#Isaac Lam, HW4, Feb 7 2024

#Old code
#Alphabet {0,1,...q-1}, Fqn(q,n): {0,1,...,q-1}^n
Fqn:=proc(q,n) local S,a,v:
option remember:
if n=0 then
 RETURN({[]}):
fi:

S:=Fqn(q,n-1):

{seq(seq([op(v),a],a=0..q-1), v in S)}:

end:

#Def. (n,M,d) q-ary code is a subset of Fqn(q,n) with M elements with
#minimal Hamming Distance d between any two members
#It can detect up to d-1 errors
#
#If d=2*t+1 correct t errors.
#
#
#HD(u,v): The Hamming distance between two words (of the same length)
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:

#SPB(q,n,d): The best you can hope for (as far as the size of C) for q-ary (n,2*t+1) code
SPB:=proc(q,n,t) local i:
trunc(q^n/add(binomial(n,i)*(q-1)^i,i=0..t)):
end:

 #2. 111010001001001110111    
  #3. Even with k=10000 my results were very poor, past n=6 for d=3, n=9 for d=5 and n=12 for d=7 I was unable to reach the bounds;
                               

#4. Since they are perfect codes, their M values are equal to the sphere packing bounds  SPB(2,2^(r)-1,3) would be the M value for these hamming codes;