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Combinatorics: Linear Code Proof

  1. Nov 26, 2013 #1
    Consider S ={0120, 1010, 2011} as a subset of codes of length four over Z3 with d = 3 By (a) Show that S is a linearly independent set.

    I am asked to show S is a linearly independent set. However, if I add 0120 + 0120, I get 0210. Since 0210 is not in the set S, is S still a linearly independent set. If so, how could I show it?
    Last edited: Nov 26, 2013
  2. jcsd
  3. Nov 26, 2013 #2


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    Yes, the property you have in mind is called "closed" - indeed S is not closed (under the operation).
    Linearly independent means that you cannot express any of the codes as a linear combination of one of the others. In other words, if you would write

    $$a [0120] + b [1010] + c [2011] = [0000]$$

    for integers ##a##, ##b## and ##c##, then the only solution to the above equation would be ##a = b = c = 0##.
  4. Nov 26, 2013 #3
    A follow up question:
    Let S be a basis for a linear code C. How many code
    words does C have? Justify your answer.

    I have no clue how to go about it
  5. Nov 26, 2013 #4


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    Have you solved your original question yet, then? Because I haven't looked at it further, but I didn't give you the full answer by far.
  6. Nov 26, 2013 #5
    I found a, b, c = 0...for the follow-up question, is it just 3^3?
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