Question regarding the connectedness of spanning 2-regular subgraphs











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If a 4-regular connected graph does not have one or more cut vertices then we can say it has a 2-regular spanning sub graph which is connected, isn't it? (A spanning sub graph with one component)



There might be several 2-regular spanning subgraphs, some which are a union of disjoint cycles, but if there is no cut vertex above type of a spanning sub graph will also be present, right?



Can someone please guide me to understand this.



Thanks a lot in advance.
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graph-theory







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    up vote
    0
    down vote

    favorite












    If a 4-regular connected graph does not have one or more cut vertices then we can say it has a 2-regular spanning sub graph which is connected, isn't it? (A spanning sub graph with one component)



    There might be several 2-regular spanning subgraphs, some which are a union of disjoint cycles, but if there is no cut vertex above type of a spanning sub graph will also be present, right?



    Can someone please guide me to understand this.



    Thanks a lot in advance.
    enter image description here






    graph-theory







    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      If a 4-regular connected graph does not have one or more cut vertices then we can say it has a 2-regular spanning sub graph which is connected, isn't it? (A spanning sub graph with one component)



      There might be several 2-regular spanning subgraphs, some which are a union of disjoint cycles, but if there is no cut vertex above type of a spanning sub graph will also be present, right?



      Can someone please guide me to understand this.



      Thanks a lot in advance.
      enter image description here






      graph-theory







      share|cite|improve this question















      If a 4-regular connected graph does not have one or more cut vertices then we can say it has a 2-regular spanning sub graph which is connected, isn't it? (A spanning sub graph with one component)



      There might be several 2-regular spanning subgraphs, some which are a union of disjoint cycles, but if there is no cut vertex above type of a spanning sub graph will also be present, right?



      Can someone please guide me to understand this.



      Thanks a lot in advance.
      enter image description here






      graph-theory




      graph-theory






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 14 hours ago

























      asked 2 days ago









      Buddhini Angelika

      1098




      1098






















          2 Answers
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          up vote
          1
          down vote













          A "2-regular spanning subgraph which is connected" is simply a Hamiltonian cycle, and in general when you have a reasonable-sounding condition for a Hamiltonian cycle to exist, it's probably not good enough.



          This MathOverflow answer gives one counter-example. (Here, a Hamiltonian cycle does not exist, because to visit every vertex, we would have to use the left and right vertices multiple times.)



          The Meredith graph is an even stronger counter-example: it is not just $2$-connected but $4$-connected (the best we can hope for a $4$-regular graph) yet is not Hamiltonian.






          share|cite|improve this answer























          • Thanks a lot @MishaLavrov in the graph in that link if we remove 2 vertices then it will be disconnected. Can't we get a good condition which will enable to decide that a connected 4 regular graph has a connected 2 regular spanning sub graph?
            – Buddhini Angelika
            yesterday










          • In general, determining if a 4-regular graph is Hamiltonian is NP-complete, so we won't have a good sufficient and necessary condition. Many sufficient conditions for Hamiltonian cycles don't work for sparse graphs, but it is a theorem of Tutte that, for example, a 4-connected planar graph is always Hamiltonian.
            – Misha Lavrov
            yesterday










          • Thank you. Can we take a 4 regular graph as a 4 connected graph?
            – Buddhini Angelika
            yesterday










          • I'm not sure what you're asking. Not all 4-regular graphs are 4-connected, but some are.
            – Misha Lavrov
            yesterday










          • Thanks @MishaLavrov. I have edited the question by adding a 4 regular graph. In that graph there exist 2 edge disjoint spanning subgraphs with one component. I.e. a Hamiltonian cycle. But I was trying to give a reason for the existence of a connected 2 regular spanning sub graph.
            – Buddhini Angelika
            14 hours ago


















          up vote
          0
          down vote













          Is this a good hint? If the graph $G $ is connected then because all degree of the vertices are even, $G $ has an eulérienne cycle. If $G $ is not connected, then each component of $G $ has an eulérienne cycle.






          share|cite|improve this answer





















          • This seems irrelevant to the question. Eulerian cycles are not $2$-regular spanning subgraphs.
            – Misha Lavrov
            yesterday










          • Thanks a lot all! Is there any good condition which will enable to decide that a 4 regular connected graph has a 2 regular spanning sub graph?
            – Buddhini Angelika
            yesterday











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          2 Answers
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          active

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          2 Answers
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          active

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          active

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          active

          oldest

          votes








          up vote
          1
          down vote













          A "2-regular spanning subgraph which is connected" is simply a Hamiltonian cycle, and in general when you have a reasonable-sounding condition for a Hamiltonian cycle to exist, it's probably not good enough.



          This MathOverflow answer gives one counter-example. (Here, a Hamiltonian cycle does not exist, because to visit every vertex, we would have to use the left and right vertices multiple times.)



          The Meredith graph is an even stronger counter-example: it is not just $2$-connected but $4$-connected (the best we can hope for a $4$-regular graph) yet is not Hamiltonian.






          share|cite|improve this answer























          • Thanks a lot @MishaLavrov in the graph in that link if we remove 2 vertices then it will be disconnected. Can't we get a good condition which will enable to decide that a connected 4 regular graph has a connected 2 regular spanning sub graph?
            – Buddhini Angelika
            yesterday










          • In general, determining if a 4-regular graph is Hamiltonian is NP-complete, so we won't have a good sufficient and necessary condition. Many sufficient conditions for Hamiltonian cycles don't work for sparse graphs, but it is a theorem of Tutte that, for example, a 4-connected planar graph is always Hamiltonian.
            – Misha Lavrov
            yesterday










          • Thank you. Can we take a 4 regular graph as a 4 connected graph?
            – Buddhini Angelika
            yesterday










          • I'm not sure what you're asking. Not all 4-regular graphs are 4-connected, but some are.
            – Misha Lavrov
            yesterday










          • Thanks @MishaLavrov. I have edited the question by adding a 4 regular graph. In that graph there exist 2 edge disjoint spanning subgraphs with one component. I.e. a Hamiltonian cycle. But I was trying to give a reason for the existence of a connected 2 regular spanning sub graph.
            – Buddhini Angelika
            14 hours ago















          up vote
          1
          down vote













          A "2-regular spanning subgraph which is connected" is simply a Hamiltonian cycle, and in general when you have a reasonable-sounding condition for a Hamiltonian cycle to exist, it's probably not good enough.



          This MathOverflow answer gives one counter-example. (Here, a Hamiltonian cycle does not exist, because to visit every vertex, we would have to use the left and right vertices multiple times.)



          The Meredith graph is an even stronger counter-example: it is not just $2$-connected but $4$-connected (the best we can hope for a $4$-regular graph) yet is not Hamiltonian.






          share|cite|improve this answer























          • Thanks a lot @MishaLavrov in the graph in that link if we remove 2 vertices then it will be disconnected. Can't we get a good condition which will enable to decide that a connected 4 regular graph has a connected 2 regular spanning sub graph?
            – Buddhini Angelika
            yesterday










          • In general, determining if a 4-regular graph is Hamiltonian is NP-complete, so we won't have a good sufficient and necessary condition. Many sufficient conditions for Hamiltonian cycles don't work for sparse graphs, but it is a theorem of Tutte that, for example, a 4-connected planar graph is always Hamiltonian.
            – Misha Lavrov
            yesterday










          • Thank you. Can we take a 4 regular graph as a 4 connected graph?
            – Buddhini Angelika
            yesterday










          • I'm not sure what you're asking. Not all 4-regular graphs are 4-connected, but some are.
            – Misha Lavrov
            yesterday










          • Thanks @MishaLavrov. I have edited the question by adding a 4 regular graph. In that graph there exist 2 edge disjoint spanning subgraphs with one component. I.e. a Hamiltonian cycle. But I was trying to give a reason for the existence of a connected 2 regular spanning sub graph.
            – Buddhini Angelika
            14 hours ago













          up vote
          1
          down vote










          up vote
          1
          down vote









          A "2-regular spanning subgraph which is connected" is simply a Hamiltonian cycle, and in general when you have a reasonable-sounding condition for a Hamiltonian cycle to exist, it's probably not good enough.



          This MathOverflow answer gives one counter-example. (Here, a Hamiltonian cycle does not exist, because to visit every vertex, we would have to use the left and right vertices multiple times.)



          The Meredith graph is an even stronger counter-example: it is not just $2$-connected but $4$-connected (the best we can hope for a $4$-regular graph) yet is not Hamiltonian.






          share|cite|improve this answer














          A "2-regular spanning subgraph which is connected" is simply a Hamiltonian cycle, and in general when you have a reasonable-sounding condition for a Hamiltonian cycle to exist, it's probably not good enough.



          This MathOverflow answer gives one counter-example. (Here, a Hamiltonian cycle does not exist, because to visit every vertex, we would have to use the left and right vertices multiple times.)



          The Meredith graph is an even stronger counter-example: it is not just $2$-connected but $4$-connected (the best we can hope for a $4$-regular graph) yet is not Hamiltonian.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited yesterday

























          answered yesterday









          Misha Lavrov

          42k555101




          42k555101












          • Thanks a lot @MishaLavrov in the graph in that link if we remove 2 vertices then it will be disconnected. Can't we get a good condition which will enable to decide that a connected 4 regular graph has a connected 2 regular spanning sub graph?
            – Buddhini Angelika
            yesterday










          • In general, determining if a 4-regular graph is Hamiltonian is NP-complete, so we won't have a good sufficient and necessary condition. Many sufficient conditions for Hamiltonian cycles don't work for sparse graphs, but it is a theorem of Tutte that, for example, a 4-connected planar graph is always Hamiltonian.
            – Misha Lavrov
            yesterday










          • Thank you. Can we take a 4 regular graph as a 4 connected graph?
            – Buddhini Angelika
            yesterday










          • I'm not sure what you're asking. Not all 4-regular graphs are 4-connected, but some are.
            – Misha Lavrov
            yesterday










          • Thanks @MishaLavrov. I have edited the question by adding a 4 regular graph. In that graph there exist 2 edge disjoint spanning subgraphs with one component. I.e. a Hamiltonian cycle. But I was trying to give a reason for the existence of a connected 2 regular spanning sub graph.
            – Buddhini Angelika
            14 hours ago


















          • Thanks a lot @MishaLavrov in the graph in that link if we remove 2 vertices then it will be disconnected. Can't we get a good condition which will enable to decide that a connected 4 regular graph has a connected 2 regular spanning sub graph?
            – Buddhini Angelika
            yesterday










          • In general, determining if a 4-regular graph is Hamiltonian is NP-complete, so we won't have a good sufficient and necessary condition. Many sufficient conditions for Hamiltonian cycles don't work for sparse graphs, but it is a theorem of Tutte that, for example, a 4-connected planar graph is always Hamiltonian.
            – Misha Lavrov
            yesterday










          • Thank you. Can we take a 4 regular graph as a 4 connected graph?
            – Buddhini Angelika
            yesterday










          • I'm not sure what you're asking. Not all 4-regular graphs are 4-connected, but some are.
            – Misha Lavrov
            yesterday










          • Thanks @MishaLavrov. I have edited the question by adding a 4 regular graph. In that graph there exist 2 edge disjoint spanning subgraphs with one component. I.e. a Hamiltonian cycle. But I was trying to give a reason for the existence of a connected 2 regular spanning sub graph.
            – Buddhini Angelika
            14 hours ago
















          Thanks a lot @MishaLavrov in the graph in that link if we remove 2 vertices then it will be disconnected. Can't we get a good condition which will enable to decide that a connected 4 regular graph has a connected 2 regular spanning sub graph?
          – Buddhini Angelika
          yesterday




          Thanks a lot @MishaLavrov in the graph in that link if we remove 2 vertices then it will be disconnected. Can't we get a good condition which will enable to decide that a connected 4 regular graph has a connected 2 regular spanning sub graph?
          – Buddhini Angelika
          yesterday












          In general, determining if a 4-regular graph is Hamiltonian is NP-complete, so we won't have a good sufficient and necessary condition. Many sufficient conditions for Hamiltonian cycles don't work for sparse graphs, but it is a theorem of Tutte that, for example, a 4-connected planar graph is always Hamiltonian.
          – Misha Lavrov
          yesterday




          In general, determining if a 4-regular graph is Hamiltonian is NP-complete, so we won't have a good sufficient and necessary condition. Many sufficient conditions for Hamiltonian cycles don't work for sparse graphs, but it is a theorem of Tutte that, for example, a 4-connected planar graph is always Hamiltonian.
          – Misha Lavrov
          yesterday












          Thank you. Can we take a 4 regular graph as a 4 connected graph?
          – Buddhini Angelika
          yesterday




          Thank you. Can we take a 4 regular graph as a 4 connected graph?
          – Buddhini Angelika
          yesterday












          I'm not sure what you're asking. Not all 4-regular graphs are 4-connected, but some are.
          – Misha Lavrov
          yesterday




          I'm not sure what you're asking. Not all 4-regular graphs are 4-connected, but some are.
          – Misha Lavrov
          yesterday












          Thanks @MishaLavrov. I have edited the question by adding a 4 regular graph. In that graph there exist 2 edge disjoint spanning subgraphs with one component. I.e. a Hamiltonian cycle. But I was trying to give a reason for the existence of a connected 2 regular spanning sub graph.
          – Buddhini Angelika
          14 hours ago




          Thanks @MishaLavrov. I have edited the question by adding a 4 regular graph. In that graph there exist 2 edge disjoint spanning subgraphs with one component. I.e. a Hamiltonian cycle. But I was trying to give a reason for the existence of a connected 2 regular spanning sub graph.
          – Buddhini Angelika
          14 hours ago










          up vote
          0
          down vote













          Is this a good hint? If the graph $G $ is connected then because all degree of the vertices are even, $G $ has an eulérienne cycle. If $G $ is not connected, then each component of $G $ has an eulérienne cycle.






          share|cite|improve this answer





















          • This seems irrelevant to the question. Eulerian cycles are not $2$-regular spanning subgraphs.
            – Misha Lavrov
            yesterday










          • Thanks a lot all! Is there any good condition which will enable to decide that a 4 regular connected graph has a 2 regular spanning sub graph?
            – Buddhini Angelika
            yesterday















          up vote
          0
          down vote













          Is this a good hint? If the graph $G $ is connected then because all degree of the vertices are even, $G $ has an eulérienne cycle. If $G $ is not connected, then each component of $G $ has an eulérienne cycle.






          share|cite|improve this answer





















          • This seems irrelevant to the question. Eulerian cycles are not $2$-regular spanning subgraphs.
            – Misha Lavrov
            yesterday










          • Thanks a lot all! Is there any good condition which will enable to decide that a 4 regular connected graph has a 2 regular spanning sub graph?
            – Buddhini Angelika
            yesterday













          up vote
          0
          down vote










          up vote
          0
          down vote









          Is this a good hint? If the graph $G $ is connected then because all degree of the vertices are even, $G $ has an eulérienne cycle. If $G $ is not connected, then each component of $G $ has an eulérienne cycle.






          share|cite|improve this answer












          Is this a good hint? If the graph $G $ is connected then because all degree of the vertices are even, $G $ has an eulérienne cycle. If $G $ is not connected, then each component of $G $ has an eulérienne cycle.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 days ago









          mathnoob

          1,133116




          1,133116












          • This seems irrelevant to the question. Eulerian cycles are not $2$-regular spanning subgraphs.
            – Misha Lavrov
            yesterday










          • Thanks a lot all! Is there any good condition which will enable to decide that a 4 regular connected graph has a 2 regular spanning sub graph?
            – Buddhini Angelika
            yesterday


















          • This seems irrelevant to the question. Eulerian cycles are not $2$-regular spanning subgraphs.
            – Misha Lavrov
            yesterday










          • Thanks a lot all! Is there any good condition which will enable to decide that a 4 regular connected graph has a 2 regular spanning sub graph?
            – Buddhini Angelika
            yesterday
















          This seems irrelevant to the question. Eulerian cycles are not $2$-regular spanning subgraphs.
          – Misha Lavrov
          yesterday




          This seems irrelevant to the question. Eulerian cycles are not $2$-regular spanning subgraphs.
          – Misha Lavrov
          yesterday












          Thanks a lot all! Is there any good condition which will enable to decide that a 4 regular connected graph has a 2 regular spanning sub graph?
          – Buddhini Angelika
          yesterday




          Thanks a lot all! Is there any good condition which will enable to decide that a 4 regular connected graph has a 2 regular spanning sub graph?
          – Buddhini Angelika
          yesterday


















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