de_F
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« on: December 05, 2018, 10:20:20 AM »



I was always wondering how people reversed algorithms from matching seed/keys and my curiosity got the better of me.
I managed to deduce the Level 1 security access key (the one used for writing a flash on the ECU for example)
Using most of the information found on this site, I gave it a shot. I made an Arduino ECU simulator that replied to the basic KWP requests from a tool. I based most of this work on information that I did find on this forum. Work done on the MED9.1 by Basano.
Most tools start by asking some basic KWP things like Hardware and software numbers and so on. I used the Arduino to feed a few messages to keep the tool happy, up until the moment the tool wants to open a development session, followed by the seed request. It then receives a seed, calculates, using the algorithm, what the reply should be, then compares it to the reply given by the tool. If this matches, access to that security level is given. There are multiple security access levels for info.
But lets stick to the reversing strategy that I tried. Given the fact that I was using an ECU simulator, I could feed the tool any seedkey I wished. so I started off with a simple seed to try and understand its working.
I got the following matching seed/keys back
seed 01 01 01 01 key 20 20 20 20
seed 02 02 02 02 key 40 40 40 40
seed 03 03 03 03 key 60 60 60 60
so there is definitely a pattern here. lets try and work this out a bit more. to start we turn the Hex into simple bits.
seed 01010101: 00000001 00000001 00000001 00000001 reply 20202020: 00100000 00100000 00100000 00100000
From this we can see what is happening. The bits are carried 5 spaces to the left.
seed 02 02 02 02 00000010 00000010 00000010 00000010 key 40 40 40 40 01000000 01000000 01000000 01000000
Same thing is happening here and on the other ones aswel. So lets try one where the bit will carry over when it moves left 5 times.
seed 08 00 00 00 00001000 00000000 00000000 00000000 shift it 5 spaces to the left and we get shifted : 00000000 00000000 00000000 00000001 reply 0A221288 00001010 00100010 00010010 10001000
Quite a bit different than what we expected. So we are dealing with some modifier. Out of reading here, I found out it was possibly a XOR so here we go :
from the above, we can conclude that the hex used for the XOR could be : 00001010 00100010 00010010 10001001 , which gives us 0A 22 12 89 in hex. A lookup in my BIN brings this up aswel. But more testing is needed.
FF FF 20 5B D9 C6 03 00 70 00 F0 00 00 01 01 00 02 00 80 00 40 07 E8 03 05 00 0F 00 02 05 40 02 02 01 00 01 01 10 00 04 24 10 68 05 81 4A 05 87 0A 22 12 89 49 4C FF 0F F0 00 B0 40 BF 4F E9 19 E6 16 A6 56 A9 59 75 85 7A 8A 3A CA 35 C5 63 93 6C 9C 2C DC 23 D3 D2 22 DD 2D 9D 6D 92 62 C4 34 CB 3B 8B 7B 84 74 58 A8 57 A7 17 E7 18 E8 4E BE
After a bit more trial and error, I found out that this was indeed correct, but that the carry over (rotating) bits are used to offset this hex value
Example, if after shifting it to the left 5 times. the result ends in xxxxxxxx xxxxxxxx xxxxxxxx xxx00001 then the hex for the XOR is 00001010 00100010 00010010 10001001 xxxxxxxx xxxxxxxx xxxxxxxx xxx00010 the hex for the XOR moves 1 to the left and becomes 00010100 01000100 00100101 00010010 xxxxxxxx xxxxxxxx xxxxxxxx xxx00100 the hex for the XOR moves 2 to the left and becomes 00101000 10001000 01001010 00100100 xxxxxxxx xxxxxxxx xxxxxxxx xxx01000 the hex for the XOR moves 3 to the left and becomes 01010001 00010000 10010100 01001000 xxxxxxxx xxxxxxxx xxxxxxxx xxx10000 the hex for the XOR moves 4 to the left and becomes 10100010 00100001 00101000 10010000
Now that we have this information we can calculate the correct response for each Seed ...
Example :
Seed 41 20 90 C8 in bits is 0100 0001 0010 0000 1001 0000 1100 1000 we rotate it 0010 0100 0001 0010 0001 1001 0000 1000 because it ends in 01000 the Hex will be 0101 0001 0001 0000 1001 0100 0100 1000
then we XOR it and we get the correct response which will be 0010 0100 0001 0010 0001 1001 0000 1000 0101 0001 0001 0000 1001 0100 0100 1000 0111 0101 0000 0010 1000 1101 0100 0000 in hex that gives us 75 02 8D 40 , which gave a correct response on my Bench ECU.
Hope this helps someone at some point.
