#OK to post homework 
#Natalya Ter-Saakov, Jan 21, 2024, Assignment 1
read "C1.txt":

#Problem 3

#A:=[a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z]:

#CCg(P,A,key): Inputs a message, a list of characters, P, an alphabet A  an integer key from 0 to 26 outputs a shifted incrypted message. 
#Note: This procedure will not work as intended if A contains 'P', 'A' or 'T'

CCg:=proc(P,A,key) local T,ind:
for ind from 1 to nops(P) do
if not (P[ind] in A) then
return FAIL:
fi:
od:

for ind from 0 to nops(A)-1 do
T[A[ind+1]]:=A[(ind+key) mod nops(A) +1]:
od:

[seq(T[P[ind]],ind=1..nops(P))]:

end:

#Problem 4

read "ENGLISH.txt":

LocFreq:=proc(startlen, endlen) local len, loc, letter, A, count, word, T, sum:
A:=[a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z]:
for letter in A do
	for len from startlen to endlen do
		for loc from 1 to len do
			T[len, loc][letter]:=0:
		od:
	od:
od:

for len from startlen to endlen do
	for word in ENG()[len] do
		for loc from 1 to len do
			T[len, loc][word[loc]]+=1:
		od:
	od:
od:

for len from startlen to endlen do
for loc from 1 to len do
T[len, loc]:=[seq(evalf(T[len, loc][A[letter]]/nops(ENG()[len])), letter=1..26)]:
print(len, loc, T[len, loc]):
od:
od:
end:


#This is not the most efficient calculation, but it works.

TotalFreq:=proc(startlen, endlen) local len, letter, loc, A, count, word, F, sum:
A:=[a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z]:
F:=[]:
for letter from 1 to 26 do
	count:=0:
	for len from startlen to endlen do
		for word in ENG()[len] do
			for loc from 1 to len do
				if word[loc]=A[letter] then count+=1: fi:
			od:
		od:
	od:
	F:=[op(F),count]:
od:
sum:=add(F):
[seq(evalf(F[letter]/sum), letter=1..26)]:
end:

print(TotalFreq(3,10)):