Research, development and trades concerning the powerful Proxmark3 device.
Remember; sharing is caring. Bring something back to the community.
"Learn the tools of the trade the hard way." +Fravia
You are not logged in.
Time changes and with it the technology
Proxmark3 @ discord
Users of this forum, please be aware that information stored on this site is not private.
Okay, I got another question now.
I snooped some communication between card and reader. Attached part of it below.
+ 27716: : 52
+ 64: 0: TAG 04 00
+ 1200: : 93 20
+ 64: 0: TAG 3b 4c ca d6 6b
+ 3022: : 93 70 3b 4c ca d6 6b fc 64
+ 64: 0: TAG 08 b6 dd
+ 2272: : 30 02 10 8b
+ 72: 0: TAG 04
+ 1376: : 52
+ 64: 0: TAG 04 00
+ 1736: : 93 70 3b 4c ca d6 6b fc 64
+ 64: 0: TAG 08 b6 dd
+ 5440: : 60 00 f5 7b
+ 112: 0: TAG 10 18 d6 ed
+ 1448: : da ff c0 a2 74 24 8f 45 !crc
+ 63: 0: TAG 0f! f3! 81! 13
+ 1408: : 01 ab d2 58 !crc
+ 72: 0: TAG a0! d2! 34 df 15! a7! 06! 4e! 15 3f! c7! d7 79! 91! 5d! 2b! 96 c4! !crc
+ 3768: : 2e d7 61 b6 !crc
+ 72: 0: TAG 0a! e5! 5a 79! 00! 73! f1 5d d5 02! df e4! 63! 17 fa! 59! 06 cd! !crcNow I want to decrypt some encrypted traces, like 01 ab d2 58 !crc.
Using simple Crapto1 application code(like test2.c in proxmark3 manual) I got Key, ks2, ks3, ks4, etc.
I can decrypt the first command after authentication(in here, 01 ab d2 58 !crc) using ks4 but,
still not clear how to decrypt 18-byte response of card, and followings of that.
I think I need a 18 -byte sized ks5, but cannot find how.
In conclusion, how can I create some codes that decrypts all commands including 18-byte sized data or 4-bit ACK/NACK?
Do I need an additional library with Crapto1?
Last edited by sikinmettugi (2010-08-03 06:48:53)
Offline
crypto1_word gets 4 byte of keystream at a time
crypto1_byte gets 1 byte of keystream at a time
crypto1_bit gets 1 bit of keystream at a time
somehow i think you should be able to manage to get 18 bytes of keystream from the cipher and xor them against the 18bytes of data you have.
Offline
So, concatenating 4 times of crypto1_word(revstate, 0, 0) and 2 crypto1_byte(revstate, 0, 0) would be the solution, right?
Last edited by sikinmettugi (2010-08-04 00:54:17)
Offline
four times four plus two is indeed eighteen
you may be a genius yet.
Offline