Find a closed walk in a graph that visits every vertex at least once.











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Prove that, for all finite connected graph, it is possible to find a closed walk that visits every vertex of the graph at least once.




I have no idea. I've try to prove using minimum spanning tree, but I need a closed walk. Any idea?










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    Prove that, for all finite connected graph, it is possible to find a closed walk that visits every vertex of the graph at least once.




    I have no idea. I've try to prove using minimum spanning tree, but I need a closed walk. Any idea?










    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite












      Prove that, for all finite connected graph, it is possible to find a closed walk that visits every vertex of the graph at least once.




      I have no idea. I've try to prove using minimum spanning tree, but I need a closed walk. Any idea?










      share|cite|improve this question














      Prove that, for all finite connected graph, it is possible to find a closed walk that visits every vertex of the graph at least once.




      I have no idea. I've try to prove using minimum spanning tree, but I need a closed walk. Any idea?







      graph-theory






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      asked Nov 22 at 17:58









      Pedro Salgado

      675




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          Enumerate you vertices $v_1,dots,v_n$. Because $G$ is connected, you can find a path from $v_i$ to $v_{i+1}$ for every $i=1,dots,n-1$. So you construct a path starting at $v_1$, visiting $v_2$, $v_3$, etc., and ending at $v_n$. Now you just go back to $v_1$ to close your tourist tour of the graph.






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            1 Answer
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            active

            oldest

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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            1
            down vote



            accepted










            Enumerate you vertices $v_1,dots,v_n$. Because $G$ is connected, you can find a path from $v_i$ to $v_{i+1}$ for every $i=1,dots,n-1$. So you construct a path starting at $v_1$, visiting $v_2$, $v_3$, etc., and ending at $v_n$. Now you just go back to $v_1$ to close your tourist tour of the graph.






            share|cite|improve this answer



























              up vote
              1
              down vote



              accepted










              Enumerate you vertices $v_1,dots,v_n$. Because $G$ is connected, you can find a path from $v_i$ to $v_{i+1}$ for every $i=1,dots,n-1$. So you construct a path starting at $v_1$, visiting $v_2$, $v_3$, etc., and ending at $v_n$. Now you just go back to $v_1$ to close your tourist tour of the graph.






              share|cite|improve this answer

























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                Enumerate you vertices $v_1,dots,v_n$. Because $G$ is connected, you can find a path from $v_i$ to $v_{i+1}$ for every $i=1,dots,n-1$. So you construct a path starting at $v_1$, visiting $v_2$, $v_3$, etc., and ending at $v_n$. Now you just go back to $v_1$ to close your tourist tour of the graph.






                share|cite|improve this answer














                Enumerate you vertices $v_1,dots,v_n$. Because $G$ is connected, you can find a path from $v_i$ to $v_{i+1}$ for every $i=1,dots,n-1$. So you construct a path starting at $v_1$, visiting $v_2$, $v_3$, etc., and ending at $v_n$. Now you just go back to $v_1$ to close your tourist tour of the graph.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 22 at 18:53

























                answered Nov 22 at 18:09









                Federico

                4,238512




                4,238512






























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