[SOLVED] 代写代考 MATH3411 Information, Codes and Ciphers (2022 T3)

Practice Test 1: Attempt review

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MATH3411 Information, Codes and Ciphers (2022 T3)

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MATH3411-5229_00252

Practice Test 1

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Started onThursday, 6 October 2022, 1:48 AM
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Completed onThursday, 6 October 2022, 1:48 AM
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Question 1
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There may be an error in the eighth digit in the following ISBN: 6-149-20074-7.

The value of the correct digit is:

ISBN-10 codes satisfy the check condition

∑i=110ixi≡0(mod11)∑i=110ixi≡0(mod11) .

A correct answer is 66, which can be typed in as follows:

Question 2
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You are given the following 7-bit ASCII codewords:

J 1001010 V 1010110 p 1110000  l 1101100
x 1111000 W 1010111 F 1000110 h 1101000

Define a 4-character 8-bit ASCII burst code by encoding characters in blocks of 3 together with a 4th character which is used as a check codeword (This is similar to the 9-character 8-bit ASCII code studied in lectures). The message

01010110 11110000 11001010 01001100

is received but contains a single error. What is the corrected and decoded message?

(No answer given)

None of these

Here, the message should, when put into rectangular format, have an even number of 1s in each row and columns. If there is a row or a column that doesn’t have an even number of 1s, then it contains an error.

A correct answer is:

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In a ternary linear code CC, the codeword x=x= 112122121010 has minimal weight among nonzero codewords.

What is the maximal numbers of errors that can always be corrected by CC?

A correct answer is 44, which can be typed in as follows:

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Let CC be a ternary linear code with parity check matrix

H=⎡⎣⎢⎢⎢1000100001000200011000012122⎤⎦⎥⎥⎥H=[1100002001210100001020000012]

Assume that the check bits correspond to columns [1,3,5,6][1,3,5,6].

The codeword xx encoding the message m=m= 011 in CC is:

Hint: Make sure that you haven’t confused the check bits and information bits.

A correct answer is 10211111021111, which can be typed in as follows:

Question 5
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Let CC be a ternary linear code with check matrix

H=H=⎡⎣⎢100012011010001102⎤⎦⎥[100001011100021012]

Suppose that the codeword xx was transmitted and received as the word y=y= 220100.
Assuming that there occurred at most one error, correct yy to find the codeword xx:

Hint: Consider the syndrome S(y)=HyTS(y)=HyT and look for it, or a scalar multiple of it, among the columns of HH.

A correct answer is 220101220101, which can be typed in as follows:

Question 6
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Let CC be the ternary linear code with generator matrix
G=G= [20011110101220][20111120110020]
What is the minimal weight w(C)w(C)?

Tip: If GG had been in standard form, then we could use the nice shortcut of quickly finding a parity check matrix HH and using the results of Tutorial Problem 19 to find our answer. However, GG is rarely of standard form here, so we need to look a little more closely at the code words and their weights. Luckily, generator matrices let us do that without too much effort, especially when the code is very small as in this question.

A correct answer is 33, which can be typed in as follows:

Question 7
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Let CC be the binary linear code with basis

B={101010,101111,110110}B={101010,101111,110110}

with information bits in positions 33, 44,and 55.
State the codeword xx that encodes the message m=m= 010:

Tip: You can solve equations and use row reductions here – but since the basis is small, it might be easiest just to look at its codewords and use intelligent trial and error.

A correct answer is 000101000101, which can be typed in as follows:

Question 8
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Let CC be the code consisting of all vectors x=x1x2x3x4∈Z47x=x1x2x3x4∈Z74 satisfying the check equation

x1x1++2x26x2+2×3+2×4≡≡0(mod7)0(mod7)x1+2×2+2×4≡0(mod7)x1+6×2+2×3≡0(mod7)

Assuming that x1x1 and x2x2 are the information bits,
find the codeword xx that encodes the message m=51m=51 :

Hint: Watch out that you don’t confuse the information bits with the check bits – that’s very easy to do.

A correct answer is 51505150, which can be typed in as follows:

Question 9
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Consider a radix 22 22-error correcting (possibly non-linear) code CC of length n=9n=9.
What is the greatest possible value of |C||C|, according to the Sphere-Packing Bound?

Tip: The code CC is not necessarily non-linear, so its size is less fixed.
Note that a code CC with the parameters that we have here might not actually exist – but we are choosing to ignore this possibility for the purposes of this question.

A correct answer is 1111, which can be typed in as follows:

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Consider a symmetric binary channel with constant bit-error probability pp, where errors in different positions are independent.
Suppose that a codeword xx is sent from the binary repetition code with codewords of length 55, and the word yy is received.
The probability that the error(s) in yy can be detected using a pure error detection strategy is:

(No answer given)

p5+5⋅(1−p)⋅p4+10⋅(1−p)2⋅p3+10⋅(1−p)3⋅p2+5⋅(1−p)4⋅pp5+5⋅(1−p)⋅p4+10⋅(1−p)2⋅p3+10⋅(1−p)3⋅p2+5⋅(1−p)4⋅p
10⋅(1−p)3⋅p2+5⋅(1−p)4⋅p10⋅(1−p)3⋅p2+5⋅(1−p)4⋅p
5⋅(1−p)⋅p4+10⋅(1−p)2⋅p3+10⋅(1−p)3⋅p2+5⋅(1−p)4⋅p+(1−p)55⋅(1−p)⋅p4+10⋅(1−p)2⋅p3+10⋅(1−p)3⋅p2+5⋅(1−p)4⋅p+(1−p)5
10⋅(1−p)3⋅p2+5⋅(1−p)4⋅p+(1−p)510⋅(1−p)3⋅p2+5⋅(1−p)4⋅p+(1−p)5
5⋅(1−p)⋅p4+10⋅(1−p)2⋅p3+10⋅(1−p)3⋅p2+5⋅(1−p)4⋅p5⋅(1−p)⋅p4+10⋅(1−p)2⋅p3+10⋅(1−p)3⋅p2+5⋅(1−p)4⋅p

Note: This question mostly just asks whether the code can detect given numbers of errors.

A correct answer is:
5⋅(1−p)⋅p4+10⋅(1−p)2⋅p3+10⋅(1−p)3⋅p2+5⋅(1−p)4⋅p5⋅(1−p)⋅p4+10⋅(1−p)2⋅p3+10⋅(1−p)3⋅p2+5⋅(1−p)4⋅p

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