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## Q：Are all minimum spanning trees optimized for fairness? |
## Q：所有的最小生成树优化公平？ |

I know by definition that a minimum spanning tree (MST) of a weighted, connected graph has the lowest global value for for sum of all edges in a path that connects all vertices. I'm curious if all MSTs are also optimized for lowest individual edge value possible. (I know that ones calculated with Kruskal's Algorithm are!) As an example, is it possible to have two MSTs in a graph (equal global sum of their edges) but have one of those MSTs contain an edge of higher value than any edge on the other MST in the graph. Thanks for any help you can give! |
我知道通过定义一个最小生成树（MST）的加权连通图有一个路径和连接所有顶点的所有边的全球最低值。 我很好奇，如果所有MSTS也是个人最低边缘值可能的优化。（我知道的Kruskal算法计算了！） 作为一个例子，它是可能的在图2 MSTS（他们的边缘等全球总和）但有一人并包含一个边缘比在图中的其他MST任何边缘。 谢谢你提供的任何帮助！ |

answer1： | 回答1： |

No, it isn't; assume it is, let $X$ be the MST with the heavy edge, and call that heavy edge $e := \{u,v\}$. Now there is some other MST $Y$ that doesn't contain edges that are at least as heavy as $e$. In $Y$, we know that there is a path from $u$ to $v$ since $X$ has such a path, so $Y$ must have one as well. Let $p$ denote that path in $Y$. There must be an edge $f$ on $p$ that isn't in $X$, since otherwise $X$ would contain the cycle formed by $p$ and $e$. Now add $f$ to $X$ and delete $e$ from it to get $X'$, which is a spanning tree as well. However, $X'$ has lower weight than $X$ since $f$ has lower weight than $e$, contradicting the assumption that $X$ is an MST. For something more general which might answer your question more succinctly, here and here are proofs that all MSTs of a graph have the same multiset of edge weights, that is: if $k$ edges of weight $w$ occur in some MST, then $k$ edges of weight $w$ occur in all MSTs for any $w$ and $k$. Something else that isn't proven in those answers that might be of interest to you: every MST only contains paths with smallest maximum weights for all pairs of vertices $\{u,v\}$, or in words: let $p$ be the unique path between $u$ and $v$ in a given MST $T$ of graph $G$; then for any $u$-$v$-path $q$ in $G$, the heaviest edge on $q$ is at least as heavy as the heaviest edge on $p$. To prove this, assume that there is some pair $\{u,v\}$ such that the heaviest edge in the unique $u$-$v$-path $p$ in some MST $T$ is heavier than the heaviest edge in some other $u$-$v$-path $q$ in $G$. Let $e$ denote the heaviest edge in $p$; now remove $e$ from $T$, which partitions $T$ into two connected components $U$ and $V$ such that $u \in U$ and $v \in V$. We know that $q$ starts in $U$ and ends in $V$, so there must be an edge $f$ that belongs to the cut $(U,V)$. Now we find that $(T \cup \{f\}) - \{e\}$ is a spanning tree again, and since the weight of $f$ is smaller than the weight of $e$ by assumption, we find that $(T \cup \{f\}) - \{e\}$ has lower weight than the MST $T$, contradicting the minimality of $T$. Thus, no such path $q$ can exist in $G$. Spanning trees with the heaviest weight being minimal are called minimum bottleneck spanning trees, and you can use the above to show that every MST is an MBST (the converse does not hold, i.e. there may be an MBST that is not an MST, specifically: an MBST might use too many heaviest edges). If an MST $T$ wasn't also an MBST, then there would be a heavier edge in $T$ than in some other spanning tree, and that edge would be on some path, and then the arguments above apply to that path. |
不，它不是；假定它是，让X是重边的MST，叫重边e：= \ { U、V } $。现在有一些其他的MST $ Y $不包含边缘，至少重如$ E $。在Y $，我们知道有一个路径从$ $ $ V美元，因为X美元有这样的路径，所以$ Y必须有一个以及。让$ $ $表示路径$ $。必须有一个边缘$ $ $ $ $ $不在$ x $，因为否则x $ $将包含由$ $ $ $ $ $的周期。现在添加$ $ $ x $和删除$ $ $从它得到$ $ $，这是一个生成树以及。然而，x美元美元，具有较低的重量比X的自F具有较低的重量比$ E $，矛盾的假设X是一个MST。 更一般的，可以更简洁地回答你的问题，这里有证据证明MSTS一个图有边的权重，这是相同的：如果multiset $k$边权$ W $则发生在MST，然后边权重K $ $ W $美元发生在任何W $和$ K $所有MSTS。 别的，不在这些问题的答案可能是您感兴趣的证明：每一个MST只包含最小最大权重的所有顶点对$ \ {你的路径，V } $，或言：让P是唯一路径之间$ U和V在给定美元美元MST $t$图G；然后任何$U$- $ V $路径$ Q $在G，在$ Q $最边缘至少重最重的边上P。 为了证明这一点，假设有一些对$ \ { U、V }这样重的边缘美元的独特$U$- $ V $路径P在MST $t$是更重的边缘在其他一些$U$- $ V $美元路径Q为G。让$ E表示最重的边缘在$ P $；现在删除E $ $ $ T，其中分区$ T $两个连接组件$ U和$ V美元，如$ U $ U和$ V \ $ V。我们知道，$ Q $开始在$ U和结束在$ V $，所以必须有一个边缘$ F $属于削减美元（U，V）$。现在我们发现$（T \杯\ { F \ }）{ } $ \ E又是一个生成树，由于F的重量小于假设$ E $的重量，我们发现$（T \杯\ \ \ { F }）{ } $ E比MST $t$重量较轻，与T最小美元美元。因此，没有这样的路径$ Q $可以存在于$ G $。 生成树的最重的是最小的被称为最小瓶颈生成树，你可以使用上面显示每个MST是MBST（反过来不成立，即有可能是MBST不是MST，具体地说：一个MBST可能使用太多的重的边缘）。如果一个MST $t$不也是MBST，那么将有$ T $重边比其他生成树，而边将一些路径，然后上述观点应用到路径。 |

graphs minimum-spanning-tree |