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MathGroup Archive 1999

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Re: Re: Langford's Problem (another solution improved)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg19644] Re: [mg19594] Re: [mg19541] Langford's Problem (another solution improved)
  • From: "Allan Hayes" <hay at haystack.demon.co.uk>
  • Date: Tue, 7 Sep 1999 00:28:38 -0400
  • References: <7qv31d$34c$1@dragonfly.wolfram.com>
  • Sender: owner-wri-mathgroup at wolfram.com

I came up with the following solution using ReplaceList.
The method seems fairly general - certainly, little thought was needed once
the idea of using ReplaceList occurred to me.

Langford[n_] :=
Module[{sol = {Table[0, {2n}]}},
    Do[sol =
        Join @@
          Map[
               ReplaceList[#1, {x___, 0, y__ /; Length[{y}] === k, 0, z___}
:> {x, k, y, k, z}] &,
               sol
           ],
      {k, n,  1, -1}]
    ; sol
]

I got the following speed comparisons:

Langford[8]; // Timing
{48.33 Second, Null}

Fred Simons: 42.4 Second

Fred Simons/Hartmut Wolf : 13.46 Second;

Andrzej Kozlowski Latest: 209.76  Second
(with my, missnamed, Backtrack2, this became 61.9 Second)


Some refinements are possible;

1) we could replace

        ReplaceList[#1, {x___, 0, y__ /; Length[{y}] === k, 0, z___} :> {x,
k, y, k, z}] &
with
        ReplaceList[#1, {x___, 0, y__ /; Length[{y}] === k, 0, z___} :> {x,
k, y, k, z}, k] &
so that ReplaceList knows to stop when k repalcements have been found

2) we could insert the k directly using With:

With[{k=k},
Join @@
          Map[
               ReplaceList[#1, {x___, 0, y__ /; Length[{y}] === k, 0, z___}
:> {x, k, y, k, z}] &,
               sol
           ]
]

3) we could use Fold instead of Do

But they have little effect on this particular computation.

Allan
---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565







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