content: update base64 post for clarity, and include janky solve script

Author: TrebledJ

content/posts/ctf/hkcert22/2022-11-14-hkcert-2022-base64-encryption.md

@@ -43,9 +43,9 @@ We’re provided with an encryption script `chall.py` (written in Python), along
 
 So how do we go about cracking this? Brute-force will be undoubtedly inefficient as we have $64! \approx 1.27 \times 10^{89}$ mapping combinations to try. It would take *years* before we have any progress! Also we’d need to look at results to determine if the English looks right (or automate it by checking a word list)—this would take even more time! Regardless, we need to find some other way.
 
-## Let’s Get Cracking
+## First Steps: Elimination by ASCII Range
 
-Here’s one idea: since the plaintext is an English article, this means that most (if not all) characters are in the printable ASCII range (32-127). This means that the most significant bit (MSB) of each byte *cannot* be 1. We can use this to create a **blacklist** of mappings. For example, originally we have 64 mappings for the letter `A`. After blacklisting, we may be left with, say, 16 mappings. This drastically reduces the search space.[^extended-ascii]
+Here’s one idea: since the plaintext is an English article, this means that most (if not all) characters are in the printable ASCII range (32-127). This means that the most significant bit (MSB) of each byte *cannot* be 1. We can use this to create a **blacklist** of mappings. For example, originally we have 64 possible mappings for the letter `A`. After blacklisting, we may be left with, say, 16 possible mappings. This drastically reduces the search space.[^extended-ascii]
 
 Since Base64 simply maps 8-bits to 6-bits, so 3 characters of ASCII would be translated to 4 characters of Base64.
 
@@ -62,23 +62,31 @@ def get_chars_with_mask(m):
     """Get Base64 chars which are masked with m."""
     return {c for i, c in enumerate(charset) if (i & m) == m}
 
+# List the 4 Base64 positions. We'll cycle through these positions (i.e. i % 4).
 msbs = [0b100000, 0b001000, 0b000010, 0b000000]
+
+# Get impossible characters for each position.
 subchars = [get_chars_with_mask(m) for m in msbs]
 
+# Create a blacklist for each Base64 char.
+# e.g. blacklist['A'] returns the set of chars which 'A' can NOT map to.
 blacklist = {c: set() for c in charset}
 
+# Loop through each char in the shuffled Base64 text.
 for i, c in enumerate(txt):
-    # Ignore char mappings which have 1 in corresponding msb.
+    # Ignore char mappings which have '1' in corresponding msb.
     # These can't map to a printable ASCII char.
     blacklist[c] |= subchars[i % 4]
 
+# Invert the blacklist to get a dictionary of possible mappings.
+# e.g. whitelist['A'] returns the set of chars which 'A' CAN map to.
 whitelist = {k: set(charset) - v for k, v in blacklist.items()}
 ```
 
 We can check the mappings we’ve eliminated:
 
 ```python
-print(''.join(sorted(blacklist['J']))
+print(''.join(sorted(blacklist['J'])))
 # '+/0123456789CDGHKLOPSTWXabefghijklmnopqrstuvwxyz'
 ```
 
@@ -97,9 +105,12 @@ We can do a similar thing on the low end. Again, since the smallest printable AS
 def get_inverted_chars_with_mask(m):
     return {c for i, c in enumerate(charset) if ((2**6 - 1 - i) & m) == m}
 
-subchars_not_in_ascii = [get_inverted_chars_with_mask(m) for m in in_ascii] # chars that don't have bits set in ascii.
+# chars that don't have bits set in ascii.
+subchars_not_in_ascii = [get_inverted_chars_with_mask(m) for m in in_ascii]
 ```
 
+## Frequency Analysis with Known Text
+
 Another idea comes to mind. Remember the plaintext is in English? Well, with English text, some letters appear more frequently than others. The same applies to words and sequences. 
 
 {% image "assets/base64-letter-frequencies.jpg", "w-65", "Frequency of English letters. But we need to be careful with letter cases." %}
@@ -120,7 +131,7 @@ V2UncmUgbm8gc3RyYW5nZXJzIHRvIGxvdmUKWW91IGtub3cgdGhlIHJ1bGVzIGFuZCBzbyBkbyBJIChk
 {% image "assets/b64-crypt-1gram.jpg", "", "dcode.fr frequency analysis for encrypted Base64." %}
 {% endimages %}
 
-<sup>Frequency analysis of plain vs. encrypted Base64.</sup>
+<sup>Frequency analysis of plain vs. encrypted Base64. Left: CNN Lite articles. Right: Encrypted challenge text.</sup>
 {.caption}
 
 From this, we can deduce that 'w' was mapped from 'G' in the original encoding (due to the gap in frequency).
@@ -132,18 +143,22 @@ One useful option is the **bigrams/n-grams** option. We can tell dcode to analys
 {% image "assets/b64-crypt-4gram.jpg", "", "dcode.fr 4-gram for encrypted Base64." %}
 {% endimages %}
 
-<sup>Frequency analysis of 4-grams in plain vs. encrypted Base64.</sup>
+<sup>Frequency analysis of 4-grams in plain vs. encrypted Base64. Left: CNN Lite articles. Right: Encrypted challenge text.</sup>
 {.caption}
 
 Observe how "YoJP0H" occurs (relatively) frequently. This corresponds to "IHRoZS", which happens to be the Base64 encoding for " the".
 
+## More Heuristics
+
 Frequency analysis is useful to group letters into buckets. But using frequency analysis alone is painful. Some guesswork is needed. Here's the complete process I went through:
 
 - Frequency Analysis: use dcode.fr to associate frequent characters.
     - We can make use of our earlier constraints to eliminate wrong guesses.[^byebye-constraints]
 
         ```python
-        guesses = { # Dictionary of guessed mappings.
+        # Dictionary of guessed mappings.
+        # key: shuffled Base64; value: plain Base64
+        guesses = {
             'w': 'G',       'Y': 'I',
             'o': 'H',       'c': 'B',