Tlemcen Car electronics - Afficher un message - [Theorie] Calcul du ISK vers PIN pour Megane 2

ebt255 · 05/06/2017, 14h16

Following the MrRorry I decided to do the same for Laguna 2ph2.

eg.1
So, we have ISK: FA2B6C787800

In binary it looks:
11111010 00101011 01101100 01111000 01111000 00000000
------FA-- -----2B--- -----6C--- -----78--- -----78--- -----00---

Now we have to count the ones (number 1 in binary): 22
Next circular shift 22 (count value) bits in right.

----------------------11111010001010110110110001 >> 1110000111100000000000
111000011110000000000011111010001010110110110001

after that bianry is: 111000011110000000000011111010001010110110110001
Next if count value (22) is even then we must xor evry bits by 1.

11100001 11100000 00000011 11101000 10101101 10110001
11111111 11111111 11111111 11111111 11111111 11111111 XOR

00011110 00011111 11111100 00010111 01010010 01001110 Result OK.
------1E-- -----1F--- -----FC--- -----17--- -----52--- -----4E--- PIN

I thought that was all but..
eg.2

ISK: FF825EA08C2E

111111111000001001011110101000001000110000101110

count value: 24
so, circular shift 24

------------------------111111111000001001011110 >> 101000001000110000101110
101000001000110000101110111111111000001001011110

then XOR

10100000 10001100 00101110 11111111 10000010 01011110
11111111 11111111 11111111 11111111 11111111 11111111 XOR

01011111 01110011 11010001 00000000 01111101 10100001 Result - WRONG!
------5F-- -----73--- -----D1--- -----00--- -----7D--- -----A1---

When count value is 24 and only then you must change it to 17.

circular shift 17
-----------------1111111110000010010111101010000 >> 01000110000101110
010001100001011101111111110000010010111101010000

Not XOR because 17 is not even.

01000110 00010111 01111111 11000001 00101111 01010000 Result OK.
------46-- -----17--- -----7F--- -----C1--- -----2F--- -----50--- PIN

I hope that is all.

05/06/2017, 14h16
ebt255 Débutant T-E Date d'inscription: juillet 2016 Localisation: Poland, Warsaw Âge: 39 Messages: 71 Thanks: 135 Thanked 167 Times in 42 Posts Pouvoir de réputation: 12	Following the MrRorry I decided to do the same for Laguna 2ph2. eg.1 So, we have ISK: FA2B6C787800 In binary it looks: 11111010 00101011 01101100 01111000 01111000 00000000 ------FA-- -----2B--- -----6C--- -----78--- -----78--- -----00--- Now we have to count the ones (number 1 in binary): 22 Next circular shift 22 (count value) bits in right. ----------------------11111010001010110110110001 >> 1110000111100000000000 111000011110000000000011111010001010110110110001 after that bianry is: 111000011110000000000011111010001010110110110001 Next if count value (22) is even then we must xor evry bits by 1. 11100001 11100000 00000011 11101000 10101101 10110001 11111111 11111111 11111111 11111111 11111111 11111111 XOR 00011110 00011111 11111100 00010111 01010010 01001110 Result OK. ------1E-- -----1F--- -----FC--- -----17--- -----52--- -----4E--- PIN I thought that was all but.. eg.2 ISK: FF825EA08C2E 111111111000001001011110101000001000110000101110 count value: 24 so, circular shift 24 ------------------------111111111000001001011110 >> 101000001000110000101110 101000001000110000101110111111111000001001011110 then XOR 10100000 10001100 00101110 11111111 10000010 01011110 11111111 11111111 11111111 11111111 11111111 11111111 XOR 01011111 01110011 11010001 00000000 01111101 10100001 Result - WRONG! ------5F-- -----73--- -----D1--- -----00--- -----7D--- -----A1--- When count value is 24 and only then you must change it to 17. circular shift 17 -----------------1111111110000010010111101010000 >> 01000110000101110 010001100001011101111111110000010010111101010000 Not XOR because 17 is not even. 01000110 00010111 01111111 11000001 00101111 01010000 Result OK. ------46-- -----17--- -----7F--- -----C1--- -----2F--- -----50--- PIN I hope that is all.