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
edited Nov 19 at 10:15
Especially Lime
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asked Nov 19 at 10:05
johntc121
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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
edited Nov 19 at 10:15
Especially Lime
21.2k22656
asked Nov 19 at 10:05
johntc121
194
add a comment |
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
edited Nov 19 at 10:15
Especially Lime
21.2k22656
asked Nov 19 at 10:05
johntc121
194
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
edited Nov 19 at 10:15
Especially Lime
21.2k22656
asked Nov 19 at 10:05
johntc121
194
edited Nov 19 at 10:15
Especially Lime
21.2k22656
asked Nov 19 at 10:05
johntc121
194
edited Nov 19 at 10:15
Especially Lime
21.2k22656
edited Nov 19 at 10:15
Especially Lime
21.2k22656
edited Nov 19 at 10:15
Especially Lime
21.2k22656
21.2k22656
asked Nov 19 at 10:05
johntc121
194
asked Nov 19 at 10:05
johntc121
194
asked Nov 19 at 10:05
johntc121
194
194
add a comment |
add a comment |
1 Answer
1
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?
answered Nov 19 at 10:14
Especially Lime
21.2k22656
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
Nov 19 at 10:30
@johntc121 yes, that's exactly right (of course you can also swap $m$ and $n$ in the second case).
– Especially Lime
Nov 19 at 10:49
Awesome. Thank you so much
– johntc121
Nov 19 at 16:13
add a comment |
1 Answer
1
active
oldest
votes
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?
answered Nov 19 at 10:14
Especially Lime
21.2k22656
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
Nov 19 at 10:30
@johntc121 yes, that's exactly right (of course you can also swap $m$ and $n$ in the second case).
– Especially Lime
Nov 19 at 10:49
Awesome. Thank you so much
– johntc121
Nov 19 at 16:13
add a comment |
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?
answered Nov 19 at 10:14
Especially Lime
21.2k22656
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
Nov 19 at 10:30
@johntc121 yes, that's exactly right (of course you can also swap $m$ and $n$ in the second case).
– Especially Lime
Nov 19 at 10:49
Awesome. Thank you so much
– johntc121
Nov 19 at 16:13
add a comment |
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?
answered Nov 19 at 10:14
Especially Lime
21.2k22656
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?
answered Nov 19 at 10:14
Especially Lime
21.2k22656
answered Nov 19 at 10:14
Especially Lime
21.2k22656
answered Nov 19 at 10:14
Especially Lime
21.2k22656
answered Nov 19 at 10:14
Especially Lime
21.2k22656
21.2k22656
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
Nov 19 at 10:30
@johntc121 yes, that's exactly right (of course you can also swap $m$ and $n$ in the second case).
– Especially Lime
Nov 19 at 10:49
Awesome. Thank you so much
– johntc121
Nov 19 at 16:13
add a comment |
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
Nov 19 at 10:30
@johntc121 yes, that's exactly right (of course you can also swap $m$ and $n$ in the second case).
– Especially Lime
Nov 19 at 10:49
Awesome. Thank you so much
– johntc121
Nov 19 at 16:13
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
Nov 19 at 10:30
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
Nov 19 at 10:30
@johntc121 yes, that's exactly right (of course you can also swap $m$ and $n$ in the second case).
– Especially Lime
Nov 19 at 10:49
@johntc121 yes, that's exactly right (of course you can also swap $m$ and $n$ in the second case).
– Especially Lime
Nov 19 at 10:49
Awesome. Thank you so much
– johntc121
Nov 19 at 16:13
Awesome. Thank you so much
– johntc121
Nov 19 at 16:13
add a comment |
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