Tags: reverse engineering
Rating:
## AsisHash 150 (re, 150p)
### PL
[ENG](#eng-version)
Dostajemy [program](./hash.elf) (elf). Analizujemy jego działanie, w dużym uproszczeniu wygląda to tak:
int main() {
char password[...];
scanf("%s", password);
char *hash = hash_password(password);
if (!strcmp(hash, good_hash)) {
puts("Congratz, you got the flag :) ");
} else {
puts("Sorry! flag is not correct!");
}
}
Funkcja hash_password jest bardzo skomplikowana i nawet nie próbowaliśmy analizować jej działania. Zamiast tego zrobiliśmy coś prostszego - ponieważ
hash jest monotoniczny (dla dłuższych haseł/wyższych znaków ascii daje wyższe wyniki), spróbowaliśmy zgadnac hash (docelowy hash flagi jest stay i równy 27221558106229772521592198788202006619458470800161007384471764) za pomocą bruteforcowania z nawrotami wszystkich możliwych flag:
import subprocess
def run(flag):
return subprocess.check_output(['./hash.elf', flag]).split('\n')[0]
def prefx(a, b):
p = 0
for ac, bc in zip(a, b):
if ac == bc:
p += 1
else:
break
return p
def plaintext(t):
return ''.join(c if 32 <= ord(c) <= 127 else '.' for c in t)
sln = '27221558106229772521592198788202006619458470800161007384471764'
charset = '0123456789abcdef}'
def tryit(f, r, n):
pp = prefx(sln, r)
print plaintext(f), r[:pp], r[pp:]
stat = '<' if sln[pp] < r[pp] else '>'
print plaintext(f), r[:pp], r[pp:], stat
for c in charset:
f2 = f[:n] + c + f[n+1:]
r2 = run(f2)
p2 = prefx(sln, r2)
if p2 > pp:
s = tryit(f2, r2, n+1)
if s == '>':
return
return stat
for c in range(256): # try to guess good initial padding
print c, '!!!!!!!!!!!' # debug info
placeholder = chr(c)
start = 'ASIS{' + placeholder * 33
start = 'ASIS{d5c808f5dc96567bda48' + placeholder * 13 # algorytm zacinał się czasami, więc ręcznie dawaliśmy mu dobre wartości początkowe przeniesione z poprzednich wykonań.
start = 'ASIS{d5c808f5dc96567bda48be9ba82fc1d' + placeholder * 2
tryit(start, run(start), len(start.replace(placeholder, '')))
Po chwili czekania i tweakowania algorytmu, dostajemy flagę:
ASIS{d5c808f5dc96567bda48be9ba82fc1d6}
### ENG version
We get a [binary](./hash.elf) (elf). We analyse its behaviour and it is doing:
int main() {
char password[...];
scanf("%s", password);
char *hash = hash_password(password);
if (!strcmp(hash, good_hash)) {
puts("Congratz, you got the flag :) ");
} else {
puts("Sorry! flag is not correct!");
}
}
The `hash_password` function is very complex and we didn't ever try to analyse it. Instead we did something simpler - since the hash is monotonous (for longer input/higher characters it gives higher results) we tried guessing correct flag (flag hash is known and equal to (27221558106229772521592198788202006619458470800161007384471764) using bruteforce with backtracing:
import subprocess
def run(flag):
return subprocess.check_output(['./hash.elf', flag]).split('\n')[0]
def prefx(a, b):
p = 0
for ac, bc in zip(a, b):
if ac == bc:
p += 1
else:
break
return p
def plaintext(t):
return ''.join(c if 32 <= ord(c) <= 127 else '.' for c in t)
sln = '27221558106229772521592198788202006619458470800161007384471764'
charset = '0123456789abcdef}'
def tryit(f, r, n):
pp = prefx(sln, r)
print plaintext(f), r[:pp], r[pp:]
stat = '<' if sln[pp] < r[pp] else '>'
print plaintext(f), r[:pp], r[pp:], stat
for c in charset:
f2 = f[:n] + c + f[n+1:]
r2 = run(f2)
p2 = prefx(sln, r2)
if p2 > pp:
s = tryit(f2, r2, n+1)
if s == '>':
return
return stat
for c in range(256): # try to guess good initial padding
print c, '!!!!!!!!!!!' # debug info
placeholder = chr(c)
start = 'ASIS{' + placeholder * 33
start = 'ASIS{d5c808f5dc96567bda48' + placeholder * 13 # the algorithm sometimes was stuck so we were starting it again with already pre-computed prefixes
start = 'ASIS{d5c808f5dc96567bda48be9ba82fc1d' + placeholder * 2
tryit(start, run(start), len(start.replace(placeholder, '')))
And after a short while and some optimizations to the algorithm we get the flag:
ASIS{d5c808f5dc96567bda48be9ba82fc1d6}
