# HW 4, Pablo Blanco
# OK to post

# old code
Help := proc(): 
print(`binomCoeff(n)`): 
end:
# outputs the binomial coefficients of N-th degree as a list
binomCoeff := proc(N) local s,j, i,y:
s:= 1:
if N<0 then
 RETURN(FAIL):
fi:

if N=0 then #not sure how to return multiple lines yet
 RETURN(s):
fi:

for i from 1 to N do
 s := 0,s,0;
 y := seq(s[j]+s[j+1], j=1..i+1):
 s:= y:
od:
[y]:
end:

# a sequence (a1,..,an) is log-concave iff s[i]^2 >= s[i-1]*s[i+1] for all 1 < i < n

# input a list and the output is whether or not it is log-concave
# if the list is log-concave: output "true"
# otherwise, output a list which shows which elements fail the condition

logConcaveCheck := proc(s) local r,i,j:
r:=seq(evalb(s[i]*s[i] >= s[i-1]*s[i+1]),i=2..nops(s)-1):
if not member(false, {r}) then
return(true):
fi:

return([true, r, true]):
end:

# given two non-negative integers, generates a matrix with prime dimensions such that there is a submatrix 7x11 which says "PB"
PBMatrix := proc(m,n) local p,q, A:
p:=ithprime(4+m);
q:=ithprime(5+n);
A:= matrix(p,q, [
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0$(q-11),
0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0$(q-11),
0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0$(q-11),
0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0$(q-11),
0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0$(q-11),
0, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0$(q-11),
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0$(q-11),
0$((p-7)*(qx))
]);
end:


# turns a matrix into a one-dimensional vector
with(linalg):
vectorize := proc(A) local i,j:
return([seq(seq(A[j,i], i=1..coldim(A)),j=1..rowdim(A))]);
end:

vectorize(PBMatrix(0,0));

# RC(2,n,d,10 000)     best:   book:
#
# RC(2,5,3,10 000);   4       4   
# RC(2,6,3,10 000);   6       8
# RC(2,7,3,10 000);   10      16
# RC(2,8,3,10 000);   18      20
# RC(2,9,3,10 000);   29      40
# RC(2,10,3,10 000);  53      72-79
# RC(2,11,3,10 000);  87      144-158
# RC(2,12,3,10 000);  161     256
# RC(2,13,3,10 000);  268     512
# RC(2,14,3,10 000);  448    1024
# RC(2,15,3,10 000);  757    2048
# RC(2,16,3,10 000);  1224   2560-3276

# RC(2,5,5,10 000);  2     2 
# RC(2,6,5,10 000);  2     2 
# RC(2,7,5,10 000);  2     2 
# RC(2,8,5,10 000);  4     4 
# RC(2,9,5,10 000);  6     6 
# RC(2,10,5,10 000); 7     12 
# RC(2,11,5,10 000); 14     24 
# RC(2,12,5,10 000); 18     32 
# RC(2,13,5,10 000); 31     64 
# RC(2,14,5,10 000); 45     128 
# RC(2,15,5,10 000); 71     256    
# RC(2,16,5,10 000); 104    256-340     

# RC(2,5,7,10 000); --
# RC(2,6,7,10 000); --
# RC(2,7,7,10 000);   2
# RC(2,8,7,10 000);   2
# RC(2,9,7,10 000);   2
# RC(2,10,7,10 000);  2
# RC(2,11,7,10 000);  4
# RC(2,12,7,10 000); 4    4
# RC(2,13,7,10 000); 6    8
# RC(2,14,7,10 000); 9    16
# RC(2,15,7,10 000); 12   32
# RC(2,16,7,10 000); 18   36-37

# using sphere packing bound and perfect, we have that
# M = (2^n)/(1 + n) 
# by applying d = 2*1 + 1 implies t=1 and use 2.18 in Hill