Mixing
Graph Theory: Partition Edges of Multigraph
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Given a connected multigraph $ G = (V,E) $ show that if $d_G(x)$ is not even $forall x
in V $ or $|E|$ is even there exists a partition of $E = A $ $cup$ $B$ such that:
$|d_A(x) - d_B(x)| leq 1 $ $forall x in V $
where $ d_A(x) $ is the number of edges in $A$ incident to a vertex $x$ in $G$.
graph-theory
asked Jan 27 at 20:17
user638879user638879
92
$endgroup$
add a comment |
$begingroup$
Given a connected multigraph $ G = (V,E) $ show that if $d_G(x)$ is not even $forall x
in V $ or $|E|$ is even there exists a partition of $E = A $ $cup$ $B$ such that:
$|d_A(x) - d_B(x)| leq 1 $ $forall x in V $
where $ d_A(x) $ is the number of edges in $A$ incident to a vertex $x$ in $G$.
graph-theory
asked Jan 27 at 20:17
user638879user638879
92
$endgroup$
add a comment |
$begingroup$
Given a connected multigraph $ G = (V,E) $ show that if $d_G(x)$ is not even $forall x
in V $ or $|E|$ is even there exists a partition of $E = A $ $cup$ $B$ such that:
$|d_A(x) - d_B(x)| leq 1 $ $forall x in V $
where $ d_A(x) $ is the number of edges in $A$ incident to a vertex $x$ in $G$.
graph-theory
asked Jan 27 at 20:17
user638879user638879
92
$endgroup$
Given a connected multigraph $ G = (V,E) $ show that if $d_G(x)$ is not even $forall x
in V $ or $|E|$ is even there exists a partition of $E = A $ $cup$ $B$ such that:
$|d_A(x) - d_B(x)| leq 1 $ $forall x in V $
where $ d_A(x) $ is the number of edges in $A$ incident to a vertex $x$ in $G$.
graph-theory
graph-theory
asked Jan 27 at 20:17
user638879user638879
92
asked Jan 27 at 20:17
user638879user638879
92
asked Jan 27 at 20:17
user638879user638879
92
asked Jan 27 at 20:17
user638879user638879
92
asked Jan 27 at 20:17
user638879user638879
92
92
add a comment |
add a comment |
1 Answer
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Let $V_o$ be a set of vertices of $G$ with odd degree. Then $|V_o|$ is even. Let $E’={(v_1,v’_1),dots, (v_k,v’_k)}$ be an arbitrary partition of $V_o$ into pairs and $G’$ be a multigraph obtained from $G$ by adding $E’$ to its set of edges. Then $G’$ has a Eulerian cycle $C$. We shall walk along $C$ starting from an arbitrary vertex $v’$ of an added edge $e’$, if there is any, or starting from an arbitrary vertex, otherwise, enumerating edges which we walked until we walk the whole $C$. Now let $A’$ (resp. $B’$) be the set of edges of $Ecup E’$ with even (resp. odd) numbers, $A=A’cap E$, and $B=B’cap E$. Since $C$ is a cycle, $d_{A’}(v)=d_{B’}(v)$ for any vertex $vne v'$ of $G$. Since at most one of edges incident to $v$ belongs to $E’$, then $|d_{A}(v')-d_{B}(v')|le 1$. If $C$ has even length then the same argument imply that $d_{A’}(v’)=d_{B’}(v’)$ and $|d_{A}(v)-d_{B}(v)|le 1$. Otherwise $d_{A’}(v’)=d_{B’}(v’)+2$. Since $| Ecup E’|=|C|$ is odd, the conditions imposed on $G$ imply that $E’$ is non-empty. Then exactly one edge of $G'$ incident to $v’$ (namely, the edge $e’$) belongs to $E’$ and $e’in A’$. Then $$d_A(v’)=d_{A’}(v’)-1=d_{B’}(v’)+1=d_{B}(v’)+1.$$
answered Jan 28 at 1:56
Alex RavskyAlex Ravsky
42.7k32383
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1 Answer
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1 Answer
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active
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$begingroup$
Let $V_o$ be a set of vertices of $G$ with odd degree. Then $|V_o|$ is even. Let $E’={(v_1,v’_1),dots, (v_k,v’_k)}$ be an arbitrary partition of $V_o$ into pairs and $G’$ be a multigraph obtained from $G$ by adding $E’$ to its set of edges. Then $G’$ has a Eulerian cycle $C$. We shall walk along $C$ starting from an arbitrary vertex $v’$ of an added edge $e’$, if there is any, or starting from an arbitrary vertex, otherwise, enumerating edges which we walked until we walk the whole $C$. Now let $A’$ (resp. $B’$) be the set of edges of $Ecup E’$ with even (resp. odd) numbers, $A=A’cap E$, and $B=B’cap E$. Since $C$ is a cycle, $d_{A’}(v)=d_{B’}(v)$ for any vertex $vne v'$ of $G$. Since at most one of edges incident to $v$ belongs to $E’$, then $|d_{A}(v')-d_{B}(v')|le 1$. If $C$ has even length then the same argument imply that $d_{A’}(v’)=d_{B’}(v’)$ and $|d_{A}(v)-d_{B}(v)|le 1$. Otherwise $d_{A’}(v’)=d_{B’}(v’)+2$. Since $| Ecup E’|=|C|$ is odd, the conditions imposed on $G$ imply that $E’$ is non-empty. Then exactly one edge of $G'$ incident to $v’$ (namely, the edge $e’$) belongs to $E’$ and $e’in A’$. Then $$d_A(v’)=d_{A’}(v’)-1=d_{B’}(v’)+1=d_{B}(v’)+1.$$
answered Jan 28 at 1:56
Alex RavskyAlex Ravsky
42.7k32383
$endgroup$
add a comment |
$begingroup$
Let $V_o$ be a set of vertices of $G$ with odd degree. Then $|V_o|$ is even. Let $E’={(v_1,v’_1),dots, (v_k,v’_k)}$ be an arbitrary partition of $V_o$ into pairs and $G’$ be a multigraph obtained from $G$ by adding $E’$ to its set of edges. Then $G’$ has a Eulerian cycle $C$. We shall walk along $C$ starting from an arbitrary vertex $v’$ of an added edge $e’$, if there is any, or starting from an arbitrary vertex, otherwise, enumerating edges which we walked until we walk the whole $C$. Now let $A’$ (resp. $B’$) be the set of edges of $Ecup E’$ with even (resp. odd) numbers, $A=A’cap E$, and $B=B’cap E$. Since $C$ is a cycle, $d_{A’}(v)=d_{B’}(v)$ for any vertex $vne v'$ of $G$. Since at most one of edges incident to $v$ belongs to $E’$, then $|d_{A}(v')-d_{B}(v')|le 1$. If $C$ has even length then the same argument imply that $d_{A’}(v’)=d_{B’}(v’)$ and $|d_{A}(v)-d_{B}(v)|le 1$. Otherwise $d_{A’}(v’)=d_{B’}(v’)+2$. Since $| Ecup E’|=|C|$ is odd, the conditions imposed on $G$ imply that $E’$ is non-empty. Then exactly one edge of $G'$ incident to $v’$ (namely, the edge $e’$) belongs to $E’$ and $e’in A’$. Then $$d_A(v’)=d_{A’}(v’)-1=d_{B’}(v’)+1=d_{B}(v’)+1.$$
answered Jan 28 at 1:56
Alex RavskyAlex Ravsky
42.7k32383
$endgroup$
add a comment |
$begingroup$
Let $V_o$ be a set of vertices of $G$ with odd degree. Then $|V_o|$ is even. Let $E’={(v_1,v’_1),dots, (v_k,v’_k)}$ be an arbitrary partition of $V_o$ into pairs and $G’$ be a multigraph obtained from $G$ by adding $E’$ to its set of edges. Then $G’$ has a Eulerian cycle $C$. We shall walk along $C$ starting from an arbitrary vertex $v’$ of an added edge $e’$, if there is any, or starting from an arbitrary vertex, otherwise, enumerating edges which we walked until we walk the whole $C$. Now let $A’$ (resp. $B’$) be the set of edges of $Ecup E’$ with even (resp. odd) numbers, $A=A’cap E$, and $B=B’cap E$. Since $C$ is a cycle, $d_{A’}(v)=d_{B’}(v)$ for any vertex $vne v'$ of $G$. Since at most one of edges incident to $v$ belongs to $E’$, then $|d_{A}(v')-d_{B}(v')|le 1$. If $C$ has even length then the same argument imply that $d_{A’}(v’)=d_{B’}(v’)$ and $|d_{A}(v)-d_{B}(v)|le 1$. Otherwise $d_{A’}(v’)=d_{B’}(v’)+2$. Since $| Ecup E’|=|C|$ is odd, the conditions imposed on $G$ imply that $E’$ is non-empty. Then exactly one edge of $G'$ incident to $v’$ (namely, the edge $e’$) belongs to $E’$ and $e’in A’$. Then $$d_A(v’)=d_{A’}(v’)-1=d_{B’}(v’)+1=d_{B}(v’)+1.$$
answered Jan 28 at 1:56
Alex RavskyAlex Ravsky
42.7k32383
$endgroup$
Let $V_o$ be a set of vertices of $G$ with odd degree. Then $|V_o|$ is even. Let $E’={(v_1,v’_1),dots, (v_k,v’_k)}$ be an arbitrary partition of $V_o$ into pairs and $G’$ be a multigraph obtained from $G$ by adding $E’$ to its set of edges. Then $G’$ has a Eulerian cycle $C$. We shall walk along $C$ starting from an arbitrary vertex $v’$ of an added edge $e’$, if there is any, or starting from an arbitrary vertex, otherwise, enumerating edges which we walked until we walk the whole $C$. Now let $A’$ (resp. $B’$) be the set of edges of $Ecup E’$ with even (resp. odd) numbers, $A=A’cap E$, and $B=B’cap E$. Since $C$ is a cycle, $d_{A’}(v)=d_{B’}(v)$ for any vertex $vne v'$ of $G$. Since at most one of edges incident to $v$ belongs to $E’$, then $|d_{A}(v')-d_{B}(v')|le 1$. If $C$ has even length then the same argument imply that $d_{A’}(v’)=d_{B’}(v’)$ and $|d_{A}(v)-d_{B}(v)|le 1$. Otherwise $d_{A’}(v’)=d_{B’}(v’)+2$. Since $| Ecup E’|=|C|$ is odd, the conditions imposed on $G$ imply that $E’$ is non-empty. Then exactly one edge of $G'$ incident to $v’$ (namely, the edge $e’$) belongs to $E’$ and $e’in A’$. Then $$d_A(v’)=d_{A’}(v’)-1=d_{B’}(v’)+1=d_{B}(v’)+1.$$
answered Jan 28 at 1:56
Alex RavskyAlex Ravsky
42.7k32383
answered Jan 28 at 1:56
Alex RavskyAlex Ravsky
42.7k32383
answered Jan 28 at 1:56
Alex RavskyAlex Ravsky
42.7k32383
answered Jan 28 at 1:56
Alex RavskyAlex Ravsky
42.7k32383
42.7k32383
add a comment |
add a comment |
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