Finished checkpoint 2

1. Writing my own code definitely helped me understand the problem better. It was kind of lucky that the most common character in each of the secret texts was in fact " ". It made me understand the work that the cipher is doing better.
2. I watched like 3 different youtube videos about the Kasiski examination process after reading the wiki page. I quickly used "the" to get a difference to the next "the" that mapped the same as 42, so I knew the keyword had to be a multiple of 2,3, or 7. I tried the process with 3 and got nowhere, so I tried it with 6 on a whim. I got the first two letters of the code to be PY, then I couldn't figure out the third so I skipped it, and got the 4th letter to be H, so I made the guess that the keyword was PYTHON and I was right!

answers.md

@@ -31,5 +31,9 @@
 
 5. What is the polyalphabetic secret word?
 
+PYTHON (this literally took me so long but it was also so fun omg)
+
 6. Decrypt this message, which was encrypted using the same secret word:
    "EbZhdaV[h^bTpchhQnhig]X[VmhhRP]ftXVnRfjVY]fgtO_X]("
+
+'The treasure is a worthless ball of aluminum foil.' (anticlimatic, lol)

caesar_cracker.py

@@ -22,10 +22,27 @@ def crack_caesar(ciphertext):
             target = char
 
     diff = Bits(target,8).int - Bits(" ",8).int
+
     cipher = CaesarCipher(diff)
-    decryptedtext = cipher.decrypt(fulltext)
+    decryptedtext = cipher.decrypt(fulltext) #I included this part to double check that the function works
 
     return diff, decryptedtext
 
 
+def polyalpha_helper(text): #used to help crack the code!
+    groups = [[],[],[],[],[],[]]
+    for i in range(len(text)):
+        for n in range(6):
+            if i%6 == n:
+                groups[n].append(text[i])
+
+    groupmax = []
+
+    for alist in groups:
+        counter = dict(Counter(alist))
+        sorter = sorted(counter.items(), key=lambda item: item[1], reverse = True)
+        max = sorter[0]
+        groupmax.append(max)
+
+    print(groupmax)