#Work on Project
#NUMBER 2:
read("/Users/kw/Documents/English.txt"):
read("/Users/kw/Documents/finalcurrent427.txt");


#Here is the transition matrix proc from our final project:
#We modify it to give counts rather than probabilities:
#_________________________________________________________
#_
MakeTrM:=proc(k1,stSize:=1) local E,F,n,m,i,j,b1,b2,Counts,colTotal,T,L,Tbl,States,st,W,Wpr:
 option remember:
 
 # Only consider words of length >=k
 if k1 = 0 or k1 = 1 then
  E:=[seq(op(L),L in ENG())]:
 else: 
  E:=ModifMtx(k1, ENG()):
  #E:=TruncMtx(k1, [seq(op(s),s in ENG())]):
 fi:
  
 E:=ConvertInd(E):          # convert all words into number lists. this is done to preserve data and for convenience.
 n:=nops(E):                # number of words

 ###
 if stSize=1 then: 
 Counts:=[[0$28]$28]:       # number of transitions
 colTotal:=[[0]$28]:        # number of appearances
 colTotal[1][1]:=n:            # every word has a start and an end
 colTotal[28][1]:=n:
 
 for i from 1 to n do:
  Counts[1][E[i][1]+1]:=Counts[1][E[i][1]+1]+1: # the first row is dedicated to "start"
  m:=nops(E[i]): # word length
  for j from 1 to m-1 do:
   colTotal[E[i][j]+1][1]:= colTotal[E[i][j]+1][1]+1: # letter E[i][j]+1 has appeared! note: "a" is colmn 2 but 1 in E.
   # terrible to read, I know. update transition count for j to j+1th letter in the ith word:
   Counts[E[i][j]+1][E[i][j+1]+1]:=Counts[E[i][j]+1][E[i][j+1]+1] + 1: 
  od:
  Counts[E[i][m]+1][28]:=Counts[E[i][m]+1][28]+1: # update "end" count
  colTotal[E[i][m]+1][1]:= colTotal[E[i][m]+1][1]+1:
 od:

 T:= [seq([seq(Counts[j][i],i=1..28)],j=1..28)]: # transition matrix!

 
 return(T):
 fi:
end:
 ## 

#__________________________________________
#END GROUP CODE
#________________________________



#words of length at least 2
M:=MakeTrM(1):

#CLEAN OFF START AND END
with(LinearAlgebra):
M:=DeleteRow(M, 27):
M:=DeleteColumn(M, 1):
M:=DeleteColumn(M, 27):
M:=DeleteRow(M,1):


#want to find number of nonzero entries 
#want to find number of entries which are greater than 10

countnonzero:=0:
countbig10:=0:

for i from 1 to 26 do
 for j from 1 to 26 do

if M(i,j)<>0 then countnonzero:=countnonzero+1: fi:
if M(i,j)>9 then countbig10:=countbig10+1: fi:


od:
od:

print(`the number of pairs which appear is`, countnonzero):
print(`the number of pairs which appear more than 10 times is`, countbig10):

#We get that 570 pairs appear, and 491 appear more than 10 times. 

#Challenge #3: This can be done with code that I have seen that my teammates wrote, but I don't want to claim it as my own for credit here.