When does the complete bipartite graph K n,m have an Euler Trail(Path)?











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So I know that an Euler trail must have no more than two odd degree vertices.



So does this mean that either $n$ or $m$ must be odd? Or is it $n = m + 1$?






proof-verification graph-theory bipartite-graph eulerian-path







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    up vote
    1
    down vote

    favorite












    So I know that an Euler trail must have no more than two odd degree vertices.



    So does this mean that either $n$ or $m$ must be odd? Or is it $n = m + 1$?






    proof-verification graph-theory bipartite-graph eulerian-path







    share|cite|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      So I know that an Euler trail must have no more than two odd degree vertices.



      So does this mean that either $n$ or $m$ must be odd? Or is it $n = m + 1$?






      proof-verification graph-theory bipartite-graph eulerian-path







      share|cite|improve this question















      So I know that an Euler trail must have no more than two odd degree vertices.



      So does this mean that either $n$ or $m$ must be odd? Or is it $n = m + 1$?






      proof-verification graph-theory bipartite-graph eulerian-path




      proof-verification graph-theory bipartite-graph eulerian-path






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 14 hours ago









      Especially Lime

      20.9k22655




      20.9k22655










      asked 15 hours ago









      johntc121

      194




      194






















          1 Answer
          1






          active

          oldest

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          up vote
          2
          down vote



          accepted










          You're right that it has an Euler trail if and only if the number of odd-degree vertices is at most $2$.



          In $K_{m,n}$ there are $m$ vertices of degree $n$ and $n$ vertices of degree $n$.



          If $m,n$ are both even then all degrees are even so there is an Euler trail.



          If exactly one of them, say $m$, is odd then there are $n$ vertices of odd degree. So when can there be an Euler trail in this case?



          If $m,n$ are both odd then there are $m+n$ odd degree vertices. When does this give an Euler trail?






          share|cite|improve this answer





















          • So does this mean that there are two possiblities for the odd occurence? Ie... n = m =1? and n =2 and m = odd number >= 1?
            – johntc121
            14 hours ago












          • @johntc121 yes, that's exactly right (of course you can also swap $m$ and $n$ in the second case).
            – Especially Lime
            14 hours ago












          • Awesome. Thank you so much
            – johntc121
            8 hours ago











          Your Answer





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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          2
          down vote



          accepted










          You're right that it has an Euler trail if and only if the number of odd-degree vertices is at most $2$.



          In $K_{m,n}$ there are $m$ vertices of degree $n$ and $n$ vertices of degree $n$.



          If $m,n$ are both even then all degrees are even so there is an Euler trail.



          If exactly one of them, say $m$, is odd then there are $n$ vertices of odd degree. So when can there be an Euler trail in this case?



          If $m,n$ are both odd then there are $m+n$ odd degree vertices. When does this give an Euler trail?






          share|cite|improve this answer





















          • So does this mean that there are two possiblities for the odd occurence? Ie... n = m =1? and n =2 and m = odd number >= 1?
            – johntc121
            14 hours ago












          • @johntc121 yes, that's exactly right (of course you can also swap $m$ and $n$ in the second case).
            – Especially Lime
            14 hours ago












          • Awesome. Thank you so much
            – johntc121
            8 hours ago















          up vote
          2
          down vote



          accepted










          You're right that it has an Euler trail if and only if the number of odd-degree vertices is at most $2$.



          In $K_{m,n}$ there are $m$ vertices of degree $n$ and $n$ vertices of degree $n$.



          If $m,n$ are both even then all degrees are even so there is an Euler trail.



          If exactly one of them, say $m$, is odd then there are $n$ vertices of odd degree. So when can there be an Euler trail in this case?



          If $m,n$ are both odd then there are $m+n$ odd degree vertices. When does this give an Euler trail?






          share|cite|improve this answer





















          • So does this mean that there are two possiblities for the odd occurence? Ie... n = m =1? and n =2 and m = odd number >= 1?
            – johntc121
            14 hours ago












          • @johntc121 yes, that's exactly right (of course you can also swap $m$ and $n$ in the second case).
            – Especially Lime
            14 hours ago












          • Awesome. Thank you so much
            – johntc121
            8 hours ago













          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          You're right that it has an Euler trail if and only if the number of odd-degree vertices is at most $2$.



          In $K_{m,n}$ there are $m$ vertices of degree $n$ and $n$ vertices of degree $n$.



          If $m,n$ are both even then all degrees are even so there is an Euler trail.



          If exactly one of them, say $m$, is odd then there are $n$ vertices of odd degree. So when can there be an Euler trail in this case?



          If $m,n$ are both odd then there are $m+n$ odd degree vertices. When does this give an Euler trail?






          share|cite|improve this answer












          You're right that it has an Euler trail if and only if the number of odd-degree vertices is at most $2$.



          In $K_{m,n}$ there are $m$ vertices of degree $n$ and $n$ vertices of degree $n$.



          If $m,n$ are both even then all degrees are even so there is an Euler trail.



          If exactly one of them, say $m$, is odd then there are $n$ vertices of odd degree. So when can there be an Euler trail in this case?



          If $m,n$ are both odd then there are $m+n$ odd degree vertices. When does this give an Euler trail?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 14 hours ago









          Especially Lime

          20.9k22655




          20.9k22655












          • So does this mean that there are two possiblities for the odd occurence? Ie... n = m =1? and n =2 and m = odd number >= 1?
            – johntc121
            14 hours ago












          • @johntc121 yes, that's exactly right (of course you can also swap $m$ and $n$ in the second case).
            – Especially Lime
            14 hours ago












          • Awesome. Thank you so much
            – johntc121
            8 hours ago


















          • So does this mean that there are two possiblities for the odd occurence? Ie... n = m =1? and n =2 and m = odd number >= 1?
            – johntc121
            14 hours ago












          • @johntc121 yes, that's exactly right (of course you can also swap $m$ and $n$ in the second case).
            – Especially Lime
            14 hours ago












          • Awesome. Thank you so much
            – johntc121
            8 hours ago
















          So does this mean that there are two possiblities for the odd occurence? Ie... n = m =1? and n =2 and m = odd number >= 1?
          – johntc121
          14 hours ago






          So does this mean that there are two possiblities for the odd occurence? Ie... n = m =1? and n =2 and m = odd number >= 1?
          – johntc121
          14 hours ago














          @johntc121 yes, that's exactly right (of course you can also swap $m$ and $n$ in the second case).
          – Especially Lime
          14 hours ago






          @johntc121 yes, that's exactly right (of course you can also swap $m$ and $n$ in the second case).
          – Especially Lime
          14 hours ago














          Awesome. Thank you so much
          – johntc121
          8 hours ago




          Awesome. Thank you so much
          – johntc121
          8 hours ago


















           

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