Re: "set" data structure in Mathematica? (speeding up graph traversal function)

*To*: mathgroup at smc.vnet.net*Subject*: [mg117595] Re: "set" data structure in Mathematica? (speeding up graph traversal function)*From*: DrMajorBob <btreat1 at austin.rr.com>*Date*: Thu, 24 Mar 2011 06:27:28 -0500 (EST)

Oops, I mangled the first timing, a bit: Clear[next, path, p] 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@paths) // Timing {1.75351, 47165} Bobby On Wed, 23 Mar 2011 02:54:36 -0500, DrMajorBob <btreat1 at austin.rr.com> wrote: > 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 > -- DrMajorBob at yahoo.com

**Re: Why Mathematica does not issue a warning when the calculations**

**Re: Why Mathematica does not issue a warning when the calculations**

**"set" data structure in Mathematica? (speeding up graph traversal function)**

**Re: "set" data structure in Mathematica? (speeding up graph traversal function)**