semaphore / Misc / Nullcon HackIM CTF Goa 2025

Challenge description:

prompt

When we follow URL, we get a single GIF:

challenge

Explore

I remember some encoding similar to Morse code that use two flags, but i forgot the exact name. So i ask ChatGPT:

Then I googled semaphore signaling and got the following table:

image

Clearly, author want us to decode this into a string with A-Z letters.

Quick view at 010editor show there are 203 frames:

image

Which is a lot for manual decoding. So let's think for a second.

How to reduce manual work extracting text from signals?

Guess 1

Maybe frames are reused and we have to decode only 26 letters?

Second result for Google search view gif by frame do the job of splitting my gif into multiple frames. So we get:

md5look4@md5look4-HP-Pavilion-Laptop-15-eg0xxx:~/Documents/hackerdom/misc$
$ ls flags | head -10
frame_000_delay-0.5s.gif
frame_001_delay-0.5s.gif
frame_002_delay-1.5s.gif
frame_003_delay-0.5s.gif
frame_004_delay-0.5s.gif
frame_005_delay-2s.gif
frame_006_delay-0.5s.gif
frame_007_delay-0.5s.gif
frame_008_delay-0.5s.gif
frame_009_delay-0.5s.gif
md5look4@md5look4-HP-Pavilion-Laptop-15-eg0xxx:~/Documents/hackerdom/misc$
$ ls flags | wc -l
203

let's test our hypnotize about resuing same PNGs for frames:

$ md5sum flags/* | sort | awk '{print $1}' | uniq -c | sort -n
      1 07c56311531859e6e7a421b367c59a80
      1 246137e88069fd3a5c00ab5cde9be9ef
      1 2f65a6b964ad828e79715e53342e2147
      1 b52f995ce9e8e2e1e4eda3d05fec6b59
      1 b718dfb5f0dbfbfb6cf9208037ed90c3
      1 d030f996f5bacd59df8c9ca02d11b144
      1 d21fbacfe32ec01a880e4b76cd1eca84
      2 1a6647e7ac8c39e058220c8e5510399e
      2 227ce1d6e3d80917bc270f39070fe60b
      2 3a6468956e8d5e570dadaa5f6934e1ce
      3 64009ef7b8978af0340a59bab80eb753
      3 d608f855e9f003b4954d9b1f1cab622b
      4 099ae99dc81fa85ea1852f97b9b9ea79
      4 2ceb567460be8331855498dfd6c212b4
      8 b54a32e827a7cc03efbb3cf6df8f9975
      9 276e02098c36e245cf74e048647b98c0
     10 11d99130fad67fe59ff43d49335018d9
     10 2c890bd907326c1b792fb6848e561069
     10 cdb69fd01a6b2f99fdb66340a62e1f88
     11 fefe09e08c1a126afb320998bfc492f8
     13 0dfbfdc9e93e6cadf5649f48104f6dee
     14 1b1a316f957972f510a55e0337bb5640
     14 646490bd6050ea6ed07bb9b7e70280b1
     14 8934acb78225e848adbddb62c0411b79
     18 c488274ddd68b681d72923bd734ebf45
     21 cf12d500f95a97cbcd8a30b3aef85acf
     24 c14e160d79cf01c70a24c597c1aadaff
md5look4@md5look4-HP-Pavilion-Laptop-15-eg0xxx:~/Documents/hackerdom/misc$
$ md5sum flags/* | sort | awk '{print $1}' | uniq -c | sort -n | wc -l
27

We see that frame are indeed reused. And we only have to manually decode 27

Now we want to generate mapping md5 → letter

I do it manually. And get something like this:

md5toletter = {
 '07c56311531859e6e7a421b367c59a80': 'Q',
 '099ae99dc81fa85ea1852f97b9b9ea79': 'M',
 '0dfbfdc9e93e6cadf5649f48104f6dee': 'B',
 '11d99130fad67fe59ff43d49335018d9': 'K',
 '1a6647e7ac8c39e058220c8e5510399e': 'AAAA',
 '1b1a316f957972f510a55e0337bb5640': 'C',
 '227ce1d6e3d80917bc270f39070fe60b': 'AAA',
 '246137e88069fd3a5c00ab5cde9be9ef': 'GGG',
 '276e02098c36e245cf74e048647b98c0': 'P',
 '2c890bd907326c1b792fb6848e561069': 'N',
 '2ceb567460be8331855498dfd6c212b4': 'L',
 '2f65a6b964ad828e79715e53342e2147': 'KK',
 '3a6468956e8d5e570dadaa5f6934e1ce': 'AAAAA',
 '64009ef7b8978af0340a59bab80eb753': 'J',
 '646490bd6050ea6ed07bb9b7e70280b1': 'H',
 '8934acb78225e848adbddb62c0411b79': 'D',
 'b52f995ce9e8e2e1e4eda3d05fec6b59': 'NN',
 'b54a32e827a7cc03efbb3cf6df8f9975': 'F',
 'b718dfb5f0dbfbfb6cf9208037ed90c3': 'EE',
 'c14e160d79cf01c70a24c597c1aadaff': 'A',
 'c488274ddd68b681d72923bd734ebf45': 'G',
 'cdb69fd01a6b2f99fdb66340a62e1f88': 'O',
 'cf12d500f95a97cbcd8a30b3aef85acf': 'E',
 'd030f996f5bacd59df8c9ca02d11b144': 'FF',
 'd21fbacfe32ec01a880e4b76cd1eca84': 'R',
 'd608f855e9f003b4954d9b1f1cab622b': 'AA',
 'fefe09e08c1a126afb320998bfc492f8': 'I'}

Some frames encode same letter, but have different duration. Most of the frames are 0.5s duration, so i try to interpret 2s as 4 repeat of letter. I also have version that does not repeat, if this is wrong path...

res_rep = 'QBAAAEFAAAAGOADBEEAAAEAAGDENPAKADABAEAKAAAAABOBIOHKLHGPOAHPKGAPKNNBMEIABIAAPLJDIOAAAAABABAIAKLPENIEAIONHODKKAEHEFFECACPGGGMGBGHCOHEHIHECAEIFEFEFACPDBCODAANAKEMGJGOGLCNEMGBHJGFHCDKCAFCEGEDCNDEDIDCDECNHDGFGNGBHAGIGPHCGFANAKANAKAAAAR'
res = 'QBAEFAGOADBEAEAGDENPAKADABAEAKABOBIOHKLHGPOAHPKGAPKNBMEIABIAPLJDIOABABAIAKLPENIEAIONHODKAEHEFECACPGMGBGHCOHEHIHECAEIFEFEFACPDBCODANAKEMGJGOGLCNEMGBHJGFHCDKCAFCEGEDCNDEDIDCDECNHDGFGNGBHAGIGPHCGFANAKANAKAR'

What encoding was used?

Guess 1

Maybe this is some substitution cipher?

let's do freq analyze:

In [2]: Counter(res_rep).most_common()
Out[2]: 
[('A', 54),
 ('E', 23),
 ('G', 21),
 ('D', 14),
 ('H', 14),
 ('C', 14),
 ('B', 13),
 ('N', 12),
 ('K', 12),
 ('I', 11),
 ('F', 10),
 ('O', 10),
 ('P', 9),
 ('L', 4),
 ('M', 4),
 ('J', 3),
 ('Q', 1),
 ('R', 1)]

This doesn't seem like substitution cipher, because there is no punctuation and two vowels are already most common.

But it's interesting that only first 18 letters are used in this ct:

In [3]: set(map(lambda c: ord(c) - ord('A'), res_rep))
Out[3]: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17}

This can suggest some baseN encoding, that by definition encode bytes using some alphabet of length N.

Guest 2.

Its looks like base18 with alphabet of prefix of standard alphabet 'A-Za-z0-9+/='

We can use this: Crypto.number.long_to_bytes(int(x, 18)) to decode x from base18:

but int use different alphabet: 0-9A-Z, so we need to translate our ct first

In [37]: translate = lambda c: '0123456789ABCDEFGH'['ABCDEFGHIJKLMNOPQRSTUVWXYZ'.index(c)]

In [38]: translate('A')
Out[38]: '0'

In [39]: res_rep_new_alpha = ''.join(map(translate, res_rep))

In [40]: res_rep_new_alpha
Out[40]: 'G10004500006E03144000400634DF0A0301040A000001E18E7AB76FE07FA60FADD1C4801800FB938E00000101080ABF4D8408ED7E3AA0474554202F666C61672E74787420485454502F312E300D0A4C696E6B2D4C617965723A205246432D343832342D73656D6170686F72650D0A0D0A0000H'

In [41]: from Crypto.Util.number import long_to_bytes

In [42]: long_to_bytes(int(res_rep_new_alpha, 18))
Out[42]: b"x\xe9\xbd\xfe\x97\xd9\xda\x006p+\xf0`\x88g*\x12&\xfa:2\x1eGA,\x89\xb4\xb7\xc9\x93\xb7\xc7\xf9\xabq\x18\xe9]\xdb\xa9\xee\xc5b\xed\x1e.\xa7\x89|)Cw5v'\x93\xa4\x12\xdc\\\xc5\x03;\xeb'\x1d]\xade\x10\xcb\x02Vc\xef\xc0\x00\xf2\xc9X\\O,\xd9&\xb0>\x1eL\xb1\xa0\x98@\xca\xcb\xcf\x14\xeeQ\xd3Z\x04\xaa|\xb7\xda\xe1\x02\xfc}B\xc5\\\xe1e\xcbY'\xe1\xd1"


In [44]: long_to_bytes(int(''.join(map(translate, res)), 18))
Out[44]: b'Pq\xb8\x8a\x96G\xc6\xe8*\x00\xe989{S\xf7\x92F\xf4\xc4\xa97\x12\xbd\xcfJi\x8f\xae~\x9bB\x00\xc6\x99\xb6U3pa`\x94\x1aEd\xa6-\x9c\xa7\x1f\xc8B\xf367\xed\x0b\xbbV\xc7dg\xf59\xe1\xb6\x00\x0fb\x8c\xab\xf4Q\xb1\xa8\\b\x88(&U\xf0zzM58csg\xf5\xe2\x08|\xc2g\xcbW\x8f\xdc\xbf\x18|\x1b\xcb\t'

this not base18

after 5 minutes staring into res_rep I notice:

Guess 3

18 unique letters is pretty close to 16, which will be well known base16 aka hex. Also 17 and 18 letters (R and Q) appears only once at start and at end. So maybe just ignore them:

we can reuse translate, and only change base in int func:

In [45]: long_to_bytes(int(''.join(map(translate, res[1:-1])), 16))
Out[45]: b'\x01\x04Pn\x03\x14\x04\x064\xdf\n\x03\x01\x04\n\x01\xe1\x8ez\xb7o\xe0\x7f\xa6\x0f\xad\x1cH\x01\x80\xfb\x93\x8e\x01\x01\x08\n\xbfM\x84\x08\xed~:\x04tT /lag.txt HTTP/1.0\xd0\xa4\xc6\x96\xe6\xb2\xd4\xc6\x17\x96W#\xa2\x05$d2\xd3C\x83#B\xd76V\xd6\x17\x06\x86\xf7&P\xd0\xa0\xd0\xa0'

In [47]: long_to_bytes(int(''.join(map(translate, res_rep[1:-1])), 16))
Out[47]: b'\x10\x00E\x00\x00n\x03\x14@\x00@\x064\xdf\n\x03\x01\x04\n\x00\x00\x01\xe1\x8ez\xb7o\xe0\x7f\xa6\x0f\xad\xd1\xc4\x80\x18\x00\xfb\x93\x8e\x00\x00\x01\x01\x08\n\xbfM\x84\x08\xed~:\xa0GET /flag.txt HTTP/1.0\r\nLink-Layer: RFC-4824-semaphore\r\n\r\n\x00\x00'

And we see part of http request:

GET /flag.txt HTTP/1.0\r\nLink-Layer: RFC-4824-semaphore

Response to this request gives us the flag

---

If u interested in how I optimize my manual parsing of frames and automate decoding frames into res and res_rep here is a whole script:

import os
import hashlib
import pprint
import subprocess

md5topath = {}
md5toletter = {
 '07c56311531859e6e7a421b367c59a80': 'Q',
 '099ae99dc81fa85ea1852f97b9b9ea79': 'M',
 '0dfbfdc9e93e6cadf5649f48104f6dee': 'B',
 '11d99130fad67fe59ff43d49335018d9': 'K',
 '1a6647e7ac8c39e058220c8e5510399e': 'AAAA',
 '1b1a316f957972f510a55e0337bb5640': 'C',
 '227ce1d6e3d80917bc270f39070fe60b': 'AAA',
 '246137e88069fd3a5c00ab5cde9be9ef': 'GGG',
 '276e02098c36e245cf74e048647b98c0': 'P',
 '2c890bd907326c1b792fb6848e561069': 'N',
 '2ceb567460be8331855498dfd6c212b4': 'L',
 '2f65a6b964ad828e79715e53342e2147': 'KK',
 '3a6468956e8d5e570dadaa5f6934e1ce': 'AAAAA',
 '64009ef7b8978af0340a59bab80eb753': 'J',
 '646490bd6050ea6ed07bb9b7e70280b1': 'H',
 '8934acb78225e848adbddb62c0411b79': 'D',
 'b52f995ce9e8e2e1e4eda3d05fec6b59': 'NN',
 'b54a32e827a7cc03efbb3cf6df8f9975': 'F',
 'b718dfb5f0dbfbfb6cf9208037ed90c3': 'EE',
 'c14e160d79cf01c70a24c597c1aadaff': 'A',
 'c488274ddd68b681d72923bd734ebf45': 'G',
 'cdb69fd01a6b2f99fdb66340a62e1f88': 'O',
 'cf12d500f95a97cbcd8a30b3aef85acf': 'E',
 'd030f996f5bacd59df8c9ca02d11b144': 'FF',
 'd21fbacfe32ec01a880e4b76cd1eca84': 'R',
 'd608f855e9f003b4954d9b1f1cab622b': 'AA',
 'fefe09e08c1a126afb320998bfc492f8': 'I'}

if not md5toletter:
    for img in os.listdir('flags'):
        img_path = os.path.join('flags', img)
        with open(img_path, 'rb') as f:
            h = hashlib.md5()
            h.update(f.read())
            md5topath[h.hexdigest()] = img_path

    pprint.pprint(md5topath)

    for md5, p in md5topath.items():
        subprocess.run(['open', p])
        print(md5, p)
        md5toletter[md5] = input()


    pprint.pprint(md5toletter)

res_rep = ""
res = ""


for img in sorted(os.listdir('flags'), key=lambda x: int(x.split('_')[1])):
    img_path = os.path.join('flags', img)
    with open(img_path, 'rb') as f:
        h = hashlib.md5()
        h.update(f.read())
        res += md5toletter[h.hexdigest()][0]
        res_rep += md5toletter[h.hexdigest()]

print(f'{res_rep = }')
print(f'{res = }')


import IPython
IPython.embed()