#Charles Wolf
#homework 11

#Feel free to share this#

#CN23(L):inputs a sequence of 2's and 3's and finds its curling number if the number is less than or equal to 4.
CN:=proc(L) local i,n,c: 
n:=nops(L):
c:=1:
if n=1 then 1:
else
for i from 1 to trunc(n/2) do
if [seq(L[n-j],j=0..i-1)]=[seq(L[n-j-i],j=0..i-1)] then c:=max(c,2):
if n>=3*i then
if [seq(L[n-j-i],j=0..i-1)]=[seq(L[n-j-2*i],j=0..i-1)] then c:=max(c,3):
if n>=4*i then
if [seq(L[n-j-2*i],j=0..i-1)]=[seq(L[n-j-3*i],j=0..i-1)] then c:=4:
fi:fi:fi:fi:fi:
od:
fi:

c:
end:

#CS(n): finds mu(n), except for possibly very high numbers where a 4 or 5 may occur.  But I don't think Maple can handle that anyway.
CS:=proc(n) local c,L,i,j,L1,mu:
mu:=n:

for i from 0 to 2^(n)-1 do

#In order to generate each sequence of length n of 2's and 3's, I start with [2$n], and then add the binary representation of a number, which is just
#a sequence of 0's and 1's.  This generates every sequence of 2's and 3's of length n.

L:=[2$n]:
L1:=convert(i,base,2):
L:=L+[op(L1),0$(n-nops(L1))]:

c:=CN(L):
if c>1 then
for j from n while (c>1) do
c:=CN(L):
L:=[op(L),c]:
mu:=max(j,mu):
od:
fi:

od:
mu:
end: