Regular connected graph with at least one bridge.












3












$begingroup$



We have a connected $k$-regular graph with at least one bridge. For which $k$ does such a graph exsist?




Clearly if $k$ is even, then this graph is Eulerian so it does not have a bridge. So $k$ must be odd.



Also if we delte this bride, then we have components $A$ and $B$ with $a$ and $b$ elements, so we have for component $A$ by handshake lemma $$(a-1)cdot k+1cdot (k-1) = 2alpha$$
where $alpha $ is the number of edges in $A$. So $a$ must be odd. The same is true for $B$.



Now, I don't know how to prove that $k$ is odd is also a sufficient condition. Any idea?






combinatorics graph-theory







share|cite|improve this question











$endgroup$

















    3












    $begingroup$



    We have a connected $k$-regular graph with at least one bridge. For which $k$ does such a graph exsist?




    Clearly if $k$ is even, then this graph is Eulerian so it does not have a bridge. So $k$ must be odd.



    Also if we delte this bride, then we have components $A$ and $B$ with $a$ and $b$ elements, so we have for component $A$ by handshake lemma $$(a-1)cdot k+1cdot (k-1) = 2alpha$$
    where $alpha $ is the number of edges in $A$. So $a$ must be odd. The same is true for $B$.



    Now, I don't know how to prove that $k$ is odd is also a sufficient condition. Any idea?






    combinatorics graph-theory







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$



      We have a connected $k$-regular graph with at least one bridge. For which $k$ does such a graph exsist?




      Clearly if $k$ is even, then this graph is Eulerian so it does not have a bridge. So $k$ must be odd.



      Also if we delte this bride, then we have components $A$ and $B$ with $a$ and $b$ elements, so we have for component $A$ by handshake lemma $$(a-1)cdot k+1cdot (k-1) = 2alpha$$
      where $alpha $ is the number of edges in $A$. So $a$ must be odd. The same is true for $B$.



      Now, I don't know how to prove that $k$ is odd is also a sufficient condition. Any idea?






      combinatorics graph-theory







      share|cite|improve this question











      $endgroup$





      We have a connected $k$-regular graph with at least one bridge. For which $k$ does such a graph exsist?




      Clearly if $k$ is even, then this graph is Eulerian so it does not have a bridge. So $k$ must be odd.



      Also if we delte this bride, then we have components $A$ and $B$ with $a$ and $b$ elements, so we have for component $A$ by handshake lemma $$(a-1)cdot k+1cdot (k-1) = 2alpha$$
      where $alpha $ is the number of edges in $A$. So $a$ must be odd. The same is true for $B$.



      Now, I don't know how to prove that $k$ is odd is also a sufficient condition. Any idea?






      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








      edited Jan 8 at 20:11







      Maria Mazur

















      asked Jan 8 at 18:57









      Maria MazurMaria Mazur

      50k1361125




      50k1361125






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          The special case $k=1$ is uninteresting and easy. So assume $kge 3$.



          Take a complete graph on $k+1$ vertices. Select and remove a perfect matching between $k-1$ of them (which is an even number). Each of those $k-1$ vertices now have degree $k-1$; connect them all to a fresh vertex to get their degree back up to $k$. The fresh vertex has exactly one valence left over, which will be your bridge when you connect the graph so far with a second copy of itself.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Great, how did you come up with this solution?
            $endgroup$
            – Maria Mazur
            Jan 8 at 20:02



















          1












          $begingroup$

          Your observation involving the handshake lemma can be strengthened a bit: $A$ must have degree sequence $(k, k, ldots, k, k-1)$, where there are $a-1$ copies of $k$. A similar statement can be made for $B$. Given two graphs $A$ and $B$ with degree sequences like this, can you construct the desired $k$-regular graph? Do you know some technique for showing that such graphs exist?






          share|cite|improve this answer









          $endgroup$














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            2 Answers
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            2 Answers
            2






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

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            3












            $begingroup$

            The special case $k=1$ is uninteresting and easy. So assume $kge 3$.



            Take a complete graph on $k+1$ vertices. Select and remove a perfect matching between $k-1$ of them (which is an even number). Each of those $k-1$ vertices now have degree $k-1$; connect them all to a fresh vertex to get their degree back up to $k$. The fresh vertex has exactly one valence left over, which will be your bridge when you connect the graph so far with a second copy of itself.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Great, how did you come up with this solution?
              $endgroup$
              – Maria Mazur
              Jan 8 at 20:02
















            3












            $begingroup$

            The special case $k=1$ is uninteresting and easy. So assume $kge 3$.



            Take a complete graph on $k+1$ vertices. Select and remove a perfect matching between $k-1$ of them (which is an even number). Each of those $k-1$ vertices now have degree $k-1$; connect them all to a fresh vertex to get their degree back up to $k$. The fresh vertex has exactly one valence left over, which will be your bridge when you connect the graph so far with a second copy of itself.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Great, how did you come up with this solution?
              $endgroup$
              – Maria Mazur
              Jan 8 at 20:02














            3












            3








            3





            $begingroup$

            The special case $k=1$ is uninteresting and easy. So assume $kge 3$.



            Take a complete graph on $k+1$ vertices. Select and remove a perfect matching between $k-1$ of them (which is an even number). Each of those $k-1$ vertices now have degree $k-1$; connect them all to a fresh vertex to get their degree back up to $k$. The fresh vertex has exactly one valence left over, which will be your bridge when you connect the graph so far with a second copy of itself.






            share|cite|improve this answer









            $endgroup$



            The special case $k=1$ is uninteresting and easy. So assume $kge 3$.



            Take a complete graph on $k+1$ vertices. Select and remove a perfect matching between $k-1$ of them (which is an even number). Each of those $k-1$ vertices now have degree $k-1$; connect them all to a fresh vertex to get their degree back up to $k$. The fresh vertex has exactly one valence left over, which will be your bridge when you connect the graph so far with a second copy of itself.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 8 at 19:54









            Henning MakholmHenning Makholm

            243k17312556




            243k17312556












            • $begingroup$
              Great, how did you come up with this solution?
              $endgroup$
              – Maria Mazur
              Jan 8 at 20:02


















            • $begingroup$
              Great, how did you come up with this solution?
              $endgroup$
              – Maria Mazur
              Jan 8 at 20:02
















            $begingroup$
            Great, how did you come up with this solution?
            $endgroup$
            – Maria Mazur
            Jan 8 at 20:02




            $begingroup$
            Great, how did you come up with this solution?
            $endgroup$
            – Maria Mazur
            Jan 8 at 20:02











            1












            $begingroup$

            Your observation involving the handshake lemma can be strengthened a bit: $A$ must have degree sequence $(k, k, ldots, k, k-1)$, where there are $a-1$ copies of $k$. A similar statement can be made for $B$. Given two graphs $A$ and $B$ with degree sequences like this, can you construct the desired $k$-regular graph? Do you know some technique for showing that such graphs exist?






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Your observation involving the handshake lemma can be strengthened a bit: $A$ must have degree sequence $(k, k, ldots, k, k-1)$, where there are $a-1$ copies of $k$. A similar statement can be made for $B$. Given two graphs $A$ and $B$ with degree sequences like this, can you construct the desired $k$-regular graph? Do you know some technique for showing that such graphs exist?






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Your observation involving the handshake lemma can be strengthened a bit: $A$ must have degree sequence $(k, k, ldots, k, k-1)$, where there are $a-1$ copies of $k$. A similar statement can be made for $B$. Given two graphs $A$ and $B$ with degree sequences like this, can you construct the desired $k$-regular graph? Do you know some technique for showing that such graphs exist?






                share|cite|improve this answer









                $endgroup$



                Your observation involving the handshake lemma can be strengthened a bit: $A$ must have degree sequence $(k, k, ldots, k, k-1)$, where there are $a-1$ copies of $k$. A similar statement can be made for $B$. Given two graphs $A$ and $B$ with degree sequences like this, can you construct the desired $k$-regular graph? Do you know some technique for showing that such graphs exist?







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 8 at 19:02









                Gregory J. PuleoGregory J. Puleo

                4,65631520




                4,65631520






























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