MATH3411 INFORMATION, CODES & CIPHERS Test 1 2015 S2 SOLUTIONS
Multiple Choice: a, b, b, d, c, c, d, e, c, a
2. (b): One or two errors.
3. (b): There are 24 = 16 codewords, namely the 16 linear combinations of the rows of G.
Copyright By PowCoder代写 加微信 powcoder
4. (d): There is just one codeword ending with 1011, namely the sum of rows 1, 3 and 4 of G. That is, the codeword is mG (which works because the information bit columns in G form the identity matrix).
7. (d): The Kraft-McMillan number K = 1 must be at most 1 for UD rli
codes. Testing values of r = 2,3,… gives us that r = 5 is the minimum radix that satifies this. (You could also draw a decision tree.)
Draw a decision tree.
One dummy symbol is needed.
Since H is in standard form (I B), we can find a generator matrix easily:
T 0 1 1 1 0 G=(−BI)= 10101
(ii) C = {00000, 01110, 10101, 11011}.
(iii) w(00000) = 0, w(01110) = 3, w(10101) = 3, w(11011) = 4.
(iv) d(00000, 01110) = 3, d(00000, 10101) = 3, d(00000, 11011) = 4, d(01110, 10101) = 4, d(01110, 11011) = 3, d(10101, 11011) = 3.
(v) TheminimumdistanceofC isd=3=2t+1wheret=1,soC is 1-error correcting and detecting. This is using the usual minimal distance decoding strategy. If we allow any strategy, then the code can correct up to 1 error and detect up to 2 errors.
Multiple choice: d, d, c/d, a, b, c, c, b, e, b
3. (c): Four errors. Update: The problem had been written so as to imply that a detection-only decoding strategy was being used, so if the receiver receives a message with three errors, then the receiver will detect the presence of errors. However, the problem could also be interpreted such that a minimal distance strategy was being used, in which case three errors would be seen as just a single error, and be (incorrectly) corrected. Under this interpretation, (d) is correct.
4. (a): There are 23 = 8 codewords, namely the 8 linear combinations of the rows of G.
5. (b): There is just one codeword ending with 011, namely the sum of rows 2 and 3 of G. That is, the codeword is mG (which works because the information bit columns in G form the identity matrix).
6. (c): 001 could either be c3 or c4c1.
7. (c): The Kraft-McMillan number K = 1 must be at most 1 for
UD codes. Testing values of r = 2,3,4,… gives us that r = 4 is the
minimum radix that satifies this. (You could also draw a decision tree.)
Draw a decision tree.
One dummy symbol is needed.
Since H is in standard form (I B), we can find a generator matrix easily:
T 1 0 1 1 0 G=(−BI)= 11101
(ii) C = {00000, 10110, 11101, 01011}.
(iii) w(00000) = 0, w(10110) = 3, w(11101) = 4, w(01011) = 3.
(iv) d(00000, 10110) = 3, d(00000, 11101) = 4, d(00000, 01011) = 3, d(10110, 11101) = 3, d(10110, 01011) = 4, d(11101, 01011) = 3.
(v) TheminimumdistanceofC isd=3=2t+1wheret=1,soC is 1-error correcting and detecting. This is using the usual minimal distance decoding strategy. If we allow any strategy, then the code can correct up to 1 error and detect up to 2 errors.
程序代写 CS代考 加微信: powcoder QQ: 1823890830 Email: powcoder@163.com