Iteratively strip off simply connected edges in graph?











up vote
6
down vote

favorite
2












Consider a set of edges composing a directed graph. For example:



edges = {DirectedEdge[1, 2], DirectedEdge[2, 3], DirectedEdge[4, 3], DirectedEdge[3, 5], DirectedEdge[5, 6], DirectedEdge[6, 7]};
Graph[edges]



enter image description here




I would like to have a function stripOff that iteratively strips off the outer edges that are simply connected to the rest, and returns them together with the remaining graph:



{incoming1, outgoing1, remains1}= stripOff[edges]
Graph[remains1]



{ {DirectedEdge[1, 2],DirectedEdge[4, 3]} ,



{DirectedEdge[6, 7]} ,



{DirectedEdge[2, 3], DirectedEdge[3, 5], DirectedEdge[5, 6]} }
enter image description here




In the next iteration step it should give



{incoming2, outgoing2, remains2}= stripOff[remains1]
Graph[remains2]



{ {DirectedEdge[2, 3]} ,



{DirectedEdge[5, 6]} ,



{DirectedEdge[3, 5]} }
enter image description here




And finally in the last iteration step



{incoming3, outgoing3, remains3}= stripOff[remains2]



{ {DirectedEdge[3, 5]} ,



{ } ,



{ } }




Is there a quick way to construct such a stripOff function in mathematica? Thanks for any suggestion!



EDIT:



Note that I am trying to iteratively strip off external legs of the graph, which are connected to a vertex only on one side, not on both.



Even though the graph



edges = {DirectedEdge[1, 2], DirectedEdge[2, 3], DirectedEdge[4, 3], DirectedEdge[5, 4]};
Graph[edges]



enter image description here




contains a sink in the middle, the function should not cut the graph in two, but only strip off outer legs:



{incoming, outgoing, remains}= stripOff[edges]



{ {DirectedEdge[1, 2], DirectedEdge[5, 4] } ,



{ } ,



{DirectedEdge[2, 3], DirectedEdge[4, 3]} }











share|improve this question
























  • shouldn't the last step give { {DirectedEdge[3, 5]} ,{DirectedEdge[3, 5]} , { } }?
    – kglr
    Nov 7 at 22:03










  • @kglr I'd like all edges to be unique, without double counting. If an edge triggers for incoming classification, it is spent and is not available to be classified as outgoing any more.
    – Kagaratsch
    Nov 7 at 22:33















up vote
6
down vote

favorite
2












Consider a set of edges composing a directed graph. For example:



edges = {DirectedEdge[1, 2], DirectedEdge[2, 3], DirectedEdge[4, 3], DirectedEdge[3, 5], DirectedEdge[5, 6], DirectedEdge[6, 7]};
Graph[edges]



enter image description here




I would like to have a function stripOff that iteratively strips off the outer edges that are simply connected to the rest, and returns them together with the remaining graph:



{incoming1, outgoing1, remains1}= stripOff[edges]
Graph[remains1]



{ {DirectedEdge[1, 2],DirectedEdge[4, 3]} ,



{DirectedEdge[6, 7]} ,



{DirectedEdge[2, 3], DirectedEdge[3, 5], DirectedEdge[5, 6]} }
enter image description here




In the next iteration step it should give



{incoming2, outgoing2, remains2}= stripOff[remains1]
Graph[remains2]



{ {DirectedEdge[2, 3]} ,



{DirectedEdge[5, 6]} ,



{DirectedEdge[3, 5]} }
enter image description here




And finally in the last iteration step



{incoming3, outgoing3, remains3}= stripOff[remains2]



{ {DirectedEdge[3, 5]} ,



{ } ,



{ } }




Is there a quick way to construct such a stripOff function in mathematica? Thanks for any suggestion!



EDIT:



Note that I am trying to iteratively strip off external legs of the graph, which are connected to a vertex only on one side, not on both.



Even though the graph



edges = {DirectedEdge[1, 2], DirectedEdge[2, 3], DirectedEdge[4, 3], DirectedEdge[5, 4]};
Graph[edges]



enter image description here




contains a sink in the middle, the function should not cut the graph in two, but only strip off outer legs:



{incoming, outgoing, remains}= stripOff[edges]



{ {DirectedEdge[1, 2], DirectedEdge[5, 4] } ,



{ } ,



{DirectedEdge[2, 3], DirectedEdge[4, 3]} }











share|improve this question
























  • shouldn't the last step give { {DirectedEdge[3, 5]} ,{DirectedEdge[3, 5]} , { } }?
    – kglr
    Nov 7 at 22:03










  • @kglr I'd like all edges to be unique, without double counting. If an edge triggers for incoming classification, it is spent and is not available to be classified as outgoing any more.
    – Kagaratsch
    Nov 7 at 22:33













up vote
6
down vote

favorite
2









up vote
6
down vote

favorite
2






2





Consider a set of edges composing a directed graph. For example:



edges = {DirectedEdge[1, 2], DirectedEdge[2, 3], DirectedEdge[4, 3], DirectedEdge[3, 5], DirectedEdge[5, 6], DirectedEdge[6, 7]};
Graph[edges]



enter image description here




I would like to have a function stripOff that iteratively strips off the outer edges that are simply connected to the rest, and returns them together with the remaining graph:



{incoming1, outgoing1, remains1}= stripOff[edges]
Graph[remains1]



{ {DirectedEdge[1, 2],DirectedEdge[4, 3]} ,



{DirectedEdge[6, 7]} ,



{DirectedEdge[2, 3], DirectedEdge[3, 5], DirectedEdge[5, 6]} }
enter image description here




In the next iteration step it should give



{incoming2, outgoing2, remains2}= stripOff[remains1]
Graph[remains2]



{ {DirectedEdge[2, 3]} ,



{DirectedEdge[5, 6]} ,



{DirectedEdge[3, 5]} }
enter image description here




And finally in the last iteration step



{incoming3, outgoing3, remains3}= stripOff[remains2]



{ {DirectedEdge[3, 5]} ,



{ } ,



{ } }




Is there a quick way to construct such a stripOff function in mathematica? Thanks for any suggestion!



EDIT:



Note that I am trying to iteratively strip off external legs of the graph, which are connected to a vertex only on one side, not on both.



Even though the graph



edges = {DirectedEdge[1, 2], DirectedEdge[2, 3], DirectedEdge[4, 3], DirectedEdge[5, 4]};
Graph[edges]



enter image description here




contains a sink in the middle, the function should not cut the graph in two, but only strip off outer legs:



{incoming, outgoing, remains}= stripOff[edges]



{ {DirectedEdge[1, 2], DirectedEdge[5, 4] } ,



{ } ,



{DirectedEdge[2, 3], DirectedEdge[4, 3]} }











share|improve this question















Consider a set of edges composing a directed graph. For example:



edges = {DirectedEdge[1, 2], DirectedEdge[2, 3], DirectedEdge[4, 3], DirectedEdge[3, 5], DirectedEdge[5, 6], DirectedEdge[6, 7]};
Graph[edges]



enter image description here




I would like to have a function stripOff that iteratively strips off the outer edges that are simply connected to the rest, and returns them together with the remaining graph:



{incoming1, outgoing1, remains1}= stripOff[edges]
Graph[remains1]



{ {DirectedEdge[1, 2],DirectedEdge[4, 3]} ,



{DirectedEdge[6, 7]} ,



{DirectedEdge[2, 3], DirectedEdge[3, 5], DirectedEdge[5, 6]} }
enter image description here




In the next iteration step it should give



{incoming2, outgoing2, remains2}= stripOff[remains1]
Graph[remains2]



{ {DirectedEdge[2, 3]} ,



{DirectedEdge[5, 6]} ,



{DirectedEdge[3, 5]} }
enter image description here




And finally in the last iteration step



{incoming3, outgoing3, remains3}= stripOff[remains2]



{ {DirectedEdge[3, 5]} ,



{ } ,



{ } }




Is there a quick way to construct such a stripOff function in mathematica? Thanks for any suggestion!



EDIT:



Note that I am trying to iteratively strip off external legs of the graph, which are connected to a vertex only on one side, not on both.



Even though the graph



edges = {DirectedEdge[1, 2], DirectedEdge[2, 3], DirectedEdge[4, 3], DirectedEdge[5, 4]};
Graph[edges]



enter image description here




contains a sink in the middle, the function should not cut the graph in two, but only strip off outer legs:



{incoming, outgoing, remains}= stripOff[edges]



{ {DirectedEdge[1, 2], DirectedEdge[5, 4] } ,



{ } ,



{DirectedEdge[2, 3], DirectedEdge[4, 3]} }








list-manipulation function-construction graphs-and-networks






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 7 at 23:59

























asked Nov 7 at 21:00









Kagaratsch

4,57731246




4,57731246












  • shouldn't the last step give { {DirectedEdge[3, 5]} ,{DirectedEdge[3, 5]} , { } }?
    – kglr
    Nov 7 at 22:03










  • @kglr I'd like all edges to be unique, without double counting. If an edge triggers for incoming classification, it is spent and is not available to be classified as outgoing any more.
    – Kagaratsch
    Nov 7 at 22:33


















  • shouldn't the last step give { {DirectedEdge[3, 5]} ,{DirectedEdge[3, 5]} , { } }?
    – kglr
    Nov 7 at 22:03










  • @kglr I'd like all edges to be unique, without double counting. If an edge triggers for incoming classification, it is spent and is not available to be classified as outgoing any more.
    – Kagaratsch
    Nov 7 at 22:33
















shouldn't the last step give { {DirectedEdge[3, 5]} ,{DirectedEdge[3, 5]} , { } }?
– kglr
Nov 7 at 22:03




shouldn't the last step give { {DirectedEdge[3, 5]} ,{DirectedEdge[3, 5]} , { } }?
– kglr
Nov 7 at 22:03












@kglr I'd like all edges to be unique, without double counting. If an edge triggers for incoming classification, it is spent and is not available to be classified as outgoing any more.
– Kagaratsch
Nov 7 at 22:33




@kglr I'd like all edges to be unique, without double counting. If an edge triggers for incoming classification, it is spent and is not available to be classified as outgoing any more.
– Kagaratsch
Nov 7 at 22:33










4 Answers
4






active

oldest

votes

















up vote
5
down vote



accepted










sourceEdges = IncidenceList[#, GeneralUtilities`GraphSources @ # ]&;
simpleSinks = Select[GeneralUtilities`GraphSinks[#],
Function[v, VertexInDegree[#, v] <= 1]] &;
sinkEdges = Complement[IncidenceList[#, simpleSinks @ #], sourceEdges @ #] &;
rest = Complement[#, sourceEdges @ #, sinkEdges @ #] &;
f = Rest @ NestWhileList[{sourceEdges @ #[[3]], sinkEdges @ #[[3]], rest @ #[[3]]}&,
{{}, {}, #}, #[[3]] =!= {}&]&;


Examples:



edges1 = {DirectedEdge[1, 2], DirectedEdge[2, 3], DirectedEdge[4, 3], DirectedEdge[3, 5],
DirectedEdge[5, 6], DirectedEdge[6, 7]};
f @ edges1



{{{1 -> 2, 4 -> 3}, {6 -> 7}, {2 -> 3, 3 -> 5, 5 -> 6}},

{{2 -> 3}, {5 -> 6}, {3 -> 5}},

{{3 -> 5}, {}, {}}}




g1 = Graph[edges1, VertexSize -> Large, 
VertexLabels -> Placed["Name", Center], ImageSize -> {200, 300}];
Row[HighlightGraph[g1, #, PlotLabel -> Column[#]] & /@ f[edges1]]


enter image description here



edges2 = {DirectedEdge[1, 2], DirectedEdge[2, 3], DirectedEdge[4, 3], 
DirectedEdge[5, 4]} ;
f @ edges2



{{{1 -> 2, 5 -> 4}, {}, {2 -> 3, 4 -> 3}},

{{2 -> 3, 4 -> 3}, {}, {}}}




g2 = Graph[edges2, VertexSize -> Large, 
VertexLabels -> Placed["Name", Center], ImageSize -> {200, 300}];
Row[HighlightGraph[g2, #, PlotLabel -> Column[#]] & /@ f[edges2]]


enter image description here



You can also use GraphComputation`SourceVertexList and GraphComputation`SinkVertexList for GeneralUtilities`GraphSources and GeneralUtilities`GraphSinks, respectively.






share|improve this answer























  • I wonder if GeneralUtilities'GraphSinks would trigger on {2->3} and {4->3} in a situation like { {1->2} , {2->3} , {4->3} , {5->4} }, where {2->3} and {4->3} do point to a sink but are not simply connected to the rest of the graph? Asking, since I'd actually like to avoid this in my case.
    – Kagaratsch
    Nov 7 at 22:26












  • @Kagaratsch, not sure I understand el = { {1->2} , {2->3} , {4->3} , {5->4} }, but GeneralUtilities`GraphSinks @Flatten[el] gives {3}.
    – kglr
    Nov 7 at 22:35












  • I see, that is what I was afraid of. In my application case I am only looking for sources and sinks which are simply connected to the rest of the graph.
    – Kagaratsch
    Nov 7 at 22:37












  • @Kagaratsch, sounds like the example in your question does not reflect your requirements accurately. Adding the example in your comment to your post with some explanation would be useful.
    – kglr
    Nov 7 at 22:50










  • Added an edit to the question.
    – Kagaratsch
    Nov 7 at 23:11


















up vote
4
down vote













g = Graph[edges, VertexLabels -> Automatic]


enter image description here



source[g_?GraphQ] := Pick[VertexList[g], VertexInDegree[g], 0]
sink[g_?GraphQ] := Pick[VertexList[g], VertexOutDegree[g], 0]

strip[g_] :=
With[{so = source[g], si = sink[g]},
{Flatten[IncidenceList[g, #] & /@ so],
Flatten[IncidenceList[g, #] & /@ si],
VertexDelete[g, Join[so, si]]}
]


enter image description here



There are minor issues, such as returning an edge twice at the last step, but that should be easy (if tedious) to fix.






share|improve this answer




























    up vote
    3
    down vote













    If you have a large graph, it will be faster to work with the vertex-edge incidence matrix instead of Graph objects. The edges you want to strip off will have either a 1 or a -1 depending on the direction of the directed edge. So, the simple edges you want to strip off will have a vertex with only a single 1 or -1 in the row. Let's take your example:



    m = IncidenceMatrix[edges];
    m //MatrixForm //TeXForm



    $left(
    begin{array}{cccccc}
    -1 & 0 & 0 & 0 & 0 & 0 \
    1 & -1 & 0 & 0 & 0 & 0 \
    0 & 1 & 1 & -1 & 0 & 0 \
    0 & 0 & -1 & 0 & 0 & 0 \
    0 & 0 & 0 & 1 & -1 & 0 \
    0 & 0 & 0 & 0 & 1 & -1 \
    0 & 0 & 0 & 0 & 0 & 1 \
    end{array}
    right)$




    The vertices that can be removed can be obtained with:



    v = Clip[Unitize[m] . ConstantArray[1, Length @ First @ m], {1, 1}, {0, 0}]



    {1, 0, 0, 1, 0, 0, 1}




    The corresponding edges can be found with:



    e = Unitize[v . Unitize[m]]



    {1, 0, 1, 0, 0, 1}




    The kind of edge can be determined using:



    v . Mod[m, 3] . DiagonalMatrix[e]



    {2, 0, 2, 0, 0, 1}




    where 1 is an outgoing edge, 2 is an incoming edge, and 3 would be both an incoming and outgoing edge.



    The matrix after removing the above vertices and edges can be found from:



    m . DiagonalMatrix[1 - e] //MatrixForm //TeXForm



    $left(
    begin{array}{cccccc}
    0 & 0 & 0 & 0 & 0 & 0 \
    0 & -1 & 0 & 0 & 0 & 0 \
    0 & 1 & 0 & -1 & 0 & 0 \
    0 & 0 & 0 & 0 & 0 & 0 \
    0 & 0 & 0 & 1 & -1 & 0 \
    0 & 0 & 0 & 0 & 1 & 0 \
    0 & 0 & 0 & 0 & 0 & 0 \
    end{array}
    right)$




    Here is a function that does one iteration:



    iter[m_] := Module[{u = Unitize[m], o, v, e},
    o = ConstantArray[1, Length @ First @ u];
    v = Clip[u . o, {1, 1}, {0, 0}];
    e = Unitize[v . Unitize[m]];
    {
    v,
    v . Mod[m, 3] . SparseArray[Band[{1,1}] -> e],
    m . SparseArray[Band[{1,1}] -> 1 - e]
    }
    ]


    For example:



    r = iter[m];
    r[[1]] (* removed vertices *)
    r[[2]] (* removed edges *)
    r[[3]] //MatrixForm //TeXForm



    {1, 0, 0, 1, 0, 0, 1}



    {2, 0, 2, 0, 0, 1}



    $left(
    begin{array}{cccccc}
    0 & 0 & 0 & 0 & 0 & 0 \
    0 & -1 & 0 & 0 & 0 & 0 \
    0 & 1 & 0 & -1 & 0 & 0 \
    0 & 0 & 0 & 0 & 0 & 0 \
    0 & 0 & 0 & 1 & -1 & 0 \
    0 & 0 & 0 & 0 & 1 & 0 \
    0 & 0 & 0 & 0 & 0 & 0 \
    end{array}
    right)$




    Putting the above together:



    res = NestWhileList[iter @* Last, iter[m], Positive @* Total @* First]


    enter image description here



    Deciding which edges are outgoing and incoming can be done with:



    KeyDrop[
    GroupBy[Thread[edges -> res[[1, 2]]], Last -> First],
    0
    ]



    <|2 -> {1 [DirectedEdge] 2, 4 [DirectedEdge] 3}, 1 -> {6 [DirectedEdge] 7}|>




    Converting the SparseArray back to a graph (the removed edges/vertices need to be eliminated from the sparse array) can be done with:



    With[
    {
    v = Pick[Range @ Length @ res[[1, 1]], res[[1, 1]], 0],
    e = Pick[Range @ Length @ res[[1, 2]], res[[1, 2]], 0]
    },

    IncidenceGraph[
    v,
    res[[1, 3]][[v, e]],
    VertexLabels->"Name"
    ]
    ]


    enter image description here



    Your second example:



    edges = {DirectedEdge[1, 2], DirectedEdge[2, 3], DirectedEdge[4, 3], DirectedEdge[5, 4]};
    NestWhileList[
    iter @* Last,
    iter @ IncidenceMatrix[edges],
    Positive @* Total @* First
    ]


    enter image description here






    share|improve this answer




























      up vote
      2
      down vote













      What you're looking for is called a "kcore" of the graph: the set of vertices with at least k edges to other vertices of the core.



      Mathematica has a function that will find this for you:
      https://reference.wolfram.com/language/ref/KCoreComponents.html
      https://reference.wolfram.com/language/example/FindTheKCoreComponentsOfAGraph.html



      To find the removed edges (if the iteration in which they are removed is not important), simply iterate over the original edge list and count those which are not attached to a vertex in the 2-core.






      share|improve this answer























        Your Answer





        StackExchange.ifUsing("editor", function () {
        return StackExchange.using("mathjaxEditing", function () {
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
        });
        });
        }, "mathjax-editing");

        StackExchange.ready(function() {
        var channelOptions = {
        tags: "".split(" "),
        id: "387"
        };
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function() {
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled) {
        StackExchange.using("snippets", function() {
        createEditor();
        });
        }
        else {
        createEditor();
        }
        });

        function createEditor() {
        StackExchange.prepareEditor({
        heartbeatType: 'answer',
        convertImagesToLinks: false,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: null,
        bindNavPrevention: true,
        postfix: "",
        imageUploader: {
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        },
        onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        });


        }
        });














        draft saved

        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f185556%2fiteratively-strip-off-simply-connected-edges-in-graph%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        5
        down vote



        accepted










        sourceEdges = IncidenceList[#, GeneralUtilities`GraphSources @ # ]&;
        simpleSinks = Select[GeneralUtilities`GraphSinks[#],
        Function[v, VertexInDegree[#, v] <= 1]] &;
        sinkEdges = Complement[IncidenceList[#, simpleSinks @ #], sourceEdges @ #] &;
        rest = Complement[#, sourceEdges @ #, sinkEdges @ #] &;
        f = Rest @ NestWhileList[{sourceEdges @ #[[3]], sinkEdges @ #[[3]], rest @ #[[3]]}&,
        {{}, {}, #}, #[[3]] =!= {}&]&;


        Examples:



        edges1 = {DirectedEdge[1, 2], DirectedEdge[2, 3], DirectedEdge[4, 3], DirectedEdge[3, 5],
        DirectedEdge[5, 6], DirectedEdge[6, 7]};
        f @ edges1



        {{{1 -> 2, 4 -> 3}, {6 -> 7}, {2 -> 3, 3 -> 5, 5 -> 6}},

        {{2 -> 3}, {5 -> 6}, {3 -> 5}},

        {{3 -> 5}, {}, {}}}




        g1 = Graph[edges1, VertexSize -> Large, 
        VertexLabels -> Placed["Name", Center], ImageSize -> {200, 300}];
        Row[HighlightGraph[g1, #, PlotLabel -> Column[#]] & /@ f[edges1]]


        enter image description here



        edges2 = {DirectedEdge[1, 2], DirectedEdge[2, 3], DirectedEdge[4, 3], 
        DirectedEdge[5, 4]} ;
        f @ edges2



        {{{1 -> 2, 5 -> 4}, {}, {2 -> 3, 4 -> 3}},

        {{2 -> 3, 4 -> 3}, {}, {}}}




        g2 = Graph[edges2, VertexSize -> Large, 
        VertexLabels -> Placed["Name", Center], ImageSize -> {200, 300}];
        Row[HighlightGraph[g2, #, PlotLabel -> Column[#]] & /@ f[edges2]]


        enter image description here



        You can also use GraphComputation`SourceVertexList and GraphComputation`SinkVertexList for GeneralUtilities`GraphSources and GeneralUtilities`GraphSinks, respectively.






        share|improve this answer























        • I wonder if GeneralUtilities'GraphSinks would trigger on {2->3} and {4->3} in a situation like { {1->2} , {2->3} , {4->3} , {5->4} }, where {2->3} and {4->3} do point to a sink but are not simply connected to the rest of the graph? Asking, since I'd actually like to avoid this in my case.
          – Kagaratsch
          Nov 7 at 22:26












        • @Kagaratsch, not sure I understand el = { {1->2} , {2->3} , {4->3} , {5->4} }, but GeneralUtilities`GraphSinks @Flatten[el] gives {3}.
          – kglr
          Nov 7 at 22:35












        • I see, that is what I was afraid of. In my application case I am only looking for sources and sinks which are simply connected to the rest of the graph.
          – Kagaratsch
          Nov 7 at 22:37












        • @Kagaratsch, sounds like the example in your question does not reflect your requirements accurately. Adding the example in your comment to your post with some explanation would be useful.
          – kglr
          Nov 7 at 22:50










        • Added an edit to the question.
          – Kagaratsch
          Nov 7 at 23:11















        up vote
        5
        down vote



        accepted










        sourceEdges = IncidenceList[#, GeneralUtilities`GraphSources @ # ]&;
        simpleSinks = Select[GeneralUtilities`GraphSinks[#],
        Function[v, VertexInDegree[#, v] <= 1]] &;
        sinkEdges = Complement[IncidenceList[#, simpleSinks @ #], sourceEdges @ #] &;
        rest = Complement[#, sourceEdges @ #, sinkEdges @ #] &;
        f = Rest @ NestWhileList[{sourceEdges @ #[[3]], sinkEdges @ #[[3]], rest @ #[[3]]}&,
        {{}, {}, #}, #[[3]] =!= {}&]&;


        Examples:



        edges1 = {DirectedEdge[1, 2], DirectedEdge[2, 3], DirectedEdge[4, 3], DirectedEdge[3, 5],
        DirectedEdge[5, 6], DirectedEdge[6, 7]};
        f @ edges1



        {{{1 -> 2, 4 -> 3}, {6 -> 7}, {2 -> 3, 3 -> 5, 5 -> 6}},

        {{2 -> 3}, {5 -> 6}, {3 -> 5}},

        {{3 -> 5}, {}, {}}}




        g1 = Graph[edges1, VertexSize -> Large, 
        VertexLabels -> Placed["Name", Center], ImageSize -> {200, 300}];
        Row[HighlightGraph[g1, #, PlotLabel -> Column[#]] & /@ f[edges1]]


        enter image description here



        edges2 = {DirectedEdge[1, 2], DirectedEdge[2, 3], DirectedEdge[4, 3], 
        DirectedEdge[5, 4]} ;
        f @ edges2



        {{{1 -> 2, 5 -> 4}, {}, {2 -> 3, 4 -> 3}},

        {{2 -> 3, 4 -> 3}, {}, {}}}




        g2 = Graph[edges2, VertexSize -> Large, 
        VertexLabels -> Placed["Name", Center], ImageSize -> {200, 300}];
        Row[HighlightGraph[g2, #, PlotLabel -> Column[#]] & /@ f[edges2]]


        enter image description here



        You can also use GraphComputation`SourceVertexList and GraphComputation`SinkVertexList for GeneralUtilities`GraphSources and GeneralUtilities`GraphSinks, respectively.






        share|improve this answer























        • I wonder if GeneralUtilities'GraphSinks would trigger on {2->3} and {4->3} in a situation like { {1->2} , {2->3} , {4->3} , {5->4} }, where {2->3} and {4->3} do point to a sink but are not simply connected to the rest of the graph? Asking, since I'd actually like to avoid this in my case.
          – Kagaratsch
          Nov 7 at 22:26












        • @Kagaratsch, not sure I understand el = { {1->2} , {2->3} , {4->3} , {5->4} }, but GeneralUtilities`GraphSinks @Flatten[el] gives {3}.
          – kglr
          Nov 7 at 22:35












        • I see, that is what I was afraid of. In my application case I am only looking for sources and sinks which are simply connected to the rest of the graph.
          – Kagaratsch
          Nov 7 at 22:37












        • @Kagaratsch, sounds like the example in your question does not reflect your requirements accurately. Adding the example in your comment to your post with some explanation would be useful.
          – kglr
          Nov 7 at 22:50










        • Added an edit to the question.
          – Kagaratsch
          Nov 7 at 23:11













        up vote
        5
        down vote



        accepted







        up vote
        5
        down vote



        accepted






        sourceEdges = IncidenceList[#, GeneralUtilities`GraphSources @ # ]&;
        simpleSinks = Select[GeneralUtilities`GraphSinks[#],
        Function[v, VertexInDegree[#, v] <= 1]] &;
        sinkEdges = Complement[IncidenceList[#, simpleSinks @ #], sourceEdges @ #] &;
        rest = Complement[#, sourceEdges @ #, sinkEdges @ #] &;
        f = Rest @ NestWhileList[{sourceEdges @ #[[3]], sinkEdges @ #[[3]], rest @ #[[3]]}&,
        {{}, {}, #}, #[[3]] =!= {}&]&;


        Examples:



        edges1 = {DirectedEdge[1, 2], DirectedEdge[2, 3], DirectedEdge[4, 3], DirectedEdge[3, 5],
        DirectedEdge[5, 6], DirectedEdge[6, 7]};
        f @ edges1



        {{{1 -> 2, 4 -> 3}, {6 -> 7}, {2 -> 3, 3 -> 5, 5 -> 6}},

        {{2 -> 3}, {5 -> 6}, {3 -> 5}},

        {{3 -> 5}, {}, {}}}




        g1 = Graph[edges1, VertexSize -> Large, 
        VertexLabels -> Placed["Name", Center], ImageSize -> {200, 300}];
        Row[HighlightGraph[g1, #, PlotLabel -> Column[#]] & /@ f[edges1]]


        enter image description here



        edges2 = {DirectedEdge[1, 2], DirectedEdge[2, 3], DirectedEdge[4, 3], 
        DirectedEdge[5, 4]} ;
        f @ edges2



        {{{1 -> 2, 5 -> 4}, {}, {2 -> 3, 4 -> 3}},

        {{2 -> 3, 4 -> 3}, {}, {}}}




        g2 = Graph[edges2, VertexSize -> Large, 
        VertexLabels -> Placed["Name", Center], ImageSize -> {200, 300}];
        Row[HighlightGraph[g2, #, PlotLabel -> Column[#]] & /@ f[edges2]]


        enter image description here



        You can also use GraphComputation`SourceVertexList and GraphComputation`SinkVertexList for GeneralUtilities`GraphSources and GeneralUtilities`GraphSinks, respectively.






        share|improve this answer














        sourceEdges = IncidenceList[#, GeneralUtilities`GraphSources @ # ]&;
        simpleSinks = Select[GeneralUtilities`GraphSinks[#],
        Function[v, VertexInDegree[#, v] <= 1]] &;
        sinkEdges = Complement[IncidenceList[#, simpleSinks @ #], sourceEdges @ #] &;
        rest = Complement[#, sourceEdges @ #, sinkEdges @ #] &;
        f = Rest @ NestWhileList[{sourceEdges @ #[[3]], sinkEdges @ #[[3]], rest @ #[[3]]}&,
        {{}, {}, #}, #[[3]] =!= {}&]&;


        Examples:



        edges1 = {DirectedEdge[1, 2], DirectedEdge[2, 3], DirectedEdge[4, 3], DirectedEdge[3, 5],
        DirectedEdge[5, 6], DirectedEdge[6, 7]};
        f @ edges1



        {{{1 -> 2, 4 -> 3}, {6 -> 7}, {2 -> 3, 3 -> 5, 5 -> 6}},

        {{2 -> 3}, {5 -> 6}, {3 -> 5}},

        {{3 -> 5}, {}, {}}}




        g1 = Graph[edges1, VertexSize -> Large, 
        VertexLabels -> Placed["Name", Center], ImageSize -> {200, 300}];
        Row[HighlightGraph[g1, #, PlotLabel -> Column[#]] & /@ f[edges1]]


        enter image description here



        edges2 = {DirectedEdge[1, 2], DirectedEdge[2, 3], DirectedEdge[4, 3], 
        DirectedEdge[5, 4]} ;
        f @ edges2



        {{{1 -> 2, 5 -> 4}, {}, {2 -> 3, 4 -> 3}},

        {{2 -> 3, 4 -> 3}, {}, {}}}




        g2 = Graph[edges2, VertexSize -> Large, 
        VertexLabels -> Placed["Name", Center], ImageSize -> {200, 300}];
        Row[HighlightGraph[g2, #, PlotLabel -> Column[#]] & /@ f[edges2]]


        enter image description here



        You can also use GraphComputation`SourceVertexList and GraphComputation`SinkVertexList for GeneralUtilities`GraphSources and GeneralUtilities`GraphSinks, respectively.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Nov 8 at 1:06

























        answered Nov 7 at 22:11









        kglr

        174k8196401




        174k8196401












        • I wonder if GeneralUtilities'GraphSinks would trigger on {2->3} and {4->3} in a situation like { {1->2} , {2->3} , {4->3} , {5->4} }, where {2->3} and {4->3} do point to a sink but are not simply connected to the rest of the graph? Asking, since I'd actually like to avoid this in my case.
          – Kagaratsch
          Nov 7 at 22:26












        • @Kagaratsch, not sure I understand el = { {1->2} , {2->3} , {4->3} , {5->4} }, but GeneralUtilities`GraphSinks @Flatten[el] gives {3}.
          – kglr
          Nov 7 at 22:35












        • I see, that is what I was afraid of. In my application case I am only looking for sources and sinks which are simply connected to the rest of the graph.
          – Kagaratsch
          Nov 7 at 22:37












        • @Kagaratsch, sounds like the example in your question does not reflect your requirements accurately. Adding the example in your comment to your post with some explanation would be useful.
          – kglr
          Nov 7 at 22:50










        • Added an edit to the question.
          – Kagaratsch
          Nov 7 at 23:11


















        • I wonder if GeneralUtilities'GraphSinks would trigger on {2->3} and {4->3} in a situation like { {1->2} , {2->3} , {4->3} , {5->4} }, where {2->3} and {4->3} do point to a sink but are not simply connected to the rest of the graph? Asking, since I'd actually like to avoid this in my case.
          – Kagaratsch
          Nov 7 at 22:26












        • @Kagaratsch, not sure I understand el = { {1->2} , {2->3} , {4->3} , {5->4} }, but GeneralUtilities`GraphSinks @Flatten[el] gives {3}.
          – kglr
          Nov 7 at 22:35












        • I see, that is what I was afraid of. In my application case I am only looking for sources and sinks which are simply connected to the rest of the graph.
          – Kagaratsch
          Nov 7 at 22:37












        • @Kagaratsch, sounds like the example in your question does not reflect your requirements accurately. Adding the example in your comment to your post with some explanation would be useful.
          – kglr
          Nov 7 at 22:50










        • Added an edit to the question.
          – Kagaratsch
          Nov 7 at 23:11
















        I wonder if GeneralUtilities'GraphSinks would trigger on {2->3} and {4->3} in a situation like { {1->2} , {2->3} , {4->3} , {5->4} }, where {2->3} and {4->3} do point to a sink but are not simply connected to the rest of the graph? Asking, since I'd actually like to avoid this in my case.
        – Kagaratsch
        Nov 7 at 22:26






        I wonder if GeneralUtilities'GraphSinks would trigger on {2->3} and {4->3} in a situation like { {1->2} , {2->3} , {4->3} , {5->4} }, where {2->3} and {4->3} do point to a sink but are not simply connected to the rest of the graph? Asking, since I'd actually like to avoid this in my case.
        – Kagaratsch
        Nov 7 at 22:26














        @Kagaratsch, not sure I understand el = { {1->2} , {2->3} , {4->3} , {5->4} }, but GeneralUtilities`GraphSinks @Flatten[el] gives {3}.
        – kglr
        Nov 7 at 22:35






        @Kagaratsch, not sure I understand el = { {1->2} , {2->3} , {4->3} , {5->4} }, but GeneralUtilities`GraphSinks @Flatten[el] gives {3}.
        – kglr
        Nov 7 at 22:35














        I see, that is what I was afraid of. In my application case I am only looking for sources and sinks which are simply connected to the rest of the graph.
        – Kagaratsch
        Nov 7 at 22:37






        I see, that is what I was afraid of. In my application case I am only looking for sources and sinks which are simply connected to the rest of the graph.
        – Kagaratsch
        Nov 7 at 22:37














        @Kagaratsch, sounds like the example in your question does not reflect your requirements accurately. Adding the example in your comment to your post with some explanation would be useful.
        – kglr
        Nov 7 at 22:50




        @Kagaratsch, sounds like the example in your question does not reflect your requirements accurately. Adding the example in your comment to your post with some explanation would be useful.
        – kglr
        Nov 7 at 22:50












        Added an edit to the question.
        – Kagaratsch
        Nov 7 at 23:11




        Added an edit to the question.
        – Kagaratsch
        Nov 7 at 23:11










        up vote
        4
        down vote













        g = Graph[edges, VertexLabels -> Automatic]


        enter image description here



        source[g_?GraphQ] := Pick[VertexList[g], VertexInDegree[g], 0]
        sink[g_?GraphQ] := Pick[VertexList[g], VertexOutDegree[g], 0]

        strip[g_] :=
        With[{so = source[g], si = sink[g]},
        {Flatten[IncidenceList[g, #] & /@ so],
        Flatten[IncidenceList[g, #] & /@ si],
        VertexDelete[g, Join[so, si]]}
        ]


        enter image description here



        There are minor issues, such as returning an edge twice at the last step, but that should be easy (if tedious) to fix.






        share|improve this answer

























          up vote
          4
          down vote













          g = Graph[edges, VertexLabels -> Automatic]


          enter image description here



          source[g_?GraphQ] := Pick[VertexList[g], VertexInDegree[g], 0]
          sink[g_?GraphQ] := Pick[VertexList[g], VertexOutDegree[g], 0]

          strip[g_] :=
          With[{so = source[g], si = sink[g]},
          {Flatten[IncidenceList[g, #] & /@ so],
          Flatten[IncidenceList[g, #] & /@ si],
          VertexDelete[g, Join[so, si]]}
          ]


          enter image description here



          There are minor issues, such as returning an edge twice at the last step, but that should be easy (if tedious) to fix.






          share|improve this answer























            up vote
            4
            down vote










            up vote
            4
            down vote









            g = Graph[edges, VertexLabels -> Automatic]


            enter image description here



            source[g_?GraphQ] := Pick[VertexList[g], VertexInDegree[g], 0]
            sink[g_?GraphQ] := Pick[VertexList[g], VertexOutDegree[g], 0]

            strip[g_] :=
            With[{so = source[g], si = sink[g]},
            {Flatten[IncidenceList[g, #] & /@ so],
            Flatten[IncidenceList[g, #] & /@ si],
            VertexDelete[g, Join[so, si]]}
            ]


            enter image description here



            There are minor issues, such as returning an edge twice at the last step, but that should be easy (if tedious) to fix.






            share|improve this answer












            g = Graph[edges, VertexLabels -> Automatic]


            enter image description here



            source[g_?GraphQ] := Pick[VertexList[g], VertexInDegree[g], 0]
            sink[g_?GraphQ] := Pick[VertexList[g], VertexOutDegree[g], 0]

            strip[g_] :=
            With[{so = source[g], si = sink[g]},
            {Flatten[IncidenceList[g, #] & /@ so],
            Flatten[IncidenceList[g, #] & /@ si],
            VertexDelete[g, Join[so, si]]}
            ]


            enter image description here



            There are minor issues, such as returning an edge twice at the last step, but that should be easy (if tedious) to fix.







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Nov 7 at 21:44









            Szabolcs

            157k13430917




            157k13430917






















                up vote
                3
                down vote













                If you have a large graph, it will be faster to work with the vertex-edge incidence matrix instead of Graph objects. The edges you want to strip off will have either a 1 or a -1 depending on the direction of the directed edge. So, the simple edges you want to strip off will have a vertex with only a single 1 or -1 in the row. Let's take your example:



                m = IncidenceMatrix[edges];
                m //MatrixForm //TeXForm



                $left(
                begin{array}{cccccc}
                -1 & 0 & 0 & 0 & 0 & 0 \
                1 & -1 & 0 & 0 & 0 & 0 \
                0 & 1 & 1 & -1 & 0 & 0 \
                0 & 0 & -1 & 0 & 0 & 0 \
                0 & 0 & 0 & 1 & -1 & 0 \
                0 & 0 & 0 & 0 & 1 & -1 \
                0 & 0 & 0 & 0 & 0 & 1 \
                end{array}
                right)$




                The vertices that can be removed can be obtained with:



                v = Clip[Unitize[m] . ConstantArray[1, Length @ First @ m], {1, 1}, {0, 0}]



                {1, 0, 0, 1, 0, 0, 1}




                The corresponding edges can be found with:



                e = Unitize[v . Unitize[m]]



                {1, 0, 1, 0, 0, 1}




                The kind of edge can be determined using:



                v . Mod[m, 3] . DiagonalMatrix[e]



                {2, 0, 2, 0, 0, 1}




                where 1 is an outgoing edge, 2 is an incoming edge, and 3 would be both an incoming and outgoing edge.



                The matrix after removing the above vertices and edges can be found from:



                m . DiagonalMatrix[1 - e] //MatrixForm //TeXForm



                $left(
                begin{array}{cccccc}
                0 & 0 & 0 & 0 & 0 & 0 \
                0 & -1 & 0 & 0 & 0 & 0 \
                0 & 1 & 0 & -1 & 0 & 0 \
                0 & 0 & 0 & 0 & 0 & 0 \
                0 & 0 & 0 & 1 & -1 & 0 \
                0 & 0 & 0 & 0 & 1 & 0 \
                0 & 0 & 0 & 0 & 0 & 0 \
                end{array}
                right)$




                Here is a function that does one iteration:



                iter[m_] := Module[{u = Unitize[m], o, v, e},
                o = ConstantArray[1, Length @ First @ u];
                v = Clip[u . o, {1, 1}, {0, 0}];
                e = Unitize[v . Unitize[m]];
                {
                v,
                v . Mod[m, 3] . SparseArray[Band[{1,1}] -> e],
                m . SparseArray[Band[{1,1}] -> 1 - e]
                }
                ]


                For example:



                r = iter[m];
                r[[1]] (* removed vertices *)
                r[[2]] (* removed edges *)
                r[[3]] //MatrixForm //TeXForm



                {1, 0, 0, 1, 0, 0, 1}



                {2, 0, 2, 0, 0, 1}



                $left(
                begin{array}{cccccc}
                0 & 0 & 0 & 0 & 0 & 0 \
                0 & -1 & 0 & 0 & 0 & 0 \
                0 & 1 & 0 & -1 & 0 & 0 \
                0 & 0 & 0 & 0 & 0 & 0 \
                0 & 0 & 0 & 1 & -1 & 0 \
                0 & 0 & 0 & 0 & 1 & 0 \
                0 & 0 & 0 & 0 & 0 & 0 \
                end{array}
                right)$




                Putting the above together:



                res = NestWhileList[iter @* Last, iter[m], Positive @* Total @* First]


                enter image description here



                Deciding which edges are outgoing and incoming can be done with:



                KeyDrop[
                GroupBy[Thread[edges -> res[[1, 2]]], Last -> First],
                0
                ]



                <|2 -> {1 [DirectedEdge] 2, 4 [DirectedEdge] 3}, 1 -> {6 [DirectedEdge] 7}|>




                Converting the SparseArray back to a graph (the removed edges/vertices need to be eliminated from the sparse array) can be done with:



                With[
                {
                v = Pick[Range @ Length @ res[[1, 1]], res[[1, 1]], 0],
                e = Pick[Range @ Length @ res[[1, 2]], res[[1, 2]], 0]
                },

                IncidenceGraph[
                v,
                res[[1, 3]][[v, e]],
                VertexLabels->"Name"
                ]
                ]


                enter image description here



                Your second example:



                edges = {DirectedEdge[1, 2], DirectedEdge[2, 3], DirectedEdge[4, 3], DirectedEdge[5, 4]};
                NestWhileList[
                iter @* Last,
                iter @ IncidenceMatrix[edges],
                Positive @* Total @* First
                ]


                enter image description here






                share|improve this answer

























                  up vote
                  3
                  down vote













                  If you have a large graph, it will be faster to work with the vertex-edge incidence matrix instead of Graph objects. The edges you want to strip off will have either a 1 or a -1 depending on the direction of the directed edge. So, the simple edges you want to strip off will have a vertex with only a single 1 or -1 in the row. Let's take your example:



                  m = IncidenceMatrix[edges];
                  m //MatrixForm //TeXForm



                  $left(
                  begin{array}{cccccc}
                  -1 & 0 & 0 & 0 & 0 & 0 \
                  1 & -1 & 0 & 0 & 0 & 0 \
                  0 & 1 & 1 & -1 & 0 & 0 \
                  0 & 0 & -1 & 0 & 0 & 0 \
                  0 & 0 & 0 & 1 & -1 & 0 \
                  0 & 0 & 0 & 0 & 1 & -1 \
                  0 & 0 & 0 & 0 & 0 & 1 \
                  end{array}
                  right)$




                  The vertices that can be removed can be obtained with:



                  v = Clip[Unitize[m] . ConstantArray[1, Length @ First @ m], {1, 1}, {0, 0}]



                  {1, 0, 0, 1, 0, 0, 1}




                  The corresponding edges can be found with:



                  e = Unitize[v . Unitize[m]]



                  {1, 0, 1, 0, 0, 1}




                  The kind of edge can be determined using:



                  v . Mod[m, 3] . DiagonalMatrix[e]



                  {2, 0, 2, 0, 0, 1}




                  where 1 is an outgoing edge, 2 is an incoming edge, and 3 would be both an incoming and outgoing edge.



                  The matrix after removing the above vertices and edges can be found from:



                  m . DiagonalMatrix[1 - e] //MatrixForm //TeXForm



                  $left(
                  begin{array}{cccccc}
                  0 & 0 & 0 & 0 & 0 & 0 \
                  0 & -1 & 0 & 0 & 0 & 0 \
                  0 & 1 & 0 & -1 & 0 & 0 \
                  0 & 0 & 0 & 0 & 0 & 0 \
                  0 & 0 & 0 & 1 & -1 & 0 \
                  0 & 0 & 0 & 0 & 1 & 0 \
                  0 & 0 & 0 & 0 & 0 & 0 \
                  end{array}
                  right)$




                  Here is a function that does one iteration:



                  iter[m_] := Module[{u = Unitize[m], o, v, e},
                  o = ConstantArray[1, Length @ First @ u];
                  v = Clip[u . o, {1, 1}, {0, 0}];
                  e = Unitize[v . Unitize[m]];
                  {
                  v,
                  v . Mod[m, 3] . SparseArray[Band[{1,1}] -> e],
                  m . SparseArray[Band[{1,1}] -> 1 - e]
                  }
                  ]


                  For example:



                  r = iter[m];
                  r[[1]] (* removed vertices *)
                  r[[2]] (* removed edges *)
                  r[[3]] //MatrixForm //TeXForm



                  {1, 0, 0, 1, 0, 0, 1}



                  {2, 0, 2, 0, 0, 1}



                  $left(
                  begin{array}{cccccc}
                  0 & 0 & 0 & 0 & 0 & 0 \
                  0 & -1 & 0 & 0 & 0 & 0 \
                  0 & 1 & 0 & -1 & 0 & 0 \
                  0 & 0 & 0 & 0 & 0 & 0 \
                  0 & 0 & 0 & 1 & -1 & 0 \
                  0 & 0 & 0 & 0 & 1 & 0 \
                  0 & 0 & 0 & 0 & 0 & 0 \
                  end{array}
                  right)$




                  Putting the above together:



                  res = NestWhileList[iter @* Last, iter[m], Positive @* Total @* First]


                  enter image description here



                  Deciding which edges are outgoing and incoming can be done with:



                  KeyDrop[
                  GroupBy[Thread[edges -> res[[1, 2]]], Last -> First],
                  0
                  ]



                  <|2 -> {1 [DirectedEdge] 2, 4 [DirectedEdge] 3}, 1 -> {6 [DirectedEdge] 7}|>




                  Converting the SparseArray back to a graph (the removed edges/vertices need to be eliminated from the sparse array) can be done with:



                  With[
                  {
                  v = Pick[Range @ Length @ res[[1, 1]], res[[1, 1]], 0],
                  e = Pick[Range @ Length @ res[[1, 2]], res[[1, 2]], 0]
                  },

                  IncidenceGraph[
                  v,
                  res[[1, 3]][[v, e]],
                  VertexLabels->"Name"
                  ]
                  ]


                  enter image description here



                  Your second example:



                  edges = {DirectedEdge[1, 2], DirectedEdge[2, 3], DirectedEdge[4, 3], DirectedEdge[5, 4]};
                  NestWhileList[
                  iter @* Last,
                  iter @ IncidenceMatrix[edges],
                  Positive @* Total @* First
                  ]


                  enter image description here






                  share|improve this answer























                    up vote
                    3
                    down vote










                    up vote
                    3
                    down vote









                    If you have a large graph, it will be faster to work with the vertex-edge incidence matrix instead of Graph objects. The edges you want to strip off will have either a 1 or a -1 depending on the direction of the directed edge. So, the simple edges you want to strip off will have a vertex with only a single 1 or -1 in the row. Let's take your example:



                    m = IncidenceMatrix[edges];
                    m //MatrixForm //TeXForm



                    $left(
                    begin{array}{cccccc}
                    -1 & 0 & 0 & 0 & 0 & 0 \
                    1 & -1 & 0 & 0 & 0 & 0 \
                    0 & 1 & 1 & -1 & 0 & 0 \
                    0 & 0 & -1 & 0 & 0 & 0 \
                    0 & 0 & 0 & 1 & -1 & 0 \
                    0 & 0 & 0 & 0 & 1 & -1 \
                    0 & 0 & 0 & 0 & 0 & 1 \
                    end{array}
                    right)$




                    The vertices that can be removed can be obtained with:



                    v = Clip[Unitize[m] . ConstantArray[1, Length @ First @ m], {1, 1}, {0, 0}]



                    {1, 0, 0, 1, 0, 0, 1}




                    The corresponding edges can be found with:



                    e = Unitize[v . Unitize[m]]



                    {1, 0, 1, 0, 0, 1}




                    The kind of edge can be determined using:



                    v . Mod[m, 3] . DiagonalMatrix[e]



                    {2, 0, 2, 0, 0, 1}




                    where 1 is an outgoing edge, 2 is an incoming edge, and 3 would be both an incoming and outgoing edge.



                    The matrix after removing the above vertices and edges can be found from:



                    m . DiagonalMatrix[1 - e] //MatrixForm //TeXForm



                    $left(
                    begin{array}{cccccc}
                    0 & 0 & 0 & 0 & 0 & 0 \
                    0 & -1 & 0 & 0 & 0 & 0 \
                    0 & 1 & 0 & -1 & 0 & 0 \
                    0 & 0 & 0 & 0 & 0 & 0 \
                    0 & 0 & 0 & 1 & -1 & 0 \
                    0 & 0 & 0 & 0 & 1 & 0 \
                    0 & 0 & 0 & 0 & 0 & 0 \
                    end{array}
                    right)$




                    Here is a function that does one iteration:



                    iter[m_] := Module[{u = Unitize[m], o, v, e},
                    o = ConstantArray[1, Length @ First @ u];
                    v = Clip[u . o, {1, 1}, {0, 0}];
                    e = Unitize[v . Unitize[m]];
                    {
                    v,
                    v . Mod[m, 3] . SparseArray[Band[{1,1}] -> e],
                    m . SparseArray[Band[{1,1}] -> 1 - e]
                    }
                    ]


                    For example:



                    r = iter[m];
                    r[[1]] (* removed vertices *)
                    r[[2]] (* removed edges *)
                    r[[3]] //MatrixForm //TeXForm



                    {1, 0, 0, 1, 0, 0, 1}



                    {2, 0, 2, 0, 0, 1}



                    $left(
                    begin{array}{cccccc}
                    0 & 0 & 0 & 0 & 0 & 0 \
                    0 & -1 & 0 & 0 & 0 & 0 \
                    0 & 1 & 0 & -1 & 0 & 0 \
                    0 & 0 & 0 & 0 & 0 & 0 \
                    0 & 0 & 0 & 1 & -1 & 0 \
                    0 & 0 & 0 & 0 & 1 & 0 \
                    0 & 0 & 0 & 0 & 0 & 0 \
                    end{array}
                    right)$




                    Putting the above together:



                    res = NestWhileList[iter @* Last, iter[m], Positive @* Total @* First]


                    enter image description here



                    Deciding which edges are outgoing and incoming can be done with:



                    KeyDrop[
                    GroupBy[Thread[edges -> res[[1, 2]]], Last -> First],
                    0
                    ]



                    <|2 -> {1 [DirectedEdge] 2, 4 [DirectedEdge] 3}, 1 -> {6 [DirectedEdge] 7}|>




                    Converting the SparseArray back to a graph (the removed edges/vertices need to be eliminated from the sparse array) can be done with:



                    With[
                    {
                    v = Pick[Range @ Length @ res[[1, 1]], res[[1, 1]], 0],
                    e = Pick[Range @ Length @ res[[1, 2]], res[[1, 2]], 0]
                    },

                    IncidenceGraph[
                    v,
                    res[[1, 3]][[v, e]],
                    VertexLabels->"Name"
                    ]
                    ]


                    enter image description here



                    Your second example:



                    edges = {DirectedEdge[1, 2], DirectedEdge[2, 3], DirectedEdge[4, 3], DirectedEdge[5, 4]};
                    NestWhileList[
                    iter @* Last,
                    iter @ IncidenceMatrix[edges],
                    Positive @* Total @* First
                    ]


                    enter image description here






                    share|improve this answer












                    If you have a large graph, it will be faster to work with the vertex-edge incidence matrix instead of Graph objects. The edges you want to strip off will have either a 1 or a -1 depending on the direction of the directed edge. So, the simple edges you want to strip off will have a vertex with only a single 1 or -1 in the row. Let's take your example:



                    m = IncidenceMatrix[edges];
                    m //MatrixForm //TeXForm



                    $left(
                    begin{array}{cccccc}
                    -1 & 0 & 0 & 0 & 0 & 0 \
                    1 & -1 & 0 & 0 & 0 & 0 \
                    0 & 1 & 1 & -1 & 0 & 0 \
                    0 & 0 & -1 & 0 & 0 & 0 \
                    0 & 0 & 0 & 1 & -1 & 0 \
                    0 & 0 & 0 & 0 & 1 & -1 \
                    0 & 0 & 0 & 0 & 0 & 1 \
                    end{array}
                    right)$




                    The vertices that can be removed can be obtained with:



                    v = Clip[Unitize[m] . ConstantArray[1, Length @ First @ m], {1, 1}, {0, 0}]



                    {1, 0, 0, 1, 0, 0, 1}




                    The corresponding edges can be found with:



                    e = Unitize[v . Unitize[m]]



                    {1, 0, 1, 0, 0, 1}




                    The kind of edge can be determined using:



                    v . Mod[m, 3] . DiagonalMatrix[e]



                    {2, 0, 2, 0, 0, 1}




                    where 1 is an outgoing edge, 2 is an incoming edge, and 3 would be both an incoming and outgoing edge.



                    The matrix after removing the above vertices and edges can be found from:



                    m . DiagonalMatrix[1 - e] //MatrixForm //TeXForm



                    $left(
                    begin{array}{cccccc}
                    0 & 0 & 0 & 0 & 0 & 0 \
                    0 & -1 & 0 & 0 & 0 & 0 \
                    0 & 1 & 0 & -1 & 0 & 0 \
                    0 & 0 & 0 & 0 & 0 & 0 \
                    0 & 0 & 0 & 1 & -1 & 0 \
                    0 & 0 & 0 & 0 & 1 & 0 \
                    0 & 0 & 0 & 0 & 0 & 0 \
                    end{array}
                    right)$




                    Here is a function that does one iteration:



                    iter[m_] := Module[{u = Unitize[m], o, v, e},
                    o = ConstantArray[1, Length @ First @ u];
                    v = Clip[u . o, {1, 1}, {0, 0}];
                    e = Unitize[v . Unitize[m]];
                    {
                    v,
                    v . Mod[m, 3] . SparseArray[Band[{1,1}] -> e],
                    m . SparseArray[Band[{1,1}] -> 1 - e]
                    }
                    ]


                    For example:



                    r = iter[m];
                    r[[1]] (* removed vertices *)
                    r[[2]] (* removed edges *)
                    r[[3]] //MatrixForm //TeXForm



                    {1, 0, 0, 1, 0, 0, 1}



                    {2, 0, 2, 0, 0, 1}



                    $left(
                    begin{array}{cccccc}
                    0 & 0 & 0 & 0 & 0 & 0 \
                    0 & -1 & 0 & 0 & 0 & 0 \
                    0 & 1 & 0 & -1 & 0 & 0 \
                    0 & 0 & 0 & 0 & 0 & 0 \
                    0 & 0 & 0 & 1 & -1 & 0 \
                    0 & 0 & 0 & 0 & 1 & 0 \
                    0 & 0 & 0 & 0 & 0 & 0 \
                    end{array}
                    right)$




                    Putting the above together:



                    res = NestWhileList[iter @* Last, iter[m], Positive @* Total @* First]


                    enter image description here



                    Deciding which edges are outgoing and incoming can be done with:



                    KeyDrop[
                    GroupBy[Thread[edges -> res[[1, 2]]], Last -> First],
                    0
                    ]



                    <|2 -> {1 [DirectedEdge] 2, 4 [DirectedEdge] 3}, 1 -> {6 [DirectedEdge] 7}|>




                    Converting the SparseArray back to a graph (the removed edges/vertices need to be eliminated from the sparse array) can be done with:



                    With[
                    {
                    v = Pick[Range @ Length @ res[[1, 1]], res[[1, 1]], 0],
                    e = Pick[Range @ Length @ res[[1, 2]], res[[1, 2]], 0]
                    },

                    IncidenceGraph[
                    v,
                    res[[1, 3]][[v, e]],
                    VertexLabels->"Name"
                    ]
                    ]


                    enter image description here



                    Your second example:



                    edges = {DirectedEdge[1, 2], DirectedEdge[2, 3], DirectedEdge[4, 3], DirectedEdge[5, 4]};
                    NestWhileList[
                    iter @* Last,
                    iter @ IncidenceMatrix[edges],
                    Positive @* Total @* First
                    ]


                    enter image description here







                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered Nov 8 at 2:21









                    Carl Woll

                    65.9k285171




                    65.9k285171






















                        up vote
                        2
                        down vote













                        What you're looking for is called a "kcore" of the graph: the set of vertices with at least k edges to other vertices of the core.



                        Mathematica has a function that will find this for you:
                        https://reference.wolfram.com/language/ref/KCoreComponents.html
                        https://reference.wolfram.com/language/example/FindTheKCoreComponentsOfAGraph.html



                        To find the removed edges (if the iteration in which they are removed is not important), simply iterate over the original edge list and count those which are not attached to a vertex in the 2-core.






                        share|improve this answer



























                          up vote
                          2
                          down vote













                          What you're looking for is called a "kcore" of the graph: the set of vertices with at least k edges to other vertices of the core.



                          Mathematica has a function that will find this for you:
                          https://reference.wolfram.com/language/ref/KCoreComponents.html
                          https://reference.wolfram.com/language/example/FindTheKCoreComponentsOfAGraph.html



                          To find the removed edges (if the iteration in which they are removed is not important), simply iterate over the original edge list and count those which are not attached to a vertex in the 2-core.






                          share|improve this answer

























                            up vote
                            2
                            down vote










                            up vote
                            2
                            down vote









                            What you're looking for is called a "kcore" of the graph: the set of vertices with at least k edges to other vertices of the core.



                            Mathematica has a function that will find this for you:
                            https://reference.wolfram.com/language/ref/KCoreComponents.html
                            https://reference.wolfram.com/language/example/FindTheKCoreComponentsOfAGraph.html



                            To find the removed edges (if the iteration in which they are removed is not important), simply iterate over the original edge list and count those which are not attached to a vertex in the 2-core.






                            share|improve this answer














                            What you're looking for is called a "kcore" of the graph: the set of vertices with at least k edges to other vertices of the core.



                            Mathematica has a function that will find this for you:
                            https://reference.wolfram.com/language/ref/KCoreComponents.html
                            https://reference.wolfram.com/language/example/FindTheKCoreComponentsOfAGraph.html



                            To find the removed edges (if the iteration in which they are removed is not important), simply iterate over the original edge list and count those which are not attached to a vertex in the 2-core.







                            share|improve this answer














                            share|improve this answer



                            share|improve this answer








                            edited Nov 8 at 15:27

























                            answered Nov 8 at 15:21









                            geofurb

                            212




                            212






























                                draft saved

                                draft discarded




















































                                Thanks for contributing an answer to Mathematica Stack Exchange!


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                Use MathJax to format equations. MathJax reference.


                                To learn more, see our tips on writing great answers.





                                Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                                Please pay close attention to the following guidance:


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                To learn more, see our tips on writing great answers.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f185556%2fiteratively-strip-off-simply-connected-edges-in-graph%23new-answer', 'question_page');
                                }
                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                這個網誌中的熱門文章

                                Tangent Lines Diagram Along Smooth Curve

                                Yusuf al-Mu'taman ibn Hud

                                Zucchini