8
\$\begingroup\$

I had a challenge a while back from a friend to try and produce the fastest possible program to compute A(3,16), where A is the Ackermann function.

He wrote it in Java, and it took ~4.4 seconds. The below (very ugly) Haskell program that I wrote took ~1.7 seconds:

{-# LANGUAGE BangPatterns #-}

import           Data.Map.Strict (Map)
import qualified Data.Map.Strict as Map

-- Main method:

main :: IO ()
main = print $ fst $ ack (I 3 16) Map.empty

-- Definitions:

data I = I {-# UNPACK #-} !Int {-# UNPACK #-} !Int
    deriving (Eq, Ord)

ack :: I -> Map I Int -> (Int, Map I Int)
ack   (I 0 n) r  = (n + 1, r)

ack i@(I m 0) r1 = maybe bak fun (Map.lookup i r1)
    where
        (!val, !r2) = ack (I (m-1) 1) r1
        bak       = (val, Map.insert i val r2)
        fun v     = (v, r1)

ack i@(I m n) r1 = maybe bak2 fun2 (Map.lookup call2 r2)
    where
        call1       = (I m (n - 1))
        (!val1, !re1) = ack call1 r1
        bak1        = (val1, Map.insert call1 val1 re1)
        fun1 v      = (v, r1)
        (!v1, !r2)    = maybe bak1 fun1 (Map.lookup call1 r1)
        call2       = (I (m - 1) v1)
        (!val2, !re2) = ack call2 r2
        bak2        = (val2, Map.insert call2 val2 re2)
        fun2 v      = (v, r2)

This is used on Ints specifically because A(3,16) is 524285, which within Ints range.

How can I make this function more memory efficient, and faster for greater Ackermann calls?

Also, the syntax is disgusting. How can I make this more readable (maybe using monads?) without having horrible performance consequences?

\$\endgroup\$
5
\$\begingroup\$

Using Data.Memocombinators

Have a look at Data.Memocombinators module. It offers combinators for memoizing functions which is essentially what you are doing with the Map.

Here is the example from the documentation on how to use it to create a memoizing fibonacci function:

import Data.MemoCombinators

fib = Memo.integral fib'
   where
     fib' 0 = 0
     fib' 1 = 1
     fib' x = fib (x-1) + fib (x-2) 
              ^^^         ^^^

Things to note:

  • fib is the memoized version of fib' which is just a helper function
  • fib' calls fib for any recursive calls

The other thing which will help is to keep in mind that the type Memo a represents a a combinator which is able to memoize a function whose argument type is a.

So:

  • integral is a combinator which can memoize functions taking an Int
  • pair integral integral is a combinator which memoize functions taking an (Int,Int)

And thus to memoize the Ackerman function:

ack = (pair integral integral) ack'
  where ack' (0,n) = n+1
        ack' (m,0) = ack (m-1,1)
        ack' (m,n) = ack (m-1, ack (m, n-1))

Again, note how ack is defined as the memoized version of ack' and ack' calls ack for recursive cases.

Using Array memoization

Faster results can be obtained by using arrays to memoize the functions ack(1,.), ack(2,.) and ack(3,.):

import Data.Array
import Data.Ix

-- memoize the function f using arrays
arrayMemo bnds f = g
  where g i = arr ! i
        arr = array bnds [ (i,f arr i) | i <- range bnds ]

a0 n = n+1

a1 = arrayMemo (0,530000) f
  where f arr 0 = a0 1
        f arr n = a0 (arr ! (n-1))

a2 = arrayMemo (0,270000) f
  where f arr 0 = a1 1
        f arr n = a1 (arr ! (n-1))

a3 = arrayMemo (0,16) f
  where f arr 0 = a2 1
        f arr n = a2 (arr ! (n-1))

The only drawback is that explicit bounds have to be determined for each level function. This approach computes a3 16 in about half a second.

\$\endgroup\$
  • 2
    \$\begingroup\$ Update: If you change all of the limits to 530000 the second solution runs in about a second. \$\endgroup\$ – ErikR Sep 13 '15 at 1:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.