Update for 15-12-21 15:00

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Tyler Perkins 2021-12-15 15:00:01 -05:00
parent f6c12a1e80
commit 276d18afa8

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@ -22,3 +22,42 @@ A *tap* is where a bit is read and fed back into itself.
An LFSR generates values based on a linear expression modulous 2, therefore we An LFSR generates values based on a linear expression modulous 2, therefore we
can reverse engineer the state of the LFSR based on a sequence we are given. can reverse engineer the state of the LFSR based on a sequence we are given.
This can be done using the Berlekamp-Massey algorithm. This can be done using the Berlekamp-Massey algorithm.
So first we will start with a simpler version. If we have a sequence and we
know the number of bits in the LSFR, we can create a matrix of the values.
If S_{i} is the _i_ th value out of an LSFR, we can solve the following
Sa = x
Where S is a matrix of the outputted values formatted below
A has the coefficents of the LFSR
and x has values of the bit string, as formatted below.
{{{
Assume 4 bits
-- ---- -- -- --
| s0 s1 s2 s3 || a0 | | s4 |
| s1 s2 s3 s4 || a1 | = | s5 |
| s2 s3 s4 s5 || a2 | | s6 |
| s3 s4 s5 s6 || a3 | | s7 |
-- ---- -- -- --
}}}
Note that
- The S matrix is _n_ bits squared
- All other matrices are _n_ tall
- You need 2*n - 1 sample bits
Given this, we can find the coefficents by solving
a = S^-1 * x
Once we do this, it will give us all of the coefficents! Everywhere there
is a 1 a tap will be located there and all of these values are XORed and placed
onto the back of the register.
To make this the Berlekamp-Massey algorithm, we first start and assume the
number of bits _n_ is 1, check if it makes the right seuqnece, and if not we
increase _n_ and try again. That all there is to it!