license-check.md

license-check (re, medium)

TL;DR

This task has a trivial XOR encoder disguised as a Verilog module.

Functions of various components:

The flag is hidden in the validator module and the input provided to encoder module needs to be decoded against bits coming from the keygen.

Below are simplified desriptions of what the modules do, with a pseudocode.

Keygen

Generates a simple (and reproducible) 8-bit key signal. Rough semantics of the code:

int_2 cnt = 0
int_8 key = 11011001b

for each clock pulse:
    key = key + 2 * cnt   // mod 256
    cnt++                 // mod 4
    out = key

Encoder

Iterate over bytes from the test flag in, using cnt as a counter. For each clock pulse:

• take the byte number cnt from the test flag
• XOR it with the key provided by the keygen
• send it to out.

... effectively, encrypting the provided flag with the dynamic key. Note that part seems not used.

int_8   key = 0
int_192 in             // Test flag for verification
int_8   cnt = 0
int_8   out = 0
int_8   part = 0       // Not used

for each clock pulse:
    part = (part + 1) & 0x03
    cnt++
    out = key ^ ((in & (0xFF << (cnt*8))) >> cnt*8)

Validator

Iterate over encrypted bytes provided via in and compare them with encrypted flag stored internally in flag register. For each clock pulse:

• if in byte is equal to the respective byte of the flag, keep out in high state
• otherwise, if it is equal to 0xEB (presumably end of string), keep out in high state too
• otherwise, switch out to low state, signaling a flag verification failure

int_8   in
int_8   out = 1
int_8   cnt = 0
int_8   flagpart = 0
int_8   flagin = 0
int_200 flag = 0x52737427253d3f7279669699ccb29a96998481808ce7b8a400

for each clock pulse:
    flagin = in
    flagpart = (flag & (0xFF << (cnt*8))) >> (cnt*8)
    if in == (flag & (0xFF << (cnt*8))) >> (cnt*8)
        out = out
    elif in == 0xeb
        out = out
    else
        out = 0
    endif
    cnt++

Getting the flag

Let's write some code, generating bytes similar to the keygen and read bytes from encrypted flag, starting with LSB:

#!/usr/bin/python3

flag = 0x52737427253d3f7279669699ccb29a96998481808ce7b8a4
result = ''
cnt = 0
key = 0b11011001

for i in range(24):
  # Keygen
  key = (key + 2 * cnt) & 0xFF
  cnt = (cnt+1) & 0x3
  # Encoder
  result=chr(key^(flag&0xFF))+result
  flag>>=8

print(result)

Result:

shc2024{peid1Eighofei8c}