On 11/2/07, Luke Palmer <[EMAIL PROTECTED]> wrote:
> On 11/1/07, Arnar Birgisson <[EMAIL PROTECTED]> wrote:
> > dnf :: LS -> DNF
> > dnf (Var s) = [[Pos s]]
> > dnf (Or l1 l2) = (dnf l1) ++ (dnf l2)
> > dnf (And l1 l2) = [t1 ++ t2 | t1 <- dnf l1, t2 <- dnf l2]
> > dnf (Not (Not d)) = dnf d
> > dnf
On 11/1/07, Arnar Birgisson <[EMAIL PROTECTED]> wrote:
> I'm learning too and found this an interesting problem. Luke, is this
> similar to what you meant?
Heh, your program is almost identical to the one I wrote to make sure
I wasn't on crack. :-)
> data LS = Var String | Not LS | And LS LS | O
Hey folks,
On Nov 1, 2007 6:41 PM, Luke Palmer <[EMAIL PROTECTED]> wrote:
> A good way to approach this is data-structure-driven programming. You
> want a data structure which represents, and can _only_ represent,
> propositions in DNF. So:
>
> data Term = Pos Var | Neg Var
> type Conj = [Term]
It's much much easier to work with n-ary than binary.
It's also easier to define disjunctive normal form by mutual recursion
with conjunctive normal form.
Jules
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Jim,
Lukes suggestion is a good one, and should help focus you on the
syntactic constraints of DNF. A property that your dnf function should
have is that the right-hand side of each case should yield a DNF
formula. Take, for example,
dnf (And s1 s2) = And (dnf s1) (dnf s2)
Does And'ing t
A good way to approach this is data-structure-driven programming. You
want a data structure which represents, and can _only_ represent,
propositions in DNF. So:
data Term = Pos Var | Neg Var
type Conj = [Term]
type DNF = [Conj]
Then write:
dnf :: LS -> DNF
The inductive definition of dnf is
I am trying to rewrite sentences in a logical language into DNF, and wonder
if someone would point out where I'm going wrong. My dim understanding of it
is that I need to move And and Not inwards and Or out, but the function
below fails, for example:
> dnf (Or (And A B) (Or (And C D) E))
