from array update algorithm to nice Haskell code
Wolfgang Jeltsch
wolfgang at jeltsch.net
Sat Jan 10 02:01:23 EST 2004
Am Freitag, 9. Januar 2004 23:45 schrieb Daan Leijen:
> > Now assume that I have a set a with O(n) elements and a set b with O(1)
> > elements.
>
> You can't have "O(1) elements" ... (A bound like O(..) talks about the worst
> case time/space of an operation.)
Well, the O calculus isn't bound to resource analysis. According to [1] O is
just something like
O :: (Natural -> Natural) -> Set (Natural -> Natural)
O f = {g | exists c. exists n0. forall n >= n0. g n <= c * f n}.
Based on this definition, I mean that I have a quantity n (e.g. the size of a
base set), and the size of the set a is less than c * n for some constant c
while the size of b is always less than a constant c'.
> > a `minusSet` b would take O(n) time and so would a `intersect` b.
> > If I'd use
> > foldr (flip delFromSet) a (setToList b)
> > and
> > [x | x <- setToList b, x `elementOf` a]
> > instead of a `minusSet` b and a `intersect` b, I would get away with
> > O(log n) time.
>
> I think that you are mixing up worst case bounds O(..) with some specific
> case. I guess that you mean by "O(1) elements" a known and constant number
> of elements.
I mean a constant maximum number of elements. An example would be that the
set b is guaranteed to have not more than one element.
> However, in the worst case, this will degenerate to some "m" number of
> elements, and your function would take O(m*log n) time, i.e. much worse than
> O(n+m).
I don't understand this. If m doesn't depend on any input values but is
really constant then O(m * log n) = O(log n)—regardless of how large m is.
Well, the specification of O(n + m) time does only say that the time is less
than c * (n + m) of some constant c which is no contradiction to an O(log n)
time. But it leaves the possibility open that the calculation really needs
linear time.
> All the best,
> Daan.
> [...]
Wolfgang
[1] Schöning, Theoretische Informatik – kurzgefasst, Spektrum Akademischer
Verlag
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