Re: Re: optimally picking one element from each list

*To*: mathgroup at smc.vnet.net*Subject*: [mg48324] Re: [mg48308] Re: [mg48297] optimally picking one element from each list*From*: DrBob <drbob at bigfoot.com>*Date*: Mon, 24 May 2004 00:45:20 -0400 (EDT)*References*: <200405220704.DAA08848@smc.vnet.net> <200405231015.GAA21155@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

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 > > -- Using M2, Opera's revolutionary e-mail client: http://www.opera.com/m2/

**Follow-Ups**:**Re: Re: Re: optimally picking one element from each list***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>

**Re: Re: Re: optimally picking one element from each list***From:*DrBob <drbob@bigfoot.com>

**References**:**optimally picking one element from each list***From:*Daniel Reeves <dreeves@umich.edu>

**Re: optimally picking one element from each list***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>

**Re: Problem with function**

**Tensor SVD?**

**Re: optimally picking one element from each list**

**Re: Re: Re: optimally picking one element from each list**