CyberSci Hardware Challenge

A copy of this writeup is written in the UNBCTF's writeups repository.

Difficulty: Medium/Hard

The objective of the challenge is to dump the firmware from the provided hardware badge and reverse engineer it. In theory this is quite easy, however it took quite some time to get familiar with the AVR architecture and instruction set.

Hints: If you're just starting out, ask Jeff Bezos for help. If you're near the end, ask Jeff Bezos for help.

Attachments: flash.hex, sram.hex, eeprom.hex

(these attachments were not provided, and had to be recovered as part of the challenge)

Solution: asd.py

The Badge

(ignore the soldering, that will come later)

Upon inital inspection we can see the board has been nicely labelled for us. There is also a small push button, a power switch, some LED's/resistors, and an IC chip present.

When powered on, the badge will cycle through different light patterns each time the small push button is pressed.

The lights didn't seem to provide a pattern - that would be too easy. So we know the challenge must be hidden in the firmware.

The chip has some small writing on the casing, and we can make out "TINY1614N". Googling around for it quickly reveals the documentation (hosted on AWS, hence the hint), which explains that we can interact with it via the UPDI (Unified Program and Debug Interface).

Dumping the firmware

What's UPDI?

Taken straight from the documentation: UPDI is a Microchip TM (formerly Atmel) proprietary interface for external programming and on-chip debugging of a device.

It operates over a one-wire interface, so other than hooking up GND and VCC (the power supply), there is minimal setup required.

Interfacing with the badge

Luckily, the challenge creator (Loudmouth Security) graciously connected the UPDI pinout into the single wire interface for us, so other than connecting GND and VCC we don't have to do much more for setup.

After conducting some research, tools such as avr-dude and updiprog quickly became of interest as they seemed to be capable of dumping the firmware.

First Attempt

Originally, it was thought that an Arduino Uno (or alike) could be converted into a USB to TTL (aka serial) adapter to enable communication with the chip.

After reading several guides of people looking to program various chips via UPDI, the following schematic seemed to be accurate.

                        VCC       (3.3v)        VCC
                         +-----------------------+
                         |                       |
 +---------------------+ |                       | +--------------------+
 | Serial Adapter      +-+      Resistor         +-+  AVR device        |
 |                     |      +----------+         |                    |
 |                  TX +------+ 1k OHM   +---------+ UPDI               |
 |                     |      +----------+    |    |                    |
 |                     |                      |    |                    |
 |                  RX +----------------------+    |                    |
 |                     |                           |                    |
 |                     +--+                     +--+                    |
 +---------------------+  |                     |  +--------------------+
                          +---------------------+
                         GND                   GND

However, communication using an Arduino was unsuccessful for various reasons.

Rather than wasting time trying to mimic the required chipset, this adapter was purchased off Amazon for around $20 CAD. Yet another reference to asking Jeff Bezos for help.

Any USB to TTL UART adapter may be equally likely to work, we simply used the adapter linked above. A search for "USB to TTL UART" on Amazon returns plenty of valid results.

Second Attempt: Success!

After an agonizing couple days wait (we were excited!), the adapter arrived and necessary drivers were installed. There were issues with avr-dude and updiprog, so a switch to pymcuprog was made.

We setup the same circuit as above on a breadboard, and are now ready to try dumping the firmware.

To test the connection, UPDI offers "ping" like functionality.

Note: pymcuprog is the recommend tool to use as many others are outdated and slow.

pymcuprog ping -d attiny1614 -t uart -u COM6
Connecting to SerialUPDI
Pinging device...
Ping response: 1E9422
Done.

Hurray! We can now talk to the device.

We want to dump both the firmware (the code), the EEPROM, and the SRAM (memory), as there may be valuable information in each.

pymcuprog read -m flash -d attiny1614 -t uart -u COM6 -f flash.hex

pymcuprog read -m eeprom -d attiny1614 -t uart -u COM8 -f eeprom.hex

pymcuprog read -m internal_sram -d attiny1614 -t uart -u COM6 -f sram.hex

Both of us were on opposite sides of Canada as we solved this. One of us (Stephen) had access to an Arduino kit with plenty of resistors. Matt, on the other hand, resorted to.. other measures.

The PCB from an old mini fridge had to be sacrificed to use a specific resistor, and made quite the interesting setup. But it worked!

Reversing the firmware

Before we can try reversing, we need to convert the files from intel hex into a binary format.

This can be done with the following tool:

avr-objcopy -I ihex -O binary flash.hex flash.bin

As mandated by CTF lore, strings was ran first and the following was spit out:

$ strings flash.bin
...
0123456789abcdef
JIGGYCIPHER{
JIGGYCIPHER{a1557d3801d7723f3d385111d85cd125625cc260ba76d3051a5de6ff24786b8fc70c1659aa4c41b9151c8eef70d438515a6642562ab9f0403f276ba556a7429829433d2572484f2585719d468964832ab1a258521f568165c3ab1bead146ecd2de65a5d812b53ca5c168473bac895d4a8bc37390f860b11e6a6eb442c249180a9c337285455ab098d47b7264ec42bed7688d59dcdfddcf3d0de674356cbee22547e087c63be52013dc319600c4a4859788ab87349c6c81ea9598127f87093c148c1d3e94edea50132f324eaf5cc88b63b8fac20cf6}

Hurrah! We got our flag! Should be some simple cipher now and we'll be done.. right? Unfortunately, no. Little did we know, we had only solved the easy part.

SRAM and EEPROM were also quickly looked at in a hex editor. SRAM contained content but it was mostly binary, no text. EEPROM was quickly ruled out of having anything important as it was mostly 1s.

$ xxd sram.bin
00000000: b98f ca8f d78f 0100 0054 6f57 de1f 93ff  .........ToW....
00000010: e0fe 2ede feb2 3f7b ad7f 73d3 df15 3eff  ......?{..s...>.
00000020: 7e2f b78b cf8a 4cde 57ae fbaa e267 bffd  ~/....L.W....g..
...

$ xxd eeprom.bin
00000000: 0081 ffff ffff ffff ffff ffff ffff ffff  ................
00000010: ffff ffff ffff ffff ffff ffff ffff ffff  ................
00000020: ffff ffff ffff ffff ffff ffff ffff ffff  ................
00000030: ffff ffff ffff ffff ffff ffff ffff ffff  ................
...

This has evolved into a Reversing/Crypto Challenge...

Any attempt to try and decode this using known ciphers fails, and performing frequency analysis yields no promising results.

Maybe there is some sort of encryption routine in the firmware!

Ghidra to the rescue?

Ghidra does support the AVR architecture, however there is no direct support for the ATTINY1614 CPU.

After reading the documentation, we found that the language was AVR8:Little Endian:16 Bit and it was compiled with GCC (uses avr-libc).

The SRAM and EEPROM dumps were added into the analysis of the flash in Ghidra via File > Import File. Using Options..., we could specify the address of the imported file, and input the real addresses of the dumps: * EEPROM: 0x1400 - 0x14FF * SRAM: 0x3800 - 0x3FFF * Flash: 0x8000 - ...

After importing the SRAM and EEPROM dumps and adjusting the memory mapping, the disassembly was still very hard to read. Ghidra's decompiler couldn't handle AVR, and after some time, we noticed some of the instructions were simply wrong (see below: Radare2 on top vs. Ghidra on bottom). Purely static analysis was also difficult, as we had no way to simulate the code. Ghidra has no emulator for AVR.

Static Analysis? Nah. Let's Emulate it!

There are few ways emulation can be done, however, Cutter was used as it was super user-friendly and only required a simple installation.

simavr was another tool we tried, but it requires you to build/define custom core for emulation. We tried, at first, but Cutter seemed much simpler so we pivoted.

The First Clue: References to JIGGYCIPHER

Despite its flaws, Ghidra did have some nice features that Cutter lacked, and it provided us with our first clue.

When looking at references to the SRAM in Ghidra, we can see a lot of references to the first eight bytes:

It took us a while to realize, but the contents of the SRAM at x3800, x3802, and x3804 were addresses to the strings we saw before in little-endian format:

• x3800: 8fb9
• x3802: 8fca
• x3804: 8fd7

Cutter mounts the flash at x0000 instead of x8000, so we need to subtract x8000 from each of these addresses in order to view the correct location. But when we do, we can see:

• 0x0fb9: 0123456789abcdef
• 0x0fca: JIGGYCIPHER{
• 0x0fd7: JIGGYCIPHER{...}

Why did we have to look up the strings in Cutter? Ghidra's addresses seemed mismatched from Cutter's, but Cutter's made more sense given the context.

Changing the data types to pointers allows us to see references to each individual string. By investigating each reference to the strings, we see a reference to JIGGYCIPHER{ (0x3802) from FUN_code_81e9.

Then, using the instructions, we can search for the function in Cutter. It turns out the function is at 0x03d2, but it is not defined as a function. Oh well, we can define it ourselves. Let's name it refs_jiggy as it references JIGGYCIPHER{.

By switching to graph view, we can get a general idea of the function's working. We can see there is some init phase, followed by loop 1, then loop 2, and finally the end of the function. Let's start analyzing what this function does.

RefsJiggy Analysis

Start

Let's start by analyzing the first block. We start by breaking it up into it's pieces.

1. This block starts by preserving some registers on the stack.
2. Allocates 0x34 (52) places on the stack.
3. Loading something from data space addresses 0x7d, 0x7e, 0x7f?? It turns out these instructions were incorrectly decompiled by Cutter (and Ghidra). The correct instructions have been commented in the version on the right.

4. Storing some registers in the stack. These are likely input parameters for the function, There are eight input registers, but AVR uses register pairs for two-byte words. So there are likely four input parameters. We will start a table, and update it as we move along and uncover some of their purposes.

   Parameter Name	Input Registers	Stack Location
   unknown	r25:r24	Y+46,Y+45
   unknown	r23:r22	Y+48,Y+47
   unknown	r21:r20	Y+50,Y+49
   unknown	r19:r18	Y+52,Y+51

5. Loads the pointer from address 0x3802 into r25:r24. We know the pointer at address 0x3802 is 0x0fca, which points to JIGGYCIPHER{. Then this part calls some function at 0x0f82, which after some quick analysis seems to be a string length function. Then some registers are stored on the stack, probably the return values from string_len.

   

6. The pointer at address 0x3802 is loaded into r25:r24 again, and is also stored on the stack.

7. r1 is stored into Y+3 and Y+4 on the stack? Turns out, r1 is used in this compilation as zero many times. So this part is simply setting those places on the stack to zero.

So far, it seems like the start phase stores some input parameters onto the stack and gets the length of the JIGGYCIPHER{ string.

Loop 1

Now moving onto Loop 1.

8. Loads some values from the stack. Referencing the start phase, we can tell Y+9,Y+8 is the length of the JIGGYCIPHER{ string. Also Y+4,Y+3 starts as zero, and is probably the incremented variable (i) of this loop.

9. Compares i against string_len('JIGGYCIPHER{'). If i is less than the string length, it continues the loop.

10. Loads some value from Y+52,Y+51, increments by one, and saves back on the stack. The value (before being incremented) is stored in r25:r24.

11. Loads some value from Y+2,Y+1, increments by one, and saves back on the stack. The value (before being incremented) is stored in r19:r18.

12. Loads a byte from the address stored in r19:r18 into register r18. Then it saves the byte in r18 into the address stored in r25:r24. This is effectively copying a byte from the address in r19:r18 to the address in r25:r24.

13. Increments i by one and return to step 8.

So it seems Loop 1 is copying the JIGGYCIPHER{ string to some target address stored in Y+52,Y+51. It is important to note that Loop 1 is copying the prefix of JIGGYCIPHER, not the full flag-like string we saw before. As its copying the prefix, we can assume this is a function generating the flag-like string.

We know the generated string is being stored in the address in Y+52,Y+51 which was one of our input parameters. Let's rename the entry in our parameter table to output_string.

Parameter Name	Input Registers	Stack Location
unknown	r25:r24	Y+46,Y+45
unknown	r23:r22	Y+48,Y+47
unknown	r21:r20	Y+50,Y+49
output_string	r19:r18	Y+52,Y+51

Loop 2

Moving onto Loop 2.

14. Set Y+5 stack value to zero.

15. This step puts the address Y+0x0c into r25:r24. This address is then passed to the init_rng function. Refer to the init_rng section to see how we reversed this function.

    Basically, init_rng takes an input address and copies the bytes 02050DFFFF0000AD8B0000050B0BFFFF00000DF000000E0D0FCAFFFFFFFFBEBAFECA to the address. These bytes act as the initial state (or seed) of the random_next function we will see later.

    Knowing now how this init_rng function works, we can see this step is copying the RNG's seed into the stack for later usage.

16. Set Y+7,Y+6 stack values to zero.

17. Loads Y+7,Y+6 from the stack. From step 16, we can see Y+7,Y+6 start as zero, so this is probably the increment value i in this loop.

18. Loads Y+46,Y+45 from the stack. From our input parameters, we can tell Y+46,Y+45 is an input parameter. So it must be some sort of count or length. Let's call it length.

    Parameter Name	Input Registers	Stack Location
    length	r25:r24	Y+46,Y+45
    unknown	r23:r22	Y+48,Y+47
    unknown	r21:r20	Y+50,Y+49
    output_string	r19:r18	Y+52,Y+51

19. Compares i against length. If i is less than length, it continues the loop.

20. Loads some value from Y+50,Y+49, increments by one, and saves back on the stack. The value (before being incremented) is stored in r25:r24.

21. Loads a byte from the address stored in r25:r24 into register r24 and then saves it on the stack at Y+10.

22. Loads some value from Y+50,Y+49, increments by one, and saves back on the stack. The value (before being incremented) is stored in r25:r24.

23. Loads a byte from the address stored in r25:r24 into register r24 and then saves it on the stack at Y+11.

    Notice that steps 20 and 21 or nearly identical to 22 and 23? The only difference is bytes being saved to Y+10 versus Y+11. This collection of steps seems to be reading some characters from the string who's address is stored in Y+50,Y+49. This is also an input parameter, so let's call it input_string.

    Parameter Name	Input Registers	Stack Location
    length	r25:r24	Y+46,Y+45
    unknown	r23:r22	Y+48,Y+47
    input_string	r21:r20	Y+50,Y+49
    output_string	r19:r18	Y+52,Y+51

24. Loads the values at Y+5 and Y+10 from the stack and XORs them. The result is stored on the stack at Y+5. Y+5 was set to zero in step 14.

25. Loads some value from Y+48,Y+47, increments by one, and saves back on the stack. The value (before being incremented) is stored in r25:r24.

26. Loads a byte from the address stored in r25:r24 into register r25. The byte at Y+11 is then loaded. These two bytes are then XORed together and the result is stored in r17.

    The first byte is loaded from Y+48,Y+47, much like bytes were loaded in the previous steps. Y+48,Y+47 is a input parameter, and seems to be another input string, so we will call it input_string_2.

    Parameter Name	Input Registers	Stack Location
    length	r25:r24	Y+46,Y+45
    input_string_2	r23:r22	Y+48,Y+47
    input_string	r21:r20	Y+50,Y+49
    output_string	r19:r18	Y+52,Y+51

27. A function, which we renamed to random_next, is called with an input parameter of the address Y+0x0c. Recall that in step 15, a string of bytes were loaded into this address. Refer to the random_next section to see how we reversed this function.

    Basically, random_next generates a random byte and stores it in r24. This random byte is dependent on its state, which is stored on the stack at Y+0x0c. The initial seed of this random number generator was loaded in step 15.

    Knowing now how this random_next function works, we can see this step is generating a random byte and storing it in r24.

28. The random byte and r17 are XORed and the result is stored in r18. Recall that r17 was set in step 26.

29. A function is called with an input parameter of the address Y+0x33 (Y+51 in decimal), which contains the address for our output string. Refer to the write_hex section to see how we reversed this function.

    Basically, write_hex receives an input byte in the register r22, converts it to its two hex characters, and writes these characters to the string's address stored in r25:r24.

    Knowing how write_hex works, we can see it is writing the byte stored in r18 to the output_string. This byte's value was determined in the previous step.

30. Loads Y+11 and Y+5 from the stack. These values are then XORed and the result is stored in r17.

31. random_next is called and the random byte is stored in r24.

32. The random byte in r24 (from step 31) and the byte in r17 (step 30) are XORed and the result is stored in r18.

33. write_hex is called and the byte stored in r18 is written the the output_string.

34. random_next is called and the random byte is stored in r24.

35. Loads a byte Y+10 from the stack into r24. Recall this byte was last written to in step 21, and is a character from input_string. This byte is then XORed against r24 (the random byte from step 34) and the result is stored in r18.

36. write_hex is called and the byte stored in r18 is written the the output_string.

37. Increments i by one and return to step 17.

A lot happened in this loop. Let's recap:

• Two characters were read from input_string.
• One character was read from input_string_2.
• write_hex was called three times, outputting three bytes (six characters) to output_string.
• Each byte that was written was encrypted in a different way, through a series of XORs against other bytes and random bytes.

End

1. Loads the output_string address from the stack, increments it by one, and stores it on the stack. The address before being incremented is stored in r25:r24.

2. A byte with value 0x7d is written the output_string address from step 38. A quick ASCII table lookup shows this byte is the } character, the end of the JIGGYCIPHER.

3. Loads the output_string address from the stack and stores r1's value into it. r1 is always zero, and so this step is putting the null-terminator onto the output_string.

4. Restores the stack and pushed registers before returning from the function.

So the end of this function adds the } character and a null-terminator to the end of the string, then returns.

Other Functions Analysis

These functions are planned to be explained, but for now we have omitted writing their analysis so we can share this writeup sooner rather than later.

init_rng

random_next

write_hex

Solution

If you don't want to read the full solution, feel free to spoil yourself: asd.py

Working our way bottom-up, we need to first mimic the random number generator so we can generate the same random numbers that were used to encrypt the plaintext.

# Build our RNG storage from the bytes recovered from the flash.
def build_triplets():
    triplets = []
    triplets_bytes = bytes.fromhex("02 05 0D FF FF 00 00 AD 8B 00 00 05 0B 0B FF FF 00 00 0D F0 00 00 0E 0D 0F FF FF FF FF BE BA FE CA")

    for i in range(3):
        triplet_bytes = triplets_bytes[i*11:i*11+11]

        shifts = [triplet_bytes[0], triplet_bytes[1], triplet_bytes[2]]
        mask = triplet_bytes[3] + (triplet_bytes[4] << 8) + (triplet_bytes[5] << 16) + (triplet_bytes[6] << 24)
        gen = triplet_bytes[7] + (triplet_bytes[8] << 8) + (triplet_bytes[9] << 16) + (triplet_bytes[10] << 24)

        triplets.append([shifts, mask, gen])

    return triplets

# Get the next bit for a given triplet
def triplet_bit(triplet):
    shifts, mask, gen = triplet
    # shifts = Z0-2
    # mask = Z3-6
    # gen = Z7-10

    next_gen = gen

    shift = shifts[2]
    next_gen = ((next_gen >> shift) ^ next_gen) & mask

    shift = shifts[0]
    next_gen = ((next_gen >> shift) ^ next_gen) & mask

    shift = shifts[1]
    next_gen = ((next_gen << shift) ^ next_gen) & mask

    triplet[2] = next_gen

    return next_gen & 1

# Get the next bit for the three triplets
def next_bit(triplets):    
    a = triplet_bit(triplets[2])
    b = triplet_bit(triplets[1])
    c = triplet_bit(triplets[0])

    # something_something
    s = (a & b) | (b & c) | (a & c)

    return s

# Get the next byte for the three triplets
def next_byte(triplets):
    s = 0
    for i in range(8):
        s = s << 1
        s = s | next_bit(triplets)
    return s

Then we can incorporate the cipher recovered from the flash, and use its length to generate all the RNG's numbers we will need.

# Set enc to integer values of the bytes in the cipher
enc = "JIGGYCIPHER{a1557d3801d7723f3d385111d85cd125625cc260ba76d3051a5de6ff24786b8fc70c1659aa4c41b9151c8eef70d438515a6642562ab9f0403f276ba556a7429829433d2572484f2585719d468964832ab1a258521f568165c3ab1bead146ecd2de65a5d812b53ca5c168473bac895d4a8bc37390f860b11e6a6eb442c249180a9c337285455ab098d47b7264ec42bed7688d59dcdfddcf3d0de674356cbee22547e087c63be52013dc319600c4a4859788ab87349c6c81ea9598127f87093c148c1d3e94edea50132f324eaf5cc88b63b8fac20cf6}"
enc = bytes.fromhex(enc[12:-1])
enc = list(enc)
## print(enc)
## [161, 85, 125, 56, ... ]

# Build the triplets (the seed) of the RNG
triplets = build_triplets()

# Generate the RNG nums, one for each encrypted byte
rng_nums = [next_byte(triplets) for i in range(len(enc))]
## print(rng_nums)
## [128, 81, 12, 36, ... ]
## These numbers were compared to the output in the emulator to ensure accuracy.

Each iteration of loop 2 in RefsJiggy appended three bytes and each of the three bytes were generated a different way. Therefore, we will need to recover each of these bytes differently. Let's start by separating each of the three bytes from enc, and their respective random numbers as well.

enc_1 = [byte for index, byte in enumerate(enc) if index % 3 == 0] # [161, 56, ...]
enc_2 = [byte for index, byte in enumerate(enc) if index % 3 == 1] # [85, ... ]
enc_3 = [byte for index, byte in enumerate(enc) if index % 3 == 2] # [125, ... ]

rng_nums_1 = [byte for index, byte in enumerate(rng_nums) if index % 3 == 0] # [128, 36, ... ]
rng_nums_2 = [byte for index, byte in enumerate(rng_nums) if index % 3 == 1] # [ 81, ... ]
rng_nums_3 = [byte for index, byte in enumerate(rng_nums) if index % 3 == 2] # [ 12, ... ]

Referring back to our reversing in Loop 2, we can deduce:

*(Y+5) = *(Y+5) ^ *(Y+10)
enc_1_byte = **(Y+48,Y+47) ^ *(Y+11) ^ random_next()
enc_2_byte = *(Y+11) ^ *(Y+5) ^ random_next()
enc_3_byte = *(Y+10) ^ random_next()

The three plaintext characters we are attempting to recover are stored in *(Y+10), *(Y+11), and **(Y+48,Y+47).

*(Y+5) is dependent on the values of *(Y+10) from the previous loops, and starts with a value of 0.

Reversing these formulas, we can get:

*(Y+10) = enc_3_byte ^ random_next()
*(Y+11) = enc_2_byte ^ *(Y+5) ^ random_next()
**(Y+48,47) = enc_1_byte ^ *(Y+11) ^ random_next()

So let's turn this pseudo-code into real Python code and get that plaintext!

*(Y+10) is a simple XOR against the random number, so we start with that:

# Recover characters from enc_3
# *(Y+10) = enc_3_byte ^ random_next()

dec_3 = []

for i in range(len(enc_3)):
    d = enc_3[i] ^ rng_nums_3[i]
    dec_3.append(d)

print("".join([chr(x) for x in dec_3]))
# qo:Gvrmnso h nutilWrd o er inso ls n te,Icm rmCbrpc,tenwhm fM~jb;hl{~ne

ASCII text! At this point I nearly had a heartattack. Over a week's effort had finally resulted in something readable.

Let's keep going.

Now that we have the values of *(Y+10), we can generate the iterative *(Y+5). With *(Y+5), we can recover *(Y+11):

# Recover characters from enc_2
# *(Y+5) = *(Y+5) ^ *(Y+10)
# *(Y+11) = enc_2_byte ^ *(Y+5) ^ random_next()

dec_2 = []
key = 0

for i in range(len(enc_2)):
    key ^= dec_3[i]
    d = enc_2[i] ^ key ^ rng_nums_2[i]
    dec_2.append(d)

print("".join([chr(x) for x in dec_2]))
# ut{oenet fteIdsra ol,yuwaygat ffehadsel  oefo yesae h e oeo i~p}ca:~edr

More ASCII text! You can infact put the above two together to recover input_string, but lets finish this out before doing that. Now with *(Y+11), we can recover **(Y+48,Y+47):

# Recover characters from enc_1
# **(Y+48,47) = enc_1_byte ^ *(Y+11) ^ random_next()
dec_1 = []

for i in range(len(enc_1)):
    d = enc_1[i] ^ dec_2[i] ^ rng_nums_1[i]
    dec_1.append(d)

print("".join([chr(x) for x in dec_1]))
# This is an Amazon(.ca?) gift card. Your claim code is ZBUH-3PYCC6-DKAG.

Well, there's input_string_2. Don't bother trying the code, we already did :)

Let's get the first input_string and wrap it all up in a nice bow now.

# Put together strings
input_string_1 = "".join([chr(x) + chr(y) for x,y in zip(dec_3, dec_2)])
input_string_2 = "".join([chr(x) for x in dec_1])

print(input_string_1)
print(input_string_2)

quot:{Governments of the Industrial World, you weary giants of flesh and steel, I come from Cyberspace, the new home of Mi~~jpb};chal:{~~ender
This is an Amazon(.ca?) gift card. Your claim code is ZBUH-3PYCC6-DKAG.

The quote above is a bit mangled, but here it is in full:

Governments of the Industrial World, you weary giants of flesh and steel, I come from Cyberspace, the new home of Mind. On behalf of the future, I ask you of the past to leave us alone. You are not welcome among us. You have no sovereignty where we gather

— John Perry Barlow, "A Declaration of the Independence of Cyberspace"

Final solve script: asd.py

Final Remarks

A big thanks to Loudmouth Security for providing the badges and this challenge. Another big thanks to CyberSci for this year's competition.

This challenge was a rollercoaster, and don't let the presentation of this writeup fool you, finding the solution was in no way a linear, clearly-defined journey. The road to solving this was bumpy, but both of us learned a lot about hardware, reversing, and maybe a bit too much about AVR in the process.

All in all, it was a lot of fun and hopefully a similar challenge will be seen next year.

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