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Cryptography 300: SpyFi

Challenge
James Brahm, James Bond’s less-franchised cousin, has left his secure communication with HQ running, but we couldn’t find a way to steal his agent identification code. Can you? Conect with nc 2018shell1.picoctf.com 31123. Source.

Solution

When we connect, we are asked for our situation report, and are given and encrypted string (different depending on our input)

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$ nc 2018shell1.picoctf.com 31123
Welcome, Agent 006!
Please enter your situation report: bla
4d6276d172d79a9b7da9098f69e9b403024a64082b1f2cdfa32d27d78ca83236e30cf5302c3f000ff2cb7b8a7106351c3fbf30cac64b016bb3fd112a28ed5478da62ea7c24a6c2956c96fbaaa3097dee75e268403189f06f1250226d6d557f9dcbadccc3485faa80ca8c04ff845819b5fb49b0414a48d3cb4993e15d5c9a05a7935360d079133605136cd91c3a85559adf26219527ad17f80a6fece062d71d6f

$ nc 2018shell1.picoctf.com 31123
Welcome, Agent 006!
Please enter your situation report: zzzzzzzzzzzzzzz
4d6276d172d79a9b7da9098f69e9b403024a64082b1f2cdfa32d27d78ca83236e30cf5302c3f000ff2cb7b8a7106351cc71a0b099baa79288b88380b6e1176093062f7a7751b1cd658a5bd1068c1e66da61b6aa76648777223b15da96a1f249e4ea96647c833ab65e5ec4ed409fe3414af4759f13ad15cc038b0084cdd2f7440456b962fdff9ae64e7a083561b1fda1a8e456864e71843a630207321a2728d26d94dd9b0159dce7368382298d639b818

Looks like the beginning is always the same.. After some experimentation, 11z’s pads the first bit of text nicely to 64 characters, then every other section are 16 character sections.

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$ python -c 'print("z" * 11 + "a" * 64 + "b" * 64)' | nc 2018shell1.picoctf.com 31123 | sed 's/Please enter your situation report: //g' | fold -w 32
Welcome, Agent 006!
4d6276d172d79a9b7da9098f69e9b403
024a64082b1f2cdfa32d27d78ca83236
e30cf5302c3f000ff2cb7b8a7106351c
c71a0b099baa79288b88380b6e117609
99908ad37adef3fb5a94680c5a64c6ca # aaaaaaaaaaaaaaaa
99908ad37adef3fb5a94680c5a64c6ca
99908ad37adef3fb5a94680c5a64c6ca
99908ad37adef3fb5a94680c5a64c6ca
5e6605d02027026603b6f00863d32bc5 # bbbbbbbbbbbbbbbb
5e6605d02027026603b6f00863d32bc5
5e6605d02027026603b6f00863d32bc5
5e6605d02027026603b6f00863d32bc5
843f702396b187b716fdfd818e0ebff1
20de53dad133d6a28195473711d21861
7577b2866bbc75dcb721fc57c1cc51ad
ad9aa6494e74fdfa8d484b6a0a895ea6
30ad4028ee72122451ffdbdad0e04c37
efc6c5598b5e33fee6d2c043ed8032e7
292a0b28f8fb2f711e633fd63f526f09

So obviously prefixed/postfixed with something and we’ll need to figure out the contents. Using $ for newline, we can get this mapping

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Agent,$Greetings   4d6276d172d79a9b7da9098f69e9b403
. My situation r   024a64082b1f2cdfa32d27d78ca83236
eport is as foll   e30cf5302c3f000ff2cb7b8a7106351c
ows:$zzzzzzzzzzz   c71a0b099baa79288b88380b6e117609

And that means the end parts maps:

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----------------   ================================

$My agent identi   843f702396b187b716fdfd818e0ebff1
fying code is: ?   20de53dad133d6a28195473711d21861
????????????????   7577b2866bbc75dcb721fc57c1cc51ad
????????????????   ad9aa6494e74fdfa8d484b6a0a895ea6
.$Down with the    30ad4028ee72122451ffdbdad0e04c37
Soviets,$006$000   efc6c5598b5e33fee6d2c043ed8032e7

Making a total guess about the last line. So let’s see if we can find the right
number of 0s to pad with.

Let’s at least confirm we’re on the right path?

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$ python -c 'print("z" * 11 + "a" * 64 +  "fying code is: p" +"a" * 64)' | nc 2018shell1.picoctf.com 31123 | sed 's/Please enter your situation report: //g' | fold -w 32
Welcome, Agent 006!
4d6276d172d79a9b7da9098f69e9b403
024a64082b1f2cdfa32d27d78ca83236
e30cf5302c3f000ff2cb7b8a7106351c
c71a0b099baa79288b88380b6e117609
99908ad37adef3fb5a94680c5a64c6ca
99908ad37adef3fb5a94680c5a64c6ca
99908ad37adef3fb5a94680c5a64c6ca
99908ad37adef3fb5a94680c5a64c6ca
20de53dad133d6a28195473711d21861 # fying code is: p
99908ad37adef3fb5a94680c5a64c6ca
99908ad37adef3fb5a94680c5a64c6ca
99908ad37adef3fb5a94680c5a64c6ca
99908ad37adef3fb5a94680c5a64c6ca
843f702396b187b716fdfd818e0ebff1
20de53dad133d6a28195473711d21861 # MATCH!!!!
7577b2866bbc75dcb721fc57c1cc51ad
ad9aa6494e74fdfa8d484b6a0a895ea6
30ad4028ee72122451ffdbdad0e04c37
efc6c5598b5e33fee6d2c043ed8032e7
292a0b28f8fb2f711e633fd63f526f09

So now we need to push it around so we can guess the characters in the middle
and check against the hash lower down.

This was semi-automated in

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import string
import sys
import time
import socket

for z in range(0, 95, 20):
    nc = Netcat('2018shell1.picoctf.com', 31123)
    nc.read() # Hello
    nc.read() # enter report
    wrap_start = 'z' * 11 + 'a' * 16

    wrap_end = 'a' * (16 + 11)  +'\n'

    # picoCTF{@g3nt6_1$_th3_c00l3$t_3355197}
    inputs = [
        'c00l3$t_3355197' + y
        for y in ['_'] + list(string.printable[z:min(z + 20, 95)])
    ]


    nc.write(wrap_start + ''.join(inputs) + wrap_end)

    resp = nc.read() # output


    def splitn(line, n=32):
        return [line[i:i+n] for i in range(0, len(line), n)]


    # split on the 'a' * 16
    # print('\n'.join(splitn(resp)))

    prefix, queries, postfix = resp.split('99908ad37adef3fb5a94680c5a64c6ca')

    pm = list(splitn(postfix))

    # ignore prefix

    for (q, i) in zip(splitn(queries), inputs):
        print(i, q in pm)
        if q in pm:
            sys.exit()

    time.sleep(1)

This would pass 20 ‘query’ strings at a time (part of the string + 1 new

letter), since the script seemed to limit the volume of input it would actually
attempt. This actualy worked surprisingly well. We started with the input string
ode is: picoCTF and have it find the {, and then just remove the left most
character, decrease the counter for the postfix padding of as, and add the
new letter to the end of the known input string.

Flag

picoCTF{@g3nt6_1$_th3_c00l3$t_3355197}
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