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Re: Re: optimally picking one element from each list
I get errors when I try Andrzej's function; maybe others had more luck?
Here's a solution using Minimize. It may be wasteful in the setup, but solution will be very fast. I'm not certain a Sort isn't needed somewhere in the last statement, but this seems OK.
An even faster method would solve a corresponding shortest path problem, if we have a solver available. I don't have to require or check for integer solutions below, because of that correspondence; basic optimal solutions will be integer automatically.
Clear[choice, optimize]
optimize[raw:{__List}] :=
Module[{sets = Union /@ raw, choice, costBounds,
choiceConstraints, pVar, costConstraints,
objective, p, cVar, vars, bounds, solution},
choice[set_, {index_}] :=
Plus @@ (p[index, #1] & ) /@ set == 1;
costBounds[{set1_, set2_}, {index_}] :=
Module[{switches},
switches = Plus @@ (p[index, #1] & ) /@
Complement[set1, set2];
{(cost[index] >= switches + (p[index, #1] -
p[index + 1, #1]) & ) /@ Intersection[
set1, set2], cost[index] >= switches}];
choiceConstraints = Flatten[MapIndexed[choice,
sets]]; pVar = Union[Cases[choiceConstraints,
_p, Infinity]]; costConstraints =
Flatten[MapIndexed[costBounds, Partition[sets, 2,
1]]]; cVar = Union[Cases[costConstraints,
_cost, {2}]]; objective = Plus @@ cVar;
vars = Flatten[{pVar, cVar}];
bounds = Thread[0 <= vars <= 1];
solution = Minimize[Flatten[{objective,
choiceConstraints, costConstraints, bounds}],
vars]; Cases[Last[solution] /. Rule -> rule,
rule[p[_, a_], 1] :> a]]
Timing@optimize@{{1,3},{2,3},{1,3},{1,3,4},{4,1}}
Timing@optimize@{{6},{2},{2},{7,8},{3,4,1},{2,8,2},{5,2},{7},{8,5,8},{2,1,
7},{5},{5}}
{0.06299999999999999*Second, {3, 3, 3, 3, 1}}
{0.015000000000000013*Second, {6, 2, 2, 7, 1, 2, 2, 7, 5, 1, 5, 5}}
Bobby
On Sun, 23 May 2004 06:15:35 -0400 (EDT), Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote:
>
> On 22 May 2004, at 16:04, Daniel Reeves wrote:
>
>> 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.
>>
>>
>>
>
> Here is an approach which, while not really fast, at least is much
> faster than yours. It will correctly minimize Length[Split[l]], but
> will not necessarily give the same answer as your approach because I
> have not bothered to make sure that ties are settled in the way you
> indicated. It should not be hard to modify it to do so but I did not
> really wish to spend time on this matter.
>
> The idea is to use the Backtrack function from
> DiscreteMath`Combinatorica`.
>
> Here is the program:
>
> <<DiscreteMath`Combinatorica`
>
> BacktrackPick[l_]:= Block[{n = Length[sp], c = Length[sp], w},
> partialQ[l_List] := Which[Length[l] == n && (w = Length[Split[
> l]]) <= c, c = w; True, Length[l] < n &&
> Length[Split[l]] <= c, True, True, False];
> solutionQ[l_List] := If[Length[l] == n && (w = Length[Split[l]]) <=
> c,
> c = w; True, False]; Last[Backtrack[sp, partialQ, solutionQ,
> All]]]
>
>
> To see how this compares with yours let's create a random test case:
>
>
>
> sp=Table[Table[Random[Integer,{1,9}],{Random[Integer,{1,3}]}],{12}]
>
>
> {{6},{2},{2},{7,8},{3,4,1},{2,8,2},{5,2},{7},{8,5,8},{2,1,7},{5},{5}}
>
> Let's try your pick function:
>
>
> Length[Split[sol1=pick[sp]]]//Timing
>
>
> {15.33 Second,9}
>
> Now BacktrackPick:
>
>
> Length[Split[sol2=BacktrackPick[sp]]]//Timing
>
>
> {0.34 Second,9}
>
> Quite a difference. This time the answers happened to be the same:
>
>
> sol1
>
> {6,2,2,8,1,2,2,7,8,7,5,5}
>
> sol2
>
> {6,2,2,8,1,2,2,7,8,7,5,5}
>
> But this will not be the case in general.
>
> I am sure that one can write a customised backtracking solution that
> will be much faster than mine. However, writing customised (that is
> without using Backtrack) backtracking programs is a bit of an art that
> requires patience and care and I have a short supply of both of these.
> So I am cc-ing this message to a well known member of this list who is
> a real master in this sort of thing ;-) so he might be interested in
> taking up the challenge.
>
> Andrzej Kozlowski
>
>
--
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