WorksheetsPart 3 (101-125)
Total questions: 26
Worksheet time: 13mins
101. Specify the error position in the string "1001110", if the initial string was encoded with Hamming
(7,4) code using the following structure (i1, i2, i3, i4, r1, r2, r3)
r1
no error
r2
i4
102. Specify the error position in the string "1001111", if the initial string was encoded with Hamming
(7,4) code using the following structure (i1, i2, i3, i4, r1, r2, r3)
r1
r3
r2
i4
103. Specify the error position in the string "1011110", if the initial string was encoded with Hamming
(7,4) code using the following structure (i1, i2, i3, i4, r1, r2, r3)
i1
i3
i2
i4
104. Specify the error position in the string "1101110", if the initial string was encoded with Hamming
(7,4) code using the following structure (i1, i2, i3, i4, r1, r2, r3)
i1
i2
i3
i4
105. Specify the formula to find the amount of information if events have different probabilities.
Hartley's formula
Shannon's formula
Fano's formula
Bayes' formula
106. Specify the formula to find the amount of information if events have the same probabilities.
Shannon's formula
Hartley's formula
Fano's formula
Bayes' formula
107. Specify the right formula if dmin is Hamming distance, s - number of correctable errors and r -
number of detecteable errors.
dmin>= s+r+1
dmin>= 2s+r+1
dmin>= s+2r+1
dmin>= s+r+2
108. Suppose the letters a, b, c, d, e, f have probabilities 1/2, 1/4, 1/8, 1/16, 1/32, 1/32 respectively.
Which of the following is the Huffman code for the letter a, b, c, d, e, f?
11, 10, 011, 010, 001, 000
0, 10, 110, 1110, 11110, 11111
11, 10, 01, 001, 0001, 0000
110, 100, 010, 000, 001, 111
109. Suppose the letters a, b, c, d, e, f have probabilities 1/2, 1/4, 1/8, 1/16, 1/32, 1/32 respectively.
What is the average length q of the Huffman code?
3,0
1,9
2,7
4,3
110. The amount of information in the message is 120 bits. Calculate the length of this message, which
is written by characters of 16-character alphabet.
30
480
120
130
111. The amount of information in the message is 60 bits. Calculate the length of this message, which is
written by characters of 4-character alphabet.
30
60
15
510
112. The basic idea behind Shannon-Fano coding is to
compress data by using more bits to encode more frequently occuring characters
compress data by using fewer bits to encode more frequently occuring characters
compress data by using fewer bits to encode fewer frequently occuring characters
expand data by using fewer bits to encode more frequently occuring characters
112. The basic idea behind Shannon-Fano coding is to
compress data by using more bits to encode more frequently occuring characters
compress data by using fewer bits to encode more frequently occuring characters
compress data by using fewer bits to encode fewer frequently occuring characters
expand data by using fewer bits to encode more frequently occuring characters
113. The efficiency of the language is 0,25 and its I average is 1 bit. Calculate the number of letters in
this language's alphabet?
32
16
8
64
114. The first code combination is 0000 and the Hamming distance of this code equals 4. Choose the
second combination.
1111
1011
0011
0000
115. The Hamming code is a method of _______.
Error control coding
Optimal coding
None of the above
116. The Hamming distance between "client" and "server" is
0
1
6
impossible to detect
117. The Hamming distance between "make" and "made" is
4
3
1
impossible to detect
118. The Hamming distance between "push" and "pull" is
0
4
2
impossible to detect
119. The Hamming distance between "starting" and "finishing" is
4
3
impossible to detect
5
120. The Hamming distance between 001111 and 010011 is
1
2
3
4
121. The Hamming distance between 010111 and 010011 is
2
3
1
4
122. The Hamming distance between 011111 and 010011 is
1
3
2
4
123. The Hamming distance between 101001 and 010011 is
1
2
4
3
124. The length of the message is 16 symbols and the message's alphabet consists of 32 symbols. Find
the amount of information in this message.
80
16
64
32
125. The length of the message is 6 symbols and the message's alphabet consists of 32 symbols. Find the
amount of information in this message.
30
6
32
24
