Task 1. The alphabet of the Multi tribe consists of 8 letters. How much information does one letter of the alphabet contain?
Problem 2. The information volume of one symbol of some message from the alphabet of the “Pulti” tribe is 6 bits. How many symbols are in the alphabet of the tribe with the help of which the Pultis composed the message?
Problem 3. A message written with letters from the 128-character alphabet contains 30 symbols. How much information does it contain?
Task 4. A message composed using a 32-character alphabet contains 80 symbols. The other message is composed using 64-character alphabet and contains 70 symbols. Compare the volumes of information contained in the messages.
Problem 5: An information message of 4 Kbytes contains 4096 characters. How many characters contains the alphabet that was used to write the message?
Problem 6. How many kilobytes does the message consist of 512 symbols of the 16-character alphabet?
Problem 7. A 256-character alphabet was used to write the text. Each page contains 30 lines of 70 characters per line. How much information do the 5 pages of text contain?
Task 8. A message is on 3 pages of 25 lines. Each line contains 60 characters. How many characters are in the used alphabet, if the whole message contains 1125 bytes?
Problem 9: A user enters text from the keyboard at a rate of 90 characters per minute. How much information is the text that he was typing for 15 minutes (using computer alphabet)?
Problem 10. A user entered text from the keyboard for 10 minutes. What is the speed of his entering the information, if the informational volume of the received text is 1 Kbyte?
Problem 11. A researcher observes the change of a parameter which can take one of seven values. The values are recorded using a minimum number of bits. The researcher recorded 120 values. Determine the information volume of the observation results.
Problem 12. If each character is encoded by two bytes, what is the information volume of the following sentence in Unicode: Today it is 35 degrees warm.
Solutions to the problems
Problem 1. Solution: 2 i = N, 2 i = 8, i= 3 bits. Answer: 3 bits.
Problem 2. Solution: N = 2 i = 26 = 64 symbols Answer: 64 symbols.
Problem 3. Given: N = 128, K = 30 Find: It – ? Solution: 1) It = K*I, where I is the volume of one symbol 2) 2 i = N, 2i = 128, i = 7 bits – the volume of one symbol 3) It = 30*7 = 210 bits – the volume of the whole message. Answer: 210 bits – the volume of the whole message.
Problem 4. Given: N1 = 32, K1 = 80, N2 = 64, K2 = 70 Find: It1, It2 Solution: 1) It = K*I, where I is the volume of one character 2) 2i = N, 2i = 32, i = 5 bits – the volume of one character of the first message; 3) 2i = N, 2i = 64, i = 6 bits – the volume of one character of the second message; 4) It1 = K1 * i = 80 * 5 = 400 bits – the volume of the first message; 5) It2 = K2 * i2 = 70 * 6 = 420 bits – the volume of the second message; Answer: there is more information in the second message than in the first.
Problem 5. Given: K = 4096, It = 4 Kb Find: N – ? Solution: 1) N = 2i; 2) It = K*I, I = It/K = 4*1024*8/4096=8 bits – the volume of one symbol; 3) N = 28 = 256 symbols – the capacity of the alphabet. Answer: the alphabet has 256 symbols.
Problem 6. Given: N = 16, K = 500 Find: It – ? Solution: 1) It = K*I, unknown I; 2) N = 2i, 16 = 2i, i = 4 bits – the volume of one character; 3) It = 4 * 512 = 2048 bits – the volume of the whole message; 4) 2048*8/1024 = 16 Kbytes. Answer: 16 Kbytes is the volume of the whole message.
Task 7. Given: N = 256, x = 30 – number of lines, y = 70 – number of characters in the line, M = 5 – number of pages. Find: It = ? Solution: 1) N = 2i, 256 = 2I, i = 8 bits = 1 byte – the volume of one character; 2) K = x*y*M = 30*70*5 = 10500 characters – in the text; 3) It = I*K = 1 * 10500 = 10500 bytes = 10 Kbytes – the volume of all text. Answer: the volume of the whole text is 10 Kbytes.
Problem 8. Given: It = 1125 bytes, x = 25 – number of lines, y = 60 – number of characters in the line, M = 3 – number of pages. Find: N – ? Solution: 1) N = 2i, unknown I; 2) It = K*I, I = It/ K; 3) K = x*y*M = 25*60*3 = 4500 characters – in text; 4) I = It/ K = 1125*8/4500 = 2 bits – volume of one character; 5) N = 22 = 4 characters – in alphabet. Answer: there are 4 symbols in the alphabet.
Problem 9. Given: V = 90zzn/min, t = 15min, N = 256. Find: It = ? Solution: 1) It = K*I; 2) K = V * t = 90*15 = 1350 symbols contains text; 3) N = 2i, 256 = 2i, I = 8 bits = 1 byte – the volume of one symbol; 4) It = 1350 * 1 = 1350 bytes = 1.3 Kbytes – the volume of all text. Answer: the text contains 1.3 Kbytes of information.
Problem 10. Given: It = 1 Kbyte, t = 10min. Find: V = ? Solution: 1) V = K/t, K is unknown; 2) K = It / I, because the capacity of the computer alphabet is 256, so I = 1 byte. Therefore, K = 1 1024/1 = 1024 characters in the text. 3) V = 1024/10 = 102 sim/min. Answer: the text input rate is 102 symbols per minute.
Problem 11. Solution.
We know the maximum number of values we need to encode with the same number of alphabet characters. It is seven.
The bit that can take only two values (0 and 1) is used as the alphabet. To determine the minimum number of bits needed to encode one value, we will use Hartley’s formula. To what degree do we have to take a 2 to get a 7? We know that 22 = 4, and 23 = 8. Hence, the value of k is between 2 and 3 and is a fraction. But the number of bits cannot be a fractional number. Therefore, in this case, it takes 3 bits to encode one value.
Since the researcher recorded 120 values, the total information volume of the observation is 3 * 120 = 360 bits or (360 / 8 =) 45 bytes.
Answer. The information volume of 120 observations taking seven different values is equal to 45 bytes.
Problem 12. Solution.
Let’s calculate the total number of characters in the sentence including spaces, digits and punctuation marks. In this case, there are 26 characters in total. Each character is encoded by two bytes. It means that the information capacity of the sentence is 26 * 2 = 52 bytes or 52 * 8 = 416 bits.
Answer. The information volume of the sentence equals 416 bits.
