#Homework 4 Due 2/4/24:

#Ok to Post

#1: Examples from p91-end:

#Matrix Operations:

with(linalg):
A:=matrix(3,3,[1,4,2,5,6,3,5,5,2]);
gaussjord(A);
det(A);
B:=inverse(A);
nullspace(A);


#Eigenvalues and Jordan Form:
jordan(A, `Q`);
print(Q);
eigenvals(A);
eigenvals(A^2);

#Loops:

for i from 1 to 10 do
	print(i^2-i+1);
od:



#---------------------------------------------

#2: 0-1 Matrix w/ #(rows), #(columns) both prime and "draw" initials such that the 1's show it and 0s are background; convert to one-dimensional vector

#NOTE THAT p>8, q>11!!!!

NameMat:=proc(p,q) local A, B, rowzero, colzero,r, c, C, F, d,K,H,v;

A:=Matrix(8,11,[1, 0, 0, 0, 1, 0, 1, 0 ,0, 0, 1,
1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1,
1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1,
1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1,
1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1,
1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1,
1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]);

A:

r:=p-8;
B:=Matrix(r, 11, 0);
c:=q-11;
d:=p;
C:=Matrix(p, c, 0);
C;

K:=<<A>,<B>>;
H:=<<K|C>>;

print(H);
v:= Vector(convert(H, list));
v:

end:



#---------------------------------------------

#3: Use RC with q=2, d=3,5,7 and 5<=n<=16 and K very big

#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:

#RV(q,n): A random word of length n in {0,1,..,q-1}
RV:=proc(q,n) local ra,i:
ra:=rand(0..q-1):
[seq( ra(), i=1..n)]:
end:

#RC(q,n,d,K): inputs q,n,d, and K and keeps picking K+1 (thanks to Nuray) random vectors
#whenver the new vector is not distance <=d-1 from all the previous one 
RC:=proc(q,n,d,K) local C,v,i,c:
C:={RV(q,n)}:
for i from 1 to K do
 v:=RV(q,n):
  if min(seq(HD(v,c), c in C))>=d then
   C:=C union {v}:
  fi:
od:
C:
end:

for n from 5 to 16 do:
print(nops(RC(2,n,3,1000)));
od:
print('c');

#The Maple output agrees with the Table in general across multiple #experiments. Typically, the Maple output value is lower than the #value given in the table, because the codes obtained by our procedure #are not necessarily optimal. In particular, for larger n, our codes #are much smaller. 

for n from 5 to 16 do:
print(nops(RC(2,n,5,1000)));
od:
print('c');

#In most experiments, the low n results are close to or equal to the #results in the table. The results for large n are still much lower #than those in the table. There are more choices of words to add for n #large, and making multiple suboptimal choices could have a more #pronounced effect on the ultimate size of the code

for n from 7 to 16 do:
print(nops(RC(2,n,7,1000)));
od:
print('c');

#


#---------------------------------------------

#4:How big are the binary Hamming codes? (What is their M?) Use the #Sphere Packing Bound. 

#We have the following parameters for the binary Hamming codes.
#n=2^r-1, q=2, d=3, t=1. Plugging these into the Sphere Packing Bound, #we obtain after minor simplification that M is at most 2^{2^{r}-1}#[1+1/(2^r-1)].