BronoCTF : No Laughing Matter

Challenge

A single file, aha.txt, containing nothing but space-separated 8-letter
“words” made up of only two characters: A and H.

$ cat aha.txt
AHHAAAHA AHHHAAHA AHHAHHHH AHHAHHHA AHHAAAHH AHHAHHHH AHHHHAHH
AHAHAHAH AHAAAHHA AHAHAHAH AHAAHHHA AHAAHHHA AHAHHAAH AHAAHHAA
...

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Spotting the pattern

Two things give this away immediately:

1. Only two distinct symbols (A and H) → almost always means binary in disguise (0/1 mapped to two letters).
2. Every “word” is exactly 8 characters long → 8 bits = 1 byte. That’s the strongest tell — this is ASCII, one character per group.

So the plan is: pick a mapping (A → 0, H → 1, or vice versa), convert
each 8-character group to a byte, and read it as ASCII.

Solving it

cipher = open("aha.txt").read().split()

decoded = ""
for byte in cipher:
    bits = byte.replace("A", "0").replace("H", "1")
    decoded += chr(int(bits, 2))

print(decoded)

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int(bits, 2) parses the 8-character "0"/"1" string as a base-2 number,
and chr() turns that number into its ASCII character. Running it against
all 35 groups in the file reconstructs the flag directly — no trial and
error needed once you notice the two-symbol/8-length pattern.

Flag

bronco{UFUNNYLMAOLOLXDIJBOLROFLHAHA}

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Takeaways

• Alphabet size is a huge clue. A ciphertext using only 2 distinct characters is very likely binary; 16 distinct characters often means hex; 64 (plus +, /, =) means base64.
• Fixed-width grouping matters. 8-character groups screamed “one byte per group” before a single line of code was written — always check group lengths against 8 (bytes), 2 (hex pairs), 4 (base64 quads), etc.
• Guessing the 0/1 mapping only has two possibilities (A=0,H=1 or A=1,H=0) — if the first guess produces garbage, just flip it and re-decode; it’s a cheap check.