2021-12-15 19:00:01 +00:00
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= Linear Feedback Shift Register =
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A LFSR is set of rules to alter a set of bits. They are useful to psudeo random
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number generators, and as key generators for stream ciphers.
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All LFSRs are cyclical in nature, and after a set amount of time will repeat
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2021-12-15 19:15:01 +00:00
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back into themselves. The initial state of the bits in the LFSR is called the
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2021-12-15 19:00:01 +00:00
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seed.
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2022-04-06 15:45:01 +00:00
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The maximum period for a _n_ bit shift register is
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2^n - 1
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2021-12-15 19:45:01 +00:00
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An LFSR can be generalized as a recurrence relationship where
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- The preceding terms are not raised to a power
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- There are no added constants
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2021-12-15 19:00:01 +00:00
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A *tap* is where a bit is read and fed back into itself.
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2021-12-15 19:30:01 +00:00
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== Reverse Engineering ==
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An LFSR generates values based on a linear expression modulous 2, therefore we
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can reverse engineer the state of the LFSR based on a sequence we are given.
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This can be done using the Berlekamp-Massey algorithm.
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2021-12-15 20:00:01 +00:00
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So first we will start with a simpler version. If we have a sequence and we
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know the number of bits in the LSFR, we can create a matrix of the values.
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If S_{i} is the _i_ th value out of an LSFR, we can solve the following
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Sa = x
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Where S is a matrix of the outputted values formatted below
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A has the coefficents of the LFSR
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and x has values of the bit string, as formatted below.
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{{{
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Assume 4 bits
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-- ---- -- -- --
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| s0 s1 s2 s3 || a0 | | s4 |
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| s1 s2 s3 s4 || a1 | = | s5 |
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| s2 s3 s4 s5 || a2 | | s6 |
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| s3 s4 s5 s6 || a3 | | s7 |
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-- ---- -- -- --
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}}}
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Note that
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- The S matrix is _n_ bits squared
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- All other matrices are _n_ tall
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- You need 2*n - 1 sample bits
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Given this, we can find the coefficents by solving
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a = S^-1 * x
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Once we do this, it will give us all of the coefficents! Everywhere there
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is a 1 a tap will be located there and all of these values are XORed and placed
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onto the back of the register.
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To make this the Berlekamp-Massey algorithm, we first start and assume the
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number of bits _n_ is 1, check if it makes the right seuqnece, and if not we
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increase _n_ and try again. That all there is to it!
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