On independent set of edges included in a perfect matching.












1












$begingroup$


Here is a problem I'm trying to solve:




Let $G$ be a graph on at least $2k+2$ vertices which has a perfect matching. Show that if every set of $k$ independent edges is included in a perfect matching then every set of $k−1$ independent edges is included in a perfect matching.




Here is my attempt:
By tutte berge theorem: $comp(G X) ≤|X|+|V(G)|−2k$ if and only if $G$ has a matching of size $k$.



Ok So I assume a perfect matching in $G$ has size at least $k$.
Let $S$ be a set of independent edges of size $k-1$, suppose it is not in a set of independent edges of size $k$, then every other edges in $G$ is incident to an end vertices of edges in $S$. But then $S$ could not be inside of a perfect matching of size at least $k$? What is wrong here?



Help, you can also give a complete solution.






proof-verification graph-theory







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Here is a problem I'm trying to solve:




    Let $G$ be a graph on at least $2k+2$ vertices which has a perfect matching. Show that if every set of $k$ independent edges is included in a perfect matching then every set of $k−1$ independent edges is included in a perfect matching.




    Here is my attempt:
    By tutte berge theorem: $comp(G X) ≤|X|+|V(G)|−2k$ if and only if $G$ has a matching of size $k$.



    Ok So I assume a perfect matching in $G$ has size at least $k$.
    Let $S$ be a set of independent edges of size $k-1$, suppose it is not in a set of independent edges of size $k$, then every other edges in $G$ is incident to an end vertices of edges in $S$. But then $S$ could not be inside of a perfect matching of size at least $k$? What is wrong here?



    Help, you can also give a complete solution.






    proof-verification graph-theory







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Here is a problem I'm trying to solve:




      Let $G$ be a graph on at least $2k+2$ vertices which has a perfect matching. Show that if every set of $k$ independent edges is included in a perfect matching then every set of $k−1$ independent edges is included in a perfect matching.




      Here is my attempt:
      By tutte berge theorem: $comp(G X) ≤|X|+|V(G)|−2k$ if and only if $G$ has a matching of size $k$.



      Ok So I assume a perfect matching in $G$ has size at least $k$.
      Let $S$ be a set of independent edges of size $k-1$, suppose it is not in a set of independent edges of size $k$, then every other edges in $G$ is incident to an end vertices of edges in $S$. But then $S$ could not be inside of a perfect matching of size at least $k$? What is wrong here?



      Help, you can also give a complete solution.






      proof-verification graph-theory







      share|cite|improve this question









      $endgroup$




      Here is a problem I'm trying to solve:




      Let $G$ be a graph on at least $2k+2$ vertices which has a perfect matching. Show that if every set of $k$ independent edges is included in a perfect matching then every set of $k−1$ independent edges is included in a perfect matching.




      Here is my attempt:
      By tutte berge theorem: $comp(G X) ≤|X|+|V(G)|−2k$ if and only if $G$ has a matching of size $k$.



      Ok So I assume a perfect matching in $G$ has size at least $k$.
      Let $S$ be a set of independent edges of size $k-1$, suppose it is not in a set of independent edges of size $k$, then every other edges in $G$ is incident to an end vertices of edges in $S$. But then $S$ could not be inside of a perfect matching of size at least $k$? What is wrong here?



      Help, you can also give a complete solution.






      proof-verification graph-theory




      proof-verification graph-theory






      share|cite|improve this question













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      asked Nov 24 '18 at 16:58









      nafhgoodnafhgood

      1,803422




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          $begingroup$

          Well, you would be done if you could extend every set $S$ of $k-1$ independent edges to a set of $k$ independent edges [make sure you see why]. And in fact you can.



          Let $M$ be a perfect matching. Then $M cup S$ contains at least one path or cycle $L$ with an odd number $ell$ of edges such that only $lfloor frac{ell}{2}rfloor$ of those edges are in $S$ and the remaining $lfloor frac{ell}{2}floor +1$ are in $M$. So let $S' = S - (Scap L) + (M cap L)$. Then (a) $S'$ is an independent set with $k-1+1 =k$ edges, and (b) every vertex covered by $S$ is also covered by $S'$ [make sure you see why for both (a) and (b)].



          Then, as $G$ has a perfect matching and at least $2k+2$ vertices, a perfect matching in $G$ has $k+1$ edges. Because $G$ also has the property that every independent set of $k$ edges can be extended to a perfect matching, this implies that there is a perfect matching containing $S'$ that has at least one edge $e$ not in $S'$. Then ${e}+S'$ is an independent set of edges. Therefore, as every vertex covered by $S$ is also covered by $S'$, it follows that ${e}+S$ is also an independent set of edges with $e$ not be in $S$ [make sure you see why]. So $S$ indeed can be extended to an independent set of $k$ edges, giving the desired result.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            a) a) edges in M and L form an alternating path. Begin and end with edges in M. So we can replace one of edge in S to two edges in M. b) since all the replaced edges are incident to each other all vertices covered by removed edges in S must be covered by the replaced edges in M. Actually I don't see b), Why is it that the end vertex of L is covered by S if l is a path?
            $endgroup$
            – nafhgood
            Nov 25 '18 at 3:02








          • 1




            $begingroup$
            $S$ may not [actually, does not] cover every vertex in $L$ but $S'$ does. And that is what you need: $S'$ to cover every vertex covered by $S$ [and perhaps two more]
            $endgroup$
            – Mike
            Nov 25 '18 at 11:25











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          1 Answer
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          1 Answer
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          0












          $begingroup$

          Well, you would be done if you could extend every set $S$ of $k-1$ independent edges to a set of $k$ independent edges [make sure you see why]. And in fact you can.



          Let $M$ be a perfect matching. Then $M cup S$ contains at least one path or cycle $L$ with an odd number $ell$ of edges such that only $lfloor frac{ell}{2}rfloor$ of those edges are in $S$ and the remaining $lfloor frac{ell}{2}floor +1$ are in $M$. So let $S' = S - (Scap L) + (M cap L)$. Then (a) $S'$ is an independent set with $k-1+1 =k$ edges, and (b) every vertex covered by $S$ is also covered by $S'$ [make sure you see why for both (a) and (b)].



          Then, as $G$ has a perfect matching and at least $2k+2$ vertices, a perfect matching in $G$ has $k+1$ edges. Because $G$ also has the property that every independent set of $k$ edges can be extended to a perfect matching, this implies that there is a perfect matching containing $S'$ that has at least one edge $e$ not in $S'$. Then ${e}+S'$ is an independent set of edges. Therefore, as every vertex covered by $S$ is also covered by $S'$, it follows that ${e}+S$ is also an independent set of edges with $e$ not be in $S$ [make sure you see why]. So $S$ indeed can be extended to an independent set of $k$ edges, giving the desired result.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            a) a) edges in M and L form an alternating path. Begin and end with edges in M. So we can replace one of edge in S to two edges in M. b) since all the replaced edges are incident to each other all vertices covered by removed edges in S must be covered by the replaced edges in M. Actually I don't see b), Why is it that the end vertex of L is covered by S if l is a path?
            $endgroup$
            – nafhgood
            Nov 25 '18 at 3:02








          • 1




            $begingroup$
            $S$ may not [actually, does not] cover every vertex in $L$ but $S'$ does. And that is what you need: $S'$ to cover every vertex covered by $S$ [and perhaps two more]
            $endgroup$
            – Mike
            Nov 25 '18 at 11:25
















          0












          $begingroup$

          Well, you would be done if you could extend every set $S$ of $k-1$ independent edges to a set of $k$ independent edges [make sure you see why]. And in fact you can.



          Let $M$ be a perfect matching. Then $M cup S$ contains at least one path or cycle $L$ with an odd number $ell$ of edges such that only $lfloor frac{ell}{2}rfloor$ of those edges are in $S$ and the remaining $lfloor frac{ell}{2}floor +1$ are in $M$. So let $S' = S - (Scap L) + (M cap L)$. Then (a) $S'$ is an independent set with $k-1+1 =k$ edges, and (b) every vertex covered by $S$ is also covered by $S'$ [make sure you see why for both (a) and (b)].



          Then, as $G$ has a perfect matching and at least $2k+2$ vertices, a perfect matching in $G$ has $k+1$ edges. Because $G$ also has the property that every independent set of $k$ edges can be extended to a perfect matching, this implies that there is a perfect matching containing $S'$ that has at least one edge $e$ not in $S'$. Then ${e}+S'$ is an independent set of edges. Therefore, as every vertex covered by $S$ is also covered by $S'$, it follows that ${e}+S$ is also an independent set of edges with $e$ not be in $S$ [make sure you see why]. So $S$ indeed can be extended to an independent set of $k$ edges, giving the desired result.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            a) a) edges in M and L form an alternating path. Begin and end with edges in M. So we can replace one of edge in S to two edges in M. b) since all the replaced edges are incident to each other all vertices covered by removed edges in S must be covered by the replaced edges in M. Actually I don't see b), Why is it that the end vertex of L is covered by S if l is a path?
            $endgroup$
            – nafhgood
            Nov 25 '18 at 3:02








          • 1




            $begingroup$
            $S$ may not [actually, does not] cover every vertex in $L$ but $S'$ does. And that is what you need: $S'$ to cover every vertex covered by $S$ [and perhaps two more]
            $endgroup$
            – Mike
            Nov 25 '18 at 11:25














          0












          0








          0





          $begingroup$

          Well, you would be done if you could extend every set $S$ of $k-1$ independent edges to a set of $k$ independent edges [make sure you see why]. And in fact you can.



          Let $M$ be a perfect matching. Then $M cup S$ contains at least one path or cycle $L$ with an odd number $ell$ of edges such that only $lfloor frac{ell}{2}rfloor$ of those edges are in $S$ and the remaining $lfloor frac{ell}{2}floor +1$ are in $M$. So let $S' = S - (Scap L) + (M cap L)$. Then (a) $S'$ is an independent set with $k-1+1 =k$ edges, and (b) every vertex covered by $S$ is also covered by $S'$ [make sure you see why for both (a) and (b)].



          Then, as $G$ has a perfect matching and at least $2k+2$ vertices, a perfect matching in $G$ has $k+1$ edges. Because $G$ also has the property that every independent set of $k$ edges can be extended to a perfect matching, this implies that there is a perfect matching containing $S'$ that has at least one edge $e$ not in $S'$. Then ${e}+S'$ is an independent set of edges. Therefore, as every vertex covered by $S$ is also covered by $S'$, it follows that ${e}+S$ is also an independent set of edges with $e$ not be in $S$ [make sure you see why]. So $S$ indeed can be extended to an independent set of $k$ edges, giving the desired result.






          share|cite|improve this answer









          $endgroup$



          Well, you would be done if you could extend every set $S$ of $k-1$ independent edges to a set of $k$ independent edges [make sure you see why]. And in fact you can.



          Let $M$ be a perfect matching. Then $M cup S$ contains at least one path or cycle $L$ with an odd number $ell$ of edges such that only $lfloor frac{ell}{2}rfloor$ of those edges are in $S$ and the remaining $lfloor frac{ell}{2}floor +1$ are in $M$. So let $S' = S - (Scap L) + (M cap L)$. Then (a) $S'$ is an independent set with $k-1+1 =k$ edges, and (b) every vertex covered by $S$ is also covered by $S'$ [make sure you see why for both (a) and (b)].



          Then, as $G$ has a perfect matching and at least $2k+2$ vertices, a perfect matching in $G$ has $k+1$ edges. Because $G$ also has the property that every independent set of $k$ edges can be extended to a perfect matching, this implies that there is a perfect matching containing $S'$ that has at least one edge $e$ not in $S'$. Then ${e}+S'$ is an independent set of edges. Therefore, as every vertex covered by $S$ is also covered by $S'$, it follows that ${e}+S$ is also an independent set of edges with $e$ not be in $S$ [make sure you see why]. So $S$ indeed can be extended to an independent set of $k$ edges, giving the desired result.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 24 '18 at 21:42









          MikeMike

          3,333311




          3,333311












          • $begingroup$
            a) a) edges in M and L form an alternating path. Begin and end with edges in M. So we can replace one of edge in S to two edges in M. b) since all the replaced edges are incident to each other all vertices covered by removed edges in S must be covered by the replaced edges in M. Actually I don't see b), Why is it that the end vertex of L is covered by S if l is a path?
            $endgroup$
            – nafhgood
            Nov 25 '18 at 3:02








          • 1




            $begingroup$
            $S$ may not [actually, does not] cover every vertex in $L$ but $S'$ does. And that is what you need: $S'$ to cover every vertex covered by $S$ [and perhaps two more]
            $endgroup$
            – Mike
            Nov 25 '18 at 11:25


















          • $begingroup$
            a) a) edges in M and L form an alternating path. Begin and end with edges in M. So we can replace one of edge in S to two edges in M. b) since all the replaced edges are incident to each other all vertices covered by removed edges in S must be covered by the replaced edges in M. Actually I don't see b), Why is it that the end vertex of L is covered by S if l is a path?
            $endgroup$
            – nafhgood
            Nov 25 '18 at 3:02








          • 1




            $begingroup$
            $S$ may not [actually, does not] cover every vertex in $L$ but $S'$ does. And that is what you need: $S'$ to cover every vertex covered by $S$ [and perhaps two more]
            $endgroup$
            – Mike
            Nov 25 '18 at 11:25
















          $begingroup$
          a) a) edges in M and L form an alternating path. Begin and end with edges in M. So we can replace one of edge in S to two edges in M. b) since all the replaced edges are incident to each other all vertices covered by removed edges in S must be covered by the replaced edges in M. Actually I don't see b), Why is it that the end vertex of L is covered by S if l is a path?
          $endgroup$
          – nafhgood
          Nov 25 '18 at 3:02






          $begingroup$
          a) a) edges in M and L form an alternating path. Begin and end with edges in M. So we can replace one of edge in S to two edges in M. b) since all the replaced edges are incident to each other all vertices covered by removed edges in S must be covered by the replaced edges in M. Actually I don't see b), Why is it that the end vertex of L is covered by S if l is a path?
          $endgroup$
          – nafhgood
          Nov 25 '18 at 3:02






          1




          1




          $begingroup$
          $S$ may not [actually, does not] cover every vertex in $L$ but $S'$ does. And that is what you need: $S'$ to cover every vertex covered by $S$ [and perhaps two more]
          $endgroup$
          – Mike
          Nov 25 '18 at 11:25




          $begingroup$
          $S$ may not [actually, does not] cover every vertex in $L$ but $S'$ does. And that is what you need: $S'$ to cover every vertex covered by $S$ [and perhaps two more]
          $endgroup$
          – Mike
          Nov 25 '18 at 11:25


















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