Update 23.1.md

docs/Chap23/23.1.md

@@ -38,7 +38,121 @@ 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)
+**Remark:** Do not assume that all edge weights are distinct.
+
+**Part 1: Proving the Forward Direction**
+
+**Goal:** Show that if every cut in the graph has a unique light edge crossing it, then the graph has exactly one MST.
+
+**Proof:**
+
+Assume, for contradiction, that the graph has **two distinct MSTs**, $T$ and $T'$.
+
+1. **Identifying an Edge in $T$ but not in $T'$:**
+
+    - Since $T$ and $T'$ are distinct, there exists at least one edge that is in $T$ but not in $T'$.
+    - Let $(u, v)$ be such an edge.
+
+2. **Creating a Cut by Removing $(u, v)$:**
+
+    - Removing $(u, v)$ from $T$ divides it into two connected components (since trees are acyclic and connected).
+    - Let $T_u$ be the set of vertices reachable from $u$ without using $(u, v)$.
+    - Let $T_v$ be the set of vertices reachable from $v$ without using $(u, v)$.
+    - The sets $T_u$ and $T_v$ form a **cut** $(T_u, T_v)$ in the graph.
+
+3. **Unique Light Edge Crossing the Cut:**
+
+    - By assumption, the cut $(T_u, T_v)$ has a **unique light edge** crossing it.
+    - Let $(x, y)$ be this unique light edge.
+    - Note that $(u, v)$ crosses this cut because $u \in T_u$ and $v \in T_v$.
+
+4. **Analyzing the Unique Light Edge:**
+
+    - **Case 1:** If $(x, y) \ne (u, v)$, then:
+        - Since $(x, y)$ is the unique light edge and not $(u, v)$, it must have a weight **less than** $w(u, v)$.
+        - **Constructing a New Spanning Tree:**
+            - Replace $(u, v)$ in $T$ with $(x, y)$ to get $T'' = T - \{ (u, v) \} \cup \{(x, y)\}$.
+            - $T''$ is connected (since $(x, y)$ connects $T_u$ and $T_v$) and spans all vertices.
+            - The total weight of $T''$ is less than that of $T$ because $w(x, y) < w(u, v)$.
+        - **Contradiction:**
+            - $T$ was assumed to be an MST, but $T''$ has a lower total weight.
+            - This contradicts the minimality of $T$.
+
+    - **Case 2:** If $(x, y) = (u, v)$, then:
+        - The unique light edge crossing the cut is $(u, v)$.
+        - **Observing $T'$:**
+            - Since $(u, v) \notin T'$, there must be a path from $u$ to $v$ in $T'$ (since $T'$ is connected).
+            - This path must cross the cut $(T_u, T_v)$ at least once.
+            - Let $e$ be an edge on this path that crosses the cut.
+        - **Comparing Edge Weights:**
+            - Since $(u, v)$ is the unique light edge crossing the cut, and $e \ne (u, v)$, it follows that $w(u, v) < w(e)$.
+        - **Constructing a New Spanning Tree:**
+            - Add $(u, v)$ to $T'$, creating a cycle.
+            - Remove $e$ from this cycle to get $T'' = T' + \{(u, v)\} - \{e\}$.
+            - $T''$ is connected and spans all vertices.
+            - The total weight of $T''$ is less than that of $T'$ because $w(u, v) < w(e)$.
+        - **Contradiction:**
+            - $T'$ was assumed to be an MST, but $T''$ has a lower total weight.
+            - This contradicts the minimality of $T'$.
+
+5. **Conclusion:**
+
+    - In both cases, assuming the existence of two distinct MSTs leads to a contradiction.
+    - Therefore, the initial assumption that there are two distinct MSTs is false.
+    - **Hence, the graph must have a unique MST.**
+
+---
+
+**Part 2: Showing the Converse is Not True**
+
+**Goal:** Provide a counterexample to show that a graph can have a unique MST even if not every cut has a unique light edge crossing it.
+
+**Counterexample:**
+
+**Graph Description:**
+
+- **Vertices:** $a$, $b$, $c$.
+- **Edges and Weights:**
+    - Edge $(a, b)$ with weight $1$.
+    - Edge $(a, c)$ with weight $1$.
+    - Edge $(b, c)$ with weight $2$.
+
+**Visualization:**
+
+```
+    (1)
+a ------- b
+ \       /
+  \     /
+   \   /
+(1) \ / (2)
+     c
+```
+
+**Analysis:**
+
+1. **Possible Spanning Trees:**
+
+    - **Tree 1:** Edges $(a, b)$ and $(a, c)$; total weight $1 + 1 = 2$.
+    - **Tree 2:** Edges $(a, b)$ and $(b, c)$; total weight $1 + 2 = 3$.
+    - **Tree 3:** Edges $(a, c)$ and $(b, c)$; total weight $1 + 2 = 3$.
+
+2. **Identifying the Unique MST:**
+
+    - The minimum total weight is $2$, achieved by Tree 1.
+    - **Therefore, the graph has a unique MST** comprising edges $(a, b)$ and $(a, c)$.
+
+3. **Examining the Cuts:**
+
+    - **Cut between $\{a\}$ and $\{b, c\}$:**
+        - Edges crossing this cut are $(a, b)$ and $(a, c)$.
+        - Both edges have the same weight $1$.
+        - **Therefore, this cut does not have a unique light edge**; it has two edges with the minimum weight.
+
+4. **Conclusion:**
+
+    - The graph has a unique MST even though at least one cut does not have a unique light edge crossing it.
+    - **This demonstrates that the converse is not true.**
 
 ## 23.1-7