[Haskell-cafe] Programming Chalenges: The 3n+1 problem
Sebastian Fischer
fischer at nii.ac.jp
Thu Apr 14 15:43:19 CEST 2011
Hi Dimitri,
> When asking "how to implement cache in Haskell" I was hopping that there
> exists some solution without using Data.Array, more "functional" approach,
> if I may say so ...
Steven's second solution is purely functional. It uses so-called tries
to cache results instead of mutable arrays. Here is an alternative
definition of Steven's `fixmemo` function:
fixmemo :: HasTrie a => ((a -> b) -> (a -> b)) -> (a -> b)
fixmemo f = fix (memo . f)
It uses the standard fixpoint combinator
Data.Function.fix :: (a -> a) -> a
and has a similar (but more restrictive) type. In order to understand
Steven's solution you need to know how to define recursive functions
using a fixpoint combinator. For example, you can define the Fibonacci
function as follows:
fibonacci :: Int -> Integer
fibonacci = fix fib
fib :: (Int -> Integer) -> (Int -> Integer)
fib fib 0 = 0
fib fib 1 = 1
fib fib n = fib (n-1) + fib (n-2)
Note how the first argument `fib` of the function `fib` shadows the
name of the global function `fib`.
The advantage of this complicated definition is that you get a
memoized version of the `fibonacci` function simply by using `fixmemo`
instead of `fix`:
memoFib :: Int -> Integer
memoFib = fixmemo fib
This definition avoids to recompute the same recursive calls over and
over again in is therefore much faster than the original version.
That said, memoization helps to solve your original problem only if
you reuse the cache for different top-level calls of the Collatz
length function. According to the Collatz conjecture no argument
appears again when computing the Collatz length of a number (otherwise
the computation would not terminate).
Ryan's solution achieves reuse of the cache by defining it at the top
level and you can do the same with tries. For the `fib` example it
would look like this:
fibCache :: Int :->: Integer
fibCache = trie (fib cachedFib)
cachedFib :: Int -> Integer
cachedFib = untrie fibCache
Now even independent calls to `cachedFib` reuse previously computed results.
In my experiments, caching did not pay off, even for computing the
maximum Collatz length of a given range. Without caching it took less
than 5 seconds to compute the maximum Collatz length of all numbers
between 1 and 1000000. With caching, it took 30 seconds.
Cheers,
Sebastian
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