The Unified State Exam in Informatics consists of 27 tasks. Task 6 tests the skills of analysis and construction of algorithms for various performers. The student must be able to compose algorithms from given commands, as well as check sequences for compliance with the algorithms. Here you can learn how to solve task 6 of the Unified State Exam in computer science, as well as study examples and solutions based on detailed tasks.

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The performer Kvadrator has two commands: add 3 and square

The performer Kvadrator has two teams, which are assigned numbers: 1 - add 3; 2 - square it. The first of them increases the number on the screen by 3, the second raises it to the second power. The performer works only with natural numbers. Create an algorithm for obtaining number B from number A, containing no more than K commands. In your answer, write down only the command numbers. If there is more than one such algorithm, then write down any of them.

The decipherer needs to recover the damaged message fragment

The decryptor needs to restore the damaged message fragment, consisting of 4 characters. There is reliable information that no more than five letters were used (A, B, C, D, E), with one of the symbols in third place... One of the letters in fourth place... One of the letters in first place ... On the second - ... Additional information has appeared that one of four options is possible. Which?

The task is included in the Unified State Examination in computer science for grade 11 under number 6.

There are two windows on the screen, each of which contains a number

There are two windows on the screen, each of which contains a number. The Adder performer has two commands, which are assigned numbers: 1 – write the sum of numbers in the first window; 2 – write the sum of numbers in the second window. By executing the first of them, the Adder adds the numbers in the windows and replaces the number in the first window with this sum, and by executing the second, it adds the numbers and replaces the number in the second window with this sum. Write down the order of commands in the program for obtaining from a pair of numbers A and B a pair of numbers C and D, containing no more than K commands, indicating only the command numbers.

The task is included in the Unified State Examination in computer science for grade 11 under number 6.

The Calculator performer has two teams, which are assigned numbers

The Calculator performer has two commands, which are assigned numbers: 1 – add 2, 2 – multiply by 3. By executing the first of them, the Calculator adds 2 to the number on the screen, and by executing the second, it triples it. Write down the order of commands in the program for obtaining number B from A, containing no more than K commands, indicating only the command numbers.

The task is included in the Unified State Examination in computer science for grade 11 under number 6.

Chains of characters (strings) are created according to the following rule

Chains of characters (strings) are created according to the following rule. The first line consists of... Each of the subsequent chains is created by such actions... Here are the first 4 lines created by this rule. What character is in the K line in the Nth place (counting from left to right)?

The task is included in the Unified State Examination in computer science for grade 11 under number 6.

Analysis of task 6 of the Unified State Exam 2017 in computer science from the demo version project. This is a task of a basic level of difficulty. Estimated task completion time is 4 minutes.

Content elements tested: formal execution of an algorithm written in natural language or the ability to create a linear algorithm for a formal performer with a limited set of commands. Content elements tested on the Unified State Exam: Formalization of the concept of an algorithm. Construction of algorithms and practical calculations.

Task 6:

The machine receives a three-digit number as input. Based on this number, a new number is constructed according to the following rules.
1. The first and second, as well as the second and third digits of the original number are added.
2. The resulting two numbers are written one after another in descending order (without separators).
Example. Original number: 348. Sums: 3 + 4 = 7; 4 + 8 = 12. Result: 127.
Specify least a number, as a result of which the machine will produce the number 1711.

Answer: ________

Analysis of task 6 of the Unified State Exam 2017:

Obviously, the result 1711 came from two numbers 17 and 11.

Now we find the smallest three-digit number.

Since we are looking for the smallest number, we will start with the smallest sum (11) in order to get the smallest first digit.

11 - 9 = 2. Thus, the number 11 is obtained as the sum of 2 and 9: 2 + 9 = 11 .

The number 17 is obtained as the sum of 9 and 8: 9 + 8 = 17 .

Now we compose the required smallest three-digit number and get 298.

Checking 2 + 9 = 11 and 9 + 8 = 17

The lesson is devoted to how to solve task 6 of the Unified State Exam in computer science

The 6th topic - “Analysis of algorithms and executors” - is characterized as tasks of a basic level of complexity, completion time - approximately 4 minutes, maximum score - 1

Performer for squaring, division, multiplication and addition

Let's take a closer look at what might be useful for solving task 6.

• in a problem for which you need to determine all possible results the work of the algorithm of any performer, the initial data can be designated by variables and the algorithm can be calculated with these variables;
• in a problem for which you need to find optimal program(or the shortest one), and which, using a given set of commands, converts a certain number to another, is better for solving build a tree of possible options; thus, calculating what results will be obtained after one step, after two steps, etc. As a result, a general solution will be found;
• if among the performer’s commands specified in the task there is an irreversible command (for example, the performer works with integers and there is a squaring command - any number can be squared, but not the square root can be extracted from any number, thereby obtaining an integer), then It’s better to build a tree of options from the end, i.e. in reverse order, moving from the final number to the starting number; whereas the resulting sequence of program commands must be written from the initial number to the final one.

Checking a numerical sequence for compliance with the algorithm

• To complete some tasks you need to repeat the topic;
• the maximum value of the sum of the digits of a decimal number is 18 , because 9 + 9 = 18 ;
• to check the correctness of the transmitted message is sometimes entered parity bit- an additional bit that complements the binary code in such a way that the number of ones becomes even as a result: i.e. if the number of units in the original message was even, then 0 is added, if odd, 1 is added:

Now we will consider specific standard exam options in computer science with an explanation of their solutions.

Analysis of task 6

Solving tasks 6 of the Unified State Exam in computer science for the topic Performers

6_1:

The performer GRASSHOPPER lives on the number line. The initial position of the GRASSHOPPER is a point 0 . GRASSHOPPER command system:

• Forward 5— The grasshopper jumps forward 5 units,
• Back 3— The grasshopper jumps back 3 units.

Which least number of times the command must appear in the program "Back 3" so that the GRASSHOPPER is at the point 21 ?

Let's consider two solutions.

✎ 1st solution:

• Let us introduce the following notation:
• let x- this is a team Forward 5
• let y- this is a team Back 3
• Since the Grasshopper moves from the beginning of the number axis (from 0 ) and eventually reaches the point 21 , then we get the equation:

Result: 3

✎ 2nd solution:

• Let's say the Grasshopper jumped to 21 (and further). He could only do this with the help of the command Forward 5. We will consider numbers > 21 And divisible by 5 without remainder(because Forward 5).
• The first number is greater 21 and divisible by 5 without a trace - this is 25 .

Result: 3

If anything remains unclear, we suggest you take a look video with solution analysis:

6_2:

There is a performer, the Grasshopper, who lives on the number line. Grasshopper command system:

• Forward N(The grasshopper jumps forward N units);
• Back M(The grasshopper jumps back M units).

Variables N And M can take any positive integer value.

It is known that the Grasshopper executed the program from 50 teams in which teams Back 2 12 more than teams Forward 3. There were no other teams in the program.
Which one command can this program be replaced with so that the Grasshopper ends up at the same point as after executing the program?

• In order to find out the number of both commands, you must enter the unknown x. Let's imagine that the number of teams Forward 3 was completed x times, then the number of commands Back 2 was x+12 once. Since there were only teams 50 and there were no other teams, then we’ll create an equation:

Result: Back 5

We invite you to watch the video analysis of task 6:

Unified State Exam 6_3:
From the performer Quad two teams assigned numbers:

1. add 1,
2. square it.

The first of these commands increases the number on the screen by 1, the second - squares it. The program for the Quad performer is a sequence of command numbers.

For example, 22111 is a program square square add 1 add 1 add 1 This program converts the number 3 V 84 .

Record a program for the performer Quad, which converts the number 5 to the number 2500 and contains at most 6 commands If there is more than one such program, then write down any of them.

• Since the number 2500 It’s quite large, so it’s difficult to figure out which commands can be used to “reach” it.
• In this kind of problem, you should start solving from the end - from the number 2500 square root of a number(since the square root is the inverse operation of squaring). If the square root cannot be extracted, we will execute the reverse command for the first command - Subtract 1(reverse for Add 1):

Result: 11212

You can watch the video of the solved 6th task of the Unified State Exam in computer science:

6_4. Option No. 11, 2019, Computer Science and ICT Model exam options, Krylov S.S., Churkina T.E.

From the performer Calculator two teams assigned numbers:

1. add 3,
2. multiply by 5.

By performing the first of them, the Calculator adds 3 to the number on the screen, and by performing the second, it multiplies it by 5.

Write down the order of commands in the program that converts number 3 to number 24 and contains no more than four commands Enter only command numbers.

• In this kind of problem, sometimes it’s easier to start the solution from the end - from the number 24 , and try to perform the action each time divide by 5(since division is the inverse operation of multiplication). If the number in question is not integer divisible by 5, then we will execute the reverse command for the first command - subtract 3(reverse for add 3):

Answer: 2111

6_5:

An executor that works with positive single-byte binary numbers has two instructions, which are assigned numbers:

1. slide right
2. add 4

By performing the first of them, the performer shifts the number one binary digit to the right, and by performing the second, he adds 4 to it.

The performer began calculations with the number 191 and executed the chain of commands 112112 . Write the result in decimal notation.

✎ 1 way:

• To execute the first command, let's convert the number to the binary number system:

Result: 16

✎ Method 2:

• When shifting to the right, a zero enters the high bit, and the low bit is sent to a special cell - the carry bit, i.e. it will be “lost”. Thus, if the number is even, then when shifted it is halved; if odd, the nearest smaller even number is halved (or the original odd integer is divided by 2 ).
• We get the results of executing a sequence of commands:

Result: 16

For a detailed explanation, watch the video:

6_6: Task 6 Unified State Exam in Computer Science 2017 FIPI option 19 (Krylov S.S., Churkina T.E.):

The Adder-Multiplier performer has two teams, which are assigned numbers:

1. Add 3
2. Multiply by x

The first one increases the number on the screen by 3 , the second multiplies it by X. The program for the performer is a sequence of command numbers. It is known that the program 12112 converts number 3 in number 120 .

Determine the value X, if it is known that it is natural.

✍ Solution:

• Let's substitute the executed commands in order according to the numbers in the sequence of commands. For convenience, we will use brackets:
  12112 :

That's right.

Result: 4

A more detailed analysis of the lesson can be seen in the video of the Unified State Exam in Computer Science 2017:

Solving tasks for the topic Checking a number sequence (Automatic)

6_7: Unified State Exam in Computer Science task 6 from the site of K. Polyakov (task number P-06):

The machine receives a four-digit number as input. Based on this number, a new number is constructed according to the following rules.

1. The first and second, as well as the third and fourth digits of the original number are added.
2. The resulting two numbers are written one after another in descending order (without separators).

Example. Original number: 3165. Sums: 3 + 1 = 4; 6 + 5 = 11. Result: 114.

Specify least number, as a result of processing which, the machine will display the number 1311.

Result: 2949

The process of solving this 6th task is presented in the video tutorial:

6_8: Task 6 Unified State Exam in Computer Science 2017 FIPI (Krylov S.S., Churkina T.E.) option 13:

The machine receives a four-digit number as input. A new number is constructed from it according to the following rules:

• The first and second digits are added, then the second and third, and then the third and fourth digits of the original number.
• The resulting three numbers are written one after another in ascending order (without separators).

Example: Original number: 7531. Sums: 7+5=12; 5+3=8; 3+1=4. Result: 4812.

Specify the largest number that the machine will produce as a result of processing 2512 .

Result: 9320

6_9: Task 6 Unified State Examination in Informatics 2017 FIPI (Ushakov D.M.) option 2:

The machine receives two two-digit hexadecimal numbers as input. These numbers contain all the digits do not exceed the number 6(if the number contains a number greater than 6, the machine refuses to work). Using these numbers, a new hexadecimal number is constructed according to the following rules:

1. Two hexadecimal numbers are calculated: the sum of the most significant digits of the resulting numbers and the sum of the least significant digits of these numbers.
2. The resulting two hexadecimal numbers are written one after another in descending order (without delimiters).

Example: Initial numbers: 25, 66. Bitwise sums: 8, B. Result: B8.

Which of the proposed numbers can be the result of the machine?
List the letters corresponding to these numbers in alphabetical order, without spaces or punctuation.

Options:
A) 127
B) C6
C) B.A.
D) E3
E) D1

Result: B.C.

A detailed solution to this 6th task can be viewed in the video:

6_10: 6 task of the Unified State Exam. Task 4 GVE Grade 11 2018 FIPI

The machine receives the input two two-digit hexadecimal numbers. These numbers contain all the digits do not exceed the number 7(if the number contains a number greater than 7, the machine refuses to work). Using these numbers, a new hexadecimal number is constructed according to the following rules.

1. Two hexadecimal numbers are calculated: the sum of the most significant digits of the resulting numbers and the sum of the least significant digits of these numbers.
2. The resulting two hexadecimal numbers are written one after another in ascending order (without delimiters).

Example. Initial numbers: 66, 43. Bitwise sums: A, 9. Result: 9A.

Determine which of the proposed numbers can be the result of the machine.

Options:
1) AD
2) 64
3)CF
4) 811

Result: 1

Solution 4 of the 11th grade GVE task, watch the video:

Solving a problem about an algorithm that constructs the number R

6_11: Task 6 Unified State Exam in Computer Science 2017 FIPI option 2 (Krylov S.S., Churkina T.E.):

N R in the following way:

1. 4N.

• all the digits of the binary notation are added, and the remainder of dividing the sum by 2 added to the end of the number (on the right). For example, record 10000 converted to record 100001 ;
• The same actions are performed on this entry - the remainder of the division of the sum of digits by is added to the right 2 .

The record obtained in this way is a binary record of the desired number R.

Enter the smallest number N, for which the result of the algorithm is greater 129 . In your answer, write this number in the decimal number system.

• Please note that after completing the second step of the task, only even numbers will be obtained! The smallest possible even number greater than 129, is the number 130 . We will work with him.
• Let's translate 130 to binary number system:

Result: 8

For a more detailed analysis, we suggest watching the video of the solution to this 6th task of the Unified State Exam in computer science:

6_12: 6 task. Demo version of the Unified State Exam 2018 computer science:

The input of the algorithm is a natural number N. The algorithm constructs a new number from it R in the following way.

1. Constructing a binary number N.
2. Two more digits are added to this entry on the right according to the following rule:
3. add up all the digits of a binary number N, and the remainder of dividing the sum by 2 is added to the end of the number (on the right). For example, record 11100 converted to record 111001 ;
4. The same actions are performed on this entry - the remainder of dividing the sum of its digits by 2 is added to the right.

The record obtained in this way (it has two digits more than in the record of the original number N) is a binary record of the desired number R.

Specify the minimum number R, which exceeds the number 83 and may be the result of this algorithm. In your answer, write this number in the decimal number system.

• Note that after the second point of the problem conditions, only even numbers are obtained (since if a number in the binary system ends with 0 , then it is even). Thus, we will only be interested in even numbers.
• The smallest possible number greater than 83, is the number 84 . We will work with him.
• Let's translate 84 to binary number system:

Result: 86

For a detailed solution to this 6th task from the demo version of the Unified State Exam 2018, watch the video:

6_13: Analysis of the 6th task of the Unified State Exam option No. 1, 2019 Computer science and ICT Typical exam options (10 options), S.S. Krylov, T.E. Churkina:

The input of the algorithm is a natural number N. The algorithm constructs a new number from it R in the following way:

1. Constructing a binary number N.
2. Two more digits are added to this entry on the right according to the following rule:
- If N divisible by 4 zero and then another one zero;
- If N when divided by 4 gives the remainder 1 zero, and then unit;
- If N when divided by 4 gives the remainder 2 , then the end of the number (on the right) is added first one, and then zero;
- If N when divided by 4 gives the remainder 3 , at the end of the number (on the right) is added first one and then another one unit.

For example, the binary notation 1001 of the number 9 would be converted to 100101, and the binary notation 1100 of the number 12 would be converted to 110000.

The record obtained in this way (it contains two digits more than in the record of the original number N) is a binary representation of a number R- the result of this algorithm.

Specify the maximum number R, which less than 100 and may be the result of this algorithm. Write this number down in your answer. in decimal number system.

• Since we need to find the largest number, we take the largest possible number that - this is the number 99 . Let's convert it to binary:

Result: 96

We invite you to watch the video solution:

Video fragment from a consultation session (consultation before the computer science exam) in preparation for the OGE. Analysis of task number 6 from the OGE on the topic Algorithm for a specific performer with a fixed set of commands. Algorithm executors. In the video clip above you will find solution to task number 6 from OGE in computer science

Task 6:

Performer The draftsman moves on the coordinate plane, leaving a trace in the form of a line. The Draftsman can issue the Move to (a, b) command (where a, b are integers), which moves the Draftsman from a point with coordinates (x, y) to a point with coordinates (x + a, y + b). If the numbers a, b are positive, the value of the corresponding coordinate increases; if negative, it decreases.

V1. The draftsman was given the following algorithm to execute:

Repeat 2 times
Command1 Move to (3, 2) Move to (2, -1)
End
Shift by (6, −4)

After completing this algorithm, the Draftsman returned to the starting point. What command should be put instead of the command Team1?

V2. The draftsman was given the following algorithm to execute:

Repeat 3 times
Shift by (-2, -3) Shift by (3, 2) Shift by (-4, 0
end

This video explains the solution to the sixth task OGE in computer science. The main accents are placed and it is shown what is worth paying attention to and what is not so important.

Options are being considered Performer Draftsman, which are most often found in the exam computer science. In practice, it is shown how to solve these problems using three simple and understandable steps.

Only three problems were solved in detail with explanations.

The first problem is classic for this assignment. It introduces the basic concepts in great detail and shows how to master it effectively and quickly.

The second task is solved in the first person. This approach brings the viewer as close as possible to the atmosphere of a real exam and helps to understand what specific actions need to be taken to get the coveted score.

The third task can be considered non-standard, although all techniques and methods remain the same.

To consolidate the material covered Don’t forget to practice in the online testing system on my website! Link under video

Happy exams!