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config.toml

@@ -30,8 +30,19 @@ enableGitInfo = true
 
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+  anchorLineNos = false
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+  guessSyntax = true
+  hl_Lines = ''
+  hl_inline = false
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+  noClasses = true
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+  tabWidth = 4
 
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content/posts/cracking_att_vvm_cipher.md

@@ -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"