Re: Complete solution to a modular System of equations

*To*: mathgroup at smc.vnet.net*Subject*: [mg60479] Re: Complete solution to a modular System of equations*From*: "Carl K. Woll" <carlw at u.washington.edu>*Date*: Sat, 17 Sep 2005 02:32:06 -0400 (EDT)*Organization*: University of Washington*References*: <dgdvp1$1r5$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

"mumat" <csarami at gmail.com> wrote in message news:dgdvp1$1r5$1 at smc.vnet.net... > In coding theory one is intersted to find the dual code of a given > code. To do this one has to solve a modular linear equation such as: > > > > eq = {a + b + d == 0, a + c + d == 0, b + c + d == 0} > > lhs = {{1, 1, 0, 1}, {1, 0, 1, 1}, {0, 1, 1, 1}}; rhs = {0, 0, 0}; > > LinearSolve[lhs,{0,0,0},Modulus->2] > > Out[52]= > {0,0,0,0} > > as you see it returns only one solution. How can I find all solutions. > The number of solutions is a power of 2. For about example there is > exacly one more solution which is > > {1,1,1,0}. > > Is there any function in Mathematica to do this? or I should start > writing my own code? > Use Reduce: In[11]:= Reduce[{Mod[lhs . {a, b, c, d} - rhs, 2] == 0, 0 <= a < 2, 0 <= b < 2, 0 <= c < 2, 0 <= d < 2}, Integers] Out[11]//OutputForm= (a == 0 && b == 0 && c == 0 && d == 0) || (a == 1 && b == 1 && c == 1 && d == 0) > thanks for your help in advance! > > regards, > > chekad > Carl Woll Wolfram Research