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~$G$.
+ is the smallest possible among 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
+mostly geometric setting. He has discovered 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
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}
+and Pettie \cite{pettie:ackermann}, to an~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.
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.
+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 vertices $x$ and $y$, let $T[x,y]$ denote the (unique) path in~$T$ joining $x$ with~$y$.
\: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{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
\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~$e$
+that $w(e')>w(e)$. Now $T-e'$ ($T$~with the edge~$e'$ removed) 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.
\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
+the 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
+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}%
\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 $e'\in T'\setminus T$.
+but as they have the same number of edges, 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
+tree $T^*:=T-e+e'$ such that $d(T^*,T')=d(T,T')-2$. Now 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}}
+\>In some cases, a~much stronger statement is true:
+
\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.
+transforming $T$ to~$T'$ such that the weight of the tree does not decrease in any step.
\proof
We improve the argument from the previous proof, refining the induction step.
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
+Let us 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
+either identical 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 we have
$w(f)\ge w(e')$ due to the choice of $e'$ and all other edges on~$C$ are lighter
than~$e'$ as $e'$ was not $T$-light.
\qed
+This lemma immediately implies that Lemma \ref{lightlemma} works in both directions:
+
\thmn{Minimality of spanning trees}\id{mstthm}%
A~spanning tree~$T$ is minimum iff there is no $T$-light edge.
\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 $n^{n-2}$
+a complete graph on~$n$ vertices with 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.
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:
+Also 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 $\mst(G-e) = \mst(G)$.
+Let~$G$ be a~graph with distinct edge weights and $e \in G\setminus\mst(G)$.
+Then $\mst(G-e) = \mst(G)$.
\proof
The tree $T=\mst(G)$ is also a~MST of~$G-e$, because every $T$-light
we will always assume distinct weights.
\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~$T$.
+If all edge weights are distinct and $T$~is an~arbitrary spanning tree, then every edge of~$G$
+is 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
+Another useful consequence of the Minimality theorem 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~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 isomorphism~$\pi$ maps spanning trees to spanning trees bijectively 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
\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.
+\:At 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}
\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
+the procedure always stops and it 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.
\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
+blue in any step of the procedure always belong to~$T_{min}$ and that the 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.
+always stops. Some parts of the proof will turn out to be useful in the upcoming chapters,
+so we will state them in a~slightly more general way.
\lemman{Blue lemma, also known as the Cut rule}\id{bluelemma}%
The lightest edge of every cut is contained in the MST.
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
+Assume that $e=xy$ is a black edge. Let us define~$M$ as 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
+with some blue path between $x$ and $y$ forms a cycle and $e$~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.
+to be in $T_{min}$ and there can be no $T_{min}$-light edges.
+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 $V\setminus M$ 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}
+\nota\id{deltanota}%
+We will use $\delta(M)$ to denote the cut separating~$M$ from its complement.
+That is, $\delta(M) = E \cap (M \times (V\setminus M))$. We will also abbreviate
+$\delta(\{v\})$ as~$\delta(v)$.
+
\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.
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~$T_{min}$.
+lemma all other (red) edges are outside~$T_{min}$. Thus the blue edges are exactly~$T_{min}$.
\qed
\rem
\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.
+The three classical MST algorithms (Bor\o{u}vka's, Jarn\'\i{}k's, and Kruskal's) 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
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}
+\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.
\endalgo
\lemma\id{boruvkadrop}%
-In each iteration of the algorithm, the number of trees in~$T$ drops at least twice.
+In each iteration of the algorithm, the number of trees in~$T$ decreases by at least
+a~factor of two.
\proof
Each tree gets merged with at least one of its neighbors, so each of the new trees
The algorithm stops in $\O(\log n)$ iterations.
\lemma\id{borcorr}%
-Bor\o{u}vka's algorithm outputs the MST of the input graph.
+The 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,
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:
-$$C=T_1[u_1v_1]\,v_1u_2\,T_2[u_2v_2]\,v_2u_3\,T_3[u_3v_3]\, \ldots \,T_k[u_kv_k]\,v_ku_1.$$
+$$C=T_1[u_1,v_1]\,v_1u_2\,T_2[u_2,v_2]\,v_2u_3\,T_3[u_3,v_3]\, \ldots \,T_k[u_k,v_k]\,v_ku_1.$$
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,
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
+trees and edges correspond to the chosen lightest inter-tree edges. We find the connected
+components of this graph, and 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 $\O(m\log n)$.
+The Bor\o{u}vka's algorithm finds the MST in time $\O(m\log n)$.
\proof
Follows from the previous lemmata.
\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
+to the 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.
+by the lightest neighboring edge until the tree spans the whole graph.
\algn{Jarn\'\i{}k \cite{jarnik:ojistem}, Prim \cite{prim:mst}, Dijkstra \cite{dijkstra:mst}}\id{jarnik}%
\algo
\endalgo
\lemma
-Jarn\'\i{}k's algorithm computes the MST of the input graph.
+The 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
+If~$G$ is connected, the algorithm always stops. In every step of
+the algorithm, $T$ is always a blue tree. because 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.}
+The most important part of the algorithm is finding the \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
+neighboring edges to the heap and delete the neighboring edges that became
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 $\O(m\log n)$.
+The 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}.
\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}
+\:$T\=\hbox{an empty spanning subgraph}$.
\:For all edges $e$ in their sorted order:
\::If $T+e$ is acyclic, add~$e$ to~$T$.
\::Otherwise drop~$e$.
\endalgo
\lemma
-Kruskal's algorithm returns the MST of the input graph.
+The Kruskal's algorithm returns the MST of the input graph.
\proof
In every step, $T$ is a forest of blue trees. Adding~$e$ to~$T$
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)}
+\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,
+the same component. 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:
+Tarjan has shown that there is a~data structure for the DSU problem
+of surprising efficiency:
-\thmn{Disjoint Set Union, Tarjan and van Leeuwen \cite{tarjan:setunion}}\id{dfu}%
+\thmn{Disjoint Set Union, Tarjan \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
This completes the following theorem:
\thm\id{kruskal}%
-Kruskal's algorithm finds the MST of a given graph in time $\O(m\log n)$.
+The 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
+We spend $\O(m\log n)$ time 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
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.)
+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
+edges, so we let each edge 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.
+Let $G$ be a multigraph and $G'$ its subgraph obtaining by removing loops
+and replacing each bundle of parallel edges by its lightest edge.
Then $G'$~has the same MST as~$G$.
\proof
\:$\ell(e)\=e$ 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 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.
+\::$T\=T\cup \{ \ell(e_1),\ldots,\ell(e_n) \}$.\hfil\break\cmt{Remember labels of all selected edges.}
+\::Contract all edges $e_k$, inheriting labels and weights.\foot{In other words, we will ask the comparison oracle for the edge $\ell(e)$ instead of~$e$.}
+\::Flatten $G$ (remove 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.
+of vertices and edges of this graph by $n_i$ and $m_i$ respectively. A~single iteration
+of the algorithm will be called a~\df{Bor\o{u}vka step}.
\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~$\O(m_i)$.
+The $i$-th 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
+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.
in all iterations is $\O(\sum_i n_i^2) = \O(\sum_i n^2/4^i) = \O(n^2)$.
\qed
+On planar graphs, the algorithm runs much faster:
+
\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 $\O(n)$.
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
+contraction. Moreover, $G_i$~is also simple, so we can use the standard bound 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
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
+to construct the planar embedding explicitly. We will show another simpler linear-time algorithm
in section~\ref{minorclosed}.
\rem
%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 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~$e$,
+(by the Minimality 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.)
+a~multigraph version of the theorem, but the direction we need is a~straightforward edge exchange,
+which obviously works in multigraphs as well as in simple graphs.)
\qed
\rem
-In the previous algorithm, the role of the mapping~$\pi^{-1}$ is of course played by the edge labels~$\ell$.
+In the Contractive Bor\o{u}vka's 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
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 $v_1,\ldots,v_n$
+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.
+\figure{distractor.eps}{\epsfxsize}{A~distractor $D_3$ and its evolution (bold edges are contracted)}
+
\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~$D_{k-1}$.
+A~single iteration of the contractive algorithm reduces the distractor~$D_k$ 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.
+selects all weight~1 edges and contracts them. This produces a~graph that is
+equal to $D_{k-1}$ 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.
+together with edges of a complete graph on the bases of these distractors. The additional edges
+have arbitrary weights that are heavier than the edges of all the distractors.
\figure{hedgehog.eps}{\epsfxsize}{A~hedgehog $H_{5,2}$ (quills bent to fit in the picture)}
\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 $H_{a,k-1}$.
+distractor (modulo an additive factor in weight) and it leaves the complete graph intact.
+The resulting graph is monotonely isomorphic to $H_{a,k-1}$.
\qed
+When we set the parameters appropriately, we get the following lower bound:
+
\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.
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