On Apr 16, 8:10 pm, David Joyner <[email protected]> wrote: > Perhaps I don't understand your program, but it appears to not address > the issue. Here is the algorithm, if I understand the question correctly: > > Let I denote the subset of range(n) which represents the erasures. > Let v denote the vector in GF(q)^n which you want to decode. > Let C denote the [n,k,d] code with generator matrix G (so the > rows of G are a basis for the vector space C over GF(q)). > For each w in GF(q)^n, let w^I denote those coordinates of w not in I. > Let L = [] be an empty list. > For each c in C > if c^I = v^I then append c to L > return L > > This gives you the list of codewords desired. > > I don't see how the output of your programs agree with this.
I may be understanding the problem wrong. Originally I thought that the problem was as follows (in this example): given a message m in GF(3)^10, we used matrix G to encode it making a vector w in GF(3)^27. During a transmission, there were 8 or so errors in known positions (erasures is the list of these positions in my code). Assuming that we still can restore the original message m, i.e. if it is unique, the procedure 'decode' gives it from w and the list of the positions of erasures, and the procedure 'correct' produces the correct codeword corresponding to it. Maybe, I still don't understand it right, but it seems as if you are saying, that the problem is to do a similar thing in cases when the solution is not unique, producing the list of all the solutions. That can be done similarly, just using the version of solve_left in 'decode1' producing the list of all solutions (plain solve_left gives only one solution even if there is more than 1), and then multiplying each of them by G, as in 'correct', If I again understood it wrong, could you, please, give a simple example (with smaller sizes), with the correct answer? Alec -- To post to this group, send email to [email protected] To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/sage-support URL: http://www.sagemath.org
