Reorganized

Cryptography/SHA/sha-256-bruteforce.py

@@ -0,0 +1,11 @@
+'''
+from asis 2013: The last crypto (binary numbers) was very puzzling. We couldn’t decipher it. But a few minutes before the CTF ending, we noticed we could brute-force the 6 missing characters offline, because in each task, there was a client-side verification with a sha-256 hash. For this task, the hash of the flag was 6307c5441ebac07051e3b90d53c3106230dd9aa128601dcd5f63efcf824ce1ba. A quick brute-force in Python revealed us the missing chars, and therefore, the final flag to submit!
+'''
+
+
+import hashlib, itertools
+hash = '6307c5441ebac07051e3b90d53c3106230dd9aa128601dcd5f63efcf824ce1ba'
+ch = 'abcdef0123456789'
+for a, b, c, d, e, f in itertools.product(ch, ch, ch, ch, ch, ch):
+    if hashlib.sha256('ASIS_a9%s00f497f2eaa4372a7fc21f0d' % (a + b + c + d + e + f)).hexdigest() == hash:
+        print 'ASIS_a9%s00f497f2eaa4372a7fc21f0d' % (a + b + c + d + e + f)

Cryptography/SHA/sha-256-bruteforce_remove_123456789x.py

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+'''
+from asis 2013
+'''
+
+from itertools import permutations
+from hashlib import sha256
+
+def test(s):
+  e = '9f2a579716af14400c9ba1de8682ca52c17b3ed4235ea17ac12ae78ca24876ef'
+  return sha256('ASIS_' + s).hexdigest() == e
+
+m = '3c6a1c371b381c943065864b95ae5546'
+s = '12456789x'
+for p in permutations(s):
+    def f(sub, c):
+        if c in sub:
+            return sub[c]
+        else:
+            return c
+    sub = {c : d for c, d in zip(s, p)}
+    z = ''.join(f(sub, c) for c in m)
+    if test(z):
+        print z
+        break