MATH3411 INFORMATION, CODES & CIPHERS Test 1, Session 2 2012, SOLUTIONS
Multiple choice: b,d,e,b,d True/False: F, T, F, F, T
(b): code can detect up to 3 errors. In fact, there is not really enough information to answer this, which was a mistake on my part, so I ac- cepted (e) as an answer.
(d): the two basis words have weight 6 and their only non-trivial com- binaiton has weight greater than 6
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(e): all the given options lead to ambiguities. With (a) c4 = c1 + c1; with(b)c4+c1 =c2+c3;with(c)c4 =c2+c2;with(d)c3 =c4+c1.
5. (d): 6. (i)
(iv) False: draw the trees. The first one has 2 non-decision nodes, the second only one so they cannot be isomorphic rooted trees.
(v) True: L = 87 98
0 1 1 1 0 0
(i)G=101010
(ii) The three rows of G will do: 011100, 101010, 110001.
(iii) 011011
(iv) See Notes for definition. Here d(C) = 3: no column is zero and clearly no two columns of H are multiples, but the first, second and last together are linearly dependent.
False Syndrome is 10.
(ii) True. in a triple error at least one row or column has an odd
number of errors
(iii) False: The Sphere Packing Bound |C| ≤ 2 for
1+n+ 12n(n−1) t = 2, n = 9, and |C| = 12 is not satisfied for n ≤ 9.
Multiple Choice: d, d, c, e, b True/False: T, T, F, F, T
(d): code can correct up to two errors.
(d): makes the code a prefix code. Also all the given options lead to ambiguities. With (a) c4 = c1 + c2; with (b) c2 = c4 + c4; with (c) c4 = c2 + c1; with (e) c3 = c4 + c2.
(e): note the code was radix 3, needs 1 dummy symbol and you need
to divide by 2 at the end. 5. (b):
(ii) True: errors could occur as the corners of a rectangle
1 0 0 0 1 0 1 1 0 1 1 1 0 0 (i)H∼0100011soG= 0111010.
True: Syndrome is 0.
(iii) False: Sphere packing bound implies |C| ≤ 2n 1+n+ 12n(n−1)
for 2-error correcting and n = 9, k = 2, |C| = 12 does not satisfy this inequality.
(iv) False: it is 6. The two basis words have weight 6 and the only non-trivial linear combination has greater weight.
(v) True: The decision trees are isomorphic, or for the two words beginning 10 flip the third bit and also flip the last bit on the word 1111.
0 0 1 0 1 1 1 0001110 1110001
(ii) 0101101
(iii) See Notes for definition. Here d(C) = 4: no column is zero, clearly no two columns of H are multiples, and for the sum of any three columns the first entry is one, but the sum of the first three columns and the last column is zero.
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