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algorithm - Finding elementary cycles of (directed) graphs


I need to enumerate all the simple cycles (i.e. elementary cycles where no vertex is repeated other than the starting) of a graph which has both directed and undirected edges, where we can treat the undirected edges as doubly directed. (Specifically, I am looking at the Cayley graphs of S3 and S4, which can be produced using CayleyGraph[SymmetricGroup[3]] and CayleyGraph[SymmetricGroup[4]] respectively.)


I have tried two ways of doing this so far. First, I have tried using ExtractCycles in the Combinatorica package, as detailed in this answer by TomD. For example, entering the ordered pairs for S3 (as "el"):


el = {{1, 2}, {1, 4}, {2, 3}, {2, 5}, {3, 1}, {3, 6}, {4, 1}, {4, 6}, {5, 4}, {5, 2}, 
{6, 5}, {6, 3}}

And then using:


Needs["Combinatorica`"]
ExtractCycles@FromOrderedPairs@el


returns:


{{5, 4, 6, 5}, {6, 3, 6}, {5, 2, 5}, {4, 1, 4}, {3, 1, 2, 3}}

However, that set is incomplete - what about e.g. {{5, 4, 1, 2, 5}, {5, 4, 1, 2, 3, 6, 5}}? These are simple cycles, so why are they not included in the list? The output for S4 is also much truncated (obviously we would expect a lot of cycles there, but the output gives only eighteen).


The second thing I tried was this answer by kguler. Taking the example of S3, using CycleGraph[3, DirectedEdges -> True] gave the right results, but CycleGraph[4, DirectedEdges -> True] did not - i.e. the cycles {5, 4, 1, 2, 5} etc. were not picked up again. Presumably this is something to do with the doubly directed edges?


Any help with this would be much appreciated!


Edit: As requested, the ordered pairs for the Cayley graph of S4 are:


el2 = {{1, 2}, {1, 9}, {2, 3}, {2, 17}, {3, 4}, {3, 13}, {4, 1}, {4, 5}, {5, 4}, {5, 6},
{6, 16}, {6, 7}, {7, 8}, {7, 22}, {8, 5}, {8, 10}, {9, 1}, {9, 10}, {10, 8}, {10, 11},
{11, 12}, {11, 21}, {12, 9}, {12, 18}, {13, 14}, {13, 3}, {14, 15}, {14, 20}, {15, 16},

{15, 23}, {16, 13}, {16, 6}, {17, 2}, {17, 18}, {18, 19}, {18, 12}, {19, 24}, {19, 20},
{20, 17}, {20, 14}, {21, 11}, {21, 22}, {22, 7}, {22, 23}, {23, 15}, {23, 24}, {24, 19},
{24, 21}}

Answer



Here is a brute force method:


cycles[el_] :=
 Module[{f, edges = Rule @@@ el // Dispatch},
  f[x_, b___, x_] := {{x, b, x}};
  f[___, x_, ___, x_] = {};
  f[c___, v_] := Join @@ (f[c, v, #] & /@ ReplaceList[v, edges]);

  Join @@ f /@ Union @@ el
 ]

In the code above the line f[___, x_, ___, x_] = {}; was used for clarity, but faster duplicate tests exist. For short cycles f[c__] /; ! UnsameQ@c = {}; should be fast(est), but if long cycles may be present you should use f[c__] /; Signature@{c} === 0 = {}; or f[c__] /; {c} =!= DeleteDuplicates@{c} = {};


Test:


el = {{1, 2}, {1, 4}, {2, 3}, {2, 5}, {3, 1}, {3, 6}, {4, 1}, {4, 6},
      {5, 4}, {5, 2}, {6, 5}, {6, 3}};

cycles[el]



{{1, 2, 3, 1}, {1, 2, 3, 6, 5, 4, 1}, {1, 2, 5, 4, 1}, {1, 2, 5, 4, 6,
   3, 1}, {1, 4, 1}, {1, 4, 6, 5, 2, 3, 1}, {1, 4, 6, 3, 1}, {2, 3, 1,
   2}, {2, 3, 1, 4, 6, 5, 2}, {2, 3, 6, 5, 4, 1, 2}, {2, 3, 6, 5, 
  2}, {2, 5, 4, 1, 2}, {2, 5, 4, 6, 3, 1, 2}, {2, 5, 2}, {3, 1, 2, 
  3}, {3, 1, 2, 5, 4, 6, 3}, {3, 1, 4, 6, 5, 2, 3}, {3, 1, 4, 6, 
  3}, {3, 6, 5, 4, 1, 2, 3}, {3, 6, 5, 2, 3}, {3, 6, 3}, {4, 1, 2, 3, 
  6, 5, 4}, {4, 1, 2, 5, 4}, {4, 1, 4}, {4, 6, 5, 4}, {4, 6, 5, 2, 3, 
  1, 4}, {4, 6, 3, 1, 2, 5, 4}, {4, 6, 3, 1, 4}, {5, 4, 1, 2, 3, 6, 
  5}, {5, 4, 1, 2, 5}, {5, 4, 6, 5}, {5, 4, 6, 3, 1, 2, 5}, {5, 2, 3, 

  1, 4, 6, 5}, {5, 2, 3, 6, 5}, {5, 2, 5}, {6, 5, 4, 1, 2, 3, 6}, {6, 
  5, 4, 6}, {6, 5, 2, 3, 1, 4, 6}, {6, 5, 2, 3, 6}, {6, 3, 1, 2, 5, 4,
   6}, {6, 3, 1, 4, 6}, {6, 3, 6}}

To remove the duplicate cycles from this output one can use:


DeleteDuplicates[RotateLeft[#, Ordering[#, 1] - 1] & /@ Most /@ #]&


{{1, 2, 3}, {1, 2, 3, 6, 5, 4}, {1, 2, 5, 4}, {1, 2, 5, 4, 6, 3}, {1, 4},
 {1, 4, 6, 5, 2, 3}, {1, 4, 6, 3}, {2, 3, 6, 5}, {2, 5}, {3, 6}, {4, 6, 5}}


Here is an alternative approach to remove the duplicates on-the-fly with a form of memoization. It seems it may be more or less efficient depending on the graph; I have not tested it well yet.


cycles2[el_] :=
 Module[{f, edges = Rule @@@ el // Dispatch},
  f[x_, b___, x_] := (
    (f[##] = {}) & @@@ NestList[RotateLeft, {x, b}, Length@{x, b} - 1];
    {{x, b}}
   );
  f[c__] /; Signature@{c} === 0 = {};
  f[c___, v_] := Join @@ (f[c, v, #] & /@ ReplaceList[v, edges]);

  Join @@ f /@ Union @@ el
  ]

cycles2[el]


{{1, 2, 3}, {1, 2, 3, 6, 5, 4}, {1, 2, 5, 4}, {1, 2, 5, 4, 6, 3}, {1, 4},
 {1, 4, 6, 5, 2, 3}, {1, 4, 6, 3}, {2, 3, 6, 5}, {2, 5}, {3, 6}, {4, 6, 5}}

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