Spelling checker strikes again.

[saga.git] / mst.tex

\ifx\endpart\undefined
\input macros.tex
\fi

\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 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.
\: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.

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.

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.

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

\section{Basic properties}\id{mstbasics}%

In this section, we will examine the basic properties of spanning trees and prove
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~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.

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 covers a~lighter edge.
\endlist

\rem
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~$T$
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 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 $e_i^\prime\in T'$.

\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 decrease 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 of spanning trees}\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{Uniqueness of MST}%
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

\nota\id{mstnota}%
When $G$ is a graph with distinct edge weights, we will use $\mst(G)$ to denote
its unique minimum spanning tree.

\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 the relative order of weights, the isomorphism applies to their MST's as well:

\defn
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 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 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}

Most MST algorithms can be described as special cases of the following procedure
(again following Tarjan \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 \rack{blue.}{red.\hfil} \cmt{Red rule}
\algout Minimum spanning tree of~$G$ consisting of edges colored blue.
\endalgo

\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
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.

\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 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

\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
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}

\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 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

\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 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.
\:$T\=$ 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\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,
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 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 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

\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 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 computes 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\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 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 computes the MST of a~given graph in time $\O(m\log n)$.

\rem
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}}
\algo
\algin A~graph~$G$ with an edge comparison oracle.
\: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~$T$.
\::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 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.
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{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
See \cite{tarjan:setunion}.
\qed

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 \<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 relabel the smaller of the two components.

We will study dynamic maintenance of connected components in more detail in Chapter~\ref{dynchap}.

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

\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 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 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~$e'\in G'$, 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
\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 $v_k$ 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 $G$ along 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}%
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 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 $\O(m_i)$ 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 $n_i$~buckets, so it takes
$\O(n_i+m_i)=\O(m_i)$.
\qed

\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 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{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 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 $m_i\le 3n_i \le 3n/2^i$.
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 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
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
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~$\ell$.

\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 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 $v_1,\ldots,v_n$
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

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

\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.

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.

\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