8 Nov
2007
8 Nov
'07
5:18 a.m.
You can make it pretty short too, if you allow yourself fix: rs=1:fix(\f p n->n++f(p++n)p)[1][0] I came up with this on the train home, but then I realised it was the same as your solution :) On 08/11/2007, at 12:57 PM, Alfonso Acosta wrote:
How about this,
infiniteRS :: [Int] infiniteRS = let acum a1 a2 = a2 ++ acum (a1++a2) a1 in 1 : acum [1] [0]
it certainly fits in one line but it's not really elegant _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe