enable syntax highlighting

This commit is contained in:
Joel Beckmeyer 2024-07-05 15:20:58 -04:00
parent 9c3d6a8748
commit b6922dcbd9
2 changed files with 33 additions and 16 deletions

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@ -30,8 +30,19 @@ enableGitInfo = true
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guessSyntax = true
hl_Lines = ''
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lineAnchors = ''
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@ -7,7 +7,7 @@ draft: false
I've been wanting to get the AT&T visual voicemail "protocol" (ADVVM) working
in the LineageOS dialer. I thought I had made a breakthrough with the discovery
of the prefix in front of the VVM mail server address:
```
```url
srv=2:vvm.mobile.att.net
```
[Another user raised an issue pointing to the same
@ -21,7 +21,7 @@ STATUS SMS.
This brings us to the subject of the mysterious data SMS coming in on port 5499
that I have wondered about ever since I discovered them in the logs when first
"implementing" ADVVM. They can be triggered by sending a message of this format:
```
```url
GET?c=ATTV:<device name>/<android short version>:<app version>&v=1.0&l=<10-digit phone number>&AD
```
These SMS seemingly contain everything *useful* that the STATUS SMS contains,
@ -54,7 +54,7 @@ how?
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:
```
```python
def get_stripped(text):
return [ord(c) & 0x0f for c in text]
```
@ -70,7 +70,7 @@ Duh! XOR is a reversible, non-destructive operator that [works great for
ciphering](https://en.m.wikipedia.org/wiki/XOR_cipher).
Let's try it:
```
```python
def xor_cipher(cipher, secret):
if isinstance(cipher[0], str):
cipher = get_stripped(cipher)
@ -92,13 +92,13 @@ Well, this doesn't quite work. When run with the ciphertext and phone number,
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:
```
```python
xor_cipher("[VW^QW\\W_X0", "00000000000")
[11, 6, 7, 14, 1, 7, 12, 7, 15, 8, 0]
```
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:
```
```python
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]
@ -107,11 +107,11 @@ xor_cipher(first_pass, "00000000000")
Now this is interesting! The output of this is the same for both of my phone
lines. Additionally, it is identical to the output from my initial decode
above. I think we've found a secondary secret, meaning the algorithm is:
```
```pseudocode
secret XOR phonenumber XOR plaintext = ciphertext
```
Let's verify:
```
```python
def decode(cipher, phonenumber):
secret = [12, 5, 3, 11, 9, 4, 5, 3, 8, 14]
@ -127,7 +127,7 @@ 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:
```
```python
lookup_table = [
['[', 'V', 'W', '^', 'Q', 'W', '\\', 'W', '_', 'X'],
['Z', 'W', 'V', '_', 'P', 'V', ']', 'V', '^', 'Y'],
@ -151,6 +151,8 @@ def validate_decode(table, phone):
print(f'Success! decode("{ciphertext}", ...) == "{plaintext}"')
validate_decode(lookup_table, "7345839476")
```
```stdout
Success! decode("[VW^QW\W_X", ...) == "0000000000"
Success! decode("ZWV_PV]V^Y", ...) == "1111111111"
Success! decode("YTU\SU^U]Z", ...) == "2222222222"
@ -170,7 +172,7 @@ 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
work for me. I verified that the number and lookup table do not work with my
decode as well:
```
```python
kop316_lookup_table = [
[ 'X', 'T', 'Q', '^', 'Z', 'S', 'U', 'U', '_', 'Y' ],
[ 'Y', 'U', 'P', '_', '[', 'R', 'T', 'T', '^', 'X' ],
@ -184,6 +186,8 @@ kop316_lookup_table = [
[ 'Q', ']', 'X', 'W', 'S', 'Z', '\\', '\\', 'V', 'P' ],
]
validate_decode(kop316_lookup_table, "2065550100")
```
```stdout
Failed on "6140620777" != decode("XTQ^ZSUU_Y", ...)
Failed on "7051731666" != decode("YUP_[RTT^X", ...)
Failed on "4362402555" != decode("ZVS\XQWW][", ...)
@ -196,7 +200,7 @@ Failed on "14912814108151515" != decode("P\YVR[]]WQ", ...)
Failed on "15813915119141414" != decode("Q]XWSZ\\VP", ...)
```
I also tried solving with both of these alternatives instead with no luck:
```
```pseudocode
secret XOR plaintext = ciphertext
phonenumber XOR plaintext = ciphertext
```
@ -204,17 +208,17 @@ 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
```
```pseudocode
ciphertext XOR plaintext = secret
```
Let's try:
```
```python
xor_cipher("XTQ^ZSUU_Y", "0000000000")
[8, 4, 1, 14, 10, 3, 5, 5, 15, 9]
```
And then validating it:
```
```python
kop316_lookup_table = [
["X", "T", "Q", "^", "Z", "S", "U", "U", "_", "Y"],
["Y", "U", "P", "_", "[", "R", "T", "T", "^", "X"],
@ -245,6 +249,8 @@ def validate_old_decode(table):
print(f'Success! decode("{ciphertext}", ...) == "{plaintext}"')
validate_old_decode(kop316_lookup_table)
```
```stdout
Success! decode("XTQ^ZSUU_Y", ...) == "0000000000"
Success! decode("YUP_[RTT^X", ...) == "1111111111"
Success! decode("ZVS\XQWW][", ...) == "2222222222"