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