#OKtoPost

#1: Read CH 5
#2: Human Exercises 5.1, 5.2, 5.3, 5.5

#5.1: This is not a binary linear code, because there is no k such #that 2^k=24. 


#5.2:Take the matrix composed of all the standard basis vectors of #length n-1 and add a column of 1s to the right side of the matrix. #This is a generator matrix and is also conveniently in standard form. 


#5.3:We simply need to show that C is a subspace of V(n,q). That is, #we show closure under vector addition and scalar multiplication. If #x,y are in C, then xH^T=0 and yH^T=0, so (x+y)H^T=xH^T+y^T=0+0=0. If #x is in C and a is a scalar, then axH^T=a(0)=0. We thus achieve the #desired result.  

#5.5: Suppose C is a binary linear code. Call E the set of even-weight 
#elements and O the set of odd-weight elements. Let v be a particular #odd-weight element. Then E+v is a set of odd-weight elements, so O is #at least as big as E. Likewise, O+v is a set of even-weight elements, #so E is at least as big as O. We conclude that if there are any odd-#weight elements in C, there are equally as many odd-weight elements #as even-weight elements. 


#3: SearchLC(q,n,k,K): inputs q and n for GF(q)^n and k (for the #dimension of the linear code, a certain subspace of GF(q)^n) and a #very large positive integer K (say 1000) and keeps generationg, K #times, random lists, M, of k members of GF(q)^n, for each determines #MinW(q,M) and outputs the one with the largest minimum weight.

#Run it, with K=1000 for q=2,3,5 ;  4 ≤ n ≤ 10   ; ; and 3 ≤ k ≤ 5 


RV:=proc(q,n) local ra,i:
ra:=rand(0..q-1):
[seq( ra(), i=1..n)]:
end:


HD:=proc(u,v) local i,co:
co:=0:
for i from 1 to nops(u) do
  if u[i]<>v[i] then
      co:=co+1:
  fi:
od:
co:
end:

LtoC:=proc(q,M) local n,k,C,c,i,M1:
option remember:
k:=nops(M):
n:=nops(M[1]):
if k=1 then
 RETURN({seq(i*M[1] mod q,i=0..q-1) }):
fi:
M1:=M[1..k-1]:
C:=LtoC(q,M1):
{seq(seq(c+i*M[k] mod q,i=0..q-1),c in C)}:
end:



#MinW(q,M): The minimal weight of the Linear code generated by M over GF(q)
MinW:=proc(q,M) local n,C,c:
n:=nops(M[1]):
C:=LtoC(q,M):

min( seq(HD(c,[0$n]), c in C minus {[0$n]} )):

end:

#--

SearchLC:=proc(q,n,k,K) local minn, largestmin, bigMinM, i, M, j:
largestmin:=0;
for i from 1 to K do:
M := {seq(RV(q,n), j = 1..k)};
minn:=MinW(q,M);
if minn>largestmin then 
 largestmin:=minn;
 bigMinM:=M;
fi:
od:
print(bigMinM);
end:


for n from 4 to 10 do:
for k from 3 to 5 do:
SearchLC(2,n,k,1000);
SearchLC(3,n,k,1000);
SearchLC(5,n,k,1000);
od:
od: