Let $e$ be an edge of minimum weight in the connected weighted graph $G$. Every minimum spanning tree of $G$...












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$begingroup$



Let $e$ be an edge of minimum weight in the connected weighted graph
$G$. Every minimum spanning tree of $G$ contains $e$.




I have been told that this is not true. But I also know this property :
Must a minimum weight spanning tree for a graph contain the least weight edge of every vertex of the graph?



Is it not true because that $e$ is not necessary the only minimum weight edge?










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    1












    $begingroup$



    Let $e$ be an edge of minimum weight in the connected weighted graph
    $G$. Every minimum spanning tree of $G$ contains $e$.




    I have been told that this is not true. But I also know this property :
    Must a minimum weight spanning tree for a graph contain the least weight edge of every vertex of the graph?



    Is it not true because that $e$ is not necessary the only minimum weight edge?










    share|cite|improve this question









    $endgroup$















      1












      1








      1


      0



      $begingroup$



      Let $e$ be an edge of minimum weight in the connected weighted graph
      $G$. Every minimum spanning tree of $G$ contains $e$.




      I have been told that this is not true. But I also know this property :
      Must a minimum weight spanning tree for a graph contain the least weight edge of every vertex of the graph?



      Is it not true because that $e$ is not necessary the only minimum weight edge?










      share|cite|improve this question









      $endgroup$





      Let $e$ be an edge of minimum weight in the connected weighted graph
      $G$. Every minimum spanning tree of $G$ contains $e$.




      I have been told that this is not true. But I also know this property :
      Must a minimum weight spanning tree for a graph contain the least weight edge of every vertex of the graph?



      Is it not true because that $e$ is not necessary the only minimum weight edge?







      graph-theory






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      share|cite|improve this question











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      share|cite|improve this question










      asked Jan 21 at 17:51









      Marine GalantinMarine Galantin

      860316




      860316






















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          $begingroup$

          If $G$ contains a cycle of edges each of the same minimum weight, then it is not possible for $G$ to have a spanning tree without omitting at least one of those edges.



          Otherwise the answer is yes: Let $T$ be a spanning tree of $G$ NOT containing $e$. Then $T+e$ contains one cycle. By assumption, there is an edge in the cycle of non-minimum weight. Remove that edge, and we have a new spanning tree $T'$ of lower weight.






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            1 Answer
            1






            active

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            active

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            active

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            3












            $begingroup$

            If $G$ contains a cycle of edges each of the same minimum weight, then it is not possible for $G$ to have a spanning tree without omitting at least one of those edges.



            Otherwise the answer is yes: Let $T$ be a spanning tree of $G$ NOT containing $e$. Then $T+e$ contains one cycle. By assumption, there is an edge in the cycle of non-minimum weight. Remove that edge, and we have a new spanning tree $T'$ of lower weight.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              If $G$ contains a cycle of edges each of the same minimum weight, then it is not possible for $G$ to have a spanning tree without omitting at least one of those edges.



              Otherwise the answer is yes: Let $T$ be a spanning tree of $G$ NOT containing $e$. Then $T+e$ contains one cycle. By assumption, there is an edge in the cycle of non-minimum weight. Remove that edge, and we have a new spanning tree $T'$ of lower weight.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                If $G$ contains a cycle of edges each of the same minimum weight, then it is not possible for $G$ to have a spanning tree without omitting at least one of those edges.



                Otherwise the answer is yes: Let $T$ be a spanning tree of $G$ NOT containing $e$. Then $T+e$ contains one cycle. By assumption, there is an edge in the cycle of non-minimum weight. Remove that edge, and we have a new spanning tree $T'$ of lower weight.






                share|cite|improve this answer









                $endgroup$



                If $G$ contains a cycle of edges each of the same minimum weight, then it is not possible for $G$ to have a spanning tree without omitting at least one of those edges.



                Otherwise the answer is yes: Let $T$ be a spanning tree of $G$ NOT containing $e$. Then $T+e$ contains one cycle. By assumption, there is an edge in the cycle of non-minimum weight. Remove that edge, and we have a new spanning tree $T'$ of lower weight.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 21 at 17:59









                PMarPMar

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