CTF Writeup - HackIM 2016 - ZorroPub (RE 100)

• Name - Zorro Pub
• Category - Reverse Engineering
• Points - 100
• Description - N/A
• Binary - Download here

Running the 64-bit ELF:

root@kali: ~/Desktop

root@kali:~/Desktop# ./zorro_bin Welcome to Pub Zorro!! Straight to the point. How many drinks you want?5 OK. I need details of all the drinks. Give me 5 drink ids:100 200 300 400 500 Looks like its a dangerous combination of drinks right there. Get Out, you will get yourself killed root@kali:~/Desktop#

The binary first accepts an integer, the number of drinks, and asks for a number of drink IDs equal to the first number. In the above example I've put '5' as the number of drinks and hence it expects 5 drink IDs. Let's load it in IDA.

Starting from the bottom, it's clear where we need to get to. Also, it shows that the flag is computed during the program's execution; "strings" will not do the job.

Let's look at the top part of the binary now:

The program scans for an integer and moves onto the next section if it is greater than 0, if not it prints "You are too drunk!! Get Out!!" and exits. The next part is the main logic of the program. I've colour-coded the boxes to make it easier (hopefully) to decipher what's happening.

The colour-scheme legend:

• Blue - Start of logic; Scans for input; call this Algorithm A
• Pink - Call this Algorithm B
• White - Final decision box
• Red - Error boxes; program exits straight after
• Green - Destination

Right after we give the program our first input (# of drinks) we arrive at the blue box at the top and Algorithm A starts. This simply scans for a digit, makes sure it's between 0x10 and 0xFFFF and XORes it with the previous input which, in the first round is 0x00. The process is repeated a number of times equal to the first parameter. The result of Alogrithm A is the XOR of all our drink IDs. When done, the program jumps to the pink boxes, Algorithm B.

Algorithm B grabs the resultant, say X, from Algorithm A and does the following:

• X AND (X-1) = new X
• Increment Counter
• If X is not 0, repeat
• If X is 0, stop

When Alogirthm B finishes, we end up in the white box in the middle. If the Counter in Algorithm B is not equal to 0xA, it displays "Looks like its a dangerous combination", else it moves onto the green box, our goal for the time being. The hurdle here is to force Algorithm B to repeat itself exactly 0xA times. This happens only when the algorithm is fed a number who's binary representation contains exactly 10 1's.

In the green box, the result from Algorithm A is used as a seed (srand()). The logic that comes after this is not of interest to us as it does not depend on our input. It's there to convolute the flag. So we know the following about the value(s) that can make it successfully to the green box function:

• It doesn't matter if we input X as the drink ID or X1 and X2, where X1 ^ X2 = X
• The value(s) should be between 0x10 and 0xFFFF
• The value(s) must contain 0xA number of 1's when represented in binary format

I've chosen to input a single number as it's easier. Since the number can be between 16 and 65,535 I've used a bash 1-liner to brute-force the answer. As the range is very small I didn't bother limiting the input to only those numbers which contain 10 1's. I personally felt that this section of the RE challenge was not well made. Most of the interesting complications of the binary's workings were invalidated by the small search-space and not even required to obtain the flag.

root@kali: ~/Desktop

root@kali:~/Desktop# for i in `seq 1 65535`; do echo $i >> answers.txt; ./zorro_bin <<< $'1\n'$i$'\n' | grep -i 'choose right mix' >> answers.txt; done

Few minutes later we end up with the flag in answers.txt : You choose right mix and here is your reward: The flag is nullcon{nu11c0n_s4yz_x0r1n6_1s_4m4z1ng}

Arduino - Welcome to the World of Microcontrollers

I've recently developed a keen interest in hardware and microcontrollers and decided to document my journey. My learning method usually involves head-diving into a big project but this time I've taken a pragmatic approach and decided to acquire a solid foundation before moving onto bigger projects.

The Arduino Environment

The Arduino environment is made up of the hardware itself, the Arduino board and shields, and the software, the Arduino IDE. Arduino Shields are add-on boards which can be stacked on top of the base Arduino board to extend its capabilites. A large range of shields exists, from Ethernet shields, to LCD shields, to RFID shields. We'll take a look at some of the shields in future posts. I'll be using the Arduino Uno in my examples but other Arduino boards can be used.

Arduino Board

The Arduino UNO board contains the following components:

• ATmega328 Microcontroller - Main microcontroller; the only one that's programmable by the user.
• ATmega16U2 Microcontoller - Handles the communication with the USB; non-programmable.
• USB Connector - Used to transfer data and provides power to the board.
• Reset Button - Resets the device.
• Power Connector - Necessary when dealing with hardware which requires more power such as motors.
• Pins - The connection to the outside world.
  • Digital I/O - Can be 1 (5V) or 0 (0V).
  • Analog Input - Accept analog input ranging from 0V to 5V. Note that they do not provide analog output.
  • Power/Reset - Provide power to a circuit.
  • ICSP (In-Cirduit Serial Programming) - Used to update the bootloader/firmware of the microcontrollers.

Arduino IDE

The Arduino IDE has everything required to get you started with coding for the Arduino. It can be downloaded from arduino.org.

It provides the following main features:

• A coding environment to write sketches (Arduino Programs) in.
• Debugs, cross-compiles and uploads sketches to the ATmega328 microcontroller.
• A Serial Monitor used to debug sketches. We'll talk about this in the next part.
• Several example sketches to get you started.

Setting up the Arduino Environment

Follow the steps below to set up a working Arduino environment:

1. Download and install the Arduino IDE.
2. Connect the Arduino board to the computer.
3. Start the IDE.
4. Select your Arduino board from Tools -> Board.
5. Select the COM port your Arduino is attached to from Tools -> Port.
6. To verify everything is working, load the Blink sketch from File -> Examples -> Basics -> Blink, and hit Upload.
7. After a few seconds, the LED next to pin13 should be blinking.

For those who do not have access to the hardware, an online Arduino simulator can be found here.

Anatomy of a Sketch

Every Arduino sketch must contain these 2 functions:

• Setup Function
  • Executes once at start, when the Arduino board is powered up
  • Used for initialisation
  • Does not take any arguments and returns void
• Loop Function
  • Iterative for as long as the Arduino is powered on
  • Executes after the Setup function
  • Contains the main logic of the program
  • Does not take any arguments and returns void

As can be seen from the Arduino IDE screenshot, these 2 functions are already in place for us to use when a new sketch is created.

The Leap to Microcontrollers

A microcontroller is a System on a Chip (SoC) on a single integrated circuit containing a processor, memory (RAM & EEPROM) and I/O pins. Essentially it's the heart and brains of the circuit; it accepts input from sensors, processes it and serves output to electronic components. Microcontrollers simplify circuits. Without them it would be very hard to create logic and changes in logic could mean drastic changes to the circuit itself. Let's start by migrating a traditional simple circuit to an Arduino simple circuit.

A simple circuit looks like this:

The closed circuit contains a power source, in our case 5V supplied by the USB, an LED and a resistor. The resistor prevents the LED from burning as the power can be overwhelming. You can solder these components together or use a solderless breadboard. The latter is a board for prototyping of electronics as it allows components to be easily connected in a non-permanent fashon. The one I have looks like this:

The holes of a breadboard are internally connected to each other in rows of 5 holes and columns along the sides. So, for example, if an end of a resistor is in hole 3F and an end of an LED is in hole 3J, they're connected. If one end is in 15A and the other is in 30A, they're disconnected. With this at hand, let's create the simple circuit:

• Connect a wire from the 5V pin to 1J
• Insert a resistor in 1I and 7I
• Insert an LED in 7H and 12H
• Close the circuit by connecting 12I with ground

The circuit should look similar to this:

As expected, the LED lights as soon as the last wire is connected. If it doesn't, make sure the LED is placed in the right direction. LEDs allow current to flow only in one direction. In this example we do not need to program anything as we don't make use of the Arduino's microcontroller. Current simply flows from the 5V pin, to the resistor, to the LED, and back. Now let's do it the Arduino way.

The wiring is very similar. Instead of connecting the circuit to the 5V pin, connect it to one of the Digital I/O pins, say pin9. This is what gives us programmatic power as we can control the voltage on any of the Digital I/O pins. The electronic schematic of the circuit now looks like this:

The LED does not light straight away like it did in the previous example. We require code which tells the microcontroller to send power to pin9. Copy the code below to the Arduino IDE:

void setup() {
    pinMode(9, OUTPUT);
}

void loop() {
    digitalWrite(9, HIGH);
}

The code initialises pin9 as an OUTPUT pin and sets it to HIGH, allowing current to flow from this pin. Let's take this opportunity to introduce some basic, indispensable functions we'll be using in nearly every sketch.

• pinMode (pin, mode) - Initialise pin to a specific mode
  • pin - The pin to be initialised
    • Digital Pins - 0 - 13
    • Analog Pins - A0 - A5
  • mode - The mode of operation
    • INPUT - Pin acts as a receiver
    • OUTPUT - Pin acts as a transmitter
    • INPUT_PULLUP - Pin acts as receiver with reverse polarity; HIGH becomes LOW and viceversa
  • Example - pinMode(9, OUTPUT)
• digitalWrite (pin, value) - Assigns a voltage to a pin
  • pin - The pin in question
  • value - Voltage to be sent: HIGH(5V) or LOW(0V)
  • Example - digitalWrite(9, LOW)
• digitalRead (pin) - Returns state of an INPUT pin.
  • pin - The pin in question
  • Return Values - HIGH(5V) or LOW(0V)
  • Example - int pinValue = digitalRead(9)

Hit the Upload button to verify the program and upload it to the microcontroller. If the LED lights up after a few seconds congrats, you've successfully built your first Arduino program. As a side node, pin 13 has a built-in LED attached to it which could be used instead without attaching any extra components.

This example might give the impression that the Arduino is just overhead as both circuits produce the same result but the 2nd example requires code. The leap to microcontrollers starts now! The LED in the 2nd example does nothing more than the LED in the 1st because we didn't program anything interesting for it to do. What if we want the LED to blink? I'll be honest, I'm not sure how this could be achieved using only electronics but, it is very easy to do with an Arduino. In fact, the circuit doesn't even need any changes; upload the following code:

void setup() {
    pinMode(9, OUTPUT);
}

void loop() {
    digitalWrite(9, HIGH);
    delay(1000);          
    digitalWrite(9, LOW); 
    delay(1000);          
}

This sketch sets pin9 to HIGH, waits for a second (delay(1000)), sets pin9 to LOW, waits for another second and repeats the process indefinitely. The result should like something like this:



Awesome isn't it ?? We now have a blinking light with no more complications than a simple circuit's.

Dealing with Digital Input

We'll now take a look at one of the most simple input devices a circuit can contain, the push-button. A push-button can be 1 of 2 states, ON, when pressed, and OFF, when released. Let's extend our previous circuit to include this and light the LED only when the button is being pressed.

The schematic of the input part looks like this:

Wire the circuit in the following manner:

• Connect a wire from the 5V pin to 20J
• Insert a push-button in 20F and 22F
• Insert a resistor in 22H and 28H
• Close the circuit by connecting 28I with ground
• Connect the sensor wire to 22G and pin10

Once again we need a resistor to avoid frying the circuit. When the button is pressed, current flows from the 5V pin to the ground, since it's a closed circuit, but also to pin 10, which is used to detect when the push-button is pressed.

At this point the LED won't light up when the push-button is pressed. You've guessed it, we need code.

void setup() {
    pinMode(10, INPUT);
    pinMode(9, OUTPUT);
}

void loop() {
    if (digitalRead(10) == HIGH)
        digitalWrite(9, HIGH);
    else
        digitalWrite(9, LOW);
}

The setup function establishes that pin9 and pin10 are used as OUTPUT and INPUT pins, respectively. When pin10 is HIGH, i.e. the push-button is being pressed, pin9 is set to HIGH, i.e. the LED lights. When the switch is off, no current flows to pin10 and hence the LED is set to LOW, i.e. it's switched off. The sketch could be condensed to the following:

void setup() {
    pinMode(9, OUTPUT);
    pinMode(10, INPUT);
}

void loop() {
    digitalWrite(9,digitalRead(10));
}

Andddd a video of it working cause why not:

CTF Writeup - TUM CTF Teaser - whitebox crypto (Rev 20)

• Name - whitebox crypto
• Category - Reverse Engineering
• Points - 20
• Description - Do not panic, it's only XTEA! I wonder what the key was...
• Binary - Download here

According to the description we're required to find the key used by the 64-bit ELF binary to encrypt the input. We're also given the algorithm used: XTEA, eXtended Tiny Encryption Algorithm. The following is a C implementation of the encrypting part of the XTEA algorithm:

    void encipher (unsigned int num_rounds, uint32_t v[2], uint32_t const key[4]) {
        unsigned int i;
        uint32_t v0=v[0], v1=v[1], sum=0, delta=0x9E3779B9;
        for (i=0; i < num_rounds; i++) {
            v0 += (((v1 << 4) ^ (v1 >> 5)) + v1) ^ (sum + key[sum & 3]);
            sum += delta;
            v1 += (((v0 << 4) ^ (v0 >> 5)) + v0) ^ (sum + key[(sum>>11) & 3]);
        }
        v[0]=v0; v[1]=v1;
    }

One cycle of the XTEA algorithm consists of 2 Feistel rounds:

A few things to notice are 1) the key is used as is, i.e. it's not used to derive a longer key and 2) the key is split into 4 equal parts. Let's look at the enciphering function in the binary:

The function has been redacted as it contains 32 cycles. From the above C implementation and the Feistel-rounds image, we know that some constant delta is being added to the key at each cycle. Like the key, the delta is also constant. The fact that there are no instructions in the binary that add 2 constants together means that, the addition of delta and the key has been pre-computed and hard coded in the program. This leaves us with the following 4 potentially-interesting instructions:

xor eax, 7B707868h
xor r10d, 1B58EA2Eh
xor r9d, 0BA9AE30h
xor r12d, 9BD661DBh

The 1st time a part of the key is used, it is used as is since the sum is 0 at this point. The 2nd and 3rd time a part of the key is used, it is incremented by delta, the 4th and 5th by (delta * 2), and so on. With this in hand let's compute the original key.

      1) 0x7B707868 => {pxh

      2) 0x1B58EA2E - Delta => 0x1B58EA2E - 0x9E3779B9 => 0x7D217075 => }!pu

      3) 0x0BA9AE30 - Delta => 0x0BA9AE30 - 0x9E3779B9 => 0x6D723477 => mr4w

      4) 0x9BD661DB - (2 * Delta) => 0x0BA9AE30 - 0x3C6EF372 => 0x5F676E69 => _gni

Combining we get: hxp{w4rming_up!}

CTF Writeup - TUM CTF Teaser - selftest (Rev 10)

• Name - selftest
• Category - Reverse Engineering
• Points - 10
• Description - Baby's 1st
• Binary - Download here

This 64-bit ELF was the easiest of the RE lot. The binary is given to us but to get the flag we need to netcat, suggesting that the key is not embedded in the binary itself. Let's connect to it and try our luck:

Command Prompt

C:\>ncat 1.ctf.link 1060 some_random_stuff :( C:\>

No surprises there. Let's look at it under IDA.

It's easy to see where we need to end up to manage to get the flag. Also, this confirms that the key is not in the binary. The following is the beginning of the program:

The first block tells us that our input is interpreted as hex and is ORed with RSI, which is 0x8000000000000000. The second block takes this number and creates a character-count map of it. For example let's say we input c0ffee, this is then ORed with 0x8000000000000000, which results in 0x8000000000c0ffee, giving us the following character-count map:

With this pinned out, we take a look at the validation routine.

The validation loop operates on the character-count map in reverse order and does the following:

1. If the byte read is 0x00, jump to the next one.
2. If the byte read is not 0x00 and is equal to the character it represents, jump to the next one.
3. If the byte read is not 0x00 and is NOT equal to the character it represents, FAIL.

Simply put, an input string is valid if the occurrences of its bytes are equal to the bytes themselves. Keep in mind that OR 0x8000000000000000 might mess up the string. This means that the following are all valid strings:

• 8888888 (There's 7 of them because RSI starts with 0x8000000000000000)
• 18888888
• 13338888888

Trying the first one out:

Command Prompt

C:\>ncat 1.ctf.link 1060 8888888 hxp{g00d_m0rning_r3v3r53r5} :) C:\>

(N)ASM101 - 0x05 - Reversing Binaries - Part II (you_are_very_good_at_this.exe)

The binary we'll be analysing in this tutorial is none other than Challenge 9 from the Flare-On challenge by Fireeye. For those of you who've never heard of it, it's a CTF-style Reverse Engineering challenge. This year, Fireeye has created 11 challenges each of which contain an e-mail address used to obtain the next binary.

Compared to the previous binary we've tackled, this will be much more involving but hang in there, the greater the obstacle the more glory in overcoming it. The operations used by the program to hide the answer are not particularly complicated but there are various anti-disassembly techniques we need to overcome. Fortunately for us there's only 2 very important ASM instructions we need to cover before we start.

Download binary here.

x86 Instructions

CMPXCHG

The Compare and Exchange operator compares the accumulator (AL/AX/EAX) to the first operand. If equal, the first operand (Destination) is loaded with the second (Source). If not, the accumulator is loaded with first operand (Destination). Programmatically it looks something like this:

    if (Accumulator == Destination) {
        ZF = 1;
        Destination = Source;
    }
    else {
        ZF = 0;
        Accumulator = Destination;
    }

Let's look at some concrete examples:

    cmpxchg ecx, edx        ; if EAX = 0x05, ECX = 0x05 & EDX = 0x04 then, ECX = 0x04
                            ; if EAX = 0x03, ECX = 0x05 & EDX = 0x04 then, EAX = 0x05
    cmpxchg cx, dx          ; same as above but for Words
    cmpxchg [edi], si       ; same as above but the Destination is a memory region

CMOVZ & CMOVNZ

CMOVZ stands for Conditional Move if Zero whereas CMOVNZ stands for Conditional Move if Not Zero. These instructions can move a value from memory to a register or from one register to another. Unlike the original MOV instruction, the source operand cannot be a memory region and immediate values are not allowed anywhere. Their description is self-explainatory but let's look at a few examples:

    cmovnz eax, ebx        ; if ZF = 0, EAX =  EBX, else nothing 
    cmovnz eax, [ebx]      ; if ZF = 0, EAX = [EBX], else nothing
    cmovz eax, ebx         ; if ZF = 1, EAX =  EBX, else nothing
    cmovz eax, [ebx]       ; if ZF = 1, EAX = [EBX], else nothing

Basically a move occurs if the condition is satisfied, else the move is not performed. Many other conditional move statements exist but since we'll only encounter 1 of them in the binary, and also because in my experience they're not that common, I won't be covering them, at least this time.

The Binary

In this tutorial I'll be using IDA once again. Originally I've solved this challenge using Immunity Debugger since IDA kept moaning and groaning about instructions changing during execution. (You'll see what I meant by that statement). Before that though, let's run the program:

Command Prompt

C:\>you_are_very_good_at_this.exe I have evolved since the first challenge. You have not. Bring it. Enter the password> lets_try_our_luck You are failure C:\>

Wow that's harsh!! The binary accepts a password, validates it in some way, and outputs a failure or success (maybe?) message. Guessing and brute-forcing is not going to get us anywhere so let's load the program in IDA and start debugging.

Your eyes can deceive you, don't trust them

In the functions window, IDA identifies a very complicated but interesting-looking function, sub_401495:

At first glance it seems that we've identified the core function of the program. Guess what? Forget about the function. The code is never reached! This wasted me quite some time during the challenge. When dealing with self-modifying code, static analysis is no good. Having said that, if there's anyone who's got an idea on how to unroll self-modifying code for static analysis, please step forward. Thanks in advance.

While single-stepping through the program there will be times where the Instruction Pointer points to a section which has been interpreted as data such as the following:

If this happens, press "c" to directly convert that part to code. IDA might prompt you for confirmation; press "YES". We can now see that the section makes much more sense:

As if it weren't bad enough, the binary is full of self-modifying code and "conditional" jumps all over the place. Consider the following:

The last line diverts the flow to another location if the Zero Flag is set, but XOR EAX, EAX is always zero, hence ZF is always set. Also what's this loc_401A54+2? Why not loc_401A56? This happens when the code pointed to by this instruction has been messed up and/or wasn't interpreted correctly. Jumping to it and pressing "c" should clear things up.

Finding our way round the maze

A good practice when debugging is to set some well-placed breakpoints. For this binary it's easier said than done since we're not guaranteed that the instructions remain consistent throughout execution. We have no choice but to start from the top. We run the program with the "Suspend on process entry point" option set, which is found under "Debugger Setup". This will force the program to halt on the very first instruction. The start of the program looks like this :

While holding F7 (single-stepping) down, we keep an eye on the stack. The program goes through several transformations before it even reaches the point where it asks us to input the password. After some time we get the following stack setup:

At the top we notice a kernel32_WriteFile system call and a pointer to the data section, .data:aIHaveEvolvedSi, which holds the string "I have evolved since the first challenge. You have not. Bring it.". This is followed by a kernel32_ReadFile call and 41 pointers to .text:00401736. The bottom 31 pointers are not shown in the image. This should look familiar as it's the real starting point. It writes the welcoming message to the console, reads our input and .. hmm .. executes some function 41 times. At this point we're not sure what this function does so let's investigate further by setting a breakpoint at address 0x00401736.

We continue the execution (F9), input a random password in the console and hit enter. The program stops at the breakpoint. This happens everytime we try and continue for 41 times after which the program displays the friendly "You are failure" message. Let's look closely at what happens at each iteration of this mysterious function.

All that glitters is not gold, it can be diamonds

It's time to start filtering the nonsense and identify the operations that matter; those that actually operate on our input, or are influenced by it. We run the program with different passwords to hopefully elicit some change in the program's execution flow. The following is a list of instructions that are affected by our input in each iteration of the function:

1. [0x00401A9C] mov al, [eax+ecx] - Loads a character from our input into AL.
2. [0x0014FDB4] mov ah, [esp+ebx+0B4h] - Loads some byte into AH.
3. [0x0014FDB8] xor al, ah - XORs our character with this byte.
4. [0x0014FDB4] mov cl, [esp+ebx+88h] - Loads some byte into CL.
5. [0x00401B14] rol al, cl - Performs a ROL on AL, CL times.
6. [0x0014FDB8] cmpxchg bl, dl - Compares and Exchanges BL with DL, if AL = BL, then ZF = 1, else 0.
7. [0x00401B3D] cmovnz esi, edx - If ZF = 1, ESI remains 0, else ESI = EDX = 1.
8. [0x00401B49] xchg eax, esi - If ESI was 1, now EAX is 1, and same for 0.
9. [0x0014FD88] cmpxchg bl, dl - At this point BL = 0 and DL = 1 so if AL = 0, BL becomes 1, else BL remains 0.
10. [0x00401B74] mov eax, ebx - EAX now holds the result, either a 1 or a 0.

At this point I suggest taking some time to go through a single iteration several times, focusing on the instructions I've listed above. The addresses might not match exactly but they should be close. At the end of the each loop, EAX is either incremented or isn't, so which one is the right path? The answer lies at the end of the program:

If EAX is 0x29, i.e. 41, i.e. the number of iterations our favourite function executes, then jump to the "You are success" section, else jump to the "You are failure" section. Excellent !! We now know that for each iteration of the function, EAX has to be incremented. Let's look again at the 10 instructions listed above and, starting from the bottom, make our way up starting from the fact that EAX has to be 1:

1. [0x00401B74] mov eax, ebx - EAX has to be 1, so EBX has to be 1.
2. [0x0014FD88] cmpxchg bl, dl - EBX has to be 1, so AL has to be 0.
3. [0x00401B49] xchg eax, esi - AL has to be 0, so ESI has to be 0.
4. [0x00401B3D] cmovnz esi, edx - ESI has to be 0, so ZF has to be 1.
5. [0x0014FDB8] cmpxchg bl, dl - ZF has to be 1, so AL has to be equal to BL. (Let BL be X)
6. [0x00401B14] rol al, cl - This operation must be equal to BL, i.e. ROL AL, CL = X
7. [0x0014FDB4] mov cl, [esp+ebx+88h] - Let CL be Y, i.e. ROL AL, Y = X
8. [0x0014FDB8] xor al, ah - ROL (XOR AL, AH), Y = X
9. [0x0014FDB4] mov ah, [esp+ebx+0B4h] - Let AH be Z, i.e. ROL (XOR AL, Z), Y = X
10. [0x00401A9C] mov al, [eax+ecx] - Let AL be I for Input, i.e. ROL (XOR I, Z), Y = X

Once again, take your time to digest it slowly and refer to the previous 10 bullet points if needs be.

Diamonds are forever

So finally we're left with a simple formula that contains 4 unknowns, 3 of which can be gathered by going through each function, and our input. Let's change the subject of the constructed formula to I since that's what we need to find:

                                 (I ^ Z) ROL Y = X 

            =>                           I ^ Z = X ROR Y

            =>                               I = ( X ROR Y ) ^ Z

Now it's a matter of going through each iteration of the function to gather X, Y and Z. Let's go through the first few together. Remember that for words, rotate operators repeat themselves every 16 (0x10) times. For example ROR AX, 56 = ROR AX, 6 and ROR AX, F3 = ROR AX, 3. So:

             +-----------------------+-----------------------------------+-------+
             |   X   |   Y   |   Z   |   X ROR Y   |   ( X ROR Y ) ^ Z   |   I   |
             +-----------------------+-----------------------------------+-------+
             |   C3  |  56   |   46  |     0F      |          49         |   I   |
             |-------------------------------------------------------------------|
             |   CC  |  F5   |   15  |     66      |          73         |   s   |
             |-------------------------------------------------------------------|
             |   BA  |  AC   |   F4  |     AB      |          5F         |   _   |
             |-------------------------------------------------------------------|
             |   4E  |  1B   |   BD  |     C9      |          74         |   t   |
             +-------------------------------------------------------------------+
             |  ...  |  ...  |  ...  |     ...     |          ...        |  ...  |
             +-------------------------------------------------------------------+

After what feels like a few years we obtain the e-mail address to advance to the next challenge:

Command Prompt

C:\>you_are_very_good_at_this.exe I have evolved since the first challenge. You have not. Bring it. Enter the password> Is_th1s_3v3n_mai_finul_foarm@flare-on.com You are success C:\>

Conclusion

I admit that this tutorial is quite a leap from the previous ones. Don't get disheartened though!!

For those of you who would like to take a go at the other challenges, Fireeye has made the binaries available for download here. Personally I quite enjoyed them and even though they can be quite frustrating at times, I've definitely learnt a lot .. それでは、また