def rotl8(v, r): return ((v << r) | (v >> (8 - r))) & 0xFF def inv_rotl8(v, r): return ((v >> r) | (v << (8 - r))) & 0xFF
T[i] = rotl8( key[i] ^ 0x5A , i % 8 ) We want Σ T[i] = 0xdeadbeef (mod 2^32) . Because the checksum is a simple sum, we can freely pick the first 63 bytes and solve for the last byte. el capo 2 cap 57
#!/usr/bin/env python3 import subprocess, os, struct def rotl8(v, r): return ((v << r) |
for i in range(SIZE-1): # let transformed byte be zero for simplicity t = 0 key[i] = inv_rotl8(t, i % 8) ^ CONST_XOR checksum = (checksum + t) & 0xffffffff The problem reduces to crafting a 64‑byte key
static const char flag[] = "ECTFel_capo_2_cap_57_success"; Because the binary is stripped, the name isn’t visible in strings , but the decompiler reveals it as a global pointer used only in the success branch. The problem reduces to crafting a 64‑byte key.bin such that the checksum after the transformation equals the required constant ( 0xdeadbeef in the example). 4.1 Deriving the Required Plain‑text Let T[i] be the transformed byte for index i . We know: