[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 the algorithm is still efficient?
The right context turns out to be the minor-closed graph classes, which are
closed under contractions and have bounded density.

\defn
A~graph~$H$ is a \df{minor} of a~graph~$G$ iff it can be obtained
from a subgraph of~$G$ by a sequence of simple graph contractions (see \ref{simpcont}).

\defn
A~class~$\cal C$ of graphs is \df{minor-closed}, when for every $G\in\cal C$ and
its every minor~$H$, the graph~$H$ lies in~$\cal C$ as well. A~class~$\cal C$ is called
\df{non-trivial} if at least one graph lies in~$\cal C$ and at least one lies outside~$\cal C$.

\example
Non-trivial minor-closed classes include planar graphs and more generally graphs
embeddable in any fixed surface. Many nice properties of planar graphs extend
to these classes, too, most notably the linearity of the number of edges.

\defn\id{density}%
Let $\cal C$ be a class of graphs. We define its \df{edge density} $\varrho(\cal C)$
to be the infimum of all~$\varrho$'s such that $m(G) \le \varrho\cdot n(G)$
holds for every $G\in\cal C$.

\thmn{Density of minor-closed classes}
A~minor-closed class of graphs has finite edge density if and only if it is
a non-trivial class.

\proof
See Theorem 6.1 in \cite{nesetril:minors}, which also lists some other equivalent conditions.
\qed

\thmn{MST on minor-closed classes \cite{mm:mst}}\id{mstmcc}%
For any fixed non-trivial minor-closed class~$\cal C$ of graphs, Algorithm \ref{contbor} finds
the MST of any graph in this class in time $\O(n)$. (The constant hidden in the~$\O$
depends on the class.)

\proof
Following the proof for planar graphs (\ref{planarbor}), we denote the graph considered
by the algorithm at the beginning of the $i$-th iteration by~$G_i$ and its number of vertices
and edges by $n_i$ and $m_i$ respectively. Again the $i$-th phase runs in time $\O(m_i)$
and $n_i \le n/2^i$, so it remains to show a linear bound for the $m_i$'s.

Since each $G_i$ is produced from~$G_{i-1}$ by a sequence of edge contractions,
all $G_i$'s are minors of~$G$.\foot{Technically, these are multigraph contractions,
but followed by flattening, so they are equivalent to contractions on simple graphs.}
So they also belong to~$\cal C$ and by the previous theorem $m_i\le \varrho({\cal C})\cdot n_i$.
\qed

\rem\id{nobatch}%
The contractive algorithm uses ``batch processing'' to perform many contractions
in a single step. It is also possible to perform contractions one edge at a~time,
batching only the flattenings. A~contraction of an edge~$uv$ can be done
in time~$\O(\deg(u))$ by removing all edges incident with~$u$ and inserting them back
with $u$ replaced by~$v$. Therefore we need to find a lot of vertices with small
degrees. The following lemma shows that this is always the case in minor-closed
classes.

\lemman{Low-degree vertices}\id{lowdeg}%
Let $\cal C$ be a graph class with density~$\varrho$ and $G\in\cal C$ a~graph
with $n$~vertices. Then at least $n/2$ vertices of~$G$ have degree at most~$4\varrho$.

\proof
Assume the contrary: Let there be at least $n/2$ vertices with degree
greater than~$4\varrho$.  Then $\sum_v \deg(v) > n/2
\cdot 4\varrho = 2\varrho n$, which is in contradiction with the number
of edges being at most $\varrho n$.
\qed

\rem
The proof can be also viewed
probabilistically: let $X$ be the degree of a vertex of~$G$ chosen uniformly at
random. Then ${\bb E}X \le 2\varrho$, hence by the Markov's inequality
${\rm Pr}[X > 4\varrho] < 1/2$, so for at least $n/2$ vertices~$v$ we have
$\deg(v)\le 4\varrho$.

\algn{Local Bor\o{u}vka's Algorithm \cite{mm:mst}}%
\algo
\algin A~graph~$G$ with an edge comparison oracle and a~parameter~$t\in{\bb N}$.
\:$T\=\emptyset$.
\:$\ell(e)\=e$ for all edges~$e$.
\:While $n(G)>1$:
\::While there exists a~vertex~$v$ such that $\deg(v)\le t$:
\:::Select the lightest edge~$e$ incident with~$v$.
\:::Contract~$G$ along~$e$.
\:::$T\=T + \ell(e)$.
\::Flatten $G$, removing parallel edges and loops.
\algout Minimum spanning tree~$T$.
\endalgo

\thm
When $\cal C$ is a minor-closed class of graphs with density~$\varrho$, the
Local Bor\o{u}vka's Algorithm with the parameter~$t$ set to~$4\varrho$ 
finds the MST of any graph from this class in time $\O(n)$. (The constant
in the~$\O$ depends on~the class.)

\proof
Let us denote by $G_i$, $n_i$ and $m_i$ the graph considered by the
algorithm at the beginning of the $i$-th iteration of the outer loop,
and the number of its vertices and edges respectively. As in the proof
of the previous algorithm (\ref{mstmcc}), we observe that all the $G_i$'s
are minors of the graph~$G$ given as the input.

For the choice $t=4\varrho$, the Lemma on low-degree vertices (\ref{lowdeg})
guarantees that at least $n_i/2$ edges get selected in the $i$-th iteration.
Hence at least a half of the vertices participates in contractions, so
$n_i\le 3/4\cdot n_{i-1}$. Therefore $n_i\le n\cdot (3/4)^i$ and the algorithm terminates
after $\O(\log n)$ iterations.

Each selected edge belongs to $\mst(G)$, because it is the lightest edge of
the trivial cut $\delta(v)$ (see the Blue Rule in \ref{rbma}).
The steps 6 and~7 therefore correspond to the operation
described by the Lemma on contraction of MST edges (\ref{contlemma}) and when
the algorithm stops, $T$~is indeed the minimum spanning tree.

It remains to analyse the time complexity of the algorithm. Since $G_i\in{\cal C}$, we have
$m_i\le \varrho n_i \le \varrho n/2^i$.
We will show that the $i$-th iteration is carried out in time $\O(m_i)$.
Steps 5 and~6 run in time $\O(\deg(v))=\O(t)$ for each~$v$, so summed
over all $v$'s they take $\O(tn_i)$, which is linear for a fixed class~$\cal C$.
Flattening takes $\O(m_i)$, as already noted in the analysis of the Contracting
Bor\o{u}vka's Algorithm (see \ref{contiter}).

The whole algorithm therefore runs in time $\O(\sum_i m_i) = \O(\sum_i n/2^i) = \O(n)$.
\qed

\rem
For planar graphs, we can get a sharper version of the low-degree lemma,
showing that the algorithm works with $t=8$ as well (we had $t=12$ as
$\varrho=3$). While this does not change the asymptotic time complexity
of the algorithm, the constant-factor speedup can still delight the hearts of
its practical users.

\lemman{Low-degree vertices in planar graphs}%
Let $G$ be a planar graph with $n$~vertices. Then at least $n/2$ vertices of~$v$
have degree at most~8.

\proof
It suffices to show that the lemma holds for triangulations (if there
are any edges missing, the situation can only get better) with at
least 3 vertices. Since $G$ is planar, $\sum_v \deg(v) < 6n$.
The numbers $d(v):=\deg(v)-3$ are non-negative and $\sum_v d(v) < 3n$,
so by the same argument as in the proof of the general lemma, for at least $n/2$
vertices~$v$ it holds that $d(v) < 6$, hence $\deg(v) \le 8$.
\qed

\rem\id{hexa}%
The constant~8 in the previous lemma is the best we can have.
Consider a $k\times k$ triangular grid. It has $n=k^2$ vertices, $\O(k)$ of them
lie on the outer face and have degrees at most~6, the remaining $n-\O(k)$ interior
vertices have degree exactly~6. Therefore the number of faces~$f$ is $6/3\cdot n=2n$,
ignoring terms of order $\O(k)$. All interior triangles can be properly colored with
two colors, black and white. Now add a~new vertex inside each white face and connect
it to all three vertices on the boundary of that face. This adds $f/2 \approx n$
vertices of degree~3 and it increases the degrees of the original $\approx n$ interior
vertices to~9, therefore about a half of the vertices of the new planar graph
has degree~9.

\figure{hexangle.eps}{\epsfxsize}{The construction from Remark~\ref{hexa}}

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

\section{Using Fibonacci heaps}
\id{fibonacci}

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

The previous implementation of the algorithm used a binary heap to store all neighboring
edges of the cut~$\delta(T)$. Instead of that, we will remember the vertices adjacent
to~$T$ and for each such vertex~$v$ we will keep the lightest edge~$uv$ such that $u$~lies
in~$T$. We will call these edges \df{active edges} and keep them in a~heap, ordered by weight.

When we want to extend~$T$ by the lightest edge of~$\delta(T)$, it is sufficient to
find the lightest active edge~$uv$ and add this edge to~$T$ together with a new vertex~$v$.
Then we have to update the active edges as follows. The edge~$uv$ has just ceased to
be active. We scan all neighbors~$w$ of the vertex~$v$. When $w$~is in~$T$, no action
is needed. If $w$~is outside~$T$ and it was not adjacent to~$T$ (there is no active edge
remembered for it so far), we set the edge~$vw$ as active. Otherwise we check the existing
active edge for~$w$ and replace it by~$vw$ if the new edge is lighter.

The following algorithm shows how these operations translate to insertions, decreases
and deletions on the heap.

\algn{Jarn\'\i{}k with active edges; Fredman and Tarjan \cite{ft:fibonacci}}\id{jarniktwo}%
\algo
\algin A~graph~$G$ with an edge comparison oracle.
\:$v_0\=$ an~arbitrary vertex of~$G$.
\:$T\=$ a tree containing just the vertex~$v_0$.
\:$H\=$ a~heap of active edges stored as pairs $(u,v)$ where $u\in T,v\not\in T$, ordered by the weights $w(vw)$, initially empty.
\:$A\=$ an~auxiliary array mapping vertices outside~$T$ to their active edges in the heap; initially all elements undefined.
\:\<Insert> all edges incident with~$v_0$ to~$H$ and update~$A$ accordingly.
\:While $H$ is not empty:
\::$(u,v)\=\<DeleteMin>(H)$.
\::$T\=T+uv$.
\::For all edges $vw$ such that $w\not\in T$:
\:::If there exists an~active edge~$A(w)$:
\::::If $vw$ is lighter than~$A(w)$, \<Decrease> $A(w)$ to~$(v,w)$ in~$H$.
\:::If there is no such edge, then \<Insert> $(v,w)$ to~$H$ and set~$A(w)$.
\algout Minimum spanning tree~$T$.
\endalgo

\thmn{Fibonacci heaps} The~Fibonacci heap performs the following operations
with the indicated amortized time complexities:
\itemize\ibull
\:\<Insert> (insertion of a~new element) in $\O(1)$,
\:\<Decrease> (decreasing value of an~existing element) in $\O(1)$,
\:\<Merge> (merging of two heaps into one) in $\O(1)$,
\:\<DeleteMin> (deletion of the minimal element) in $\O(\log n)$,
\:\<Delete> (deletion of an~arbitrary element) in $\O(\log n)$,
\endlist
\>where $n$ is the maximum number of elements present in the heap at the time of
the operation.

\proof
See Fredman and Tarjan \cite{ft:fibonacci} for both the description of the Fibonacci
heap and the proof of this theorem.
\qed

\thm
Algorithm~\ref{jarniktwo} with a~Fibonacci heap finds the MST of the input graph in time~$\O(m+n\log n)$.

\proof
The algorithm always stops, because every edge enters the heap~$H$ at most once.
As it selects exactly the same edges as the original Jarn\'\i{}k's algorithm,
it gives the correct answer.

The time complexity is $\O(m)$ plus the cost of the heap operations. The algorithm
performs at most one \<Insert> or \<Decrease> per edge and exactly one \<DeleteMin>
per vertex and there are at most $n$ elements in the heap at any given time,
so by the previous theorem the operations take $\O(m+n\log n)$ time in total.
\qed

\cor
For graphs with edge density at least $\log n$, this algorithm runs in linear time.

\rem
We can consider using other kinds of heaps which have the property that inserts
and decreases are faster than deletes. Of course, the Fibonacci heaps are asymptotically
optimal (by the standard $\Omega(n\log n)$ lower bound on sorting by comparisons, see
for example \cite{clrs}), so the other data structures can improve only
multiplicative constants or offer an~easier implementation.

A~nice example is a~\df{$d$-regular heap} --- a~variant of the usual binary heap
in the form of a~complete $d$-regular tree. \<Insert>, \<Decrease> and other operations
involving bubbling the values up spend $\O(1)$ time at a~single level, so they run
in~$\O(\log_d n)$ time. \<Delete> and \<DeleteMin> require bubbling down, which incurs
comparison with all~$d$ sons at every level, so they run in~$\O(d\log_d n)$.
With this structure, the time complexity of the whole algorithm
is $\O(nd\log_d n + m\log_d n)$, which suggests setting $d=m/n$, giving $\O(m\log_{m/n}n)$.
This is still linear for graphs with density at~least~$n^{1+\varepsilon}$.

Another possibility is to use the 2-3-heaps \cite{takaoka:twothree} or Trinomial
heaps \cite{takaoka:trinomial}. Both have the same asymptotic complexity as Fibonacci
heaps (the latter even in worst case, but it does not matter here) and their
authors claim implementation advantages.

\FIXME{Mention Thorup's Fibonacci-like heaps for integers?}

\para
As we already noted, the improved Jarn\'\i{}k's algorithm runs in linear time
for sufficiently dense graphs. In some cases, it is useful to combine it with
another MST algorithm, which identifies a~part of the MST edges and contracts
the graph to increase its density. For example, we can perform several
iterations of the Contractive Bor\o{u}vka's algorithm and find the rest of the
MST by the above version of Jarn\'\i{}k's algorithm.

\algn{Mixed Bor\o{u}vka-Jarn\'\i{}k}
\algo
\algin A~graph~$G$ with an edge comparison oracle.
\:Run $\log\log n$ iterations of the Contractive Bor\o{u}vka's algorithm (\ref{contbor}),
  getting a~MST~$T_1$.
\:Run the Jarn\'\i{}k's algorithm with active edges (\ref{jarniktwo}) on the resulting
  graph, getting a~MST~$T_2$.
\:Combine $T_1$ and~$T_2$ to~$T$ as in the Contraction lemma (\ref{contlemma}).
\algout Minimum spanning tree~$T$.
\endalgo

\thm
The Mixed Bor\o{u}vka-Jarn\'\i{}k algorithm finds the MST of the input graph in time $\O(m\log\log n)$.

\proof
Correctness follows from the Contraction lemma and from the proofs of correctness of the respective algorithms.
As~for time complexity: The first step takes $\O(m\log\log n)$ time
(by Lemma~\ref{contiter}) and it gradually contracts~$G$ to a~graph~$G'$ of size
$m'\le m$ and $n'\le n/\log n$. The second step then runs in time $\O(m'+n'\log n') = \O(m)$
and both trees can be combined in linear time, too.
\qed

\para
Actually, there is a~much better choice of the algorithms to combine: use the
improved Jarn\'\i{}k's algorithm multiple times, each time stopping after a~while.
The good choice of the stopping condition is to place a~limit on the size of the heap.
Start with an~arbitrary vertex, grow the tree as usually and once the heap gets too large,
conserve the current tree and start with a~different vertex and an~empty heap. When this
process runs out of vertices, it has identified a~sub-forest of the MST, so we can
contract the graph along the edges of~this forest and iterate.

\algn{Iterated Jarn\'\i{}k; Fredman and Tarjan \cite{ft:fibonacci}}
\algo
\algin A~graph~$G$ with an edge comparison oracle.
\:$T\=\emptyset$. \cmt{edges of the MST}
\:$\ell(e)\=e$ for all edges~$e$. \cmt{edge labels as usually}
\:$m_0\=m$.
\:While $n>1$: \cmt{We will call iterations of this loop \df{phases}.}
\::$F\=\emptyset$. \cmt{forest built in the current phase}
\::$t\=2^{2m_0/n}$. \cmt{the limit on heap size}
\::While there is a~vertex $v_0\not\in F$:
\:::Run the improved Jarn\'\i{}k's algorithm (\ref{jarniktwo}) from~$v_0$, stop when:
\::::all vertices have been processed, or
\::::a~vertex of~$F$ has been added to the tree, or
\::::the heap had more than~$t$ elements.
\:::Denote the resulting tree~$R$.
\:::$F\=F\cup R$.
\::$T\=T\cup \ell[F]$. \cmt{Remember MST edges found in this phase.}
\::Contract~$G$ along all edges of~$F$ and flatten it.
\algout Minimum spanning tree~$T$.
\endalgo

\nota
For analysis of the algorithm, let us denote the graph entering the $i$-th
phase by~$G_i$ and likewise with the other parameters. The trees from which
$F_i$~has been constructed will be called $R_i^1, \ldots, R_i^{z_i}$. The
non-indexed $G$, $m$ and~$n$ will correspond to the graph given as~input.

\para
However the choice of the parameter~$t$ can seem mysterious, the following
lemma makes the reason clear:

\lemma\id{ijphase}%
The $i$-th phase of the Iterated Jarn\'\i{}k's algorithm runs in time~$\O(m)$.

\proof
During the phase, the heap always contains at most~$t_i$ elements, so it takes
time~$\O(\log t_i)=\O(m/n_i)$ to delete an~element from the heap. The trees~$R_i^j$
are disjoint, so there are at most~$n_i$ \<DeleteMin>'s over the course of the phase.
Each edge is considered at most twice (once per its endpoint), so the number
of the other heap operations is~$\O(m_i)$. Together, it equals $\O(m_i + n_i\log t_i) = \O(m_i+m) = \O(m)$.
\qed

\lemma
Unless the $i$-th phase is final, the forest~$F_i$ consists of at most $2m_i/t_i$ trees.

\proof
As every edge of~$G_i$ is incident with at most two trees of~$F_i$, it is sufficient
to establish that there are at least~$t_i$ edges incident with the vertices of every
such tree~(*).

The forest~$F_i$ evolves by additions of the trees~$R_i^j$. Let us consider the possibilities
how the algorithm could have stopped growing the tree~$R_i^j$:
\itemize\ibull
\:the heap had more than~$t_i$ elements (step~10): since the elements stored in the heap correspond
  to some of the edges incident with vertices of~$R_i^j$, the condition~(*) is fulfilled;
\:the algorithm just added a~vertex of~$F_i$ to~$R_i^j$ (step~9): in this case, an~existing
  tree of~$F_i$ is extended, so the number of edges incident with it cannot decrease;\foot{%
  To make this true, we counted the edges incident with the \em{vertices} of the tree
  instead of edges incident with the tree itself, because we needed the tree edges
  to be counted as well.}
\:all vertices have been processed (step~8): this can happen only in the final phase.
\qeditem
\endlist

\thm\id{itjarthm}%
The Iterated Jarn\'\i{}k's algorithm finds the MST of the input graph in time
$\O(m\timesbeta(m,n))$, where $\beta(m,n):=\min\{ i: \log^{(i)}n < m/n \}$.

\proof
Phases are finite and in every phase at least one edge is contracted, so the outer
loop is eventually terminated. The resulting subgraph~$T$ is equal to $\mst(G)$, because each $F_i$ is
a~subgraph of~$\mst(G_i)$ and the $F_i$'s are glued together according to the Contraction
lemma (\ref{contlemma}).

Let us bound the sizes of the graphs processed in individual phases. As the vertices
of~$G_{i+1}$ correspond to the components of~$F_i$, by the previous lemma $n_{i+1}\le
2m_i/t_i$. Then $t_{i+1} = 2^{2m/n_{i+1}} \ge 2^{2m/(2m_i/t_i)} = 2^{(m/m_i)\cdot t_i} \ge 2^{t_i}$,
therefore:
$$
\left. \vcenter{\hbox{$\displaystyle t_i \ge 2^{2^{\scriptstyle 2^{\scriptstyle\vdots^{\scriptstyle m/n}}}} $}}\;\right\}
\,\hbox{a~tower of~$i$ exponentials.}
$$
As soon as~$t_i\ge n$, the $i$-th phase must be final, because at that time
there is enough space in the heap to process the whole graph. So~there are
at most~$\beta(m,n)$ phases and we already know (Lemma~\ref{ijphase}) that each
phase runs in linear time.
\qed

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

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

\cor
When we use the Iterated Jarn\'\i{}k's algorithm on graphs with edge density
at least~$\log^{(k)} n$ for some $k\in{\bb N}^+$, it runs in time~$\O(km)$.

\proof
If $m/n \ge \log^{(k)} n$, then $\beta(m,n)\le k$.
\qed

\rem
Gabow et al.~\cite{gabow:mst} have shown how to speed this algorithm up 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.}

\FIXME{Reference to Q-Heaps.}

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

\section{Verification of minimality}


\endpart