Notes from CTF tasks (math/programming-related)

Table of Contents


Who: tope of mode13h (two-man team from Norway)

Contact: tope#9134@Discord (I am friendly to the point of being flirtatious, so feel free to message me)

Audience: I am writing these notes to some imagined fellow CTFer who looked at the same tasks that I did, but missed some critical step for the solve, or perhaps someone who did solve them and just wants to compare notes. I'll rarely re-post the full problem as given in the CTF, nor do very detailed expositions. For the time being this isn't intended as a pedagogical resource. The target audience ("you") are not complete beginners, some facility with math is assumed.

Content: My favorite tasks are of the "puzzle"1 variety. In CTFs these would normally be found under misc and crypto. My safe space is the intersection between programming and mathematics.

Motivation: I am fascinated by the generalities behind puzzles–i.e. their mathematical expression–and I love solving problems with programming. Fascination gives off a glow, and this glow-energy is what I'd like to call motivation. The competitive element of CTFs has little bearing, and I tend to avoid CTFs with uninteresting problems.


General Comments

An A+ CTF! An incredible amount of tasks, from mid-tier to top-tier difficulty level. Objectively the task set alone deserves a ~100 rating on CTFTime3.

I did the crypto tasks and will outline solves for those below. I also glanced at some of the "easier" tasks in misc and web, but even the entry-level tasks in those categories were beyond me.

Personal downside #1: very biased toward web, rev, and pwn, which I am pretty clueless about, though those are the "true" staples of CTF. I realize this is more like a pure upside for most people.

On the end of the first day there were only three crypto tasks released yet more than fifteen(!?) web, pwn, and rev tasks out. I sat around twiddling my thumbs a bit, feeling pretty useless.

The only misc-like task I saw involved PHP, which is more of a mental plague than a programming language4. I played around with it a bit, but finally just went to bed feeling a little bit dumber and somewhat frustrated because I thought all tasks had been released and that I'd be useless for the rest of the CTF.

Personal downside #2: Two more crypto were released just after I'd gone to sleep on day two, so I only had a few hours for them when I woke up. Although the new tasks were a nice surprise, the timing was very unfortunate for me.

Plea for organizers in general: please consider having all tasks out by the mid-way point of the CTF. If not, then communicate your planned task release schedule so it's possible to better manage our time?

We ended up placing 13th5.


You're given some code which generates a QR-code image of the flag and uses that image to create two new randomized images which–when combined–would reconstruct the original (albeit transposed? I am guessing, I didn't actually run the code). You also receive one of the images generated. It struck me as a little cryptic.

I ignored the whole visual threshold part of the task and noted it uses Python's random module to generate the pixels in the output images. That's a lot of random data, and the QR-code image it's based on has a lot of known fixed output. After double-checking that the QR-image had enough known pixels (you get to reverse one bit of state per pixel) and where it was (it would be hell if it wasn't contiguous), it reduces to a "standard" reverse-the-RNG-state task.

For the reversing Python's MT the easy way you need \(32 \cdot 672\) contiguous bits of output. That is if you don't have to worry about missing state. You need to undo the tempering that MT does to its output words:

def untemper(x):
  x ^= (x >> 18)
  x ^= (x << 15) & 0xefc60000
  x ^= ((x << 7) & 0x9d2c5680) ^ ((x << 14) & 0x94284000) ^ ((x << 21) & 0x14200000) ^ ((x << 28) & 0x10000000)
  x ^= (x >> 11) ^ (x >> 22)
  return x

You combine words i, i+1, and i+M to get word i+N:

tmp = w[i][31] ^ w[i+1][:31]  # using slice notation to select bits
w[i+N] = tmp>>1 ^ w[i+M] ^ (A if tmp[0] else 0)

M, N, A are constants from _randommodule.c in CPython source code. Then you reapply the tempering (see aforementioned .c source) and it should match Python's output. That's the basic idea.

It was a task which is very easy to realize the solution but rather painful to implement. I struggled with infinite little indexing bugs. The headache I got from trying to directly indexing into bin(random.getrandbits(...)) was not worth it. (Bits within words run from bit \(2^{31}\) to \(1\), but the words are in little-endian order.) I even had bugs in the final solution as some randomness was leaking through, but fortunately QR-codes are highly redundant, so I didn't care. Then again I probably did things needlessly complicated by reverse-generating the original QR-code instead of simply generating the "companion" image to the one shared.

Apart from headaches it gave me with my own bugs, it's actually a fairly clever task, because there aren't any obvious "hints" that points you in the right direction, so it might take a while to notice. I was pretty lucky.


Several random 256-bit elliptic curves are generated and used to do a standard key exchange to AES-encrypt the flag. The curves are all asserted to be non-smooth and resistant to MOV attack. You get the public output of the exchanges as well as the encrypted messages. It's a very nice and clean task with a likewise straightforward implementation. It was a pure blessing after the indexical spaghetti that was VSS.

The key to realize is that the secret is reused (for both client and server). The generated random curves have a lot of small random subgroups, so you can solve the discrete log in those subgroups (multiply the generator and public keys by \((p-1)/q\) to put it in the subgroup of size \(q\)), get constraints like \(secret = x_i \pmod{q_i}\), and then do Chinese Remainder when you have enough. A partial Pohlig-Hellman, if you will.

I think I did most of this task in REPL, but the pseudo-code sketch would be:

crt_r, crt_s = [], []
for ec,pk_r,pk_s in data:
  order = ec.order()
  for f in small_factors(order):
    crt_r.append( (babystepgiantstep(f, pk_r*(order/f), g*(order/f)), f) )
    crt_s.append( (babystepgiantstep(f, pk_s*(order/f), g*(order/f)), f) )

r,s = crt(*crt_r)[0], crt(*crt_s)[0]


In this task (against a live server) you're asked for elliptic curve parameters \((p, a, b)\) and two points \((G_1, G_2)\) and then have to solve 30 rounds of distinguishing \(G_1 \cdot r \cdot s\) from \(G_1 \cdot x\) when given \(G_1 \cdot r\) and \(G_2 \cdot s\) (for random secret integers \(r, s, x\)). It will throw an exception if you give it a singular curve or points not on the curve.

At first glance I thought this was too similar to FlagBot, because there are no checks against the points being in a small subgroup. I knew you could also use trace 1 curves on which the much-copy-pasted Smart attack works, but I only have that code in some Sage docker; I wanted to use my own comfortable code and homemade Python libraries. Besides, I got this: I thought it would just be a matter of trivially putting them into small subgroups and solving the CRT again. A bit too easy, maybe…

Yeah, after a while I realized my oops: it required a very large prime modulus \(p\) and calls E.order() on the curve – and there's a timer on the program. The E.order() call takes well over a minute, sometimes several minutes, and so there's no time to do the loop. I wasted some time trying to find random curves for which E.order() was smooth or took less time but…

Finally I relented and tested E.order() on a curve with \(|E_p| = p\). It was instant, of course, so… sigh Copy-pasted Sage code it is, then.

Now the problem was to generate curves with trace 1, which I didn't know how to do, but poiko talked of a DEFCON task which had involved generating such curves and gave me a paper:

From the paper I figured I wanted curves under primes \(p = 11 m (m+1) + 3\) with \((a,b)\) which gives j-invariant equal to \(-2^{15}\), I quickly found plenty such curves. Then copy-paste Smart_Attack() and blah, blah, the usual stuff. (I really despise copy-pasting.) A bit unrewarding due to my stubbornness, but I have to admit it was a good task in the end, even if the "hidden" constraint of "the given curve must also have a fast E.order()" was a bit devious6.

crypto:easy RSA?

A fun task which had two stages.

It generates some RSA-modulus \(n = p*q\) and encrypts a large vector with lots of numbers. These numbers are the vector (dot) products between the flag and random numbers, offset by some small error, and then modulo a small prime. You are given the random numbers used for free, but the errors are secret.

The primes are generated in a truly bizarre way:

mark = 3**66
def get_random_prime():
  total = 0
  for i in range(5):
    total += mark**i * getRandomNBitInteger(32)
  fac = str(factor(total)).split(" * ")
  return int(fac[-1])

It generates a number that has "small" digits in base-\(3^{66}\) and returns the largest prime in the factorization of this number. That's one of the oddest ways to generate primes I've seen.

But (the "a-ha" moment) it means that \(n\cdot x\) also has "small" digits in base-\(3^{66}\) for some \(x\) which is itself also "small" (compared to \(n\)).

I used a lattice like

[   n * HH, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[   1 * HH, HL, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[B**1 * HH, 0, HL, 0, 0, 0, 0, 0, 0, 0, 0],
[B**2 * HH, 0, 0, HL, 0, 0, 0, 0, 0, 0, 0],
[B**3 * HH, 0, 0, 0, HL, 0, 0, 0, 0, 0, 0],
[B**4 * HH, 0, 0, 0, 0, HL, 0, 0, 0, 0, 0],
[B**5 * HH, 0, 0, 0, 0, 0, HL, 0, 0, 0, 0],
[B**6 * HH, 0, 0, 0, 0, 0, 0, HL, 0, 0, 0],
[B**7 * HH, 0, 0, 0, 0, 0, 0, 0, HL, 0, 0],
[B**8 * HH, 0, 0, 0, 0, 0, 0, 0, 0, HL, 0],

to find \(x\). There might be better ways, but it worked. Now to factor. Trivially you can find the first and last digits of the primes. There's probably some more clever way to do the whole thing, but I just did the dumb thing and used the base-\(B\) digits of \(n\cdot x\) as coefficients for a polynomial over \(\mathbb{Z}_p\) for some large \(p\) and factored that. That gave me two numbers back (when multiplied so the least/most significant digits match), each of which shared one of the big primes with \(n\). This was all a bit rushed because it took me too long to discover the "trick" of finding \(n \cdot x\) and I just did everything in the REPL at this point, using my own utility libraries and NTL wrapper.

So now I had \(n = p \cdot q\) and could decrypt the vector to get a big system of linear relations with "small errors" modulo some small prime. The lattice is life, the lattice is love. Maybe there's a more clever way to solve this as well, but did I mention I was a little pressed for time at this point? The system in its entirely was too big for my poor, slow LLL, but a small random subset of equations worked just fine (every equation involved the entire flag).


I "solved" this task but was a little too late. I had already realized the solution by the last hour or two, but underestimated how many datasets I needed and wasted some time with self-doubt, double-checking, and massaging uncooperative lattices. When the competition ended I went to eat and by the time I got back there was enough data and I got the (bittersweet) flag easily.

Like FlagBot it's a deceptively clean and simple task, where you don't realize the problem until that sweet "a-ha!" moment. The server uses generated DSA constants (a (huge) prime \(p\), a generator \(g\) for a large prime (\(q\)) subgroup of \(\mathbb{Z}_p\)) to prove that it knows \(x\) in \(y=g^x \pmod p\) by giving you \(y\), \(t=y^v \pmod p\) (for some secret ephemeral key \(v\)) and \(r = v - c \cdot x \pmod q\) with \(c\) being a SHA256 constant the recepient can derive from \((g, y, t)\). It looks respectable and above the board at first glance…

But basically you need to forget everything I just said and ignore the whole business with \(p\) and discrete logs. With \(r\) you're given a system of inequalities like \(c_i \cdot flag + r_i < flag \pmod{q_i}\), because the ephemeral key is erroneously generated to be less than \(x\) which was asserted to be 247 bits in the code (while \(q\) is 256 bits), so each equation ought to reveal a little bit more about \(x\). It is clearly another lattice problem.

I'm not sure if there are any better or more clever ways to set up the lattice, but what worked for me in the end was the most obvious and straightforward:

[[1,     0, c0, c1, ...,  cn],
 [0, 2^248, r0, r2, ...,  rn],
 [0,     0, q0,  0, ...,   0]
 [0,     0,  0, q1, ...,   0]
 [0,     0,  0,  0, ...,  qn]]

b'n1ctf{S0me_kn0wl3dg3_is_leak3d}' Since I can no longer enter it on the site, I'll leave it here to be sad and alone. :( 2020

General Comments

An unfortunate step down from last weekend's N1CTF2020 even though this one has a much higher rating on CTFTime.

DISCLAIMER: I'm judging with very biased goggles here; I'm sure web/rev/pwn were all kinds of excellent and top notch woo-hoo, but misc and crypto was very disappointing and demotivating for this rating bracket and I tapped out early. It felt too easy and/or uninspired and/or unoriginal.

I am also going to sound incredibly negative. It wasn't that bad, but it definitely wasn't a 100-point CTF, and that's the stance I'm critiquing from. The tasks I looked at felt like they belonged in the 50-70 range.

crypto:Pwnhub Collection which was (unnecessarily) gated behind rev would probably be my pick for "best" crypto-task, because it actively combines two related techniques in a sensible way. crypto:Conquering Premium Access had a copy-pastable solution through Google, which was not fun to discover. crypto:Bad Primes was somewhat trivial7. misc:BabyJS was cute but that's about it. More info about them below. (As usual I will just outline the solves, these are just notes that should allow anyone who attempted or failed the problems to solve them.)

crypto:Bad Primes8

You're given \(c = flag^e \pmod{p\cdot q}\) with all numbers known except \(flag\). The caveat is that \(e | p-1\) so there's no \(d = e^{-1} \pmod{p-1}\) for recovering the flag directly. (Finding the flag \(\pmod q\) works trivially, so we ignore that part.)

So finding the e-th root of \(c \pmod p\) is the interesting part. You can do it in two different ways:

The I-don't-want-to-learn-anything Way (also known as I'm-a-professional-I-don't-have-time-for-this): press the [Skip This Step] button and do GF(p)(c).nth_root(e) in Sage.

The Problem Child Way:

So we want a number \(x\) such that \(x^e = c \pmod p\) with \(p-1 = e\cdot s\) and \(gcd(s,e) = 1\).

The a-ha comes after exploring the hunch that \(s\) is important here. Look at \(j = e^{-1} \pmod s\). This means means \(j\cdot e = 1 + k\cdot s\) for some \(k\). In fact, it means that \(j\cdot e\cdot e = e + k\cdot (p-1)\). A-ha! So now \((c^j)^e = c^{e \cdot j} = c^{k\cdot s + 1} = x^{e\cdot (k\cdot s + 1)} = x^{k\cdot (p-1) + e} = x^e \pmod p\). Voila! \(c^j\) is a root.

Other reasoning lines probably exist too, but this was the one I took.

Next step we iteratively multiply with any e-th root of unity (= \(r^{(p-1)/e}\) for any primitive element \(r\)) to cycle through the rest of the roots, reconstructing the number in under mod \(p\cdot q\) with CRT to check if it is the flag.


A misc task that was misclassified as web(?). It's an exposition on the various ways JavaScript tries very hard to be an awful programming language9. It's still not as awful as PHP, but we can't all be the champ.

is(a, 'number');
is(b, 'number');
assert('1.1', a === b);
assert('1.2', 1337 / a !== 1337 / b);

[0.0, -0.0] works because the division gives [+Infinity, -Infinity]. A pimple, no big deal. Floats are tricky anyway.

isnt(c, 'undefined');
isnt(d, 'undefined');
const cast = (f, ...a) =>;
[c, d] = cast(Number, c, d);
assert('2.1', c !== d);
[c, d] = cast(String, c, d);
assert('2.2', c === d);

Probably a billion solutions here. I did [{}, "[Object object]"]. What's that weird rash…

let { e } = json;
is(e, 'number');
const isCorrect = e++<e--&&!++e<!--e&&--e>e++;
assert('3', isCorrect);

Up to boils now. Actual boils. <!-- is a comment. \(e=-1\) works.

const { f } = json;
isnt(f, 'undefined');
assert('4', f == !f);

f=[] works, I don't know why and I don't want to know. I fear that knowing certain things actually make you less knowledgeable.

const { g } = json;
isnt(g, 'undefined');
// what you see:
function check(x) {
    return {
        value: x * x
// what the tokenizer sees:
assert('5', g == check(g));

This one was a little cute, like a magician's clever misdirection. The original check() is replaced by this check(x) { return; ... } function. So null works. Why does it return undefined? Haha!

Blood-shot eyes, trichotillomania, psychosis.

const { h } = json;
is(h, 'number');
try {
} catch(e){}

Something like 1e1000 is converted to the string "Infinity". Makes sense, n'est-ce pas?

const { i } = json;
isnt(i, 'undefined');
assert('7', i in [,,,...'"',,,Symbol.for("'"),,,]);

This unending dream. 3 is in the array because array index 3 is defined? I don't know the logic. Any language which tries to pretend lists and maps are somehow isomorphic data structures or algebras is insane. These languages were invented by insane people.

const js = eval(`(${clean})`);
assert('8', Object.keys(json).length !== Object.keys(js).length);

Put __proto__ in the object and it will get hidden by the eval, because :jazzhands-OO:.

const { y, z } = json;
isnt(y, 'undefined');
isnt(z, 'undefined');

When I looked over the task I figured I could pass all the other checks, but this one seemed a bit need-to-actually-stop-and-think, "huh?" It had me puzzled for a while, I don't really know JavaScript all that well. (I had to ask Google if JavaScript has apply-overriding and stuff like that.) I also didn't realize that everything has a .constructor. But eventually I discovered it just by playing around in nodejs and from that a-ha the pieces fell into place.

y = "constructor" so y[y] becomes a function, i.e. y.constructor, and y[y][y] becomes the function constructor, which takes its body as a string argument (?!) to be eval'd. So z = "return console.log;" for example.

crypto:Conquering Premium Access

AES is used to encrypt the flag. You're handed the ciphertext and "power traces" (voltage measurement?) of some hardware using AES to encrypt 10,000 known 16-byte plaintexts with the same key/IV as it did the flag.

The task hints about "T-tables", "aligned" and that masking isn't employed. But the task also said "thank you to professor so-and-so for the data" and links to a university course, which is the biggest clue in my eyes. From that I figured it's probably very "textbook" and intended to be solved with least effort.

So, textbook reading material:

Indeed: almost disappointingly so. Finding a correlation based on the 1-bits-equals-more-power assumption turned out to work directly. You can ignore the rounds, ignore the hints10, ignore everything, because there's so much data and it's so artificially clean. Find the key (and iv) such that the weight of sbox[key^iv^plaintext] has the highest correlation with (overall) power consumption. sbox[key^iv^plaintext] is what the state gets set to in the first round (in CBC mode), before ShiftRows etc. (Note that ShiftRows doesn't change the weight.) Technically I ignored IV too because I simply forgot about it, but that was fine too. You can simply use the full traces, and don't have to target any point in time at all.

See also: which I also came across and seems to copy a lot of text/material from the above link, but is a much more thorough write-up than anything I can produce.

Notice how it's also a verrrry similar problem? Yep: copy-pasting that code should just work out of the box here too, though it is awful and slow so I ended up rewriting it and doing most of my playing in REPL to pretend I wasn't a fraud, trying to learn something in spite of everything.

And yeah, AES ECB to decrypt, so no IV was used, which I as stated implicitly assumed. Can't recall if the task hinted to that or not, maybe it did?

misc:P*rn Protocol

A PDF described a very simple data protocol. Implement the protocol, request login, log in with the given username and password, and get the flag.


This felt more like a "socket programming for beginners" tutorial than anything else. Why was this even part of the task set?

crypto:Pwnhub Collection

Labelled as hard crypto but really just easy-ish crypto gated behind ASAN rev. poiko reversed it for me because I'm a newbie, so can't really say much about the rev part.

So from what poiko told me, the server does something roughly equivalent of the following pseudocode:

# coll = list of pairs (category, link)
coll = [t.strip().split(' ') for t in open('collection')]
coll.append( (input(), input()) )

txt = ' '.join(x + '{' + y + '}' for x,y in sorted(coll, key=lambda x:x[0]))
# NB: sorted only on category element

print(aes_cbc(fixed_key, fixed_iv, txt).hexdigest())

So there's a string like "blah{bleh} foo{bar} qux{crux}" where we can add an element that gets sorted in.

Finding the "categories" (the strings outside the curly braces) can be done very efficiently with a binary search. Start with a string like byte(127)*n that gets sorted last, observe the ciphertext output. Then for each byte keep a high,low that you bisect and observe if we were flipped to another position in the string (an earlier ciphertext block will change). This finds all the categories very quickly.

They turned out to be crypto flag misc etc. (Which was something poiko already guessed from doing the rev part, but I just double-checked.) Next step is discovering the stuff inside the curly braces. Because I was being dumb, it took me longer than it should have to realize it's the even-more-textbook method of byte-by-byte brute forcing.

Input a category that gets sorted before flag with a long arbitrary link that aligns things like so:

# |---aes block---||---aes block---||---aes block---|
# xxxx f{aaaaaaaaaaaaaaaaaaaa} flag{xxxxxxxxxxxxxxxxx

Now pop off one a so the first x gets shifted in and note what that block becomes in the output. Then discover the byte by cycling through and check the output of encrypting with link = "aaaa...aaaa} flag{" + b for unknown byte b. I.e. the string that's encrypted becomes:

# |---aes block---||---aes block---||---aes block---|
# xxxx f{aaaaaaaaaaaaaaaaaaa} flag{*} flag{xxxxxxxxxx
# with '*' being the variable byte.

Once there's a match you add the byte to what you know is there already ("} flag{") and repeat for the next byte, until the entire link has been discovered. Print it, ship it, done.

CyberSecurityRumble 2020

General Comments

Overall a pretty good CTF. Decent number of tasks, variety, difficulty level and so forth. The stuff I looked at seemed alright, though I got entangled in some web business toward the end that soured things but that was my own fault.

We placed a cool 10th tho (much thanks to poiko spamming flags left and right the last day).

There wasn't an abundance of crypto or mathematical misc tasks, but then I don't really come to expect that much, it felt like there was enough. I kinda think of crypto/math as the awkward little sister that's there only because the organizers' mother forced them to bring her along. The cool kids are talking about real world stuff, hacking the Gibson, phreaking the phone lines, doing bug bounties, whatever it is that real hackermen do, while she's just sitting there arranging her fries to count out the Stirling numbers. Most of the time I'm just glad she's invited at all; if CTFs were to reflect real world industry, I suspect it'd be like 30 web-tasks and 2 revs.

I solved most of the easy crypto on the first evening, then did crypto:dtls the following morning, but the two remaining tasks were labelled web. Ugh. I looked at the task named blow and just thought "fuck no." Some kind of hellish JWS/JWT JavaScript task they kept adding hints to because it got no solves. From the hints alone I surmised that it has some trivial solution in crypto terms, but it was probably a pain in the ass (aka the "web" part) to even get started or progress to where it's relevant? So I worked on web:Yanote instead, but failed that task miserably and eventually went off to nurse my imposter syndrome in the corner.

crypto:Pady McPadface

A live server that knows the flag traverses it bit by bit and gives \((r^2 + bit)^e \pmod n\) for a fixed RSA modulus \(n\) using the standard \(e = 65537\). \(r\) is an ephemeral random number a few bits than \(log_2 \sqrt{n}\) so \(r^2 < n\).

So basically I wanted to discover if the ciphertexts were quadratic residues or not under the given (composite) modulus. I knew what to look up, and it turned out to be pretty easy, but I was actually surprised to learn the Jacobi symbol is easily calculable without knowing the factorization of \(n\). Huh! (It's also surprising that this has never come up before.) I'm glad to have learnt it, it does seem like a neat trick.

def trailing_zeros(a):
  return (a ^ a - 1).bit_length() - 1

def jacobi(a,n):
  assert n & 1 # n must be odd
  sign = 1

  a = a % n
  while a > 0:
    r = trailing_zeros(a)

    # 2 divides an odd number of times into a, then flip the sign if n is not
    # on the form 8k±1
    if n % 8 in (3,5) and r & 1:
      sign = -sign

    a,n = n, a >> r
    if a % 4 == 3 and n % 4 == 3:
      sign = -sign
    a = a % n

  if n != 1:
    # a divides into n
    return 0

  return sign

So then it was a matter of connecting to the server and getting several ciphertext instances and calculating their Jacobi symbols \(\left(\frac{c}{n}\right)\). If you find an instance where it is -1 then you know the number is not a quadratic residue, so \(bit\) must be 1 in that position. After some collected data you set the others to 0 and get the flag.


The task implements some basic code for working with the secp256k1 elliptic curve (\(y^2 = x^3 + 7\)). The ECC implementation is "home-made" with code mostly taken from Wikipedia. It asks for a point, multiplies it with a random secret number and asks you to recover this secret. No check is done on whether or not the point you give is actually on the secp256k1 curve though, so it's a classic invalid curve attack.

Basically you can set the \(b\) parameter in \(y^2 = x^3 + b\) to whatever you want. (If you give the point \((s,t)\) it will use the curve \(y^2 = x^3 + (t^2-s^3)\).) I've solved some task before using a singular curve but couldn't remember how the mapping worked off the top of my head and didn't find my old code for it, so instead I idly looked for other \(b\) parameters where I could solve the discrete log with Pohlig-Hellman as I had the tools for that more readily available. I think there's 7 (?) different Frobenius traces among these curves, corresponding to different orders of the resulting elliptic curve, but each had a very large factor (133-135 bits) so this turned out to be a dead end.

I went back to pen and paper to see if I could manually work out the singular mapping, partly as self-punishment for forgetting. Given one of the simplest non-trivial points, \((1,1)\), the server will use the singular non-elliptic curve \(y^2 = x^3\). The point-doubling formula then does the following: \(x \mapsto \left(\frac{3x^2}{2y}\right)^2 - 2x = \frac{1}{4}x\) and similarly \(y \mapsto \frac{1}{8}y\). Hm! That gave me the needed deja vu: indeed it seems that \(\frac{(n * (1,1))_x}{(n * (1,1))_y} = n\), so it worked out nicely, and I didn't have to wade through the full algebra. So yeah, just feed it (1,1) and then do \(\frac{x}{y} \pmod p\) to get the secret.

It's probably supposed to be a pretty trivial task, but I made it overly complicated by investigating all the curve-stuff.


A server is signing messages using the Python ecdsa library. I don't remember the exact details but I think you were supposed to forge a signature. Immediately there's was a glaring red flag with a entropy=sony_rand parameter where sony_rand was a function that returned random bytes using Python's random module. ecdsa uses this for generating the ephemeral \(k\) in its DSA signature generation.

At first I thought this was going to be a very challenging task, because even though the randomness of getrandbits isn't of cryptographic quality, it's very hard to actually mirror it unless given at least some white (fully revealed) output. I know it's seeded in Python from PID and two different clock functions, so it might be bruteable, but it's hard to predict how far into the stream the server is and so on and so forth; it seemed very daunting to brute all that. I wondered if you could get startup time or restart the server, or maybe the Mersenne Twister output had some cool linear relationship I wasn't aware of…

I was just being dumb. I didn't notice it was a forking TCP server… It will clone the state and basically just re-use the same ephemeral \(k\) every single time. This becomes immediately obvious when you set to actually collect some signatures and see the \(r\) being repeated. I got pretty lucky that it was so obvious once you start collecting data, if not it'd be screwed.

So from two signatures \((r, \frac{H_1 + r x}{k})\) and \((r, \frac{H_2 + r x}{k})\) where you can calculate the hashes \(H_1, H_2\), recovering the private multiplier \(x\) is trivial and you can sign anything. I think I just set sk.privkey.secret_multiplier = x and used ecdsa to do it in REPL.


A server will run any Python code that's a valid palindrome (x == x[::-1]) in a Python subprocess (of a user that can't read the flag file), but then exec the stdout from that process (in the process that can read the flag file). There's a limited attempt at preventing the task from becoming trivial: it will remove any of the substrings "#", '"""', "'''" from the input. The removal happens only once, in order. So, the a-ha: you can actually use """ by giving something like "'''"". The server will remove the single quotes and keep the double ones.

Unfortunately I don't have the actual solution at hand since I can't find it in my REPL history, but the basic idea was to just use a simple payload like print("__import__('os').system('sh')") and then figure out how to make that into a palindrome. I think I settled on something like """;";""";<code>;""";<reverse of code>;""";";""". Notice how """;";""" is interpreted to be the string ';";' when evaluated normally, but also provides an end for the triple-quote: ..."""; ";" "".

(Another idea would be to use some trickery involving the escape character r"\" which could potentially escape/unescape strings depending on order, but I didn't think of that at the time.)


Trivial welcome-task I had missed until poiko reminded me. It basically does print(flag ^ flag[4:]) (if one were able to do that in Python). You know flag starts with CSR{ which gives you the next four bytes, and those gives the next four, and so on.


The task gives you a pcap file and this shell script:

# /usr/lib/|head -n 1
# GNU C Library (GNU libc) stable release version 2.31.

# clone
git clone dtls

# build
cd dtls && autoconf && autoheader && ./configure && make && cd ..

# listen
tcpdump -i lo  port 31337 -w dump.pcap &

# Wait for tcpdump...
sleep 2

# serve
dtls/tests/dtls-server -p 31337 &

# send flag
cat flag.txt|dtls/tests/dtls-client -p 31337

I got the repo and looked around. I know it implements basic TLS over UDP, but ignored all that. The first thing I searched for was RNG. It doesn't use cryptographic source of random numbers but instead will just use srand() and rand() from libc (as hinted to in the script above). That's a problem in and by itself.

But it also turns out that the code is buggy, for example it initializes the RNG with dtls_prng_init((unsigned long)*buf), where buf was read to from /dev/urandom. It's probably intending to use a long for the seed, but guess what? buf is a char array, so it's actually just using a single byte for the seed to generate all those important crypto bytes.

Now, how to actually attack it. I knew I didn't want to actually interact with TLS, because TLS is web and web is a hell run by committees, so instead I did the following:

  • Make it print out the random bytes it generates. Run the script and compare it with the given pcap to figure out where those numbers go and what they should actually be.
  • Trivially brute the seed.
  • Modify the library to load its seed from an environment variable so I could run the client and server with fixed seeds.
  • Force its internal time function to return 0, as I can see from the pcap that's what it used there. (It actually tries to use integer seconds since the start of the program, thus 0 since the transfer happens immediately.)

I ran the script again and compared my pcap with the one given. It's generating and using the same numbers, uses the same cookie, etc. So then I simply copied the final encrypted packet from the pcap file and put it into the source code with an ugly hack like this:

static char bufzzz[] = "...stuff...";

static int
decrypt_verify(dtls_peer_t *peer, uint8 *packet, size_t length,
         uint8 **cleartext)
  if (packet[0] == 0x17) {
    printf("pulling the old switcharoo\n");
    packet = bufzzz;
    length = 84;

  // ...

For some reason (that I don't care about) this causes TinyDTLS' own debug printing of the message to get messed up, so I also had to print the cleartext myself at the end of the function. Out comes the flag. No TLS needed (thank God).

This was a pretty cool task. It was practical and "real-world" but without feeling contrived and without there being JavaScript or any web frameworks involved. It was possibly the "cleanest" real-world-y task I've seen. Well done to the task author.


I didn't solve this one. I failed it. Not only did I fail it technically, but I failed it mentally, which is probably worse. But I worked on it for a while so I'm still going to write about it. It triggered all sorts of feelings of inadequacy and failure that you wouldn't believe.

All because of this "web" tag.

The "web" tag usually means "use educated guesses for any blackbox behavior" but to me it usually reads as "realize that nobody agrees with what you think is reasonable." It reads "smell that? Take a deep whiff, that's the smell of learned helplessness, baby."

Clue: the server says "invalid CRC-HMAC" if you give it an invalid cookie.

Ignore the pedantic fact that it would normally be called HMAC-<name> and "CRC" is very non-specific, kind of like just saying "HASH."

However CRCs are more fun than hashes in that they're easier to play around with computationally. So for a second I dared to dream about some sort of semi-fun task of trying to figure out the modulus and key and all the other constants of some unknown HMAC(key, ipad, opad, H=CRC(modulus, a, b)) given an oracle(x) = HMAC(key, A+x+B). That would have been pretty nice actually.

So my approach was all wrong. Because I started to hear this "looOOooOOoOool" coming from a dark corner of my mind: the "web" tag coming back to haunt me. So maybe none of that cool stuff. Maybe it's just…straight up CRC-32…? Well, yes and no.

See, I hate this. Instead of "figuring something out" you have to guess what they mean about what that something is. Since it's "web" and in web they tend to not be too particular about things, HMAC might mean something else. Potentially it could be a homegrown custom HMAC, potentially it could be something else entirely. It's not even given that it does HMAC(<key>, <pickle-data>). For all you know it could do F(<key>, <username> + ":" + <permission>) or whatever. Hell, maybe it's not even a CRC! Maybe it's just adler32(<data> + <secret>)

Okay, relax. Start testing some basic stuff.

I figured out it had normal CRC behavior at least. It was, as they say, affine-ish. Working with polynomials over \(\mathbb{Z}_2\), \(CRC(x + y) = CRC(x) + CRC(y) + K_{b}\) where \(K_{b}\) is a constant that depends on the byte-lengths of x and y (technically on their xor difference?). You'd expect this to work for any HMAC based on a CRC too. This allows you to (easily) log in as admin (but on the admin page I just found people doing "test post pls ignore" and being generally "????????").

But when I actually tried to "solve" the blackbox function completely, I kept getting wrong constants everywhere. And I didn't know what was wrong, didn't know if it's because I had bugs in the old code I was using (some GF2[X] implementation from way back), or had some wrong assumption.

After a while I suspected there was no HMAC and maybe just a straight up unknown CRC, which frustrated me even more because if so, then this should be easy, no? What if something else is different? I spammed random stuff trying to figure out the modulus for the CRC from the constants I was getting out, double-checked my math… I even wondered if you were supposed to birthday-hammer user-create to find a duplicate checksum, and maybe that would finally reveal what was going on, got completely lost…

Getting full control over the cookie is trickier, but in my more clear-headed state I think it's possible without solving the black-box function as I was trying to do. I believe all you need is in the relationship. You can get all the constant K terms you need by making three users that xor to zero for any given length. For an unknown CRC, for three constants \(K_i\), \(K_{i+1}\) and \(K_{i+2}\) (that you'd get from ("0","q","A") for example, since they xor to 0), then you can find the modulus as a factor of \(x^8 K_{i+2} + (x^8+1) K_{i+1} + K_{i}\) (does that work? Now I'm not sure, I'm copying a note…), then the relationship between the various \(K_i\) as \(K_n = K_0 \frac{x^n-1}{(x-1) x^{n-1}} \pmod M\) (TODO: math might be wrong), and you should have everything you need.

Then you make usernames that xor together to be your pickle payload (the server accepted non-printables in username) and change the byte for the username length (this last part is a pen-and-paper mess, so it's not something I'd actually try unless I was sure). The stuff after the byte 0x2e (".") in the pickle data is ignored, so it's fine if the cookie ends with garbage. This was probably the intended solution, though it's a bug-prone approach, maybe a bit painful to implement.

However I basically just got so frustrated with not getting the math to work out (I mean, I couldn't even figure out the damn modulus!11) and debugging my own code that I gave up. I have an issue with committing to guesses unless I have a reasonable idea it will work. Failure-sensitive. Besides, engaging with guessy stuff feels dirty unless you vibe some level of mental harmony with the task author (which I didn't).

Dirty confession: I even debased myself and tried to brute force a 32-bit key against HMAC-CRC32 using the usual constants, but it didn't work. That was the last straw. After that I signed off and went for a long walk to hang out with the ducks by the river (not a metaphor–there are actual ducks).

Addendum: it was plain CRC-32, and the "HMAC" just a salt being added to the data bytestring. Prefix or postfix salt? Length of salt? Well, it's web, so the answer could be to just guess and try it, and if that doesn't work, guess something else.

BalsnCTF 2020

General Comments

Another great CTF!

But yet again I have some Debbie Downer comments.

Disclaimer: this is clearly a top-tier CTF, close to ~100 points, so my main criticisms below are more subjective and personal than objective. I will be playing the part of Wolfgang, so I might seem a lot more negative than what is deserved.

The difficulty of certain tasks seemed to come from them simply being the composition of several sub-tasks that in some cases weren't even related. I am not a fan of that at all. It's exhausting and it feels like a really cheap way to increase difficulty. Happy Farm is the obvious example12. Instead of having tasks T1, T2, T3 that are all fairly independent problems, they're composed or chained in some way to make a much more work-intensive task. This joined task can even appear more difficult than the most difficult in the set13, yet it won't feel that rewarding.

That's just a strong personal preference, perhaps other people think it's cool and the bee's knees. I preferred N1CTF2020 for example, which had a similar difficulty level, but where none of the tasks seemed "artificially" difficult by composition or layering.

import random

assert \
b"""As an extreme example, what if every category was just a single chained
monster-task? No flag until you do them all. I personally am /much/ more
comfortable on the other end of this spectrum, where it's a series of smaller
bite-sized puzzles. Composing small but hard-ish puzzles isn't /that/ difficult.
Here's one, for example.""" == random.getrandbits(8*328).to_bytes(328, 'big')

Another consequence is one ends up with less number of tasks total, so there's less to choose from or pick what order to do things in.

A lot of the frustration behind the above rant comes from all the time I wasted on Patience I, which I only got a partial solution to, and the exhaustion that set in from solving the three-in-one task Happy Farm. As it were I felt really burned out and didn't do anything at all productive on Sunday, except to assist poiko with one of his solves. Well. I also discovered several mice in my apartment this weekend, which was…not fun.

Anyway, aeshash and IEAIE were both really cool, and I look forward to actually doing them outside of the CTF.

We ended up placing 12th, so we just missed the mark for the coveted 13th position.

Happy Farm

A layered crypto task. It's simply the concatenation of three totally different crypto tasks. Unlike some other such problems (I seem to remember some chained RSA problem from either a Dragon CTF or a DefCon qualifier?) the systems here aren't even related. In my opinion it really should have been given as three separate tasks, even if they were capped at a third of the points.

But as it were, this took too much time and exhausted most of my energy/hype.

Also, the task had the added fun flavour of the code being about seeds and fertilizers to grow onions, drawing numbers as ASCII-art bulbs and so forth. It stopped being fun after the first subtask.

Task 1: AES CBC

After spending some time reading the code (which is a hot mess) this task becomes pretty simple. All you need to know is how CBC mode works with AES.

Your have an oracle you can query twice. You give it (n, iv, data) with n=0..8999 and it gives back \(\mathtt{aescbc}_{k,iv}^n(data)\) for an unknown key \(k\). That is: AES is initialized in CBC mode with the unknown k and the given iv. It then iteratively encrypt data n times. The goal is to find \(\mathtt{aescbc}_{k,siv}^{9000}(sdata)\) for some known \((siv,sdata)\). You may not give it \(sdata\) to encrypt directly.

So first you give it something like (8999, siv[0]^sdata[0] || siv[1:], b'0' || sdata[1:]). The xor is just to make the data not match sdata, but it will be undone by the CBC mode. Then you give it (1, response[-16:], response) (last block is the IV for next block) and the result is the answer.

Task 2: RSA and Euler hacks

The server sets \(e = 3\), \(s = (2^{1023})^e \pmod n\), and \(d = 1/3 \pmod{\phi(n)}\). \(n\) is composed of two 512-bit primes. Note that modulus \(n\) is initially unknown. Note also that \(x^d \pmod n\) is "sort-of" like \(\sqrt[3]{x} \pmod n\).

The serves gives you \(s\) and asks for an integer \(k \in [0,8999]\). It then gives you \(r = s^{d^k}\), which is "sort-of" like taking the cube root \(k\) times.

Finally you can feed it a full pair \((x,j)\) and it will calculate \(y = x^{d^j} \pmod n\) but with some of the low-order bits hidden. This whole thing was a bit convoluted. There's ASCII drawings of onions that you use to figure out where the missing bits are and blah blah. You basically get \(y\) minus some "small" random number on the order of 2332-ish (if I recall correctly). And a final caveat: the \(j\) you give in this step has to be less than or equal to the \(k\) you gave initially.

The goal is to find \(s^{d^{9000}} \pmod n\).

First, to do anything, we need to find \(n\), which we can do with what I call Euler hacks. We know that \(n' = 2^{3\cdot 1023} - s\) contains \(n\) as a factor. This number is too big to try to factor out \(n\) directly, but we have another number that also has \(n\) as a factor. Because \(r\) is "sort-of" like iteratively taking the cube root of \(s\) (which was itself a cube, because \(e=3\)), if we repeatedly cube it back up and subtract the known remainder we'll get a factor of \(n\), for example in \(r^{3^{k-1}} - 2^{1023}\). This last number can't be calculated directly if the \(k\) is large, but we can do it under \(\pmod{n'}\) which will preserve the factor of \(n\). I suspect the caveat mentioned above with \(j \le k\) is simply there to force you to give a large initial \(k\) to make this step a little bit harder? Finally we take the gcd of these numbers and get n (possibly multiplied by some small factor).

nk = 2**(3*3*11*31) - s
nj = pow(r, 3**8997, nk) - 2**(11*31) # I gave j=8999
n = gcd(nk, nj)

# The exponents are factored becaue I thought maybe the fact that 1023
# contained a factor of 3 would be relevant, but it wasn't.

Anyway, with \(n\) known this reduces to a lattice/Coppersmith-type problem. In the second step I gave \(k=8999\) and got \(r = s^{d^{8999}} \pmod n\). In the last step I gave \(j=8999\) again but now using \(x=2^{1023}\) as the base number (i.e. a cube root of \(s\)), which gives me back \(y\) as an "approximation" of the target number \(s^{d^{9000}} \pmod n\).

The mask for the unknown bits in this approximation is

mask = 0xf00000000ffff0000ff00ffff00ffff00fffffffffffff000ffffffffffffffff00fffffffffffff0ff

Because \(y^3 = r \pmod n\) I can use this equation and Sage's .small_roots(). I didn't really think about doing anything else, I just wanted it to end. However, the missing bits are not contiguous: if treated as a contiguous block then it's maybe a little bit over what a standard out-of-the-box Howgrave-Coppersmith implementation can handle. I wanted to avoid introducing more variables for the bit pattern and I also didn't want to sit there and have to fiddle with magic parameters hoping to get lucky. But "brute forcing" the first hex character seemed to be enough. E.g. in sage:

print("small rootsin': ")
for i in range(16):
  print(i, end=' ', flush=True)
  rewt = ((c + i*2^328 + x)^3 - b).small_roots(X=2**300, epsilon=0.05)
  if rewt:
    print('thank god... You may now have a snack')

So y + i*2^328 + rewt[0] is the solution.

Task 3: Borked RCA

Thankfully, this last layer was very easy. By now I was really sick of this whole task.

Long story short, you can query an oracle four times, giving it a k each time (k has the usual constraints). It then does something like this:

def blah(secret, k):
  L,R = secret[:len(secret)//2], secret[len(secret)//2:]
  for _ in range(k):
    L,R = R, bytes_xor(L, self.rc4_encrypt(R))
  return L+R

And gives you the result. The goal is to find the result of blah(secret, 9000**3).

rc4_encrypt is supposed to be a standard RC4 implementation initialized with an unknown key. (I.e. it generates len(plaintext) bytes of stream cipher output and xors it into the plaintext.) But if you're lucky like me, you notice this little nugget right away, before you've even parsed what the task is about:

def swap(self, a, b):
  a, b = b, a

I think I actually laughed when I saw this innocent little method. It's trying so hard not to be noticed. Aww, it's so cute.

And but so. The RC4 implementation never really swaps elements around. Which means it will have a really short order. Let's look at it:

i = (i + 1) % 256
j = (j + s[i]) % 256
self.swap(s[i], s[j]) # lol!
output.append(s[(s[i] + s[j]) % 256])

Every 256 iterations, the i will be 0 and j will be sum(range(256))%256 == 128. Meaning that every 512 iterations we'll have \(j=i=0\) and it will repeat.

OK. So. It's trivial to find secret: just give it k=0 and it won't encrypt it. I think secret was 240 bytes long, so to find the order of blah we just need to align things…lcm with the order…hm divide by 2…blah blah boring details. Long story short, blah(secret, 9000**3) == secret. It's just what came out of the first oracle query. Almost a bit of an anti-climax, really.

The Last Bitcoin

Proof-of-work Python script that asks you to find a string X such that sha256(random_token + X) starts with 200 zero bits.

That part is straightforward and clearly impossible.

Then you notice that if you give it such a string it exits with error code 1 indicating success. Python also exists with 1 if it gets an uncaught exception. And it will give such an exception if you give it non-ascii input. So I simply typed æ and got the flag.

The Danger of Google's Omnipotence

This is really poiko's solve, he did the reverse, provided all the data, and so on, I just helped with the last part in figuring out how to reduce the magic operation to something tractable.

The problem at bird's eye is, I was told, something like this:

k = 0xfad9c53c828be5dafc765d4a52a54168442b6f57569db5f320a45d0d0e39d92d04284087fe2e36da1375d55e6e4e9f746cf9d9916c791e0467dc0aedf77581d7e1ab342f99e49f4c684fd7424e806cc2fb1dd54c614487b6a3909dc469f76eb8df050f3928d4c371d8aace5c81fbb1e467b987ec5ae1f5ecd0b8ffe69369edc9
flag = <some 2^16 array> # secret
A = <some 2^16 array> # known

B = flag
for _ in range(k):
  B = strange_matmul(B, A)

# B is output here

Reversing strange_matmul and figuring out what it does is the main problem, which poiko already did. But as a last step we'd need to strength-reduce strange_matmul to some actual math, like a real matrix multiplication for example, and then we have B = flag * A^k which can hopefully be inverted.

poiko gave me sort-of pseudocode for strange_matmul:

def strange_matmul(m, n):
    out = []
    for i in range(256):
        for j in range(256):
            val = 0
            for k in range(256):
                v1 = m[i*256+k]
                v2 = n[k*256+j]
                if v1 == 0 or v2 == 0:
                val = strange_add(val, a_inv[(a[v1] + a[v2]) % 7**3])
    return out

Where strange_add adds the numbers as if they were in base-7 but without using carry, and a and a_inv were some lookup tables he also shared. I don't remember if they were in the code or something he constructed.

Anyway, so it looks like a normal matrix multiplication, but the element-wise operations aren't carried out using normal arithmetic. So we both played around a little with a and a_inv and there were definite group-like properties. I even wrote a little utility class so I could do arithmetic with such numbers directly and play with them in the REPL. At first it "felt" like a commutative additive group, with 0 almost being the identity except for 0 + 0 = 1 which meant it wasn't really associative either, because \((x+0)+0 \ne x+(0+0)\). a was a permulation, but a_inv was not as a_inv[0] == 1 when it seemed like it should have been 0. This makes sense from the above code where 0 is avoided. Anyway, from the strange_add I strongly suspected it was a mapping into the åker14 \(\mathbb{F}_{7^3}\) where the numbers represented the coefficients of a qudratic polynomial over \(\mathbb{F}_{7}\). If so, what's the modulus?

\(x^3\) in \(\mathbb{F}_{7}[X]/(x^3 + P)\) should become \(-P\) where \(deg P < 3\).

>>> -fux(7)**3
[6 0 4]

fux was my utility-class, and fux(7) should represent x, so the modulus is \(x^3 + 6 x^2 + 4\). (It turns out to also be the default one that sage gives when constructing GF(7^3), so that's probably its origin.) With this intuition behind it the math seems to work out.

So, we had 256x256 matrices over \(\mathbb{F}_{7^3}\), and calculating the inverse is straightforward but boring sage stuff. It would likely have taken many minutes though, but one thing I lucked out15 on was I did .multiplicative_order() on A^-1 which sage found quickly as 342. So I could shave several minutes off the calculation by doing flagimg = B*(A^-1)^(k % 342).

Anyway, flagimg was a 256x256 noisy image showing a readable flag.

aeshash & IEAIE & Patience I

I feel these deserve an entry even though I failed all of them, but I did spend time on them. I probably didn't praise the first two tasks in this list enough. They are what I consider actually good crypto tasks of high difficulty (unlike Happy Farm). Especially aeshash! aeshash's task author deserves some real kudos.

For either task I had no prior experience and tooling, so I knew it'd be a lot of work for me. I made poiko waste some of his rev-time to make sure I got a aeshash into Python, which I felt sort of bad about… Because I only really started to scratch the surface of them. And then for most of Sunday I didn't really have any energy left to continue. However they're the kind of tasks that makes me want to solve them in my leisure time, which is a real treat.

IEAIE was some image encryption scheme using a logistic map and permuations. It was very sensitive to differential analysis. I did a few things, was able to reveal a bunch of key-state, but ran into some floating point rounding headaches in my offline tests. I hate floating point numbers, so I took a sanity break from it and never really got back. For once the relevant papers are given in the task description so it's not really a Google-the-Paper challenge.

aeshash involved leaking state in a three-round pseudo-AES (each round using initial state as round-key) and then targeting a specific output. Literally: an AES hash. I have no idea about this one, which is all the more exciting. Despite me usually solving the crypto challenges I've never worked with crypto profesionally or academically, so I lack a ton of basic knowledge and experience, such as breaking weakened AES, which is probably like Crypto 101. But I figured it will be a good place to start learning.

I also sunk a ton of time into Patience I doing proper OpenCV for the first time in my life… God, this task sapped me. The composed text which I suspect is one third of the flag(?) was easy, I also found this strange Morse-ish sequence but was unable to guess what it really was, noted the spikes in the lighting but had no idea what they meant either. In the end I even started to fear the minute little camera shakes were also part of it. Please God. Maybe all of these things played together in some beautiful and harmonious way in the end, but really it just felt like three concatenated subtasks that just all had to do with video?

Addendum: having seen spoilers on how Patience I is solved (but not the others), I can at least say I don't feel bad anymore for not figuring out this task. It's a bad one. Though I was wrong about the "subtasks" being separate.

The first string you find, the easy part, overlay all the LEDs, was _q!fitlboEc. I mistakenly thought it might be flag[x::3] or something like that. (Thus why I thought the three parts were separate.) But apparently it's an alphabet of sorts, which you index using…I don't know, it was unclear exactly what. And the timing pattern was Morse, but the international extended version with punctuation, making P=B_=_Q=D which is apparently supposed to mean "replace 'b' with 'p', and 'q' with 'q'"?… The logic or motivation behind any of these things totally escapes me.

I even said to poiko while I was working on this task, thinks like "hm, there's a timing pattern here, it feels a lot like Morse, but I'm actually going to be a little bit disappointed if that turns out to be true…" and "it can't be too escape room-ish, right? It's Balsn, they're good, they wouldn't make something like that" and so on. Disappointed!


Again, poiko solved this one, but I helped out a tiny little bit at the end, so I feel comfortable writing what it was about.

Apparently a scala program that seemed to construct an infinite stream by doing what I would write in Haskell as:

w = [[0,1,2,3], [0], [0], [0]]
stream = [0] : map (concat . map (w !!)) stream

It makes an infinite stream of lists like [0] : [0,1,2,3] : [0,1,2,3,0,0,0] : [0,1,2,3,0,0,0,0,1,2,3,0,1,2,3,0,1,2,3] : .... It sums all the sublists in this stream, extracts index 60107, and takes its remainder under \(2^{62}\). It converts the resulting number to 4 bytes which becomes an xor-key for flag data.

Immediately I smelled a recurrence relation for the sum. Tried and true OEIS tells the truth: \(a_{n+2} = a_{n+1} + 3 a_{n}\) 16, and so, starting from \(a_0 = 0\) and \(a_1 = 6\) (for the sum of these lists) gives all that's needed. It can be calculated linearly in a list, or logarithmically by exponentiation on a 2x2 matrix.

DragonCTF 2020

General Comments

My expectations were a bit high, DragonCTF has had some of the best CTFs in the past. Yet I had the thought that even last year's 24-hour "teaser" CTF was better? So a bit mixed feelings. I'm not saying this wasn't good but…? It's hard not to vote with one's heart.

The difficulty was alright, but it felt like very few tasks in total. Thinking partly of my poor teammate poiko this time, who seemed to get only a single pure rev task, and an easy one at that. I know he was hyped and had looked forward to fiendish VM-in-VMs and unknown architectures. We kept waiting for more to be released, but alas.

For my own sake, Frying in motion was an cool idea, tho a bit draining. Bit Flip 1 was easy. Bit Flip 2 & 3 were puzzling, but not the good kind of puzzling (which motivates me to write code or learn stuff outside of the CTF), but "I feel I just missing some (probably) trivial trick here, oh well next time I guess"-puzzling.

Sunday I ended up watching Gilmore Girls and poiko was reduced to doing Linux stuff and actual hacking. Imagine having to do actual hacking. What's that about? Sheesh.

Anyway, we placed like dumpster tier anyway. Wasn't our weekend, I guess.

Frying in motion

I put off this task while looking at the Bit Flip tasks because I kind of expected implementing it would be an unfun hellhole of tiny bugs. It's one of those tasks where the idea is simple and/or obvious but it's a pain in the ass to implement17. Not a fan of that.

But so: the task involves reversing a strfry() given the output of another strfry(). Fine, sounds like an RNG task.

strfry() does this:

char *
strfry (char *string)
  static int init;
  static struct random_data rdata;

  if (!init)
      static char state[32];
      rdata.state = NULL;
      __initstate_r (random_bits (),
                     state, sizeof (state), &rdata);
      init = 1;

  size_t len = strlen (string);
  if (len > 0)
    for (size_t i = 0; i < len - 1; ++i)
        int32_t j;
        __random_r (&rdata, &j);
        j = j % (len - i) + i;

        char c = string[i];
        string[i] = string[j];
        string[j] = c;

  return string;

(Aside: is it just me or is the GNU style one of the ugliest code styles in existence?)

So strfry() just uses glibc's random_r() stuff, using a 32-byte state buffer (of which only 28 bytes are used), initialized with a 32-bit seed based on some CPU clock counter. (random_bits() actually gives less than 32-bits of actual bits but let's ignore that.)

It's clearly a "brute the seed" sort of task, because even partially reversing the internal state directly (let's say if it had initialized with 28 bytes of random data from urandom) just given strfry() output alone is very hard unless many such calls can be queried in a series. (And even then it would be a finicky as all hell.)

Another thing the task does is it calls strfry() a lot of times on a temporary string before it does the challenge-response step. The number of resulting random_r() calls is predictable, but you don't get any output. So, it's "brute the seed" and then "seek in the RNG stream."

Seeking in glibc's random_r() stream turns out to be easy, as it seems to be just a linear recurrence. It does varying things depending on how many bytes of state you give it, but all of them are very simple. It's sort of like a more flexible Mersenne Twister without the output scrambling. For 32-byte state it will use \(a_{n+7} = a_{n+4} + a_n\) which can be modelled most simply (to me, anyway) with matrix exponentiation:

A = ZZ[2**32].matrix([[0, 0, 1, 0, 0, 0, 1],
                      [1, 0, 0, 0, 0, 0, 0],
                      [0, 1, 0, 0, 0, 0, 0],
                      [0, 0, 1, 0, 0, 0, 0],
                      [0, 0, 0, 1, 0, 0, 0],
                      [0, 0, 0, 0, 1, 0, 0],
                      [0, 0, 0, 0, 0, 1, 0]])

# now you can model state after n steps as A**n * state
# modulo indexing errors.

This is the same for seeking in plenty of RNGs whose internals are linear or "affine" like xoshiro and MT. Note that the state has the last 7 outputs from the RNG, which will come in handy. Note also that the RNG discard the lowest bit when giving actual output. (But due to the lack of scrambling the low order bits will still be pretty bad by most RNG standards.)

Now about the seed. The state is initialized by initstate_r() with \(state_i = seed \cdot 16807^i \pmod {2^{31}-1}\).

There's plenty of sources of bugs here. One is that it actually starts outputting at \(state_3\) or something like that, and then iterates in reverse order from how I set up my vectors, which I should probably have fixed–so that's on me. It also discards the 70 (in our case) first outputs from random_r() so these have to be taken into account. And of course the mixed moduli need to be kept separate. Notably care must be taken so numpy doesn't overflow the seed-modulus.

Now to match the strfry() output. For example, given out = strfry(in) with unique characters, I know that the first random call (i=0) satisfies random_r() % (len(in) - i) == in.index(out[i]). Then re-do the swap, get a similar constraint from the next character, and so forth. I did this for the 7 first characters for a 90+ char string to make sure I had enough information to uniquely determine any seed from a given state. (A potentially more clever idea would be to use a longer string and only consider even indices, taking only 1 bit from each character, to get constraints that are all mod 2? I didn't think of that at the time.)

I think in the end I had some idiotic thing like:

\(( A^{70+n+7} \left[ seed \cdot 16807^i \pmod {2^{31}-1} \right]_{i=0\ldots 6} \pmod {2^{32}})_{7-j} \gg 1 = index_{j} \pmod {length - j} \pmod {\mathtt{random\_indexing\_errors}}\)

I mean… It's not pretty. I knew it wouldn't be. Doing this sort of thing 90% of my time is spent fixing off-by-one errors or putting indices in the wrong order or something like that. For my test code I had something like:

# precomputed stuff
start = np.array([282475249, 16807, 1, 470211272, 1144108930, 984943658, 1622650073], np.uint32)
A = np.array([
  [0xe9373f80, 0xdc7dfb8c, 0x9a732dfd, 0x8e41234d, 0x45478090, 0xe4134087, 0x78b11946],
  [0x78b11946, 0xe9373f80, 0xdc7dfb8c, 0x21c214b7, 0x8e41234d, 0x45478090, 0xe4134087],
  [0xe4134087, 0x78b11946, 0xe9373f80, 0xf86abb05, 0x21c214b7, 0x8e41234d, 0x45478090],
  [0x45478090, 0xe4134087, 0x78b11946, 0xa3efbef0, 0xf86abb05, 0x21c214b7, 0x8e41234d],
  [0x8e41234d, 0x45478090, 0xe4134087, 0xea6ff5f9, 0xa3efbef0, 0xf86abb05, 0x21c214b7],
  [0x21c214b7, 0x8e41234d, 0x45478090, 0xc2512bd0, 0xea6ff5f9, 0xa3efbef0, 0xf86abb05],
  [0xf86abb05, 0x21c214b7, 0x8e41234d, 0x4cdcc58b, 0xc2512bd0, 0xea6ff5f9, 0xa3efbef0],
], np.uint32)

def find_seed(A, start, m):
  seedvec = start.copy()
  for i in range(1,2**32):
    out = (A*seedvec).sum(1).astype(np.uint32)
    out //= 2
    out %= np.array(range(89,96), np.uint32)
    if np.all(out == m):
      return i
    seedvec += start
    seedvec %= 2147483647

However, doing a full brute like this wasn't fast enough against the live server. It was really close but it required me to get lucky, and DragonCTF has this super annoying PoW blocker on all their tasks… I probably should have picked apart random_bits() but at this point I just wanted to move on with my life. I think it might also be possible to thread the various moduli and simply calculate the seed directly for each index-modulus and then CRT it (esp. if using the idea noted above), but Lord almighty, the paper work involved. Yeah, brute forcing is dirty, it doesn't feel rewarding or clean, but it's a dirty real-worldsy task, so I didn't feel too bad about it.

Because this CTF had like 1 rev task, poiko was free to help out, so he wrote a utility in big boy code to do the above for me while I fixed the remaining indexical bugs, and I could just use:

s = int(subprocess.getoutput('./BLAS ' + ' '.join(str(x) for x in M)))
# s = find_seed(M)

Out comes the seed quick as mercury rain. Then seek the stream as above, then seek the stream again because you forgot to add +1, then do the final strfry() on the given string, then realize you're an idiot and reverse the final strfry() because you misremembered the code, then finally get the flag.

Bit Flip 1

Diffie-Hellman between Alice and Bob to get a shared key which is used as an AES key.

The stream of numbers \(2^{256} \mathtt{sha256}(x) + \mathtt{sha256}(x+1)\) for \(x = r, r+2, r+4, \ldots\) is scanned to find a prime. When a prime is found at \(x\), then the lower 64 bits of \(\mathtt{sha256}(x+1)\) becomes Alice's secret. The numbers here are big-endian strings of 32 bytes.

\(r\) is os.urandom(16) and unknown, but you get to flip arbitrary bits in it so once it's known you can set it to whatever you want. You're told how many steps it took in the above sequence to find a prime (simulating a timing attack), so finding the original \(r\) bit-by-bit is easy, something like:

def find_next_bit(oracle, num_known, bits_known):
  next_bit = 0
  while True:
    next_bit = 2 * next_bit + (1 << num_known)
    x = oracle(next_bit - bits_known - 2)  # fetch ABC1111X
    # if unlucky
    if x != 0:
  y = oracle(next_bit + bits_known)        # fetch ABx0000X

  # if C was 0, then oracle(y) fetched at offset +2 closer to the prime
  # if C was 1, there's a tiny chance for false positive
  return y != x - 1

(Could be made a lot more efficient by making use of past ranges. Probably only \(1 + \epsilon\) queries is needed for most bits instead of fixed 2 I'm doing here?)

But anyway, this basically solves the task, because now you get the prime, Alice's secret, and Bob's public key is given.


I'm a closeted Haskell advocate, I've played with other dependent type things like Agda 2, and there's some community overlap, so I recognized it immediately. However.

I've never used Coq before. I've never used it before, but I now know that I hate it. I hate its syntax, its IDE, its logo, I hate all the people who think making puns on its name is smirk-worthy, and the people who pretend the name doesn't lend itself to puns. I hate how all the blogs make proving stupid toy theorems about Peano numbers seem like a fun adventure. I hate myself.

I have renewed understanding for beginners when they do stuff like:

if i == 0:
  s = "a"
elif i == 2:
  s = "b"
elif i == 3:
  s = "c"
elif ...


sum = i = 0
for _ in range(100000):
  log("increasing the index")
  i += 1
  log("accumulating sum...")
  sum += av[i]**2 + av[2*i + av[i]%2]

log("solving langlands")

log("printing sum")

I get it. That's me in Coq.

I solved the trivial ones easily. The math_theorem one I just pulled from some tutorial thing. (Super unrewarding way to solve it, but there you go.) I was stuck on the last problem for a long time, reading tutorials, blog posts, looking at other proofs, and so on. I also infected poiko and made him waste time on it. Both of us wasted several combined hours trying to learn Coq's syntax and usage for this dumb, pseudo-trivial theorem.

In the end we both came up with different proofs for the final theorem at nearly the same time. Mine was:

intro n. intro l. revert n. induction l as [|? ? IHl]; intro n; destruct n; simpl.
- intro. apply PeanoNat.Nat.nlt_0_r in H. contradiction.
- intro. apply PeanoNat.Nat.nlt_0_r in H. contradiction.
- destruct 1. exists a. auto. exists a. auto.
- auto with arith.

It probably makes no sense and looks ridiculous to anyone who knows anything about Coq.

poiko's solution might make more sense (I couldn't say):

unfold lt; intro n; induction n as [| n hn]; intro l.
- destruct l; simpl. inversion 1. inversion 1. exists a. auto. exists a. auto.
- destruct l. simpl. * inversion 1. * intros. apply hn. auto with arith.

I'm having a hard time deciding whether this is a good task or not. I mean it made me despise Coq, and that's not good; I'm usually all for ivory tower type theory stuff. On the one hand, motivating CTF players to learn some higher-order type theory and theorem proving sounds great, but on the other hand it feels very arbitrary and polarizing the way this task was put together, like putting up a poem in Finnish that tells you how to make the flag through puns–free points for anyone who speaks the language, a frustrating Google hammer experience for the rest.

Bit Flip 2 & 3

Similar to Bit Flip 1 but Bob's key isn't printed. This ruins the easy solution. You can still reverse \(r\) as above, so you get a little bit of control over what the prime becomes as well as Alice's secret, both of which are of course known, but there's not much else to do.

These tasks had me super puzzled. Here's just various notes I made to myself mentally.

The brute forcing options included:

  • Brute force Bob's 64-bit secret. Nope, not gonna happen.
  • Brute force hashes to make Alice's secret 0 like a 64-bit PoW task. Also not gonna happen.
  • Search the SHA256-stream for primes under which 5 (the generator used) has a very low order, for example by making \(p-1\) smooth. Given that Bit Flip 3 is just like Bit Flip 2 but with the added constraint of strong primes, it seemed to heavily suggest this was the path. But finding smooth numbers of this size without sieving you control is near impossible. So this was also not gonna happen.

I wondered about the (p % 5 == 4) chech in Bit Flip 3 – like are there any other constraints for random primes where a given number such as 5 might have low order in the associated field?

The task also specifically imports is_prime() from gmpy2, even though it uses Crypto.Util.numbers (which has isPrime()). It's the only thing it uses from gmpy2 too. That was curious, but I thought it was just some author idiosyncracy. Besides, you can't really construct the primes being generated, you can only pick one from some set of random ones, so again it's not like you can even begin to think about fooling a standard prime check.

All tasks also xors the shared secret with 1337 for seemingly no reason. No idea why. The shared secret will in "almost all" cases be a 500+ bit number, of which the 128 most significant bits are used, so it doesn't matter. (Addendum: I was wrong here, I totally missed the "bug" in Bit Flip 3.)

IV isn't reused, there doesn't seem to be a meet-in-the-middle, data collection probably doesn't help.

I failed to even find the problem I was supposed to work on, so I just put it out of my mind.

Post-CTF Addendum: after the CTF ended I found out that Bit Flip 3 can be solved trivially because long_to_bytes(x, 16) does something quite unexpected:

>>> long_to_bytes(random.getrandbits(513), 16)[:16]

This was actually really clever! Ever since I started joining CTFs I've had a low-key distaste for long_to_bytes() and how pervasive it is. It feels so Python 2-y. It's in every crypto task under the sun, so its usage didn't stand out to me at all, and I never bothered to check up on it in this case. I'm sorry I missed this a-ha moment. Deceptive!

However it seems that several people solved Bit Flip 2 by "brute force the hashes modulo OSINT." Namely: the BitCoin blockchain has a lot of such hashes, so you can find hashes there that match the requirements. Although that's a clever idea, it sort of lowers the task in my estimation greatly if that is the intended solution. It's just a really unfun way to solve problems. It feels a lot like "enter the MD5 hash in Google" or "look up the factors on" but version 2.0. Big boy OSINT is still OSINT. If anyone has a programmatic or mathematical solution to Bit Flip 2 that doesn't look up dirty money-laundered power-guzzling hashes from other sources I'm very interested to see what I missed?

Not that anyone reads these notes, but hey.

HITCON & perfectblue

Reserving this space. Skipped both of these in favor of reading postmodern fiction, curling up in bed awash with emo poetry, and going to couples therapy in plays acted out in my own head.

But it seems they had cool tasks which I plan to solve off-line between whatever bouts of depression. PBCTF in particular (got poiko to scrape the tasks as I was even offline at that point) seemed to have tons of tasty goodies.

ASIS 2020

General Comments

I was bribed with pizza to rejoin this CTF after two weeks of being largely offline. I didn't play on Friday, as I was lost in the braindance of Cyberpunk 2077, but was able to quit and reboot into Linux on Saturday.

But oof, this CTF had some actual hiccups, not like the fake critiques I've unfairly levied against other CTFs above. A bunch of things seemed to have slipped through Q&A, there were a couple of broken problems and questionable code in the crypto/ppc sections; poiko also indicated that the rev challenges were really uninspired, everything boiling down to obfuscated x86 with guessy elements(?).

However, looking at CTFTime now, especially the voting section, it seems almost like there's reverse group-think going on, like suddenly it's Socially Acceptable to be Critical or to "offset" high votes with exaggeratedly low ones, so everyone's letting off steam. Does this CTF really only deserve 20 points? That's goddamn vicious. Was it really Ekoparty or e-Jornadas level? No way.

Their heart was clearly in the right place. The difficulty level was alright, and the intent seemed good. Plenty of the tasks were interesting, but sure, there were Problems and a couple of tasks with bad "quality assurance." I still feel this CTF was leagues better than OSINT-riddled pub-quiz or escape-room CTFs. It tasted like at least 50-60 points to me, even given the fuckups for a few of the tasks.


(Aside: both this and Coffeehouse were coded in a certain style that I dislike. It seems by a Python 2 programmer that's afraid of using anything but the trusty old str datatype, and insists on using them as if they were bytes. This programmer also enjoys converting X-bit numbers to text strings of "0" and "1" to do binary operations as if working with actual numbers is a bit scary.)

So Chloe involved a fair bit of "reversing" the bad code to get data into types that are actually sane.

What you get is cipher + byte(r)*16 (48 bytes) xored with a 16-byte key. r is a random byte.

So modulo an 8-bit brute force, key (originally from os.urandom) is already given in the last 16 bytes of the output.

Now for cipher which is calculated in some Feistel-like manner with an added permutation at the end:

L,R = flag[:8],flag[8:]

for i in range(16):
  L,R = R, L ^ R ^ roundKey[i] ^ (2**64-1 if i%2 else 0)

cipher = permute(L + R)

The Feistel function is just a simple xor and predictable negation, so the whole "Feistel network" is easily collapsible, as you can simply check the xor keys that end up being used in left and right side:

# precalculate: (the magic numbers are a[i] = a[i-1]^a[i-2]^2**(i-1) starting
# from [0,1])
xorL = reduce(op.xor, [roundKey[i] for i in bits_by_idx(14043)])
xorR = reduce(op.xor, [roundKey[i] for i in bits_by_idx(28086)])

# and the above Feistel-loop simply collapses to:
cipher = permute(flag ^ (xorL << 64) ^ xorR)

The roundKey array is given by the key found earlier.

The last permute() step here performs a fixed random permutation on the bits in each byte. There's only \(8!\) such permutations so finding the reverse one is also easily brutable (~15-16 bits). To brute force the permutation I just did it the simple thing by turning each permutation into a 256-byte translation table, so I could simply do cipher.translate(Pinv[i]) to test a permutation.

So: cycle through the bytes, reverse permutation, and do a simple xor and check if it gives flag data.

Coffeehouse (cafe house)

A 32-bit block-cipher which uses a xor, add, and shift in a loop for diffusion. I.e. it takes the flag and encrypts 4-byte blocks by looping over the block 32 times and doing various add-and-xor-shift with the key and a counter.

I really wish the cipher wasn't 32-bits because it's just too tempting to brute force…

You only get the encrypted flag as data, so I guess that's clue #1 that at least a partial brute-force was expected. You guess that no utf-8 is used in the flag so there's N bits that you know were 0 in the plaintext, and from there you can filter keys by writing the trivial decrypt() function which was missing. Technically this dumb way ends up being around 37-bit(?) brute force.

This is what I did and I'm not particularly proud of it. My program (C++) found the solution in seconds, but a numpy equivalent with parallel keys would probably not be much slower.

My guess is that maybe the task author intended you to find some way to strength-reduce the loop? Hmmm. Like, what if it was 64-bit or 128-bit instead? Or the inner loop had several orders of magnitude more rounds? Then one would have to solve it by finding invariants… And suddenly it's a much cooler task–given that there is a solution to these strengthened problems. I don't really know, I didn't look into it, I just did the bad thing and moved on. But my intuition tells me that the diffusion step is really weak and there's plenty of invariants to be found, so I wouldn't have complained if I was forced to look into it more, forced to not be lazy. (Again, given that there actually is a solution?)


I spent the better half of Saturday evening looking at this task. Spoiler: I didn't solve it. (I don't think anyone did? I think many people were in the same boat as me, spent a lot of time on it, and maybe ended up frustrated?)

It seemed magical, a promise of real beauty! So simple, so clean, pure mathematics, yet I had no idea how to solve it. That's the best kind of feeling, isn't it? When you feel stupid because of math. It drew me in immediately.

Secrets: \(flag\), \(p\) (a fixed prime modulus of length ~512 bits), and \(s\) (presumably a very small non-negative integer).

You're given five different pairs of numbers \(\lfloor a_{i} / 256 \rfloor\) and \(c_{i} = a_{i}^e \pmod p\). Each of the numbers \(a_{i}\) are constructed from ostensibly random alphanumeric bytestring of the same length as flag. You're also given \(flag^e \pmod p\). So far so good. The coup-de-grâce is the exponent: \(e = 7 \cdot 37 \cdot 191 \cdot 337 \cdot 31337 \cdot 2^s\).

Hmmm…? The usual Euler hack of finding \(p\) from \(p\cdot k = gcd(a_i^e - c_i, a_j^e - c_j)\) (for some small \(k\)) doesn't work because \(e\) is too large, even if \(s = 0\). The numbers involved would be hundreds of gigabytes. Oh my!

The fact that the \(a_{i}\) numbers get their last byte censored seems to indicate that a solution has to be easily calculable, because the task will necessarily(?) involve a small brute-force step to test a given solution.

So, some ideas I went over:

  • is there some way to calculate remainder(a^e, b^e - c) iteratively, without fully reifying the numbers involved?
  • sub-question: is there some way to calculate the remainder or gcd directly from vectors of residues modulo sane primes? Prime-factor FFT?
  • can continued fractions be used somehow, as continued fractions and gcd are very similar algorithms.
  • can some clever algebra be used to relieve \(e\) of some of its odd exponents? In particular: has \(p\) been carefully picked to be revealed in one of the low-term expansions of some \(a^e - b^e\) expression?
  • is \(p\) of a special form? E.g. does \(e\) divide into \(p-1\) giving a power-of-two quotient? (I can't recall if I even tested these last two, because it would have made the task ugly and idiotic; I didn't want it to be idiotic, I really wanted to preserve the task's beauty.)
  • in the relations \(a_i^e - c_i = k_i\cdot p\), we can recover \(k_i\cdot p \pmod q\) for saner numbers \(q\) (in particular we can find small factors of \(k_i\)), but how does that help? All the \(k_i\) integers are too big to actually be used or reconstructed.
  • is the special form of the \(a_i\) numbers relevant?
  • the \(a_i\) plaintext numbers are ostensibly random, but are they really? Could a seed have been fixed such that it can be used to reveal \(p\) or \(flag\) in a manner ordinarily impossible? Hmm, unlikely, right?
  • does it help to find vectors (e.g. LLL) combining the given terms to 0?

Most of these avenues I never really tested. What kept stopping me was the brute force step on the \(a_i\) terms: I felt I had to be damn sure about an implementation lest I waste a lot of time, something I could actually verify before brute-forcing.

So I ended up really disliking the brute force step added to this task actually. (Unless it wasn't intended as a brute-force step, but some kind of magic fairy-land lattice problem? I don't see how that's possible, though?) It seemed unnecessary. I also don't know why \(s\) was hidden as well, that just seemed overly sadistic.

But in the end I never really became sure of anything. If there is a solution, I would be very happy if someone could enlighten me! So far my best intuition is in the direction of some novel FFT application?

Trio couleurs

Spoiler: I didn't solve this one either, but noting my thoughts.

I think this task was released during the day on Saturday. The server instantly disconnected you at first, so I ignored it for a while, and only came back to it when I finally gave up on congruence above. By then it was late and I found that this was a really work-intensive task…

(Again this style of converting numbers to text strings of "0" and "1", so I take it it's the same task author as Chloe and Coffeehouse, who does seem to have a particular phobia.)

Code is given which looks like DES, used in some broken 3DES-ish manner. It didn't encrypt the same as a reference DES implementation, so then I have to waste time to figure out the difference. Turns out it was regular DES reduced to 8 rounds. (Unless this kind of debugging is relevant for the task, I wish the task made these things clearer, and simply stated sincerely that it was "DES reduced to 8 rounds" or similar.)

But OK, so it seemed like a linear cryptoanalysis task. The 3DES part of the task is broken since it trivially leaks the penultimate ciphertext.

It doesn't actually seem like a bad task, it's "true" crypto, though I believe it requires some prior experience to be doable in hours as opposed to days. Key discovery from 8-round DES is, as far as I know, not trivial, and require a lot of plaintexts. I haven't done it before, nor do I have any linear analysis code, so this contributed to my decision to skip the task, as it would potentially be a big time sink, I needed sleep, and not many hours remained of the CTF.

It's something I could have fun solving off-line though.

Baby MD5

I'm embarrassed to admit I also wasted a long time staring at this stupid task doing exactly what the task encouraged me not to do: overthink it. I wrote a script to solve ASIS' proof-of-work using all available cores to make reconnects fast, and I tried to sample the parameters to see if it would sometimes give \(n = m\) in which case you can make an MD5 collision and… Well, I suppose this was the "trap" set up to trick idiots like me.

While I was doing this, poiko came along and remarked "hm, isn't 'dead' hex, though? Couldn't you–" and I instantly cringed and facepalmed pretty hard. Yeah, indeed… Dumb, dumb, dumb.

Connect, assert that \(m > n\) and then find a string X such that iterating the function lambda x: md5(x.encode()).hexdigest() starting with prefix + X produces a hex string that starts with dead after \(m - n\) iterations; that hex string is the second input. Could also do this trick for \(m < n\) if the random prefix turns out to be (lowercase) hex.

גל התקפה (Attack Wave)

I solved this one while taking a break from congruence above.

The boring part: 5.1 WAV file, two channels was a synthetic voice saying what sounded like "RGBA" and the other 4 channels contained random noise. The noise data were all in [0,255] so I extracted that, treated is as image data, and that's where the fun18 begins.

It's what I like to call a "numpy massaging" task. In my mind these tasks exists in a grey area between the guessy and the mathematical19. You don't have the source, you don't know what the task authors did or their reasons for doing it, so in that sense it's guessy, but what you do have is pure, cold data and with data you can "use the force, Luke." There are certain intuitions to follow in seeking low entropy, seeking order in the chaos. As long as you're given some breadcrumb trail of patterns to start with, it's usually fine.

For example, displaying the byte data as images of whatever dimension shows semi-repeated patterns on the top and bottom (low entropy), whereas the middle of the image appeared much more random (high entropy), which makes it "feel" like an image with maybe uniform coloring around the edge with text in the center, like this:


Now we seek to lower this entropy with various basic operations. Xoring all the channels together direct only increases the entropy, but I noticed that the three "RGB" channels were very similar (whereas the "A" channel was more chaotic). Looking at the differential of where certain numbers or bits occur is a good way to get a feel for the repeated patterns. They start out with this very cyclic pattern, but each is offset a little. So, rotating the arrays to by small offsets aligns them to produce very low-entropy patterns, differing in very few bits, so you know you're closer. I combined this trick with the more mysterious "A" layer as well, and combining "A" with any of the other channels (or all three) quickly produced an image with the flag clearly readable:


Unfortunately I have no idea what the author's intent was. From the periodic artifacts still present in the above image, my guess is that there was indeed some "intended" way where you get an actual clean image, possibly with the letters all black and clearly defined. If I were to guess further it might have something to do with the fact that the data size makes it feel like there's a row/column missing from the image, that the data given is actually just differentials between the pixels… But I didn't investigate further after I got the flag. It was a fairly quick task.

Election & Galiver

I don't exactly know when these tasks were released, but I didn't see them until I joined up on Sunday, with like 4-5 hours left of the CTF. Oof.

Why, oh why, do most CTFs insist on these staggered releases? Every damn time. Is it intended to "keep things exciting" in the competitive sense until the very end? Because I don't think it's working20.

Galiver looked cool, seemed to be DH with gaussian integers, but I made the terrible choice of focusing on the PPC task Election instead because it "seemed easy."

Then I invested too much into Election and I was sunk by my own sunk-cost fallacy. The cherry on top was that this task was actually bugged server-side.

Election was a PPC where you are given a string of base-10 digits and have to give some binary arithmetic expression like X op Y (where op can be one of {+,-,*,/,**}) in which these digits can be found as a substring. The expression you give cannot be more than 8 characters long. You can also just enter a single number so for levels up to strings of length 8, you can just feed it back the string given granted it doesn't start with a 0.

Each level the base-10 string you get has length +1. It has an unknown number of levels, which is the first guessy part, because at first you don't really know what you're up against. Making a small in-memory database of the shorter expressions and their digits gets you to length 12+ strings and you realize you'll need to make a full database of all strings producible by these arithmetic expressions. I dumped these to external files and just used grep in my solve-script. These files were several gigabytes each.

The addition, subtraction, and multiplications are trivial and can be ignored.

All the guessy elements stem from the server-side code being hidden. That's always a bad sign. First you have to assume (pray) that it actually picks valid substrings. OK, fine, I'll allow it, but it doesn't feel great to trust the author like this, especially when knowing that the author is merely human, and the code might not have been tested by others.

The exponentiation expressions are the simplest to dump, but most time-consuming in CPU time (gmpy2 and not Python's slow bigints), so I started with those. Constructing the database of digit expansions from the divisions are the more brain-intensive and tricky if you want to do it "properly." From some pen and paper and Wikipedia I re-figured the repetend in the expansion of, for example, a/b is of length \(k\) where \(k\) is the least number such that \(10^k = 1 (mod b)\). Thus for prime \(b\) it will be up to \(b-1\) digits – so I figured this part of the database migth require a magnitude more disk space, because the numbers involved are larger (D**DDDDD vs D/DDDDDD), and thus wanted to take some extra care to only store what is necessary. I knew that for the cyclic numbers the numerator can be fixed to 1, but it was unclear to me whether \(a/b\) for composite numbers could give new "unique" decimal sequences. I don't have that much SSD disk space…

But bottom line: expressions on the form D/DDDDDD can give up to \(999999+\epsilon\) digits before repeating. And this is where the server-side bug comes in. As my database kept filling up I started encountering this:

[DEBUG] Sent 0x9 bytes: b'1/102451\n'
[DEBUG] Received 0x60 bytes: b'sorry, the result of your expression does not contain the given substr: 07718811919\n'


>>> getcontext().prec = 200000
>>> '07718811919' in str(Decimal(1)/102451)

Hmmm. I kept getting more and more of these errors, barring me from further levels. I didn't really know what was going wrong. I thought maybe I had a bug somewhere making it skip certain "easy" expressions and giving the server fractions it didn't expect (like if the server erroneously expected some fixed low-term expression for the later levels). It felt really bad to be in the dark. Finally I went on IRC and tried to contact admin. It turned out the server actually only calculates fractions to 99999 digits:

15:56 <factoreal> see this:
15:56 <factoreal> if len(formula) <= 8:
15:56 <factoreal> IIif re.match(regex, formula):
15:56 <factoreal> IIItry:
15:56 <factoreal> IIIIif '/' in formula:
15:56 <factoreal> IIIIIa, b = formula.split('/')
15:56 <factoreal> IIIIIres = mpfr(Fraction(int(a), int(b)), 99999)
15:56 <factoreal> IIIIelse:
15:56 <factoreal> IIIIIres = eval(formula)
15:56 <factoreal> IIIIif substr in str(res):
15:56 <factoreal> IIIIIreturn True
15:57 <franksh> hm, yeah, so bugged. :/ 99999 isn't enough to cover all repetends/decimal expansions of some fractions.
15:58 <franksh> but i didn't find a way to get or see server source?
15:59 <factoreal> the cycle of 1/p where p is prime is p-1
15:59 <factoreal> franksh I will share it after CTF, ok?
16:00 <franksh> yeah, but 6-digit numbers are valid in d/dddddd so need an order of magnitude more to cover all
16:00 <franksh> and sure
16:01 <factoreal> you are right

The classic off-by-one-order-of-magnitude bug. But by then, as is shown by the timestamps above (CTF ended 16:00 for me), it was too late to do anything. That felt pretty shitty, and a sour note to end it on.

From IRC, I think the one solver of this task said something about only searching fractions 1/p with (fixed) 30000 precision(?), which seemed like pure luck (or they're just a very good guesser). That also gives a clue to a third problem with the task, in that it didn't select the base-10 strings randomly at all. You could "guess" that exponentiation was irrelevant, and only search 1/p up to 100000 in an on-line fashion? Urk.

Pretty disappointing. Should probably have done Galiver instead. Oh well, offline I guess.



by "puzzles" I mean problems that have a natural translation into mathematics and reveal there a certain simplicity and elegance in generalized form. It stands in marked contrast to "riddles"2, which tend to resist mathematical patterns, and are usually dependent on culture, language, or specific trivia.


I am pretty bad at escape-room CTFs where you hold up the message under a full moon to reveal a bunch of points that turn out to align with certain zodiac signs in the sky, and then after translating the zodiacs into numbers using the provided plastic spy-codex, you get a number which turns out to be the Dewey decimal index where you can find the book that was used for the book cipher. (See example.) That kind of Sherlock Holmes stuff is completely orthogonal with the kind of puzzles I enjoy.


Given appropriate adjustment of over-rated CTFs that feature stego and guess-tasks.


I firmly believe the inventors of PHP ought to be retroactively punished for having made the world a worse place. Every time I had to read a PHP doc page and look at the comments by PHP users I could feel myself losing IQ points as if breathing in neurotoxic mold.


We're only a two-man team, so naturally we don't stand a chance for a competitive placement, but it was a lot of fun with good, challenging tasks.


Maybe I'm only critical because I'm embarrassed it fooled me!


which is fine, but it could have been a sub-problem in another task for example. One could also say it tries to teach you some math–but it's the sort of stuff with a trivial Sage/Google escape hatch. The trademark of low-effort tasks that get ground up by the "point mill" into bland paste, rather than offering any fun/engaging problem solving.


Also: this was my first impression: #!/usr/bin/env python2. :shudder: This has nothing to do with the task itself, but it always makes my heart sink and I lose a bit of faith in the problem author and the world in general.


The question "why is JavaScript?" is ostensibly answered with "because fuck you, because we're engineers and we get the job done, that's why." But the question "why is JavaScript the way it is?" can only be answered with a shrug and an embarrassed silence. Indeed, why is any scatological piece of art the way it is.


OK, actually the fact that it uses T-tables probably helps, as high-weight input will likely lead to high-weight output from the lookups there? I don't know, I've never done any power-analysis before.


I was assuming the wrong byte length of course, because it wasn't HMAC, nor was it plain CRC(x), but it was salted. And I hadn't figured out the byte-length independent approach.


I suspect Show your Patience and Intelligence I is another one, but since I only partially solved it, I couldn't really say outright. See also from last year's BalsnCTF, which–don't get me wrong–is a really cool and mind-blowing task, but you need three fairly independent key insights to solve it. Although I guess in web this kind of stuff is more like the standard way of doing things?


because the domain knowledge required for the entire thing grows linearly with unrelated sub-tasks added. This is another reason why I don't like mixed challenges. Crypto behind web, misc behind rev, etc. It feels like a sort of "miserly" way of forcing the solve count to stay low even though the subtasks are themselves easy or unoriginal. Generalists and people with "broad" CTF-knowledge are already greatly rewarded and have a (fair!) advantage over more specialized noobs (like me), but this doubles down on that advantage on the task level.


Norwegian for 'field'


I say I lucked out because my sage seems to hit a bug when I do A.multiplicative_order() directly. It just hangs and churns CPU for over a minute, so most likely I would have given up on this avenue. Who knows why sage does what it does sometimes. :its_a_mystery:


looking up on OEIS felt like cheating but it saved a couple of minutes. It's clear in hindsight, as the list goes: A, B, BAAA, BAAABBB, BAAABBBBAAABAAABAAA and so on.


There's only two hard problems in computer science: naming things, cache invalidation, and off-by-one errors. (Recycled Twitter joke.)


for some definition of fun. I must admit I sort of enjoy data/numpy massaging, especially when it's precise binary data (as opposed to analog data where you spend most of the time on cleanup or noise filtering).


"True" guessy challenges for me are more like OSINT riddles or web challs where you have no data or feedback–you simply have to "get" the core idea, and it's a binary yes/no. "Numpy massaging" tasks become a sort-of subgame where you play with combining arrays or matrices while trying to make them "fit," and there is some indication or feeling whether you're "closer" to a solution or not, e.g. by seeing what patterns you're able to produce. Xoared from BalCCon2k CTF was another example like this, where you do simple transforms of some blob of data, seeking lower and lower entropy, and eventually it suddenly becomes a readable image.


Besides, online CTFs are completely broken in terms of the "competitive element," since forming super-teams ad infinitum and rolling up with a 20-man group is all fair play, given the right communication tools and politics. The only "competitive" element that exists is between antisocial dissidents that have refused to merge into super-teams (yet)?

Author: franksh

Created: 2020-12-18 fr. 23:17