Do the minimum spanning trees of a weighted graph have the same number of edges with a given weight?

Question

If a weighted graph $G$ has two different minimum spanning trees $T_1 = (V_1, E_1)$ and $T_2 = (V_2, E_2)$, then is it true that for any edge $e$ in $E_1$, the number of edges in $E_1$ with the same weight as $e$ (including $e$ itself) is the same as the number of edges in $E_2$ with the same weight as $e$? If the statement is true, then how can we prove it?

Answer 1

Claim: Yes, that statement is true.

Proof Sketch: Let $T_1,T_2$ be two minimal spanning trees with edge-weight multisets $W_1,W_2$. Assume $W_1 \neq W_2$ and denote their symmetric difference with $W = W_1 \mathop{\Delta} W_2$.

Choose edge $e \in T_1 \mathop{\Delta} T_2$ with $w(e) = \min W$, that is $e$ is an edge that occurs in only one of the trees and has minimum disagreeing weight. Such an edge, that is in particular $e \in T_1 \mathop{\Delta} T_2$, always exists: clearly, not all edges of weight $\min W$ can be in both trees, otherwise $\min W \notin W$. W.l.o.g. let $e \in T_1$ and assume $T_1$ has more edges of weight $\min W$ than $T_2$.

Now consider all edges in $T_2$ that are also in the cut $C_{T_1}(e)$ that is induced by $e$ in $T_1$. If there is an edge $e'$ in there that has the same weight as $e$, update $T_1$ by using $e'$ instead of $e$; note that the new tree is still a minimal spanning tree with the same edge-weight multiset as $T_1$. We iterate this argument, shrinking $W$ by two elements and thereby removing one edge from the set of candidates for $e$ in every step. Therefore, we get after finitely many steps to a setting where all edges in $T_2 \cap C_{T_1}(e)$ (where $T_1$ is the updated version) have weights other than $w(e)$.

Now we can always choose $e' \in C_{T_1}(e) \cap T_2$ such that we can swap $e$ and $e'$¹, that is we can create a new spanning tree

$\qquad \displaystyle T_3 = \begin{cases} (T_1 \setminus \{e\}) \cup \{e'\}, &w(e') \lt w(e) \\[.5em] (T_2 \setminus \{e'\}) \cup \{e\}, &w(e') \gt w(e) \end{cases}$

which has smaller weight than $T_1$ and $T_2$; this contradicts the choice of $T_1,T_2$ as minimal spanning trees. Therefore, $W_1 = W_2$.

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1. The nodes incident of $e$ are in $T_2$ connected by a path $P$; $e'$ is the unique edge in $P \cap C_{T_1}(e)$.

Answer 2

Here is a slightly simpler argument that also works for other matroids. (I saw this question from another one.)

Suppose that $G$ has $m$ edges. Without loss of generality, assume that the weight function $w$ takes on values in $[m]$, so we have a partition of $E$ into sets $E_i := w^{-1}(i)$ for $i\in [m]$. We can do induction on the number $j$ of non-empty $E_i$ and number of vertices $n$ in $G$; for $j=1$ and any $n$, the statement is obvious.

A standard fact about matroids is that for every MST $T$ there is a linear extension of the ordering induced by $w$ so that the greedy algorithm produces $T$.

To close the induction, take $t$ to be the largest number so that $E_t$ is not empty. Set $E' = E_1 \cup \cdots \cup E_{t-1}$. Observe that any linear extension of $w$ puts every edge in $E'$ before any edge in $E_t$. According to the fact, any MST consists of a spanning forest $F$ of the subgraph induced by $E'$ and some edges from $E_t$. By the inductive hypothesis, each connected component of $F$ has the same number of edges from each $E_i$ for $i < t$. Since all the choices of $F$ have the same size, the number of edges from $E_t$ required to complete $F$ to a spanning tree is independent of the choice of $F$ and we are done.