TnSb/blog
1
0
Fork 0
Code Issues Releases Activity

revise code examples

Browse Source
This commit is contained in:
Joel Beckmeyer 2024-06-29 13:06:23 -04:00
parent 6afec009c4
commit a42bc2bd42
1 changed files with 87 additions and 42 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

125
content/posts/cracking_att_vvm_cipher.md
View File

@ -1,7 +1,8 @@
---
title: "Cracking the AT&T VVM cipher"
tags: ["Hacking"]
date: 2024-06-19T20:16:00-04:00
date: 2024-06-19
lastmod: 2024-06-29
draft: false
---
I've been wanting to get the AT&T visual voicemail "protocol" (ADVVM) working
@ -55,15 +56,8 @@ Our dictionary is self-contained in an upper 4-bit prefix. Let's focus on the
bottom four bits (or [nibble](https://en.wikipedia.org/wiki/Nibble)) by
removing the upper bits:
```
def strip_prefix(char):
return ord(char)&0x0f
```
Now we can create a bytearray containing the stripped versions of the passcode:
```
def get_stripped(text):
stripped = [strip_prefix(c) for c in text]
return bytearray(stripped)
return [ord(c) & 0x0f for c in text]
```
Now, how do we actually figure out the transform? There are a number of ciphers
@ -78,19 +72,20 @@ ciphering](https://en.m.wikipedia.org/wiki/XOR_cipher).
Let's try it:
```
def decode(cipher, secret):
def xor_cipher(cipher, secret):
if isinstance(cipher[0], str):
cipher = get_stripped(cipher)
if isinstance(secret[0], str):
secret = get_stripped(secret)
# remember that the cipher "passes through" digits past the 10th, so we
# just overwrite the first 10
# remember that the cipher "passes through" digits past the length of the
# secret, so we just take the rest unciphered
text = [i^j for i, j in zip(cipher, secret)]
if len(cipher) > 10:
text += cipher[10:]
if len(cipher) > len(secret):
text += cipher[len(secret):]
return text
decode("[VW^QW\\W_X0", "7345839476")
xor_cipher("[VW^QW\\W_X0", "7345839476")
[12, 5, 3, 11, 9, 4, 5, 3, 8, 14, 0]
```
@ -99,16 +94,15 @@ it doesn't output the plaintext password that I am expecting. Not to worry,
because we can also use the plaintext instead of the actual secret to gain some
insight. Below is the one with my throwaway number:
```
decode("[VW^QW\\W_X0", "00000000000")
[11, 6, 7, 14, 1, 7, 12, 7, 15, 8, 0, 0]
xor_cipher("[VW^QW\\W_X0", "00000000000")
[11, 6, 7, 14, 1, 7, 12, 7, 15, 8, 0]
```
Ignore the slight bug due to an assumption about the length of the secret :|
The output of this is different for my throwaway and my actual phone number. So
there is a unique 10-digit secret. Let's try a two-step decode:
```
first_pass = ''.join(chr(i) for i in decode("[VW^QW\\W_X0", "7345839476"))
decode(first_pass, "00000000000")
[12, 5, 3, 11, 9, 4, 5, 3, 8, 14, 0, 0]
first_pass = ''.join(chr(i) for i in xor_cipher("[VW^QW\\W_X0", "7345839476"))
xor_cipher(first_pass, "00000000000")
[12, 5, 3, 11, 9, 4, 5, 3, 8, 14, 0]
```
Now this is interesting! The output of this is the same for both of my phone
@ -119,25 +113,19 @@ secret XOR phonenumber XOR plaintext = ciphertext
```
Let's verify:
```
def decode(cipher, phonenumber, secret):
cipher = get_stripped(cipher)
phonenumber = get_stripped(phonenumber)
def decode(cipher, phonenumber):
secret = [12, 5, 3, 11, 9, 4, 5, 3, 8, 14]
# remember that the cipher "passes through" digits past the 10th, so we
# just overwrite the first 10
text = [i^j^k for i, j, k in zip(cipher, phonenumber, secret)]
if len(cipher) > 10:
text += cipher[10:]
return text
first_pass = ''.join(chr(i) for i in xor_cipher(cipher, phonenumber))
return xor_cipher(first_pass, secret)
decode("[VW^QW\\W_X0", "7345839476")
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
decode("[WU]URZPWQ", "7345839476")
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
```
Beatiful! This yields the same result for both phone numbers.
Beatiful! This yields the same result for both phone numbers. Our secret is
[12, 5, 3, 11, 9, 4, 5, 3, 8, 14].
Let's validate this more completely:
```
@ -153,17 +141,17 @@ lookup_table = [
['S', '^', '_', 'V', 'Y', '_', 'T', '_', 'W', 'P'],
['R', '_', '^', 'W', 'X', '^', 'U', '^', 'V', 'Q'],
]
def validate_decode(table):
def validate_decode(table, phone):
for plaintext_char in range(10):
expected_plaintext = str(plaintext_char) * 10
ciphertext = ''.join([table[plaintext_char][i] for i in range(10)])
plaintext = ''.join([str(i) for i in decode(ciphertext, "7345839476")])
ciphertext = "".join([table[plaintext_char][i] for i in range(10)])
plaintext = "".join([str(i) for i in decode(ciphertext, phone)])
if plaintext != expected_plaintext:
print(f"Failed on \"{plaintext}\" != decode(\"{ciphertext}\", ...)")
print(f'Failed on "{plaintext}" != decode("{ciphertext}", ...)')
else:
print(f"Success! decode(\"{ciphertext}\", ...) == \"{plaintext}\"")
print(f'Success! decode("{ciphertext}", ...) == "{plaintext}"')
validate_decode(lookup_table)
validate_decode(lookup_table, "7345839476")
Success! decode("[VW^QW\W_X", ...) == "0000000000"
Success! decode("ZWV_PV]V^Y", ...) == "1111111111"
Success! decode("YTU\SU^U]Z", ...) == "2222222222"
@ -176,7 +164,8 @@ Success! decode("S^_VY_T_WP", ...) == "8888888888"
Success! decode("R_^WX^U^VQ", ...) == "9999999999"
```
Addendum: After writing the majority of this, [a user pointed
**Addendum**:
After writing the majority of this, [a user pointed
out](https://gitlab.com/LineageOS/issues/android/-/issues/6964#note_1961619585)
that there is already some documentation on this ciphering, so I took a look.
Unfortunately, it looks like the cipher method has changed as this does not
@ -213,3 +202,59 @@ secret XOR plaintext = ciphertext
phonenumber XOR plaintext = ciphertext
```
But neither worked.
There is a secret that works for decoding this, and we can find it by following
the same method from above, using the fact that
```
ciphertext XOR plaintext = secret
```
Let's try:
```
xor_cipher("XTQ^ZSUU_Y", "0000000000")
[8, 4, 1, 14, 10, 3, 5, 5, 15, 9]
```
And then validating it:
```
kop316_lookup_table = [
["X", "T", "Q", "^", "Z", "S", "U", "U", "_", "Y"],
["Y", "U", "P", "_", "[", "R", "T", "T", "^", "X"],
["Z", "V", "S", "\\", "X", "Q", "W", "W", "]", "["],
["[", "W", "R", "]", "Y", "P", "V", "V", "\\", "Z"],
["\\", "P", "U", "Z", "^", "W", "Q", "Q", "[", "]"],
["]", "Q", "T", "[", "_", "V", "P", "P", "Z", "\\"],
["^", "R", "W", "X", "\\", "U", "S", "S", "Y", "_"],
["_", "S", "V", "Y", "]", "T", "R", "R", "X", "^"],
["P", "\\", "Y", "V", "R", "[", "]", "]", "W", "Q"],
["Q", "]", "X", "W", "S", "Z", "\\", "\\", "V", "P"],
]
def old_decode(cipher):
cipher = get_stripped(cipher)
secret = [8, 4, 1, 14, 10, 3, 5, 5, 15, 9]
return xor_cipher(cipher, secret)
def validate_old_decode(table):
for plaintext_char in range(10):
expected_plaintext = str(plaintext_char) * 10
ciphertext = "".join([table[plaintext_char][i] for i in range(10)])
plaintext = "".join([str(i) for i in old_decode(ciphertext)])
if plaintext != expected_plaintext:
print(f'Failed on "{plaintext}" != decode("{ciphertext}", ...)')
else:
print(f'Success! decode("{ciphertext}", ...) == "{plaintext}"')
validate_old_decode(kop316_lookup_table)
Success! decode("XTQ^ZSUU_Y", ...) == "0000000000"
Success! decode("YUP_[RTT^X", ...) == "1111111111"
Success! decode("ZVS\XQWW][", ...) == "2222222222"
Success! decode("[WR]YPVV\Z", ...) == "3333333333"
Success! decode("\PUZ^WQQ[]", ...) == "4444444444"
Success! decode("]QT[_VPPZ\", ...) == "5555555555"
Success! decode("^RWX\USSY_", ...) == "6666666666"
Success! decode("_SVY]TRRX^", ...) == "7777777777"
Success! decode("P\YVR[]]WQ", ...) == "8888888888"
Success! decode("Q]XWSZ\\VP", ...) == "9999999999"
```
To reiterate, the secret used here was `[8, 4, 1, 14, 10, 3, 5, 5, 15, 9]`.