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In trying to prove that every tree, T, has at most one perfect matching, I came across this idea:

Since the matchings are perfect, each vertex has degree $0$ or $2$ in the symmetric difference, so every component is an isolated vertex or a cycle.

Why is this true? Why is it true that since each vertex is either of degree $0$ or of degree $2$, then all components are either isolated or a cycle?

hululu
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larry
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4 Answers4

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First, why the claim is true. Take any vertex. If it's matches to the same vertex in both perfect matching, then it had degree zero on the symmetric difference. Otherwise it has degree two.

Second, after removing all isolated vertices in the symmetric difference, all vertices have degree two. Take any such vertex and follow its two edges. What you get is a growing path that eventually closed to a cycle since the graph is finite.

Since trees have no cycles, this implies that any two perfect matching are equal, by consisting their symmetric difference.

A different proof is by induction. The idea is that every leaf must be matched to its unique neighbor.

Yuval Filmus
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Let $M$ and $M_0$ be perfect matchings in a tree. Form the symmetric difference of the edge sets, $M \bigtriangleup M_0$. Since the matchings are perfect, each vertex has degree $0$ or $2$ in the symmetric difference, so every component is an isolated vertex or a cycle. Since the tree has no cycle, every vertex must have degree $0$ in the symmetric difference, which means that the two matchings are the same.

Gassa
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user521739
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    Can you please explain why there is no vertex with degree 1? I don't really understand that. – Sofia Apr 04 '22 at 15:44
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    @Sofia Consider a vertex $v$. Since the matchings $M$ and $M_0$ are perfect, $v$ is matched in both them. Either $v$ has the same match in both, in which case it doesn't appear in the symmetric difference giving degree $0$, or they are different, say ${v, u} \in M$ and ${v, w} \in M_0$, then both ${v, u}, {v, w} \in M \triangle M_0$ giving degree $2$. – zaxioms Dec 03 '23 at 17:17
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Proof By Induction: For the basis step, a tree with one vertex has no perfect matching; a tree with two vertices has one. We will take a set of two vertices to prove the statement.

For the induction step, consider an arbitrary tree $T$ on $n > 2$ vertices, and consider a leaf $v$. In any perfect matching, $v$ must be matched to its neighbor $u$. Since each perfect matching in $T$ must contain the edge $uv$, the number of perfect matchings in $T$ equals the number of perfect matchings in $T - \{u, v\}$.

Each component of $T − \{u, v\}$ is a tree; by the induction hypothesis, each component has at most one perfect matching. (More generally, a forest has at most one perfect matching.)

Thành Nguyễn
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Baski_Flex
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Since every tree of two or more vertices is two chromatic. Tree with even no of vertices will have the perfect matching as all the vertices with same color can be grouped together and a matching can be established between two groups. But any tree with odd no of vertex will have no perfect matching for obvious reason. Hence proved.

Palash Kumar Mallik
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  • Be careful when using the word "obvious" in math proofs. Just because something may be obvious to one person doesn't mean it is obvious to another. Please elaborate on the line "But any tree odd no of vertex will have no perfect matching for obvious reason." – Xoque55 Dec 04 '15 at 03:44
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    There are two problems with this answer. First, the question is to show that every tree has at most one perfect matching. You have claimed that there is one. Why are there not others? Second, you claim that you can create a perfect matching between the nodes of one color and the nodes of another color. But consider any “star” graph with $2k$ vertices ($k>1$), one of which is adjacent to every other (the others being leaves). The partite sets of a two-color of this graph have different sizes, $1$ and $2n-1$, so they can’t be perfectly matched. – Steve Kass Dec 04 '15 at 03:47