For a given graph~$G$ with weights $w:E(G)\rightarrow {\bb R}$:
\itemize\ibull
\: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.
+\:A~\df{spanning tree} of~$G$ is any spanning subgraph of~$G$ that is a tree.
\:For any subgraph $H\subseteq G$ we define its \df{weight} $w(H):=\sum_{e\in E(H)} w(e)$.
When comparing two weights, we will use the terms \df{lighter} and \df{heavier} in the
obvious sense.
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
-$\O(m\beta(m,n))$-time algorithm by Fredman and Tarjan \cite{ft:fibonacci} and
-algorithms with inverse-Ackermann type complexity by Chazelle \cite{chazelle:ackermann}
-and Pettie \cite{pettie:ackermann}.
+In the next 50 years, several significantly faster algorithms were discovered, ranging
+from the $\O(m\timesbeta(m,n))$ time algorithm by Fredman and Tarjan \cite{ft:fibonacci},
+over algorithms with inverse-Ackermann type complexity by Chazelle \cite{chazelle:ackermann}
+and Pettie \cite{pettie:ackermann}, to another algorithm by Pettie \cite{pettie:optimal}
+whose time complexity is provably optimal.
-\FIXME{Write the rest of the history.}
+In the upcoming chapters, we will explore this colorful universe of MST algorithms.
+We will meet the canonical works of the classics, the clever ideas of their successors,
+various approaches to the problem including randomization and solving of important
+special cases. At several places, we will try to contribute our little stones to this
+mosaic.
-This chapter attempts to survery the important algorithms for finding the MST and it
-also presents several new ones.
+%--------------------------------------------------------------------------------
-\section{Basic Properties}
+\section{Basic properties}\id{mstbasics}%
In this section, we will examine the basic properties of spanning trees and prove
-several important theorems to base the algorithms upon. We will follow the theory
-developed by Tarjan in~\cite{tarjan:dsna}.
+several important theorems which will serve as a~foundation for our MST algorithms.
+We will mostly 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
+For the whole section, we will fix a~connected 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.
the edges of this path \df{edges covered by~$e$}.
\: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.
+\:An edge~$e$ is called \df{$T$-heavy} if it covers a~lighter edge.
\endlist
\rem
-Please note that the above properties also apply to tree edges
-which by definition cover only themselves and therefore they are always heavy.
+Edges of the tree~$T$ cover only themselves and thus they are neither heavy nor light.
+The same can happen if an~edge outside~$T$ covers only edges of the same weight,
+but this will be rare because all edge weights will be usually distinct.
\lemman{Light edges}\id{lightlemma}%
Let $T$ be a spanning tree. If there exists a $T$-light edge, then~$T$
\lemman{Exchange property for trees}\id{xchglemma}%
Let $T$ and $T'$ be spanning trees of a common graph. Then there exists
-a sequence of edge exchanges which transforms $T$ to~$T'$. More formally,
+a sequence of edge exchanges that transforms $T$ to~$T'$. More formally,
there exists a sequence of spanning trees $T=T_0,T_1,\ldots,T_k=T'$ such that
$T_{i+1}=T_i - e_i + e_i^\prime$ where $e_i\in T_i$ and $e_i^\prime\in T'$.
\lemman{Monotone exchanges}\id{monoxchg}%
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.
+transforming $T$ to~$T'$ such that the weight does not decrease in any step.
\proof
We improve the argument from the previous proof, refining the induction step.
than~$e'$ as $e'$ was not $T$-light.
\qed
-\thmn{Minimality by order}\id{mstthm}%
+\thmn{Minimality of spanning trees}\id{mstthm}%
A~spanning tree~$T$ is minimum iff there is no $T$-light edge.
\proof
However, as the following theorem shows, this is possible only if the weight
function is not injective.
-\thmn{MST uniqueness}%
+\thmn{Uniqueness of MST}%
If all edge weights are distinct, then the minimum spanning tree is unique.
\proof
$T_1$ and $T_2$ must be identical.
\qed
-\rem\id{edgeoracle}%
-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.
-
\nota\id{mstnota}%
When $G$ is a graph with distinct edge weights, we will use $\mst(G)$ to denote
its unique minimum spanning tree.
+The following trivial lemma will be often invaluable:
+
+\lemman{Edge removal}
+Let~$G$ be a~graph with distinct edge weights and $e$ any its edge
+which does not lie in~$\mst(G)$. Then $\mst(G-e) = \mst(G)$.
+
+\proof
+The tree $T=\mst(G)$ is also a~MST of~$G-e$, because every $T$-light
+edge in~$G-e$ is also $T$-light in~$G$. Then we apply the uniqueness of
+the MST of~$G-e$.
+\qed
+
+\paran{Comparison oracles}\id{edgeoracle}%
+To simplify the description of MST algorithms, we will assume that the weights
+of all edges are distinct and that instead of numeric weights we are given a~comparison oracle.
+The oracle is a~function that answers questions of type ``Is $w(e)<w(f)$?'' in
+constant time. This will conveniently shield us from problems with representation
+of real numbers in algorithms and in the few cases where we need a more concrete
+input, we will explicitly state so.
+
+In case the weights are not distinct, we can easily break ties by comparing some
+unique identifiers of edges. According to our characterization of minimum spanning
+trees, the unique MST of the new graph will still be a~MST of the original graph.
+
+\obs
+If all edge weights are distinct and $T$~is an~arbitrary tree, then for every tree~$T$ all edges are
+either $T$-heavy, or $T$-light, or contained in~$T$.
+
+\paran{Monotone isomorphism}%
Another useful consequence is that whenever two graphs are isomorphic and the
-isomorphism preserves weight order, the isomorphism applies to their MST's
-as well:
+isomorphism preserves the relative order of weights, the isomorphism applies to their MST's as well:
\defn
-A~\df{monotone isomorphism} of two weighted graphs $G_1=(V_1,E_1,w_1)$ and
+A~\df{monotone isomorphism} between two weighted graphs $G_1=(V_1,E_1,w_1)$ and
$G_2=(V_2,E_2,w_2)$ is a bijection $\pi: V_1\rightarrow V_2$ such that
for each $u,v\in V_1: uv\in E_1 \Leftrightarrow \pi(u)\pi(v)\in E_2$ and
for each $e,f\in E_1: w_1(e)<w_1(f) \Leftrightarrow w_2(\pi[e]) < w_2(\pi[f])$.
\lemman{MST of isomorphic graphs}\id{mstiso}%
-Let~$G_1$ and $G_2$ be two weighted graphs with unique edge weights and $\pi$
-their monotone isomorphism. Then $\mst(G_2) = \pi[\mst(G_1)]$.
+Let~$G_1$ and $G_2$ be two weighted graphs with distinct edge weights and $\pi$
+a~monotone isomorphism between them. Then $\mst(G_2) = \pi[\mst(G_1)]$.
\proof
The isomorphism~$\pi$ maps spanning trees onto spanning trees and it preserves
the relation of covering. Since it is monotone, it preserves the property of
being a light edge (an~edge $e\in E(G_1)$ is $T$-light $\Leftrightarrow$
-the edge $\pi[e]\in E(G_2)$ is~$f[T]$-light). Therefore by Theorem~\ref{mstthm}, $T$
-is the MST of~$G_1$ if and only if $\pi[T]$ is the MST of~$G_2$.
+the edge $\pi[e]\in E(G_2)$ is~$f[T]$-light). Therefore by the Minimality Theorem
+(\ref{mstthm}), $T$ is the MST of~$G_1$ if and only if $\pi[T]$ is the MST of~$G_2$.
\qed
-\section{The Red-Blue Meta-Algorithm}
+%--------------------------------------------------------------------------------
+
+\section{The Red-Blue meta-algorithm}
Most MST algorithms can be described as special cases of the following procedure
-(again following \cite{tarjan:dsna}):
+(again following Tarjan \cite{tarjan:dsna}):
\algn{Red-Blue Meta-Algorithm}\id{rbma}%
\algo
\:In the beginning, all edges are colored black.
\: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}
+\::or pick a cycle~$C$ such that its heaviest edge is not red \hfil\break and color this edge \rack{blue.}{red.\hfil} \cmt{Red rule}
\algout Minimum spanning tree of~$G$ consisting of edges colored blue.
\endalgo
-\rem
+\para
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
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.
+\paran{Correctness}%
+Let us prove that the meta-algorithm is correct. First we show that the edges colored
+blue in any step of the procedure always belong to~$T_{min}$ and that edges colored
+red are guaranteed to be outside~$T_{min}$. Then we demonstrate that the procedure
+always stops. We will prefer a~slightly more general formulation of the lemmata, which will turn out
+to be useful in the future chapters.
+
+\lemman{Blue lemma, also known as the Cut rule}\id{bluelemma}%
+The lightest edge of every cut is contained in the MST.
\proof
-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
+By contradiction. Let $e$ be the lightest edge of a cut~$C$.
+If $e\not\in T_{min}$, then there must exist an edge $e'\in T_{min}$ that 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
+$w(e)<w(e')$.
+\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.
+\lemman{Red lemma, also known as the Cycle rule}\id{redlemma}%
+An~edge~$e$ is not contained in the MST iff it is the heaviest on some cycle.
\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 $T_{min}$.
+The implication from the left to the right follows directly from the Minimality
+theorem: if~$e\not\in T_{min}$, then $e$~is $T_{min}$-heavy and so it is the heaviest
+edge on the cycle $T_{min}[e]+e$.
+
+We will prove the other implication again by contradiction. Suppose that $e$ is 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~spanning tree lighter than
+$T_{min}$.
\qed
\figure{mst-rb.eps}{289pt}{Proof of the Blue (left) and Red (right) lemma}
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~\ref{mstthm}).
-In this case we can apply the red rule.
+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.
+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}
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~$T_{min}$
+a different color would mean that the 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
lemma all other (red) edges are outside~$T_{min}$, so the blue edges are exactly~$T_{min}$.
\qed
-\section{Classical algorithms}
+\rem
+The MST problem is a~special case of the problem of finding the minimum basis
+of a~weighted matroid. Surprisingly, when we modify the Red-Blue procedure to
+use the standard definitions of cycles and cuts in matroids, it will always
+find the minimum basis. Some of the other MST algorithms also easily generalize to
+matroids and in some sense matroids are exactly the objects where ``the greedy approach works''. We
+will however not pursue this direction in our work, referring the reader to the Oxley's monograph
+\cite{oxley:matroids} instead.
+
+%--------------------------------------------------------------------------------
+
+\section{Classical algorithms}\id{classalg}%
The three classical MST algorithms can be easily stated in terms of the Red-Blue meta-algorithm.
-For each of them, we first state the general version of the algorithm, then we prove that
+For each of them, we first show the general version of the algorithm, then we prove that
it gives the correct result and finally we discuss the time complexity of various
implementations.
+\paran{Bor\o{u}vka's algorithm}%
+The oldest MST algorithm is based on a~simple idea: grow a~forest in a~sequence of
+iterations until it becomes connected. We start with a~forest of isolated
+vertices. In each iteration, we let each tree of the forest select the lightest
+edge of those having exactly one endpoint in the tree (we will call such edges
+the \df{neighboring edges} of the tree). We add all such edges to the forest and
+proceed with the next iteration.
+
\algn{Bor\o{u}vka \cite{boruvka:ojistem}, Choquet \cite{choquet:mst}, Sollin \cite{sollin:mst} and others}
\algo
\algin A~graph~$G$ with an edge comparison oracle.
\algout Minimum spanning tree~$T$.
\endalgo
-\lemma
-Bor\o{u}vka's algorithm returns the MST of the input graph.
+\lemma\id{boruvkadrop}%
+In each iteration of the algorithm, the number of trees in~$T$ drops at least twice.
+
+\proof
+Each tree gets merged with at least one of its neighbors, so each of the new trees
+contains two or more original trees.
+\qed
+
+\cor
+The algorithm stops in $\O(\log n)$ iterations.
+
+\lemma\id{borcorr}%
+Bor\o{u}vka's algorithm outputs the MST of the input graph.
\proof
In every iteration of the algorithm, $T$ is a blue subgraph,
we do not need the Red rule to explicitly exclude edges.
It remains to show that adding the edges simultaneously does not
-produce a cycle. Consider the first iteration of the algorithm where $T$ contains some cycle~$C$. Without
-loss of generality we can assume that $C=T_1[u_1v_1]\,v_1u_2\,T_2[u_2v_2]\,v_2u_3\,T_3[u_3v_3]\, \ldots \,T_k[u_kv_k]\,v_ku_1$.
+produce a cycle. Consider the first iteration of the algorithm where $T$ contains a~cycle~$C$. Without
+loss of generality we can assume that:
+$$C=T_1[u_1v_1]\,v_1u_2\,T_2[u_2v_2]\,v_2u_3\,T_3[u_3v_3]\, \ldots \,T_k[u_kv_k]\,v_ku_1.$$
Each component $T_i$ has chosen its lightest incident edge~$e_i$ as either the edge $v_iu_{i+1}$
-or $v_{i-1}u_i$ (indexing cyclically). Assume that $e_1=v_1u_2$ (otherwise we reverse the orientation
-of the cycle). Then $e_2=v_2u_3$ and $w(e_2)<w(e_1)$ and so on, giving $w(e_1)>w(e_2)>\ldots>w(e_k)>w(e_1)$,
-which is a contradiction. (Note that distinctness of edge weights was crucial here.)
-\qed
-
-\lemma\id{boruvkadrop}%
-In each iteration of the algorithm, the number of trees in~$T$ drops at least twice.
-
-\proof
-Each tree gets merged with at least one neighboring trees, so each of the new trees
-consists of at least two original trees.
+or $v_{i-1}u_i$ (indexing cyclically). Suppose that $e_1=v_1u_2$ (otherwise we reverse the orientation
+of the cycle). Then $e_2=v_2u_3$ and $w(e_2)<w(e_1)$ and we can continue in the same way,
+getting $w(e_1)>w(e_2)>\ldots>w(e_k)>w(e_1)$, which is a contradiction.
+(Note that distinctness of edge weights was crucial here.)
\qed
-\cor
-The algorithm stops in $\O(\log n)$ iterations.
-
\lemma\id{boruvkaiter}%
Each iteration can be carried out in time $\O(m)$.
\proof
-Following \cite{mm:mst},
-we assign a label to each tree and we keep a mapping from vertices to the
+We assign a label to each tree and we keep a mapping from vertices to the
labels of the trees they belong to. We scan all edges, map their endpoints
to the particular trees and for each tree we maintain the lightest incident edge
so far encountered. Instead of merging the trees one by one (which would be too
-slow), we build an auxilliary graph whose vertices are labels of the original
+slow), we build an auxiliary graph whose vertices are the labels of the original
trees and edges correspond to the chosen lightest inter-tree edges. We find connected
-components of this graph, these determine how the original labels are translated
+components of this graph, these determine how are the original labels translated
to the new labels.
\qed
Follows from the previous lemmata.
\qed
-\algn{Jarn\'\i{}k \cite{jarnik:ojistem}, Prim \cite{prim:mst}, Dijkstra \cite{dijkstra:mst}}
+\paran{Jarn\'\i{}k's algorithm}%
+The next algorithm, discovered independently by Jarn\'\i{}k, Prim and Dijkstra, is similar
+to Bor\o{u}vka's algorithm, but instead of the whole forest it concentrates on
+a~single tree. It starts with a~single vertex and it repeatedly extends the tree
+by the lightest neighboring edge until it spans the whole graph.
+
+\algn{Jarn\'\i{}k \cite{jarnik:ojistem}, Prim \cite{prim:mst}, Dijkstra \cite{dijkstra:mst}}\id{jarnik}%
\algo
\algin A~graph~$G$ with an edge comparison oracle.
-\:$T\=$ a single-vertex tree containing any vertex of~$G$.
+\:$T\=$ a single-vertex tree containing an~arbitrary vertex of~$G$.
\:While there are vertices outside $T$:
\::Pick the lightest edge $uv$ such that $u\in V(T)$ and $v\not\in V(T)$.
\::$T\=T+uv$.
\endalgo
\lemma
-Jarn\'\i{}k's algorithm returns the MST of the input graph.
+Jarn\'\i{}k's algorithm computes the MST of the input graph.
\proof
-During the course of the algorithm, $T$ is always a blue tree. Step~4 corresponds to applying
-the Blue rule to a cut between~$T$ and the rest of the given graph. We need not care about
+If~$G$ is connected, the algorithm always stops. Let us prove that in every step of
+the algorithm, $T$ is always a blue tree. Step~4 corresponds to applying
+the Blue rule to the cut $\delta(T)$ separating~$T$ from the rest of the given graph. We need not care about
the remaining edges, since for a connected graph the algorithm always stops with the right
number of blue edges.
\qed
-\impl
-The most important part of the algorithm is finding \em{neighboring edges,} i.e., edges
-going between $T$ and $V(G)\setminus T$. In the straightforward implementation,
-searching for the lightest neighboring edge takes $\Theta(m)$ time, so the whole
-algorithm runs in time $\Theta(mn)$.
+\impl\id{jarnimpl}%
+The most important part of the algorithm is finding \em{neighboring edges.}
+In a~straightforward implementation, searching for the lightest neighboring
+edge takes $\Theta(m)$ time, so the whole algorithm runs in time $\Theta(mn)$.
We can do much better by using a binary
heap to hold all neighboring edges. In each iteration, we find and delete the
minimum edge from the heap and once we expand the tree, we insert the newly discovered
-neighboring edges to the heap while deleting the neighboring edges which become
+neighboring edges to the heap while deleting the neighboring edges that become
internal to the new tree. Since there are always at most~$m$ edges in the heap,
each heap operation takes $\O(\log m)=\O(\log n)$ time. For every edge, we perform
at most one insertion and at most one deletion, so we spend $\O(m\log n)$ time in total.
From this, we can conclude:
\thm
-Jarn\'\i{}k's algorithm finds the MST of the graph in time $\O(m\log n)$.
+Jarn\'\i{}k's algorithm computes the MST of a~given graph in time $\O(m\log n)$.
\rem
-We will show several faster implementations in section \ref{fibonacci}.
+We will show several faster implementations in section \ref{iteralg}.
+
+\paran{Kruskal's algorithm}%
+The last of the three classical algorithms processes the edges of the
+graph~$G$ greedily. It starts with an~empty forest and it takes the edges of~$G$
+in order of their increasing weights. For every edge, it checks whether its
+addition to the forest produces a~cycle and if it does not, the edge is added.
+Otherwise, the edge is dropped and not considered again.
-\algn{Kruskal \cite{kruskal:mst}, the Greedy algorithm}
+\algn{Kruskal \cite{kruskal:mst}}
\algo
\algin A~graph~$G$ with an edge comparison oracle.
-\:Sort edges of~$G$ by their increasing weight.
+\:Sort edges of~$G$ by their increasing weights.
\:$T\=\emptyset$. \cmt{an empty spanning subgraph}
\:For all edges $e$ in their sorted order:
\::If $T+e$ is acyclic, add~$e$ to~$T$.
\proof
In every step, $T$ is a forest of blue trees. Adding~$e$ to~$T$
-in step~4 applies the Blue rule on the cut separating two components of~$T$ ($e$ is the lightest,
+in step~4 applies the Blue rule on the cut separating some pair of components of~$T$ ($e$ is the lightest,
because all other edges of the cut have not been considered yet). Dropping~$e$ in step~5 corresponds
-to the red rule on the cycle found ($e$~must be the heaviest, since all other edges of the
-cycle have been already processed). At the end of the algorithm, all edges have been colored,
+to the Red rule on the cycle found ($e$~must be the heaviest, since all other edges of the
+cycle have been already processed). At the end of the algorithm, all edges are colored,
so~$T$ must be the~MST.
\qed
\impl
-Except for the initial sorting, which in general takes $\Theta(m\log n)$ time, the only
-other non-trivial operation is detection of cycles. What we need is a data structure
+Except for the initial sorting, which in general requires $\Theta(m\log m)$ time, the only
+other non-trivial operation is the detection of cycles. What we need is a~data structure
for maintaining connected components, which supports queries and edge insertion.
-The following theorem shows that it can be done with a surprising efficiency.
+This is closely related to the well-known Disjoint Set Union problem:
+
+\problemn{Disjoint Set Union (DSU)}
+Maintain an~equivalence relation on a~finite set under a~sequence of operations \<Union>
+and \<Find>. The \<Find> operation tests whether two elements are equivalent and \<Union>
+joins two different equivalence classes into one.
+
+\para
+We can maintain the connected components of our forest~$T$ as equivalence classes. When we want
+to add an~edge~$uv$, we first call $\<Find>(u,v)$ to check if both endpoints of the edge lie in
+the same components. If they do not, addition of this edge connects both components into one,
+so we perform $\<Union>(u,v)$ to merge the equivalence classes.
+
+Tarjan and van Leeuwen have shown that there is a~data structure for the DSU problem
+with surprising efficiency:
-\thmn{Incremental connectivity}%
-When only edge insertions and queries are allowed, connected components
-can be maintained in $\O(\alpha(n))$ time amortized per operation.
+\thmn{Disjoint Set Union, Tarjan and van Leeuwen \cite{tarjan:setunion}}\id{dfu}%
+Starting with a~trivial equivalence with single-element classes, a~sequence of operations
+comprising of $n$~\<Union>s intermixed with $m\ge n$~\<Find>s can be processed in time
+$\O(m\timesalpha(m,n))$, where $\alpha(m,n)$ is a~certain inverse of the Ackermann's function
+(see Definition \ref{ackerinv}).
\proof
-Proven by Tarjan and van Leeuwen in \cite{tarjan:setunion}.
+See \cite{tarjan:setunion}.
\qed
-\FIXME{Define Ackermann's function. Use $\alpha(m,n)$?}
+This completes the following theorem:
+
+\thm\id{kruskal}%
+Kruskal's algorithm finds the MST of a given graph in time $\O(m\log n)$.
+If the edges are already sorted by their weights, the time drops to
+$\O(m\timesalpha(m,n))$.
+
+\proof
+We spend $\O(m\log n)$ on sorting, $\O(m\timesalpha(m,n))$ on processing the sequence
+of \<Union>s and \<Find>s, and $\O(m)$ on all other work.
+\qed
\rem
-The cost of the operations on components is of course dwarfed by the complexity
+The cost of the \<Union> and \<Find> operations is of course dwarfed by the complexity
of sorting, so a much simpler (at least in terms of its analysis) data
structure would be sufficient, as long as it has $\O(\log n)$ amortized complexity
per operation. For example, we can label vertices with identifiers of the
-corresponding components and always recolor the smaller of the two components.
+corresponding components and always relabel the smaller of the two components.
-\thm
-Kruskal's algorithm finds the MST of a given graph in time $\O(m\log n)$
-or $\O(m\alpha(n))$ if the edges are already sorted by their weights.
+We will study dynamic maintenance of connected components in more detail in Chapter~\ref{dynchap}.
-\proof
-Follows from the above analysis.
-\qed
+%--------------------------------------------------------------------------------
-\section{Contractive algorithms}
+\section{Contractive algorithms}\id{contalg}%
While the classical algorithms are based on growing suitable trees, they
can be also reformulated in terms of edge contraction. Instead of keeping
-a forest of trees, we can keep each tree contracted to a single vertex.
+a~forest of trees, we can keep each tree contracted to a single vertex.
This replaces the relatively complex tree-edge incidencies by simple
vertex-edge incidencies, potentially speeding up the calculation at the
expense of having to perform the contractions.
in which these costs are carefully balanced, leading for example to
a linear-time algorithm for MST in planar graphs.
-There are two definitions of edge contraction which differ when an edge of a
-triangle is contracted. Either we unify the other two edges to a single edge
+There are two definitions of edge contraction that differ when an edge of
+a~triangle is contracted. Either we unify the other two edges to a single edge
or we keep them as two parallel edges, leaving us with a~multigraph. We will
-use the multigraph version and show that we can easily reduce the multigraph
+use the multigraph version and we will show that we can easily reduce the multigraph
to a simple graph later. (See \ref{contract} for the exact definitions.)
We only need to be able to map edges of the contracted graph to the original
-edges, so each edge will carry a unique label $\ell(e)$ which will be preserved by
+edges, so each edge will carry a unique label $\ell(e)$ that will be preserved by
contractions.
\lemman{Flattening a multigraph}\id{flattening}%
Then $G'$~has the same MST as~$G$.
\proof
-Loops can be never used in a spanning tree. If there is a spanning tree~$T$
-containing a removed edge~$e$ parallel to an edge~$e'\in G'$, exchaning $e'$
-for~$e$ in~$T$ makes it lighter. \qed
-
-\rem Removal of the heavier of a pair of parallel edges can be also viewed
-as an application of the Red rule on a two-edge cycle. And indeed it is, the
-Red-Blue procedure works on multigraphs as well as on simple graphs and all the
-classical algorithms also do. We only would have to be more careful in the
-formulations and proofs, which we preferred to avoid. We also note that most of
-the algorithms can be run on disconnected multigraphs with little or no
-modifications.
+Every spanning tree of~$G'$ is a spanning tree of~$G$. In the other direction:
+Loops can be never contained in a spanning tree. If there is a spanning tree~$T$
+containing a~removed edge~$e$ parallel to an edge~$e'\in G'$, exchanging $e'$
+for~$e$ makes~$T$ lighter. (This is indeed the multigraph version of the Red
+lemma applied to a~two-edge cycle, as we will see in \ref{multimst}.)
+\qed
\algn{Contractive version of Bor\o{u}vka's algorithm}\id{contbor}
\algo
\:$T\=\emptyset$.
\:$\ell(e)\=e$ for all edges~$e$. \cmt{Initialize the labels.}
\:While $n(G)>1$:
-\::For each vertex $v_i$ of~$G$, let $e_i$ be the lightest edge incident to~$v_i$.
-\::$T\=T\cup \{ \ell(e_i) \}$. \cmt{Remember labels of all selected edges.}
-\::Contract $G$ along all edges $e_i$, inheriting labels and weights.
+\::For each vertex $v_k$ of~$G$, let $e_k$ be the lightest edge incident to~$v_k$.
+\::$T\=T\cup \{ \ell(e_k) \}$. \cmt{Remember labels of all selected edges.}
+\::Contract all edges $e_k$, inheriting labels and weights.\foot{In other words, we ask the comparison oracle for the edge $\ell(e)$ instead of~$e$.}
\::Flatten $G$, removing parallel edges and loops.
\algout Minimum spanning tree~$T$.
\endalgo
+\nota
+For the analysis of the algorithm, we will denote the graph considered by the algorithm
+at the beginning of the $i$-th iteration by $G_i$ (starting with $G_0=G$) and the number
+of vertices and edges of this graph by $n_i$ and $m_i$ respectively.
+
\lemma\id{contiter}%
-Each iteration of the algorithm can be carried out in time~$\O(m)$.
+The $i$-th iteration of the algorithm (also called the \df{Bor\o{u}vka step}) can be carried
+out in time~$\O(m_i)$.
\proof
The only non-trivial parts are steps 6 and~7. Contractions can be handled similarly
-to the unions in the original Bor\o{u}vka's algorithm (see \ref{boruvkaiter}).
-We build an auxillary graph containing only the selected edges~$e_i$, find
+to the unions in the original Bor\o{u}vka's algorithm (see \ref{boruvkaiter}):
+We build an auxiliary graph containing only the selected edges~$e_k$, find
connected components of this graph and renumber vertices in each component to
-the identifier of the component. This takes $\O(m)$ time.
+the identifier of the component. This takes $\O(m_i)$ time.
Flattening is performed by first removing the loops and then bucket-sorting the edges
(as ordered pairs of vertex identifiers) lexicographically, which brings parallel
-edges together. The bucket sort uses two passes with $n$~buckets, so it takes
-$\O(n+m)=\O(m)$.
+edges together. The bucket sort uses two passes with $n_i$~buckets, so it takes
+$\O(n_i+m_i)=\O(m_i)$.
\qed
-\thm
-The Contractive Bor\o{u}vka's algorithm finds the MST in time $\O(m\log n)$.
+\thm\id{contborthm}%
+The Contractive Bor\o{u}vka's algorithm finds the MST of the input graph in
+time $\O(\min(n^2,m\log n))$.
\proof
-As in the original Bor\o{u}vka's algorithm, the number of phases is $\O(\log n)$.
-Then apply the previous lemma.
+As in the original Bor\o{u}vka's algorithm, the number of iterations is $\O(\log n)$.
+When combined with the previous lemma, it gives an~$\O(m\log n)$ upper bound.
+
+To get the $\O(n^2)$ bound, we observe that the number of trees in the non-contracting
+version of the algorithm drops at least by a factor of two in each iteration (Lemma \ref{boruvkadrop})
+and the same must hold for the number of vertices in the contracting version.
+Therefore $n_i\le n/2^i$. While the number of edges need not decrease geometrically,
+we still have $m_i\le n_i^2$ as the graphs~$G_i$ are simple (we explicitly removed multiple
+edges and loops at the end of the previous iteration). Hence the total time spent
+in all iterations is $\O(\sum_i n_i^2) = \O(\sum_i n^2/4^i) = \O(n^2)$.
\qed
-\thmn{\cite{mm:mst}}\id{planarbor}%
+\thmn{Contractive Bor\o{u}vka on planar graphs, \cite{mm:mst}}\id{planarbor}%
When the input graph is planar, the Contractive Bor\o{u}vka's algorithm runs in
time $\O(n)$.
\proof
-Let us denote the graph considered by the algorithm at the beginning of the $i$-th
-iteration by $G_i$ (starting with $G_0=G$) and its number of vertices and edges
-by $n_i$ and $m_i$ respectively. As we already know from the previous lemma,
-the $i$-th iteration takes $\O(m_i)$ time. We are going to prove that the
-$m_i$'s are decreasing exponentially.
-
-The number of trees in the non-contracting version of the algorithm decreases
-at least twice in each iteration (Lemma \ref{boruvkadrop}) and the
-same must hold for the number of vertices in the contracting version.
-Therefore $n_i\le n/2^i$.
-
-However, every $G_i$ is planar, because the class of planar graphs is closed
-under edge deletion and contraction. The~$G_i$ is also simple as we explicitly removed multiple edges and
-loops at the end of the previous iteration. So we can use the standard theorem on
+Let us refine the previous proof. We already know that $n_i \le n/2^i$. We will
+prove that when~$G$ is planar, the $m_i$'s are decreasing geometrically. We know that every
+$G_i$ is planar, because the class of planar graphs is closed under edge deletion and
+contraction. Moreover, $G_i$~is also simple, so we can use the standard theorem on
the number of edges of planar simple graphs (see for example \cite{diestel:gt}) to get $m_i\le 3n_i \le 3n/2^i$.
-From this we get that the total time complexity is $\O(\sum_i m_i)=\O(\sum_i n/2^i)=\O(n)$.
+The total time complexity of the algorithm is therefore $\O(\sum_i m_i)=\O(\sum_i n/2^i)=\O(n)$.
\qed
\rem
-There are other possibilities how to find the MST of a planar graph in linear time.
-Matsui \cite{matsui:planar} has described an algorithm based on simultaneously
-processing the graph and its dual. The advantage of our approach is that we do not
-need to construct the planar embedding first.
-
-\rem
-To achieve the linear time complexity, the algorithm needs a very careful implementation.
-Specifically, when we represent the graph using adjacency lists, whose heads are stored
-in an array indexed by vertex identifiers, we must renumber the vertices in each iteration.
-Otherwise, unused identifiers could end up taking most of space in the arrays and scans of these
-arrays would have super-linear cost with respect to the size of the current graph~$G_i$.
+There are several other possibilities how to find the MST of a planar graph in linear time.
+For example, Matsui \cite{matsui:planar} has described an algorithm based on simultaneously
+working on the graph and its topological dual. The advantage of our approach is that we do not need
+to construct the planar embedding explicitly. We will show one more linear algorithm
+in section~\ref{minorclosed}.
\rem
-The algorithm can be also implemented on the pointer machine. Representation of graphs
-by pointer structures easily avoids the aforementioned problems with sparse arrays,
-but we need to handle the bucket sorting somehow. We can create a small data structure
-for every vertex and use a pointer to this structure as a unique identifier of the vertex.
-We will also keep a list of all vertex structures. During the bucket sort, each vertex
-structure will contain a pointer to the corresponding bucket and the vertex list will
-define the order of vertices (which can be arbitrary, but has to be fixed).
+To achieve the linear time complexity, the algorithm needs a very careful implementation,
+but we defer the technical details to section~\ref{bucketsort}.
+\paran{General contractions}%
Graph contractions are indeed a~very powerful tool and they can be used in other MST
algorithms as well. The following lemma shows the gist:
\lemman{Contraction of MST edges}\id{contlemma}%
Let $G$ be a weighted graph, $e$~an arbitrary edge of~$\mst(G)$, $G/e$ the multigraph
-produced by contracting $G$ along~$e$, and $\pi$ the bijection between edges of~$G-e$ and
+produced by contracting~$e$ in~$G$, and $\pi$ the bijection between edges of~$G-e$ and
their counterparts in~$G/e$. Then: $$\mst(G) = \pi^{-1}[\mst(G/e)] + e.$$
\proof
%in a MST.
The right-hand side of the equality is a spanning tree of~$G$, let us denote it by~$T$ and
the MST of $G/e$ by~$T'$. If $T$ were not minimum, there would exist a $T$-light edge~$f$ in~$G$
-(according to Theorem \ref{mstthm}). If the path $T[f]$ covered by~$f$ does not contain~$e$,
+(by Theorem \ref{mstthm}). If the path $T[f]$ covered by~$f$ does not contain~$e$,
then $\pi[T[f]]$ is a path covered by~$\pi(f)$ in~$T'$. Otherwise $\pi(T[f]-e)$ is such a path.
In both cases, $f$ is $T'$-light, which contradicts the minimality of~$T'$. (We do not have
-a~multigraph version of the theorem, but this direction is a straightforward edge exchange,
-which of course works in multigraphs as well.)
+a~multigraph version of the theorem, but the side we need is a~straightforward edge exchange,
+which obviously works in multigraphs as well.)
\qed
\rem
In the previous algorithm, the role of the mapping~$\pi^{-1}$ is of course played by the edge labels~$\ell$.
-Finally, we will show a family of graphs where the $\O(m\log n)$ bound on time complexity
+\paran{A~lower bound}%
+Finally, we will show a family of graphs for which the $\O(m\log n)$ bound on time complexity
is tight. The graphs do not have unique weights, but they are constructed in a way that
the algorithm never compares two edges with the same weight. Therefore, when two such
-graphs are monotonely isomorphic (see~\ref{mstiso}), the algorithm processes them in the same way.
+graphs are monotonically isomorphic (see~\ref{mstiso}), the algorithm processes them in the same way.
\defn
A~\df{distractor of order~$k$,} denoted by~$D_k$, is a path on $n=2^k$~vertices $v_1,\ldots,v_n$
representation of the number~$i$. The vertex $v_1$ is called a~\df{base} of the distractor.
\rem
-Alternatively, we can use a recursive definition: $D_0$ is a single vertex, $D_k$ consists
-of two disjoint copies of $D_{k-1}$ joined by an edge of weight~$k$.
+Alternatively, we can use a recursive definition: $D_0$ is a single vertex, $D_{k+1}$ consists
+of two disjoint copies of~$D_k$ joined by an edge of weight~$k$.
\figure{distractor.eps}{\epsfxsize}{A~distractor $D_3$ and its evolution (bold edges are contracted)}
\defn
A~\df{hedgehog}~$H_{a,k}$ is a graph consisting of $a$~distractors $D_k^1,\ldots,D_k^a$ of order~$k$
together with edges of a complete graph on the bases of the distractors. These additional edges
-have unique weights larger than~$k$, i.e., they are heavier than the edges of the distractors.
+have arbitrary weights, but heavier than the edges of all distractors.
\figure{hedgehog.eps}{\epsfxsize}{A~hedgehog $H_{5,2}$ (quills bent to fit in the picture)}
\thmn{Lower bound for Contractive Bor\o{u}vka}%
For each $n$ there exists a graph on $\Theta(n)$ vertices and $\Theta(n)$ edges
-such that the Contractive Bor\o{u}vka's algorithm spends time $\Omega(m\log n)$ on it.
+such that the Contractive Bor\o{u}vka's algorithm spends time $\Omega(n\log n)$ on it.
\proof
Consider the hedgehog $H_{a,k}$ for $a=\lceil\sqrt n\rceil$ and $k=\lceil\log_2 a\rceil$.
edges as every $H_{a,k}$ contains a complete graph on~$a$ vertices.
\qed
-\section{Minor-closed graph classes}
-
-The contracting algorithm given in the previous section has been found to perform
-well on planar graphs, but in the general case its time complexity was not linear.
-Can we find some broader class of graphs where the algorithm is still efficient?
-The right context turns out to be the minor-closed graph classes, which are
-closed under contractions and have bounded density.
-
-\defn
-A~graph~$H$ is a \df{minor} of a~graph~$G$ iff it can be obtained
-from a subgraph of~$G$ by a sequence of graph contractions (see \ref{simpcont}).
-
-\defn
-A~class~$\cal C$ of graphs is \df{minor-closed}, when for every $G\in\cal C$ and
-its minor~$H$, the graph~$H$ lies in~$\cal C$ as well. A~class~$\cal C$ is called
-\df{non-trivial} if at least one graph lies in~$\cal C$ and at least one lies outside~$\cal C$.
-
-\example
-Non-trivial minor-closed classes include planar graphs and more generally graphs
-embeddable in any fixed surface. Many nice properties of planar graphs extend
-to these classes, too, most notable the linearity of the number of edges.
-
-\defn\id{density}%
-Let $\cal C$ be a class of graphs. We define its \df{edge density} $\varrho(\cal C)$
-to be the infimum of all~$\varrho$'s such that $m(G) \le \varrho\cdot n(G)$
-holds for every $G\in\cal C$.
-
-\thmn{Density of minor-closed classes}
-A~minor-closed class of graphs has finite edge density if and only if it is
-a non-trivial class.
-
-\proof
-See Theorem 6.1 in \cite{nesetril:minors}, which also lists some other equivalent conditions.
-\qed
-
-\thmn{MST on minor-closed classes \cite{mm:mst}}\id{mstmcc}%
-For any fixed non-trivial minor-closed class~$\cal C$ of graphs, Algorithm \ref{contbor} finds
-the MST of any graph from this class in time $\O(n)$. (The constant hidden in the~$\O$
-depends on the class.)
-
-\proof
-Following the proof for planar graphs (\ref{planarbor}), we denote the graph considered
-by the algorithm at the beginning of the $i$-th iteration by~$G_i$ and its number of vertices
-and edges by $n_i$ and $m_i$ respectively. Again the $i$-th phase runs in time $\O(m_i)$
-and $n_i \le n/2^i$, so it remains to show a linear bound for the $m_i$'s.
-
-Since each $G_i$ is produced from~$G_{i-1}$ by a sequence of edge contractions,
-all $G_i$'s are minors of~$G$.\foot{Technically, these are multigraph contractions,
-but followed by flattening, so they are equivalent to contractions on simple graphs.}
-So they also belong to~$\cal C$ and by the previous theorem $m_i\le \varrho({\cal C})n_i$.
-\qed
-
-\rem\id{nobatch}%
-The contractive algorithm uses ``batch processing'' to perform many contractions
-in a single step. It is also possible to perform contractions one edge at a~time,
-batching only the flattenings. A~contraction of an edge~$uv$ can be performed
-in time~$\O(\deg(u))$ by removing all edges incident with~$u$ and inserting them back
-with $u$ replaced by~$v$. Therefore we need to find a lot of vertices with small
-degrees. The following lemma shows that this is always the case in minor-closed
-classes.
-
-\lemman{Low-degree vertices}\id{lowdeg}%
-Let $\cal C$ be a graph class with density~$\varrho$ and $G\in\cal C$ a~graph
-with $n$~vertices. Then at least $n/2$ vertices of~$G$ have degree at most~$4\varrho$.
-
-\proof
-Assume the contrary: let there be at least $n/2$ vertices with degree
-greater than~$4\varrho$. Then $\sum_v \deg(v) > n/2
-\cdot 4\varrho = 2\varrho n$, which is in contradiction with the number
-of edges being at most $\varrho n$.
-\qed
-
-\rem
-The proof can be also viewed
-probabilistically: let $X$ be the degree of a vertex of~$G$ chosen uniformly at
-random. Then ${\bb E}X \le 2\varrho$, hence by the Markov's inequality
-${\rm Pr}[X > 4\varrho] < 1/2$, so for at least $n/2$ vertices~$v$ we have
-$\deg(v)\le 4\varrho$.
-
-\algn{Local Bor\o{u}vka's Algorithm \cite{mm:mst}}%
-\algo
-\algin A~graph~$G$ with an edge comparison oracle and a~parameter~$t$.
-\:$T\=\emptyset$.
-\:$\ell(e)\=e$ for all edges~$e$.
-\:While $n(G)>1$:
-\::While there exists a~vertex~$v$ such that $\deg(v)\le t$:
-\:::Select the lightest edge~$e$ incident with~$v$.
-\:::Contract~$G$ along~$e$.
-\:::$T\=T\cup \{ \ell(e_i) \}$.
-\::Flatten $G$, removing parallel edges and loops.
-\algout Minimum spanning tree~$T$.
-\endalgo
-
-\thm
-When $\cal C$ is a minor-closed class of graphs with density~$\varrho$, the
-Local Bor\o{u}vka's Algorithm with the parameter~$t$ set to~$4\varrho$
-finds the MST of any graph from this class in time $\O(n)$. (The constant
-in the~$\O$ depends on~the class only.)
-
-\proof
-Let us denote by $G_i$, $n_i$ and $m_i$ the graph considered by the
-algorithm at the beginning of the $i$-th iteration of the outer loop,
-and the number of its vertices and edges respectively. As in the proof
-of the previous algorithm (\ref{mstmcc}), we observe that all the $G_i$'s
-are minors of the graph~$G$ given as the input.
-
-For the choice $t=4\varrho$, the Lemma on low-degree vertices (\ref{lowdeg})
-guarantees that at least $n_i/2$ edges get selected in the $i$-th iteration.
-Hence at least a half of the vertices participates in contractions, so
-$n_i\le 3/4\cdot n_{i-1}$. Therefore $n_i\le n\cdot (3/4)^i$ and the algorithm terminates
-after $\O(\log n)$ iterations.
-
-Each selected edge belongs to $\mst(G)$, because it is the lightest edge of
-the trivial cut separating~$v$ from the rest of the graph (see the Blue
-Rule in \ref{rbma}). The steps 6 and~7 therefore correspond to the operation
-described by the Lemma on contraction of MST edges (\ref{contlemma}) and when
-the algorithm stops, $T$~is indeed the minimum spanning tree.
-
-It remains to analyse the time complexity of the algorithm. Since $G_i\in{\cal C}$, we have
-$m_i\le \varrho n_i \le \varrho n/2^i$.
-We will show that the $i$-th iteration is carried out in time $\O(m_i)$.
-Steps 5 and~6 run in time $\O(\deg(v))=\O(t)$ for each~$v$, so summed
-over all $v$'s they take $\O(tn_i)$, which is linear for a fixed class~$\cal C$.
-Flattening takes $\O(m_i)$, as already noted in the analysis of the Contracting
-Bor\o{u}vka's Algorithm (see \ref{contiter}).
-
-The whole algorithm therefore runs in time $\O(\sum_i m_i) = \O(\sum_i n/2^i) = \O(n)$.
-\qed
-
-\rem
-For planar graphs, we can get a sharper version of the low-degree lemma,
-showing that the algorithm works with $t=8$ as well (we got $t=12$ from the
-general version). While this does not change the asymptotic time complexity
-of the algorithm, the constant-factor speedup can still delight the hearts of
-its practical users.
-
-\lemman{Low-degree vertices in planar graphs}%
-Let $G$ be a planar graph with $n$~vertices. Then at least $n/2$ vertices of~$v$
-have degree at most~8.
-
-\proof
-It suffices to show that the lemma holds for triangulations (if there
-are any edges missing, the situation can only get better) with at
-least 3 vertices. Since $G$ is planar, $\sum_v \deg(v) < 6n$.
-The numbers $d(v):=\deg(v)-3$ are non-negative and $\sum_v d(v) < 3n$,
-hence by the same argument as in the previous proof, for at least $n/2$
-vertices~$v$ it holds that $d(v) < 6$, hence $\deg(v) \le 8$.
-\qed
+%--------------------------------------------------------------------------------
+\section{Lifting restrictions}
+In order to have a~simple and neat theory, we have introduced several restrictions
+on the graphs in which we search for the MST. As in some rare cases we are going to
+meet graphs that do not fit into this simplified world, let us quickly examine what
+happens when the restrictions are lifted.
+\paran{Disconnected graphs}\id{disconn}%
+The basic properties of minimum spanning trees and the algorithms presented in
+this chapter apply to minimum spanning forests of disconnected graphs, too.
+The proofs of our theorems and the steps of our algorithms are based on adjacency
+of vertices and existence of paths, so they are always local to a~single
+connected component. The Bor\o{u}vka's and Kruskal's algorithm need no changes,
+the Jarn\'\i{}k's algorithm has to be invoked separately for each component.
-\section{Using Fibonacci heaps}
-\id{fibonacci}
+We can also extend the notion of light and heavy edges with respect
+to a~tree to forests: When an~edge~$e$ connects two vertices lying in the same
+tree~$T$ of a~forest~$F$, it is $F$-heavy iff it is $T$-heavy (similarly
+for $F$-light). Edges connecting two different trees are always considered
+$F$-light. Again, a~spanning forest~$F$ is minimum iff there are no $F$-light
+edges.
-% G has to be connected, so m=O(n)
-% mention Steiner trees
-% mention matroids
-% sorted weights
-% \O(...) as a set?
-% impedance mismatch in terminology: contraction of G along e vs. contraction of e.
-% use \delta(X) notation
+\paran{Multigraphs}\id{multimst}%
+All theorems and algorithms from this chapter work for multigraphs as well,
+only the notation sometimes gets crabbed, which we preferred to avoid. The Minimality
+theorem and the Blue rule stay unchanged. The Red rule is naturally extended to
+self-loops (which are never in the MST) and two-edge cycles (where the heavier
+edge can be dropped) as already suggested in the Flattening lemma (\ref{flattening}).
+
+\paran{Multiple edges of the same weight}\id{multiweight}%
+In case when the edge weights are not distinct, the characterization of minimum
+spanning trees using light edges is still correct, but the MST is no longer unique
+(as already mentioned, there can be as much as~$n^{n-2}$ MST's).
+
+In the Red-Blue procedure, we have to avoid being too zealous. The Blue lemma cannot
+guarantee that when a~cut contains multiple edges of the minimum weight, all of them
+are in the MST. It will however tell that if we pick one of these edges, an~arbitrary
+MST can be modified to another MST that contains this edge. Therefore the Blue rule
+will change to ``Pick a~cut~$C$ such that it does not contain any blue edge and color
+one of its lightest edges blue.'' The Red lemma and the Red rule can be handled
+in a~similar manner. The modified algorithm will be then guaranteed to find one of
+the possible MST's.
+
+The Kruskal's and Jarn\'\i{}k's algorithms keep working. This is however not the case of the
+Bor\o{u}vka's algorithm, whose proof of correctness in Lemma \ref{borcorr} explicitly referred to
+distinct weights and indeed, if they are not distinct, the algorithm will occasionally produce
+cycles. To avoid the cycles, the ties in edge weight comparisons have to be broken in a~systematic
+way. The same applies to the contractive version of this algorithm.
\endpart