rediscovered several times.
Recently, several significantly faster algorithms were discovered, most notably the
-$\O(m\beta(m,n))$-time algorithm by Fredman and Tarjan \cite{ft:fibonacci} and
+$\O(m\timesbeta(m,n))$-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}.
%--------------------------------------------------------------------------------
-\section{Basic Properties}
+\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
%--------------------------------------------------------------------------------
-\section{The Red-Blue Meta-Algorithm}
+\section{The Red-Blue meta-algorithm}
Most MST algorithms can be described as special cases of the following procedure
(again following \cite{tarjan:dsna}):
When an edge is colored red in any step of the procedure, it is not contained in the minimum spanning tree.
\proof
-Again by contradiction. Assume that $e$ is an edge painted red as the heaviest edge
+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
produce a cycle. Consider the first iteration of the algorithm where $T$ contains a~cycle~$C$. Without
loss of generality we can assume that $C=T_1[u_1v_1]\,v_1u_2\,T_2[u_2v_2]\,v_2u_3\,T_3[u_3v_3]\, \ldots \,T_k[u_kv_k]\,v_ku_1$.
Each component $T_i$ has chosen its lightest incident edge~$e_i$ as either the edge $v_iu_{i+1}$
-or $v_{i-1}u_i$ (indexing cyclically). Assume that $e_1=v_1u_2$ (otherwise we reverse the orientation
+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.)
%--------------------------------------------------------------------------------
-\section{Contractive algorithms}
+\section{Contractive algorithms}\id{contalg}%
While the classical algorithms are based on growing suitable trees, they
can be also reformulated in terms of edge contraction. Instead of keeping
\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.
+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.
-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).
+To achieve the linear time complexity, the algorithm needs a very careful implementation,
+but we defer the technical details to section~\ref{bucketsort}.
\para
Graph contractions are indeed a~very powerful tool and they can be used in other MST
edges as every $H_{a,k}$ contains a complete graph on~$a$ vertices.
\qed
-%--------------------------------------------------------------------------------
-
-\section{Minor-closed graph classes}
-
-The contracting algorithm given in the previous section has been found to perform
-well on planar graphs, but in the general case its time complexity was not linear.
-Can we find any broader class of graphs where the algorithm is still efficient?
-The right context turns out to be the minor-closed graph classes, which are
-closed under contractions and have bounded density.
-
-\defn
-A~graph~$H$ is a \df{minor} of a~graph~$G$ iff it can be obtained
-from a subgraph of~$G$ by a sequence of simple graph contractions (see \ref{simpcont}).
-
-\defn
-A~class~$\cal C$ of graphs is \df{minor-closed}, when for every $G\in\cal C$ and
-its every minor~$H$, the graph~$H$ lies in~$\cal C$ as well. A~class~$\cal C$ is called
-\df{non-trivial} if at least one graph lies in~$\cal C$ and at least one lies outside~$\cal C$.
-
-\example
-Non-trivial minor-closed classes include planar graphs and more generally graphs
-embeddable in any fixed surface. Many nice properties of planar graphs extend
-to these classes, too, most notably the linearity of the number of edges.
-
-\defn\id{density}%
-Let $\cal C$ be a class of graphs. We define its \df{edge density} $\varrho(\cal C)$
-to be the infimum of all~$\varrho$'s such that $m(G) \le \varrho\cdot n(G)$
-holds for every $G\in\cal C$.
-
-\thmn{Density of minor-closed classes}
-A~minor-closed class of graphs has finite edge density if and only if it is
-a non-trivial class.
-
-\proof
-See Theorem 6.1 in \cite{nesetril:minors}, which also lists some other equivalent conditions.
-\qed
-
-\thmn{MST on minor-closed classes \cite{mm:mst}}\id{mstmcc}%
-For any fixed non-trivial minor-closed class~$\cal C$ of graphs, Algorithm \ref{contbor} finds
-the MST of any graph in this class in time $\O(n)$. (The constant hidden in the~$\O$
-depends on the class.)
-
-\proof
-Following the proof for planar graphs (\ref{planarbor}), we denote the graph considered
-by the algorithm at the beginning of the $i$-th iteration by~$G_i$ and its number of vertices
-and edges by $n_i$ and $m_i$ respectively. Again the $i$-th phase runs in time $\O(m_i)$
-and $n_i \le n/2^i$, so it remains to show a linear bound for the $m_i$'s.
-
-Since each $G_i$ is produced from~$G_{i-1}$ by a sequence of edge contractions,
-all $G_i$'s are minors of~$G$.\foot{Technically, these are multigraph contractions,
-but followed by flattening, so they are equivalent to contractions on simple graphs.}
-So they also belong to~$\cal C$ and by the previous theorem $m_i\le \varrho({\cal C})\cdot n_i$.
-\qed
-
-\rem\id{nobatch}%
-The contractive algorithm uses ``batch processing'' to perform many contractions
-in a single step. It is also possible to perform contractions one edge at a~time,
-batching only the flattenings. A~contraction of an edge~$uv$ can be done
-in time~$\O(\deg(u))$ by removing all edges incident with~$u$ and inserting them back
-with $u$ replaced by~$v$. Therefore we need to find a lot of vertices with small
-degrees. The following lemma shows that this is always the case in minor-closed
-classes.
-
-\lemman{Low-degree vertices}\id{lowdeg}%
-Let $\cal C$ be a graph class with density~$\varrho$ and $G\in\cal C$ a~graph
-with $n$~vertices. Then at least $n/2$ vertices of~$G$ have degree at most~$4\varrho$.
-
-\proof
-Assume the contrary: Let there be at least $n/2$ vertices with degree
-greater than~$4\varrho$. Then $\sum_v \deg(v) > n/2
-\cdot 4\varrho = 2\varrho n$, which is in contradiction with the number
-of edges being at most $\varrho n$.
-\qed
-
-\rem
-The proof can be also viewed
-probabilistically: let $X$ be the degree of a vertex of~$G$ chosen uniformly at
-random. Then ${\bb E}X \le 2\varrho$, hence by the Markov's inequality
-${\rm Pr}[X > 4\varrho] < 1/2$, so for at least $n/2$ vertices~$v$ we have
-$\deg(v)\le 4\varrho$.
-
-\algn{Local Bor\o{u}vka's Algorithm \cite{mm:mst}}%
-\algo
-\algin A~graph~$G$ with an edge comparison oracle and a~parameter~$t\in{\bb N}$.
-\:$T\=\emptyset$.
-\:$\ell(e)\=e$ for all edges~$e$.
-\:While $n(G)>1$:
-\::While there exists a~vertex~$v$ such that $\deg(v)\le t$:
-\:::Select the lightest edge~$e$ incident with~$v$.
-\:::Contract~$G$ along~$e$.
-\:::$T\=T + \ell(e)$.
-\::Flatten $G$, removing parallel edges and loops.
-\algout Minimum spanning tree~$T$.
-\endalgo
-
-\thm
-When $\cal C$ is a minor-closed class of graphs with density~$\varrho$, the
-Local Bor\o{u}vka's Algorithm with the parameter~$t$ set to~$4\varrho$
-finds the MST of any graph from this class in time $\O(n)$. (The constant
-in the~$\O$ depends on~the class.)
-
-\proof
-Let us denote by $G_i$, $n_i$ and $m_i$ the graph considered by the
-algorithm at the beginning of the $i$-th iteration of the outer loop,
-and the number of its vertices and edges respectively. As in the proof
-of the previous algorithm (\ref{mstmcc}), we observe that all the $G_i$'s
-are minors of the graph~$G$ given as the input.
-
-For the choice $t=4\varrho$, the Lemma on low-degree vertices (\ref{lowdeg})
-guarantees that at least $n_i/2$ edges get selected in the $i$-th iteration.
-Hence at least a half of the vertices participates in contractions, so
-$n_i\le 3/4\cdot n_{i-1}$. Therefore $n_i\le n\cdot (3/4)^i$ and the algorithm terminates
-after $\O(\log n)$ iterations.
-
-Each selected edge belongs to $\mst(G)$, because it is the lightest edge of
-the trivial cut $\delta(v)$ (see the Blue Rule in \ref{rbma}).
-The steps 6 and~7 therefore correspond to the operation
-described by the Lemma on contraction of MST edges (\ref{contlemma}) and when
-the algorithm stops, $T$~is indeed the minimum spanning tree.
-
-It remains to analyse the time complexity of the algorithm. Since $G_i\in{\cal C}$, we have
-$m_i\le \varrho n_i \le \varrho n/2^i$.
-We will show that the $i$-th iteration is carried out in time $\O(m_i)$.
-Steps 5 and~6 run in time $\O(\deg(v))=\O(t)$ for each~$v$, so summed
-over all $v$'s they take $\O(tn_i)$, which is linear for a fixed class~$\cal C$.
-Flattening takes $\O(m_i)$, as already noted in the analysis of the Contracting
-Bor\o{u}vka's Algorithm (see \ref{contiter}).
-
-The whole algorithm therefore runs in time $\O(\sum_i m_i) = \O(\sum_i n/2^i) = \O(n)$.
-\qed
-
-\rem
-For planar graphs, we can get a sharper version of the low-degree lemma,
-showing that the algorithm works with $t=8$ as well (we had $t=12$ as
-$\varrho=3$). While this does not change the asymptotic time complexity
-of the algorithm, the constant-factor speedup can still delight the hearts of
-its practical users.
-
-\lemman{Low-degree vertices in planar graphs}%
-Let $G$ be a planar graph with $n$~vertices. Then at least $n/2$ vertices of~$v$
-have degree at most~8.
-
-\proof
-It suffices to show that the lemma holds for triangulations (if there
-are any edges missing, the situation can only get better) with at
-least 3 vertices. Since $G$ is planar, $\sum_v \deg(v) < 6n$.
-The numbers $d(v):=\deg(v)-3$ are non-negative and $\sum_v d(v) < 3n$,
-so by the same argument as in the proof of the general lemma, for at least $n/2$
-vertices~$v$ it holds that $d(v) < 6$, hence $\deg(v) \le 8$.
-\qed
-
-\rem\id{hexa}%
-The constant~8 in the previous lemma is the best we can have.
-Consider a $k\times k$ triangular grid. It has $n=k^2$ vertices, $\O(k)$ of them
-lie on the outer face and have degrees at most~6, the remaining $n-\O(k)$ interior
-vertices have degree exactly~6. Therefore the number of faces~$f$ is $6/3\cdot n=2n$,
-ignoring terms of order $\O(k)$. All interior triangles can be properly colored with
-two colors, black and white. Now add a~new vertex inside each white face and connect
-it to all three vertices on the boundary of that face. This adds $f/2 \approx n$
-vertices of degree~3 and it increases the degrees of the original $\approx n$ interior
-vertices to~9, therefore about a half of the vertices of the new planar graph
-has degree~9.
-
-\figure{hexangle.eps}{\epsfxsize}{The construction from Remark~\ref{hexa}}
-
-%--------------------------------------------------------------------------------
-
-\section{Using Fibonacci heaps}
-\id{fibonacci}
-
-We have seen that the Jarn\'\i{}k's Algorithm \ref{jarnik} runs in $\O(m\log n)$ time
-(and this bound can be easily shown to be tight). Fredman and Tarjan have shown a~faster
-implementation in~\cite{ft:fibonacci} using their Fibonacci heaps. In this section,
-we convey their results and we show several interesting consequences.
-
-The previous implementation of the algorithm used a binary heap to store all neighboring
-edges of the cut~$\delta(T)$. Instead of that, we will remember the vertices adjacent
-to~$T$ and for each such vertex~$v$ we will keep the lightest edge~$uv$ such that $u$~lies
-in~$T$. We will call these edges \df{active edges} and keep them in a~heap, ordered by weight.
-
-When we want to extend~$T$ by the lightest edge of~$\delta(T)$, it is sufficient to
-find the lightest active edge~$uv$ and add this edge to~$T$ together with a new vertex~$v$.
-Then we have to update the active edges as follows. The edge~$uv$ has just ceased to
-be active. We scan all neighbors~$w$ of the vertex~$v$. When $w$~is in~$T$, no action
-is needed. If $w$~is outside~$T$ and it was not adjacent to~$T$ (there is no active edge
-remembered for it so far), we set the edge~$vw$ as active. Otherwise we check the existing
-active edge for~$w$ and replace it by~$vw$ if the new edge is lighter.
-
-The following algorithm shows how these operations translate to insertions, decreases
-and deletions on the heap.
-
-\algn{Jarn\'\i{}k with active edges; Fredman and Tarjan \cite{ft:fibonacci}}\id{jarniktwo}%
-\algo
-\algin A~graph~$G$ with an edge comparison oracle.
-\:$v_0\=$ an~arbitrary vertex of~$G$.
-\:$T\=$ a tree containing just the vertex~$v_0$.
-\:$H\=$ a~heap of active edges stored as pairs $(u,v)$ where $u\in T,v\not\in T$, ordered by the weights $w(vw)$, initially empty.
-\:$A\=$ an~auxiliary array mapping vertices outside~$T$ to their active edges in the heap; initially all elements undefined.
-\:\<Insert> all edges incident with~$v_0$ to~$H$ and update~$A$ accordingly.
-\:While $H$ is not empty:
-\::$(u,v)\=\<DeleteMin>(H)$.
-\::$T\=T+uv$.
-\::For all edges $vw$ such that $w\not\in T$:
-\:::If there exists an~active edge~$A(w)$:
-\::::If $vw$ is lighter than~$A(w)$, \<Decrease> $A(w)$ to~$(v,w)$ in~$H$.
-\:::If there is no such edge, then \<Insert> $(v,w)$ to~$H$ and set~$A(w)$.
-\algout Minimum spanning tree~$T$.
-\endalgo
-
-\thmn{Fibonacci heaps} The~Fibonacci heap performs the following operations
-with the indicated amortized time complexities:
-\itemize\ibull
-\:\<Insert> (insertion of a~new element) in $\O(1)$,
-\:\<Decrease> (decreasing value of an~existing element) in $\O(1)$,
-\:\<Merge> (merging of two heaps into one) in $\O(1)$,
-\:\<DeleteMin> (deletion of the minimal element) in $\O(\log n)$,
-\:\<Delete> (deletion of an~arbitrary element) in $\O(\log n)$,
-\endlist
-\>where $n$ is the maximum number of elements present in the heap at the time of
-the operation.
-
-\proof
-See Fredman and Tarjan \cite{ft:fibonacci} for both the description of the Fibonacci
-heap and the proof of this theorem.
-\qed
-
-\thm
-Algorithm~\ref{jarniktwo} with a~Fibonacci heap finds the MST of the input graph in time~$\O(m+n\log n)$.
-
-\proof
-The algorithm always stops, because every edge enters the heap~$H$ at most once.
-As it selects exactly the same edges as the original Jarn\'\i{}k's algorithm,
-it gives the correct answer.
-
-The time complexity is $\O(m)$ plus the cost of the heap operations. The algorithm
-performs at most one \<Insert> or \<Decrease> per edge and exactly one \<DeleteMin>
-per vertex and there are at most $n$ elements in the heap at any given time,
-so by the previous theorem the operations take $\O(m+n\log n)$ time in total.
-\qed
-
-\cor
-For graphs with edge density at least $\log n$, this algorithm runs in linear time.
-
-\rem
-We can consider using other kinds of heaps which have the property that inserts
-and decreases are faster than deletes. Of course, the Fibonacci heaps are asymptotically
-optimal (by the standard $\Omega(n\log n)$ lower bound on sorting by comparisons, see
-for example \cite{clrs}), so the other data structures can improve only
-multiplicative constants or offer an~easier implementation.
-
-A~nice example is a~\df{$d$-regular heap} --- a~variant of the usual binary heap
-in the form of a~complete $d$-regular tree. \<Insert>, \<Decrease> and other operations
-involving bubbling the values up spend $\O(1)$ time at a~single level, so they run
-in~$\O(\log_d n)$ time. \<Delete> and \<DeleteMin> require bubbling down, which incurs
-comparison with all~$d$ sons at every level, so they run in~$\O(d\log_d n)$.
-With this structure, the time complexity of the whole algorithm
-is $\O(nd\log_d n + m\log_d n)$, which suggests setting $d=m/n$, giving $\O(m\log_{m/n}n)$.
-This is still linear for graphs with density at~least~$n^{1+\varepsilon}$.
-
-Another possibility is to use the 2-3-heaps \cite{takaoka:twothree} or Trinomial
-heaps \cite{takaoka:trinomial}. Both have the same asymptotic complexity as Fibonacci
-heaps (the latter even in worst case, but it does not matter here) and their
-authors claim implementation advantages.
-
-\FIXME{Mention Thorup's Fibonacci-like heaps for integers?}
-
-\para
-As we already noted, the improved Jarn\'\i{}k's algorithm runs in linear time
-for sufficiently dense graphs. In some cases, it is useful to combine it with
-another MST algorithm, which identifies a~part of the MST edges and contracts
-the graph to increase its density. For example, we can perform several
-iterations of the Contractive Bor\o{u}vka's algorithm and find the rest of the
-MST by the above version of Jarn\'\i{}k's algorithm.
-
-\algn{Mixed Bor\o{u}vka-Jarn\'\i{}k}
-\algo
-\algin A~graph~$G$ with an edge comparison oracle.
-\:Run $\log\log n$ iterations of the Contractive Bor\o{u}vka's algorithm (\ref{contbor}),
- getting a~MST~$T_1$.
-\:Run the Jarn\'\i{}k's algorithm with active edges (\ref{jarniktwo}) on the resulting
- graph, getting a~MST~$T_2$.
-\:Combine $T_1$ and~$T_2$ to~$T$ as in the Contraction lemma (\ref{contlemma}).
-\algout Minimum spanning tree~$T$.
-\endalgo
-
-\thm
-The Mixed Bor\o{u}vka-Jarn\'\i{}k algorithm finds the MST of the input graph in time $\O(m\log\log n)$.
-
-\proof
-Correctness follows from the Contraction lemma and from the proofs of correctness of the respective algorithms.
-As~for time complexity: The first step takes $\O(m\log\log n)$ time
-(by Lemma~\ref{contiter}) and it gradually contracts~$G$ to a~graph~$G'$ of size
-$m'\le m$ and $n'\le n/\log n$. The second step then runs in time $\O(m'+n'\log n') = \O(m)$
-and both trees can be combined in linear time, too.
-\qed
-
-\para
-Actually, there is a~much better choice of the algorithms to combine: use the
-improved Jarn\'\i{}k's algorithm multiple times, each time stopping after a~while.
-The good choice of the stopping condition is to place a~limit on the size of the heap.
-Start with an~arbitrary vertex, grow the tree as usually and once the heap gets too large,
-conserve the current tree and start with a~different vertex and an~empty heap. When this
-process runs out of vertices, it has identified a~sub-forest of the MST, so we can
-contract the graph along the edges of~this forest and iterate.
-
-\algn{Iterated Jarn\'\i{}k; Fredman and Tarjan \cite{ft:fibonacci}}
-\algo
-\algin A~graph~$G$ with an edge comparison oracle.
-\:$T\=\emptyset$. \cmt{edges of the MST}
-\:$\ell(e)\=e$ for all edges~$e$. \cmt{edge labels as usually}
-\:$m_0\=m$.
-\:While $n>1$: \cmt{We will call iterations of this loop \df{phases}.}
-\::$F\=\emptyset$. \cmt{forest built in the current phase}
-\::$t\=2^{2m_0/n}$. \cmt{the limit on heap size}
-\::While there is a~vertex $v_0\not\in F$:
-\:::Run the improved Jarn\'\i{}k's algorithm (\ref{jarniktwo}) from~$v_0$, stop when:
-\::::all vertices have been processed, or
-\::::a~vertex of~$F$ has been added to the tree, or
-\::::the heap had more than~$t$ elements.
-\:::Denote the resulting tree~$R$.
-\:::$F\=F\cup R$.
-\::$T\=T\cup \ell[F]$. \cmt{Remember MST edges found in this phase.}
-\::Contract~$G$ along all edges of~$F$ and flatten it.
-\algout Minimum spanning tree~$T$.
-\endalgo
-
-\nota
-For analysis of the algorithm, let us denote the graph entering the $i$-th
-phase by~$G_i$ and likewise with the other parameters. The trees from which
-$F_i$~has been constructed will be called $R_i^1, \ldots, R_i^{z_i}$. The
-non-indexed $G$, $m$ and~$n$ will correspond to the graph given as~input.
-
-\para
-However the choice of the parameter~$t$ can seem mysterious, the following
-lemma makes the reason clear:
-
-\lemma\id{ijphase}%
-The $i$-th phase of the Iterated Jarn\'\i{}k's algorithm runs in time~$\O(m)$.
-
-\proof
-During the phase, the heap always contains at most~$t_i$ elements, so it takes
-time~$\O(\log t_i)=\O(m/n_i)$ to delete an~element from the heap. The trees~$R_i^j$
-are disjoint, so there are at most~$n_i$ \<DeleteMin>'s over the course of the phase.
-Each edge is considered at most twice (once per its endpoint), so the number
-of the other heap operations is~$\O(m_i)$. Together, it equals $\O(m_i + n_i\log t_i) = \O(m_i+m) = \O(m)$.
-\qed
-
-\lemma
-Unless the $i$-th phase is final, the forest~$F_i$ consists of at most $2m_i/t_i$ trees.
-
-\proof
-As every edge of~$G_i$ is incident with at most two trees of~$F_i$, it is sufficient
-to establish that there are at least~$t_i$ edges incident with the vertices of every
-such tree~(*).
-
-The forest~$F_i$ evolves by additions of the trees~$R_i^j$. Let us consider the possibilities
-how the algorithm could have stopped growing the tree~$R_i^j$:
-\itemize\ibull
-\:the heap had more than~$t_i$ elements (step~10): since the elements stored in the heap correspond
- to some of the edges incident with vertices of~$R_i^j$, the condition~(*) is fulfilled;
-\:the algorithm just added a~vertex of~$F_i$ to~$R_i^j$ (step~9): in this case, an~existing
- tree of~$F_i$ is extended, so the number of edges incident with it cannot decrease;\foot{%
- To make this true, we counted the edges incident with the \em{vertices} of the tree
- instead of edges incident with the tree itself, because we needed the tree edges
- to be counted as well.}
-\:all vertices have been processed (step~8): this can happen only in the final phase.
-\qeditem
-\endlist
-
-\thm\id{itjarthm}%
-The Iterated Jarn\'\i{}k's algorithm finds the MST of the input graph in time
-$\O(m\mathop\beta(m,n))$, where $\beta(m,n):=\min\{ i: \log^{(i)}n < m/n \}$.
-
-\proof
-Phases are finite and in every phase at least one edge is contracted, so the outer
-loop is eventually terminated. The resulting subgraph~$T$ is equal to $\mst(G)$, because each $F_i$ is
-a~subgraph of~$\mst(G_i)$ and the $F_i$'s are glued together according to the Contraction
-lemma (\ref{contlemma}).
-
-Let us bound the sizes of the graphs processed in individual phases. As the vertices
-of~$G_{i+1}$ correspond to the components of~$F_i$, by the previous lemma $n_{i+1}\le
-2m_i/t_i$. Then $t_{i+1} = 2^{2m/n_{i+1}} \ge 2^{2m/(2m_i/t_i)} = 2^{(m/m_i)\cdot t_i} \ge 2^{t_i}$,
-therefore:
-$$
-\left. \vcenter{\hbox{$\displaystyle t_i \ge 2^{2^{\scriptstyle 2^{\scriptstyle\vdots^{\scriptstyle m/n}}}} $}}\;\right\}
-\,\hbox{a~tower of~$i$ exponentials.}
-$$
-As soon as~$t_i\ge n$, the $i$-th phase must be final, because at that time
-there is enough space in the heap to process the whole graph. So~there are
-at most~$\beta(m,n)$ phases and we already know (Lemma~\ref{ijphase}) that each
-phase runs in linear time.
-\qed
-
-\cor
-The Iterated Jarn\'\i{}k's algorithm runs in time $\O(m\log^* n)$.
-
-\proof
-$\beta(m,n) \le \beta(1,n) = \log^* n$.
-\qed
-
-\cor
-When we use the Iterated Jarn\'\i{}k's algorithm on graphs with edge density
-at least~$\log^{(k)} n$ for some $k\in{\bb N}^+$, it runs in time~$\O(km)$.
-
-\proof
-If $m/n \ge \log^{(k)} n$, then $\beta(m,n)\le k$.
-\qed
-
-\FIXME{Reference to Q-Heaps.}
-
-%--------------------------------------------------------------------------------
-
-\section{Verification of minimality}
-
-\FIXME{\cite{pettie:onlineverify} online lower bound}
-
-% use \para
-% G has to be connected, so m=O(n)
-% mention Steiner trees
-% mention matroids
-% sorted weights
-% \O(...) as a set?
-% impedance mismatch in terminology: contraction of G along e vs. contraction of e.
-% use \delta(X) notation
-% mention disconnected graphs
-% unify use of n(G) vs. n
-% Euclidean MST
-
\endpart
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