 \u003d 243. Answer: 243. 36 Options 5 Task 10 No. 4797. There are 32 pencils in the closed box, some of them are blue. At random is taken out one pencil. The message "This pencil is not blue" carries 4 bits of information. How many blue pencils in the box? 37 Shenonna formula: where x is the amount of information in the message about the event P, P is the probability of the event P. The likelihood that it was not blue where - the number of blue pencils. Usessed by the Schuenonna formula, we get that \u003d 30-year-old training session Self-training Video Tutorial LinkslinkHttps: / /www.youtube.com/watch?v\u003dbobnzjwlsnu Literature LiteratureHttp: //kpolyakov.narod.ru/ Krylov S.S., Churkina Tue. EGE 2015. Informatics and ICT. Typical exam options. - M.: "National Education", 2015. Leschinner V.R. Ege 2015. Informatics. Typical test tasks. - M.: Exam, 2015.Evich L.N., Kulabukh S.Yu. Informatics and ICT. Preparation for the EEG-2015. - Rostov-on-Don: Legion, 2014. Ushakov D.M., Yakushkin P.A. Computer science. The most complete publication of typical options for the tasks of the EE 2. - M.: Astrel, 2014. Evich L.N., Kulabukhov S.Yu. Informatics and ICT. Preparation for the EEG-2015. - Rostov-on-Don: Legion, 2014. Ostrovskaya E.M., Satykina N.N. Ege 2015. Informatics. We rent without problems! - M.: Eksmo, 2014. Satykina N.N., Ostrovsky E.M. Ege 2015. Informatics. Thematic training tasks. - M.: Eksmo, 2014.Sorina E.M., Zorin M.V. Ege 2015. Informatics. Collection of tasks. - M.: Eksmo, 2015.39 Useful sites for preparing for the exam! 40informatics - it's just http://easyinformatics.ru/Videosulation Problems of EGE-2013 http: //www.agechev.rf/ege.htm Temporary portal for preparing for Exams http://inf.reshuege.ru/?redir\u003d1Egue on computer science 2013 http://infogehelp.ru/40