List of publications.

[saga.git] / adv.tex

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\chapter{Advanced MST Algorithms}

\section{Minor-closed graph classes}\id{minorclosed}%

The contractive algorithm given in Section~\ref{contalg} 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 this 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\id{minordef}%
A~graph~$H$ is a \df{minor} of a~graph~$G$ (written as $H\minorof G$) 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
every minor~$H$ of~$G$, 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:
\itemize\ibull
\:planar graphs,
\:graphs embeddable in any fixed surface (i.e., graphs of bounded genus),
\:graphs embeddable in~${\bb R}^3$ without knots or without interlocking cycles,
\:graphs of bounded tree-width or path-width.
\endlist

\para
Many of the nice structural properties of planar graphs extend to
minor-closed classes, too (see Lov\'asz \cite{lovasz:minors} for a~nice survey
of this theory and Diestel \cite{diestel:gt} for some of the deeper results).
The most important property is probably the characterization
of such classes in terms of their forbidden minors.

\defn
For a~class~$\cal H$ of graphs we define $\Forb({\cal H})$ as the class
of graphs that do not contain any of the graphs in~$\cal H$ as a~minor.
We will call $\cal H$ the set of \df{forbidden (or excluded) minors} for this class.
We will often abbreviate $\Forb(\{M_1,\ldots,M_n\})$ to $\Forb(M_1,\ldots,M_n)$.

\obs
For every~${\cal H}\ne\emptyset$, the class $\Forb({\cal H})$ is non-trivial
and closed on minors. This works in the opposite direction as well: for every
minor-closed class~$\cal C$ there is a~class $\cal H$ such that ${\cal C}=\Forb({\cal H})$.
One such~$\cal H$ is the complement of~$\cal C$, but smaller ones can be found, too.
For example, the planar graphs can be equivalently described as the class $\Forb(K_5, K_{3,3})$
--- this follows from the Kuratowski's theorem (the theorem speaks of forbidden
subdivisions, but while in general this is not the same as forbidden minors, it
is for $K_5$ and $K_{3,3}$). The celebrated theorem by Robertson and Seymour
guarantees that we can always find a~finite set of forbidden minors:

\thmn{Excluded minors, Robertson \& Seymour \cite{rs:wagner}}
For every non-trivial minor-closed graph class~$\cal C$ there exists
a~finite set~$\cal H$ of graphs such that ${\cal C}=\Forb({\cal H})$.

\proof
This theorem has been proven in a~long series of papers on graph minors
culminating with~\cite{rs:wagner}. See this paper and follow the references
to the previous articles in the series.
\qed

\para
For analysis of the contractive algorithm,
we will make use of another important property --- the bounded density of
minor-closed classes. The connection between minors and density dates back to
Mader in the 1960's and it can be proven without use of the Robertson-Seymour
theory.

\defn\id{density}%
Let $G$ be a~graph and $\cal C$ be a class of graphs. We define the \df{edge density}
$\varrho(G)$ of~$G$ as the average number of edges per vertex, i.e., $m(G)/n(G)$. The
edge density $\varrho(\cal C)$ of the class is then defined as the infimum of $\varrho(G)$ over all $G\in\cal C$.

\thmn{Mader \cite{mader:dens}}\id{maderthm}%
For every $k\in{\bb N}$ there exists $h(k)\in{\bb R}$ such that every graph
of average degree at least~$h(k)$ contains a~subdivision of~$K_{k}$ as a~subgraph.

\proofsketch
(See Lemma 3.5.1 in \cite{diestel:gt} for a~complete proof in English.)

Let us fix~$k$ and prove by induction on~$m$ that every graph of average
degree at least~$2^m$ contains a~subdivision of some graph with $k$~vertices
and $m$~edges (for $k\le m\le {k\choose 2}$). When we reach $m={k\choose 2}$, the theorem follows
as the only graph with~$k$ vertices and~$k\choose 2$ edges is~$K_k$.

The base case $m=k$: Let us observe that when the average degree
is~$a$, removing any vertex of degree less than~$a/2$ does not decrease the
average degree. A~graph with $a\ge 2^k$ therefore has a~subgraph
with minimum degree $\delta\ge a/2=2^{k-1}$. Such subgraph contains
a~cycle on more than~$\delta$ vertices, in other words a~subdivision of
the cycle~$C_k$.

Induction step: Let~$G$ be a~graph with average degree at least~$2^m$ and
assume that the theorem already holds for $m-1$. Without loss of generality,
$G$~is connected. Consider a~maximal set $U\subseteq V$ such that the subgraph $G[U]$
induced by~$U$ is connected and the graph $G\sgc U$ ($G$~with $U$~contracted to
a~single vertex) has average degree at least~$2^m$ (such~$U$ exists, because
$G=G\sgc U$ whenever $\vert U\vert=1$). Now consider the subgraph~$H$ induced
in~$G$ by the neighbors of~$U$. Every $v\in V(H)$ must have $\deg_H(v) \ge 2^{m-1}$,
as otherwise we can add this vertex to~$U$, contradicting its
maximality. By the induction hypothesis, $H$ contains a~subdivision of some
graph~$R$ with $k$~vertices and $m-1$ edges. Any two non-adjacent vertices
of~$R$ can be connected in the subdivision by a~path lying entirely in~$G[U]$,
which reveals a~subdivision of a~graph with $m$~edges. \qed

\thmn{Density of minor-closed classes, Mader~\cite{mader:dens}}
Every non-trivial minor-closed class of graphs has finite edge density.

\proof
Let~$\cal C$ be any such class, $X$~its excluded minor with the smallest number
of vertices~$x$.
As $X\minorof K_x$, the class $\cal C$ is entirely contained in ${\cal C}'=\Forb(K_x)$, so
$\varrho({\cal C}) \le \varrho({\cal C}')$ and therefore it suffices to prove the
theorem for classes excluding a~single complete graph~$K_x$.

We will show that $\varrho({\cal C})\le 2h(x)$, where $h$~is the function
from the previous theorem. If any $G\in{\cal C}$ had more than $2h(x)\cdot n(G)$
edges, its average degree would be at least~$h(x)$, so by the previous theorem
$G$~would contain a~subdivision of~$K_x$ and hence $K_x$ as a~minor.
\qed

Let us return to the analysis of our algorithm.

\thmn{MST on minor-closed classes, Mare\v{s} \cite{mm:mst}}\id{mstmcc}%
For any fixed non-trivial minor-closed class~$\cal C$ of graphs, the Contractive Bor\o{u}vka's
algorithm (\ref{contbor}) finds the MST of any graph of 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 Bor\o{u}vka step 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 we have $n_i \le n/2^i$, 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 the input graph.\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 Density theorem $m_i\le \varrho({\cal C})\cdot n_i$.
The time complexity is therefore $\sum_i \O(m_i) = \sum_i \O(n_i) = \O(\sum_i n/2^i) = \O(n)$.
\qed

\paran{Local contractions}\id{nobatch}%
The contractive algorithm uses ``batch processing'' to perform many contractions
in a single step. It is also possible to perform them 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 $\E X \le 2\varrho$, 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, Mare\v{s} \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~$e$.
\:::$T\=T + \ell(e)$.
\::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 the beginning of the $i$-th iteration, at least $n_i/2$ vertices
have degree at most~$t$. Each selected edge removes one such vertex and
possibly increases the degree of another one, so at least $n_i/4$ edges get selected.
Hence $n_i\le 3/4\cdot n_{i-1}$ and $n_i\le n\cdot (3/4)^i$, so 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, Lemma \ref{rbma}).
The steps 6 and~7 therefore correspond to the operation
described by the Contraction Lemma (\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 know that
$m_i\le \varrho n_i \le \varrho n/2^i$.
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 $\O(n_i)$ 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

\paran{Back to planar graphs}%
For planar graphs, we can obtain a sharper version of the low-degree lemma
showing that the algorithm works with $t=8$ as well (we had $t=12$ from
$\varrho=3$). 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 4 vertices. Since $G$ is planar, we have $\sum_v \deg(v) < 6n$.
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 $d(v) < 6$, and thus $\deg(v) \le 8$.
\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 they have degree 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 (see the picture). 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}}

\rem
The observation in~Theorem~\ref{mstmcc} was also independently made by Gustedt \cite{gustedt:parallel},
who studied a~parallel version of the Contractive Bor\o{u}vka's algorithm applied
to minor-closed classes.

\rem
The bound on the average degree needed to enforce a~$K_k$ minor, which we get from Theorem \ref{maderthm},
is very coarse. Kostochka \cite{kostochka:lbh} and independently Thomason \cite{thomason:efc}
have proven that an~average degree $\Omega(k\sqrt{\log k})$ is sufficient and that this
is the best what we can get.

\rem
Minor-closed classes share many other interesting properties, for example bounded chromatic
numbers of various kinds, as shown by Theorem 6.1 of \cite{nesetril:minors}. We can expect
that many algorithmic problems will turn out to be easy for them.

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

\section{Iterated algorithms}\id{iteralg}%

We have seen that the Jarn\'\i{}k's Algorithm \ref{jarnik} runs in $\Theta(m\log n)$ time.
Fredman and Tarjan \cite{ft:fibonacci} have shown a~faster implementation
using their Fibonacci heaps. In this section, we will convey their results and we
will show several interesting consequences.

The previous implementation of the algorithm used a binary heap to store all edges
separating the current tree~$T$ from the rest of the graph, i.e., 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 maintain the lightest edge~$uv$ such that $u$~lies in~$T$. We will call these edges \df{active edges}
and keep them in a~Fibonacci 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 the 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 already 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 in the heap.

\algn{Active Edge Jarn\'\i{}k; 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~Fibonacci heap of active edges stored as pairs $(u,v)$ where $u\in T,v\not\in T$, ordered by the weights $w(uv)$, and initially empty.
\:$A\=$ a~mapping of 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

\paran{Analysis}%
To analyse the time complexity of this algorithm, we will use the standard
theorem on~complexity of the Fibonacci heap:

\thmn{Fibonacci heaps, Fredman and Tarjan \cite{ft:fibonacci}} 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 the 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 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 the 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. There are at most $n$ elements in the heap at any given time,
thus by the previous theorem the operations take $\O(m+n\log n)$ time in total.
\qed

\cor
For graphs with edge density $\Omega(\log n)$, this algorithm runs in linear time.

\remn{Other heaps}%
We can consider using other kinds of heaps that 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 the \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 spend $\O(d\log_d n)$.
With this structure, the time complexity of the whole algorithm
is $\O(nd\log_d n + m\log_d n)$, which suggests setting $d=m/n$, yielding $\O(m\log_{m/n}n)$.
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 the worst case, but it does not matter here) and their
authors claim faster implementation. For integer weights, we can use Thorup's priority
queues described in \cite{thorup:sssp} which have constant-time \<Insert> and \<Decrease>
and $\O(\log\log n)$ time \<DeleteMin>. (We will however omit the details since we will
show a~faster integer algorithm soon.)

\paran{Combining MST algorithms}%
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
them to increase the density of the graph. For example, we can perform several Bor\o{u}vka
steps  and then find the rest of the MST by the Active Edge 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 $\log\log n$ Bor\o{u}vka steps (\ref{contbor}), getting a~MST~$T_1$.
\:Run the Active Edge Jarn\'\i{}k's algorithm (\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

\paran{Iterating Jarn\'\i{}k's algorithm}%
Actually, there is a~much better choice of the algorithms to combine: use the
Active Edge Jarn\'\i{}k's algorithm multiple times, each time stopping it after a~while.
A~good choice of the stopping condition is to place a~limit on the size of the heap.
We start with an~arbitrary vertex, grow the tree as usually and once the heap gets too large,
we 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 edges of~this forest and iterate.

\algn{Iterated Jarn\'\i{}k; Fredman and Tarjan \cite{ft:fibonacci}}\id{itjar}%
\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}
\::$t\=2^{\lceil 2m_0/n \rceil}$. \cmt{the limit on heap size}
\::While there is a~vertex $v_0\not\in F$:
\:::Run the Active Edge 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 has grown to 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 all edges of~$F$ and flatten~$G$.
\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. Let the trees from which
$F_i$~has been constructed 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}%
Each phase of the Iterated Jarn\'\i{}k's algorithm runs in time~$\O(m)$.

\proof
During the $i$-th 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 edge-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\id{ijsize}%
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 every such tree, including
edges connecting two vertices of the same 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 each elements stored in the heap
  corresponds to a~unique edge incident with~$R_i^j$, we have enough such edges;
\: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{%
  This is the place where we needed to count the interior edges 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
$\O(m\timesbeta(m,n))$, where $\beta(m,n):=\min\{ i \mid \log^{(i)}n \le m/n \}$.

\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 the 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 $t_{i+1} = 2^{\lceil 2m/n_{i+1} \rceil} \ge 2^{2m/n_{i+1}} \ge 2^{2m/(2m_i/t_i)} = 2^{(m/m_i)\cdot t_i} \ge 2^{t_i}$,
therefore:
$$
\left. \vcenter{\hbox{$\displaystyle t_i \ge 2^{2^{\scriptstyle 2^{\scriptstyle\rddots^{\scriptstyle m/n}}}} $}}\;\right\}
\,\hbox{a~tower of~$i$ exponentials.}
$$
As soon as~$t_i\ge n$, the $i$-th phase is final, because at that time
there is enough space in the heap to process the whole graph without stopping. So~there are
at most~$\beta(m,n)$ phases and we already know that each phase runs in linear
time (Lemma~\ref{ijphase}).
\qed

\cor
The Iterated Jarn\'\i{}k's algorithm runs in time $\O(m\log^* n)$.

\proof
$\beta(m,n) \le \beta(1,n) \le \log^* n$.
\qed

\cor\id{ijdens}%
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

\paran{Integer weights}%
The algorithm spends most of the time in phases which have small heaps. Once the
heap grows to $\Omega(\log^{(k)} n)$ for any fixed~$k$, the graph gets dense enough
to guarantee that at most~$k$ phases remain. This means that if we are able to
construct a~heap of size $\Omega(\log^{(k)} n)$ with constant time per operation,
we can get a~linear-time algorithm for MST. This is the case when the weights are
integers:

\thmn{MST for integer weights, Fredman and Willard \cite{fw:transdich}}\id{intmst}%
MST of a~graph with integer edge weights can be found in time $\O(m)$ on the Word-RAM.

\proof
We will combine the Iterated Jarn\'\i{}k's algorithm with the Q-heaps from Section \ref{qheaps}.
We modify the first pass of the algorithm to choose $t=\log n$ and use the Q-heap tree instead
of the Fibonacci heap. From Theorem \ref{qh} and Remark \ref{qhtreerem} we know that the
operations on the Q-heap tree run in constant time, so the modified first phase takes time~$\O(m)$.
Following the analysis of the original algorithm in the proof of Theorem \ref{itjarthm} we obtain
$t_2\ge 2^{t_1} = 2^{\log n} = n$, so the algorithm stops after the second phase.\foot{%
Alternatively, we can use the Q-heaps directly with $k=\log^{1/4}n$ and then the algorithm stops
after the third phase.}
\qed

\paran{Further improvements}%
Gabow et al.~\cite{gabow:mst} have shown how to speed up the Iterated Jarn\'\i{}k's algorithm to~$\O(m\log\beta(m,n))$.
They split the adjacency lists of the vertices to small buckets, keep each bucket
sorted and consider only the lightest edge in each bucket until it is removed.
The mechanics of the algorithm is complex and there is a~lot of technical details
which need careful handling, so we omit the description of this algorithm.
A~better algorithm will be shown in Chapter~\ref{optchap}.

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

\section{Verification of minimality}\id{verifysect}%

Now we will turn our attention to a~slightly different problem: given a~spanning
tree, how to verify that it is minimum? We will show that this can be achieved
in linear time and it will serve as a~basis for a~randomized linear-time
MST algorithm in Section~\ref{randmst}.

MST verification has been studied by Koml\'os \cite{komlos:verify}, who has
proven that $\O(m)$ edge comparisons are sufficient, but his algorithm needed
super-linear time to find the edges to compare. Dixon, Rauch and Tarjan \cite{dixon:verify}
have later shown that the overhead can be reduced
to linear time on the RAM using preprocessing and table lookup on small
subtrees. Later, King has given a~simpler algorithm in \cite{king:verifytwo}.

In this section, we will follow Koml\'os's steps and study the comparisons
needed, saving the actual efficient implementation for later.

\para
To verify that a~spanning tree~$T$ is minimum, it is sufficient to check that all
edges outside~$T$ are $T$-heavy (by the Minimality Theorem, \ref{mstthm}). In fact, we will be
able to find all $T$-light edges efficiently. For each edge $uv\in E\setminus T$,
we will find the heaviest edge of the tree path $T[u,v]$ and compare its weight
to $w(uv)$. It is therefore sufficient to solve the following problem:

\problem
Given a~weighted tree~$T$ and a~set of \df{query paths} $Q \subseteq \{ T[u,v] \mid u,v\in V(T) \}$
specified by their endpoints, find the heaviest edge \df{(peak)} of every path in~$Q$.

\paran{Bor\o{u}vka trees}%
Finding the peaks can be burdensome if the tree~$T$ is degenerated,
so we will first reduce it to the same problem on a~balanced tree. We run
the Bor\o{u}vka's algorithm on~$T$, which certainly produces $T$ itself, and we
record the order, in which the subtrees have been merged, in another tree~$B(T)$.
The peak queries on~$T$ can be then easily translated to peak queries on~$B(T)$.

\defn
For a~weighted tree~$T$ we define its \df{Bor\o{u}vka tree} $B(T)$ as a~rooted tree which records
the execution of the Bor\o{u}vka's algorithm run on~$T$. The leaves of $B(T)$
are all the vertices of~$T$, an~internal vertex~$v$ at level~$i$ from the bottom
corresponds to a~component tree~$C(v)$ formed in the $i$-th iteration of the algorithm. When
a~tree $C(v)$ selects an adjacent edge~$e$ and gets merged with some other trees to form
a~component $C(u)$, we add an~edge $uv$ to~$B(T)$ and set its weight to $w(e)$.

\figure{bortree.eps}{\epsfxsize}{An octipede and its Bor\o{u}vka tree}

\obs
As the algorithm finishes with a~single component in the last phase, the Bor\o{u}vka tree
is really a~tree. All its leaves are on the same level and each internal vertex has
at least two sons. Such trees will be called \df{complete branching trees.}

\lemma
For every tree~$T$ and every pair of its vertices $x,y\in V(T)$, the peak
of the path $T[x,y]$ has the same weight as the peak of~the path $B(T)[x,y]$.

\proof
Let us denote the path $T[x,y]$ by~$P$ and its heaviest edge by~$h=ab$. Similarly,
let us use $P'$ for $B(T)[x,y]$ and $h'$ for the heaviest edge of~$P'$.

We will first prove that~$h$ has its counterpart of the same weight in~$P'$,
so $w(h') \ge w(h)$. Consider the lowest vertex $u$ of~$B(T)$ such that the
component $C(u)$ contains both $a$ and~$b$, and consider the sons $v_a$ and $v_b$ of~$u$
for which $a\in C(v_a)$ and $b\in C(v_b)$. As the edge~$h$ must have been
selected by at least one of these components, we assume without loss of generality that
it was $C(v_a)$, and hence we have $w(uv_a)=w(h)$. We will show that the
edge~$uv_a$ lies in~$P'$, because exactly one of the vertices $x$, $y$ lies
in~$C(v_a)$. Both cannot lie there, since it would imply that $C(v_a)$,
being connected, contains the whole path~$P$, including~$h$. On the other hand,
if $C(v_a)$ contained neither~$x$ nor~$y$, it would have to be incident with
another edge of~$P$ different from~$h$, so this lighter edge would be selected
instead of~$h$.

In the other direction: for any edge~$uv\in P'$, the tree~$C(v)$ is incident
with at least one edge of~$P$, so the selected edge must be lighter or equal
to this edge and hence also to~$h$.
\qed

\para
We will simplify the problem even further: For an~arbitrary tree~$T$, we split each
query path $T[x,y]$ to two half-paths $T[x,a]$ and $T[a,y]$ where~$a$ is the
\df{lowest common ancestor} of~$x$ and~$y$ in~$T$. It is therefore sufficient to
consider only paths that connect a~vertex with one of its ancestors.

When we combine the two transforms, we get:

\lemman{Balancing of trees}\id{verbranch}%
For each tree~$T$ on $n$~vertices and a~set~$Q$ of $q$~query paths on~$T$, it is possible
to find a~complete branching tree~$T'$, together with a~set~$Q'$ of paths on~$T'$,
such that the weights of the heaviest edges of the paths in~$Q$ can be deduced from
the same of the paths in~$Q'$. The tree $T'$ has at most $2n$ vertices and $\O(\log n)$
levels. The set~$Q'$ contains at most~$2q$ paths and each of them connects a~vertex of~$T'$
with one of its ancestors. The construction of~$T'$ involves $\O(n)$ comparisons
and the transformation of the answers takes $\O(q)$ comparisons.

\proof
The tree~$T'$ will be the Bor\o{u}vka tree for~$T$, obtained by running the
contractive version of the Bor\o{u}vka's algorithm (Algorithm \ref{contbor})
on~$T$. The algorithm runs in linear time, for example because trees are planar
(Theorem \ref{planarbor}). We therefore spend $\O(n)$ comparisons in it.

As~$T'$ has~$n$ leaves and it is a~complete branching tree, it has at most~$n$ internal vertices,
so~$n(T')\le 2n$ as promised. Since the number of iterations of the Bor\o{u}vka's
algorithm is $\O(\log n)$, the depth of the Bor\o{u}vka tree must be logarithmic as well.

For each query path $T[x,y]$ we find the lowest common ancestor of~$x$ and~$y$
and split the path by the two half-paths. This produces a~set~$Q'$ of at most~$2q$ half-paths.
The peak of every original query path is then the heavier of the peaks of its halves.
\qed

\paran{Bounding comparisons}%
We will now describe a~simple variant of the depth-first search which finds the
peaks of all query paths of the balanced problem. As we promised,
we will take care of the number of comparisons only, as long as all other operations
are well-defined and they can be performed in polynomial time.

\defn
For every edge~$e=uv$, we consider the set $Q_e$ of all query paths containing~$e$.
The vertex of a~path, that is closer to the root, will be called the \df{top} of the path,
the other vertex its \df{bottom.}
We define arrays $T_e$ and~$P_e$ as follows: $T_e$~contains
the tops of the paths in~$Q_e$ in order of their increasing depth (we
will call them \df{active tops} and each of them will be stored exactly once). For
each active top~$t=T_e[i]$, we define $P_e[i]$ as the peak of the path $T[v,t]$.

\obs
As for every~$i$ the path $T[v,T_e[i+1]]$ is contained within $T[v,T_e[i]]$,
the edges of~$P_e$ must have non-increasing weights, that is $w(P_e[i+1]) \le
w(P_e[i])$.
This leads to the following algorithm:

\alg $\<FindPeaks>(u,p,T_p,P_p)$ --- process all queries located in the subtree rooted
at~$u$ entered from its parent via an~edge~$p$.
\id{findpeaks}

\algo
\:Process all query paths whose bottom is~$u$ and record their peaks.
This is accomplished by finding the index~$i$ of each path's top in~$T_p$ and reading
the desired edge from~$P_p[i]$.

\:For every son~$v$ of~$u$, process the edge $e=uv$:

\::Construct the array of tops~$T_e$ for the edge~$e$: Start with~$T_p$, remove
   the tops of the paths that do not contain~$e$ and add the vertex~$u$ itself
   if there is a~query path which has~$u$ as its top and whose bottom lies somewhere
   in the subtree rooted at~$v$.

\::Prepare the array of the peaks~$P_e$: Start with~$P_p$, remove the entries
   corresponding to the tops that are no longer active. If $u$ became an~active
   top, append~$e$ to the array.

\::Finish~$P_e$:
   Since the paths leading to all active tops have been extended by the
   edge~$e$, compare $w(e)$ with weights of the edges recorded in~$P_e$ and replace
   those edges which are lighter by~$e$.
   Since $P_p$ was sorted, we can use binary search
   to locate the boundary between the lighter and heavier edges in~$P_e$.

\::Recurse on~$v$: call $\<FindPeaks>(v,e,T_e,P_e)$.
\endalgo

\>As we need a~parent edge to start the recursion, we add an~imaginary parent
edge~$p_0$ of the root vertex~$r$, for which no queries are defined. We can
therefore start with $\<FindPeaks>(r,p_0,\emptyset,\emptyset)$.

Let us account for the comparisons:

\lemma\id{vercompares}%
When the procedure \<FindPeaks> is called on the balanced problem, it
performs $\O(n+q)$ comparisons, where $n$ is the size of the tree and
$q$ is the number of query paths.

\proof
We will calculate the number of comparisons~$c_i$ performed when processing the edges
going from the $(i+1)$-th to the $i$-th level of the tree.
The levels are numbered from the bottom, so leaves are at level~0 and the root
is at level $\ell\le \lceil \log_2 n\rceil$. There are $n_i\le n/2^i$ vertices
at the $i$-th level, so we consider exactly $n_i$ edges. To avoid taking a~logarithm\foot{All logarithms are binary.}
of zero, we define $\vert T_e\vert=1$ for $T_e=\emptyset$.
\def\eqalign#1{\null\,\vcenter{\openup\jot
  \ialign{\strut\hfil$\displaystyle{##}$&$\displaystyle{{}##}$\hfil
      \crcr#1\crcr}}\,}
$$\vcenter{\openup\jot\halign{\strut\hfil $\displaystyle{#}$&$\displaystyle{{}#}$\hfil&\quad#\hfil\cr
c_i &\le \sum_e \left( 1 + \log \vert T_e\vert \right)&(Total cost of the binary searches.)\cr
    &\le n_i + \sum_e \log\vert T_e\vert&(We sum over $n_i$ edges.)\cr
    &\le n_i + n_i \cdot {\sum_e \log\vert T_e\vert \over n_i}&(Consider the average of logarithms.) \cr
    &\le n_i + n_i \cdot \log{\sum_e \vert T_e\vert \over n_i}&(Logarithm is concave.) \cr
    &\le n_i + n_i \cdot \log{q+n\over n_i}&(Bound the number of tops by queries.) \cr
    &= n_i \cdot \left( 1 + \log\left({q+n\over n}\cdot{n\over n_i}\right) \right)\cr
    &= n_i + n_i\log{q+n\over n} + n_i\log{n\over n_i}.\cr
}}$$
Summing over all levels, we estimate the total number of comparisons as:
$$
c = \sum_i c_i = \left( \sum_i n_i \right) + \left( \sum_i n_i \log{q+n\over n}\right) + \left( \sum_i n_i \log{n\over n_i}\right).
$$
The first part is equal to~$n$, the second one to $n\log((q+n)/n)\le q+n$. For the third
one, we would like to plug in the bound $n_i \le n/2^i$, but we unfortunately have one~$n_i$
in the denominator. We save the situation by observing that the function $f(x)=x\log(n/x)$
is decreasing\foot{We can easily check the derivative: $f(x)=(x\ln n-x\ln x)/\ln 2$, so $f'(x)\cdot \ln2 =
\ln n - \ln x - 1$. We want $f'(x)<0$ and therefore $\ln x > \ln n - 1$, i.e., $x > n/e$.}
for $x > n/e$, so for $i\ge 2$ it holds that:
$$
n_i\log{n\over n_i} \le {n\over 2^i}\cdot\log{n\over n/2^i} = {n\over 2^i} \cdot i.
$$
We can therefore rewrite the third part as:
$$\eqalign{
\sum_i n_i\log{n\over n_i} &\le n_0\log{n\over n_0} + n_1\log{n\over n_1} + n\cdot\sum_{i\ge 2}{i\over 2^i} \le\cr
&\le n\log1 + n_1\cdot {n\over n_1} + n\cdot\O(1) = \O(n).\cr
}$$
Putting all three parts together, we conclude that:
$$
c \le n + (q+n) + \O(n) = \O(n+q). \qedmath
$$

\para
When we combine this lemma with the above reduction from general trees to balanced trees, we get the following theorem:

\thmn{Verification of the MST, Koml\'os \cite{komlos:verify}}\id{verify}%
For every weighted graph~$G$ and its spanning tree~$T$, it is sufficient to
perform $\O(m)$ comparisons of edge weights to determine whether~$T$ is minimum
and to find all $T$-light edges in~$G$.

\proof
We first transform the problem to finding all peaks of a~set
of query paths in~$T$ (these are exactly the paths covered by the edges
of $G\setminus T$). We use the reduction from Lemma \ref{verbranch} to get
an~equivalent problem with a~full branching tree and a~set of parent-descendant
paths. The reduction costs $\O(m+n)$ comparisons.
Then we run the \<FindPeaks> procedure (Algorithm \ref{findpeaks}) to find
the tops of all query paths. According to Lemma \ref{vercompares}, this spends another $\O(m+n)$
comparisons. Since we (as always) assume that~$G$ is connected, $\O(m+n)=\O(m)$.
\qed

\paran{Other applications}%
The problem of computing path maxima or minima in a~weighted tree has several other interesting
applications. One of them is computing minimum cuts separating given pairs of vertices in a~given
weighted undirected graph~$G$. We construct a~Gomory-Hu tree~$T$ for the graph as described in \cite{gomoryhu}
(see also \cite{bhalgat:ght} for a~more efficient algorithm running in time
$\widetilde\O(mn)$ for unit-cost graphs). The crucial property of this tree is that for every two
vertices $u$, $v$ of the graph~$G$, the minimum-cost edge on $T[u,v]$ has the same cost
as the minimum cut separating $u$ and~$v$ in~$G$. Since the construction of~$T$ generally
takes $\Omega(n^2)$ time, we could of course invest this time in precomputing the minima for
all pairs of vertices. This would however require quadratic space, so we can better use
the method of this section which fits in $\O(n+q)$ space for $q$~queries.

\paran{Dynamic verification}%
A~dynamic version of the problem is also often considered. It calls for a~data structure
representing a~weighted forest with operations for modifying the structure of the forest
and querying minima or maxima on paths. Sleator and Tarjan have shown in \cite{sleator:trees}
how to do this in $\O(\log n)$ time amortized per operation, which leads to
an~implementation of the Dinic's maximum flow algorithm \cite{dinic:flow}
in time $\O(mn\log n)$.

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

\section{Verification in linear time}\id{verifysect}%

We have proven that $\O(m)$ edge weight comparisons suffice to verify minimality
of a~given spanning tree. Now we will show an~algorithm for the RAM
which finds the required comparisons in linear time. We will follow the idea
of King from \cite{king:verifytwo}, but as we have the power of the RAM data structures
from Section~\ref{bitsect} at our command, the low-level details will be easier,
especially the construction of vertex and edge labels.

\para
First of all, let us make sure that the reduction to fully branching trees
in the Balancing lemma (\ref{verbranch}) can be made run in linear time. As already noticed
in the proof, the Bor\o{u}vka's algorithm runs in linear time. Constructing
the Bor\o{u}vka tree in the process adds at most a~constant overhead to every
step of the algorithm.

Finding the common ancestors is not trivial, but Harel and Tarjan have shown
in \cite{harel:nca} that linear time is sufficient on the RAM. Several more
accessible algorithms have been developed since then (see the Alstrup's survey
paper \cite{alstrup:nca} and a~particularly elegant algorithm described by Bender
and Falach-Colton in \cite{bender:lca}). Any of them implies the following
theorem:

\thmn{Lowest common ancestors}\id{lcathm}%
On the RAM, it is possible to preprocess a~tree~$T$ in time $\O(n)$ and then
answer lowest common ancestor queries presented online in constant time.

\cor
The reductions in Lemma \ref{verbranch} can be performed in time $\O(m)$.

\para
Having the balanced problem at hand, it remains to implement the procedure \<FindPeaks>
of Algorithm \ref{findpeaks} efficiently. We need a~compact representation of
the arrays $T_e$ and~$P_e$, which will allow to reduce the overhead of the algorithm
to time linear in the number of comparisons performed. To achieve
this goal, we will encode the arrays in RAM vectors (see Section \ref{bitsect}
for the vector operations).

\defn

\em{Vertex identifiers:} Since all vertices processed by the procedure
lie on the path from the root to the current vertex~$u$, we modify the algorithm
to keep a~stack of these vertices in an~array. We will often refer to each vertex by its
index in this array, i.e., by its depth. We will call these identifiers \df{vertex
labels} and we note that each label requires only $\ell=\lceil \log\lceil\log n\rceil\rceil$
bits. As every tree edge is uniquely identified by its bottom vertex, we can
use the same encoding for \df{edge labels.}

\em{Slots:} As we are going to need several operations which are not computable
in constant time on the RAM, we precompute tables for these operations
like we did in the Q-heaps (cf.~Lemma \ref{qhprecomp}). A~table for word-sized
arguments would take too much time to precompute, so we will generally store
our data structures in \df{slots} of $s=\lceil 1/3\cdot\log n\rceil$ bits each.
We will soon show that it is possible to precompute a~table of any reasonable
function whose arguments fit in two slots.

\em{Top masks:} The array~$T_e$ will be represented by a~bit mask~$M_e$ called the \df{top mask.} For each
of the possible tops~$t$ (i.e., the ancestors of the current vertex), we store
a~single bit telling whether $t\in T_e$. Each top mask fits in $\lceil\log n\rceil$
bits and therefore in a~single machine word. If needed, it can be split to three slots.
Unions and intersections of sets of tops then translate to $\band$/$\bor$
on the top masks.

\em{Small and big lists:} The heaviest edge found so far for each top is stored
by the algorithm in the array~$P_e$. Instead of keeping the real array,
we store the labels of these edges in a~list encoded in a~bit string.
Depending on the size of the list, we use one of two possible encodings:
\df{Small lists} are stored in a~vector that fits in a~single slot, with
the unused fields filled by a~special constant, so that we can easily infer the
length of the list.

If the data do not fit in a~small list, we use a~\df{big list} instead. It
is stored in $\O(\log\log n)$ words, each of them containing a~slot-sized
vector. Unlike the small lists, we use the big lists as arrays. If a~top~$t$ of
depth~$d$ is active, we keep the corresponding entry of~$P_e$ in the $d$-th
field of the big list. Otherwise, we keep that entry unused.

We want to perform all operations on small lists in constant time,
but we can afford spending time $\O(\log\log n)$ on every big list. This
is true because whenever we use a~big list, $\vert T_e\vert = \Omega(\log n/\log\log n)$,
hence we need $\log\vert T_e\vert = \Omega(\log\log n)$ comparisons anyway.

\em{Pointers:} When we need to construct a~small list containing a~sub-list
of a~big list, we do not have enough time to see the whole big list. To handle
this, we introduce \df{pointers} as another kind of edge identifiers.
A~pointer is an~index to the nearest big list on the path from the small
list containing the pointer to the root. As each big list has at most $\lceil\log n\rceil$
fields, the pointer fits in~$\ell$ bits, but we need one extra bit to distinguish
between regular labels and pointers.

\lemman{Precomputation of tables}
When~$f$ is a~function of up to two arguments computable in polynomial time, we can
precompute a~table of the values of~$f$ for all values of arguments that fit
in a~single slot. The precomputation takes $\O(n)$ time.

\proof
Similar to the proof of Lemma \ref{qhprecomp}. There are $\O(2^{2s}) = \O(n^{2/3})$
possible values of arguments, so the precomputation takes time $\O(n^{2/3}\cdot\poly(s))
= \O(n^{2/3}\cdot\poly(\log n)) = \O(n)$.
\qed

\example
As we can afford spending $\O(n)$ time on preprocessing,
we can assume that we can compute the following functions in constant time:

\itemize\ibull
\:$\<Weight>(x)$ --- the Hamming weight of a~slot-sized number~$x$
(we already considered this operation in Algorithm \ref{lsbmsb}, but we needed
quadratic word size for it). We can easily extend this function to $\log n$-bit numbers
by splitting the number in three slots and adding their weights.

\:$\<FindKth>(x,k)$ --- the $k$-th set bit from the top of the slot-sized
number~$x$. Again, this can be extended to multi-slot numbers by calculating
the \<Weight> of each slot first and then finding the slot containing the
$k$-th~\1.

\:$\<Bits>(m)$ --- for a~slot-sized bit mask~$m$, it returns a~small list
of the positions of the bits set in~$\(m)$.

\:$\<Select>(x,m)$ --- constructs a~slot containing the substring of $\(x)$
selected by the bits set in~$\(m)$.

\:$\<SubList>(x,m)$ --- when~$x$ is a~small list and~$m$ a bit mask, it returns
a~small list containing the elements of~$x$ selected by the bits set in~$m$.
\endlist

\para
We will now show how to perform all parts of the procedure \<FindPeaks>
in the required time. We will denote the size of the tree by~$n$ and the
number of query paths by~$q$.

\lemma
Depths of all vertices and all top masks can be computed in time $\O(n+q)$.

\proof
Run depth-first search on the tree, assign the depth of a~vertex when entering
it and construct its top mask when leaving it. The top mask can be obtained
by $\bor$-ing the masks of its sons, excluding the level of the sons and
including the tops of all query paths that have their bottoms at the current vertex
(the depths of the tops are already assigned).
\qed

\lemma\id{verth}%
The arrays $T_e$ and~$P_e$ can be indexed in constant time.

\proof
Indexing~$T_e$ is exactly the operation \<FindKth> applied on the corresponding
top mask~$M_e$.

If $P_e$ is stored in a~big list, we calculate the index of the particular
slot and the position of the field inside the slot. This field can be then
extracted using bit masking and shifts.

If it is a~small list, we extract the field directly, but we have to
dereference it in case it is a pointer. We modify the recursion in \<FindPeaks>
to pass the depth of the lowest edge endowed with a~big list and when we
encounter a~pointer, we index this big list.
\qed

\lemma\id{verhe}%
For an~arbitrary active top~$t$, the corresponding entry of~$P_e$ can be
extracted in constant time.

\proof
We look up the precomputed depth~$d$ of~$t$ first.
If $P_e$ is stored in a~big list, we extract the $d$-th entry of the list.
If the list is small, we find the position of the particular field
by counting bits of the top mask~$M_e$ at position~$d$ and higher
(this is \<Weight> of $M_e$ with the lower bits masked out).
\qed

\lemma\id{verfh}%
\<FindPeaks> processes an~edge~$e$ in time $\O(\log \vert T_e\vert + q_e)$,
where $q_e$~is the number of query paths having~$e$ as its bottom edge.

\proof
The edge is examined in steps 1, 3, 4 and~5 of the algorithm. We will show how to
perform each of these steps in constant time if $P_e$ is a~small list or
$\O(\log\log n)$ if it is big.

\em{Step~1} looks up $q_e$~tops in~$P_e$ and we already know from Lemma \ref{verhe}
how to do that in constant time per top.

\em{Step~3} is trivial as we have already computed the top masks and we can
reconstruct the entries of~$T_e$ in constant time according to Lemma \ref{verth}.

\em{Step~5} involves binary search on~$P_e$ in $\O(\log\vert T_e\vert)$ comparisons,
each of them indexes~$P_e$, which is $\O(1)$ again by Lemma \ref{verth}. Rewriting the
lighter edges is $\O(1)$ for small lists by replication and bit masking, for a~big
list we do the same for each of its slots.

\em{Step~4} is the only non-trivial one. We already know which tops to select
(we have the top masks $M_e$ and~$M_p$ precomputed), but we have to carefully
extract the sublist.
We need to handle these four cases:

\itemize\ibull
\:\em{Small from small:} We use $\<Select>(T_e,T_p)$ to find the fields of~$P_p$
that shall be deleted by a~subsequent call to \<SubList>. Pointers
can be retained as they still refer to the same ancestor list.

\:\em{Big from big:} We can copy the whole~$P_p$, since the layout of the
big lists is fixed and the items, which we do not want, simply end up as unused
fields in~$P_e$.

\:\em{Small from big:} We use the operation \<Bits> to construct a~list
of pointers (we use bit masking to add the ``this is a~pointer'' flags).

\:\em{Big from small:} First we have to dereference the pointers in the
small list~$S$. For each slot~$B_i$ of the ancestor big list, we construct
a~subvector of~$S$ containing only the pointers referring to that slot,
adjusted to be relative to the beginning of the slot (we use \<Compare>
and \<Replicate> from Algorithm \ref{vecops} and bit masking). Then we
use a~precomputed table to replace the pointers by the fields of~$B_i$
they point to. We $\bor$ together the partial results and we again have
a~small list.

Finally, we have to spread the fields of this small list to the whole big list.
This is similar: for each slot of the big list, we find the part of the small
list keeping the fields we want (we call \<Weight> on the sub-words of~$M_e$ before
and after the intended interval of depths) and we use a~tabulated function
to shift the fields to the right locations in the slot (controlled by the
sub-word of~$M_e$ in the intended interval).
\qeditem
\endlist

\>We now have all the necessary ingredients to prove the following theorem
and thus conclude this section:

\thmn{Verification of MST on the RAM}\id{ramverify}%
There is a~RAM algorithm which for every weighted graph~$G$ and its spanning tree~$T$
determines whether~$T$ is minimum and finds all $T$-light edges in~$G$ in time $\O(m)$.

\proof
Implement the Koml\'os's algorithm from Theorem \ref{verify} with the data
structures developed in this section.
According to Lemma \ref{verfh}, the algorithm runs in time $\sum_e \O(\log\vert T_e\vert + q_e)
= \O(\sum_e \log\vert T_e\vert) + \O(\sum_e q_e)$. The second sum is $\O(m)$
as there are $\O(1)$ query paths per edge, the first sum is $\O(\#\hbox{comparisons})$,
which is $\O(m)$ by Theorem \ref{verify}.
\qed

\>In Section \ref{kbestsect}, we will need a~more specialized statement:

\cor\id{rampeaks}%
There is a~RAM algorithm which for every weighted tree~$T$ and a~set~$P$ of
paths in~$T$ calculates the peaks of these paths in time $\O(n(T) + \vert P\vert)$.

\paran{Verification on the Pointer Machine}\id{pmverify}%
Buchsbaum et al.~\cite{buchsbaum:verify} have recently shown that linear-time
verification can be achieved even on the Pointer Machine. They first solve the
problem of finding the lowest common ancestors for a~set of pairs of vertices
by batch processing: They combine an~algorithm of time complexity $\O(m\timesalpha(m,n))$
based on the Disjoint Set Union data structure with the framework of topological graph
computations described in Section \ref{bucketsort}. Then they use a~similar
technique for finding the peaks themselves.

\paran{Online verification}%
The online version of this problem has turned out to be more difficult. It calls for an~algorithm
that preprocesses the tree and then answers queries for peaks of paths presented online. Pettie
\cite{pettie:onlineverify} has proven an~interesting lower bound based on the inverses of the
Ackermann's function. If we want to answer queries within $t$~comparisons, we
have to invest $\Omega(n\log\lambda_t(n))$ time into preprocessing.\foot{$\lambda_t(n)$ is the
$t$-th row inverse of the Ackermann's function, $\alpha(n)$ is its diagonal inverse. See
\ref{ackerinv} for the exact definitions.} This implies that with
preprocessing in linear time, the queries require $\Omega(\alpha(n))$ time.

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

\section{A randomized algorithm}\id{randmst}%

When we analysed the Contractive Bor\o{u}vka's algorithm in Section~\ref{contalg},
we observed that while the number of vertices per iteration decreases exponentially,
the number of edges generally does not, so we spend $\Theta(m)$ time on every phase.
Karger, Klein and Tarjan \cite{karger:randomized} have overcome this problem by
combining the Bor\o{u}vka's algorithm with filtering based on random sampling.
This leads to a~randomized algorithm which runs in linear expected time.

The principle of the filtering is simple: Let us consider any spanning tree~$T$
of the input graph~$G$. Each edge of~$G$ that is $T$-heavy is the heaviest edge
of some cycle, so by the Red lemma (\ref{redlemma}) it cannot participate in
the MST of~$G$. We can therefore discard all $T$-heavy edges and continue with
finding the MST on the reduced graph. Of course, not all choices of~$T$ are equally
good, but it will soon turn out that when we take~$T$ as the MST of a~randomly selected
subgraph, only a~small expected number of edges remains.

Selecting a~subgraph at random will unavoidably produce disconnected subgraphs
at occasion, so we will drop the implicit assumption that all graphs are
connected for this section and we will always search for the minimum spanning forest.
As we already noted (\ref{disconn}), with a~little bit of care our
algorithms and theorems keep working.

Since we need the MST verification algorithm for finding the $T$-heavy edges,
we will assume that we are working on the RAM.

\lemman{Random sampling, Karger \cite{karger:sampling}}\id{samplemma}%
Let $H$~be a~subgraph of~$G$ obtained by including each edge independently
with probability~$p$. Let further $F$~be the minimum spanning forest of~$H$. Then the
expected number of $F$-nonheavy\foot{That is, $F$-light edges and also edges of~$F$ itself.}
edges in~$G$ is at most $n/p$.

\proof
Let us observe that we can obtain the forest~$F$ by running the Kruskal's algorithm
(\ref{kruskal}) combined with the random process producing~$H$ from~$G$. We sort all edges of~$G$
by their weights and we start with an~empty forest~$F$. For each edge, we first
flip a~biased coin (that gives heads with probability~$p$) and if it comes up
tails, we discard the edge. Otherwise we perform a~single step of the Kruskal's
algorithm: We check whether $F+e$ contains a~cycle. If it does, we discard~$e$, otherwise
we add~$e$ to~$F$. At the end, we have produced the subgraph~$H$ and its MSF~$F$.

When we  exchange the check for cycles with flipping the coin, we get an~equivalent
algorithm which will turn out to be more convenient to analyse:
\algo
\:If $F+e$ contains a~cycle, we immediately discard~$e$ (we can flip
the coin, but we need not to, because the edge will be discarded regardless of
the outcome). We note that~$e$ is $F$-heavy with respect to both the
current state of~$F$ and the final MSF.
\:If $F+e$ is acyclic, we flip the coin:
\::If it comes up heads, we add~$e$ to~$F$. In this case, $e$~is neither $F$-light
   nor $F$-heavy.
\::If it comes up tails, we discard~$e$. Such edges are $F$-light.
\endalgo

The number of $F$-nonheavy edges is therefore equal to the total number of the coin
flips in step~2 of this algorithm. We also know that the algorithm stops before
it adds $n$~edges to~$F$. Therefore it flips at most as many coins as a~simple
random process that repeatedly flips until it gets~$n$ heads. As waiting for
every occurrence of heads takes expected time~$1/p$, waiting for~$n$ heads
must take $n/p$. This is the bound we wanted to achieve.
\qed

\para
We will formulate the algorithm as a~doubly-recursive procedure. It alternatively
performs steps of the Bor\o{u}vka's algorithm and filtering based on the above lemma.
The first recursive call computes the MSF of the sampled subgraph, the second one
finds the MSF of the original graph, but without the heavy edges.

As in all contractive algorithms, we use edge labels to keep track of the
original locations of the edges in the input graph. For the sake of simplicity,
we do not mention it in the algorithm explicitly.

\algn{MSF by random sampling --- the KKT algorithm}\id{kkt}%
\algo
\algin A~graph $G$ with an~edge comparison oracle.
\:Remove isolated vertices from~$G$. If no vertices remain, stop and return an~empty forest.
\:Perform two Bor\o{u}vka steps (iterations of Algorithm \ref{contbor}) on~$G$ and
  remember the set~$B$ of the edges having been contracted.
\:Select a~subgraph~$H\subseteq G$ by including each edge independently with
  probability $1/2$.
\:$F\=\msf(H)$ calculated recursively.
\:Construct $G'\subseteq G$ by removing all $F$-heavy edges of~$G$.
\:$R\=\msf(G')$ calculated recursively.
\:Return $R\cup B$.
\algout The minimum spanning forest of~$G$.
\endalgo

\nota
Let us analyse the time complexity of this algorithm by studying properties of its \df{recursion tree.}
This tree describes the subproblems processed by the recursive calls. For any vertex~$v$
of the tree, we denote the number of vertices and edges of the corresponding subproblem~$G_v$
by~$n_v$ and~$m_v$ respectively.
If $m_v>0$, the recursion continues: the left son of~$v$ corresponds to the
call on the sampled subgraph~$H_v$, the right son to the reduced graph~$G^\prime_v$.
(Similarly, we use letters subscripted with~$v$ for the state of the other variables
of the algorithm.)
The root of the recursion tree is obviously the original graph~$G$, the leaves are
trivial graphs with no edges.

\obs
The Bor\o{u}vka steps together with the removal of isolated vertices guarantee that the number
of vertices drops at least by a~factor of four in every recursive call. The size of a~subproblem~$G_v$
at level~$i$ is therefore at most $n/4^i$ and the depth of the tree is at most $\lceil\log_4 n\rceil$.
As there are no more than~$2^i$ subproblems at level~$i$, the sum of all~$n_v$'s
on that level is at most $n/2^i$, which is at most~$2n$ when summed over the whole tree.

We are going to show that the worst case of the KKT algorithm is not worse than
of the plain contractive algorithm, while the average case is linear.

\lemma
For every subproblem~$G_v$, the KKT algorithm spends $\O(m_v+n_v)$ time plus the cost
of the recursive calls.

\proof
We know from Lemma \ref{contiter} that each Bor\o{u}vka step takes time $\O(m_v+n_v)$.\foot{We
need to add $n_v$, because the graph could be disconnected.}
The selection of the edges of~$H_v$ is straightforward.
Finding the $F_v$-heavy edges is not, but we have already shown in Theorem \ref{ramverify}
that linear time is sufficient on the RAM.
\qed

\thmn{Worst-case complexity of the KKT algorithm}
The KKT algorithm runs in time $\O(\min(n^2,m\log n))$ in the worst case on the RAM.

\proof
The argument for the $\O(n^2)$ bound is similar to the analysis of the plain
contractive algorithm. As every subproblem~$G_v$ is a~simple graph, the number
of its edges~$m_v$ is less than~$n_v^2$. By the previous lemma, we spend time
$\O(n_v^2)$ on it. Summing over all subproblems yields $\sum_v \O(n_v^2) =
\O((\sum_v n_v)^2) = \O(n^2)$.

In order to prove the $\O(m\log n)$ bound, it is sufficient to show that the total time
spent on every level of the recursion tree is $\O(m)$. Suppose that $v$~is a~vertex
of the recursion tree with its left son~$\ell$ and right son~$r$. Some edges of~$G_v$
are removed in the Bor\o{u}vka steps, let us denote their number by~$b_v$.
The remaining edges fall either to~$G_\ell = H_v$, or to $G_r = G^\prime_v$, or possibly
to both.

We can observe that the intersection $G_\ell\cap G_r$ cannot be large: The edges of~$H_v$ that
are not in the forest~$F_v$ are $F_v$-heavy, so they do not end up in~$G_r$. Therefore the
intersection can contain only the edges of~$F_v$. As there are at most $n_v/4$ such edges,
we have $m_\ell + m_r + b_v \le m_v + n_v/4$.

On the other hand, the first Bor\o{u}vka step selects at least $n_v/2$ edges,
so $b_v \ge n_v/2$. The duplication of edges between $G_\ell$ and~$G_r$ is therefore
compensated by the loss of edges by contraction and $m_\ell + m_r \le m_v$. So the total
number of edges per level does not decrease and it remains to apply the previous lemma.
\qed

\thmn{Expected complexity of the KKT algorithm}\id{kktavg}%
The expected time complexity of the KKT algorithm on the RAM is $\O(m)$.

\proof
The structure of the recursion tree depends on the random choices taken,
but as its worst-case depth is at most~$\lceil \log_4 n\rceil$, the tree
is always a~subtree of the complete binary tree of that depth. We will
therefore prove the theorem by examining the complete tree, possibly with
empty subproblems in some vertices.

The left edges of the tree (edges connecting a~parent with its left
son) form a~set of \df{left paths.} Let us consider the expected time spent on
a~single left path. When walking the path downwards from its top vertex~$r$,
the expected size of the subproblems decreases exponentially: for a~son~$\ell$
of a~vertex~$v$, we have $n_\ell \le n_v/4$ and $\E m_\ell = \E m_v/2$. The
expected total time spend on the path is therefore $\O(n_r+m_r)$ and it remains
to sum this over all left paths.

With the exception of the path going from the root of the tree,
the top~$r$ of a~left path is always a~right son of a~unique parent vertex~$v$.
Since the subproblem~$G_r$ has been obtained from its parent subproblem~$G_v$
by filtering out all heavy edges, we can use the Sampling lemma (\ref{samplemma}) to show that
$\E m_r \le 2n_v$. The sum of the expected sizes of all top subproblems is
then $\sum_r n_r + m_r \le \sum_v 3n_v = \O(n)$. After adding the exceptional path
from the root, we get $\O(m+n)=\O(m)$.
\qed

\paran{High probability}%
There is also a~high-probability version of the above theorem. According to
Karger, Klein and Tarjan \cite{karger:randomized}, the time complexity
of the algorithm is $\O(m)$ with probability $1-\exp(-\Omega(m))$. The proof
again follows the recursion tree and it involves applying the Chernoff bound
\cite{chernoff} to bound the tail probabilities.

\paran{Different sampling}%
We could also use a~slightly different formulation of the Sampling lemma
suggested by Chan \cite{chan:backward}. He changes the selection of the subgraph~$H$
to choosing an~$mp$-edge subset of~$E(G)$ uniformly at random. The proof is then
a~straightforward application of the backward analysis method. We however preferred
the Karger's original version, because generating a~random subset of a~given size
requires an~unbounded number of random bits in the worst case.

\paran{On the Pointer Machine}%
The only place where we needed the power of the RAM is finding the heavy edges,
so we can employ the pointer-machine verification algorithm mentioned in \ref{pmverify}
to bring the results of this section to the~PM.

\endpart