Is it possible that the BMA detects an irreducible polynomial from a sequence that was not generated by an LFSR? I am feeding a sequence into the BMA under the assumption that it was generated by an LFSR. It detects a polynomial of a certain length, but the sequence can’t be reconstructed from that polynomial. I don’t want to assume that the implementation of the BMA has a bug. If the question above cannot be answered with yes in any case, the implementation must be incorrect.


1 Answer 1



Given an arbitrarily long sequence $(x_1,\ldots,x_t,\ldots)$ consider its initial segments.

Any (finite) sequence $$x^{(n)}:=(x_1,\ldots,x_n)$$can be generated by a circulating LFSR of the same length. Just cycle the bits, no taps.

BMA will detect this if no shorter LFSR exists that generates $x^{(n)}$. And will converge to correctly detect the shortest LFSR that uniquely generates $x^{(n)}$ if you feed it twice as many initial bits $x^{(2n)}.$

Fix your code.

  • $\begingroup$ Thanks. At least I don’t have to work on assumptions now $\endgroup$
    – neolith
    Commented Aug 7, 2021 at 21:00

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