CRC reverse engineering
in [Cryptography]
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From: Peter Pearson on 3 May 2007 14:00 On Wed, 2 May 2007 17:16:06 +0000 (UTC), David Wagner wrote: > mvrpfswe wrote: >>There must be a more elegant way of figuring this out rather then a >>brute force, > > One conceptually simple thing you can try is linear algebra. > > Assume that each bit of the presumed-CRC-output can be written an > (unknown) linear function data bits; write down a system of linear > equations; and then use linear algrebra (Gaussian elimination) to > solve the system of linear equations and find the linear function > (if there is one) that describes how to compute that output bit as > a function of the input bits. Indeed, the 28 sample values you presented can all be written as linear combinations of 8 rows, each of which has just one 1-bit in the byte that varies: .. . . 01 . . . 5fd6 .. . . 02 . . . 5aab .. . . 04 . . . 5951 .. . . 08 . . . 5fa2 .. . . 10 . . . 5a43 .. . . 20 . . . 5a86 .. . . 40 . . . f4c7 .. . . 80 . . . b0da -- To email me, substitute nowhere->spamcop, invalid->net.
From: mvrpfswe on 3 May 2007 21:00 On May 3, 11:00 am, Peter Pearson <ppear...(a)nowhere.invalid> wrote: > On Wed, 2 May 2007 17:16:06 +0000 (UTC), David Wagner wrote: > > mvrpfswe wrote: > >>There must be a more elegant way of figuring this out rather then a > >>brute force, > > > One conceptually simple thing you can try is linear algebra. > > > Assume that each bit of the presumed-CRC-output can be written an > > (unknown) linear function data bits; write down a system of linear > > equations; and then use linear algrebra (Gaussian elimination) to > > solve the system of linear equations and find the linear function > > (if there is one) that describes how to compute that output bit as > > a function of the input bits. > > Indeed, the 28 sample values you presented can all be written as > linear combinations of 8 rows, each of which has just one 1-bit > in the byte that varies: > > . . . 01 . . . 5fd6 > . . . 02 . . . 5aab > . . . 04 . . . 5951 > . . . 08 . . . 5fa2 > . . . 10 . . . 5a43 > . . . 20 . . . 5a86 > . . . 40 . . . f4c7 > . . . 80 . . . b0da > > -- > To email me, substitute nowhere->spamcop, invalid->net. This certainly solves the problem for the LSB of the serial to the output. However since the rank of the system is somewhat poor only 8 bits changing out of 48 it will not catch any other bit-errors then in the LSB (serial). Now the problem reduces (?) to how the data CRC changes in each state such that the solution can be expanded to the entire data set. Any ideas? The starting CRC will probably be constant and non zero at the end of byte 3 and also byte 5 and 6 will most likely alter the data in a constant way I believe? I also did manage to find a marker with a different type then the previous ones: 0x00, 0x00, 0x3C, 0x17, 0x20, 0x65, 0x11, 0xCE -Johan
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