Update 23.1.md

docs/Chap23/23.1.md

@@ -38,7 +38,26 @@ Let $A$ be any cut that causes some vertices in the cycle on once side of the cu
 
 > Show that a graph has a unique minimum spanning tree if, for every cut of the graph, there is a unique light edge crossing the cut. Show that the converse is not true by giving a counterexample.
 
-(Removed)
+### The Proof
+
+Assume for a contradiction that the graph has two distinct minimum spanning trees $T$ and $T'$. Let $(u, v)$ be an edge in $T$ which is not in $T'$. Removing edge $(u, v)$ cuts the tree $T$ into two components. Let $T_u$ and $T_v$ be the sets of vertices in the components containing $u$ and $v$, respectively.
+
+Consider the cut $(T_u, T_v)$, and let $(x, y)$ be the unique light edge crossing the cut. Since $(u, v)$ crosses this cut and $(x, y)$ is unique, there are two possibilities:
+
+1. If $(x, y) \ne (u, v)$, then $w(x, y) < w(u, v)$ (since $(x, y)$ is the unique light edge). Replacing $(u, v)$ with $(x, y)$ in $T$ gives a spanning tree $T'' = T - \{(u, v)\} \cup \{(x, y)\}$ with total weight less than $T$, contradicting the minimality of $T$.
+2. If $(x, y) = (u, v)$, then $(u, v)$ is the unique light edge crossing the cut $(T_u, T_v)$. Since $(u, v) \notin T'$, there must be an alternative path from $u$ to $v$ in $T'$, which must cross the cut $(T_u, T_v)$ via some edge $e \ne (u, v)$. Because $(u, v)$ is the unique light edge, we have $w(u, v) < w(e)$. Replacing $e$ with $(u, v)$ in $T'$ gives a spanning tree $T'' = T' - \{e\} \cup \{(u, v)\}$ with total weight less than $T'$, again contradicting the minimality of $T'$.
+
+In both cases, we reach a contradiction. Therefore, the graph must have a unique minimum spanning tree.
+
+### Counterexample for the Converse
+
+Consider a graph with three vertices $a, b, c$ and edge weights:
+
+$$
+w(a, b) = w(a, c) = 1, \quad w(b, c) = 2.
+$$
+
+The graph has a unique minimum spanning tree containing edges $(a, b)$ and $(a, c)$. However, the cut $\{\{a\}, \{b, c\}\}$ does not have a unique light edge crossing it, since both edges $(a, b)$ and $(a, c)$ cross the cut and have the same minimal weight of 1.
 
 ## 23.1-7