[Haskell-cafe] lazy evaluation

[Josef Svenningsson]

On Feb 6, 2008 3:06 PM, Miguel Mitrofanov <miguelimo38 at yandex.ru> wrote:
>
> On 6 Feb 2008, at 16:32, Peter Padawitz wrote:
>
> > Can anybody give me a simple explanation why the second definition
> > of a palindrome checker does not terminate, although the first one
> > does?
> >
> > pal :: Eq a => [a] -> Bool
> > pal s = b where (b,r) = eqrev s r []
> >
> > eqrev :: Eq a => [a] -> [a] -> [a] -> (Bool,[a])
> > eqrev (x:s1) ~(y:s2) acc = (x==y&&b,r) where (b,r) = eqrev s1 s2
> > (x:acc)
> > eqrev _ _ acc                  = (True,acc)
>
> I.    eqrev "" (_|_) acc = (True, acc)
> II.a. eqrev "1" (_|_) "" = ('1' == (_|_) && b, r) where (b,r) = eqrev
> "" (_|_) "1"
>        By (I), (b,r) = (True, "1"), so eqrev "1" (_|_) "" = ((_|_),"1")
> II.b. eqrev "1" "1" "" = ('1' == '1' && b, r) where (b,r) = eqrev ""
> "" "1"
>        (b,r) = (True,"1"), so eqrev "1" "1" "" = (True,"1")
>
> Therefore, the least fixed point of \r -> eqrev "1" r "" is "1" and
> the answer is True.
>
> > pal :: Eq a => [a] -> Bool
> > pal s = b where (b,r) = eqrev' s r
> >
> > eqrev' :: Eq a => [a] -> [a] -> (Bool,[a])
> > eqrev' (x:s1) ~(y:s2) = (x==y&&b,r++[y]) where (b,r) = eqrev' s1 s2
> > eqrev' _ _                   = (True,[])
>
> I.  eqrev' "" (_|_) = (True,[])
> II.a. eqrev' "1" (_|_) = ('1' == (_|_) && b, r ++ [(_|_)]) where (b,r)
> = eqrev' "" (_|_)
>        By (I), (b,r) = (True,[]), so eqrev' "1" (_|_) = ((_|_),[(_|_)])
> II.b. eqrev' "1" [(_|_)] = ('1' == (_|_) && b, r ++ [(_|_)]) where
> (b,r) = eqrev' "" []
>        (b,r) = (True,[]), so eqrev' "1" [(_|_)] = ((_|_),[(_|_)])
> Therefore, the least fixed point of \r -> eqrev' "1" r is [(_|_)] and
> the answer is (_|_). No wonder it hangs.
>
This proof also shows us where the problem lies and how to fix it. It
turns out to be really easy: change 'r++[y]' to 'r++[x]' and the
program works.

Cheers,

Josef