SBOX = [0xb9, 0xb3, 0x49, 0x94, ...] # 略 def pad(data): if len(data) == 0: return b"\x00" * 8 while len(data) % 8 != 0: data += b"\x00" return data def sub(state): return [SBOX[x] for x in state] def mix(block, state): for i in range(8): state[i] ^= block[7 - i] & 0x1f state[i] ^= block[i] & 0xe0 return state def shift(state): t = 0 for s in state: t ^= s u = state[0] for i in range(7): state[i] ^= t ^ state[i] ^ state[i+1] state[7] ^= t ^ state[7] ^ u return state def hash(data): assert len(data) % 8 == 0 state = [2**i-1 for i in range(1, 9)] for i in range(0, len(data), 8): block = data[i: i+8] state = sub(state) state = mix(block, state) state = shift(state) state = sub(state) return bytes(state).hex() Banner = """ ____ __ __ ____ _ _ _ ____ _ _ / ___|| \/ / ___| | | | | / \ / ___|| | | | \___ \| |\/| \___ \ _____| |_| | / _ \ \___ \| |_| | ___) | | | |___) |_____| _ |/ ___ \ ___) | _ | |____/|_| |_|____/ |_| |_/_/ \_\____/|_| |_| """ print(Banner, end= "\n\n") print("Can you even Collide?") MSG1 = bytes.fromhex(input("First message : ").strip()) MSG2 = bytes.fromhex(input("Second message : ").strip()) MSG1 = pad(MSG1) MSG2 = pad(MSG2) H1 = hash(MSG1) H2 = hash(MSG2) print("H(MSG1) = {}".format(H1)) print("H(MSG2) = {}".format(H2)) if MSG1 == MSG2: print("Really ?") elif H1 == H2 : if H1 == "0000000000000000": print("Good job, Here's your reward: ") print(open("flag.txt","r").read()) else: print("So close, yet so far :(") else: print("Not even close :(")
overview
独自hash が定義されている
目的は
Hash(m1) = 0, Hash(m2) = 0, m1 != m2を満たすm1,m2を見つけること。すなわち2種類の 現像攻撃 を両方やれ、と言っているhashはsubmixshiftの3種類の操作を組み合わせて行われる。問題名のSMSはこの操作の頭文字をとったものだろうこれらの操作のうち、
subは簡単に逆算できるrev_sbox = {SBOX[i]:i for i in range(len(SBOX))} def unsub(state): return [rev_sbox[x] for x in state]
また、
shiftも z3 を使った逆算が可能def unshift(state): shifted = state from z3 import BitVec, Solver, sat def shift(state): t = 0 for s in state: t ^= s u = state[0] for i in range(7): state[i] ^= t ^ state[i] ^ state[i+1] state[7] ^= t ^ state[7] ^ u return state state = [BitVec("state{}".format(i), 8) for i in range(8)] shifted_bits = shift(state[:]) solver = Solver() for i in range(8): solver.add(shifted_bits[i] == shifted[i]) assert solver.check() == sat m = solver.model() return [m[state[i]].as_long() for i in range(8)]
mixだけは2入力あるが、一方を固定することで、もう一方の値が決まる。これも z3 を使ってできるdef unmix(state, prev_state): from z3 import * def mix(block, state): for i in range(8): state[i] ^= block[7 - i] & 0x1f state[i] ^= block[i] & 0xe0 return state block = [BitVec("block{}".format(i), 8) for i in range(8)] mixed = mix(block, prev_state[:]) solver = Solver() solver.add([mixed[i] == state[i] for i in range(8)]) assert solver.check() == sat m = solver.model() return [m[block[i]].as_long() for i in range(8)]
solution
ということですべての操作が逆算できる。愚直に入力を求めていけば第一種現像攻撃が実現できる
↑ができたら、ブロックを一つ増やすなどして初期状態を別のものに変えたあと、もう一度同様のことをすると、出力は同じで入力は異なる、ということが実現できる
import os SBOX = [ 0xb9, 0xb3, 0x49, 0x94, 0xf9, 0x3, 0xd0, 0xfc, 0x67, 0xa3, 0x72, 0xb5, 0x45, 0x82, 0x54, 0x93, 0x5b, 0x88, 0x5c, 0xe0, 0x96, 0x41, 0xc7, 0xa, 0xdb, 0x7f, 0x77, 0x29, 0x9, 0xb, 0x8d, 0x80, 0x2d, 0xaf, 0xe1, 0x4a, 0x38, 0x73, 0x3a, 0x6a, 0xf2, 0xb6, 0xdc, 0xbd, 0x79, 0x2a, 0xcb, 0x55, 0x10, 0x61, 0x63, 0x68, 0x13, 0x95, 0x9f, 0x1c, 0x4f, 0x35, 0x5f, 0xae, 0x37, 0xb8, 0xfe, 0xea, 0x7a, 0x4b, 0xc3, 0xe8, 0xc6, 0x44, 0x60, 0xb2, 0x5a, 0x2e, 0xeb, 0x47, 0x1e, 0x4d, 0x9a, 0x98, 0x36, 0xe7, 0x48, 0x3e, 0x42, 0x6b, 0xa1, 0x65, 0xb1, 0x57, 0x6c, 0x4, 0xff, 0xfd, 0x34, 0x40, 0x31, 0x8c, 0xbe, 0xda, 0x2c, 0x1b, 0x7c, 0x64, 0x3f, 0xd1, 0xc9, 0x9b, 0x25, 0x87, 0xaa, 0xd, 0x15, 0x1f, 0xce, 0x30, 0xfb, 0xd5, 0xef, 0xbb, 0x24, 0x28, 0x90, 0x2f, 0x85, 0xc5, 0x4c, 0x97, 0xa8, 0x16, 0x43, 0xac, 0x74, 0xc0, 0x8b, 0xc4, 0xe9, 0x7e, 0xf5, 0xd2, 0xab, 0x12, 0xd8, 0xdd, 0xa9, 0xad, 0x21, 0xd7, 0xed, 0x1, 0x32, 0xbf, 0xa6, 0x8a, 0xe3, 0x6f, 0xde, 0x84, 0xc8, 0x6d, 0x92, 0x99, 0x51, 0x39, 0xe5, 0x46, 0x9c, 0xf0, 0x0, 0x8e, 0xbc, 0xa2, 0x22, 0x9d, 0xc2, 0xfa, 0xb0, 0x33, 0x56, 0xec, 0xdf, 0x89, 0x52, 0x8, 0x62, 0x7, 0x59, 0xb7, 0xe4, 0x14, 0x9e, 0x70, 0xd9, 0xe, 0x3d, 0x26, 0x1d, 0x66, 0x71, 0xe2, 0x5, 0x6e, 0x5d, 0xf6, 0x18, 0xf, 0xcf, 0xd6, 0xe6, 0xba, 0x1a, 0x78, 0xf8, 0x76, 0xd3, 0x50, 0xf7, 0x58, 0x17, 0x91, 0x11, 0x86, 0xf1, 0xa4, 0x19, 0x4e, 0x6, 0xa0, 0xca, 0xa5, 0xf3, 0xee, 0xcd, 0x53, 0x5e, 0xa7, 0xc, 0xb4, 0x2, 0xc1, 0x3b, 0x27, 0x69, 0x7d, 0x8f, 0xcc, 0x20, 0x7b, 0x81, 0x2b, 0x83, 0x23, 0xd4, 0x3c, 0xf4, 0x75 ] def sub(state): return [SBOX[x] for x in state] rev_sbox = {SBOX[i]:i for i in range(len(SBOX))} def unsub(state): return [rev_sbox[x] for x in state] def mix(block, state): for i in range(8): state[i] ^= block[7 - i] & 0x1f state[i] ^= block[i] & 0xe0 return state def unmix(state, prev_state): from z3 import BitVec, Solver, sat block = [BitVec("block{}".format(i), 8) for i in range(8)] mixed = mix(block, prev_state[:]) solver = Solver() solver.add([mixed[i] == state[i] for i in range(8)]) assert solver.check() == sat m = solver.model() return [m[block[i]].as_long() for i in range(8)] def shift(state): t = 0 for s in state: t ^= s u = state[0] for i in range(7): state[i] ^= t ^ state[i] ^ state[i+1] state[7] ^= t ^ state[7] ^ u return state def unshift(state): from z3 import BitVec, Solver, sat shifted = state state = [BitVec("state{}".format(i), 8) for i in range(8)] shifted_bits = shift(state[:]) solver = Solver() for i in range(8): solver.add(shifted_bits[i] == shifted[i]) assert solver.check() == sat m = solver.model() return [m[state[i]].as_long() for i in range(8)] def unhash(state, prev_state): return unmix(unshift(state), sub(prev_state)) # MSG1 init_state = [2**i-1 for i in range(1, 9)] output = bytes.fromhex("0000000000000000") last_state = unsub(output) block1 = unhash(last_state, init_state) # MSG2 init_state = [2**i-1 for i in range(1, 9)] state = sub(init_state) state = mix(block1, state[:]) state = shift(state[:]) init_state = state[:] output = bytes.fromhex("0000000000000000") last_state = unsub(output) block2 = unhash(last_state, init_state) from ptrlib import Socket sock = Socket("localhost", 19999) sock.sendlineafter(": ", bytes(block1).hex()) sock.sendlineafter(": ", bytes(block1 + block2).hex()) sock.interactive()
