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

...

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)

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}

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.