The lesson considers the solution of the 13th task of the exam in computer science
Topic 13 - "The amount of information" - is characterized as tasks of an increased level of complexity, the execution time is about 3 minutes, the maximum score is 1
when working with text
- By using K bit can be encoded Q = 2 K different characters:
- Q- the power of the alphabet
- K Q character options
- 2 - binary number system (data is stored in binary form)
- I, you need to multiply the number of characters N by the number of bits to store one character K:
- I
- N- message length (number of characters),
- K- the number of bits to store one character.
- These two formulas use the same variable:
N = 2 i
Q = 2 K I = N * K
Consider an example using two formulas at the same time:
Example:
Message volume - 7.5 kB 7680 characters. What is the power of the alphabet?
✍ Solution:
- Let's use the formula:
- Let's find the number of bits required to store 1 character (first, we translate the value into bits):
- Next, let's use the formula:
- 8 bits per character allow you to encode:
I = N * K;
I- message size = 7.5 Kbytes;
N- number of characters = 7680;
K- the number of bits per character
\ [K = \ frac (7.5 * 2 ^ (13)) (7680) = \ frac (7.5 * 2 ^ (13)) (15 * 2 ^ 9) = \ frac (7.5 * 16 ) (15) = 8 \]
those. K = 8 bits per character
Q = 2 K
K- the number of bits to store one character from Q character options (= 8)
Q Is the power of the alphabet, i.e. number of character options
2 8 = 256 different characters
256 characters is power
Answer: 256
Measuring the amount of information
when working with various systems
- By using K bit can be encoded Q = 2 K different (numbers) of objects of a certain system:
- Q- the total number of objects in a certain system, data about which are stored in a computer or transmitted in a message,
- K- the number of bits to store one object out of the total Q,
- 2 - binary number system (data is stored in binary form).
- To find the information volume of a message I, you need to multiply the number of objects in the message - N- by the number of bits K to store one object:
- I- information volume of the message,
- N- the number of objects in the message
- K- the number of bits for storing one object of the system.
* other designations are also accepted: N = 2 i
Example:
There is an automatic system for informing the warehouse about the need to deliver certain groups of consumables to the workshop. The system is designed so that through the communication channel to the warehouse a conditional number of consumables is transmitted(in this case, the same, but the minimum possible number of bits in the binary representation of this number is used). A delivery request is known to have been sent 9 groups materials from 19 used in production. Determine the size of the sent message
(Give the answer in bits)
✍ Solution:
- Let's use the formula:
- to store the number of one group, you need a bit:
K- the number of bits for storing one material group number
Q- the total number of numbers for various groups of consumables = 19
I = N * K;
I- message volume =? bit;
N- the number of transmitted group numbers (= 9);
K- the number of bits per 1 number (= 5)
Answer: 45
The solution of tasks 13 Unified State Exam in Informatics
Unified State Exam in Informatics 2017 task 13 FIPI option 1 (Krylov S.S., Churkina T.E.):
7 33 -symbol alphabet. The database for storing information about each user has the same and minimum possible integer byte bit... In addition to their own password, the system stores additional information for each user, for which an integer number of bytes is allocated; this number is the same for all users.
To store information about 60 users needed 900 byte.
How many bytes are allocated to store additional information about one user?
In response, write down only an integer - the number of bytes.
✍ Solution:
- First, let's define a password. According to the formula Q = M N we get:
Result: 9
A step-by-step solution to this 13 task of the exam in computer science is also available in the video lesson:
Unified State Exam 2017 collection by D.M. Ushakov's "10 training options ..." option 1:
The cable network is voting among viewers on which of the four films they would like to watch in the evening. The cable network is used by 2000
human. Participated in the voting 1200
human.
What is the amount of information ( in bytes) recorded by an automated voting system?
✍ Solution:
- Since the four movie numbers are stored in the computer system, you can find the number of bits needed to store the movie number:
Result: 300
Unified State Exam 2017 collection by D.M. Ushakov's "10 training options ..." option 6:
When registering in a computer system, each user is given a password consisting of 15 characters and containing only characters from 12 -character set A, B, C, D, E, F, G, H, I, K, L, M, N... The database for storing information about each user has the same and minimum possible integer byte... In this case, character-by-character passwords are used, all characters are encoded with the same and minimum possible number bit... In addition to the password itself, additional information is stored in the system for each user, for which 12 bytes per user.
Determine the amount of memory ( in bytes) required to store information about 30
users.
In the answer, write down only an integer - the number of bytes.
✍ Solution:
Result: 600
An example of solving this task of the exam is available in the video lesson:
Unified State Exam 2017 collection by D.M. Ushakov's "10 training options ..." option 10:
The school rehearsal exam is taken 105 human. Each of them is assigned a special number that identifies it in the automatic response checking system. When registering a participant to record his number, the system uses the minimum possible number of bit, the same for each participant.
What is the amount of information in bits recorded by the device after registration 60
participants?
✍ Solution:
Result: 420
An example of solving this task of the exam is available in the video lesson:
13 task. Demo version of the exam 2018 informatics:
10 characters. Capital letters of the Latin alphabet are used as symbols, i.e. 26 various symbols. In the database for storing each password, the same and minimum possible integer is allocated byte... In this case, character-by-character passwords are used, all characters are encoded with the same and minimum possible number bit.
Determine the amount of memory ( in bytes) required for storing data about 50
users.
In the answer, write down only an integer - the number of bytes.
✍ Solution:
- The main formula for solving this problem is:
- To find the number of bits required to store one password, you first need to find the number of bits required to store 1 character in the password. By the formula we get:
where Q- the number of character options that can be encoded with N bit.
Result: 350
For a detailed solution to the 13th task of the demo version of the exam in 2018, see the video:
Solution 13 of the USE task in informatics (diagnostic version of the exam paper, USE simulator 2018, S. S. Krylov, D. M. Ushakov):
In some countries, the license plate number consists of 7 characters... Each character can be one of 18 different letters or decimal digit.
Each such number in a computer program is recorded in the minimum possible and the same whole number byte, in this case, character-by-character coding is used and each character is encoded with the same and minimum possible amount bit.
Determine the amount of memory in bytes allocated by this program for writing 50
numbers.
✍ Solution:
- Since the number can be used either one letter from 18 , or one digit from 10 , then only one of the following can be used as one character in the number 28 characters:
Result: 250
Video parsing:
Solution 13 of the USE task in computer science (control option No. 1 of the examination paper, Simulator 2018, S. S. Krylov, D. M. Ushakov):
The rehearsal exam is passed 9
streams across 100
a person in everyone. Each of them is assigned a special code consisting of a stream number and a stream number. When coding these participant numbers, the checking system uses the minimum possible number of bit, the same for each participant, separately for the number of the stream and the number in the stream. In this case, to write the code, the minimum possible and equally integer amount is used bytes.
What is the amount of information in bytes recorded by the device after registration 80
participants?
Please provide only the number in your answer.
✍ Solution:
- The code consists of two components: 1. stream number (in bits) and 2. sequential number (in bits). Let's find the number of bits needed to store them:
Result: 160
Video analysis of the task:
Solution 13 of the USE task in informatics (K. Polyakov, v. 4):
Message volume - 7.5 kB... This message is known to contain 7680 characters. What is the power of the alphabet?
✍ Solution:
- Let's use the formula:
I = 7.5 KB = 7.5 * 2 13 bits
\ [K = \ frac (7.5 * 2 ^ (13)) (7680) = \ frac (7.5 * 2 ^ (13)) (15 * 2 ^ 9) = \ frac (7.5 * 16 ) (15) = 8 \]
2 8 = 256
different characters
(by the formula Q = 2 N)
Result: 256
Video analysis of the task is presented after the next task.
Message encoding (text):
Solution 13 of the exam in computer science (K. Polyakov, v. 6):
The power of the alphabet is 256
. How many KB of memory will be required to save 160 pages of text containing on average 192 characters on every page?
✍ Solution:
- Let's find the total number of characters on all pages (for convenience, we will use powers of two):
\ [I = (15 * 2 ^ (11)) * 2 ^ 3 bits = \ frac (15 * 2 ^ (14)) (2 ^ (13)) KB = 30 KB \]
I = 30 Kbyte
Result: 30
See a detailed analysis of tasks for coding text: from 1 to 2100), month number (day from 1 to 12) and the number of the day in the month (number from 1 to 31). Each field is recorded separately from other fields using the smallest possible number of bits.
Determine the minimum number of bits required to encode one record.
✍ Solution:
- Formula required Q = 2 n.
- Let's calculate the required number of bits to store each item of the entire record:
Solution 13 of the exam in computer science (K. Polyakov, v. 33):
The license plate consists of several letters (the number of letters is the same in all license plates), followed by three numbers. This uses 10 digits but only 5 letters: H, O, M, E and R... You need to have at least 100 000 different numbers.
What is the smallest number of letters in a license plate?
✍ Solution:
- Formula required Q = m n.
Result: 3
We offer you to watch a video analysis of the task:
Solution 13 of the exam in computer science (K. Polyakov, v. 58):
When registering in a computer system, each user is given a password consisting of 9 characters... Use as symbols uppercase and lowercase letters of the Latin alphabet (in it 26 characters), and decimal digits... The database for storing information about each user is allocated the same and minimum possible integer number of bytes. In this case, a character-by-character encoding of passwords is used, all characters are encoded with the same and minimum possible number of bits. In addition to the password itself, additional information is stored in the system for each user, for which the 18 bytes per user. Highlighted in the computer system 1 Kb to store information about users.
What is the largest number of users that can be stored in the system? In the answer, write down only an integer - the number of users.
✍ Solution:
- Since both uppercase and lowercase letters are used, we get all the options for characters for encoding:
Result: 40
Watch the video with the solution to the task:
Task 13 2018 in the Russian language. Theory.
Since the writing very much depends on what part of speech is in front of us, then the theory should also be studied for task 23.
So together, separately or with a hyphen? Here are some life hacks.
1) Distinguish between prepositions, conjunctions and particles. Prepositions put the next word in the desired case, connect words in a sentence. Conjunctions do not change adjacent words and connect homogeneous members or sentences in a complex. Particles add a touch of meaning to a word or create a form of imperative, subjunctive, or conditional mood.
2) Words so that, too, also, but they have both continuous and separate spellings. If this is a union and it can be replaced with other unions (To = in order to. Also, also = and), then write together. Example: I came to (in order) to win. (And) I also want to go to the park. (And) He was also there. He was not very handsome, but (but) good.
3) Union IE is written separately.
4) Particles would, whether, but (if they are particles) are always written separately. (What_would_same) How to recognize particles? You can just omit them. Example: What (should) me read?
Particles THAT, EITHER, ANYTHING, SO, KA, SOME are written with a hyphen.
The particles are even, perhaps, really written together.
5) Prepositions in the course, in continuation, in the conclusion, in contrast, (at the end there may be AND, if it is not just a preposition, but a preposition with a noun) for purposes, in force, in a measure, in the area, during, in relation , except, at the expense, not counting are always written together.
6) Prepositions in spite of and in spite of are written in two words, and if we have a verbal participle in front of us, in spite of, in spite of, then we write in three words.
7) It should be distinguished: keep in mind, in view of (due to) bad weather, in the form.
8) In view of, instead of, it seems, due to, like, about, over, after, towards, we write together, if these are prepositions (can be replaced with other prepositions). If this is a preposition + noun, write separately. How to check: Try to insert a word in between. Example: Talk about (about) work. Put money into (your) account.
9) Prepositions from behind, from under, over, we write a series of hyphens.
10) Gender with nouns is written together (half a tomato) if the noun begins with a consonant. Through a hyphen, when the noun begins with l (half a lemon), a capital letter (half a Moscow) and a vowel (half a watermelon). And if between gender and noun there is still an adjective written separately (half a teaspoon)
11) Remember, in most cases, adverbs are still written LITTELY. We write through a hyphen if there are prefixes PO, B, VO in combination with the suffixes OMU, HIM, YH, THEIR, AND (in an amicable way, firstly, in a wolfish way, in the third)
Spelling prepositions |
||
Together | Apart | Hyphenated |
In view of Instead of Like As a consequence Like About In excess of In spite of Regardless of Followed by Towards against | As In connection with Continuing During Finally In contrast In order to By virtue of Moderately In the area of Throughout In a relationship With the exception of At the expense of Not to mention | Because of From under Over Pose |
Examples of |
||
In view of the difficulties Instead of you Like a deep hole Due to bad weather Like an umbrella About the lesson Over the measure Downstairs (doors) Above (letters) In the middle (of the road) Despite mistakes Despite the storm Following the summer Towards the sun Opposite the school | As an exception Due to heavy rain Throughout the day During the summer At the end of the meeting | Because of him From under the closet Fog spreads over the river |
Homonymous parts of speech |
||
In view of (preposition) - in view of (preposition + n.) City Like (preposition) him - in the genus (preposition + noun) of the Romanovs Due to (pretext) bad weather - in the investigation (preposition + noun) mistakes were made in the case During, in continuation, in conclusion - derivative prepositions for the designation of time, at the end we write E - BUT! In the course of the river, in the continuation of the novel, in the conclusion of the abstract (a simple preposition B + noun, at the end of I) Talk about a lesson (about a preposition) - put money into an account (preposition + noun) Below, above, in the middle in the absence of a controlled noun. - adverbs, but with controlled nouns. - prepositions (A man stood at the bottom (pl.) - at the bottom of the door there was a hole (preposition) |
||
Spelling of unions |
||
Together | Apart |
|
To Too Also Moreover And But Why Then From what Because of this Why That's why so If Insofar as | That is (A thousand years, that is, many) I mean (yesterday, I mean the third) for now almost wherein as if whereas because so not that ... not that that is because of because though due to the fact that then ... then |
|
Homonymous parts of speech |
||
Particle pronouns are spelled separately I listened to the same as you What should I do? No matter what Combinations of conjunctions and adverbs are written separately And so and so I read it many times and remember it this way. Combination of prepositions and pronouns What will you stay with? With that text there are tasks Not far from that house It takes a long time to walk through that forest |
||
Particle Spelling |
||
Together | Hyphenated | Apart |
Even, really, really | That, -or, -white, -ka, -tka, -s, -de, some- (coy-), -that- with verbs, adverbs and with words all the same, so | same (w), would (b), whether (e), as if, they say, as if, almost some, if it is followed by a preposition (some have) |
Spelling of numbers |
||
Together | Apart |
|
ten, hundred, hundred formed from them: fifty, eighty, four hundred, five hundred, seven hundred, seven hundred. on the-hundredth, -thousandth, -millionth: two hundredth, five thousandth, one hundred millionth | Compound numbers: five thousand seven hundred fifty three Fractional numbers: five-eighths |
|
| Spelling compound words
|
|
| Together
| Hyphenated
|
| Noun with connecting vowels "o" and "e": steamer. Exs. with elements board- and -meter: flight attendant. Exs. and adj. with foreign language elements: anti-, aviation-, auto-, bio-, bicycle-, helio-, hydro-, zoo-, inter-, counter-, macro-, micro-, mono-, moto-, neo-, radio-, stereo-, television-, ultra-, photo-, extra-: antivirus, zoologist, bike rental, etc. But! Rear admiral Compound nouns, the first part of which is an imperative verb in -i: hoarder, daredevil. But! Tumbleweed Composite words and abbreviations: university, USA, special correspondent. The names of residents, nationalities, tribes, people by their occupation, interests, affiliation with the organization: Mexican, yacht club. Complex noun, adj., Narc. adverbs, the first part of which is the numeral in the genitive case: five-volume, double. Adj., Formed from compound nouns, written without a hyphen: paintwork (paintwork). App., One of the parts of which is not used independently: omnivorous. App., Formed from two words, one of which is independent, and the other is subordinate: agricultural (agriculture). Adj., Formed by the combination of “adverb + adjective (participle): evergreen. BUT! a combination of an adverb and an adjective (participle) is written separately, if the first part answers the question how? how? and if the adverb ends in n-ski: easily amenable, friendly aimed. Appendix, the first part of which is a numeral: thirty degrees, forty minutes. Words with a half-: crescent. Words with a half-consonant, except l: half a century, half a day | Sush. consisting of two words without connecting vowels: king-bell, prime minister. Exs. with foreign language elements: vice, life, chief, non-commissioned, headquarters, ex: vice president, ex-husband. The names of the intermediate cardinal points: northwest, southeast. Names of plants with conjunction and or verb: coltsfoot. Adj., Formed from compound hyphenated nouns: southwestern. App., Formed from equal words: Russian-English (Russian and English). App., Denoting colors: gray-blue, yellow-green. Adj., Denoting quality with an additional touch: sweet and sour. Words with half-before a vowel, capital letter or consonant l: half-alphabet, half-Africa, half-lemon |
In the 13th task of the profile level of the USE in mathematics, it is necessary to solve the equation, but already of an increased level of complexity, since the tasks of the former level C begin from the 13th task, and this task can be called C1. Let's move on to examining examples of typical tasks.
Analysis of typical options for tasks No. 13 of the USE in mathematics of the profile level
The first variant of the task (demo version 2018)
a) Solve the equation cos2x = 1-cos (n / 2-x)
b) Find all the roots of this equation belonging to the interval [-5n / 2; -n].
Solution algorithm:
- t
- We do the reverse change and solve the simplest trigonometric equations.
- We build a number axis.
- We put roots on it.
- We mark the ends of the segment.
- We select those values that lie within the interval.
- We write down the answer.
Solution:
1. We transform the right-hand side of the equality using the reduction formula cos ( π/ 2−x) = sin x... We have:
cos2x = 1 - sin x.
Transform the left side of the equation using the double argument cosine formula using sine:
cos (2x) = 1−2sin 2x
We get the following equation: 1 − sin 2 x= 1− sin x
Now the equation contains only one trigonometric function sin x.
2. We introduce a replacement: t= sin x... We solve the resulting quadratic equation:
1−2t 2 =1−t,
−2t 2 +t=0,
t(−2t+1)=0,
t = 0 or -2t + 1 = 0,
t 1 = 0 t 2 = 1/2.
3. We do the reverse replacement:
sin x= 0 or sin x = ½
We solve these equations:
sin x =0↔x=πn, nЄZ
sin ( x)=1/2↔x= (-1) n ∙ ( π / 6)+πn, nЄZ.
Therefore, we obtain two families of solutions.
1. In the previous section, two families were obtained, each of which contains infinitely many solutions. It is necessary to find out which of them are in the given interval. To do this, we build a number line.
2. We put on it the roots of both families, marking them with green (first) and blue (second).
3. Mark the ends of the gap in red.
4. In the indicated interval there are three roots that are three roots: −2 π ;−11π/ 6 and −7 π/ 6.
a) πn, nЄZ;(-1) n ∙ ( π / 6)+πn, nЄZ
b) −2 π ;−11π 6;−7π 6
The second variant of the task (from Yashchenko, no. 1)
a) Solve the equation.
Solution algorithm:
- Replace this function with a variable t and solve the resulting quadratic equation.
- We make the reverse substitution and solve the simplest exponential, then trigonometric equations.
- We build a coordinate plane and a circle of unit radius on it.
- Mark the points that are the ends of the segment.
- We select those values that lie inside the segment.
- We write down the answer.
Solution:
1. Introduce the replacement t = 4 cos x. then the equation will take the form:
We solve the quadratic equation using the discriminant and root formulas:
D = b 2 - c = 81 - 4 ∙ 4 ∙ 2 = 49,
t 1 = (9 - 7) / 8 = ¼, t 2 = (9 + 7) / 8 = 2.
1. Build a coordinate plane and a circle of unit radius on it.
2. Mark the points that are the ends of the segment.
3. Select the values that lie inside the segment.
These are the roots. There are two of them.
a) 
The third variant of the task (from Yashchenko, no. 6)
a) Solve the equation 
.
b) Find all the roots of this equation that belong to the segment.
Solution algorithm:
- Using trigonometric formulas, we bring the equation to a form containing only one trigonometric function.
- Replace this function with a variable t and solve the resulting quadratic equation.
- We make the reverse substitution and solve the simplest exponential, and then trigonometric equations.
- We solve inequalities for each case.
- We write down the answer.
Solution:
1. According to the reduction formulas 
.
2. Then this equation will take the form:
3. Introducing a replacement 
... We get:
We solve the usual quadratic equation using the discriminant and root formulas:
Both roots are positive.
3. Back to the variable x:
