HKCERT CTF 2022 Postmortem (II): Harder Crypto Challenges

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HKCERT CTF 2022 Postmortem (II): Harder Crypto Challenges
Regarding the thumbnail. This is my Discord avatar combined with Komi from the anime Komi can’t communicate, where the challenge Mystiz can’t codeis referred to. Made by @byronwai.

We will continue walking through the remaining crypto challenges I wrote for HKCERT CTF 2022: Mystiz can't code, Slow keystream and King of Rock, Paper, Scissors.

亞洲協會香港中心 / Mystiz can't code (Crypto)

Challenge Summary

Mystiz can't code! He wanted to implement Advanced Encryption Standard (AES) in batch but he couldn't do it correctly.

Anyway, he decided to give up and uploads what he has for now. Since this is implemented based on AES, it should be almost as secure as AES as well?

We are given a batch script which attempts to implement AES. We are also given two ciphertexts encrypted with the same key (and one of the plaintexts is given).

Microsoft Windows [Version 10.0.19044.2006]
(c) Microsoft Corporation. All rights reserved.

C:\Users\mystiz>aes.bat [**REDACTED_KEY**] 68656c6c6f20776f726c6421
2740f489df8449453fd87f075a648e94

C:\Users\mystiz>aes.bat [**REDACTED_KEY**] [**REDACTED_FLAG**]
9a3538b25faf70f2654e7816df540acedb753d319f76311d95a4ad5e797fff3e13f7dcbde563baf8f7ac62580196b5ca911789fedade0fd6fb40642413d521992311f9bc01d127db4bbcf257ee5deb8fcd49b23aadd12f52fa7829e7e281373f

Solution

😒 Objection! I can actually code. I actually built a wrote a working AES then broke it in the way I wanted.

Part I: Background story

I am a newbie to batch scripts. The only batch script I wrote was to shut down a machine. Back in the day MSN still exists (an application in the stone age, huh?), I used to rename those batch scripts into files, changed their icons and sent them to my friends. It was pretty satisfying when you see they became offline as soon as they receive the files.

Well, after uncountably many years, I am asked to write batch scripts for my work to maintain Windows machines. I was stuck into an issue for a few hours... and it was pretty amusing. Anyway, I thought that is worth making it into a CTF challenge.

Part II: Understanding the cipher

Denote the flawed AES by AES†. When we compare the batch script with the below figure showing the implementation, it seems that everything matches.

HKCERT CTF 2022 Postmortem (II): Harder Crypto Challenges
A brief flow for the correct AES implementation.

For instance, the EncryptBlock and the AddRoundKey functions below looked legit (except that the state is a global variable).

:EncryptBlock
    set block_id=%1

    call :LoadState %block_id%

    set round_key=0
    call :AddRoundKey %round_key%

    for /l %%r in (1, 1, 9) do (
        set round_key=%%r
        call :SubBytes
        call :ShiftRows
        call :MixColumns
        call :AddRoundKey %round_key%
    )

    set round_key=10
    call :SubBytes
    call :ShiftRows
    call :AddRoundKey %round_key%
    
    call :SaveState %block_id%
exit /b 0
:AddRoundKey
    set round_id=%1
    for /l %%i in (0, 1, 15) do (
        set /a j=16*%round_id%+%%i
        set /a STATE[%%i]="STATE[%%i]^KEY[%j%]"
    )
exit /b 0

If you are convinced, then unfortunately you fell into the same pitfall as I did. In reality, there is a bizarre behaviour that I couldn't understand while writing the challenge. In short, both EncryptBlock and AddRoundKey are flawed so that AES† is not working as intended. The below figure shows how the messages are encrypted using AES†. n in the figure is the number of blocks the script has encrypted.

HKCERT CTF 2022 Postmortem (II): Harder Crypto Challenges
The actual encryption flow for the given batch file.

We can see from above that only two bytes from the subkeys are used when encrypting a block, which is k33 from the 0-th and the n-th round keys. The other parts are performed in the same way that AES does. Up to here, we can already write a script to exhaust the 216 subkey candidates for each block.

hkcert22{pr09r4mm1ng_in_b47ch_1s_s0_d1fficul7_4nd_why_d03s_th1n9s_1n_br4ck3t_run5_in_p4ral13l}

Part III: Batch, why?

🧠 RIP brain cells. I am unable to figure out all the details… It didn’t work as expected. Please let me know if you know what’s going on.

In this section, we will study the causes of the problem. In short, the commands in the parentheses are executed simultaneously1.

For batch scripts, all function variables are global unless we specify SETLOCAL in the function. Let's consider how the first block is encrypted. It calls LoadStateat the very beginning, which eventually sets j←15. Since j is global, its value will be shared. After that, AddRoundKey, which is defined below, is called with the argument round_id = 0.

:AddRoundKey
    set round_id=%1
    for /l %%i in (0, 1, 15) do (
        set /a j=16*%round_id%+%%i
        set /a STATE[%%i]="STATE[%%i]^KEY[%j%]"
    )
exit /b 0

Since the commands are executed simultaneously, j was unable to update timely when it is used to update the state. The value of j in the line that sets the value of STATE[i] is fixed for all i=0,1,...,15. Therefore, instead of XORing the message with the round key, every byte of the message is XORed with KEY[15] (i.e., k33 from the 0-th round key).

:EncryptBlock
    set block_id=%1

    call :LoadState %block_id%

    set round_key=0
    call :AddRoundKey %round_key%

    for /l %%r in (1, 1, 9) do (
        set round_key=%%r
        call :SubBytes
        call :ShiftRows
        call :MixColumns
        call :AddRoundKey %round_key%
    )

    set round_key=10
    call :SubBytes
    call :ShiftRows
    call :AddRoundKey %round_key%
    
    call :SaveState %block_id%
exit /b 0

Let's have a look at the EncryptBlock function. The parameter for the AddRoundKeyfunction, %round_key%, is actually zero for all the rounds. This sets j←15 when AddRoundKey is called, which will be used in the final AddRoundKey call. With that said, when n=0, only KEY[15] will be used for encryption.

On the other hand for n>0, LoadState sets j←16n+15. This j will only be used on the first AddRoundKey (and the subsequent js for AddRoundKey are 15), which implies that KEY[16*n+15] (i.e., k33 of the n-th round key) will be used for the first AddRoundKey and k33 of the 0-th round key.

澳洲牛奶公司 / Slow keystream (Crypto)

Challenge Summary

Want a challenge in Hong Kong?

Enter the Australia Dairy Company and execute the 379-byte flag generator written in Golang. I bet you will be expelled from the restaurant before the second character of the flag appears.

We are given a binary written in Golang (with source code given below), and flag.enc. The objective is to recover the input.

package main

import (
	"fmt"
	"math/rand"
	"os"
)

func main() {
	rand.Seed(1337)

	flag, err := os.ReadFile("flag.enc")
	if err != nil {
		fmt.Println("cannot open flag.enc")
		os.Exit(1)
	}

	for i, j := uint64(0), 0; j < len(flag); i++ {
		rand.Uint64()
		if i == uint64(1)<<j {
			x := byte(rand.Uint64())
			fmt.Print(string(flag[j] ^ x))
			j += 1
		}
	}
	fmt.Println()
}

Solution

Part I: What is in Golang's PRNG?

In this challenge, the characters of the encrypted flag is XORed with the outputs from rand.Uint64. With that said, we have to look deeper into the internals of the PRNG. We can start with the documentation - with that, we can read the source code and see where the actual code is. Tracing in the codebase, eventually, we end up at its implementation (snippets below):

// https://cs.opensource.google/go/go/+/refs/tags/go1.19:src/math/rand/rng.go

// Copyright 2009 The Go Authors. All rights reserved.
// Use of this source code is governed by a BSD-style
// license that can be found in the LICENSE file.

package rand

/*
 * Uniform distribution
 *
 * algorithm by
 * DP Mitchell and JA Reeds
 */

const (
	rngLen   = 607
	rngTap   = 273
	rngMax   = 1 << 63
	rngMask  = rngMax - 1
	int32max = (1 << 31) - 1
)

var (
	// rngCooked used for seeding. See gen_cooked.go for details.
	rngCooked [rngLen]int64 = [...]int64{
		-4181792142133755926, -4576982950128230565, 1395769623340756751, 5333664234075297259,
		// ...600 numbers snipped...
		8382142935188824023, 9103922860780351547, 4152330101494654406,
	}
)

type rngSource struct {
	tap  int           // index into vec
	feed int           // index into vec
	vec  [rngLen]int64 // current feedback register
}

// seed rng x[n+1] = 48271 * x[n] mod (2**31 - 1)
func seedrand(x int32) int32 {
	const (
		A = 48271
		Q = 44488
		R = 3399
	)

	hi := x / Q
	lo := x % Q
	x = A*lo - R*hi
	if x < 0 {
		x += int32max
	}
	return x
}

// Seed uses the provided seed value to initialize the generator to a deterministic state.
func (rng *rngSource) Seed(seed int64) {
	rng.tap = 0
	rng.feed = rngLen - rngTap

	seed = seed % int32max
	if seed < 0 {
		seed += int32max
	}
	if seed == 0 {
		seed = 89482311
	}

	x := int32(seed)
	for i := -20; i < rngLen; i++ {
		x = seedrand(x)
		if i >= 0 {
			var u int64
			u = int64(x) << 40
			x = seedrand(x)
			u ^= int64(x) << 20
			x = seedrand(x)
			u ^= int64(x)
			u ^= rngCooked[i]
			rng.vec[i] = u
		}
	}
}

// Uint64 returns a non-negative pseudo-random 64-bit integer as an uint64.
func (rng *rngSource) Uint64() uint64 {
	rng.tap--
	if rng.tap < 0 {
		rng.tap += rngLen
	}

	rng.feed--
	if rng.feed < 0 {
		rng.feed += rngLen
	}

	x := rng.vec[rng.feed] + rng.vec[rng.tap]
	rng.vec[rng.feed] = x
	return uint64(x)
}

Part II: Fibonnnnacccci and its matrix representation

Note. The operations below are taken modulo 264.

We can see that rng.vec is an array of 607 64-bit numbers. When Uint64 is called, two numbers from the array are retrieved, summed and returned. The sum will be stored in the array for future calls. Let v(n):=(v0(n),v1(n),...,v606(n)) be the vector after n Uint64 calls. Then the first two v(n)s being:

vk(1):={v333(0)+v606(0)if k=333vk(0)otherwise,vk(2):={v332(1)+v605(1)if k=332vk(1)otherwise,...

It is obvious that Uint64 changes only one entry at a time. Let's define

(a0,a1,...,a606):=(v333(0),v332(0),...,v0(0),v606(0),v605(0),...,v334(0)).

Furthermore, let also a607,a608,... be the only changing entry in each round, i.e., a607:=v333(1), a608:=v332(2) and so on. Therefore a607=a0+a334,a608=a1+a335,....

This is simply a lagged Fibonacci generator. We can represent it as a matrix multiplication, which will be useful when we compute the values efficiently:

HKCERT CTF 2022 Postmortem (II): Harder Crypto Challenges

Part III: The final steps

Finally, let's revisit flag.go:

func main() {
	rand.Seed(1337)

	flag, err := os.ReadFile("flag.enc")
	if err != nil {
		fmt.Println("cannot open flag.enc")
		os.Exit(1)
	}

	for i, j := uint64(0), 0; j < len(flag); i++ {
		rand.Uint64()
		if i == uint64(1)<<j {
			x := byte(rand.Uint64())
			fmt.Print(string(flag[j] ^ x))
			j += 1
		}
	}
	fmt.Println()
}

Let n be the length of the flag. The below flow explains which rand.Uint64() are used after the random is seeded:

  1. Generate two unused rand.Uint64() and set k←0.
  2. Generate a rand.Uint64() to decrypt the k-th byte from flag.enc.
  3. Generate 2k+1 unused rand.Uint64().
  4. If k<n−1, set k←k+1 and jump to step 2. Otherwise, we are done.

Equivalently, this is the draft of the solve script:

  1. Initialize M be the 607×607 matrix, set a←(a0,a1,...,a607) and k←0.
  2. Set a←M2⋅a.
  3. Set a←T⋅a.
  4. Decrypt the k-th character of the flag from the last entry of a.
  5. Set a←S⋅a and S←S2.
  6. If k<n−1, set k←k+1 and jump to step 3. Otherwise, we are done.

In particular, step 5 is used for "fast forwarding". If S=Tm, then a←S⋅a is a way to skip m steps at a time. Also, S←S2 sets S to T2m, which makes subsequent calls to skip 2m steps.

Solve script

RNGCOOKED = [
    -4181792142133755926, -4576982950128230565, 1395769623340756751, 5333664234075297259,
    # ...Snipped. Please copy directly from Golang's source code 
    8382142935188824023, 9103922860780351547, 4152330101494654406
]
RNGLEN = 607
RNGTAP = 273

class GoRng:
    def __init__(self, seed):
        self.tap = 0
        self.feed = RNGLEN - RNGTAP

        self.vec = vector(Zmod(2^8), [0 for _ in range(RNGLEN)])

        seed %= (1<<31) - 1
        if seed == 0: seed = 89482311

        x = seed
        for i in range(-20, 0):
            x = self.__seedrand(x)

        for i in range(RNGLEN):
            x = self.__seedrand(x)
            u = x<<40
            x = self.__seedrand(x)
            u ^^= x<<20
            x = self.__seedrand(x)
            u ^^= x
            u ^^= RNGCOOKED[i]
            self.vec[i] = u

    def __seedrand(self, x):
        A, Q, R = 48271, 44488, 3399
        hi = x // Q
        lo = x % Q
        x = A*lo - R*hi
        return x % ((1<<31) - 1)


rng = GoRng(1337)

T = Matrix(Zmod(2^8), 607, 607)

for i in range(606):
    T[i, i+1] = 1
T[606, 0]     = 1
T[606, 334]   = 1

S = T

a = vector(Zmod(2^8), list(rng.vec[333::-1]) + list(rng.vec[:333:-1]))

# Skip the first two randoms
a = T^2 * a

with open('flag.enc', 'rb') as f: flag = f.read()
output = []
for i, f in enumerate(flag):
    a = T*a
    output.append(int(a[-1]) ^^ f)
    a = S*a
    S = S*S
    print(bytes(output))
💭 Why modulo 256 instead of 264? I was using modulo 264 all the time during testing, but @CrazyManArmy showed me that we can use 256 because 256 is a factor of 264, and we only care the lower 8 bits from rand.Uint64. This decreases the running time of the solve script from 1.5 hours to… 3 seconds.

海帆道 / King of Rock, Paper, Scissors (Crypto)

Challenge Summary

Let's play rock paper scissors securely. Since we are using a commitment scheme, none of us can cheat!

In this challenge, all messages are transmitted via protocol buffers (protobuf). Messages related to the protocol are given below:

message ServerRoundInitMessage {
  bytes nonce = 1; // server's nonce
}

message ClientRoundInitMessage {
  bytes hash = 1;  // MD5 digest of ClientMoveMessage
  bytes nonce = 2; // client's nonce
}

message ServerMoveMessage {
  Move move = 1;
}

message ClientMoveMessage {
  bytes nonce_server = 1;
  bytes nonce_client = 2;
  Move move = 3;
}

message ServerRoundFinalMessage {
  Player winner = 1;
};

Define a commitment scheme C between two parties (the server and the client) to play rock paper scissors, define the protocol below:

  1. The server generates a nonce and sends the ServerRoundInitMessage to the client.
  2. The client
    • determines a move (rock, scissors or stone),
    • generates a nonce,
    • constructs a ClientMoveMessage and computes its MD5 digest,
    • constructs a ClientRoundInitMessage, and
    • sends the ClientRoundInitMessage to the server.
  3. The server determines a move and sends the ServerMoveMessage to the client.
  4. The client sends the ClientMoveMessage to the server.
  5. The server sends the ServerRoundFinalMessage to the client on who's the winner.

A game consists of 128 rounds. The objective is to win more than 95% of the rounds and lose none of them.

Solution

Part I: A short demo of the protocol

📝 Note. For simplicity, I will use JSON instead of protobuf encoding to show how the protocol works. I used the latter for the challenge because it is easier to exploit… We will talk about that later.
Step 1: ServerRoundInitMessage (Server → Client)

Initially, the server generates a nonce 15f0...5273, constructs and sends the below ServerRoundInitMessage to the client.

// ServerRoundInitMessage
{"nonce": "15f0...5273"}
Step 2: ClientRoundInitMessage (Client → Server)

The client wants to play scissors this time. It generates a nonce d7b5...1ff1. The client constructs a ClientMoveMessage message and its MD5 digest is 03e0...a148.

// ClientMoveMessage
{"nonce_server": "15f0...5273", "nonce_client": "d7b5...1ff1", "move": "scissors"}

Afterwards, the client constructs ClientRoundInitMessage with the client's nonce and the hash, and sends it to the server:

// ClientRoundInitMessage
{"hash": "03e0...a148", "nonce": "d7b5...1ff1"}
Step 3: ServerMoveMessage (Server → Client)

At this stage, the server does not know what the client's move is. The server wants to play rock. It constructs the ServerMoveMessage and sends it to the player:

// ServerMoveMessage
{"move": "rock"}
Step 4: ClientMoveMessage (Client → Server)

The client reveals the commitment by sending the ClientMoveMessage to the server (which is already constructed in step 2).

// ClientMoveMessage
{"nonce_server": "15f0...5273", "nonce_client": "d7b5...1ff1", "move": "scissors"}
Step 5: ServerRoundFinalMessage (Server → Client)

The server can verify the validity of ClientMoveMessage, by checking whether

  1. [client nonce] its client nonce equals the nonce in ClientRoundInitMessage,
  2. [server nonce] its server nonce equals the nonce in ServerRoundInitMessage, and
  3. [digest] its MD5 digest equals the hash specified in ClientRoundInitMessage.
🎯 Motivation. The validations above can detect a forged move. The client can understand from ServerMoveMessage that they are losing. However, if they change the move, the ClientMoveMessage and its digest will be updated as well. In that case, (3) is no longer valid.

Since everything checks out, the message is valid. The server finally knows that the client played scissors. Since the server plays rock, it wins the round. The server also sends a ServerRoundFinalMessage announcing their win:

// ServerRoundFinalMessage
{"winner": "server"}
🤏 TL;DR. Can I skip this? If you know Unicoll and know what it could do, yes.

MD5 is insecure. It is very easy to find an MD5 collision in 2022. corkami/collisions is a repository explaining all sorts of hash collisions. There are various attacks, FastColl, UniColl and HashClash, and each of them has a different running speed and capability.

In short, we can forge a move without changing the MD5 digest. An idea would be to inject some "junk" fields that no one cared. They are not used to validate anything while affecting the hashed outcome.

To achieve this, we will use UniColl. Given a payload p of length 64n+4k (for some small k's), it will generate a pair of outputs m1 and m2 such that

  1. The lengths of m1 and m2 are 64n+128 bytes,
  2. MD5(m1)=MD5(m2),
  3. m1 begins with p, and
  4. Only two bytes of m1 and m2 are different, which are the (64n+10)-th byte and the (64n+74)-th byte (one-indexed). Also, m1[64n+10]−m2[64n+10]=1 and m1[64n+74]−m2[64n+74]=−1.

This is an example of a MD5 collision that UniColl finds. Suppose I want to find a pair of colliding messages starting with hello world. (and the other starts with hello wormd!, which is also 12 bytes long). We will use HashClash as a demo.

# Inside hashclash/scripts/demo
echo -n "hello world." > prefix.txt
../poc_no.sh prefix.txt
MD5 differential path toolbox
Copyright (C) 2009 Marc Stevens
http://homepages.cwi.nl/~stevens/
(trimmed)
Found collision!
468115c46b73eac261b6d9f5284b68e3  collision1.bin
468115c46b73eac261b6d9f5284b68e3  collision2.bin
31dda9646624a15c6078582c9af60cf597949a18  collision1.bin
87fbd7c43d1ed226d13a478f0b89f55a744864d9  collision2.bin
4 -rw-rw-r-- 1 mystiz mystiz 128 Dec 18 14:48 collision1.bin
4 -rw-rw-r-- 1 mystiz mystiz 128 Dec 18 14:48 collision2.bin
hd collision1.bin            
00000000  68 65 6c 6c 6f 20 77 6f  72 6c 64 2e 12 d9 a6 74  |hello world....t|
00000010  d7 a8 55 aa c5 66 63 f8  58 98 89 71 e8 b2 49 52  |..U..fc.X..q..IR|
00000020  f9 e0 de f1 a0 62 e4 0d  ba a9 57 b1 96 ed 91 eb  |.....b....W.....|
00000030  1c b7 69 1d 36 89 e7 1a  a2 41 a4 fd 7f bb 98 1f  |..i.6....A......|
00000040  f7 72 24 73 6b 78 6b 20  14 db d0 a2 36 66 2a bc  |.r$skxk ....6f*.|
00000050  36 fe cb 1e 32 1b 35 5b  b7 ce d7 1b 0f e5 3f 98  |6...2.5[......?.|
00000060  1b 8b ed ab 73 14 fb f7  57 d2 2d ed c9 45 8a 1c  |....s...W.-..E..|
00000070  2c f0 ab 4e fb 47 96 81  75 4c 25 34 ad ec 01 db  |,..N.G..uL%4....|
00000080
hd collision2.bin
00000000  68 65 6c 6c 6f 20 77 6f  72 6d 64 2e 12 d9 a6 74  |hello wormd....t|
00000010  d7 a8 55 aa c5 66 63 f8  58 98 89 71 e8 b2 49 52  |..U..fc.X..q..IR|
00000020  f9 e0 de f1 a0 62 e4 0d  ba a9 57 b1 96 ed 91 eb  |.....b....W.....|
00000030  1c b7 69 1d 36 89 e7 1a  a2 41 a4 fd 7f bb 98 1f  |..i.6....A......|
00000040  f7 72 24 73 6b 78 6b 20  14 da d0 a2 36 66 2a bc  |.r$skxk ....6f*.|
00000050  36 fe cb 1e 32 1b 35 5b  b7 ce d7 1b 0f e5 3f 98  |6...2.5[......?.|
00000060  1b 8b ed ab 73 14 fb f7  57 d2 2d ed c9 45 8a 1c  |....s...W.-..E..|
00000070  2c f0 ab 4e fb 47 96 81  75 4c 25 34 ad ec 01 db  |,..N.G..uL%4....|
00000080
md5sum collision*.bin
468115c46b73eac261b6d9f5284b68e3  collision1.bin
468115c46b73eac261b6d9f5284b68e3  collision2.bin
🐌 Is it slow? No. The above script took only three minutes on my laptop.

We can see the properties we mentioned - m1 and m2 have the same digest, and m1and m2 differs by two bytes. Before we solve the challenge with UniColl, let's look at the protobuf message format.

Part III: Walk-through on protobuf message format

🤏 TL;DR. Can I skip again? If you know how protobuf works… probably.

A protobuf message consists of a series of records, which is composed of three items: the field number, the wire type and the content. Let's look at the wire type first.

📖 Documentation? The developer guide pretty much covers everything. Please refer to the guide for more details.
The wire types

We will only mention two wire types: VARINT and LEN because they are the only types that are used in ClientMoveMessage. It is probably not what you expect -- we used bytes and Move (which is an enum) instead of those types!

In reality, VARINT is a wire type that is used when transmitting the data types int32, int64, enum and so on. It uses a k-byte buffer to represent an unsigned integer not longer than 7k-bit long. A one-byte VARINT can express up to 127, and a two-byte VARINT can express up to 16383=214−1. Also, the most significant bit from each byte is considered to be the "continuation bit", which determines whether the subsequent byte belongs to the VARINT.

Below is an example of a VARINT of three bytes long (represented as binary):

[byte 1] [byte 2] [byte 3]
11000000 10100000 00010000
*(   64) *(   32)  (   16)

We can see that the MSB of the first two bytes are 1 and the MSB of the last byte is 0. This happens on every VARINT: All MSBs are 1 except the last byte. After that, we will interpret the integer from its base 128 expressions:

64⋅1280+32⋅1281+16⋅1282=266304.

Now two exercises.

🧮 Exercise 1. Which of the following is/are valid VARINTs?

  1. 11001001 10010001 11111111 10000110 01001101
  2. 10101101 00101110 11010110 10100100 10000100
  3. 11110111 11010001 11000011 10000101 11101100
  4. 01011000 00011110 01011100 00001010 00001111
  5. 10001011 11101011 11110010 01111001 00110010
🧮 Exercise 2. What does the value of the VARINT 10111001 00001010?

For exercise 1, only the first entry is a valid VARINT. The answer for exercise 2 is 1337=0001010 01110012.

Now onto the wire type LEN. One data type it can handle is bytes. Its expression is straightforward -- it begins with a VARINT consisting of the length l, followed by l bytes being the buffer. For example, the below LEN is equivalent to hello:

00000101 01101000 01100101 01101100 01101100 01101111
********                                              -- an unsigned integer 5
         ******** ******** ******** ******** ******** -- 5-byte buffer "hello" 

There are two more types, I64 and I32, which are self-explanatory and I will not cover them. Each of the four wire types is assigned with a type ID:

Type ID 0 1 2 4
Type Name VARINT I64 LEN I32
Formation of messages

Recall that a record is composed of a field number, a wire type and the content. We described the latter two since they are related. What about the field number?

enum Move {
  ROCK = 0;
  PAPER = 1;
  SCISSORS = 2;
}

message ClientMoveMessage {
  bytes nonce_server = 1;
  bytes nonce_client = 2;
  Move move = 3;
}

From the ClientMoveMessage above, the field numbers of nonce_server, nonce_clientand move are 1, 2 and 3 respectively.

Now we understand everything and can finally build a field. A field begins with a VARINT given by 8f+t (f and t are the field number and the wire type), followed by the content. The 7-byte field below represents nonce_server to hello.

0a 05 68 65 6c 6c 6f
**                   -- an unsigned integer 10 = 8*1 + 2
                          field number = 1 (nonce_server)
                          wire type = 2 (LEN)
   **                -- an unsigned integer 5 (the length)
      ** ** ** ** ** -- 5-byte buffer "hello" 

Finally, a message is just a concatenation of records. Let's use ClientMoveMessageas an example:

HKCERT CTF 2022 Postmortem (II): Harder Crypto Challenges

The three red bytes 0A16=8⋅1+2, 1216=8⋅2+2 and 1816=8⋅3+0 respectively represent a LEN with field number 1 (nonce_server), a LEN with field number 2 (nonce_client) and a VARINT with field number 3 (move).

nonce_server: 1a0ab7c21d1c0200ffe48b8e0e37d369 (hex-encoded)
nonce_client: 152b631ddef3cd401dc0b730739460a8 (hex-encoded)
move: SCISSORS

There are quite some details that I did not cover yet. Since this subsection is overly long, I will talk about it when we need them.

Part IV: Generating the first pair of collision

🎉 Congratulations! We finally arrive at the important part. Congratulations on bearing all the background information in the previous sections.

Now the goal is clear: For any nonce_server, find three ClientMoveMessages such that their nonce_clients are all the same, their moves are all different and the MD5 digests are the same. Here is the first property on protobuf:

💡 Property on protobuf (1). If the wire type does not match the data type, then the record will be dropped.

For instance, type wire type I64 does not handle enums. Thus the below record is omitted when parsing ClientMoveMessage.

19 01 0a 05 04 05 06 07 08
**                         -- an unsigned integer 0x19 = 8*3 + 1
                                field number = 3 (move)
                                wire type = 1 (I64)
   ** ** ** ** ** ** ** ** -- the 64-bit integer 0x0807060504050a01

Besides, if we change the first byte to 18, it breaks down into two valid records:

18 01 0a 05 04 05 06 07 08
**                         -- an unsigned integer 0x18 = 8*3 + 0
                                field number = 3 (move)
                                wire type = 0 (VARINT)
   **                      -- an unsigned integer 1
      **                   -- an unsigned integer 0x0a = 8*1 + 2
                                field number = 1 (nonce_server)
                                wire type = 2 (LEN)
         **                -- an unsigned integer 5 (the length)
            ** ** ** ** ** -- 5-byte buffer "04 05 06 07 08"
nonce_server: 0405060708 (hex-encoded)
move: PAPER

Let's make a pair of ClientMoveMessages with the same MD5 digest and the moves being ROCK and PAPER, and will tackle SCISSORS later. Also, since nonce_serverand nonce_client can be handled easily, we will not deal with them now.

This is a 20-byte prefix I made for Unicoll and it is expected that the size of the outputs is 128 bytes long.

1a 07 00 00 00 00 00 00 00 18 01 0a 05 04 05 06 07 08 1a 6c

These are the outputs from Unicoll:

1a 07 00 00 00 00 00 00 00 18 01 0a 05 04 05 06 07 08 1a 6c 11 e1 00 b5 d6 e0 e8 59 d2 1a ba f6
45 c7 64 92 91 43 f4 6c 46 11 cd 78 ac 72 d5 49 7f dd ec 8b 6f 18 a7 9f 5e 55 1a 49 90 0b 29 6e
23 8b 8e 23 06 62 2a 07 ec be 56 14 39 02 91 60 ce c6 60 be 16 35 74 a3 b5 78 11 21 3b e6 2d d8
2a bb cc aa 9b 55 89 bc 43 a8 cc 5c 97 0f 92 25 f6 72 0b 99 0f 47 2e e0 71 15 4e 97 85 e0 23 86
1a 07 00 00 00 00 00 00 00 19 01 0a 05 04 05 06 07 08 1a 6c 11 e1 00 b5 d6 e0 e8 59 d2 1a ba f6
45 c7 64 92 91 43 f4 6c 46 11 cd 78 ac 72 d5 49 7f dd ec 8b 6f 18 a7 9f 5e 55 1a 49 90 0b 29 6e
23 8b 8e 23 06 62 2a 07 ec bd 56 14 39 02 91 60 ce c6 60 be 16 35 74 a3 b5 78 11 21 3b e6 2d d8
2a bb cc aa 9b 55 89 bc 43 a8 cc 5c 97 0f 92 25 f6 72 0b 99 0f 47 2e e0 71 15 4e 97 85 e0 23 86 

Both of the messages have MD5 digest 9c456db27e76b061617914bcb8d20f74. Now let's introduce one more property before we illustrate them.

💡 Property on protobuf (2). There is a default for every data type, which is used when the corresponding field is missing. For instance, it uses the zeroth item from the enum if the field is not specified.
HKCERT CTF 2022 Postmortem (II): Harder Crypto Challenges
The first message is from Unicoll which plays paper.
HKCERT CTF 2022 Postmortem (II): Harder Crypto Challenges
The second message is from Unicoll which plays rock.

The first message composes of four records:

move = bytes.fromhex('00 00 00 00 00 00 00')   # bytes
move = 1                                       # enum, equivalent to "PAPER"
nonce_client = bytes.fromhex('04 05 06 07 08') # bytes
move = bytes.fromhex('11 e1 00 ... e0 23 86')  # bytes

Since it is illegal to assign move as bytes (it is an enum), only the second and the third records will be kept. Therefore, only two fields are assigned.

nonce_server: 0405060708 (hex-encoded)
move: PAPER

Besides, the second message composes of three records but none of them is valid. Since no moves are assigned, it uses the default value, ROCK.

move = bytes.fromhex('00 00 00 00 00 00 00')   # bytes
move = 0x08_07_06_05_04_05_0a_01               # fixed64
move = bytes.fromhex('11 e1 00 ... e0 23 86')  # bytes

Part V: Let's collide with scissors, too

MD5 uses Merkle-Damgard scheme. In short, we can concatenate the same string after the collision pair, and their MD5 digests would still be the same. We can make use of this to generate multiple messages yielding the same digest. Let m1, m2, m1′ and m2′ be 128-byte messages such that:

  1. MD5(m1)=MD5(m2) and
  2. MD5(m1 ∣∣ m1′)=MD5(m1 ∣∣ m2′).

Then the below equality holds:

MD5(m1 ∣∣ m1′)=MD5(m1 ∣∣ m2′)=MD5(m2 ∣∣ m1′)=MD5(m2 ∣∣ m2′).

What can be done? One more property on protobuf!

💡 Property on protobuf (3). If there are multiple valid fields, only the last will be used.

Essentially we can generate m1′ and m2′ which respectively "overwrites the move with scissors" and "does nothing". This is the same as what we did previously: m1overwrote the move with paper and m2 did nothing... and we have a 148-byte prefix for Unicoll!

1a 07 00 00 00 00 00 00 00 18 01 0a 05 04 05 06 07 08 1a 6c 11 e1 00 b5 d6 e0 e8 59 d2 1a ba f6
45 c7 64 92 91 43 f4 6c 46 11 cd 78 ac 72 d5 49 7f dd ec 8b 6f 18 a7 9f 5e 55 1a 49 90 0b 29 6e
23 8b 8e 23 06 62 2a 07 ec be 56 14 39 02 91 60 ce c6 60 be 16 35 74 a3 b5 78 11 21 3b e6 2d d8
2a bb cc aa 9b 55 89 bc 43 a8 cc 5c 97 0f 92 25 f6 72 0b 99 0f 47 2e e0 71 15 4e 97 85 e0 23 86

1a 07 00 00 00 00 00 00 00 18 02 0a 05 04 05 06 07 08 1a 6c

There is the collision pair generated, and both of the messages have the MD5 digest d65c76e692c2b4c81b05a20928375fc3:

1a 07 00 00 00 00 00 00 00 18 01 0a 05 04 05 06 07 08 1a 6c 11 e1 00 b5 d6 e0 e8 59 d2 1a ba f6
45 c7 64 92 91 43 f4 6c 46 11 cd 78 ac 72 d5 49 7f dd ec 8b 6f 18 a7 9f 5e 55 1a 49 90 0b 29 6e
23 8b 8e 23 06 62 2a 07 ec be 56 14 39 02 91 60 ce c6 60 be 16 35 74 a3 b5 78 11 21 3b e6 2d d8
2a bb cc aa 9b 55 89 bc 43 a8 cc 5c 97 0f 92 25 f6 72 0b 99 0f 47 2e e0 71 15 4e 97 85 e0 23 86

1a 07 00 00 00 00 00 00 00 18 02 0a 05 04 05 06 07 08 1a 6c 37 41 76 c6 67 9c 61 22 b6 c0 7c e8
47 80 c9 13 9b 01 66 db 88 46 2e 6a cc fe ea d5 5a 12 ff f1 4d a6 c4 9d ef c4 cb 40 62 b4 90 cf
92 3f 36 ec 96 0f c9 54 43 00 e5 a9 3e 4d df f8 99 92 77 54 1a 27 64 00 79 f2 52 2c 48 39 6f b9
a6 ef 7d c3 07 4d 40 c3 8b d5 73 42 37 3f 6d e1 b5 50 67 87 ed 32 cd 23 75 f5 15 05 51 75 51 70
1a 07 00 00 00 00 00 00 00 18 01 0a 05 04 05 06 07 08 1a 6c 11 e1 00 b5 d6 e0 e8 59 d2 1a ba f6
45 c7 64 92 91 43 f4 6c 46 11 cd 78 ac 72 d5 49 7f dd ec 8b 6f 18 a7 9f 5e 55 1a 49 90 0b 29 6e
23 8b 8e 23 06 62 2a 07 ec be 56 14 39 02 91 60 ce c6 60 be 16 35 74 a3 b5 78 11 21 3b e6 2d d8
2a bb cc aa 9b 55 89 bc 43 a8 cc 5c 97 0f 92 25 f6 72 0b 99 0f 47 2e e0 71 15 4e 97 85 e0 23 86

1a 07 00 00 00 00 00 00 00 19 02 0a 05 04 05 06 07 08 1a 6c 37 41 76 c6 67 9c 61 22 b6 c0 7c e8
47 80 c9 13 9b 01 66 db 88 46 2e 6a cc fe ea d5 5a 12 ff f1 4d a6 c4 9d ef c4 cb 40 62 b4 90 cf
92 3f 36 ec 96 0f c9 54 43 ff e4 a9 3e 4d df f8 99 92 77 54 1a 27 64 00 79 f2 52 2c 48 39 6f b9
a6 ef 7d c3 07 4d 40 c3 8b d5 73 42 37 3f 6d e1 b5 50 67 87 ed 32 cd 23 75 f5 15 05 51 75 51 70

And they are accordingly when decoded,

move = bytes.fromhex('00 00 00 00 00 00 00')   # bytes
move = 1                                       # enum, equivalent to "PAPER"
nonce_client = bytes.fromhex('04 05 06 07 08') # bytes
move = bytes.fromhex('11 e1 00 ... e0 23 86')  # bytes

move = bytes.fromhex('00 00 00 00 00 00 00')   # bytes
move = 2                                       # enum, equivalent to "SCISSORS"
nonce_client = bytes.fromhex('04 05 06 07 08') # bytes
move = bytes.fromhex('37 41 76 ... 75 51 70')  # bytes

# As a result:
move = 2 # SCISSORS
nonce_client = bytes.fromhex('04 05 06 07 08')
move = bytes.fromhex('00 00 00 00 00 00 00')   # bytes
move = 1                                       # enum, equivalent to "PAPER"
nonce_client = bytes.fromhex('04 05 06 07 08') # bytes
move = bytes.fromhex('11 e1 00 ... e0 23 86')  # bytes

move = bytes.fromhex('00 00 00 00 00 00 00')   # bytes
move = 0x08_07_06_05_04_05_0a_02               # fixed64
move = bytes.fromhex('37 41 76 ... 75 51 70')  # bytes

# As a result:
move = 1 # PAPER
nonce_client = bytes.fromhex('04 05 06 07 08')

Part VI: The final messages

From above, we have m1 ∣∣ m1′ and m1 ∣∣ m2′. We can replace m1 with m2 to get four messages with the same digest. We can pick any move by concatenating different message chunks, as shown below:

Startm1: move ← 📄m2: do nothing✂️📄✂️🪨m'1: move ← ✂️m'2: do nothingm'1: move ← ✂️m'2: do nothing

Finally, we can concatenate the nonce_server and nonce_client to whatever we want. These are the final messages, with the assumption that nonce_server = ff...ff and nonce_client = 00...00:

HKCERT CTF 2022 Postmortem (II): Harder Crypto Challenges
m2 ∣∣ m2′ ∣∣ suffix. Plays 🪨.
HKCERT CTF 2022 Postmortem (II): Harder Crypto Challenges
m1 ∣∣ m2′ ∣∣ suffix. Plays 📄.
HKCERT CTF 2022 Postmortem (II): Harder Crypto Challenges
m1 ∣∣ m1′ ∣∣ suffix. Plays ✂️.

We can now use m1, m2, m1′ and m2′ to win every single game and get the flag: hkcert22{bu7_wh0_u53s_md5_1n_2022}. Congratulations to @hoifanrd (short for Hoi Fan Road, or 海帆道) being the only solver of the challenge.

版权声明:admin 发表于 2022年12月27日 下午8:40。
转载请注明:HKCERT CTF 2022 Postmortem (II): Harder Crypto Challenges | CTF导航

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