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I have a list of pairs, for example:

pairs={{13, 10}, {12, 14}, {10, 36}, {35, 11}, {3, 5}, {1, 6},
{20, 24}, {21, 22}, {33, 7}, {31, 8}, {31, 27}, {32, 25}, 
{21, 35}, {34, 19}, {18, 15}, {14, 16}, {9, 5}, {4, 7}, 
{1, 13}, {15, 2}, {6, 36}, {4, 34}, {8, 2}, {9, 3}, {25, 20},
{19, 26}, {22, 11}, {23, 12}, {32, 28}, {30, 33}, {23, 16},
{24, 17}, {29, 27}, {26, 30}, {17, 28}, {18, 29}};

pairs can be seen as the definition of a relation $R$. $x$ and $y$ satisfy the relation if and only if {x,y} $\in$ pairs. I need to compute the equivalence classes of the symmetric transitive closure of $R$.

In other words, I need to compute a list eqvclss. The elements of eqvclss are lists themselves. For example, 13, 10, 36, 6, 1, ... should all be in the same list in eqvclss. (If you understand that, then I explained the question properly; if you don't, say so in the comments so I can try to improve).

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  • $\begingroup$ Related: mathematica.stackexchange.com/q/3234/121 $\endgroup$
    – Mr.Wizard
    Jun 29, 2012 at 5:29
  • $\begingroup$ From {13,10} and {10, 36}, you conclude that 10 and 13 belong together. But 13 also belongs with 1 because {1, 13} exists. However, there is no pair {1, 10} or {10, 1}, so 1 shouldn't be in the same group as 10, which is in the same group as 13 which belongs with 1. So the problem for Gather is that your identity relation is not transitive. As a result, you have to give up some conditions. Maybe you want all gathered groups to be numbers that are at least indirectly connected? $\endgroup$
    – Jens
    Jun 29, 2012 at 5:29
  • $\begingroup$ @Jens Thanks for your comment. I see now. I did some drastic edits to the question (prior to seeing your comment). Now it's more to the point of what I want. Your comment now seems out of context, but it does answer the original question. $\endgroup$
    – a06e
    Jun 29, 2012 at 5:32

5 Answers 5

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ConnectedComponents

Using Daniel Lichtblau's answer to a related question 

ConnectedComponents[pairs] //Sort /@ # & //Sort (* thanks: CarlWoll *)

{{3, 5, 9},
{11, 21, 22, 35},
{12, 14, 16, 23},
{1, 6, 10, 13, 36},
{17, 20, 24, 25, 28, 32},
{2, 8, 15, 18, 27, 29, 31},
{4, 7, 19, 26, 30, 33, 34}}

In versions prior to 10.3 use

 ConnectedComponents[Graph[UndirectedEdge @@@ pairs]] //Sort /@ # & //Sort

MatrixPower

Implementing transitive closure using MatrixPower:

m = Max@pairs;

(*the adjacency matrix of atomic elements in pairs:*)
SparseArray[pairs ~Append~ {i_, i_} -> 1, {m, m}];

(*symmetrize the adjacency matrix:*)
% + %\[Transpose] // Sign;

(*find the transitive closure:*)
Sign @ MatrixPower[N@%, m];

(*eliminate duplicate rows,and extract the atomic elements of pairs in each row:*)
Select[DeleteDuplicates @ Normal @ %, Tr@# > 1 &];
Join @@ Position[#, 1] & /@ %;

(*organize:*)
Sort[Sort /@ %]

{{3, 5, 9},
{11, 21, 22, 35},
{12, 14, 16, 23},
{1, 6, 10, 13, 36},
{17, 20, 24, 25, 28, 32},
{2, 8, 15, 18, 27, 29, 31},
{4, 7, 19, 26, 30, 33, 34}}

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5
  • $\begingroup$ +1 I think like this better than my answer. $\endgroup$
    – a06e
    Jun 29, 2012 at 5:43
  • $\begingroup$ @becko, thanks for the vote. (Does the edit answer your request re horizontal scrollbar?) $\endgroup$
    – kglr
    Jun 29, 2012 at 6:05
  • $\begingroup$ Yes. The annoying scrollbar is gone. Thanks. $\endgroup$
    – a06e
    Jun 29, 2012 at 6:11
  • $\begingroup$ Slightly simpler is ConnectedComponents[pairs] $\endgroup$
    – Carl Woll
    Jan 26, 2018 at 15:54
  • $\begingroup$ Thank you @Carl. Updated with the simpler version. $\endgroup$
    – kglr
    Jan 26, 2018 at 16:03
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Adapting Heike's fine answer from the prior question:

pairs //. x_ :> Union @@@ Gather[x, # ⋂ #2 =!= {} &]

{{1, 6, 10, 13, 36},
 {12, 14, 16, 23},
 {11, 21, 22, 35},
 {3, 5,  9},
 {17, 20, 24, 25, 28, 32},
 {4, 7, 19, 26, 30, 33, 34},
 {2, 8, 15, 18, 27, 29, 31}}
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3
  • $\begingroup$ I had not seen that. I can see is almost the same to what I did. $\endgroup$
    – a06e
    Jun 29, 2012 at 6:01
  • $\begingroup$ @becko and I did not see your answer before posting this, nevertheless I think it is different enough to be of interest. Do you agree? $\endgroup$
    – Mr.Wizard
    Jun 29, 2012 at 6:05
  • $\begingroup$ Yes. Besides, your way is more elegantly coded. $\endgroup$
    – a06e
    Jun 29, 2012 at 6:10
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Here's code for version 7:

Needs["Combinatorica`"]

gr = FromUnorderedPairs @ pairs;

ConnectedComponents @ gr

{{1, 6, 10, 13, 36},
 {2, 8, 15, 18, 27, 29, 31},
 {3, 5, 9},
 {4, 7, 19, 26, 30, 33, 34},
 {11, 21, 22, 35},
 {12, 14, 16, 23},
 {17, 20, 24, 25, 28, 32}}

GraphPlot[gr, VertexLabeling -> True]

Mathematica graphics

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I have separately posted this method as an answer to this question

If you want to preserve the order in which the vertices are connected within each cycle,

ExtractCycles (in the Combinatorica package) may also be of interest.

For example

Needs["Combinatorica`"]
ExtractCycles@FromUnorderedPairs@pairs

gives

(* { {28, 17, 24, 20, 25, 32, 28},

{23, 12, 14, 16, 23},

{35, 11, 22, 21, 35},

{34, 4, 7, 33, 30, 26, 19, 34},

{9, 3, 5, 9},

{15, 2, 8, 31, 27, 29, 18, 15},

{13, 1, 6, 36, 10, 13} } *)

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I just found a way to do it:

SymmetricTransitiveClosure[pairs : {{_, _} ..}] := 
    FixedPoint[DeleteDuplicates /@ 
        Flatten /@ Gather[#, Intersection[#1, #2] =!= {} &] &, pairs]

I won't acccept this yet. Perhaps someone comes up with something better.

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