[Haskell-cafe] Boolean formula: part2
Dmitry Bogatov
KAction at gnu.org
Wed Sep 17 20:00:06 UTC 2014
* Dmitry Bogatov <KAction at gnu.org> [2014-09-17 23:48:49+0400]
Sorry for followup, I found that problem is not in this function per-se,
but in filtering one. Since I require formula to be complicated enough,
I have to filter out simple ones, and simple ones gets constructed
before any more complicated. I get all formulas only about P before
it even touches Q. How can I speed things up?
-- Set of hacks, that tries to formalize my opinion what
-- good formula is.
goodFormula :: Formula -> Bool
goodFormula = \case
(Function2 t x y) -> x /= y && goodFormula x && goodFormula y
&& case (x, y) of
(Negation u, Negation v) -> u /= v
(Negation u, y) -> u /= y
(x, Negation v) -> x /=v
(Function2 t' u' v', _) -> t' /= t' || ((y /= u') && (y /= v'))
_ -> True
(Negation x) -> goodFormula x
_ -> True
flenght :: Formula -> Int
flenght fmt = case value fmt of
Right _ -> 0
Left l -> length (toList l)
main = do
-- mapM_ print $ take (buildFormulas [P, Q, R] 7)
print $ filter (\f -> goodFormula f && flenght f == 3) (buildFormulas [P, Q, R] 4) !! 1
-- print $ (buildFormulas [P, Q, R] 5) !! 1
> Hello! I have following code:
>
> buildFormulas :: [Variable] -> Int -> [Formula]
> buildFormulas vars 0 = map Var vars
> buildFormulas vars n = join $
> forM [0 .. n - 1 ] $ \i -> do
> x <- buildFormulas vars i
> y <- buildFormulas vars (n - i - 1)
> z <- buildFormulas vars (n - 1)
> let t1 = case z of
> (Negation _) -> []
> _ -> [Negation z]
> t1 ++ [conj x y, impl x y, disj x y]
>
> It is correct, typechecks, but it is insanely slow (calculating
> take 35 (buildFormulas [P, Q, R] 5) hangs my computer).
>
> My guess is that it builds too much before returning first results.
> Can you suggest more efficent way enumerate formulas of set of
> variables?
--
Best regards, Dmitry Bogatov <KAction at gnu.org>,
Free Software supporter, esperantisto and netiquette guardian.
GPG: 54B7F00D
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