Polictf 2017 – Lucky Consecutive Guessing

This is a short snippet to solve the LCG on Polictf’17. The basic idea is to reduce the problem aX_i + b = X_{i+1} \bmod M to aX_i = X_{i+1} \bmod M. This problem is quite easy to solve.

The basic idea is to view the problem as the latter and correct the discrepancies (\delta = (b, ab, a^2b,\dots)). So, for instance, if we sample

V = \texttt{1310585136, 1634517111, 2548394614, 745784911},

then these values contain the implicit discrepancies. By performing correction of the values (subtracting \sum_k^i \delta_k from the corresponding V_i we obtain a corrected value).

So, now we can use LLL to solve the simpler problem. Once a solution is found, we use the discrepancies to correct it to the case aX_i + b = X_{i+1} \bmod M.

import random

class LinearCongruentialGenerator:

    def __init__(self, a, b, nbits):
        self.a = a
        self.b = b
        self.nbits = nbits
        self.state = random.randint(0, 1 << nbits)

    def nextint(self):
        self.state = ((self.a * self.state) + self.b) % (1 << self.nbits)
        return self.state >> (self.nbits - 32)

    def break_lcg(a,b,p,i,j,outputs):
        deltas = [b % p, (a*b + b) % p, (a^2*b + a*b + b) % p, (a^3*b + a^2*b + a*b + b) % p]
        Y = [((val << (i - j)) ) for val, delta in zip(outputs, deltas)]
        L = matrix([
                   [p,    0,  0, 0 ],
                   [a^1, -1,  0, 0 ],
                   [a^2,  0, -1, 0 ],
                   [a^3,  0,  0, -1]
        B = L.LLL()
        Y = vector([x - y for x, y in zip(Y, deltas)])
        target = vector([ round(RR(w) / p) * p - w for w in B * vector(Y) ])
        states = list(B.solve_right(target))
        return [x + y + z for x, y, z in zip(Y, states, deltas)]

# parameters
p = 2^85
a = 0x66e158441b6995
b = 0xb
n = 85
r = 32
sequence = [1310585136, 1634517111, 2548394614, 745784911]

# break the LCG
start_seed = break_lcg(a,b,p,n,r,sequence)

# run it to sample required values
lcg = LinearCongruentialGenerator(a, b, n)
lcg.state = start_seed[0]
for i in range(0, 104):
    print lcg.nextint()

Generating wallpaper-consistent terminal colors with K-means++

K-means algorithm

Assume that we have to following image:


We can represent the image in 3D-space as follows:


Suppose that we want to find eight dominant colors in the image. We could create a histogram and take the eight most common values. This would be fine, but often, very similar colors would repeat. Instead, we can try to find eight centroids in the image and find clusters of points beloning to each centroid (say, a point p belongs to a centroid p_{\text{centroid}} if \|p - p_{\text{centroid}}\|_2 < R for some radius R). All points not belonging to a centroid will be penalized using some heuristic. This is basically the K-means clustering algorithm.


The algorithm can be used to generate set of dominant colors to be used for instance in the terminal. Running the above algorithm on the image, we get


or as list

['#321331', '#e1a070', '#621d39', '#e05a40', '#9ed8aa', '#8b3860', '#f7d188', '#ab3031']

with some minor adjustment

"color": [
"foreground": "#c5c8c6",
"background": "#282a2e"

In iTerm, it might look like this


This can be achieved using the following code:

#!/usr/bin/env python
import sys, os
from sklearn.cluster import KMeans
from PIL import Image

nbrcentroids = 10
# Constant increase in colors
beta = 10
# Multplicative factor in background
gamma = 0.4
rgb2hex = lambda rgb: '#%s' % ''.join(('%02x' % min(p + beta, 255) for p in rgb))
darken = lambda rgb : (p * gamma for p in rgb)

def getcentroids(filename, n=8):
    img = Image.open(filename)
    img.thumbnail((100, 100))
    # Run K-means algorithm on image
    kmeans = KMeans(init='k-means++', n_clusters=n)
    # Get centroids from solution
    rgbs = [map(int, c) for c in kmeans.cluster_centers_]
    return rgbs

def set_colors_gnome(centroids):
    centroids = sorted(centroids, key=lambda rgb: sum(c**2 for c in rgb))
    prefix = 'gsettings set org.pantheon.terminal.settings '
    # Set background and foreground
    os.system(prefix + 'background \"%s\"' % rgb2hex(darken(centroids[0])))
    os.system(prefix + 'foreground \"%s\"' % rgb2hex(centroids[-1]))
    # Set ANSI colors
    colors = ':'.join(rgb2hex(centroid) for centroid in centroids[1:-1])
    os.system(prefix + 'palette \"' + colors + ':' + colors + '\"')

def bar(mode):
    write = sys.stdout.write
    for i in range(0, nbrcentroids):
        write('\033[0;3%dmBAR ' % i)

centroids = getcentroids(sys.argv[1], n=nbrcentroids)

It is on Github too!

PlaidCTF’17 – Multicast

We are given a challenge which contains a Sage script, which holds the following lines of code

nbits = 1024
e = 5
flag = open("flag.txt").read().strip()
assert len(flag) <= 64
m = Integer(int(flag.encode('hex'),16))
out = open("data.txt","w")

for i in range(e):
    while True:
        p = random_prime(2^floor(nbits/2)-1, lbound=2^floor(nbits/2-1), proof=False)
        q = random_prime(2^floor(nbits/2)-1, lbound=2^floor(nbits/2-1), proof=False)
        ni = p*q
        phi = (p-1)*(q-1)
        if gcd(phi, e) == 1:

    while True:
        ai = randint(1,ni-1)
        if gcd(ai, ni) == 1:

    bi = randint(1,ni-1)
    mi = ai*m + bi
    ci = pow(mi, e, ni)

…along with a text file containing the encrypted flag. Let us first parse the file and put in a JSON structure

data = {'n' : [n0, n1, ...], 'a' : [a0, a1, ...], ...}

We see that the flag is encrypted several times, up to affine transformation. If we define a polynomial g_i(x) = (a_ix + b_i)^5 - c_i \mod N_i, this has a root g_i(m) = 0 \mod N_i. We could try to find this root using Coppersmith, but it turns out it not possible since the size of (am + b)^5 is larger than N_i. However, due to a publication by Håstad, we can construct the following:

g(x) = \sum \phi_i(N_i) g_i(x)


\phi_i \mod N_j = \begin{cases}0 & \quad \text{if } i \neq j\\1 & \quad \text{when } i = j\\\end{cases}

It is easy to construct \phi_i as

\phi_i = \frac{1}{N_i}\cdot \prod_k N_k \cdot [\prod_k N_k / N_i]^{-1}_{N_i}

using the Chinese Remainder Theorem. Clearly, g(x) also has a root g(m) = 0 \mod (\prod N_i) (this is easy to check!). Since m^5 is strictly smaller than \prod_k N_k, the polynomial roots of g(x) can be found using Coppersmith!

p = data['n'][0] * data['n'][1] * data['n'][2] * data['n'][3] * data['n'][4]
PR. = PolynomialRing(Zmod(p))
f = 0

# compute the target polynomial
for i in range(0, 5):
    q = p / data['n'][i]
    qinv = inverse_mod(q, data['n'][i])
    f = f + q * qinv * ((data['a'][i]*x + data['b'][i])^5 - data['c'][i])

# make f monic
f = f * inverse_mod(f[5], p)

print f.small_roots(X=2^512, beta=1)[0]

The code outputs the long integer representation of



F4 Phantom

With F-4 Phantom II, we want to break the encryption! Please help us!!

‍‍‍nc 54979

We get a key-generation function, as seen below.

def gen_pubkey(e, p):
    assert gmpy.is_prime(p) != 0
    B = bin(p).strip('0b')
    k = random.randrange(len(B))
    k, l = min(k, len(B) - k), max(k, len(B) - k)
    assert k != l
    BB = B[:k] + str(int(B[k]) ^ 1) + B[k+1:l] + str(int(B[l]) ^ 1) + B[l+1:]
    q = gmpy.next_prime(int(BB, 2))
    assert p != q
    n = p*q
    key = RSA.construct((long(n), long(e)))
    pubkey = key.publickey().exportKey("PEM")
    return n, p, q, pubkey

So, pq = p \cdot (p \pm 2^k \pm 2^l + \Delta), where \Delta = \texttt{next\_prime}(p \pm 2^k \pm 2^l) - (p \pm 2^k \pm 2^l). The density of primes (well, asymptotically) is \pi(n) \sim \frac{x}{log x}, so we expect \Delta to be quite small.

We define a := p \pm 2^k \pm 2^l. Now, we solve the following equation

x \cdot (x \pm 2^k \pm 2^l + \Delta) = N \iff x \cdot (x + a) - N = 0 \iff x = \frac{a}{2} \pm \sqrt{\frac{a^2}{4} + N}

Now, we know that the roots should be numerically close to p and q. So, we may simply call \texttt{next\_prime} on them and check if they divide N. Note that we need to do this for different hypotheses (different k).

import gmpy, base64
from Crypto.PublicKey import RSA

def solve(a):
   a2 = a ** 2/4
   L = gmpy.sqrt(a2 + N) # L > a/2
   C_1 = gmpy.next_prime(-a/2 + L)
   if N % C_1 == 0: return C_1
   C_2 = gmpy.next_prime(a/2 + L)
   if N % C_2 == 0: return C_2
   return False

# just some public key
pem_data1 = '''

pub =  RSA.importKey(base64.b64decode(pem_data1))
N = pub.n
m = (len(bin(N))+1)/2-2
print '[ ] Solving equations...'

for mm in range(m, m+3):
  for j in range(2, m/2+1):
    k = mm - j
    l = j
    retval = solve(2**(k-1) + 2**(l-1))
    if retval: 
      q = retval
    retval = solve(2**(k-1) - 2**(l-1))
    if retval: 
      q = retval

p = N / q
print '[+] Found p = {}... and q = {}...'.format(str(p)[:40], str(q)[:40])
λ python solve.py
[ ] Solving equations...
[+] Found p = 1381282812140256921334162381757305071089... and q = 1381282811302268925712750063033928508701...

Now, it turns out that \text{gcd}(e, (p-1)(q-1)) \neq 1. We cannot decrypt the message! If we look closely, we see that \text{gcd}(e, (p-1)) = 1. Hmm… so, we can decrypt the message c^d = m \mod p. And the message is the flag, which is constant. What if we sample two ciphertexts encrypting the same message but under different keys? Yes… so, basically, we have m \mod p_1 and m \mod p_2. Then, we can combine it using CRT! This is easy!

import gmpy
from Crypto.Util.number import long_to_bytes

enc1 = 63188108518214820361083256140053967663112132356420859206347143811869148973386950682507343981284848159232100220605963292020722612854139075311063776086523677522160515093598202087755508512714885329251980275085813640578839753523295579661559881983237388178475574713319340090857313275483787001349560782781513895950569634984688753665
q1 = 33452043035349425454164058954054458228134102234436666511159820871022348004023966976017694615018111901825497223286811050478588079083387098695346723120647425399335947
p1 = 33452043035349425454164058954054813129854949698738692548175391185736387867969615080539316436404430573352896343366560167202569408949342999951142479436993639588965567
e1 = 3562731839

enc2 = 162331112890791758781057932826106636167735138703054666826574266304486608255768782351247144197937186145526374008317633308191215438756014724244242639042178681790803086016105986729819920057318114565244866632716401408212076926367974085730803964312385570202673351687846963322223962280999264618753873289802828995706526584379102468
q2 = 1381282812140256921334162381757305071089685391539884775199049048751523002134390007428038278188586333291047282664366863789424963612326970767160301759006158962675449
p2 = 1381282811302268925712750063033928508701820008572424411412024462643800411901779755548441592138469189655615818433739872652769585433967353091413641137354055362924329
e2 = 197840663

d1 = gmpy.invert(e1, p1-1)
d2 = gmpy.invert(e2, p2-1)

assert(gmpy.gcd(e1, p1-1) == 1)
assert(gmpy.gcd(e2, p2-1) == 1)

m = (pow(enc1, d1, p1) * p2 * gmpy.invert(p2, p1) + pow(enc2, d2, p2) * p1 * gmpy.invert(p1, p2)) % (p1*p2)
print long_to_bytes(m)


********** great job! the flag is: ASIS{Still____We_Can_Solve_Bad_F4!}


A fine OTP server

Connect to OTP generator server, and try to find one OTP.

nc 35156

We get the code for generating OTPs:

def gen_otps():
    template_phrase = 'Welcome, dear customer, the secret passphrase for today is: '

    OTP_1 = template_phrase + gen_passphrase(18)
    OTP_2 = template_phrase + gen_passphrase(18)

    otp_1 = bytes_to_long(OTP_1)
    otp_2 = bytes_to_long(OTP_2)

    nbit, e = 2048, 3
    privkey = RSA.generate(nbit, e = e)
    pubkey  = privkey.publickey().exportKey()
    n = getattr(privkey.key, 'n')

    r = otp_2 - otp_1
    if r < 0:
        r = -r
    IMP = n - r**(e**2)
    if IMP > 0:
    	c_1 = pow(otp_1, e, n)
    	c_2 = pow(otp_2, e, n)
    return pubkey, OTP_1[-18:], OTP_2[-18:], c_1, c_2

We note that \texttt{otp\_1}^3 < N. Because of this, the task is really trivial. We can simply compute the cubic root without taking any modulus into consideration. \texttt{gmpy2.iroot(arg, 3)} can solve this efficiently! This gives us:


Secured OTP server

Connect to OTP generator server, and try to find one OTP.
This is secure than first server 🙂

nc 12431

This is almost identical to the previous, but the OTP is chosen such that it will overflow the modulus when cubed.

    template_phrase = '*************** Welcome, dear customer, the secret passphrase for today is: '
    A = bytes_to_long(template_phrase + '00' * 18)

Note that we can write an OTP as m = A \cdot 2^k + B, where B < 2^k. So, we have m^3 = A^3 + 3A^2D + 3AD^2 + D^3. The observant reader have noticed that if we remove A^3, it will not wrap around N. Now, we can mod out the terms containing A, i.e., find m - A^3 \mod A. Now we are back in the scenario of previous challenge… so, we can use the method from before! Hence, (m - A^3 \bmod A)^{1/3}. Finally, we get



You should solve a DLP challenge, but how? Of course , you don’t expect us to give you a regular and boring DLP problem!

nc 28416

The ciphertext is generated using the following function:

def encrypt(nbit, msg):
    msg = bytes_to_long(msg)
    p = getPrime(nbit)
    q = getPrime(nbit)
    n = p*q
    s = getPrime(4)
    enc = pow(n+1, msg, n**(s+1))
    return n, enc

We have that e = (n+1)^m \mod n^s = n^m + {m \choose {m-1}}n^{m-1} + \cdots + + {m \choose 2}n^2 + {m \choose 1}n + 1. We can write this as {m \choose 1}n + 1 + \mathcal{O}(n^2). Now, we can eliminate the ordo term by taking e \mod n^2 = {m \choose 1}n + 1 \iff e - 1 \mod n^2 = mn. Assuming that m < n, we can determine m by simple division. So, \frac{e - 1 \mod n^2}{n} = m. In Python, we have that

Turns out this is the case, and, hence, we obtain


unsecure ASIS sub-d

ASIS has many insecure sub-domains, but we think they are over HTTPS and attackers can’t leak the private data, what do you think?

So, we get a PCAP. The first thing we do is to extract data binwalk -e a.pcap. This generates a folder with all the certificates used.

Then, we look at the moduli.

for f in $FILES
  openssl x509 -inform der -in $f -noout -text -modulus 

This generates a file with all moduli. Let us try something simple! Common moduli! For each pair, we check if \text{gcd}(N_i, N_j) \neq 1. If so, we have found a factor. Turns out two moduli have a common factor, so we can factor each of them and decrypt their traffic:

p1 = 146249784329547545035308340930254364245288876297216562424333141770088412298746469906286182066615379476873056564980833858661100965014105001127214232254190717336849507023311015581633824409415804327604469563409224081177802788427063672849867055266789932844073948974256061777120104371422363305077674127139401263621 
q1 = 136417036410264428599995771571898945930186573023163480671956484856375945728848790966971207515506078266840020356163911542099310863126768355608704677724047001480085295885211298435966986319962418547256435839380570361886915753122740558506761054514911316828252552919954185397609637064869903969124281568548845615791

p2 = 159072931658024851342797833315280546154939430450467231353206540935062751955081790412036356161220775514065486129401808837362613958280183385111112210138741783544387138997362535026057272682680165251507521692992632284864412443528183142162915484975972665950649788745756668511286191684172614506875951907023988325767 
q2 = 136417036410264428599995771571898945930186573023163480671956484856375945728848790966971207515506078266840020356163911542099310863126768355608704677724047001480085295885211298435966986319962418547256435839380570361886915753122740558506761054514911316828252552919954185397609637064869903969124281568548845615791

We can now generate two PEM-keys

d1 = gmpy.invert(e, (p1 - 1)*(q1 - 1))
key = RSA.construct((long(p1*q1), long(e), long(d1)))
f = open('privkey.pem','w')

Putting it into Wireshark, we obtain two images:


I totally agree.

Alice, Bob and Rob

We have developed a miniature of a crypto-system. Can you break it?
We only want to break it, don’t get so hard on our system!

This is McElice PKC. The ciphertexts are generated by splitting each byte in blocks of four bits. The following matrix is used as public key:

G = \begin{pmatrix}1& 1& 0& 0& 0& 1& 1& 0\\ 1& 1& 1& 1& 1& 1& 1& 1\\ 0& 1& 1& 0& 1& 1& 0& 0 \\ 0& 1& 1& 1& 0& 0& 1& 0 \end{pmatrix}

The ciphertext is generated as \mathbf{m}G + \mathbf{e}, which is a function from 4 bits to a byte. \mathbf{e} is an error (or pertubation) vector with only one bit set. This defines a map f: \mathbb{F}_4 \rightarrow \mathbb{F}_8. So, each plaintext byte is two ciphertext bytes.

We can first create a set of codewords

P = numpy.matrix([[1, 1, 0, 0, 0, 1, 1, 0], [1, 1, 1, 1, 1, 1, 1, 1], [0, 1, 1, 0, 1, 1, 0, 0],[0, 1, 1, 1, 0, 0, 1, 0]])
image = {} # set of codewords
for i in range(0, 2**4):
    C = (numpy.array([int(b) for b in (bin(i))[2:].zfill(4)]) * P % 2).tolist()[0]
    image[int(''.join([str(c) for c in C]), 2)] = i

Then, go through each symbol in the ciphertext, flip all possible bits (corresponding to zeroing out \mathbf{e}) and perform lookup in the set of codewords P (compute the intersection between the Hamming ball of the ciperhext block and P).

f = open('flag.enc','r')
out = ''
for i in xrange(18730/2):
    blocks = f.read(2)
    j = ord(blocks[0])
    C1 = 0
    C2 = 0
    for i in range(0, 8):
        if j ^ 2**i in image:
            C1 = image[j ^ 2**i] << 4
    j = ord(blocks[1])
    for i in range(0, 8):
        if j ^ 2**i in image:
            C2 = image[j ^ 2**i]
    out += chr(C1+C2)

Turns out it is a PNG:


Confidence DS CTF ’17 – Public Key Infrastructure

Log in as admin to get the flag.

nc pki.hackable.software 1337

This was a fun challenge, because it required many steps to complete. The first thing we notice is that MD5 is being used:

def h(x):
    return int(hashlib.md5(x).hexdigest(), 16)

def makeMsg(name, n):
    return 'MSG = {n: ' + n + ', name: ' + name + '}'

def makeK(name, n):
    return 'K = {n: ' + n + ', name: ' + name + ', secret: ' + SECRET + '}'

def sign(name, n):
    k = h(makeK(name, n))
    r = pow(G, k, P) % Q
    s = (modinv(k, Q) * (h(makeMsg(name, n)) + PRIVATE * r)) % Q
    return (r*Q + s)

Naturally, we cannot register as \texttt{admin}. An immediate conclusion is that we need to some kind of collision attack. But how? We cannot exploit any collision in the \texttt{admin} account. What if we can create a collision in r while keeping s distinct?

from coll import Collider, md5pad, filter_disallow_binstrings
import os

def nullpad(payload):
    return payload + b'\x00' * (64 - len(payload) % 64)

def left_randpad(payload):
    random_string = os.urandom(64)
    return random_string[:len(payload) % 64] + payload

# nullpad the prefix so that it does not interfere
# with the two collision blocks
prefix = nullpad(b'K = {n: ')
suffix = b', name: groci, secret:'

collider = Collider()
c1, c2 = collider.get_last_coll()
cols = collider.get_collisions()

# here are our collisions!
A = next(cols)
B = next(cols)

The above code creates a collision such that k = h(makeK(name, n)) generates the same value. The problem is that the server does not return the signature, but str(pow(sign(name, n), 65537, int(n.encode('hex'), 16))).

To get the signature, which is computed as Z = (r \cdot Q + s)^{65537} \mod M, we need to perform some additional computations. Obviously, we could factor the modulus M and compute \phi(M) and then obtain the signature as easy as Z^{65537^{-1} \mod \phi(M)}… but factoring…


The probability that it is smooth enough is not very high. Also, the number is around 300 digits so not a good candidate for msieve or other factoring software… so what then?

Well… what if 2^{96} \cdot A + c and 2^{96} \cdot B + c both are prime? Then, we can easily recover the signature as Z^{65537^{-1} \mod (M-1)}! No need for time-consuming factoring! Embodied in Python, it could look like this:

import hashlib
from Crypto.Util.number import isPrime

def num2b(i):
    c = hex(i).strip('L')
    c = (len(c) % 2) * '0' + c[2:]
    return c.decode('hex')

# get the payloads
A = int(A[64:64*3].encode('hex'), 16)
B = int(B[64:64*3].encode('hex'), 16)

# append the same data to both payload until
# they resulting numbers are BOTH prime
mul = 2 ** 96
for i in range(1, 2**20, 2):
    if isPrime(AA + i) and isPrime(BB + i):
        print AA+i
        print BB+i

As an example, we obtain

\begin{array}{rcl} n_1 &= &  105830311539247723296176540884596952305269885491629657210626981367998\\ &&347855559972583535368860650411949121405830046936667132754993342040282\\ &&396579458814796410615711380733886667185822027819749278255976029130752\\ &&166929087394904457033345839480235723175883533744993499706397792250975\\ &&3538392909994754306314762383722990106537161406818023\\ \end{array}


\begin{array}{rcl} n_2 &=&  105830311539247723296176540884596952305269885520672956204333681813273\\ &&006022550217898599044674310459977554450027911751595002340337873829980\\ &&015343362188266640170082456168713800027076364038815247707214102650115\\ &&472228269311796278082939271876115926467296299772272093195558205398667\\ &&4873073960726016011965433749481512129478713571026663\\ \end{array}

Putting this into action, we might do something like:

def connect(name, n):
   name_encoded = base64.b64encode(name)
   n_encoded = base64.b64encode(n)
   server = remote('pki.hackable.software', 1337)
   payload = 'register:' + name_encoded + ',' + n_encoded
   server.send(payload + '\n')
   return pow(int(server.recvline()), modinv(65537, n1-1), n1)

payload1 = nullpad('K = {n: ') + num2b(n1)
payload2 = nullpad('K = {n: ') + num2b(n2)

name = 'groci'
PORT = 1331

assert(h(makeK(name, payload1)) == h(makeK(name, payload2)))
assert(h(makeMsg(name, payload1)) != h(makeMsg(name, payload2)))

# Get first signature
sig = connect(name, payload1)
r1, s1 = sig / Q, sig % Q
z1 = h(makeMsg(name, payload1))

# Get second signature
connect(name, payload2)
sig = pow(int(server.recvline()), modinv(65537, n2-1), n2)
r2, s2 = sig / Q, sig % Q
z2 = h(makeMsg(name, payload2))

# Make sure we got a nonce re-use
assert(r1 == r2)

# OK, use standard nonce re-use technique...
delta = modinv(((s1 - s2) % Q), Q)
k = ( ((z1 - z2) % Q) * delta) % Q
r_inv = modinv(r1, Q)
PRIVATE = (((((s1 * k) % Q) - z1) % Q) * r_inv) % Q

# Now that we know the private key, 
# lets forge/sign the admin account!
name = 'admin'
n = 'snelhest'
k = h(makeK(name, n))

r = pow(G, k, P) % Q
s = (modinv(k, Q) * (h(makeMsg(name, n)) + PRIVATE * r)) % Q
sig = r*Q + s

# connect and login
name_encoded = base64.b64encode(name)
n_encoded = base64.b64encode(n)
server = remote('pki.hackable.software', 1337)
payload = 'login:' + name_encoded + ',' + n_encoded + ',' + base64.b64encode(num2b(sig))
server.send(payload + '\n')
print server.recvline()

We obtain the following signatures:

r_1 = 455070882754437660339639715779025537190883529105 s_1 = 719850042802710537609987071764516609303635832220
r_2 = 455070882754437660339639715779025537190883529105 s_2 = 584220801769294572739392594733340817152152143208

Now, with our forced nonce re-use, we can use standard techniques to obtain the private key and sign our \texttt{admin} account, which gives the flag


Note to reader: I apologize for the excessive use of memes.

0ctf’17 – All crypto tasks


Just a simple scheme.
nc 8221

The encrypt function in the scheme takes the username as input. It hashes the username with MD5, appends the name to the hash and encrypts with a secret key k, i.e. \textsf{Enc}_k(H_\text{MD5}(\text{username}) \|\text{username}). Then, the secret becomes \text{IV} \| C_1 \| C_2 .... Notably, the first block C_1 contains the hash, but encrypted.

Recall how the decryption is defined:

C_i = \textsf{Dec}_k(C_i) \oplus C_{i-1}, C_0 = \text{IV}

So, what would happen if we input u = H_\text{MD5}(\texttt{admin}) \| \texttt{admin} as username? Then, we have encrypted H_\text{MD5}(u) \| u, but we want only u. As mentioned before, the hash fits perfectly into a single block. So, by removing the \text{IV}C_1 becomes the new \text{IV} (which has no visible effect on the plaintext anymore!). Then, we have \textsf{Enc}_k(H_\text{MD5}(\texttt{admin}) \| \texttt{admin}), which is all what we need.

The flag is flag{Easy_br0ken_scheme_cann0t_keep_y0ur_integrity}.


I swear that the safest cryptosystem is used to encrypt the secret!

We start off analyzing the code. Seeing process(m, k) function, we note that this is actually something performed in \mathbb F_2[x]/P(x) with the mapping that, for instance, an integer 11 is \texttt{1011} in binary, which corresponds to a polynomial x^3 + x + 1. The code is doing (m \oplus k)^2 in GF(2^{256}).

The keygen function repeatedly calls key = process(key, seed). The first value for key is random, but the remaining ones does not. seed remains the same. Define K to be the key and S the seed. Note that all elements are in GF(2^{256}). The first stream value Z_0 is unknown.

Z_0 = K
Z_1 = (Z_0 \oplus S)^2 = K^2 \oplus S^2
Z_2 = (Z_1 \oplus S)^2 = Z_1^2 \oplus S^2

So, we can compute the seed and key as S^2 = Z_2 \oplus Z_1^2 and K^2 = Z_1 \oplus S^2 = Z_1 \oplus Z_2 \oplus Z_1^2. The individual square roots exist and are unique.

def num2poly(num):
    poly = R(0)
    for i, v in enumerate(bin(num)[2:][::-1]):
        if (int(v)):
            poly += x ** i
    return poly

def poly2num(poly):
    bin = ''.join([str(i) for i in poly.list()])
    return int(bin[::-1], 2)

def gf2num(ele):
    return ele.polynomial().change_ring(ZZ)(2)

P = 0x10000000000000000000000000000000000000000000000000000000000000425L

fake_secret1 = "I_am_not_a_secret_so_you_know_me"
fake_secret2 = "feeddeadbeefcafefeeddeadbeefcafe"
secret = str2num(urandom(32))

R = PolynomialRing(GF(2), 'x')
x = R.gen()
GF2f = GF(2**256, name='a', modulus=num2poly(P))

f = open('ciphertext', 'r')
A = GF2f(num2poly(int(f.readline(), 16)))
B = GF2f(num2poly(int(f.readline(), 16)))
C = GF2f(num2poly(int(f.readline(), 16)))

b = GF2f(num2poly(str2num(fake_secret1)))
c = GF2f(num2poly(str2num(fake_secret2)))

# Retrieve partial key stream using known plaintexts
Y = B + b
Z = C + c

Q = (Z + Y**2)
K = (Y + Q).sqrt()

print 'flag{%s}' % hex(gf2num(A + K)).decode('hex')

This gives the flag flag{t0_B3_r4ndoM_en0Ugh_1s_nec3s5arY}.


Well, maybe the previous one is too simple. So I designed the ultimate one to protect the top secret!

There are some key insights:

  • The process1(m, k) function is basically the same as in previous challenge, but it computes the multiplication m \otimes k with the exception that elements are in GF(2^{128}) this time.  We omitt the multiplication symbol from now on.
  • The process2(m, k) function might look involved, but all that it does is to compute the matrix multplication between two 2 \times 2 matrices (with elements in GF(2^{128})), i.e., XY = \begin{pmatrix}x_0 & x_1 \\ x_2 & x_3\end{pmatrix}\begin{pmatrix}y_0 & y_1 \\ y_2 & y_3\end{pmatrix}
  • We start with matrices X = \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} and Y = \begin{pmatrix}A & B \\ 0 & 1\end{pmatrix}.
  • Raising A to a power yields has a closed form formula: A^s = \begin{pmatrix}A^s & B(A^{s-1} \oplus A \oplus 1 )\\ 0 & 1\end{pmatrix} =\begin{pmatrix}A^s & B\frac{A^{s} \oplus 1 }{A \oplus 1}\\ 0 & 1\end{pmatrix}.
  • The nextrand(rand) function takes the integral value of N, we call this s and computes Y^s = \begin{pmatrix}A & B \\ 0 & 1\end{pmatrix}^s via a square-and-multiply type algorithm. In python, it would be
    def proc2(key):
        AN = A**gf2num(N)
        return key*AN+(AN+1)/(A+1)*B

Let us look at the nextrand(rand) function a little more. Let R be the random value fed to the function. Once Y^s is computed, it returns

Q = R A^s \oplus  \frac{A^s \oplus 1}{A\oplus 1} B

Define U = \frac{B}{A\oplus 1}. Adding this to the above yields

Q\oplus  U = R A^s \oplus  \frac{A^s \oplus 1 \oplus 1}{A\oplus 1} B = A^s(R \oplus \frac{B}{A\oplus 1}) = A^s(R\oplus U).

So, A^s = \frac{Q\oplus U}{R \oplus U}. Note that given two elements of the key stream, all these elements are known. Once determined, we compute the (dicrete) \log \frac{Q\oplus U}{R\oplus U} to find s. And once we have s, we also have N. Then, all secrets have been revealed!

From the plaintext, we can immediately get the key K by XORing the first part of the plaintext with the corresponding part of the ciphertext. This gives K =\texttt{0x2fe7878d67cdbb206a58dc100ad980ef}.

R = PolynomialRing(GF(2), 'x')
x = R.gen()

GF2f = GF(2**128, name='a', modulus=num2poly(0x100000000000000000000000000000087))

A = GF2f(num2poly(0xc6a5777f4dc639d7d1a50d6521e79bfd))
B = GF2f(num2poly(0x2e18716441db24baf79ff92393735345))
G1 = GF2f(num2poly(G[1]))
G0 = GF2f(num2poly(0x2fe7878d67cdbb206a58dc100ad980ef))

U = B/(A+1)
Z = (G1+U)/(G0+U)
N = discrete_log(Z, A, K.order()-1)

We can then run the encryption (the default code) with the parameters N,K fixed to obtain the flag flag{LCG1sN3ver5aFe!!}.

Boston Key Party 2017 – Sponge

In this challenge, we are given a hash function, which essentially splits the input into chunks of ten bytes. It then appends six bytes of null data to it and XORs with the current state. The state is encrypted with the all-zero key and updated. This process is repeated until all blocks have been exhausted.

A simple case is to consider what happens when we hash ten bytes of null data.

\begin{array}{lcll} \textbf{State}_{i} & & \textbf{State}_{i+1} & \textbf{Input}\\ \texttt{00000000000000000000~000000000000}&\rightarrow&\texttt{66e94bd4ef8a2c3b884c~fa59ca342b2e}&\texttt{00000000000000000000}\\ \texttt{e6e94bd4ef8a2c3b884d~fa59ca342b2e} &\rightarrow& \texttt{cccb674e90ee226bea81~557bff1e7123} & \texttt{80000000000000000001} \end{array}

If we can bring it back to the same state, we have a local collision. Here is an example of such a collision:

\begin{array}{lcll} \textbf{State}_{i} & & \textbf{State}_{i+1} & \textbf{Input}\\ \texttt{00000000000000000000~000000000000}&\rightarrow&\texttt{66e94bd4ef8a2c3b884c~fa59ca342b2e}&\texttt{00000000000000000000}\\ \texttt{210e7f3dc8b624d41038 fa59ca342b2e} &\rightarrow& \texttt{26e1e20ccd3ce8e975a6 33c3d2824408} &\texttt{47e734e9273c08ef9874}\\ \texttt{0f809b9375310b0656cf 33c3d2824408}& & \texttt{d979b8a852674734c855 000000000000} & \texttt{2961799fb80de3ef2369}\\ \texttt{00000000000000000000 000000000000} & \rightarrow &\texttt{66e94bd4ef8a2c3b884c~fa59ca342b2e}&\texttt{d979b8a852674734c855}\\ \texttt{e6e94bd4ef8a2c3b884d~fa59ca342b2e} &\rightarrow& \texttt{cccb674e90ee226bea81~557bff1e7123} & \texttt{80000000000000000001} \end{array}

So, how did we create the above collision? Well, actually, it is not too complicated… first, note that we cannot control the last six bytes. Recall that \textsf{Enc}(\textsf{Enc}(x \oplus a \| \texttt{0x00}^6) \oplus b \| \texttt{0x00}^6) \oplus c \| \texttt{0x00}^6 = y. Let us reorder the function as follows:

\textsf{Enc}(x \oplus a \| \texttt{0x00}^6)  = \textsf{Dec}(y \oplus c \| \texttt{0x00}^6) \oplus b \| \texttt{0x00}^6

If we force the trailing six bytes to null and then decrypt that block for different values on y \oplus c \| \texttt{0x00}^6, since we can control c. Equivalently, we can encrypt for different values of x \oplus a \| \texttt{0x00}^6, where the trailing six bits will be the trailing six bits of \textsf{Enc}(\texttt{0x00}^{16}).

Utilizing the birthday paradox, we can find a collision in the six bytes in \sim \sqrt{256^6} time and space. This is done by putting about 2^{24} values of the encryption (or decryption) in a table. Then, we generate the decryptions (or encryptions, respectively) and look in the table.


In Python, it could look something like:

trailing_bytes_first = AES.new('\x00'*16).encrypt('\x00'*16)[-6:]

for i in range(0, 2**24):
    plaintext = os.urandom(10) + trailing_bytes_first
    data[AES.new('\x00'*16).encrypt(plaintext)[-6:]] = plaintext

for i in range(0, 2**24):
    ciphertext = os.urandom(10) + '\x00\x00\x00\x00\x00\x00'
    if AES.new('\x00'*16).decrypt(ciphertext)[-6:] in data:
        print [data[AES.new('\x00'*16).decrypt(ciphertext)[-6:]]], [ciphertext]

Two such values are

local_collision_blocks = ['!\x0e\x7f=\xc8\xb6$\xd4\x108\xfaY\xca4+.', '\xd9y\xb8\xa8RgG4\xc8U\x00\x00\x00\x00\x00\x00']

Finally, we generate a collision as follows

GIVEN = 'I love using sponges for crypto'
A = AES.new('\x00'*16).encrypt(local_collision_blocks[0])
B = AES.new('\x00'*16).decrypt(local_collision_blocks[1])
local_collision = '\x00'*10 + xorstring(local_collision_blocks[0][:10], AES.new('\x00'*16).encrypt('\x00'*16)) + xorstring(A,B)[:10] + xorstring(local_collision_blocks[1][:10], GIVEN[:10])