Number of spanning trees in a subgraph












0












$begingroup$


A spanning tree of a connected graph is identified with any acyclic subgraph that contains all vertices of this graph. How to formally prove or where to find a proof that any subgraph of a connected graph contains equal or less numer of spanning trees than the original graph ?






combinatorics graph-theory







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$endgroup$












  • $begingroup$
    Every spanning tree of the subgraph is a spanning tree of the original graph.
    $endgroup$
    – SmileyCraft
    Dec 23 '18 at 22:17






  • 1




    $begingroup$
    @SmileyCraft Only if the subgraph is spanning, which it may not be.
    $endgroup$
    – Misha Lavrov
    Dec 23 '18 at 22:34










  • $begingroup$
    If the subgraph is not spanning (I assume you mean connected) then there is no spanning tree in the first place. But then still every spanning tree of the subgraph is a spanning tree of the original graph.
    $endgroup$
    – SmileyCraft
    Dec 23 '18 at 22:36










  • $begingroup$
    I mean that if the subgraph does not have all the vertices that the original graph has, then a spanning tree of the subgraph extends to a spanning tree of the original graph, but you have to add more vertices and edges to get there
    $endgroup$
    – Misha Lavrov
    Dec 23 '18 at 22:57
















0












$begingroup$


A spanning tree of a connected graph is identified with any acyclic subgraph that contains all vertices of this graph. How to formally prove or where to find a proof that any subgraph of a connected graph contains equal or less numer of spanning trees than the original graph ?






combinatorics graph-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    Every spanning tree of the subgraph is a spanning tree of the original graph.
    $endgroup$
    – SmileyCraft
    Dec 23 '18 at 22:17






  • 1




    $begingroup$
    @SmileyCraft Only if the subgraph is spanning, which it may not be.
    $endgroup$
    – Misha Lavrov
    Dec 23 '18 at 22:34










  • $begingroup$
    If the subgraph is not spanning (I assume you mean connected) then there is no spanning tree in the first place. But then still every spanning tree of the subgraph is a spanning tree of the original graph.
    $endgroup$
    – SmileyCraft
    Dec 23 '18 at 22:36










  • $begingroup$
    I mean that if the subgraph does not have all the vertices that the original graph has, then a spanning tree of the subgraph extends to a spanning tree of the original graph, but you have to add more vertices and edges to get there
    $endgroup$
    – Misha Lavrov
    Dec 23 '18 at 22:57














0












0








0





$begingroup$


A spanning tree of a connected graph is identified with any acyclic subgraph that contains all vertices of this graph. How to formally prove or where to find a proof that any subgraph of a connected graph contains equal or less numer of spanning trees than the original graph ?






combinatorics graph-theory







share|cite|improve this question









$endgroup$




A spanning tree of a connected graph is identified with any acyclic subgraph that contains all vertices of this graph. How to formally prove or where to find a proof that any subgraph of a connected graph contains equal or less numer of spanning trees than the original graph ?






combinatorics graph-theory




combinatorics graph-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 23 '18 at 22:11









Piotr WilczekPiotr Wilczek

86




86












  • $begingroup$
    Every spanning tree of the subgraph is a spanning tree of the original graph.
    $endgroup$
    – SmileyCraft
    Dec 23 '18 at 22:17






  • 1




    $begingroup$
    @SmileyCraft Only if the subgraph is spanning, which it may not be.
    $endgroup$
    – Misha Lavrov
    Dec 23 '18 at 22:34










  • $begingroup$
    If the subgraph is not spanning (I assume you mean connected) then there is no spanning tree in the first place. But then still every spanning tree of the subgraph is a spanning tree of the original graph.
    $endgroup$
    – SmileyCraft
    Dec 23 '18 at 22:36










  • $begingroup$
    I mean that if the subgraph does not have all the vertices that the original graph has, then a spanning tree of the subgraph extends to a spanning tree of the original graph, but you have to add more vertices and edges to get there
    $endgroup$
    – Misha Lavrov
    Dec 23 '18 at 22:57


















  • $begingroup$
    Every spanning tree of the subgraph is a spanning tree of the original graph.
    $endgroup$
    – SmileyCraft
    Dec 23 '18 at 22:17






  • 1




    $begingroup$
    @SmileyCraft Only if the subgraph is spanning, which it may not be.
    $endgroup$
    – Misha Lavrov
    Dec 23 '18 at 22:34










  • $begingroup$
    If the subgraph is not spanning (I assume you mean connected) then there is no spanning tree in the first place. But then still every spanning tree of the subgraph is a spanning tree of the original graph.
    $endgroup$
    – SmileyCraft
    Dec 23 '18 at 22:36










  • $begingroup$
    I mean that if the subgraph does not have all the vertices that the original graph has, then a spanning tree of the subgraph extends to a spanning tree of the original graph, but you have to add more vertices and edges to get there
    $endgroup$
    – Misha Lavrov
    Dec 23 '18 at 22:57
















$begingroup$
Every spanning tree of the subgraph is a spanning tree of the original graph.
$endgroup$
– SmileyCraft
Dec 23 '18 at 22:17




$begingroup$
Every spanning tree of the subgraph is a spanning tree of the original graph.
$endgroup$
– SmileyCraft
Dec 23 '18 at 22:17




1




1




$begingroup$
@SmileyCraft Only if the subgraph is spanning, which it may not be.
$endgroup$
– Misha Lavrov
Dec 23 '18 at 22:34




$begingroup$
@SmileyCraft Only if the subgraph is spanning, which it may not be.
$endgroup$
– Misha Lavrov
Dec 23 '18 at 22:34












$begingroup$
If the subgraph is not spanning (I assume you mean connected) then there is no spanning tree in the first place. But then still every spanning tree of the subgraph is a spanning tree of the original graph.
$endgroup$
– SmileyCraft
Dec 23 '18 at 22:36




$begingroup$
If the subgraph is not spanning (I assume you mean connected) then there is no spanning tree in the first place. But then still every spanning tree of the subgraph is a spanning tree of the original graph.
$endgroup$
– SmileyCraft
Dec 23 '18 at 22:36












$begingroup$
I mean that if the subgraph does not have all the vertices that the original graph has, then a spanning tree of the subgraph extends to a spanning tree of the original graph, but you have to add more vertices and edges to get there
$endgroup$
– Misha Lavrov
Dec 23 '18 at 22:57




$begingroup$
I mean that if the subgraph does not have all the vertices that the original graph has, then a spanning tree of the subgraph extends to a spanning tree of the original graph, but you have to add more vertices and edges to get there
$endgroup$
– Misha Lavrov
Dec 23 '18 at 22:57










1 Answer
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$begingroup$

In general, the combinatorial strategy to prove that $|X| le |Y|$ is to find an injective function from $X$ to $Y$.



In this case, you have a subgraph $H$ of a connected graph $G$; to prove that $G$ has at least as many spanning trees as $H$, you want to find an injective function which turns spanning trees of $H$ into spanning trees of $G$.



Given a spanning tree $T$ of $H$, we can extend it to a spanning tree of $G$ as follows. For as long as $T$ is not a spanning tree of $G$ (that is, for as long as it does not contain the vertices of $G$) we can make $T$ larger by




  1. Picking an arbitrary vertex $v$ of $G$ which is not yet in $T$.

  2. Finding the shortest path from $v$ to $T$, ending at a vertex $w$ in $T$.

  3. Creating a larger tree $T'$ consisting of the tree $T$ and the $v,w$-path found in step 2.


Repeat this with the new tree $T'$ until you have a tree that spans all the vertices of $G$.



You should check that (a) this process is always possible to carry out, (b) it always ends with a spanning tree of $G$, and (c) if we start with different spanning trees of $H$, we are guaranteed to end up with different spanning trees of $П$.



(Statement (c) is necessary to guarantee that the function we construct is injective, which is vital to get the inequality we want.)






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    $begingroup$

    In general, the combinatorial strategy to prove that $|X| le |Y|$ is to find an injective function from $X$ to $Y$.



    In this case, you have a subgraph $H$ of a connected graph $G$; to prove that $G$ has at least as many spanning trees as $H$, you want to find an injective function which turns spanning trees of $H$ into spanning trees of $G$.



    Given a spanning tree $T$ of $H$, we can extend it to a spanning tree of $G$ as follows. For as long as $T$ is not a spanning tree of $G$ (that is, for as long as it does not contain the vertices of $G$) we can make $T$ larger by




    1. Picking an arbitrary vertex $v$ of $G$ which is not yet in $T$.

    2. Finding the shortest path from $v$ to $T$, ending at a vertex $w$ in $T$.

    3. Creating a larger tree $T'$ consisting of the tree $T$ and the $v,w$-path found in step 2.


    Repeat this with the new tree $T'$ until you have a tree that spans all the vertices of $G$.



    You should check that (a) this process is always possible to carry out, (b) it always ends with a spanning tree of $G$, and (c) if we start with different spanning trees of $H$, we are guaranteed to end up with different spanning trees of $П$.



    (Statement (c) is necessary to guarantee that the function we construct is injective, which is vital to get the inequality we want.)






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      In general, the combinatorial strategy to prove that $|X| le |Y|$ is to find an injective function from $X$ to $Y$.



      In this case, you have a subgraph $H$ of a connected graph $G$; to prove that $G$ has at least as many spanning trees as $H$, you want to find an injective function which turns spanning trees of $H$ into spanning trees of $G$.



      Given a spanning tree $T$ of $H$, we can extend it to a spanning tree of $G$ as follows. For as long as $T$ is not a spanning tree of $G$ (that is, for as long as it does not contain the vertices of $G$) we can make $T$ larger by




      1. Picking an arbitrary vertex $v$ of $G$ which is not yet in $T$.

      2. Finding the shortest path from $v$ to $T$, ending at a vertex $w$ in $T$.

      3. Creating a larger tree $T'$ consisting of the tree $T$ and the $v,w$-path found in step 2.


      Repeat this with the new tree $T'$ until you have a tree that spans all the vertices of $G$.



      You should check that (a) this process is always possible to carry out, (b) it always ends with a spanning tree of $G$, and (c) if we start with different spanning trees of $H$, we are guaranteed to end up with different spanning trees of $П$.



      (Statement (c) is necessary to guarantee that the function we construct is injective, which is vital to get the inequality we want.)






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        In general, the combinatorial strategy to prove that $|X| le |Y|$ is to find an injective function from $X$ to $Y$.



        In this case, you have a subgraph $H$ of a connected graph $G$; to prove that $G$ has at least as many spanning trees as $H$, you want to find an injective function which turns spanning trees of $H$ into spanning trees of $G$.



        Given a spanning tree $T$ of $H$, we can extend it to a spanning tree of $G$ as follows. For as long as $T$ is not a spanning tree of $G$ (that is, for as long as it does not contain the vertices of $G$) we can make $T$ larger by




        1. Picking an arbitrary vertex $v$ of $G$ which is not yet in $T$.

        2. Finding the shortest path from $v$ to $T$, ending at a vertex $w$ in $T$.

        3. Creating a larger tree $T'$ consisting of the tree $T$ and the $v,w$-path found in step 2.


        Repeat this with the new tree $T'$ until you have a tree that spans all the vertices of $G$.



        You should check that (a) this process is always possible to carry out, (b) it always ends with a spanning tree of $G$, and (c) if we start with different spanning trees of $H$, we are guaranteed to end up with different spanning trees of $П$.



        (Statement (c) is necessary to guarantee that the function we construct is injective, which is vital to get the inequality we want.)






        share|cite|improve this answer











        $endgroup$



        In general, the combinatorial strategy to prove that $|X| le |Y|$ is to find an injective function from $X$ to $Y$.



        In this case, you have a subgraph $H$ of a connected graph $G$; to prove that $G$ has at least as many spanning trees as $H$, you want to find an injective function which turns spanning trees of $H$ into spanning trees of $G$.



        Given a spanning tree $T$ of $H$, we can extend it to a spanning tree of $G$ as follows. For as long as $T$ is not a spanning tree of $G$ (that is, for as long as it does not contain the vertices of $G$) we can make $T$ larger by




        1. Picking an arbitrary vertex $v$ of $G$ which is not yet in $T$.

        2. Finding the shortest path from $v$ to $T$, ending at a vertex $w$ in $T$.

        3. Creating a larger tree $T'$ consisting of the tree $T$ and the $v,w$-path found in step 2.


        Repeat this with the new tree $T'$ until you have a tree that spans all the vertices of $G$.



        You should check that (a) this process is always possible to carry out, (b) it always ends with a spanning tree of $G$, and (c) if we start with different spanning trees of $H$, we are guaranteed to end up with different spanning trees of $П$.



        (Statement (c) is necessary to guarantee that the function we construct is injective, which is vital to get the inequality we want.)







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 24 '18 at 17:25

























        answered Dec 23 '18 at 23:46









        Misha LavrovMisha Lavrov

        46.8k657107




        46.8k657107






























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