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Re: [PrimeNumbers] Does Fourier transform help with exhaustive subtractions?

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  • Peter Kosinar
    Hello Kaveh, ... Consider the sets A = { N, 2N, 3N, ... N^2 }, B = { 0, 1, 2, ... N-1 }. The set obtained by subtracting each element of B from each element of
    Message 1 of 5 , Nov 9, 2006
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      Hello Kaveh,

      > I have two sets (A and B) of n positive integers each, and I need to
      > subtract each element of B from each element of A. Obviously, the
      > computation runs in O(n^2).
      >
      > I was wondering if fast Fourier transfrom can be used to reduce the
      > time complexity of this task.

      Consider the sets A = { N, 2N, 3N, ... N^2 }, B = { 0, 1, 2, ... N-1 }.
      The set obtained by subtracting each element of B from each element of A
      is { 1, 2, 3, ... N^2 }. In other words, the size of the resulting set can
      be quadratic and thus the worst-case runtime of any algorithm producing
      it is at least quadratic [*].

      Thus, without having some additional restrictions on the structure of the
      elements in those sets, the best you can hope for is a quadratic
      algorithm.

      Or did I misunderstood your problem statement?

      Peter

      [*] Assuming you're representing the set as an array of numbers.

      --
      [Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ] 134813278
    • Joshua Zucker
      ... I think he s talking about making the nxm table of all the differences of (an element of A) - (an element of B). What I wonder about is what you want to do
      Message 2 of 5 , Nov 9, 2006
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        On 11/9/06, David Cleaver <wraithx@...> wrote:
        > Anyway, wouldn't this be an O(2n-1) = O(2n) = O(n) operation? You could
        > add up all elements of B and then subtract that sum from each element of
        > A. Or are there additional restrictions you're not telling us about?

        I think he's talking about making the nxm table of all the differences
        of (an element of A) - (an element of B).

        What I wonder about is what you want to do with that table: sometimes
        it's O(n+m) if you just, say, want the total; O(nm) if you want all
        the elements; O(1) (but O(min(n,m)) for lookup) if you just want to
        check whether a few particular numbers occur or not ...

        --Joshua Zucker
      • Joshua Zucker
        ... Hi Kaveh, I think you misunderstood Peter s point (quoted above). With fast multiplication, n digits times m digits, the result will only have n+m digits,
        Message 3 of 5 , Nov 9, 2006
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          On 11/9/06, Peter Kosinar <goober@...> wrote:
          > Consider the sets A = { N, 2N, 3N, ... N^2 }, B = { 0, 1, 2, ... N-1 }.
          > The set obtained by subtracting each element of B from each element of A
          > is { 1, 2, 3, ... N^2 }. In other words, the size of the resulting set can
          > be quadratic and thus the worst-case runtime of any algorithm producing
          > it is at least quadratic [*].

          Hi Kaveh,
          I think you misunderstood Peter's point (quoted above).

          With fast multiplication, n digits times m digits, the result will
          only have n+m digits, so Peter's argument says it takes at least
          O(n+m) time just to store the resulting bignum! And indeed FFT mult
          takes O(n ln n) which in general is way more than 2n.

          With your subtraction problem, n numbers minus each of m numbers, the
          result can have nm numbers, so he argues that since storing each
          number takes O(1) time, the whole computation must take at least O(nm)
          time. I think that argument sounds very convincing to me!

          My solution, which takes O(1) time, depends on not storing all the
          numbers, but rather on doing an O(n+m) lookup each time you want to
          check if a number is in the set "A-B" that you want. Peter already
          anticipated that, though, in his footnote which reminds us that the
          O(nm) minimum is only true if you want to store all the nm
          differences.

          So, if you really need all the numbers in a list, Peter is right and
          the minimum time is O(nm). If you can use some other way of
          remembering the numbers, or if you know some other properties of the
          numbers (like, say, the maximum is a lot less than nm, so you know
          there can't be nm different differences among them) then there might
          be an algorithm that exploits those other possibiities.

          --Joshua Zucker
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