From 1536a91e2fd6172813b0d67c02a26a0a7c22e65d Mon Sep 17 00:00:00 2001 From: Xing Shi Cai Date: Mon, 18 Nov 2024 10:56:37 +0800 Subject: [PATCH 1/2] Update 23.1.md Add a proof for 23.1-6 --- docs/Chap23/23.1.md | 22 +++++++++++++++++++++- 1 file changed, 21 insertions(+), 1 deletion(-) diff --git a/docs/Chap23/23.1.md b/docs/Chap23/23.1.md index 5c300c5f04..d653fe1983 100755 --- a/docs/Chap23/23.1.md +++ b/docs/Chap23/23.1.md @@ -38,7 +38,27 @@ 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 $(x, y) \neq (u, v)$, 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 $e \neq (u, v)$. 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 From 5c45ec3a7ffcb03da53819e5499e21262c054858 Mon Sep 17 00:00:00 2001 From: Peng-Yu Chen Date: Wed, 11 Dec 2024 23:54:06 -0500 Subject: [PATCH 2/2] Update 23.1.md --- docs/Chap23/23.1.md | 5 ++--- 1 file changed, 2 insertions(+), 3 deletions(-) diff --git a/docs/Chap23/23.1.md b/docs/Chap23/23.1.md index d653fe1983..7dccc9b57c 100755 --- a/docs/Chap23/23.1.md +++ b/docs/Chap23/23.1.md @@ -44,9 +44,8 @@ Assume for a contradiction that the graph has two distinct minimum spanning tree 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 $(x, y) \neq (u, v)$, 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 $e \neq (u, v)$. 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 $T'$. +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.