\section{Minor-closed graph classes}\id{minorclosed}%
-The contractive algorithm given in section~\ref{contalg} has been found to perform
+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 the algorithm is still efficient?
+Can we find any broader class of graphs where this algorithm is still linear?
The right context turns out to be the minor-closed graph classes, which are
closed under contractions and have bounded density.
\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
+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
\para
Many of the nice structural properties of planar graphs extend to
-minor-closed classes, too (see \cite{lovasz:minors} for a~nice survey
-of this theory and \cite{diestel:gt} for some of the deeper results).
+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.
--- 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.
+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 ${\cal C}=\Forb({\cal H})$.
+\qed
-\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
-theorem.
+theory.
\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 $G\in\cal C$.
+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}}
+\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.
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 ${k\choose 2}\ge m\ge k$~edges. For $m={k\choose 2}$ the theorem follows
+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
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 $G.U$ ($G$~with $U$~contracted to
+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
-$G=G.U$ whenever $\vert U\vert=1$). Now consider the subgraph~$H$ induced
+$G=G\sgc U$ 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 $r$~vertices and $m-1$ edges. Any two non-adjacent vertices
+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
Every non-trivial minor-closed class of graphs has finite edge density.
\proof
-Let~$\cal C$ be any such class, $X$~its smallest excluded minor and $x=n(X)$.
-As $H\minorof K_x$, the class $\cal C$ entirely lies in ${\cal C}'=\Forb(K_x)$, so
+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~$K_x$.
$G$~would contain a~subdivision of~$K_x$ and hence $K_x$ as a~minor.
\qed
-\rem
-Minor-closed classes share many other interesting properties, as shown for
-example by Theorem 6.1 of \cite{nesetril:minors}.
+Let us return to the analysis of our algorithm.
-\thmn{MST on minor-closed classes \cite{mm:mst}}\id{mstmcc}%
+\thmn{MST on minor-closed classes, Tarjan \cite{tarjan:dsna}}\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 iteration by~$G_i$ and its number of vertices
+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 $n_i \le n/2^i$, so it remains to show a linear bound for the $m_i$'s.
+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~$G$.\foot{Technically, these are multigraph 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 previous theorem $m_i\le \varrho({\cal C})\cdot n_i$.
+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
-\rem\id{nobatch}%
+\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 contractions one edge at a~time,
+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
${\rm Pr}[X > 4\varrho] < 1/2$, 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}}%
+\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~$t\in{\bb N}$.
\:$T\=\emptyset$.
\: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~$e$.
+\:::Contract~$e$.
\:::$T\=T + \ell(e)$.
\::Flatten $G$, removing parallel edges and loops.
\algout Minimum spanning tree~$T$.
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, so at least $n_i/4$ edges get selected.
-Hence $n_i\le 3/4\cdot n_{i-1}$ and therefore $n_i\le n\cdot (3/4)^i$ and the
+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 in \ref{rbma}).
+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 Lemma on contraction of MST edges (\ref{contlemma}) and when
+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 have
+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 linear for a fixed class~$\cal C$.
-Flattening takes $\O(m_i)$, as already noted in the analysis of the Contracting
+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
-\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
+\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.
\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, $\sum_v \deg(v) < 6n$.
+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 $d(v) < 6$, hence $\deg(v) \le 8$.
+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 have degrees at most~6, the remaining $n-\O(k)$ interior
+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. This adds $f/2 \approx n$
+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
+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 made by Gustedt in~\cite{gustedt:parallel},
-who studied a~parallel version of the contractive Bor\o{u}vka's algorithm applied
+The observation in~Theorem~\ref{mstmcc} was also used by Gustedt \cite{gustedt:parallel},
+to construct 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{Using Fibonacci heaps}
-\id{fibonacci}
+\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 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.
+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~$\delta(T)$.
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 in~$T$, no action
+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 on the heap.
+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 $w(uv)$, initially empty.
+\:$H\=$ 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:
\algout Minimum spanning tree~$T$.
\endalgo
-\para
-To analyze the time complexity of this algorithm, we will use the standard
+\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} The~Fibonacci heap performs the following operations
+\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 value of an~existing 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 $\O(\log n)$,
\qed
\cor
-For graphs with edge density at least $\log n$, this algorithm runs in linear time.
+For graphs with edge density $\Omega(\log n)$, this algorithm runs in linear time.
-\rem
+\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 a~\df{$d$-regular heap} --- a~variant of the usual binary heap
+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
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:pqsssp} which have constant-time \<Insert> and \<Decrease>
+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.)
-\para
+\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
-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 Active Edge Jarn\'\i{}k's algorithm.
+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 $\log\log n$ iterations of the Contractive Bor\o{u}vka's algorithm (\ref{contbor}),
- getting a~MST~$T_1$.
+\: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}).
and both trees can be combined in linear time, too.
\qed
-\para
+\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 after a~while.
+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 graph along the edges of~this forest and iterate.
+contract the edges of~this forest and iterate.
-\algn{Iterated Jarn\'\i{}k; Fredman and Tarjan \cite{ft:fibonacci}}
+\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}
-\:$m_0\=m$.
+\:$m_0\=m$. \cmt{in the following, $n$ and $m$ will change with the graph}
\: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}
\:::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.
+\::Contract all edges of~$F$ and flatten~$G$.
\algout Minimum spanning tree~$T$.
\endalgo
lemma makes the reason clear:
\lemma\id{ijphase}%
-The $i$-th phase of the Iterated Jarn\'\i{}k's algorithm runs in time~$\O(m)$.
+Each 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
+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
+\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
-connecting two vertices of the tree.
+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 edges incident with~$R_i^j$, we have enough such edges;
+ 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.}
\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 $\beta(m,n):=\min\{ i: \log^{(i)}n \le m/n \}$.
+$\O(m\timesbeta(m,n))$, 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
\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 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.
+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) = \log^* n$.
+$\beta(m,n) \le \beta(n,n) \le \log^* n$.
\qed
-\cor
+\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)$.
If $m/n \ge \log^{(k)} n$, then $\beta(m,n)\le k$.
\qed
-\obs
+\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
we can get a~linear-time algorithm for MST. This is the case when the weights are
integers:
-\thmn{MST for graphs with integer weights, Fredman and Willard \cite{fw:transdich}}\id{intmst}%
+\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 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 stop
+Alternatively, we can use the Q-heaps directly with $k=\log^{1/4}n$ and then the algorithm stops
after the third phase.}
\qed
-\rem
+\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.
-
-\FIXME{Reference to Chazelle.}
+A~better algorithm will be shown in Chapter~\ref{optchap}.
%--------------------------------------------------------------------------------
-\section{Verification of minimality}
+\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
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
-superlinear time to find the edges to compare. Dixon, Rauch and Tarjan
-have later shown in \cite{dixon:verify} that the overhead can be reduced
+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}.
needed, saving the actual efficient implementation for later.
\para
-To verify that a~spanning~$T$ is minimum, it is sufficient to check that all
-edges outside~$T$ are $T$-heavy (by Theorem \ref{mstthm}). In fact, we will be
+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] ; u,v\in V(T) \}$
-specified by their endpoints, find the heaviest edge (\df{peak}) for every path in~$Q$.
+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$.
-\para
+\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)$.
+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 phase of the algorithm. When
+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)$.
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 endpoints of~$h$ lies
-in~$C(v_a)$. Both endpoints cannot lie there, since it would imply that $C(v_a)$,
+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
When we combine the two transforms, we get:
-\lemma\id{verbranch}%
+\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
(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 passes of the Bor\o{u}vka's
+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$
The peak of every original query path is then the heavier of the peaks of its halves.
\qed
-\para
+\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 transformed problem. As we promised,
+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, which is closer to the root, will be called its \df{top,}
+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
+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]$.
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 in the subtree rooted
+\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}
\::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 which has bottom somewhere
+ 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
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 lighter and heavier edges in~$P_e$.
+ 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
Let us account for the comparisons:
\lemma\id{vercompares}%
-When the procedure \<FindPeaks> is called on the transformed problem, it
+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.
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
+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
$$
\para
-When we combine this lemma with the above reduction, we get the following theorem:
+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
comparisons. Since we (as always) assume that~$G$ is connected, $\O(m+n)=\O(m)$.
\qed
-\rem
+\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}
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.
-\rem
+\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}
%--------------------------------------------------------------------------------
-\section{Verification in linear time}
+\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,
+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:verify}, but as we have the power of the RAM data structures
+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 Lemma \ref{verbranch} can be made run in linear time. As already noticed
+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.
The reductions in Lemma \ref{verbranch} can be performed in time $\O(m)$.
\para
-Having the reduced problem at hand, it remains to implement the procedure \<FindPeaks>
+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 will be linear in the number of comparisons performed. To achieve
+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).
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 will need several operations which are not computable
+\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 show soon that it is possible to precompute a~table of any reasonable
+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 calling $\band$/$\bor$
+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 which fits in a~single slot, with
+\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, which
+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
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 normal labels and pointers.
+between regular labels and pointers.
\lemman{Precomputation of tables}
-When~$f$ is a~function of two arguments computable in polynomial time, we can
+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.
\qed
\example
-As we can afford spending spending $\O(n)$ time on preprocessing,
+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 to $\log n$-bit numbers
+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
\qed
\lemma\id{verfh}%
-The procedure \<FindPeaks> processes an~edge~$e$ in time $\O(\log \vert T_e\vert + q_e)$,
+\<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
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 we do not want simply end up as unused
+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
\qeditem
\endlist
-\>We are now ready to combine these steps and get the following theorem:
+\>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~$T$
+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}, it runs in time $\sum_e \O(\log\vert T_e\vert + q_e)
+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
-\rem\id{pmverify}%
-Buchsbaum et al.~have recently shown in \cite{buchsbaum:verify} 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 $\O(m\alpha(m,n))$
-using the Union-Find data structure with table lookup for small subtrees. Then they use a~similar
-technique for finding the peaks themselves. The tricky part is of course the table
-lookup, which they handle by radix-sorting pointer-based codes of the subtrees.
+\>In Section \ref{kbestsect}, we will need a~more specialized statement:
-\rem
-The online version of this problem (build a~data structure for a~weighted tree
-in linear time and then answer queries for individual paths in constant time)
-is still open even for the RAM.
+\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}%
+\section{A randomized algorithm}\id{randmst}%
-When we analysed the contractive Bor\o{u}vka's algorithm in Section~\ref{contalg},
+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
subgraph, only a~small expected number of edges remains.
Selecting a~subgraph at random will unavoidably produce disconnected subgraphs
-at occassion, so we will drop the implicit assumption that all graphs are
+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 (Remark \ref{disconn}), with a~little bit of care our
+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}}
+\lemman{Random sampling, Karger \cite{karger:sampling}}\id{samplemma}%
Let $H$~be a~subgraph of~$G$ obtained by including each edge independently
-with probability~$p$ and $F$~the minimum spanning forest of~$H$. Then the
-expected number of $F$-nonheavy edges in~$G$ is at most $n/p$.
+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 (which gives heads with probability~$p$) and if it comes up
+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
-algoritm: We check whether $F+e$ contains a~cycle. If it does, we discard~$e$, otherwise
+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
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 which repeatedly flips until it gets~$n$ heads. As waiting for
-every occurence of heads takes expected time~$1/p$, waiting for~$n$ heads
-must take $n/p$. This is the bound we wanted to achieve.
+random process that repeatedly flips until it gets~$n$ heads. As waiting for
+every occurrence of heads takes expected time~$1/p$ (the distribution is geometric),
+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
-peforms steps of the Bor\o{u}vka's algorithm and filtering based on the above lemma.
+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 graph without the heavy edges.
+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.
+we do not mention it in the algorithm explicitly.
-\algn{MSF by random sampling --- the KKT algorithm}
+\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 edges contracted.
-\:Select subgraph~$H\subseteq G$ by including each edge independently with
+ 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~$G$.
\nota
Let us analyse the time complexity of this algorithm by studying properties of its \df{recursion tree.}
-The tree describes the subproblems processed by the recursive calls. For any vertex~$t$
-of the tree, we denote the number of vertices and edges of the corresponding subproblem~$G_t$
-by~$n_t$ and~$m_t$ respectively.
-If $m_t>0$, the recursion continues: the left son of~$t$ corresponds to the
-call on the sampled subgraph~$H_t$, the right son to the reduced graph~$G^\prime_t$.
-(Similarly, we use letters subscripted with~$t$ for the state of the other variables
+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~$G_t$
+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~$n_t$'s
-on that level is at most $n/2^i$, which is at most~$2n$ for the whole tree.
+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~$G_t$, the KKT algorithm spends time $\O(m_t+n_t)$ plus the time
-spent on the recursive calls.
+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 $\O(m_t+n_t)$.\foot{We
-add $n_t$ as the graph could be disconnected.}
-The selection of the edges of~$H_t$ is straightforward.
-Finding the $F_t$-heavy edges is not, but we have already shown in Theorem \ref{ramverify}
+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
\proof
The argument for the $\O(n^2)$ bound is similar to the analysis of the plain
-contractive algorithm. As every subproblem~$G_t$ is a~simple graph, the number
-of its edges~$m_t$ is less than~$n_t^2$. By the previous lemma, we spend time
-$\O(n_t^2)$ on it. Summing over all subproblems yields $\sum_t \O(n_t^2) =
-\O((\sum_t n_t)^2) = \O(n^2)$.
+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 $t$~is a~vertex
-of the recursion tree with its left son~$\ell$ and right son~$r$. Some edges of~$G_t$
-are removed in the Bor\o{u}vka steps, let us denote their number by~$b_t$.
-The remaining edges fall either to~$G_\ell = H_t$, or to $G_r = G^\prime_t$, or possibly
+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~$H_t$ that
-are not in the forest~$F_t$ are $F_t$-heavy, so they do not end up in~$G_r$. Therefore the
-intersection can contain only the edges of~$F_t$. As there are at most $n_t/4$ such edges,
-we have $m_\ell + m_r + b_t \le m_t + n_t/4$.
+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 $n_t/2$ edges,
-so $b_t \ge n_t/2$. The duplication of edges between $G_\ell$ and~$G_r$ is therefore
-compensated by the loss of edges by contraction and $m_\ell + m_r \le m_t$. So the total
+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{Average-case complexity of the KKT algorithm}
+\thmn{Expected complexity of the KKT algorithm}\id{kktavg}%
The expected time complexity of the KKT algorithm on the RAM is $\O(m)$.
\proof
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 at some vertices.
+empty subproblems in some vertices.
-The set of all left edges in the tree (edges connecting a~parent with its left
+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~$t$, we have $n_\ell \le n_t/4$ and $\E m_\ell = \E m_t/2$. The
+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~$t$.
-Since the subproblem~$G_r$ has been obtained from its parent subproblem~$G_t$
-by filtering out all heavy edges, we can use the Sampling lemma to show that
-$\E m_r \le 2n_t$. The sum of the expected sizes of all top subproblems is
-then $\sum_r n_r + m_r \le \sum_t 3n_t = \O(n)$. After adding the exceptional path
+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
-\rem
+\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.
-\rem
-We could also use a~slightly different formulation of the sampling lemma
-suggested by Chan \cite{chan:backward}. It changes the selection of the subgraph~$H$
+\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 prefered
+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.
-\rem
-The only place where we needed the power of the RAM is the verification algorithm,
-so we can use the pointer-machine verification algorithm mentioned in Remark \ref{pmverify}
+\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.
-%--------------------------------------------------------------------------------
-
-\section{Special cases and related problems}
-
-Finally, we will focus our attention on various special cases of the minimum
-spanning tree problem which frequently arise in practice.
-
-\examplen{Graphs with sorted edges}
-When the edges are already sorted by their weights, we can use the Kruskal's
-algorithm to find the MST in time $\O(m\timesalpha(n))$ (Theorem \ref{kruskal}).
-We however can do better: As the minimality of a~spanning tree depends only on the
-order of weights and not on the actual values (Theorem \ref{mstthm}), we can
-renumber the weights to $1, \ldots, m$ and find the MST using the Fredman-Willard
-algorithm for integer weights. According to Theorem \ref{intmst} it runs in
-time $\O(m)$ on the Word-RAM.
-
-\examplen{Graphs with a~small number of distinct weights}
-When the weights of edges are drawn from a~set of a~fixed size~$U$, we can
-sort them in linear time and so reduce the problem to the previous case.
-A~more practical way is to use the Jarn\'\i{}k's algorithm (\ref{jarnimpl}),
-but replace the heap by an~array of $U$~buckets. As the number of buckets
-is constant, we can find the minimum in constant time and hence the whole
-algorithm runs in time $\O(m)$, even on the Pointer Machine. For large
-values of~$U,$ we can build a~binary search tree or the van Emde-Boas
-tree (see Section \ref{ramdssect} and \cite{boas:vebt}) on the top of the buckets to bring the complexity
-of finding the minimum down to $\O(\log U)$ or $\O(\log\log U)$ respectively.
-
-\examplen{Graphs with floating-point weights}
-A~common case of non-integer weights are rational numbers in floating-point (FP)
-representation. Even in this case we will be able to find the MST in linear time.
-The most common representation of binary FP numbers specified by the IEEE
-standard 754-1985 \cite{ieee:binfp} has a~useful property: When the
-bit strings encoding non-negative FP numbers are read as ordinary integers,
-the order of these integers is the same as of the original FP numbers. We can
-therefore once again replace the edge weights by integers and use the linear-time
-integer algorithm. While the other FP representations (see \cite{dgoldberg:fp} for
-an~overview) need not have this property, the corresponding integers can be adjusted
-in $\O(1)$ time to the format we need. (More advanced tricks of this type have been
-employed by Thorup in \cite{thorup:floatint} to extend his linear-time algorithm
-for single-source shortest paths to FP edge lengths.)
-
-\examplen{Graphs with bounded degrees}
-For graphs with vertex degrees bounded by a~constant~$\Delta$, the problem is either
-trivial (if $\Delta<3$) or as hard as for arbitrary graphs. There is a~simple linear-time
-transform of arbitrary graphs to graphs with maximum degree~3 which preserves the MST:
-
-\lemman{Degree reduction}\id{degred}%
-For every graph~$G$ there exists a~graph~$G'$ with maximum degree at most~3 and
-a~function $\pi: E(G)\rightarrow E(G')$ such that $\mst(G) = \pi^{-1}(\mst(G'))$.
-The graph $G'$ and the embedding~$\pi$ can be constructed in time $\O(m)$.
-
-\figure{french.eps}{\epsfxsize}{Degree reduction in Lemma~\ref{degred}}
-
-\proof
-We show how to eliminate a~single vertex~$v$ of degree $d>3$ and then apply
-induction.
-
-Assume that $v$~has neighbors $w_1,\ldots,w_d$. We replace~$v$ and the edges~$vw_i$
-by $d$~new vertices $v_1,\ldots,v_d$, joined by a~path $v_1v_2\ldots v_d$, and
-edges~$v_iw_i$. Each edge of the path will receive a~weight smaller than all
-original weights, the other edges will inherit the weights of the edges $vw_i$
-they replace. The edges of the path will therefore lie in the MST (this is
-obvious from the Kruskal's algorithm) and as~$G$ can be obtained from the
-new~$G'$ by contracting the path, the rest follows from the Contraction lemma
-(\ref{contlemma}).
-
-This step can be carried out in time $\O(d)$. As it replaces a high-degree
-vertex by vertices of degree~3, the whole procedure stops in at most~$n$ such
-steps, so it takes time $\O(\sum_{v\in V}\deg(v)) = \O(m)$ including the
-time needed to find the high-degree vertices at the beginning.
-\qed
-
-\examplen{Euclidean MST}
-The MST also has its counterparts in the realm of geometric algorithms. Suppose
-that we have $n$~points $x_1,\ldots,x_n$ in the plane and we want to find the
-shortest system of segments connecting these points. If we want the segments to
-touch only in the given points, this is equivalent to finding the MST of the
-complete graph on the vertices $V=\{x_1,\ldots,x_n\}$ with edge weights
-defined as the Euclidean distances of the points. Since the graph is dense, many of the MST
-algorithms discussed run in linear time with the size of the graph, hence
-in time $\O(n^2)$.
-
-There is a~more efficient method based on the observation that the MST
-is always a~subgraph of the Delaunay's tesselation for the given points
-(this was first noted by Shamos and Hoey in~\cite{shamos:closest}). The
-tesselation is a~planar graph, which guarantees that it has $\O(n)$ edges,
-and it is a~dual graph of the Voronoi diagram of the given points, which can
-be constructed in time $\O(n\log n)$ using for example the Fortune's
-algorithm \cite{fortune:voronoi}. We can therefore reduce the problem
-to finding the MST of the tesselation for which $\O(n\log n)$ time
-is more than sufficient.
-
-This approach fails for non-Euclidean metrics, but in some cases
-(in particular for the rectilinear metric) the $\O(n\log n)$ time bound
-is also achievable by the algorithm of Zhou et al.~\cite{zhou:nodel}
-based on the sweep-line technique and the Red rule. For other
-variations on the geometric MST, see Eppstein's survey paper
-\cite{eppstein:spanning}.
-
-\examplen{Steiner trees}
-The constraint that the segments in the previous example are allowed to touch
-each other only in the given points looks artificial and it is indeed uncommon in
-practical applications (including the problem of designing electrical transmission
-lines originally studied by Bor\o{u}vka). If we lift this restriction, we get
-the problem known by the name Steiner tree.\foot{It is named after the Swiss mathematician
-Jacob Steiner who studied a~special case of this problem in the 19th century.}
-We can also define it in terms of graphs:
-
-\defn A~\df{Steiner tree} of a~weighted graph~$(G,w)$ with a~set~$M\subseteq V$
-of \df{mandatory notes} is a~tree~$T\subseteq G$ that contains all the mandatory
-vertices and its weight is minimum possible.
-
-For $M=V$ the Steiner tree is identical to the MST, but if we allow an~arbitrary
-choice of the mandatory vertices, it is NP-hard. This has been proven by Garey and Johnson
-\cite{garey:steiner,garey:rectisteiner} for not only the graph version with
-weights $\{1,2\}$, but also for the planar version with Euclidean or rectilinear
-metric. There is a~polynomial approximation algorithm with ratio $5/3$ for
-graphs due to Pr\"omel and Steger \cite{proemel:steiner} and a~polynomial-time
-approximation scheme for the Euclidean Steiner tree in an~arbitrary dimension
-by Arora \cite{arora:tspapx}.
-
-\examplen{Approximating the weight of the MST}
-Sometimes we are not interested in the actual edges forming the MST and only
-the weight matters. If we are willing to put up with a~randomized approximation,
-we can even achieve sub-linear complexity. Chazelle et al.~\cite{chazelle:mstapprox}
-have shown an~algorithm which, given $0 < \varepsilon < 1/2$, approximates
-the weight of the MST of a~graph with average degree~$d$ and edge weights from the set
-$\{1,\ldots,w\}$ in time $\O(dw\varepsilon^{-2}\cdot\log(dw/\varepsilon))$,
-producing a~weight which has relative error at most~$\varepsilon$ with probability
-at least $3/4$. They have also proven an~almost matching lower bound $\Omega(dw\varepsilon^{-2})$.
-
-For the $d$-dimensional Euclidean case, there is a~randomized approximation
-algorithm by Czumaj et al.~\cite{czumaj:euclidean} which with high probability
-produces a~spanning tree within relative error~$\varepsilon$ in $\widetilde\O(\sqrt{n}\cdot \poly(1/\varepsilon))$\foot{%
-$\widetilde\O(f) = \O(f\cdot\log^{\O(1)} f)$ and $\poly(n)=n^{\O(1)}$.}
-queries to a~data structure containing the points. The data structure is expected
-to answer orthogonal range queries and cone approximate nearest neighbor queries.
-There is also a~$\widetilde\O(n\cdot \poly(1/\varepsilon))$ time approximation
-algorithm for the MST weight in arbitrary metric spaces by Czumaj and Sohler \cite{czumaj:metric}.
-(This is still sub-linear since the corresponding graph has roughly $n^2$ edges.)
-
\endpart
projects / saga.git / blobdiff