Graph Theory: How to prove this statement?












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$begingroup$


Assume $T$ is a weighted tree and we want to find a spanning tree that it's maximum edge-weight is minimum. Prove that each MST (Minimum Spanning Tree) is an answer to this problem, but not all answers are MST.



Any hints how to prove the statement?






discrete-mathematics graph-theory trees







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  • $begingroup$
    What do you mean by that $T$'s maximum edge-weight is minimum?
    $endgroup$
    – Santana Afton
    Apr 11 '17 at 11:07
















0












$begingroup$


Assume $T$ is a weighted tree and we want to find a spanning tree that it's maximum edge-weight is minimum. Prove that each MST (Minimum Spanning Tree) is an answer to this problem, but not all answers are MST.



Any hints how to prove the statement?






discrete-mathematics graph-theory trees







share|cite|improve this question









$endgroup$












  • $begingroup$
    What do you mean by that $T$'s maximum edge-weight is minimum?
    $endgroup$
    – Santana Afton
    Apr 11 '17 at 11:07














0












0








0





$begingroup$


Assume $T$ is a weighted tree and we want to find a spanning tree that it's maximum edge-weight is minimum. Prove that each MST (Minimum Spanning Tree) is an answer to this problem, but not all answers are MST.



Any hints how to prove the statement?






discrete-mathematics graph-theory trees







share|cite|improve this question









$endgroup$




Assume $T$ is a weighted tree and we want to find a spanning tree that it's maximum edge-weight is minimum. Prove that each MST (Minimum Spanning Tree) is an answer to this problem, but not all answers are MST.



Any hints how to prove the statement?






discrete-mathematics graph-theory trees




discrete-mathematics graph-theory trees






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Apr 11 '17 at 10:51









Ju BcJu Bc

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30918












  • $begingroup$
    What do you mean by that $T$'s maximum edge-weight is minimum?
    $endgroup$
    – Santana Afton
    Apr 11 '17 at 11:07


















  • $begingroup$
    What do you mean by that $T$'s maximum edge-weight is minimum?
    $endgroup$
    – Santana Afton
    Apr 11 '17 at 11:07
















$begingroup$
What do you mean by that $T$'s maximum edge-weight is minimum?
$endgroup$
– Santana Afton
Apr 11 '17 at 11:07




$begingroup$
What do you mean by that $T$'s maximum edge-weight is minimum?
$endgroup$
– Santana Afton
Apr 11 '17 at 11:07










2 Answers
2






active

oldest

votes


















2












$begingroup$

I assume $T$ is a weighted graph, rather than a tree, since otherwise there is only one spanning tree.



Kruskal's MST algorithm processes the edges of the graph in order of weight. So it will only consider an edge of weight $w$ if there is no spanning tree available using lower-weight edges.



For the other bit, try to construct a graph where all spanning trees have to use a particular edge. Now give that edge very large weight (and other edges all different smaller weights). Then any spanning tree will minimise the maximum weight, but there is only one minimum spanning tree.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    As the question asked for any MST, why is it enough to only consider the one Kruskal's algorithm computes?
    $endgroup$
    – Dirk
    Apr 11 '17 at 11:17










  • $begingroup$
    Because Kruskal's algorithm can compute any MST.
    $endgroup$
    – Especially Lime
    Apr 11 '17 at 13:02



















0












$begingroup$

I will assume you meant $T$ to be a weighted graph, and for notational convenience I'll rename $T$ to $G$. Let's call a spanning tree min-max spanning tree if the maximum edge weight in it is minimum over all spanning trees. So, we want to show that every minimum spanning tree is a min-max spanning tree, but a min-max spanning tree need not be a minimum spanning tree.



The second is easier to prove, so I'll start with that. Consider an undirected graph $G = (V,E)$, with $V = {v_1, v_2, v_3}$ and edge weights between $v_1$ and $v_2$ as $2$, between $v_2$ and $v_3$ as $1$, between $v_3$ and $v_1$ as $1$. Then the tree with edges between $v_1$ and $v_2$ and between $v_1$ and $v_3$ is a min-max spanning tree but it isn't a minimum spanning tree.



To prove that every minimum spanning tree is a min-max spanning tree, I'll make use of Cycle property. For the sake of contradiction, assume there exists a minimum spanning tree $T$ and a min-max spanning tree $T'$ for a graph $G$ such that there is an edge $e$ in $T$ whose weight is larger than any edge in $T'$. If we add edge $e$ to $T'$ we get a cycle $C$ containing $e$. Since edge $e$ is the heaviest edge in the cycle $C$, by the cycle property it cannot be present in a minimum spanning tree contradicting our initial assumption.






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    I assume $T$ is a weighted graph, rather than a tree, since otherwise there is only one spanning tree.



    Kruskal's MST algorithm processes the edges of the graph in order of weight. So it will only consider an edge of weight $w$ if there is no spanning tree available using lower-weight edges.



    For the other bit, try to construct a graph where all spanning trees have to use a particular edge. Now give that edge very large weight (and other edges all different smaller weights). Then any spanning tree will minimise the maximum weight, but there is only one minimum spanning tree.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      As the question asked for any MST, why is it enough to only consider the one Kruskal's algorithm computes?
      $endgroup$
      – Dirk
      Apr 11 '17 at 11:17










    • $begingroup$
      Because Kruskal's algorithm can compute any MST.
      $endgroup$
      – Especially Lime
      Apr 11 '17 at 13:02
















    2












    $begingroup$

    I assume $T$ is a weighted graph, rather than a tree, since otherwise there is only one spanning tree.



    Kruskal's MST algorithm processes the edges of the graph in order of weight. So it will only consider an edge of weight $w$ if there is no spanning tree available using lower-weight edges.



    For the other bit, try to construct a graph where all spanning trees have to use a particular edge. Now give that edge very large weight (and other edges all different smaller weights). Then any spanning tree will minimise the maximum weight, but there is only one minimum spanning tree.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      As the question asked for any MST, why is it enough to only consider the one Kruskal's algorithm computes?
      $endgroup$
      – Dirk
      Apr 11 '17 at 11:17










    • $begingroup$
      Because Kruskal's algorithm can compute any MST.
      $endgroup$
      – Especially Lime
      Apr 11 '17 at 13:02














    2












    2








    2





    $begingroup$

    I assume $T$ is a weighted graph, rather than a tree, since otherwise there is only one spanning tree.



    Kruskal's MST algorithm processes the edges of the graph in order of weight. So it will only consider an edge of weight $w$ if there is no spanning tree available using lower-weight edges.



    For the other bit, try to construct a graph where all spanning trees have to use a particular edge. Now give that edge very large weight (and other edges all different smaller weights). Then any spanning tree will minimise the maximum weight, but there is only one minimum spanning tree.






    share|cite|improve this answer









    $endgroup$



    I assume $T$ is a weighted graph, rather than a tree, since otherwise there is only one spanning tree.



    Kruskal's MST algorithm processes the edges of the graph in order of weight. So it will only consider an edge of weight $w$ if there is no spanning tree available using lower-weight edges.



    For the other bit, try to construct a graph where all spanning trees have to use a particular edge. Now give that edge very large weight (and other edges all different smaller weights). Then any spanning tree will minimise the maximum weight, but there is only one minimum spanning tree.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Apr 11 '17 at 11:15









    Especially LimeEspecially Lime

    22.9k23059




    22.9k23059








    • 1




      $begingroup$
      As the question asked for any MST, why is it enough to only consider the one Kruskal's algorithm computes?
      $endgroup$
      – Dirk
      Apr 11 '17 at 11:17










    • $begingroup$
      Because Kruskal's algorithm can compute any MST.
      $endgroup$
      – Especially Lime
      Apr 11 '17 at 13:02














    • 1




      $begingroup$
      As the question asked for any MST, why is it enough to only consider the one Kruskal's algorithm computes?
      $endgroup$
      – Dirk
      Apr 11 '17 at 11:17










    • $begingroup$
      Because Kruskal's algorithm can compute any MST.
      $endgroup$
      – Especially Lime
      Apr 11 '17 at 13:02








    1




    1




    $begingroup$
    As the question asked for any MST, why is it enough to only consider the one Kruskal's algorithm computes?
    $endgroup$
    – Dirk
    Apr 11 '17 at 11:17




    $begingroup$
    As the question asked for any MST, why is it enough to only consider the one Kruskal's algorithm computes?
    $endgroup$
    – Dirk
    Apr 11 '17 at 11:17












    $begingroup$
    Because Kruskal's algorithm can compute any MST.
    $endgroup$
    – Especially Lime
    Apr 11 '17 at 13:02




    $begingroup$
    Because Kruskal's algorithm can compute any MST.
    $endgroup$
    – Especially Lime
    Apr 11 '17 at 13:02











    0












    $begingroup$

    I will assume you meant $T$ to be a weighted graph, and for notational convenience I'll rename $T$ to $G$. Let's call a spanning tree min-max spanning tree if the maximum edge weight in it is minimum over all spanning trees. So, we want to show that every minimum spanning tree is a min-max spanning tree, but a min-max spanning tree need not be a minimum spanning tree.



    The second is easier to prove, so I'll start with that. Consider an undirected graph $G = (V,E)$, with $V = {v_1, v_2, v_3}$ and edge weights between $v_1$ and $v_2$ as $2$, between $v_2$ and $v_3$ as $1$, between $v_3$ and $v_1$ as $1$. Then the tree with edges between $v_1$ and $v_2$ and between $v_1$ and $v_3$ is a min-max spanning tree but it isn't a minimum spanning tree.



    To prove that every minimum spanning tree is a min-max spanning tree, I'll make use of Cycle property. For the sake of contradiction, assume there exists a minimum spanning tree $T$ and a min-max spanning tree $T'$ for a graph $G$ such that there is an edge $e$ in $T$ whose weight is larger than any edge in $T'$. If we add edge $e$ to $T'$ we get a cycle $C$ containing $e$. Since edge $e$ is the heaviest edge in the cycle $C$, by the cycle property it cannot be present in a minimum spanning tree contradicting our initial assumption.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      I will assume you meant $T$ to be a weighted graph, and for notational convenience I'll rename $T$ to $G$. Let's call a spanning tree min-max spanning tree if the maximum edge weight in it is minimum over all spanning trees. So, we want to show that every minimum spanning tree is a min-max spanning tree, but a min-max spanning tree need not be a minimum spanning tree.



      The second is easier to prove, so I'll start with that. Consider an undirected graph $G = (V,E)$, with $V = {v_1, v_2, v_3}$ and edge weights between $v_1$ and $v_2$ as $2$, between $v_2$ and $v_3$ as $1$, between $v_3$ and $v_1$ as $1$. Then the tree with edges between $v_1$ and $v_2$ and between $v_1$ and $v_3$ is a min-max spanning tree but it isn't a minimum spanning tree.



      To prove that every minimum spanning tree is a min-max spanning tree, I'll make use of Cycle property. For the sake of contradiction, assume there exists a minimum spanning tree $T$ and a min-max spanning tree $T'$ for a graph $G$ such that there is an edge $e$ in $T$ whose weight is larger than any edge in $T'$. If we add edge $e$ to $T'$ we get a cycle $C$ containing $e$. Since edge $e$ is the heaviest edge in the cycle $C$, by the cycle property it cannot be present in a minimum spanning tree contradicting our initial assumption.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        I will assume you meant $T$ to be a weighted graph, and for notational convenience I'll rename $T$ to $G$. Let's call a spanning tree min-max spanning tree if the maximum edge weight in it is minimum over all spanning trees. So, we want to show that every minimum spanning tree is a min-max spanning tree, but a min-max spanning tree need not be a minimum spanning tree.



        The second is easier to prove, so I'll start with that. Consider an undirected graph $G = (V,E)$, with $V = {v_1, v_2, v_3}$ and edge weights between $v_1$ and $v_2$ as $2$, between $v_2$ and $v_3$ as $1$, between $v_3$ and $v_1$ as $1$. Then the tree with edges between $v_1$ and $v_2$ and between $v_1$ and $v_3$ is a min-max spanning tree but it isn't a minimum spanning tree.



        To prove that every minimum spanning tree is a min-max spanning tree, I'll make use of Cycle property. For the sake of contradiction, assume there exists a minimum spanning tree $T$ and a min-max spanning tree $T'$ for a graph $G$ such that there is an edge $e$ in $T$ whose weight is larger than any edge in $T'$. If we add edge $e$ to $T'$ we get a cycle $C$ containing $e$. Since edge $e$ is the heaviest edge in the cycle $C$, by the cycle property it cannot be present in a minimum spanning tree contradicting our initial assumption.






        share|cite|improve this answer









        $endgroup$



        I will assume you meant $T$ to be a weighted graph, and for notational convenience I'll rename $T$ to $G$. Let's call a spanning tree min-max spanning tree if the maximum edge weight in it is minimum over all spanning trees. So, we want to show that every minimum spanning tree is a min-max spanning tree, but a min-max spanning tree need not be a minimum spanning tree.



        The second is easier to prove, so I'll start with that. Consider an undirected graph $G = (V,E)$, with $V = {v_1, v_2, v_3}$ and edge weights between $v_1$ and $v_2$ as $2$, between $v_2$ and $v_3$ as $1$, between $v_3$ and $v_1$ as $1$. Then the tree with edges between $v_1$ and $v_2$ and between $v_1$ and $v_3$ is a min-max spanning tree but it isn't a minimum spanning tree.



        To prove that every minimum spanning tree is a min-max spanning tree, I'll make use of Cycle property. For the sake of contradiction, assume there exists a minimum spanning tree $T$ and a min-max spanning tree $T'$ for a graph $G$ such that there is an edge $e$ in $T$ whose weight is larger than any edge in $T'$. If we add edge $e$ to $T'$ we get a cycle $C$ containing $e$. Since edge $e$ is the heaviest edge in the cycle $C$, by the cycle property it cannot be present in a minimum spanning tree contradicting our initial assumption.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 26 '18 at 7:41









        AdityaAditya

        19116




        19116






























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