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:

>
> 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

```

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