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

*To*: mathgroup at smc.vnet.net*Subject*: [mg117600] Re: "set" data structure in Mathematica? (speeding up graph traversal function)*From*: DrMajorBob <btreat1 at austin.rr.com>*Date*: Thu, 24 Mar 2011 06:28:23 -0500 (EST)

> Another change was the use of Cases in place of ReplaceList, which is > mostly responsible for the great difference in timing you get. No, it isn't. "next" is computed exactly ONCE for each node, so this is all the time involved: Clear[next] nodes = Union@Flatten[graph /. Rule -> List]; next[i_Integer] := next[i] = Cases[graph, Rule[i, j_] :> j] next /@ nodes; // Timing {0.001473, Null} Here's the same thing with ReplaceList: Clear[next] nodes = Union@Flatten[graph /. Rule -> List]; next[i_Integer] := next[i] = ReplaceList[i, graph] next /@ nodes; // Timing {0.002368, Null} And here it is with Dispatch, in two different ways: Clear[next] nodes = Union@Flatten[graph /. Rule -> List]; next[i_Integer] := next[i] = ReplaceList[i, Dispatch@graph] next /@ nodes; // Timing {0.001981, Null} Clear[next] d = Dispatch@graph; nodes = Union@Flatten[graph /. Rule -> List]; next[i_Integer] := next[i] = ReplaceList[i, d] next /@ nodes; // Timing {0.000291, Null} The last is a lot faster RELATIVELY, but not in absolute terms -- unless your plan is to eliminate "next" entirely: Clear[path, p] d = Dispatch@graph; path[i_Integer] := path@p@i path[p[a___, last_]] := Module[{c = Complement[ReplaceList[last, d], {a, last}]}, If[c == {}, p[a, last], path@p[a, last, #] & /@ c ] ] (paths = (path /@ nodes); Length@Flatten@paths) // Timing {1.7552, 47165} But that's no better than this: 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], path@p[a, last, #] & /@ c ] ] ((paths = Flatten[path /@ nodes] /. p -> List) // Length) // Timing {1.74741, 47165} or this: Clear[next, path, p] d = Dispatch@graph; nodes = Union@Flatten[graph /. Rule -> List]; next[i_Integer] := next[i] = ReplaceList[i, d] path[i_Integer] := path@p@i path[p[a___, last_]] := Module[{c = Complement[ReplaceList[last, d], {a, last}]}, If[c = {}, p[a, last], path@p[a, last, #] & /@ c]] (paths = (path /@ nodes); Length@Flatten@paths) // Timing {1.74914, 47165} I like that solution best. Bobby On Wed, 23 Mar 2011 13:58:37 -0500, Szabolcs Horv=E1t <szhorvat at gmail.com> wrote: > Thank you for reading the lengthy post and replying, Bobby! > > Essentially what you are doing is replacing several calls to MemberQ > by a single call to Complement. This indeed works well! > > Another change was the use of Cases in place of ReplaceList, which is > mostly responsible for the great difference in timing you get. > However, using Dispatch[graph] instead of just graph in ReplaceList > appears to be faster than Cases. This can be combined with your idea > to use Complement to get another speedup. > > Cheers, > Szabolcs > > 2011/3/22 DrMajorBob <btreat1 at austin.rr.com>: >> 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: "set" data structure in Mathematica? (speeding up graph traversal function)**

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

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

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