# OK to post
# RDB, 2024-01-21, HW1

# 1. Exempt.

# 2. I've read through RSA details before, but admittedly I don't remember them
# that well! I promise to remember them well soon.

# 3.

CCg := proc(msg, alphabet, k)
    chars := {op(msg)}:
    dict := table():

    for c in chars do
        idx := ListTools[Search](c, alphabet):
        if idx = 0 then
            error("message contains ", c);
            return FAIL:
        fi:

        dict[c] := idx:
    od:

    get_shift := x -> (dict[x] + k - 1) mod nops(alphabet) + 1:
    [seq(alphabet[get_shift(x)], x in msg)]:
end:

# To check.
alph := [seq(0..9)];
outs := [seq(CCg([0], alph, k), k=0..10)];
ins := [seq(CCg(outs[k + 1], alph, -k), k=0..10)];

# 4.

# I'd like to raise the objection that we should really be working with strings
# here, e.g. "a", "i", and so on. If I happened to use the variable `k` above,
# then we would get nonsense when reading ENGLISH.txt! This would make the
# below code much easier as well, because we could use more of StringTools.

# Nontheless, I've done this using the current variables-as-letters format.

makeFreqLists := proc(list_of_words)
    res := table():

    # Make a 0'd out table.
    for i1 from 3 to 10 do
        for j1 from 1 to i1 do
            res[i1, j1] := [0 $ 26]:
        od:
    od:

    denoms := [seq(nops(length_words), length_words in list_of_words)]:
    F := [0 $ 26]:

    for len from 3 to 10 do
        for word in list_of_words[len] do
            for pos from 1 to len do
                # Going to cheat here and use strings.
                # lowercase letters are 97-122 in ASCII, and StringTools[Ord]
                # converts letters to ASCII values.
                sym := word[pos]:
                idx := StringTools[Ord](convert(sym, string)) - 96:
                res[len, pos][idx] += 1 / denoms[len]:
                F[idx] += 1:
            od:
        od:
    od:

    n_letters := add(k1 * denoms[k1], k1=3..10):
    op(res), F / n_letters, F:
end:

read "ENGLISH.txt":
A, B, C1 := makeFreqLists(ENG()):

# Lots of data here.
# I'm most interested in knowing when "e" is the most common letter (as it is
# across all letters in all positions).

e_winners := []:
e_losers := []:
for len from 3 to 10 do
    for pos from 1 to len do
        if max[index](A[len, pos]) = 5 then
            e_winners := [op(e_winners), [len, pos]]:
        else
            e_losers := [op(e_losers), [len, pos]]:
        fi:
    od:
od: