Computing the Möbius function

  1. CRGreathouse

    CRGreathouse 3,682
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    From the Mathematica documentation:
    SquareFreeQ[n] checks to see if n has a square prime factor. This is done by computing MoebiusMu[n] and seeing if the result is zero; if it is, then n is not squarefree, otherwise it is. Computing MoebiusMu[n] involves finding the smallest prime factor q of n. If n has a small prime factor (less than or equal to 1223), this is very fast. Otherwise, FactorInteger is used to find q.

    How does this work? I can't think of a way to compute mu(n) faster than:
    * finding a small p for which p^2 | n, or
    * finding that n is prime, or
    * checking that for all p < (n)^(1/3), p^2 does not divide n
    * factoring n, which can be faster than the third option for large n


    Edit: I checked my copy of "Open problems in number theoretic complexity, II", which has O8: Is SQUAREFREES in P? Assuming that this is still open, any algorithm known would either be superpolynomial or of unproven complexity. But even an exponential-time improvement on the naive algorithm would be of interest.
     
    Last edited: Sep 17, 2008
  2. jcsd
  3. CRGreathouse

    CRGreathouse 3,682
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    The interesting line, to me, is this:
    If n has a small prime factor (less than or equal to 1223), this is very fast.

    This seems a fairly strong claim. In particular, it would mean that the "very fast" algorithm could distinguish between n = 2pq and n = 2pq^2 for large p and q.

    I saw a reference to this at http://www.marmet.org/louis/sqfgap/ :
     
  4. Hurkyl

    Hurkyl 16,090
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    The implication cannot be true, because it would mean that if N doesn't have a small factor, that we could compute [itex]\mu(N) = -\mu(3N)[/itex], where the latter can be done rapidly.
     
  5. CRGreathouse

    CRGreathouse 3,682
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    Yeah, I figured as much, thanks.

    So factoring n is pretty much the fastest way to find mu(n), right? Except in the special cases I outlined above?
     
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