// [rev] arxelerated (21 solves)

// Writeup author: @kristoj
// Challenge author: @Rising

Challenge description

oh no, my device starts shredding everything I put in it...

qemu-system-arm -cpu cortex-m3 -machine lm3s6965evb -kernel arxelerated -semihosting -semihosting-config enable=on,target=native -serial mon:stdio

Table of contents


A really fun challenge about an obfuscated ARM kernel. It's an engaging puzzle for those familiar with firmware analysis.

Note: You can skip to the summary if you are not interested in the details.

First impression

By examining the provided materials, we encounter two main files:

  • arxelerated: A firmware file for ARM Cortex-M3.
  • out.enc: An encrypted file, presumably containing data we need to decrypt.

Executing the firmware using the specified QEMU command reveals a beautifully rendered cat image.


Upon entering a key, numerous dots begins to appear on the console while the image is being covered with random pixels. Once the image is fully obscured, a lenghty hexadecimal string is outputted on the console.


The length of the hex string matches that found in the out.enc file, and its prefix coincides with the beginning of the encrypted file. Repeated executions of the QEMU command confirm the hex string's consistency, suggesting deterministic encryption. It's plausible to infer that the out.enc file encrypts data related to the challenge's objective, possibly the flag.

Understanding the flow

Initial static analysis

By executing the file command on the arxelerated file, we ascertain that it is an ARM ELF binary, characterized by the following properties:

$ file arxelerated arxelerated: ELF 32-bit LSB executable, ARM, EABI5 version 1 (SYSV), statically linked, not stripped

Further examination with the strings command reveals that the binary was compiled with the Sourcery G++ 4.2-68 compiler and utilizes functions related to an OSRAM display:

$ strings arxelerated | grep GCC GCC: (Arm GNU Toolchain 13.2.rel1 (Build arm-13.7)) 13.2.1 20231009 GCC: (Sourcery G++ 4.2-68) 4.2.1 GCC: (Sourcery G++ 4.2-68) 4.2.1 GCC: (Sourcery G++ 4.2-68) 4.2.1 GCC: (Sourcery G++ 4.2-68) 4.2.1 $ strings arxelerated | grep OSRAM OSRAMWriteCommand OSRAMWriteData g_pucOSRAM128x64x4VerticalInc g_pucOSRAM128x64x4HorizontalInc g_pucOSRAM128x64x4Init OSRAM128x64x4Clear OSRAM128x64x4ImageDraw OSRAM128x64x4Enable

Opening the binary in IDA PRO with the ARM little-endian processor type provides a closer look at its structure:


The start function primarily initializes the system and then calls the sub_504 function, which performs several critical operations:

  1. sub_474: Initializes the clock and display.
  2. OSRAM128x64x4ImageDraw(&unk_20000004, 0, 0, 128, 64): which, we can assume, draws the cat image on the display.
  3. UARTCharPut(0x4000C000, '>'): Outputs the > character on the console.
  4. UARTCharGet(0x4000C000): Awaits a character input from the console.
  5. __und(0): Triggers an undefined (UDF) instruction, leading to an exception.

Because of the UDF instruction, we can not understand what the program does only by a static analysis. Let's use GDB to debug the binary and see what is happening.

Dynamic analysis

Due to the UDF instruction, static analysis alone is insufficient. We proceed with dynamic analysis using GDB connected to a QEMU instance configured to halt at startup, by adding the -S -s flags to the QEMU command.

qemu-system-arm -cpu cortex-m3 -machine lm3s6965evb -kernel arxelerated -semihosting -semihosting-config enable=on,target=native -serial mon:stdio -S -s
$ gdb-multiarch > file arxelerated > target remote localhost:1234 > break *0x54A > continue > x/i $pc => 0x54a <frame_dummy+290>: udf #0 > next > x/i $pc => 0x6cc <frame_dummy+676>: tst.w lr, #4

We can see that the program jumps to the sub_6CC function that handles the interrupt, like an Interrupt Service Routine. Here, the Current Program Status Register (CPSR) is saved and it is used to call a function pointed by off_F80. We can assume that this is the IRQ vector table.


In this case, when the execution reaches the BLX instruction at 0x708, the program jumps to the sub_600 function, which is the first function in the IRQ vector table:

> break *0x708 > continue > x/i $pc => 0x708 <frame_dummy+736>: blx r5 > next > x/i $pc => 0x600 <frame_dummy+472>: push {r4}

Because of the UDF instruction was called with the parameter #0, we can guess that this argument represents the index of the function in the IRQ vector table that will be called by the handler.

Undefined instruction handling

By using IDA python, we can extract all the addresses where there is a UDF instruction to see if the guess is true.

> def get_und_addresses(): addresses = [] for i in range(0x0, 0x1000): if get_bytes(i, 1) == b'\xde': addresses.append(i-1) print("UND addresses:") print(list(map(hex, addresses))) return addresses def get_und_arguments(): addresses = get_und_addresses() arguments = [] for address in addresses: arguments.append(get_operand_value(address, 0)) return arguments def print_und_handlers(): arguments = get_und_arguments() print("UND arguments:") print(list(map(hex, arguments))) > print_und_handlers() UND addresses: ['0x4c0', '0x4ce', '0x4d8', '0x4f0', '0x4fa', '0x54a', '0x6e0', '0x774'] UND arguments: ['0x1', '0x3', '0x4', '0x3', '0x2', '0x0', '0xff', '0xff']

If we put a breakpoint to this addresses inside GDB, we can see that the program jumps to the functions pointed by the IRQ vector table at the corresponding index (except for 0xff arguments, whose instruction are never reached). So the guess was correct!
Furthermore, we can notice that when the program jumps to the functions in the IRQ vector table, the registers are the same as the ones during the undefined instruction.


To see some comprehensible pseudo-code from function callers, we aim to replace the UDF instructions with calls to their corresponding handler functions. However, there's a problem: the Branch Link (BL) instruction spans 4 bytes, while UDF instructions are only 2 bytes long.
By double-clicking the calculated addresses in the IDA python console, we can inspect the UDF instructions and see that most of the time they are preceded by a redundant instruction MOV R3, R3. This extra space provides an opportunity to insert the BL instruction.

BL instruction offsets

The BL instruction is PC-relative, so we need to calculate the offset between the UDF instruction and the function we want to call.

> def calculate_BL_bytes(source, destination): source = source + 4 offset = destination - source offset = (offset >> 1) & 0x00FFFFFF imm11 = offset & 0x07FF S = (offset >> 24) & 0x1 J1 = not(S ^ ((offset >> 22) & 0x1)) J2 = not(S ^ ((offset >> 23) & 0x1)) high_halfword = 0xF000 | (S << 10) low_halfword = 0xD000 | ((J1 & 0x1) << 13) | ((J2 & 0x1) << 11) | imm11 return high_halfword.to_bytes(2, byteorder='little') + low_halfword.to_bytes(2, byteorder='little')

After some trial and error, we can see that it returns the correct bytes for the BL instruction that will call the function pointed by the destination address.

Patching the binary

We can automatize the patching process:

> IRQ_table_addr = 0xF80 > IRQ_table = [0x600 + int.from_bytes(get_bytes(IRQ_table_addr + i*4, 1), 'big') - 1 for i in range(0, 5)] > UND_addresses = get_und_addresses() > MOV_R3_R3_bytes = get_bytes(0x4CC, 2) > patchable_UND_addrs = [addr for addr in UND_addresses if get_bytes(addr-2, 2) == MOV_R3_R3_bytes] > patchable_UND_args = [get_operand_value(addr, 0) for addr in patchable_UND_addrs] > def patch_UNDs(): for addr, arg in zip(patchable_UND_addrs, patchable_UND_args): print("Patching UND at", hex(addr), "with argument", hex(arg)) print("Calling function at", hex(IRQ_table[arg])) BL_bytes = calculate_BL_bytes(addr-2, IRQ_table[arg]) for i in range(0, 4): patch_byte(addr-2+i, BL_bytes[i]) > patch_UNDs()

While most UDF instructions can be patched using the above method, the one at 0x4C0 lacks a preceding redundant instruction to repurpose. Reviewing the surrounding instructions reveals a potential optimization to get some space for the patch:

.text:000004AC PUSH {R4-R6,LR} .text:000004AE MOV R6, R0 .text:000004B0 MOV R5, R1 .text:000004B2 LDR R2, [R0] .text:000004B4 LDR R3, [R1] .text:000004B6 MOV R0, R2 .text:000004B8 MOV R1, R3 .text:000004BA MOVS R4, #0 .text:000004BC MOV R2, R4 .text:000004BE MOV R3, R4 .text:000004C0 UND #1

As we can see, the registers R2 and R3 are used to store values pointed by R0 and R1, but then they are overwritten with R4. We can load those values directly into R0 and R1 without wasting instructions, by patching the binary like this:

.text:000004AC PUSH {R4-R6,LR} .text:000004AE MOV R6, R0 .text:000004B0 MOV R5, R1 .text:000004B2 LDR R0, [R0] ; a1 .text:000004B4 LDR R1, [R1] ; a2 .text:000004B6 MOVS R4, #0 .text:000004B8 MOV R2, R4 .text:000004BA MOV R3, R4 .text:000004BC NOP .text:000004BE BL sub_614

After updating the interested code with the shortcuts U, C and P, we can see the pseudo-code of the functions sub_504 and sub_4AC:

ida_patched ida_patched

And now the challenge begins.

Reverse engineering

We can give a better look to the sub_504 function, which is the function called by the start function:

  • Initialization:

  • Main Loop:

  • Outputing:

Main loop

The loop operates by taking two consecutive 4-byte values from the screen buffer and XORing them with two previously calculated values (initially set to zero). It then passes the result to the sub_4AC function and updates the previous values. Subsequently, the loop induces a sleep period, redraws the screen buffer, and updates the indexes.


By analyzing this function, we can notice it is an encryption routine that manipulates the input values through XOR and ROR operations. These operations are orchestrated by calling functions from the IRQ vector table, including sub_660, while simultaneously outputting ten dots on the console.


At this point, I was not able to guess the CRAX Block Cypher algorithm, so I decided to rewrite the encryption function in Python and decrypt it by using z3.

Solving the encryption

Dependency analysis

The encryption function's behavior indicates that each pair of new values depends solely on the two previously encrypted values and specific lookup values (located at 0x20001034 and 0xF98), which likely function as part of an encryption key. From our initial observations noted in the overview, the first 800 hexadecimal characters of the output and the attached file are identical. This allows us to use the last 8 common bytes as a starting point for our encryption emulation.

Z3 solver

To manage z3's memory constraints effectively, we opt to decrypt the data 8 bytes at a time. This strategy is efficient because knowing the previous 8 bytes enables z3 to decrypt the next 8 bytes within a reasonable timeframe.

[...] # from z3_solve_next_two.py # I only need the next 2 VALUES LN = 2 screen_plain = [BitVec(f"screen_plain_{i}", 32) for i in range(LN)] screen_cp = [x for x in screen_plain] screen_enc = main_loop(screen_cp) s = Solver() for i in range(LN): s.add(screen_enc[i] == screen_out[i + OFFSET]) if s.check() == sat: print("SAT!") m = s.model() print(", ".join([hex(m[screen_plain[i]].as_long()) for i in range(LN)]))

Given the slow performance of the decryption script and the prevalence of zeros in the decrypted data, we can refine the script to first check if the decrypted 8 bytes were zeros by encrypting them and comparing the result to the encrypted data. If they did not match, only then would we employ z3 for decryption.

[...] # from solve.py # first 800 hex_char (= 400 bytes = 100 _DWORD) are the same OFFSET = 100 result = screen_in[:OFFSET] # sub_600 VALUES = [140050, 6267691, 5949456, 4919] # last known values after OFFSET prev = [3983684781, 1999312935] for i in range(0, 1024-OFFSET, 2): if execute_check_next_zeros(): next_two = ['0x0', '0x0'] else: next_two = execute_z3_solve_next_two(OFFSET+i, prev + VALUES) result += next_two prev = [screen_out[OFFSET+i], screen_out[OFFSET+i+1]] [...]

Patching the screen buffer

After approximately one hour of execution, the decrypted data was stored in the screen_pl.py file. Using IDA Python, we patch the binary to display what was on the screen prior to encryption, instead of the default cat image:

# paste the screen_pl.py content here > screen_pl = [...] > SCREEN_ADDR = 0x20000004 > for i in range(0, len(screen_pl)): patch_byte(SCREEN_ADDR+i, screen_pl[i])

Upon executing the patched binary in QEMU, the display shows the flag:



This challenge revolves around an ARM kernel that encrypts the screen buffer, displaying encrypted data similar to the provided out.enc file on the console.

The initial step involves deobfuscating the binary. This is accomplished by tracing its execution using gdb-multiarch combined with QEMU's -S -s options, allowing for an in-depth analysis of the binary's operations. Key to this process is patching the undefined (UDF) instructions with appropriate handlers from the IRQ vector table.

Through detailed examination of the pseudo-code, the encryption is identified as a CRAX Block Cipher algorithm.

To reverse the encryption, we can employ the CRAX Block Cipher's decryption function or, alternatively, we can reimplement the encrypt function in Python and decrypt the screen buffer with the z3 solver.

With the decryption complete, we patch the binary and run it in QEMU, revealing the flag on the screen buffer.

// Attachments
python3//check_next_are_zeros.pyDownload download file
python3//out.pyDownload download file
python3//result.pyDownload download file
python3//screen_in.pyDownload download file
python3//screen_pl.pyDownload download file
python3//solve.pyDownload download file
python3//z3_solve_next_two.pyDownload download file