On Sun, 2008-07-13 at 20:24 +0200, Logesh Pillay wrote: > I know we can perform memoization in Haskell. The well known recursive > Fibonacci example works v. well. f(10000) returns a virtually instant > answer which would not be possible otherwise. > > My (probably naive) function to give the number of partitions of a number :- > p = ((map p' [0 .. ]) !!) > where > p' 0 = 1 > p' 1 = 1 > p' n = sum [(((-1) ^ (k+1)) * ( p' (n-((k*(3*k-1)) `div` 2)) + p' > (n-((k*(3*k+1)) `div` 2)))) | k <- [1 .. n]] > > It is an attempt to apply the Euler recurrence formula (no 11 in > http://mathworld.wolfram.com/PartitionFunctionP.html ) > > It works but it is shockingly slow. It is orders of magnitude slower > than the Python memoized version which runs very fast. > parts = {0:1, 1:1} > def P(n): > if not n in parts: > parts[n] = sum ([( ((-1) ** (k+1)) * ( P(n-((k*(3*k-1))//2)) + > P(n-((k*(3*k+1))//2)) ) ) for k in xrange (1, n+1)]) > return parts[n] > > Why? Its as if memoization is being ignored in the haskell version.
That's because you aren't using it. > How to fix? Use your memoized function. _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe