ASIS CTF’17

F4 Phantom

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

‍‍‍nc 66.172.27.77 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 = '''
MIGmMA0GCSqGSIb3DQEBAQUAA4GUADCBkAKBhyW0uQDhy7/eShVn6VQWQisC8sda
zl8xJmCe8eEPGNfDVMFgN7Ll4WdRLYE4T8dOvoMb99Cmjhym7Orb/qfP5q/BpTQy
5hr1w5RZVtYMqQ5I+M4qg7E8kkVhGJh/13bNwqEYvV6VNSCK+bgFWVGMelTQam31
Fiq6wsTLbIBCrLie52Cil1584QIEC8rPFw==
'''

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
      break
    retval = solve(2**(k-1) - 2**(l-1))
    if retval: 
      q = retval
      break

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)

…prints:

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

Alright!

A fine OTP server

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

nc 66.172.27.77 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:

ASIS{0f4ae19fefbb44b37f9012b561698d19}

Secured OTP server

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

nc 66.172.33.77 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

ASIS{gj____Finally_y0u_have_found_This_is_Franklin-Reiter’s_attack_CongratZ_ZZzZ!_!!!}

DLP

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 146.185.143.84 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

ASIS{Congratz_You_are_Discrete_Logarithm_Problem_Solver!!!}

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.

#!/bin/bash
FILES=*.crt
for f in $FILES
do
  openssl x509 -inform der -in $f -noout -text -modulus 
done

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')
f.write(key.exportKey('PEM'))
f.close()

Putting it into Wireshark, we obtain two images:

flag
flag2

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)
f=open('decrypted','w')
f.write(out)

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()
collider.bincat(prefix)
collider.safe_diverge()
c1, c2 = collider.get_last_coll()
collider.bincat(suffix)
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…

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)

# PURE BRUTEFORCE, NOTHING FANCY HERE!
# 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
        break

As an example, we obtain

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

and

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

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
DrgnS{ThisFlagIsNotInterestingJustPasteItIntoTheScoreboard}!

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

0ctf’17 – All crypto tasks

Integrity

Just a simple scheme.
nc 202.120.7.217 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}.

OTP1

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}.

OTP2

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 , we call this and computes 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.

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.

lookup

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])

BSides’17 – Delphi

In this challenge, we are given a server which accepts encrypted commands and returns the resulting output. First we define our oracle go(cmd).

import urllib2

def go(cmd):
	return urllib2.urlopen('http://delphi-status-e606c556.ctf.bsidessf.net/execute/' + cmd).read()

This simply return the status from the server. It is common for this kind of CTF challenges to use some block-cipher variant such as some of the AES modes.

The first guess I had was that AES-CBC was being used. ~~That would mean that if we try to flip some bit in a block somewhere in the middle of the ciphertext, the first decrypted part would remain intact, whilst the trailing blocks would get scrambled.~~

~~Assume that we have four ciphertext blocks~~ $C_0, C_1, C_2, C_3$ ~~and the decryption is~~ $\textsf{dec}_k : C_0\| C_1\| C_2\| C_3 \mapsto P_0\|P_1\|P_2\|P_3$ . ~~Now, we flip a bit in~~ $C_1$ ~~so that we get~~ $C_1'$ , ~~then we have~~ $\textsf{dec}_k : C_0\| C'_1\| C_2\| C_3 \mapsto P_0\|P'_1\|P'_2\|P'_3$ . (This is not true, thanks to hellman for pointing that out in the comments).

Turns out this is not the case. In fact, the error did only propagate one block and not further, i.e., $\textsf{dec}_k : C_0\| C'_1\| C_2\| C_3 \mapsto P_0\|P'_1\|P'_2\|P_3$ . Having a look at the Wikipedia page, I found that this is how AES-CFB/(CBC) would behave (image from Wikipedia):

1202px-cfb_decryption-svg

601px-cbc_decryption-svg

Since $\textsf{dec}_k(C_0) \oplus C_1 = P_1$ , we can inject some data into the decrypted ciphertext! Assume that we want $P'_1 = Q$ . Then, we can set $C'_1 = C_1 \oplus P_1 \oplus Q$ , since then $\textsf{dec}_k(C_0) \oplus C'_1 = P_1\oplus P_1 \oplus Q = Q$ . Embodying the above in Python, we might get something like

def xor(a, b):
    return ''.join(chr(ord(x) ^ ord(y))
              for x, y in zip(a, b))

cmd = '8d40ab447609a876f9226ba5983275d1ad1b46575784725dc65216d1739776fdf8ac97a8d0de4b7dd17ee4a33f85e71d5065a02296783e6644d44208237de9175abed53a8d4dc4b5377ffa268ea1e9af5f1eca7bb9bfd93c799184c3e0546b3ad5e900e5045b729de2301d66c3c69327'
response = ' to test multiple-block patterns' # the block we attack

split_blocks = [cmd[i * 32: i * 32 + 32]
                for i in range(len(cmd) / 32)]

block = 3 # this is somewhat arbitrary

# get command and pad it with blank space
append_cmd = '  some command'
append_cmd = append_cmd + '\x20' * (16 - len(append_cmd))

new_block = xor(split_blocks[block].decode("hex"),
                response).encode('hex')
new_block = xor(new_block.decode("hex"),
                append_cmd).encode('hex')

split_blocks[block] = new_block
cmd = ''.join(split_blocks)
#print cmd
print go(cmd)

We can verify that this works. Running the server, we get

This is a longer string th\x8a\r\xe4\xd9.\n\xde\x86\xb6\xbd*\xde\xf8X\x15I  some command  e-block patterns\n

OK, so the server accepts it. Nice. Can we exploit this? Obviously — yes. We can guess that the server does something like

echo "{input string}";

First, we break off the echo statement. Then we try to cat the flag and comment out the rest. We can do this in one block! Here is how:

append_cmd = '\"; cat f* #'

Then, the server code becomes

echo "{partial + garbage}"; cat f* #{more string}";

The server gives the following response:

This is a longer string th:\xd7\xb1\xe8\xc2Q\xd7\xe8*\x02\xe8\xe8\x9c\xa6\xf71\n
FLAG:a1cf81c5e0872a7e0a4aec2e8e9f74c3\n

Indeed, this is the flag. So, we are done!

Some icons for macOS Sierra.

Today, I drew some replacement icons for some apps I use daily (i.e., Terminal, Wireshark, Hex Fiend and Hopper) since I did not like default ones. I release them under the GPL v.3.0 License, so feel free to use them in your OS 🙂

Terminal black

Terminal dark grey

Terminal light grey

Wireshark

Hex Editor

Hopper Disassembler

In action (with the dark grey and light grey terminal icons, respectively):

Dark icon on Finder light theme

Light icon on Finder dark theme

Here are some additional icons:

Slack

Tor Browser

VLC

Skype

Covert communication over a noisy channel

Consider a situation in which one part Alice wants to communicate with another part Bob over a discrete memoryless channel with crossover probability $p_1$ while ensuring a low level of detection in the presence of a warden Willie, who observes the communication through another discrete memoryless channel with crossover probability $p_2$ . We assume that all parts are equally computationally bounded.

Skärmavbild 2016-08-23 kl. 22.44.47

In this situation, we require that $p_1 < p_2$ for non-zero $p_1,p_2$ . Let $\mathbf{u} \in \mathbb{F}_2^l$ be a sequence of secret bits she wants to transmit.

NOTE: Alice and Bob may on forehand agree upon any strategy they desire, but the strategy is known also to Willie.

Communicating with a secret key

If Alice and Bob share a secret key $k$ , they may use it to pick a common encoding, i.e., an error-correcting code. This encoding in turn is used to encode the message, which of course must be able to correct the errors produced by the channel. Assuming that the correction capability is below the error rate of Willie’s channel, he cannot decode. Let $N$ bits be the length of the publicly transmitted sequence. An established result states that if $\sqrt N \cdot \log N$ bits are shared between Alice and Bob, they can transmit $\sqrt N$ secret bits without Willie seeing them (with high probability).

Communicating without a secret key

Now, assume that Alice and Bob do not share a common key. Alice performs the following steps:

Picks a cryptographically secure random vector $\mathbf r$ .
Computes the scalar product $\mathbf r \cdot \mathbf u = k$ .
Sends the bits $\mathbf r \| k$ over the channel.

This reduces to the problem of $\textsc{Learning Parity with Noise}$ (LPN) and can be solved with the BKW algorithm (or similar) if $p_1$ is sufficiently low. In particular, if Bob receives

$N = 4 \cdot \log 2 \cdot l \cdot \epsilon_1^{-2 (l+1)}$

such sequences, or equivalently,

$N \cdot l = 4 \cdot \log 2 \cdot l^2 \cdot \epsilon_1^{-2 (l+1)}$

bits, he will be able to decode with high probability. Here we have exploited the piling-up lemma disregrading the fact that some bits in $\mathbf{u}$ are zero and does not contribute. For some probabilities $p_1<p_2$ and natural number $N$ , the information is hidden from Willie. The information rate is determined as follows: $N = \mathcal{O}(l^2\cdot \epsilon_1^{-2(l+1)})$ , so

$\mathcal{O}(\sqrt N)=\mathcal{O}(l \cdot\epsilon_1^{-l-1})\iff\mathcal{O}(\sqrt N\cdot\epsilon_1^{l+1})=\mathcal{O}(l)$ .

This bound can be improved upon by an increase in the number of parities.

Here is a more detailed paper on a similar topic. Another paper is here.

grocid.net

scribbles from past, present and possibly future