Re: "set" data structure in Mathematica? (speeding up

*To*: mathgroup at smc.vnet.net*Subject*: [mg117576] Re: "set" data structure in Mathematica? (speeding up*From*: DrMajorBob <btreat1 at austin.rr.com>*Date*: Wed, 23 Mar 2011 02:54:36 -0500 (EST)

Here's an improvement, I think. For the test graph you listed below, I get Clear[next, path, p, listPath] nodes = Union@Flatten[graph /. Rule -> List]; next[i_Integer] := next[i] = Cases[graph, Rule[i, j_] :> j] path[i_Integer] := path@p@i path[p[a___, last_]] := Module[{c = Complement[next@last, {a, last}]}, If[c == {}, p[a, last], Flatten[path@p[a, last, #] & /@ c] ] ] (paths = (path /@ nodes); Length@Flatten[pathways]) // Timing {1.73881, 405496} Omitting Flatten and using p -> List can get a structure like yours (mostly): Clear[next, path, p, listPath] nodes = Union@Flatten[graph /. Rule -> List]; next[i_Integer] := next[i] = Cases[graph, Rule[i, j_] :> j] path[i_Integer] := path@p@i path[p[a___, last_]] := Module[{c = Complement[next@last, {a, last}]}, If[c == {}, p[a, last], Flatten[path@p[a, last, #] & /@ c] ] ] ((paths = Flatten[path /@ nodes] /. p -> List) // Length) // Timing {1.85056, 47165} Compare the above with Clear[traverse] traverse[path_][node_] := With[{l = ReplaceList[node, graph], p = Append[path, node]}, If[l === {}, {p}, If[MemberQ[path, node], {}, Join @@ (traverse[p] /@ l)]]] pathways = traverse[{}] /@ nodes; // Timing {8.09828, Null} paths == Cases[pathways, {__Integer}, Infinity] True Bobby On Tue, 22 Mar 2011 05:06:19 -0500, Szabolcs Horv=E1t <szhorvat at gmail.com> wrote: > Dear MathGroup, > > I am trying to speed up a function. > > I'm looking for an efficient set-like data structure, to be used as > follows: > > I have a recursive function f, which works like this (omitting the > details): > > f[set_][value_] := > if value in set then > return nothing > otherwise > newSet = insert value into set > return f[newSet] /@ (list of computed values) > > In other words, the function calculates a list of values, then calls > itself recursively with each of them. These values are collected into a > set during the recursive calls, and one of the stopping conditions is > that a value gets repeated. > > Note: of course there's another stopping condition as well, otherwise > the function wouldn't return anything, but that is irrelevant here. > > Also note: each branch of the recursion will have its own collection > (set) of values. > > The trivial solution would be just using a List of values, inserting new > values with Append, and testing whether they're already in the list with > MemberQ. I found that the function spends most of its time in MemberQ, > so I am looking for a more efficient solution than a plain list that > needs to be iterated over to find elements in it. > > I tried using newSet = Union[set, {value}] instead of append, and > testing whether Length[newSet] === Length[set], but this turned out to > be even slower. > > I wonder if there's a better solution. > > One way would be using "function definitions", i.e. "inserting" 'value' > into 'set' as set[value] = True, and testing membership simply as > TrueQ[set[value]], but this object can't easily be passed around in a > recursive function because it is essentially a global variable. Note > again that the recursion has several branches as the function is mapped > over a whole list of values. > > Does anyone have an idea how to solve this problem? > > -- Szabolcs > > P.S. The problem I'm trying to solve is to find (actually just count) > ALL acyclic paths in a directed graph, starting from a given node. > Starting from a node, I descend along the connections until there's no > way to go, or until I encounter a node that I have already visited. > > Perhaps it's better if I post the function ... > > traverse[graph_][path_][node_] := > With[ > {l = ReplaceList[node, graph], > p = Append[path, node]}, > If[l === {}, > {p}, > If[ > MemberQ[path, node], > {}, > Join @@ (traverse[graph][p] /@ l) > ] > ] > ] > > graph is a list of rules representing a simple directed graph, path is > the list of already visited nodes (initially {}), node is the starting > node. It's generally a good idea to pass the graph as a dispatch table > (Dispatch[]), but then MemberQ[] takes even longer to evaluate than > ReplaceList[]. > > If there are cycles in the graph, there will be a large number of paths, > and the function will be slow. > > P.P.S > > Here's a small graph with some loops to test with: > > {1 -> 2, 3 -> 4, 5 -> 1, 5 -> 2, 6 -> 2, 6 -> 5, 7 -> 2, 7 -> 6, > 8 -> 1, 8 -> 2, 8 -> 5, 8 -> 6, 9 -> 2, 9 -> 5, 9 -> 6, 9 -> 7, > 10 -> 2, 10 -> 6, 10 -> 7, 11 -> 6, 11 -> 7, 12 -> 6, 12 -> 7, > 13 -> 6, 13 -> 7, 14 -> 2, 14 -> 5, 14 -> 6, 14 -> 7, 15 -> 3, > 15 -> 6, 15 -> 7, 15 -> 8, 16 -> 3, 16 -> 6, 16 -> 7, 16 -> 13, > 17 -> 3, 17 -> 6, 17 -> 7, 17 -> 9, 17 -> 11, 17 -> 13, 18 -> 3, > 18 -> 6, 18 -> 7, 18 -> 9, 18 -> 11, 18 -> 12, 18 -> 13, 18 -> 14, > 18 -> 15, 18 -> 19, 18 -> 20, 19 -> 6, 19 -> 7, 19 -> 9, 19 -> 10, > 19 -> 11, 19 -> 12, 19 -> 13, 19 -> 14, 19 -> 16, 19 -> 17, 19 -> 18, > 19 -> 20, 20 -> 3, 20 -> 7, 20 -> 9, 20 -> 10, 20 -> 11, 20 -> 12, > 20 -> 14, 20 -> 16, 20 -> 18, 20 -> 19, 21 -> 3, 21 -> 9, 21 -> 10, > 21 -> 11, 21 -> 12, 21 -> 13, 21 -> 14, 21 -> 15, 21 -> 16, 21 -> 17, > 21 -> 19, 21 -> 20, 22 -> 3, 22 -> 7, 22 -> 9, 22 -> 10, 22 -> 11, > 22 -> 12, 22 -> 13, 22 -> 14, 22 -> 16, 22 -> 19, 22 -> 20, 23 -> 3, > 23 -> 7, 23 -> 9, 23 -> 10, 23 -> 14, 23 -> 16, 23 -> 17, 23 -> 18, > 23 -> 19, 23 -> 20, 24 -> 7, 24 -> 9, 24 -> 10, 24 -> 11, 24 -> 14, > 24 -> 15, 24 -> 16, 24 -> 17, 24 -> 18, 24 -> 20, 25 -> 3, 25 -> 9, > 25 -> 10, 25 -> 11, 25 -> 14, 25 -> 15, 25 -> 16, 25 -> 18, 25 -> 19, > 25 -> 20, 26 -> 7, 26 -> 9, 26 -> 10, 26 -> 11, 26 -> 13, 26 -> 14, > 26 -> 15, 26 -> 16, 26 -> 19, 26 -> 20, 27 -> 3, 27 -> 6, 27 -> 7, > 27 -> 9, 27 -> 10, 27 -> 11, 27 -> 12, 27 -> 13, 27 -> 14, 27 -> 15, > 27 -> 16, 27 -> 17, 27 -> 18, 27 -> 19, 27 -> 20, 27 -> 22, 27 -> 28, > 28 -> 3, 28 -> 9, 28 -> 10, 28 -> 11, 28 -> 12, 28 -> 13, 28 -> 14, > 28 -> 15, 28 -> 16, 28 -> 17, 28 -> 18, 28 -> 19, 28 -> 20, 28 -> 22, > 28 -> 27, 28 -> 29, 29 -> 3, 29 -> 6, 29 -> 7, 29 -> 8, 29 -> 9, > 29 -> 10, 29 -> 11, 29 -> 12, 29 -> 13, 29 -> 14, 29 -> 15, 29 -> 16, > 29 -> 17, 29 -> 18, 29 -> 19, 29 -> 20, 29 -> 21, 29 -> 22, > 29 -> 23, 29 -> 24, 29 -> 25, 29 -> 26, 29 -> 27, 29 -> 28} > -- DrMajorBob at yahoo.com

**Problem with DateListPlot Aspect Ratio**

**Mathematica debugger**

**Re: "set" data structure in Mathematica? (speeding up**

**Re: "set" data structure in Mathematica? (speeding up**