\section{The Problem}
The problem of finding a minimum spanning tree of a weighted graph is one of the
-best studied problems in the area of combinatorial optimization and it can be said
-that it stood at the cradle of this discipline. Its colorful history (see \cite{graham:msthistory}
-and \cite{nesetril:history} for the full account) begins in~1926 with
-the pioneering work of Bor\accent23uvka
+best studied problems in the area of combinatorial optimization since its birth.
+Its colorful history (see \cite{graham:msthistory} and \cite{nesetril:history} for the full account)
+begins in~1926 with the pioneering work of Bor\accent23uvka
\cite{boruvka:ojistem}\foot{See \cite{nesetril:boruvka} for an English translation with commentary.},
who studied primarily an Euclidean version of the problem related to planning
of electrical transmission lines (see \cite{boruvka:networks}), but gave an efficient
-algorithm for the general version of the problem. As it was well before the birth of graph
-theory, the language of his paper was complicated, so we will rather state the problem
+algorithm for the general version of the problem. As it was well before the dawn of graph
+theory, the language of his paper was complicated, so we will better state the problem
in contemporary terminology:
\proclaim{Problem}Given an undirected graph~$G$ with weights $w:E(G)\rightarrow {\bb R}$,
-find its minimum spanning tree, where:
+find its minimum spanning tree, defined as follows:
\defn\thmid{mstdef}%
For a given graph~$G$ with weights $w:E(G)\rightarrow {\bb R}$:
\itemize\ibull
-\:A~tree $T$ is a \df{spanning tree} of~$G$ if and only if $V(T)=V(G)$ and $E(T)\subseteq E(G)$.
+\:A~subgraph $H\subseteq G$ is called a \df{spanning subgraph} if $V(H)=V(G)$.
+\:A~\df{spanning tree} of $G$ is any its spanning subgraph which is a tree.
\:For any subgraph $H\subseteq G$ we define its \df{weight} $w(H):=\sum_{e\in E(H)} w(e)$.
-\:A~spanning tree~$T$ is \df{minimal} iff $w(T)$ is the smallest possible of all spanning trees.
- We use an abbreviation \df{MST} for such trees.
-\:For a disconnected graph, a \df{(minimal) spanning forest (MSF)} is defined as
- a union of (minimal) spanning trees of its connected components.
+\:A~\df{minimum spanning tree (MST)} of~$G$ is a spanning tree~$T$ such that its weight $w(T)$
+ is the smallest possible of all the spanning trees of~$G$.
+\:For a disconnected graph, a \df{(minimum) spanning forest (MSF)} is defined as
+ a union of (minimum) spanning trees of its connected components.
\endlist
Bor\accent23uvka's work was further extended by Jarn\'\i{}k \cite{jarnik:ojistem}, again in
-mostly geometric setting, giving another polynomial algorithm. However, when
+mostly geometric setting, giving another efficient algorithm. However, when
computer science and graph theory started forming in the 1950's and the
spanning tree problem was one of the central topics of the flourishing new
-disciplines, the previous work was not well known and the algorithms have been
+disciplines, the previous work was not well known and the algorithms had to be
rediscovered several times.
Recently, several significantly faster algorithms were discovered, most notably the
several important theorems to base the algorithms upon. We will follow the theory
developed by Tarjan in~\cite{tarjan:dsna}.
-For the whole section, we will fix a graph~$G$ with edge weights~$w$ and all other
-graphs will be subgraphs of~$G$ containing all of its vertices. We will use the
-same notation for the subgraph and for the corresponding set of edges.
+For the whole section, we will fix a graph~$G$ with edge weights~$w$ and all
+other graphs will be spanning subgraphs of~$G$. We will use the same notation
+for the subgraphs as for the corresponding sets of edges.
First of all, let us show that the weights on edges are not necessary for the
definition of the MST. We can formulate an equivalent characterization using
\:For vertices $x$ and $y$, let $T[x,y]$ denote the (unique) path in~$T$ joining $x$ and~$y$.
\:For an edge $e=xy$ we will call $T[e]:=T[x,y]$ the \df{path covered by~$e$} and
the edges of this path \df{edges covered by~$e$}.
-\:An edge~$e$ is called \df{$T$-light} if it covers a heavier edge, i.e., if there
+\:An edge~$e$ is called \df{light with respect to~$T$} (or just \df{$T$-light}) if it covers a heavier edge, i.e., if there
is an edge $f\in T[e]$ such that $w(f) > w(e)$.
\:An edge~$e$ is called \df{$T$-heavy} if it is not $T$-light.
\endlist
\lemman{Light edges}\thmid{lightlemma}%
Let $T$ be a spanning tree. If there exists a $T$-light edge, then~$T$
-is not minimal.
+is not minimum.
\proof
-If there is a $T$-light edge~$e$, then there exists an edge $f\in T[e]$ such
-that $w(f)>w(e)$. Now $T-f$ is a forest of two trees with endpoints of~$e$
+If there is a $T$-light edge~$e$, then there exists an edge $e'\in T[e]$ such
+that $w(e')>w(e)$. Now $T-e'$ is a forest of two trees with endpoints of~$e$
located in different components, so adding $e$ to this forest must restore
-connectivity and $T':=T-f+e$ is another spanning tree with weight $w(T')
-= w(T)-w(f)+w(e) < w(T)$. Hence $T$ could not have been minimal.
+connectivity and $T':=T-e'+e$ is another spanning tree with weight $w(T')
+= w(T)-w(e')+w(e) < w(T)$. Hence $T$ could not have been minimum.
\qed
\figure{mst2.eps}{278pt}{An edge exchange as in the proof of Lemma~\thmref{lightlemma}}
$T_{i+1}=T_i - e_i + e_i^\prime$ where $e_i\in T_i$ and $e_i^\prime\in T'$.
\proof
-By induction on $d(T,T'):=\vert T\symdiff T'\vert$. When $d(T,T')=0$, then
-both trees are identical and an empty sequence suffices. Otherwise, the trees are different,
-but they are of the same size, so there must exist an edge $e'\in T'\setminus T$.
+By induction on $d(T,T'):=\vert T\symdiff T'\vert$. When $d(T,T')=0$,
+both trees are identical and no exchanges are needed. Otherwise, the trees are different,
+but as they are of the same size, there must exist an edge $e'\in T'\setminus T$.
The cycle $T[e']+e'$ cannot be wholly contained in~$T'$, so there also must
exist an edge $e\in T[e']\setminus T'$. Exchanging $e$ for~$e'$ yields a spanning
tree $T^*:=T-e+e'$ such that $d(T^*,T')=d(T,T')-2$ and we can apply the induction
\figure{mst1.eps}{295pt}{One step of the proof of Lemma~\thmref{xchglemma}}
\lemman{Monotone exchanges}\thmid{monoxchg}%
-Let $T$ be a spanning tree such that there are no $T$-light edges and $T$
+Let $T$ be a spanning tree such that there are no $T$-light edges and $T'$
be an arbitrary spanning tree. Then there exists a sequence of edge exchanges
transforming $T$ to~$T'$ such that the weight does not increase in any step.
We improve the argument from the previous proof, refining the induction step.
When we exchange $e\in T$ for $e'\in T'\setminus T$ such that $e\in T[e']$,
the weight never drops, since $e'$ is not a $T$-light edge and therefore
-$w(e') \ge w(e)$, so $w(T^*)=w(T)-w(e)+w(e')\le w(T)$.
-
-To allow the induction to proceed, we have to make sure that there are still
-no light edges with respect to~$T^*$. In fact, it is enough to avoid $T^*$-light
-edges in $T'\setminus T^*$, since these are the only edges considered by the
-induction step. Instead of picking $e'$ arbitrarily, we will pick the lightest
-edge available. Now consider an edge $f\in T'\setminus T^*$. We want to show
-that $f$ is heavier than all edges on $T^*[f]$.
-
-The path $T^*[f]$ is either the original path $T[f]$ (if $e\not\in T[f]$)
-or $T[f] \symdiff C$, where $C$ is the cycle $T[e']+e$. The first case is
-trivial, in the second case $w(f)\ge w(e')$ and all other edges on~$C$
-are lighter than~$e'$.
+$w(e') \ge w(e)$, so $w(T^*)=w(T)-w(e)+w(e')\ge w(T)$.
+
+To keep the induction going, we have to make sure that there are still no light
+edges with respect to~$T^*$. In fact, it is enough to avoid such edges in
+$T'\setminus T^*$, since these are the only edges considered by the induction
+steps. To accomplish that, we replace the so far arbitrary choice of $e'\in T'\setminus T$
+by picking the lightest such edge.
+
+Now consider an edge $f\in T'\setminus T^*$. We want to show that $f$ is not
+$T^*$-light, i.e., that it is heavier than all edges on $T^*[f]$. The path $T^*[f]$ is
+either equal to the original path $T[f]$ (if $e\not\in T[f]$) or to $T[f] \symdiff C$,
+where $C$ is the cycle $T[e']+e'$. The former case is trivial, in the latter one
+$w(f)\ge w(e')$ due to the choice of $e'$ and all other edges on~$C$ are lighter
+than~$e'$ as $e'$ was not $T$-light.
\qed
-\theorem
-A~spanning tree~$T$ is minimal iff there is no $T$-light edge.
+\theorem\thmid{mstthm}%
+A~spanning tree~$T$ is minimum iff there is no $T$-light edge.
\proof
-If~$T$ is minimal, then by Lemma~\thmref{lightlemma} there are no $T$-light
+If~$T$ is minimum, then by Lemma~\thmref{lightlemma} there are no $T$-light
edges.
Conversely, when $T$ is a spanning tree without $T$-light edges
-and $T_{min}$ is an arbitrary minimal spanning tree, then according to the Monotone
+and $T_{min}$ is an arbitrary minimum spanning tree, then according to the Monotone
exchange lemma (\thmref{monoxchg}) there exists a non-decreasing sequence
of exchanges transforming $T$ to $T_{min}$, so $w(T)\le w(T_{min})$
-and thus $T$~is also minimal.
+and thus $T$~is also minimum.
\qed
-In general, a single graph can have many minimal spanning trees (for example
+In general, a single graph can have many minimum spanning trees (for example
a complete graph on~$n$ vertices and unit edge weights has $n^{n-2}$
minimum spanning trees according to the Cayley's formula \cite{cayley:trees}).
-However, as the following lemma shows, this is possible only if the weight
+However, as the following theorem shows, this is possible only if the weight
function is not injective.
-\lemman{MST uniqueness}
+\theoremn{MST uniqueness}
If all edge weights are distinct, then the minimum spanning tree is unique.
\proof
-Consider two minimal spanning trees $T_1$ and~$T_2$. According to the previous
+Consider two minimum spanning trees $T_1$ and~$T_2$. According to the previous
theorem, there are no light edges with respect to neither of them, so by the
-Monotone exchange lemma (\thmref{monoxchg}) there exists a non-decreasing sequence
-of edge exchanges going from $T_1$ to $T_2$. Each exchange in this sequence is
-strictly increasing, because all edge weights all distinct. On the other hand,
+Monotone exchange lemma (\thmref{monoxchg}) there exists a sequence of non-decreasing
+edge exchanges going from $T_1$ to $T_2$. As all edge weights all distinct,
+these edge exchanges must be in fact strictly increasing. On the other hand,
we know that $w(T_1)=w(T_2)$, so the exchange sequence must be empty and indeed
$T_1$ and $T_2$ must be identical.
\qed
\rem\thmid{edgeoracle}%
-To simplify the description of MST algorithms, we will expect that the input
-graph has all edge weights distinct. We will also assume that instead of explicit
-edge weights we will be given a comparison oracle, that is a function which answers
-questions ``$w(e)<w(f)$?'' in constant time. Please note that this is without loss
-of generality, because when some edges have identical weights, we can determine their
-relative order by comparing some unique identifiers of these edges and every MST
-of the new graph will be also a MST of the original one.
-In the few cases where we need a more concrete input, we will explicitly say so.
+To simplify the description of MST algorithms, we will expect that the weights
+of all edges are distinct and that instead of numeric weights (usually accompanied
+by problems with representation of real numbers in algorithms) we will be given
+a comparison oracle, that is a function which answers questions ``$w(e)<w(f)$?'' in
+constant time. In case the weights are not distinct, we can easily break ties by
+comparing some unique edge identifiers and according to our characterization of
+minimum spanning trees, the unique MST of the new graph will still be a MST of the
+original graph. In the few cases where we need a more concrete input, we will
+explicitly state so.
\section{The Red-Blue Meta-Algorithm}
\algo
\algin A~graph $G$ with an edge comparison oracle (see \thmref{edgeoracle})
\:In the beginning, all edges are colored black.
-\:While possible, use one of the following rules:
-\::Pick a cut~$C$ such that its lightest edge is not blue \hfil\break and color this edge blue. \cmt{Blue rule}
-\::Pick a cycle~$C$ such that its heaviest edge is not red \hfil\break and color this edge \hphantas{red.}{blue.} \cmt{Red rule}
+\:Apply rules as long as possible:
+\::Either pick a cut~$C$ such that its lightest edge is not blue \hfil\break and color this edge blue, \cmt{Blue rule}
+\::Or pick a cycle~$C$ such that its heaviest edge is not red \hfil\break and color this edge \hphantas{red.}{blue.} \cmt{Red rule}
\algout Minimum spanning tree of~$G$ consisting of edges colored blue.
\endalgo
\rem
-This is not a proper algorithm, since the selection of rules to apply is not specified.
-We will however prove that for any such selection the procedure stops with the correct result.
-Also, it will turn out that each of the classical MST algorithms can be stated as a
-specific rule selection strategy of this procedure, which justifies the name
-meta-algorithm for it.
+This procedure is not a proper algorithm, since it does not specify how to choose
+the rule to apply. We will however prove that no matter how the rules are applied,
+the procedure always stops and gives the correct result. Also, it will turn out
+that each of the classical MST algorithms can be described as a specific way
+of choosing the rules in this procedure, which justifies the name meta-algorithm.
+
+\proclaim{Notation}%
+We will denote the unique minimum spanning tree of the input graph by~$T_{min}$.
+We intend to prove that this is also the output of the procedure.
\lemman{Blue lemma}
When an edge is colored blue in any step of the procedure, it is contained in the minimum spanning tree.
\proof
-\qed
+By contradiction. Let $e$ be an edge painted blue as the lightest edge of a cut~$C$.
+If $e\not\in T_{min}$, then there must exist an edge $e'\in T_{min}$ which is
+contained in~$C$ (take any pair of vertices separated by~$C$, the path
+in~$T_{min}$ joining these vertices must cross~$C$ at least once). Exchanging
+$e$ for $e'$ in $T_{min}$ yields an even lighter spanning tree since
+$w(e)<w(e')$. \qed
\lemman{Red lemma}
When an edge is colored red in any step of the procedure, it is not contained in the minimum spanning tree.
\proof
+Again by contradiction. Assume that $e$ is an edge painted red as the heaviest edge
+of a cycle~$C$ and that $e\in T_{min}$. Removing $e$ causes $T_{min}$ to split to two
+components, let us call them $T_x$ and $T_y$. Some vertices of~$C$ now lie in $T_x$,
+the others in $T_y$, so there must exist in edge $e'\ne e$ such that its endpoints
+lie in different components. Since $w(e')<w(e)$, exchanging $e$ for~$e'$ yields
+a lighter spanning tree than $T_{min}$.
\qed
\figure{mst-rb.eps}{289pt}{Proof of the Blue (left) and Red (right) lemma}
As long as there exists a black edge, at least one rule can be applied.
\proof
+Assume that $e=xy$ be a black edge. Let us denote $M$ the set of vertices
+reachable from~$x$ using only blue edges. If $y$~lies in~$M$, then $e$ together
+with some blue path between $x$ and $y$ forms a cycle and it must be the heaviest
+edge on this cycle. This holds because all blue edges have been already proven
+to be in $T_{min}$ and there can be no $T_{min}$-light edges (see Theorem~\thmref{mstthm}).
+In this case we can apply the red rule.
+
+On the other hand, if $y\not\in M$, then the cut formed by all edges between $M$
+and $V(G)\setminus M$ contains no blue edges, therefore we can use the blue rule.
\qed
\figure{mst-bez.eps}{295pt}{Configurations in the proof of the Black lemma}
the minimum spanning tree of the input graph.
\proof
+To prove that the procedure stops, let us notice that no edge is ever recolored,
+so we must run out of black edges after at most $m$ steps. Recoloring
+to the same color is avoided by the conditions built in the rules, recoloring to
+a different color would mean that the an edge would be both inside and outside~$T_{min}$
+due to our Red and Blue lemmata.
+
+When no further rules can be applied, the Black lemma guarantees that all edges
+are colored, so by the Blue lemma all blue edges are in~$T_{min}$ and by the Red
+lemma all other (red) edges are outside~$T_{min}$, so the blue edges are exactly~$T_{min}$.
\qed
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