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MathGroup Archive 2004

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Re: optimally picking one element from each list

  • To: mathgroup at smc.vnet.net
  • Subject: [mg48348] Re: optimally picking one element from each list
  • From: "Carl K. Woll" <carlw at u.washington.edu>
  • Date: Tue, 25 May 2004 07:17:15 -0400 (EDT)
  • Organization: University of Washington
  • References: <c8mugt$a3l$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Daniel,

Here is a solution, which I think is correct, and which ought to be
considerably faster than the others that have been proposed. I couldn't get
the other solutions to work, so I did not bother to compare timings.

pick[data_]:=Module[{common,tmp},
 common={};

tmp=Reverse[If[(common=Intersection[common,#])=={},common=#,common]&/@data];
 common=.;
 Reverse[If[MemberQ[#,common],common,common=First[#]]&/@tmp]
 ]

Basically, you start at the beginning, and find the element which gives you
the longest string of common elements. Once the string can no longer be
extended, start a new string. It seems to me that this algorithm ought to
give you a correct answer (there are many correct answers).

Carl Woll

"Daniel Reeves" <dreeves at umich.edu> wrote in message
news:c8mugt$a3l$1 at smc.vnet.net...
> Suppose you have a list of lists and you want to pick one element from
> each and put them in a new list so that the number of elements that are
> identical to their next neighbor is maximized.
>   (in other words, for the resulting list l, minimize Length[Split[l]].)
>   (in yet other words, we want the list with the fewest interruptions of
> identical contiguous elements.)
>
> EG, pick[{ {1,2,3}, {2,3}, {1}, {1,3,4}, {4,1} }]
>      --> {    2,      2,    1,     1,      1   }
>
> Here's a preposterously brute force solution:
>
> pick[x_] := argMax[-Length[Split[#]]&, Distribute[x, List]]
>
>    where argMax can be defined like so:
>
>    (* argMax[f,domain] returns the element of domain for which f of
>       that element is maximal -- breaks ties in favor of first occurrence.
>     *)
>    SetAttributes[argMax, HoldFirst];
>    argMax[f_, dom_List] := Fold[If[f[#1] >= f[#2], #1, #2] &,
>                                 First[dom], Rest[dom]]
>
> Below is an attempt at a search-based approach, which is also way too
> slow.  So the gauntlet has been thrown down.  Anyone want to give it a
> shot?
>
>
> (* Used by bestFirstSearch. *)
> treeSearch[states_List, goal_, successors_, combiner_] :=
>     Which[states=={}, $Failed,
>           goal[First[states]], First[states],
>           True, treeSearch[
>                   combiner[successors[First[states]], Rest[states]],
>                   goal, successors, combiner]]
>
> (* Takes a start state, a function that tests whether a state is a goal
>    state, a function that generates a list of successors for a state, and
>    a function that gives the cost of a state.  Finds a goal state that
>    minimizes cost.
> *)
> bestFirstSearch[start_, goal_, successors_, costFn_] :=
>   treeSearch[{start}, goal, successors,
>              Sort[Join[#1,#2], costFn[#1] < costFn[#2] &]&]
>
> (* A goal state is one for which we've picked one element of every list
>    in l.
> *)
> goal[l_][state_] := Length[state]==Length[l]
>
> (* If in state s we've picked one element from each list in l up to list
>    i, then the successors are all the possible ways to extend s to pick
>    elements thru list i+1.
> *)
> successors[l_][state_] := Append[state,#]& /@ l[[Length[state]+1]]
>
> (* Cost function g: higher cost for more neighbors different
>    (Length[Split[state]]) and then breaks ties in favor of longer
>    states to keep from unnecessarily expanding the search tree.
> *)
> g[l_][state_] :=
Length[Split[state]]*(Length[l]+1)+Length[l]-Length[state]
>
> (* Pick one element from each of the lists in l so as to minimize the
>    cardinality of Split, ie, maximize the number of elements that are
>    the same as their neighbor.
> *)
> pick[l_] := bestFirstSearch[{}, goal[l], successors[l], g[l]]
>
>
> -- 
> http://ai.eecs.umich.edu/people/dreeves  - -  google://"Daniel Reeves"
>
>              If you must choose between two evils,
>              pick the one you've never tried before.
>



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