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Re: Eliminating intermediate results

  • To: mathgroup at smc.vnet.net
  • Subject: [mg91539] Re: [mg91515] Eliminating intermediate results
  • From: Daniel Lichtblau <danl at wolfram.com>
  • Date: Tue, 26 Aug 2008 03:29:02 -0400 (EDT)
  • References: <200808250905.FAA24382@smc.vnet.net>

Mariano Suárez-Alvarez wrote:
> Hi,
> 
> Suppose you have a function 'generate' which given a parameter p
> produces a list of expressions, and you have a predicate predQ which
> looks at those expressions and does some test. To get the list of
> elements in the list which satisfy the predicate, one says:
> 
>   Select[generate[p], predQ]
> 
> Now, if the list generated by 'generate[p]' is big with respect to the
> resulting filtered list, this tends to become impracticable soon.
> 
> For a concrete example: suppose you want to compute the length of
> the list of compositions of a number n in k parts with no zeros. One
> simpleminded way to do that is:
> 
>   Timing@  Length@ Select[Compositions[25, 6], (! MemberQ[#, 0]) &]
> 
> which returns
> 
>   {10.1025, 118755}
> 
> An alternative, considerably more clumsy way is to eliminate the
> intermediate list, and count. For example, using the very cool
> Sow/Reap pair of functions, one can do:
> 
> Timing@ Length@ Module[{n = 30, k = 6, bound},
>    bound = NumberOfCompositions[n, k];
>    Reap[
>      For[i = 0; p = Append[Table[0, {k - 1}], n], i < bound, i++;
>       p = NextComposition[p],
>       If[! MemberQ[p, 0], Sow[p]]
>       ]
>      ][[2, 1]]
>    ]
> 
> which in turns gives
> 
>   {5.9171, 118755}
> 
> The difference here might very well be due to the different ways
> in which NextComposition and Compositions are impemented
> in Combinatorica (I haven't looked). 

I think this is the case. Mathematica is often better suited, in terms 
of speed, for processing in an "all at once" fashion rather than "one at 
a time".


> But code like
> 
>   Length@ Select[Compositions[25, 6], (! MemberQ[#, 0]) &]
> 
> which constructs huge intermediate data structures seems to
> be quite inefficient in Mathematica.

I surmise you mean inefficient in memory usage rather than speed? Or is 
speed here also an issue?


> Is there a standard approach to aleviate this issue? (preferably
> which does not involved ugly code as the one above,
> preserving the nice compositional style)

I am not aware of one. You can improve speed by using just a counter 
instead of Sow/Reap/Length, and by using Compile to rewrite 
Combinatorica's NextComposition. Here is some code to do this.

nextCompC = Compile[{{l, _Integer, 1}}, Module[{c = l, h = 1, t},
    While[c[[h]] == 0, h++];
    If[h == Length[l],
     c = Join[{Last[c]}, Table[0, {Length[l] - 1}]]
     ,
     t = c[[h]] - 1;
     c[[{h, h + 1}]] = {0, c[[h + 1]] + 1};
     c[[1]] = t;
     ];
    c
    ]]

Now you could do:

Timing@Module[{n = 30, j = 0, k = 6, bound},
   bound = NumberOfCompositions[n, k];
   p = Append[Table[0, {k - 1}], n];
   For[i = 0, i < bound, i++,
    p = nextCompC[p];
    If[! MemberQ[p, 0], j++]];
   j]

I get a factor of 3 or so speed improvement in this way. Nothing great, 
but neither is it useless.


> In the context of functional languages, one speaks of
> 'deforestation',
> which is the effort to eliminate intermediate trees (ie, data
> structures)
> which appear frequently when programming in a compositional
> style. The Haskell compiler, for example, iirc, is smart enough
> to *not* construct an intermediate list in the Haskell
> analogue of
> 
>   Select[Range[1000], EvenQ]
> 
> Coming from Haskell, it be nice to be able to rely on
> Mathematica to be smart, so I don't have to ;)
> 
> -- m

There might be ways to emulate this in Mathematica, but not using 
Select. The semantics are that it will evaluate its first argument and 
only then apply the predicate argument to cull from that first argument.

One approach might be to generate the compositions in a tree-like 
fashion, and prune branches for which all leaves (that is, individual 
compositions) will of necessity contain zeros. Maybe set it up as a 
recursion wherein we have

compositions[n,k] = Join[Table[Prepend[compositions[n-j,k-1],j], {j,1,n}]

(I have not tried this and offer no guarantees as to how close or far it 
is from workable.)

Daniel Lichtblau
Wolfram Research


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