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infrom theory-1

Total questions: 100

Worksheet time: 50mins

Name
Class
Date
1.

______ indicate(s) an error in a received combination.

a)

Parity bits

b)

Error syndrome

c)

Data bits

d)

None of the given

2.

... is a measure of uncertainty

a)

Encoding

b)

Entropy

c)

Information

d)

Redundancy

3.

A code has two allowable combinations 101 and 010. What is the allowable combination for the error combination 001?

a)

101

b)

010

c)

001

d)

None

4.

A code has two allowable combinations 101 and 010. What is the allowable combination for the error combination 100?

a)

101

b)

010

c)

100

d)

None

5.

A code has two allowable combinations 101 and 010. What is the allowable combination for the error combination 000?

a)

010

b)

101

c)

000

d)

None

6.

A code has two allowable combinations 101 and 010. What is the allowable combination for the error combination 111?

a)

101

b)

010

c)

111

d)

None

7.

A code has two allowable combinations 101 and 010. What is the allowable combination for the error combination 011?

a)

010

b)

101

c)

011

d)

None

8.

A code has two allowable combinations 101 and 010. What is the allowable combination for the error combination 110?

a)

010

b)

101

c)

110

d)

None

9.

A codeword of the Hamming code consists of _______ and _______ bits.

a)

data; parity

b)

with errors; without errors

c)

allowable; not allowable

d)

none of the given

10.

A Huffman code is a = 1, b = 000, c = 001, d = 01. Probabilities are p(a) = 0.4, p(b) = 0.1, p(c) = 0.2, p(d) = 0.3. The average length of codewords q is

a)

2.1 bit

b)

1.9 bit

c)

2.0 bit

d)

8.0 bit

11.

A redundancy of a code S = ...

a)

1 - lavr/Imax

b)

lavr/Imax

c)

1 + lavr/Imax

d)

Imax/lavr

12.

An average length of codewords qavr = ...

a)

Σ (pi*qi)

b)

Σ (pi/qi)

c)

Σpi / n

d)

Σqi / n

13.

An efficiency of a code E = ...

a)

lavr/Imax

b)

Imax/lavr

c)

lavr/100

d)

Imax - lavr

14.

ASCII code is a

a)

Variable length code

b)

Fixed length code

c)

Error-correction code

d)

None of the given

15.

By the Bayes' rule for conditional entropy H(Y|X) = ...

a)

H(X|Y) - H(X) + H(Y)

b)

[P(A)] /P(B)

c)

H(X|Y) - H(X)

d)

H(X|Y)+ H(Y)

16.

By the Bayes' theorem ...

a)

P(B|A) = P(A and B)/P(A)

b)

P(A|B) = [P(B|A)][P(A)] /P(B)

c)

P(B|A) = P(A and B)*P(A)

d)

P(A|B) = [P(B|A)][P(A)] * P(B)

17.

By the Chain rule H(X,Y) = H(Y|X) + ...

a)

H(X)

b)

H(Y)

c)

H(Y|X)

d)

H(X|Y)

18.

By the Hartley's formula the amount of information I = ...

a)

I = n*log m

b)

I = m*n

c)

I = log (m/n)

d)

I = log (m*n)

19.

By the Hartley's formula the entropy H = ...

a)

H = - Σ(pi*log pi)

b)

H = - Σ (log pi)

c)

H = log m

d)

H = - Σ (pi/log pi)

20.

By the property of joint entropy H(X,Y) <= ...

a)

H(X)

b)

H(Y)

c)

H(X) + H(Y)

d)

None of the given

21.

By the property of joint entropy H(X,Y) ...

a)

H(X,Y) >= H(X) and H(X,Y) <= H(Y)

b)

H(X,Y) <= H(X) and H(X,Y) >= H(Y)

c)

H(X,Y) >= H(X) and H(X,Y) >= H(Y)

d)

H(X,Y) >= H(X) + H(Y)

22.

By the Shannon's formula the amount of information I = ...

a)

H = - n * Σ(pi*log pi)

b)

H = - n * Σ (log pi)

c)

H = - n * Σ pi

d)

H = - n * Σ (pi/log pi)

23.

By the Shannon's formula the entropy H = ...

a)

H = - Σ(pi*log pi)

b)

H = - Σ (log pi)

c)

H = - Σ pi

d)

H = - Σ (pi/log pi)

24.

Calculate the code rate for Hamming (15,11) code

a)

1

b)

0,733

c)

0,571

d)

0,839

25.

Calculate the code rate for Hamming (31,26) code

a)

1

b)

0.839

c)

0.733

d)

0.571

26.

Calculate the code rate for Hamming (7,4) code

a)

1

b)

0.571

c)

0.733

d)

0.839

27.

Calculate the efficiency of the language if it has 32 letters and its I average is 1 bit.

a)

0.8

b)

0.2

c)

5

d)

1

28.

Calculate the redundancy of the language if it has 32 letters and its I average is 1 bit.

a)

0.8

b)

0.2

c)

5

d)

1

29.

Choose an example of block code

a)

Shannon-Fano code

b)

Huffman code

c)

Hamming code

d)

None of the given

30.

Choose conditions of an optimal coding (p – probability, l – length of a code word)

a)

pi < pj and li=lj

b)

pi > pj and li

c)

pi > pj and li>lj

d)

none of the given

31.

Choose the formula to create the Hamming code

a)

(n, k) = (2r - 1, 2r - 1 - r)

b)

(n, k) = (2r, 2r - 1 - r)

c)

(n, k) = (2r - 1, 2r - r)

d)

(n, k) = (2r - 1, 2r - 1 + r)

32.

Choose the formula to determine the number N of possible messages with length n if the message source alphabet consists of m characters, each of which can be an element of the message.

a)

N = mn

b)

N = nm

c)

N = m * n

d)

N = log m

33.

Code has dmin = 1. How many errors can be corrected by this code?

a)

2

b)

3

c)

0

d)

1

34.

Code has dmin = 1. How many errors can be detected by this code?

a)

2

b)

3

c)

0

d)

1

35.

Code has dmin = 10. How many errors can be detected by this code?

a)

4

b)

8

c)

9

d)

10

36.

Code has dmin = 11. How many errors can be corrected by this code?

a)

11

b)

7

c)

5

d)

10

37.

Code has dmin = 11. How many errors can be detected by this code?

a)

5

b)

9

c)

10

d)

11

38.

Code has dmin = 12. How many errors can be detected by this code?

a)

5

b)

10

c)

11

d)

12

39.

Code has dmin = 2. How many errors can be corrected by this code?

a)

2

b)

3

c)

0

d)

1

40.

Code has dmin = 2. How many errors can be detected by this code?

a)

2

b)

3

c)

1

d)

0

41.

Code has dmin = 3. How many errors can be corrected by this code?

a)

2

b)

3

c)

1

d)

4

42.

Code has dmin = 3. How many errors can be detected by this code?

a)

1

b)

3

c)

2

d)

4

43.

Code has dmin = 4. How many errors can be detected by this code?

a)

5

b)

1

c)

3

d)

4

44.

Code has dmin = 5. How many errors can be corrected by this code?

a)

5

b)

3

c)

2

d)

4

45.

Code has dmin = 5. How many errors can be detected by this code?

a)

6

b)

2

c)

4

d)

5

46.

Code has dmin = 6. How many errors can be detected by this code?

a)

6

b)

2

c)

5

d)

4

47.

Code has dmin = 7. How many errors can be corrected by this code?

a)

5

b)

6

c)

3

d)

4

48.

Code has dmin = 7. How many errors can be detected by this code?

a)

7

b)

3

c)

6

d)

5

49.

Code has dmin = 8. How many errors can be detected by this code?

a)

8

b)

6

c)

7

d)

3

50.

Code has dmin = 9. How many errors can be corrected by this code?

a)

5

b)

7

c)

4

d)

8

51.

Code has dmin = 9. How many errors can be detected by this code?

a)

7

b)

9

c)

8

d)

4

52.

Code rate R (k information bits and n total bits) is defined as

a)

k = n/R

b)

R = k * n

c)

R = k/n

d)

n = R * k

53.

Conditional entropy H(Y|X) lies between

a)

- H(Y) and 0

b)

0 and H(Y)

c)

- H(Y) and H(Y)

d)

0 and 1

54.

Convert the message into a signal suitable for transmission over the channel of communication, referred to as ...

a)

Encoding

b)

Decoding

c)

Entropy

d)

Redundancy

55.

Determine the Hamming distance for code that can detect 3 errors and correct 2 errors.

a)

6

b)

5

c)

7

d)

9

56.

Determine the Hamming distance for code that can detect 3 errors and correct 1 errors.

a)

5

b)

4

c)

6

d)

8

57.

Determine the Hamming distance for code that can detect 5 errors and correct 3 errors.

a)

9

b)

8

c)

10

d)

14

58.

Encode a string "0000" with Hamming (7,4) code using the following structure (i1, i2, i3, i4, r1, r2, r3)

a)

0000001

b)

0000111

c)

0000000

d)

0000101

59.

Encode a string "0001" with Hamming (7,4) code using the following structure (i1, i2, i3, i4, r1, r2, r3)

a)

0001010

b)

0001001

c)

0001011

d)

0001111

60.

Encode a string "0010" with Hamming (7,4) code using the following structure (i1, i2, i3, i4, r1, r2, r3)

a)

0010010

b)

0010111

c)

0010110

d)

0010100

61.

Encode a string "0011" with Hamming (7,4) code using the following structure (i1, i2, i3, i4, r1, r2, r3)

a)

0011100

b)

0011001

c)

0011101

d)

0011111

62.

Encode a string "0100" with Hamming (7,4) code using the following structure (i1, i2, i3, i4, r1, r2, r3)

a)

0100011

b)

0100110

c)

0100111

d)

0100101

63.

Encode a string "0101" with Hamming (7,4) code using the following structure (i1, i2, i3, i4, r1, r2, r3)

a)

0101101

b)

0101000

c)

0101100

d)

0101110

64.

Encode a string "0110" with Hamming (7,4) code using the following structure (i1, i2, i3, i4, r1, r2, r3)

a)

0110101

b)

0110011

c)

0110001

d)

0110000

65.

Encode a string "0111" with Hamming (7,4) code using the following structure (i1, i2, i3, i4, r1, r2, r3)

a)

0111110

b)

0111000

c)

0111010

d)

0111011

66.

Encode a string "1000" with Hamming (7,4) code using the following structure (i1, i2, i3, i4, r1, r2, r3)

a)

1000111

b)

1000100

c)

1000101

d)

1000001

67.

Encode a string "1001" with Hamming (7,4) code using the following structure (i1, i2, i3, i4, r1, r2, r3)

a)

1001111

b)

1001010

c)

1001110

d)

1001100

68.

Encode a string "1010" with Hamming (7,4) code using the following structure (i1, i2, i3, i4, r1, r2, r3)

a)

1010111

b)

1010001

c)

1010011

d)

1010010

69.

Encode a string "1011" with Hamming (7,4) code using the following structure (i1, i2, i3, i4, r1, r2, r3)

a)

1011100

b)

1011010

c)

1011000

d)

1011001

70.

Encode a string "1100" with Hamming (7,4) code using the following structure (i1, i2, i3, i4, r1, r2, r3)

a)

1100110

b)

1100000

c)

1100010

d)

1100011

71.

Encode a string "1101" with Hamming (7,4) code using the following structure (i1, i2, i3, i4, r1, r2, r3)

a)

1101101

b)

1101011

c)

1101001

d)

1101000

72.

Encode a string "1110" with Hamming (7,4) code using the following structure (i1, i2, i3, i4, r1, r2, r3)

a)

1110000

b)

1110101

c)

1110100

d)

1110110

73.

Encode a string "1111" with Hamming (7,4) code using the following structure (i1, i2, i3, i4, r1, r2, r3)

a)

1111110

b)

1111011

c)

1111111

d)

1111101

74.

The length of the message is 16 symbols and the message’s alphabet consists of 4 symbols. Find the amount of information in this message.

a)

A) 16

b)

B) 80

c)

C) 64

d)

D) 32

75.

The efficiency of the language is 0.5 and its I average is equal to 1 bit. Calculate the number of letters in this language’s alphabet?

a)

64

b)

32

c)

4

d)

16

76.

Elements of alphabets X and Y are statistically related. It is known that H(X)=4 bits and H(Y)=10 bits. What are a range of variation for a conditional entropy H(Y|X) when H(X|Y) changes from its min to max?

a)

(from 7 to 11)

b)

(from 4 to 12)

c)

(from 6 to 10)

d)

(from 6 to 11)

77.

For Hamming distance dmin and r errors in the received word, the condition to be able to detect the errors is

a)

dmin>= r+1

b)

dmin>= 2r+1

c)

dmin>= 2r+2

d)

dmin>= r+2

78.

For Hamming distance dmin and s errors in the received word, the condition to be able to correct the errors is

a)

dmin >= s+1

b)

dmin >= 2s+1

c)

dmin >= 2s+2

d)

dmin >= s+2

79.

Hamming (7,4) code can correct ___ error(s)

a)

2

b)

3

c)

1

d)

0

80.

Hamming distance can easily be found with ...

a)

XNOR operation

b)

XOR operation

c)

OR operation

d)

AND operation

81.

How does a noise affect the data?

a)

change only the 0 to 1

b)

change only the 1 to 0

c)

change the 0 to 1 and the 1 to 0

d)

None of the above

82.

How many data bits are in the (15, 11) Hamming code?

a)

11

b)

4

c)

15

d)

5

83.

How many data bits are in the (31, 26) Hamming code?

a)

26

b)

31

c)

5

d)

4

84.

How many data bits are in the (7, 4) Hamming code?

a)

4

b)

3

c)

7

d)

10

85.

How many parity bits are in the (15, 11) Hamming code?

a)

4

b)

15

c)

11

d)

5

86.

How many parity bits are in the (31, 26) Hamming code?

a)

26

b)

31

c)

5

d)

4

87.

How many parity bits are in the (7, 4) Hamming code?

a)

3

b)

4

c)

7

d)

11

88.

A Huffman code is a = 0, b = 10, c = 110, d = 1110, e = 1111. Probabilities are p(a) = 0.50, p(b) = 0.30, p(c) = 0.15, p(d) = 0.03, p(e) = 0.02. The average length of a code words is

a)

1.75 bit

b)

2.0 bit

c)

1.3 bit

d)

1.7 bit

89.

Which letter will get the shortest codeword after Huffman coding of the word «bbaacccabaac»?

a)

a

b)

b

c)

c

d)

none

90.

An alphabet consist of the letters a, b, c, d, e and f. The probability of occurrence is p(a) = 0.06, p(b) = 0.15, p(c) = 0.4 and p(d) = 0.18, p(e)=0.17, p(f)=0.04. The Huffman code is

a)

c=1,d=000,e=001,b=010,a=0110,f=0111

b)

c=0,d=111,e=110,b=101,a=1001,f=1000

c)

c=1,d=01,e=001,b=000,a=00010,f=00011

d)

c=1,d=01,e=001,b=000,a=0010,f=00011

e)

c=0,d=101,e=110,b=101,a=1000,f=1001

91.

If k - number of bits before Hamming encoding and n - number of bits after Hamming encoding then

a)

k > n

b)

k < n

c)

k = n

d)

k = 1/2 n

92.

In digital communication system, smaller the code rate, ... are the redundant bits.

a)

less

b)

equal

c)

more

d)

unpredictable

93.

Main idea of error control codes is

a)

To add some redundancy

b)

To delete some redundancy

c)

To double all bits

d)

None of the given

94.

Noise affects ...

a)

information source

b)

receiver

c)

channel

d)

transmitter

95.

Shannon-Fano and Huffman codes are an encoding algorithms used for

a)

lossy data compression

b)

lossless data compression

c)

error correction

d)

error detection

96.

Specify the case when entropy is maximum

a)

p1=0.5 and p2=0.5

b)

p1=1 and p2=0

c)

p1=0 and p2=1

d)

p1=0.9 and p2=0.1

97.

Specify the error position in the string "0001110", if the initial string was encoded with Hamming (7,4) code using the following structure (i1, i2, i3, i4, r1, r2, r3)

a)

i4

b)

i1

c)

i2

d)

r2

98.

Specify the error position in the string "1000110", if the initial string was encoded with Hamming (7,4) code using the following structure (i1, i2, i3, i4, r1, r2, r3)

a)

i1

b)

i4

c)

i2

d)

i3

99.

Specify the error position in the string "1001010", if the initial string was encoded with Hamming (7,4) code using the following structure (i1, i2, i3, i4, r1, r2, r3)

a)

r2

b)

r1

c)

r3

d)

i3

100.

Specify the error position in the string "1001100", if the initial string was encoded with Hamming (7,4) code using the following structure (i1, i2, i3, i4, r1, r2, r3)

a)

r3

b)

r2

c)

r1

d)

no error

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