[saga.git] / mst.tex

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\chapter{Minimum Spanning Trees}

\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 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\o{u}vka
\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 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, defined as follows:

\defn\id{mstdef}%
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.
\: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.
\: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\o{u}vka's work was further extended by Jarn\'\i{}k \cite{jarnik:ojistem}, again in
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 had to be
rediscovered several times.

Recently, several significantly faster algorithms were discovered, most notably the
$\O(m\timesbeta(m,n))$-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}.

\FIXME{Write the rest of the history.}

This chapter attempts to survery the important algorithms for finding the MST and it
also presents several new ones.

%--------------------------------------------------------------------------------

\section{Basic properties}

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}.

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
an ordering of edges instead.

\defnn{Heavy and light edges}\id{heavy}%
Let~$T$ be a~spanning tree. Then:
\itemize\ibull
\: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{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

\rem
Please note that the above properties also apply to tree edges
which by definition cover only themselves and therefore they are always heavy.

\lemman{Light edges}\id{lightlemma}%
Let $T$ be a spanning tree. If there exists a $T$-light edge, then~$T$
is not minimum.

\proof
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 $T':=T-e'+e$ 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~\ref{lightlemma}}

The converse of this lemma is also true and to prove it, we will once again use
technique of transforming trees by \df{exchanges} of edges. In the proof of the
lemma, we have made use of the fact that whenever we exchange an edge~$e$ of
a spanning tree for another edge~$f$ covered by~$e$, the result is again
a spanning tree. In fact, it is possible to transform any spanning tree
to any other spanning tree by a sequence of exchanges.

\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,
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 $e_i^\prime\in T'$.

\proof
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
hypothesis to $T^*$ and $T'$ to get the rest of the exchange sequence.
\qed

\figure{mst1.eps}{295pt}{One step of the proof of Lemma~\ref{xchglemma}}

\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.

\proof
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 $w(T^*)=w(T)-w(e)+w(e')\ge w(T)$.

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

\thmn{Minimality by order}\id{mstthm}%
A~spanning tree~$T$ is minimum iff there is no $T$-light edge.

\proof
If~$T$ is minimum, then by Lemma~\ref{lightlemma} there are no $T$-light
edges.
Conversely, when $T$ is a spanning tree without $T$-light edges
and $T_{min}$ is an arbitrary minimum spanning tree, then according to the Monotone
exchange lemma (\ref{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 minimum.
\qed

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 theorem shows, this is possible only if the weight
function is not injective.

\thmn{MST uniqueness}%
If all edge weights are distinct, then the minimum spanning tree is unique.

\proof
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 (\ref{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\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.

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:

\defn
A~\df{monotone isomorphism} of 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 $\mst(G_2) = \pi[\mst(G_1)]$.

\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~$G_2$.
\qed

%--------------------------------------------------------------------------------

\section{The Red-Blue meta-algorithm}

Most MST algorithms can be described as special cases of the following procedure
(again following \cite{tarjan:dsna}):

\algn{Red-Blue Meta-Algorithm}\id{rbma}%
\algo
\algin A~graph $G$ with an edge comparison oracle (see \ref{edgeoracle})
\: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}
\algout Minimum spanning tree of~$G$ consisting of edges colored blue.
\endalgo

\rem
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.

\nota
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
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. Suppose 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}

\lemman{Black 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~\ref{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}

\thmn{Red-Blue correctness}%
For any selection of rules, the Red-Blue procedure stops and the blue edges form
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

%--------------------------------------------------------------------------------

\section{Classical algorithms}

The three classical MST algorithms can be easily stated in terms of the Red-Blue meta-algorithm.
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.

\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.
\:$T\=$ a forest consisting of vertices of~$G$ and no edges.
\:While $T$ is not connected:
\::For each component $T_i$ of~$T$, choose the lightest edge $e_i$ from the cut
   separating $T_i$ from the rest of~$T$.
\::Add all $e_i$'s to~$T$.
\algout Minimum spanning tree~$T$.
\endalgo

\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
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,
because every addition of some edge~$e_i$ to~$T$ is a straightforward
application of the Blue rule. We stop when the blue subgraph is connected, so
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 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). 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

\lemma\id{boruvkaiter}%
Each iteration can be carried out in time $\O(m)$.

\proof
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 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 are the original labels translated
to the new labels.
\qed

\thm
Bor\o{u}vka's algorithm finds the MST in time $\O(m\log n)$.

\proof
Follows from the previous lemmata.
\qed

\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 an~arbitrary vertex of~$G$.
\:While there are vertices outside $T$:
\::Pick the lightest edge $uv$ such that $u\in V(T)$ and $v\not\in V(T)$.
\::$T\=T+uv$.
\algout Minimum spanning tree~$T$.
\endalgo

\lemma
Jarn\'\i{}k's algorithm computers the MST of the input graph.

\proof
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
of the cut $\delta(T)$. 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
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 a~given graph in time $\O(m\log n)$.

\rem
We will show several faster implementations in section \ref{fibonacci}.

\algn{Kruskal \cite{kruskal:mst}, the Greedy algorithm}
\algo
\algin A~graph~$G$ with an edge comparison oracle.
\:Sort edges of~$G$ by their increasing weight.
\:$T\=\emptyset$. \cmt{an empty spanning subgraph}
\:For all edges $e$ in their sorted order:
\::If $T+e$ is acyclic, add~$e$ to~$T$.
\::Otherwise drop~$e$.
\algout Minimum spanning tree~$T$.
\endalgo

\lemma
Kruskal's algorithm returns the MST of the input graph.

\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 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 are colored,
so~$T$ must be the~MST.
\qed

\impl
Except for the initial sorting, which in general takes $\Theta(m\log m)$ 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.
(This is also known under the name Disjoint Set Union problem, i.e., maintenance
of an~equivalence relation on a~set with queries on whether two elements are equivalent
and the operation of joining two equivalence classes into one.)
The following theorem shows that it can be done with surprising efficiency.

\thmn{Incremental connectivity}%
When only edge insertions and connectivity queries are allowed, connected components
can be maintained in $\O(\alpha(n))$ time amortized per operation.

\proof
Proven by Tarjan and van Leeuwen in \cite{tarjan:setunion}.
\qed

\FIXME{Define Ackermann's function. Use $\alpha(m,n)$?}

\rem
The cost of the operations on components 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.

\thm
Kruskal's algorithm finds the MST of a given graph in time $\O(m\log n)$
or $\O(m\timesalpha(n))$ if the edges are already sorted by their weights.

\proof
Follows from the above analysis.
\qed

%--------------------------------------------------------------------------------

\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.
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.

We will show a contractive version of the Bor\o{u}vka's algorithm
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
or we keep them as two parallel edges, leaving us with a~multigraph. We will
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 $\ell(e)$ that will be preserved by
contractions.

\lemman{Flattening a multigraph}\id{flattening}%
Let $G$ be a multigraph and $G'$ its subgraph such that all loops have been
removed and each bundle of parallel edges replaced by its lightest edge.
Then $G'$~has the same MST as~$G$.

\proof
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~$e'\in G'$, exchaning $e'$
for~$e$ makes~$T$ 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 would only have to be more careful in the
formulations and proofs, which we preferred to avoid.

\algn{Contractive version of Bor\o{u}vka's algorithm}\id{contbor}
\algo
\algin A~graph~$G$ with an edge comparison oracle.
\:$T\=\emptyset$.
\:$\ell(e)\=e$ 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.\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

\lemma\id{contiter}%
Each iteration of the algorithm can be carried out in time~$\O(m)$.

\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
connected components of this graph and renumber vertices in each component to
the identifier of the component. This takes $\O(m)$ 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
$\O(n+m)=\O(m)$.
\qed

\thm
The Contractive Bor\o{u}vka's algorithm finds the MST in time $\O(m\log n)$.

\proof
As in the original Bor\o{u}vka's algorithm, the number of iterations is $\O(\log n)$.
Then apply the previous lemma.
\qed

\thmn{\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 geometrically.

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 $n_i\le n/2^i$.

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. Hence 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 $\O(\sum_i m_i)=\O(\sum_i n/2^i)=\O(n)$.
\qed

\rem
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. We will show one more linear algorithm soon. The advantage
of our approach is that we do not need to construct the planar embedding explicitly.

\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 elements could end up taking most of the space in the arrays and the scans of these
arrays would have super-linear cost with respect to the size of the current graph~$G_i$.

\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).

\para
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
their counterparts in~$G/e$. Then: $$\mst(G) = \pi^{-1}[\mst(G/e)] + e.$$

\proof
% We seem not to need this lemma for multigraphs...
%If there are any loops or parallel edges in~$G$, we can flatten the graph. According to the
%Flattening lemma (\ref{flattening}), the MST stays the same and if we remove a parallel edge
%or loop~$f$, then $\pi(f)$ would be removed when flattening~$G/e$, so $f$ never participates
%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~$G$
(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 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
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.

\defn
A~\df{distractor of order~$k$,} denoted by~$D_k$, is a path on $n=2^k$~vertices $v_1,\ldots,v_n$
where each edge $v_iv_{i+1}$ has its weight equal to the number of trailing zeroes in the binary
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, $D_{k+1}$ 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)}

\lemma
A~single iteration of the contractive algorithm reduces~$D_k$ to a graph isomorphic with~$D_{k-1}$.

\proof
Each vertex~$v$ of~$D_k$ is incident with a single edge of weight~1. The algorithm therefore
selects all weight~1 edges and contracts them. This produces a graph which is
exactly $D_{k-1}$ with all weights increased by~1, which does not change the relative order of edges.
\qed

\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 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)}

\lemma
A~single iteration of the contractive algorithm reduces~$H_{a,k}$ to a graph isomorphic with $H_{a,k-1}$.

\proof
Each vertex is incident with an edge of some distractor, so the algorithm does not select
any edge of the complete graph. Contraction therefore reduces each distractor to a smaller
distractor (modulo an additive factor in weight) and leaves the complete graph intact.
This is monotonely isomorphic to $H_{a,k-1}$.
\qed

\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 $\Omega(n\log n)$ on it.

\proof
Consider the hedgehog $H_{a,k}$ for $a=\lceil\sqrt n\rceil$ and $k=\lceil\log_2 a\rceil$.
It has $a\cdot 2^k = \Theta(n)$ vertices and ${a \choose 2} + a\cdot 2^k = \Theta(a^2) + \Theta(a^2) = \Theta(n)$ edges
as we wanted.

By the previous lemma, the algorithm proceeds through a sequence of hedgehogs $H_{a,k},
H_{a,k-1}, \ldots, H_{a,0}$ (up to monotone isomorphism), so it needs a logarithmic number of iterations plus some more
to finish on the remaining complete graph. Each iteration runs on a graph with $\Omega(n)$
edges as every $H_{a,k}$ contains a complete graph on~$a$ vertices.
\qed

\endpart