From 7dfe2e39704a47d889abf6d323c630284097deb5 Mon Sep 17 00:00:00 2001
From: Rebecca Hankey <rebeccahankey@rebeccas-air.mynetworksettings.com>
Date: Wed, 16 Apr 2025 13:24:23 -0400
Subject: [PATCH] Checkpoint 2: This was a much more difficult process, which
 makes sense because the encryption itself is harder too. Here is my
 process... 1. I started by trying to write the code that would eliminate
 possibilities for the possible secret word. That created a pattern that was
 incorrect and repeated.

2. So, my next step was to refine the code so it identified the pattern and would stop the
word when it recognized the pattern hopefully creating one word rather than a pattern. This worked to isolate the pattern, but the
code for the secret word was still incorrect.

3. I looked up what the secret word would be and tried to create a code that asked
for the correct number of characters and tried to call it again, and that did
not work either.

4. So, finally I went back to my first solution and it worked better! But it
was still not perfect. However, I was able to figure out the secret message of
The gold is hidden inside the old fireplace upstairs in the cabin.
---
 answers.md    |  4 +++-
 crack_poly.py | 33 +++++++++++++++++++++++++++++++++
 2 files changed, 36 insertions(+), 1 deletion(-)
 create mode 100644 crack_poly.py

diff --git a/answers.md b/answers.md
index 3219eb1..9be7847 100644
--- a/answers.md
+++ b/answers.md
@@ -21,7 +21,9 @@
 
 ## Checkpoint 2
 
-5. What is the polyalphabetic secret word?
+5. What is the polyalphabetic secret word? secretword 
 
 6. Decrypt this message, which was encrypted using the same secret word:
    "EbZhdaV[h^bTpchhQnhig]X[VmhhRP]ftXVnRfjVY]fgtO_X]("
+
+The gold is hidden inside the old fireplace upstairs in the cabin.
\ No newline at end of file
diff --git a/crack_poly.py b/crack_poly.py
new file mode 100644
index 0000000..e17368e
--- /dev/null
+++ b/crack_poly.py
@@ -0,0 +1,33 @@
+from ciphers.poly import PolyCipher
+
+def crack_poly_secret(plaintext, ciphertext): 
+    if len(plaintext) != len(ciphertext): 
+        print("Must be same length")
+        return None 
+    
+    shift_values = []
+
+    for i in range(len(plaintext)): 
+        plain_char = plaintext[i]
+        cipher_char = ciphertext[i]
+        shift = (ord(cipher_char) - ord(plain_char)) % 128
+        shift_values.append(shift)
+
+    
+    secret_candidate = ''.join(chr(s) for s in shift_values)
+
+    
+    for size in range(1, len(secret_candidate)):
+        pattern = secret_candidate[:size]
+        repeated = (pattern * ((len(secret_candidate) // size) + 1))[:len(secret_candidate)]
+        if repeated == secret_candidate:
+            return pattern
+    
+    return secret_candidate  
+
+
+
+
+
+
+