5 \chapter{Advanced MST Algorithms}
7 \section{Minor-closed graph classes}\id{minorclosed}%
9 The contractive algorithm given in section~\ref{contalg} has been found to perform
10 well on planar graphs, but in the general case its time complexity was not linear.
11 Can we find any broader class of graphs where the algorithm is still efficient?
12 The right context turns out to be the minor-closed graph classes, which are
13 closed under contractions and have bounded density.
16 A~graph~$H$ is a \df{minor} of a~graph~$G$ (written as $H\minorof G$) iff it can be obtained
17 from a~subgraph of~$G$ by a sequence of simple graph contractions (see \ref{simpcont}).
20 A~class~$\cal C$ of graphs is \df{minor-closed}, when for every $G\in\cal C$ and
21 its every minor~$H$, the graph~$H$ lies in~$\cal C$ as well. A~class~$\cal C$ is called
22 \df{non-trivial} if at least one graph lies in~$\cal C$ and at least one lies outside~$\cal C$.
25 Non-trivial minor-closed classes include:
28 \:graphs embeddable in any fixed surface (i.e., graphs of bounded genus),
29 \:graphs embeddable in~${\bb R}^3$ without knots or without interlocking cycles,
30 \:graphs of bounded tree-width or path-width.
34 Many of the nice structural properties of planar graphs extend to
35 minor-closed classes, too (see \cite{lovasz:minors} for a~nice survey
36 of this theory and \cite{diestel:gt} for some of the deeper results).
37 The most important property is probably the characterization
38 of such classes in terms of their forbidden minors.
41 For a~class~$\cal H$ of graphs we define $\Forb({\cal H})$ as the class
42 of graphs which do not contain any of the graphs in~$\cal H$ as a~minor.
43 We will call $\cal H$ the set of \df{forbidden (or excluded) minors} for this class.
44 We will often abbreviate $\Forb(\{M_1,\ldots,M_n\})$ to $\Forb(M_1,\ldots,M_n)$.
47 For every~${\cal H}\ne\emptyset$, the class $\Forb({\cal H})$ is non-trivial
48 and closed on minors. This works in the opposite direction as well: for every
49 minor-closed class~$\cal C$ there is a~class $\cal H$ such that ${\cal C}=\Forb({\cal H})$.
50 One such~$\cal H$ is the complement of~$\cal C$, but smaller ones can be found, too.
51 For example, the planar graphs can be equivalently described as the class $\Forb(K_5, K_{3,3})$
52 --- this follows from the Kuratowski's theorem (the theorem speaks of forbidden
53 subdivisions, but while in general this is not the same as forbidden minors, it
54 is for $K_5$ and $K_{3,3}$). The celebrated theorem by Robertson and Seymour
55 guarantees that we can always find a~finite set of forbidden minors.
57 \thmn{Excluded minors, Robertson \& Seymour \cite{rs:wagner}}
58 For every non-trivial minor-closed graph class~$\cal C$ there exists
59 a~finite set~$\cal H$ of graphs such that ${\cal C}=\Forb({\cal H})$.
62 This theorem has been proven in a~long series of papers on graph minors
63 culminating with~\cite{rs:wagner}. See this paper and follow the references
64 to the previous articles in the series.
68 For analysis of the contractive algorithm,
69 we will make use of another important property --- the bounded density of
70 minor-closed classes. The connection between minors and density dates back to
71 Mader in the 1960's and it can be proven without use of the Robertson-Seymour
75 Let $\cal C$ be a class of graphs. We define its \df{edge density} $\varrho(\cal C)$
76 to be the infimum of all~$\varrho$'s such that $m(G) \le \varrho\cdot n(G)$
77 holds for every $G\in\cal C$.
79 \thmn{Mader \cite{mader:dens}}
80 For every $k\in{\bb N}$ there exists $h(k)\in{\bb R}$ such that every graph
81 of average degree at least~$h(k)$ contains a~subdivision of~$K_{k}$ as a~subgraph.
84 (See Lemma 3.5.1 in \cite{diestel:gt} for a~complete proof in English.)
86 Let us fix~$k$ and prove by induction on~$m$ that every graph of average
87 degree at least~$2^m$ contains a~subdivision of some graph with $k$~vertices
88 and ${k\choose 2}\ge m\ge k$~edges. For $m={k\choose 2}$ the theorem follows
89 as the only graph with~$k$ vertices and~$k\choose 2$ edges is~$K_k$.
91 The base case $m=k$: Let us observe that when the average degree
92 is~$a$, removing any vertex of degree less than~$a/2$ does not decrease the
93 average degree. A~graph with $a\ge 2^k$ therefore has a~subgraph
94 with minimum degree $\delta\ge a/2=2^{k-1}$. Such subgraph contains
95 a~cycle on more than~$\delta$ vertices, in other words a~subdivision of
98 Induction step: Let~$G$ be a~graph with average degree at least~$2^m$ and
99 assume that the theorem already holds for $m-1$. Without loss of generality,
100 $G$~is connected. Consider a~maximal set $U\subseteq V$ such that the subgraph $G[U]$
101 induced by~$U$ is connected and the graph $G.U$ ($G$~with $U$~contracted to
102 a~single vertex) has average degree at least~$2^m$ (such~$U$ exists, because
103 $G=G.U$ whenever $\vert U\vert=1$). Now consider the subgraph~$H$ induced
104 in~$G$ by the neighbors of~$U$. Every $v\in V(H)$ must have $\deg_H(v) \ge 2^{m-1}$,
105 as otherwise we can add this vertex to~$U$, contradicting its
106 maximality. By the induction hypothesis, $H$ contains a~subdivision of some
107 graph~$R$ with $r$~vertices and $m-1$ edges. Any two non-adjacent vertices
108 of~$R$ can be connected in the subdivision by a~path lying entirely in~$G[U]$,
109 which reveals a~subdivision of a~graph with $m$~edges. \qed
111 \thmn{Density of minor-closed classes, Mader~\cite{mader:dens}}
112 Every non-trivial minor-closed class of graphs has finite edge density.
115 Let~$\cal C$ be any such class, $X$~its smallest excluded minor and $x=n(X)$.
116 As $H\minorof K_x$, the class $\cal C$ entirely lies in ${\cal C}'=\Forb(K_x)$, so
117 $\varrho({\cal C}) \le \varrho({\cal C}')$ and therefore it suffices to prove the
118 theorem for classes excluding a~single complete graph~$K_x$.
120 We will show that $\varrho({\cal C})\le 2h(x)$, where $h$~is the function
121 from the previous theorem. If any $G\in{\cal C}$ had more than $2h(x)\cdot n(G)$
122 edges, its average degree would be at least~$h(x)$, so by the previous theorem
123 $G$~would contain a~subdivision of~$K_x$ and hence $K_x$ as a~minor.
127 Minor-closed classes share many other interesting properties, as shown for
128 example by Theorem 6.1 of \cite{nesetril:minors}.
130 \thmn{MST on minor-closed classes \cite{mm:mst}}\id{mstmcc}%
131 For any fixed non-trivial minor-closed class~$\cal C$ of graphs, the Contractive Bor\o{u}vka's
132 algorithm (\ref{contbor}) finds the MST of any graph of this class in time
133 $\O(n)$. (The constant hidden in the~$\O$ depends on the class.)
136 Following the proof for planar graphs (\ref{planarbor}), we denote the graph considered
137 by the algorithm at the beginning of the $i$-th iteration by~$G_i$ and its number of vertices
138 and edges by $n_i$ and $m_i$ respectively. Again the $i$-th phase runs in time $\O(m_i)$
139 and $n_i \le n/2^i$, so it remains to show a linear bound for the $m_i$'s.
141 Since each $G_i$ is produced from~$G_{i-1}$ by a sequence of edge contractions,
142 all $G_i$'s are minors of~$G$.\foot{Technically, these are multigraph contractions,
143 but followed by flattening, so they are equivalent to contractions on simple graphs.}
144 So they also belong to~$\cal C$ and by the previous theorem $m_i\le \varrho({\cal C})\cdot n_i$.
148 The contractive algorithm uses ``batch processing'' to perform many contractions
149 in a single step. It is also possible to perform contractions one edge at a~time,
150 batching only the flattenings. A~contraction of an edge~$uv$ can be done
151 in time~$\O(\deg(u))$ by removing all edges incident with~$u$ and inserting them back
152 with $u$ replaced by~$v$. Therefore we need to find a lot of vertices with small
153 degrees. The following lemma shows that this is always the case in minor-closed
156 \lemman{Low-degree vertices}\id{lowdeg}%
157 Let $\cal C$ be a graph class with density~$\varrho$ and $G\in\cal C$ a~graph
158 with $n$~vertices. Then at least $n/2$ vertices of~$G$ have degree at most~$4\varrho$.
161 Assume the contrary: Let there be at least $n/2$ vertices with degree
162 greater than~$4\varrho$. Then $\sum_v \deg(v) > n/2
163 \cdot 4\varrho = 2\varrho n$, which is in contradiction with the number
164 of edges being at most $\varrho n$.
168 The proof can be also viewed
169 probabilistically: let $X$ be the degree of a vertex of~$G$ chosen uniformly at
170 random. Then ${\bb E}X \le 2\varrho$, hence by the Markov's inequality
171 ${\rm Pr}[X > 4\varrho] < 1/2$, so for at least $n/2$ vertices~$v$ we have
172 $\deg(v)\le 4\varrho$.
174 \algn{Local Bor\o{u}vka's Algorithm \cite{mm:mst}}%
176 \algin A~graph~$G$ with an edge comparison oracle and a~parameter~$t\in{\bb N}$.
178 \:$\ell(e)\=e$ for all edges~$e$.
180 \::While there exists a~vertex~$v$ such that $\deg(v)\le t$:
181 \:::Select the lightest edge~$e$ incident with~$v$.
182 \:::Contract~$G$ along~$e$.
183 \:::$T\=T + \ell(e)$.
184 \::Flatten $G$, removing parallel edges and loops.
185 \algout Minimum spanning tree~$T$.
189 When $\cal C$ is a minor-closed class of graphs with density~$\varrho$, the
190 Local Bor\o{u}vka's Algorithm with the parameter~$t$ set to~$4\varrho$
191 finds the MST of any graph from this class in time $\O(n)$. (The constant
192 in the~$\O$ depends on~the class.)
195 Let us denote by $G_i$, $n_i$ and $m_i$ the graph considered by the
196 algorithm at the beginning of the $i$-th iteration of the outer loop,
197 and the number of its vertices and edges respectively. As in the proof
198 of the previous algorithm (\ref{mstmcc}), we observe that all the $G_i$'s
199 are minors of the graph~$G$ given as the input.
201 For the choice $t=4\varrho$, the Lemma on low-degree vertices (\ref{lowdeg})
202 guarantees that at the beginning of the $i$-th iteration, at least $n_i/2$ vertices
203 have degree at most~$t$. Each selected edge removes one such vertex and
204 possibly increases the degree of another, so at least $n_i/4$ edges get selected.
205 Hence $n_i\le 3/4\cdot n_{i-1}$ and therefore $n_i\le n\cdot (3/4)^i$ and the
206 algorithm terminates after $\O(\log n)$ iterations.
208 Each selected edge belongs to $\mst(G)$, because it is the lightest edge of
209 the trivial cut $\delta(v)$ (see the Blue Rule in \ref{rbma}).
210 The steps 6 and~7 therefore correspond to the operation
211 described by the Lemma on contraction of MST edges (\ref{contlemma}) and when
212 the algorithm stops, $T$~is indeed the minimum spanning tree.
214 It remains to analyse the time complexity of the algorithm. Since $G_i\in{\cal C}$, we have
215 $m_i\le \varrho n_i \le \varrho n/2^i$.
216 We will show that the $i$-th iteration is carried out in time $\O(m_i)$.
217 Steps 5 and~6 run in time $\O(\deg(v))=\O(t)$ for each~$v$, so summed
218 over all $v$'s they take $\O(tn_i)$, which is linear for a fixed class~$\cal C$.
219 Flattening takes $\O(m_i)$, as already noted in the analysis of the Contracting
220 Bor\o{u}vka's Algorithm (see \ref{contiter}).
222 The whole algorithm therefore runs in time $\O(\sum_i m_i) = \O(\sum_i n/2^i) = \O(n)$.
226 For planar graphs, we can get a sharper version of the low-degree lemma,
227 showing that the algorithm works with $t=8$ as well (we had $t=12$ as
228 $\varrho=3$). While this does not change the asymptotic time complexity
229 of the algorithm, the constant-factor speedup can still delight the hearts of
232 \lemman{Low-degree vertices in planar graphs}%
233 Let $G$ be a planar graph with $n$~vertices. Then at least $n/2$ vertices of~$v$
234 have degree at most~8.
237 It suffices to show that the lemma holds for triangulations (if there
238 are any edges missing, the situation can only get better) with at
239 least 3 vertices. Since $G$ is planar, $\sum_v \deg(v) < 6n$.
240 The numbers $d(v):=\deg(v)-3$ are non-negative and $\sum_v d(v) < 3n$,
241 so by the same argument as in the proof of the general lemma, for at least $n/2$
242 vertices~$v$ it holds that $d(v) < 6$, hence $\deg(v) \le 8$.
246 The constant~8 in the previous lemma is the best we can have.
247 Consider a $k\times k$ triangular grid. It has $n=k^2$ vertices, $\O(k)$ of them
248 lie on the outer face and have degrees at most~6, the remaining $n-\O(k)$ interior
249 vertices have degree exactly~6. Therefore the number of faces~$f$ is $6/3\cdot n=2n$,
250 ignoring terms of order $\O(k)$. All interior triangles can be properly colored with
251 two colors, black and white. Now add a~new vertex inside each white face and connect
252 it to all three vertices on the boundary of that face. This adds $f/2 \approx n$
253 vertices of degree~3 and it increases the degrees of the original $\approx n$ interior
254 vertices to~9, therefore about a half of the vertices of the new planar graph
257 \figure{hexangle.eps}{\epsfxsize}{The construction from Remark~\ref{hexa}}
260 The observation in~Theorem~\ref{mstmcc} was also made by Gustedt in~\cite{gustedt:parallel},
261 who studied a~parallel version of the contractive Bor\o{u}vka's algorithm applied
262 to minor-closed classes.
264 %--------------------------------------------------------------------------------
266 \section{Using Fibonacci heaps}
269 We have seen that the Jarn\'\i{}k's Algorithm \ref{jarnik} runs in $\Theta(m\log n)$ time.
270 Fredman and Tarjan have shown a~faster implementation in~\cite{ft:fibonacci}
271 using their Fibonacci heaps. In this section, we convey their results and we
272 show several interesting consequences.
274 The previous implementation of the algorithm used a binary heap to store all edges
275 separating the current tree~$T$ from the rest of the graph, i.e., edges of the cut~$\delta(T)$.
276 Instead of that, we will remember the vertices adjacent to~$T$ and for each such vertex~$v$ we
277 will maintain the lightest edge~$uv$ such that $u$~lies in~$T$. We will call these edges \df{active edges}
278 and keep them in a~Fibonacci heap, ordered by weight.
280 When we want to extend~$T$ by the lightest edge of~$\delta(T)$, it is sufficient to
281 find the lightest active edge~$uv$ and add this edge to~$T$ together with the new vertex~$v$.
282 Then we have to update the active edges as follows. The edge~$uv$ has just ceased to
283 be active. We scan all neighbors~$w$ of the vertex~$v$. When $w$~is in~$T$, no action
284 is needed. If $w$~is outside~$T$ and it was not adjacent to~$T$ (there is no active edge
285 remembered for it so far), we set the edge~$vw$ as active. Otherwise we check the existing
286 active edge for~$w$ and replace it by~$vw$ if the new edge is lighter.
288 The following algorithm shows how these operations translate to insertions, decreases
289 and deletions on the heap.
291 \algn{Active Edge Jarn\'\i{}k; Fredman and Tarjan \cite{ft:fibonacci}}\id{jarniktwo}%
293 \algin A~graph~$G$ with an edge comparison oracle.
294 \:$v_0\=$ an~arbitrary vertex of~$G$.
295 \:$T\=$ a tree containing just the vertex~$v_0$.
296 \:$H\=$ 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.
297 \:$A\=$ a~mapping of vertices outside~$T$ to their active edges in the heap; initially all elements undefined.
298 \:\<Insert> all edges incident with~$v_0$ to~$H$ and update~$A$ accordingly.
299 \:While $H$ is not empty:
300 \::$(u,v)\=\<DeleteMin>(H)$.
302 \::For all edges $vw$ such that $w\not\in T$:
303 \:::If there exists an~active edge~$A(w)$:
304 \::::If $vw$ is lighter than~$A(w)$, \<Decrease> $A(w)$ to~$(v,w)$ in~$H$.
305 \:::If there is no such edge, then \<Insert> $(v,w)$ to~$H$ and set~$A(w)$.
306 \algout Minimum spanning tree~$T$.
310 To analyze the time complexity of this algorithm, we will use the standard
311 theorem on~complexity of the Fibonacci heap:
313 \thmn{Fibonacci heaps} The~Fibonacci heap performs the following operations
314 with the indicated amortized time complexities:
316 \:\<Insert> (insertion of a~new element) in $\O(1)$,
317 \:\<Decrease> (decreasing value of an~existing element) in $\O(1)$,
318 \:\<Merge> (merging of two heaps into one) in $\O(1)$,
319 \:\<DeleteMin> (deletion of the minimal element) in $\O(\log n)$,
320 \:\<Delete> (deletion of an~arbitrary element) in $\O(\log n)$,
322 \>where $n$ is the number of elements present in the heap at the time of
326 See Fredman and Tarjan \cite{ft:fibonacci} for both the description of the Fibonacci
327 heap and the proof of this theorem.
331 Algorithm~\ref{jarniktwo} with the Fibonacci heap finds the MST of the input graph in time~$\O(m+n\log n)$.
334 The algorithm always stops, because every edge enters the heap~$H$ at most once.
335 As it selects exactly the same edges as the original Jarn\'\i{}k's algorithm,
336 it gives the correct answer.
338 The time complexity is $\O(m)$ plus the cost of the heap operations. The algorithm
339 performs at most one \<Insert> or \<Decrease> per edge and exactly one \<DeleteMin>
340 per vertex. There are at most $n$ elements in the heap at any given time,
341 thus by the previous theorem the operations take $\O(m+n\log n)$ time in total.
345 For graphs with edge density at least $\log n$, this algorithm runs in linear time.
348 We can consider using other kinds of heaps which have the property that inserts
349 and decreases are faster than deletes. Of course, the Fibonacci heaps are asymptotically
350 optimal (by the standard $\Omega(n\log n)$ lower bound on sorting by comparisons, see
351 for example \cite{clrs}), so the other data structures can improve only
352 multiplicative constants or offer an~easier implementation.
354 A~nice example is a~\df{$d$-regular heap} --- a~variant of the usual binary heap
355 in the form of a~complete $d$-regular tree. \<Insert>, \<Decrease> and other operations
356 involving bubbling the values up spend $\O(1)$ time at a~single level, so they run
357 in~$\O(\log_d n)$ time. \<Delete> and \<DeleteMin> require bubbling down, which incurs
358 comparison with all~$d$ sons at every level, so they spend $\O(d\log_d n)$.
359 With this structure, the time complexity of the whole algorithm
360 is $\O(nd\log_d n + m\log_d n)$, which suggests setting $d=m/n$, yielding $\O(m\log_{m/n}n)$.
361 This is still linear for graphs with density at~least~$n^{1+\varepsilon}$.
363 Another possibility is to use the 2-3-heaps \cite{takaoka:twothree} or Trinomial
364 heaps \cite{takaoka:trinomial}. Both have the same asymptotic complexity as Fibonacci
365 heaps (the latter even in the worst case, but it does not matter here) and their
366 authors claim faster implementation. For integer weights, we can use Thorup's priority
367 queues described in \cite{thorup:pqsssp} which have constant-time \<Insert> and \<Decrease>
368 and $\O(\log\log n)$ time \<DeleteMin>. (We will however omit the details since we will
369 show a~faster integer algorithm soon.)
372 As we already noted, the improved Jarn\'\i{}k's algorithm runs in linear time
373 for sufficiently dense graphs. In some cases, it is useful to combine it with
374 another MST algorithm, which identifies a~part of the MST edges and contracts
375 the graph to increase its density. For example, we can perform several
376 iterations of the Contractive Bor\o{u}vka's algorithm and find the rest of the
377 MST by the Active Edge Jarn\'\i{}k's algorithm.
379 \algn{Mixed Bor\o{u}vka-Jarn\'\i{}k}
381 \algin A~graph~$G$ with an edge comparison oracle.
382 \:Run $\log\log n$ iterations of the Contractive Bor\o{u}vka's algorithm (\ref{contbor}),
384 \:Run the Active Edge Jarn\'\i{}k's algorithm (\ref{jarniktwo}) on the resulting
385 graph, getting a~MST~$T_2$.
386 \:Combine $T_1$ and~$T_2$ to~$T$ as in the Contraction lemma (\ref{contlemma}).
387 \algout Minimum spanning tree~$T$.
391 The Mixed Bor\o{u}vka-Jarn\'\i{}k algorithm finds the MST of the input graph in time $\O(m\log\log n)$.
394 Correctness follows from the Contraction lemma and from the proofs of correctness of the respective algorithms.
395 As~for time complexity: The first step takes $\O(m\log\log n)$ time
396 (by Lemma~\ref{contiter}) and it gradually contracts~$G$ to a~graph~$G'$ of size
397 $m'\le m$ and $n'\le n/\log n$. The second step then runs in time $\O(m'+n'\log n') = \O(m)$
398 and both trees can be combined in linear time, too.
402 Actually, there is a~much better choice of the algorithms to combine: use the
403 Active Edge Jarn\'\i{}k's algorithm multiple times, each time stopping after a~while.
404 A~good choice of the stopping condition is to place a~limit on the size of the heap.
405 We start with an~arbitrary vertex, grow the tree as usually and once the heap gets too large,
406 we conserve the current tree and start with a~different vertex and an~empty heap. When this
407 process runs out of vertices, it has identified a~sub-forest of the MST, so we can
408 contract the graph along the edges of~this forest and iterate.
410 \algn{Iterated Jarn\'\i{}k; Fredman and Tarjan \cite{ft:fibonacci}}
412 \algin A~graph~$G$ with an edge comparison oracle.
413 \:$T\=\emptyset$. \cmt{edges of the MST}
414 \:$\ell(e)\=e$ for all edges~$e$. \cmt{edge labels as usually}
416 \:While $n>1$: \cmt{We will call iterations of this loop \df{phases}.}
417 \::$F\=\emptyset$. \cmt{forest built in the current phase}
418 \::$t\=2^{\lceil 2m_0/n \rceil}$. \cmt{the limit on heap size}
419 \::While there is a~vertex $v_0\not\in F$:
420 \:::Run the Active Edge Jarn\'\i{}k's algorithm (\ref{jarniktwo}) from~$v_0$, stop when:
421 \::::all vertices have been processed, or
422 \::::a~vertex of~$F$ has been added to the tree, or
423 \::::the heap has grown to more than~$t$ elements.
424 \:::Denote the resulting tree~$R$.
426 \::$T\=T\cup \ell[F]$. \cmt{Remember MST edges found in this phase.}
427 \::Contract~$G$ along all edges of~$F$ and flatten it.
428 \algout Minimum spanning tree~$T$.
432 For analysis of the algorithm, let us denote the graph entering the $i$-th
433 phase by~$G_i$ and likewise with the other parameters. Let the trees from which
434 $F_i$~has been constructed be called $R_i^1, \ldots, R_i^{z_i}$. The
435 non-indexed $G$, $m$ and~$n$ will correspond to the graph given as~input.
438 However the choice of the parameter~$t$ can seem mysterious, the following
439 lemma makes the reason clear:
442 The $i$-th phase of the Iterated Jarn\'\i{}k's algorithm runs in time~$\O(m)$.
445 During the phase, the heap always contains at most~$t_i$ elements, so it takes
446 time~$\O(\log t_i)=\O(m/n_i)$ to delete an~element from the heap. The trees~$R_i^j$
447 are edge-disjoint, so there are at most~$n_i$ \<DeleteMin>'s over the course of the phase.
448 Each edge is considered at most twice (once per its endpoint), so the number
449 of the other heap operations is~$\O(m_i)$. Together, it equals $\O(m_i + n_i\log t_i) = \O(m_i+m) = \O(m)$.
453 Unless the $i$-th phase is final, the forest~$F_i$ consists of at most $2m_i/t_i$ trees.
456 As every edge of~$G_i$ is incident with at most two trees of~$F_i$, it is sufficient
457 to establish that there are at least~$t_i$ edges incident with every such tree, including
458 connecting two vertices of the tree.
460 The forest~$F_i$ evolves by additions of the trees~$R_i^j$. Let us consider the possibilities
461 how the algorithm could have stopped growing the tree~$R_i^j$:
463 \:the heap had more than~$t_i$ elements (step~10): since the each elements stored in the heap
464 corresponds to a~unique edges incident with~$R_i^j$, we have enough such edges;
465 \:the algorithm just added a~vertex of~$F_i$ to~$R_i^j$ (step~9): in this case, an~existing
466 tree of~$F_i$ is extended, so the number of edges incident with it cannot decrease;\foot{%
467 This is the place where we needed to count the interior edges as well.}
468 \:all vertices have been processed (step~8): this can happen only in the final phase.
473 The Iterated Jarn\'\i{}k's algorithm finds the MST of the input graph in time
474 $\O(m\timesbeta(m,n))$, where $\beta(m,n):=\min\{ i: \log^{(i)}n \le m/n \}$.
477 Phases are finite and in every phase at least one edge is contracted, so the outer
478 loop is eventually terminated. The resulting subgraph~$T$ is equal to $\mst(G)$, because each $F_i$ is
479 a~subgraph of~$\mst(G_i)$ and the $F_i$'s are glued together according to the Contraction
480 lemma (\ref{contlemma}).
482 Let us bound the sizes of the graphs processed in the individual phases. As the vertices
483 of~$G_{i+1}$ correspond to the components of~$F_i$, by the previous lemma $n_{i+1}\le
484 2m_i/t_i$. Then $t_{i+1} = 2^{\lceil 2m/n_{i+1} \rceil} \ge 2^{2m/n_{i+1}} \ge 2^{2m/(2m_i/t_i)} = 2^{(m/m_i)\cdot t_i} \ge 2^{t_i}$,
487 \left. \vcenter{\hbox{$\displaystyle t_i \ge 2^{2^{\scriptstyle 2^{\scriptstyle\rddots^{\scriptstyle m/n}}}} $}}\;\right\}
488 \,\hbox{a~tower of~$i$ exponentials.}
490 As soon as~$t_i\ge n$, the $i$-th phase must be final, because at that time
491 there is enough space in the heap to process the whole graph. So~there are
492 at most~$\beta(m,n)$ phases and we already know (Lemma~\ref{ijphase}) that each
493 phase runs in linear time.
497 The Iterated Jarn\'\i{}k's algorithm runs in time $\O(m\log^* n)$.
500 $\beta(m,n) \le \beta(1,n) = \log^* n$.
504 When we use the Iterated Jarn\'\i{}k's algorithm on graphs with edge density
505 at least~$\log^{(k)} n$ for some $k\in{\bb N}^+$, it runs in time~$\O(km)$.
508 If $m/n \ge \log^{(k)} n$, then $\beta(m,n)\le k$.
512 The algorithm spends most of the time in phases which have small heaps. Once the
513 heap grows to $\Omega(\log^{(k)} n)$ for any fixed~$k$, the graph gets dense enough
514 to guarantee that at most~$k$ phases remain. This means that if we are able to
515 construct a~heap of size $\Omega(\log^{(k)} n)$ with constant time per operation,
516 we can get a~linear-time algorithm for MST. This is the case when the weights are
519 \thmn{MST for graphs with integer weights, Fredman and Willard \cite{fw:transdich}}\id{intmst}%
520 MST of a~graph with integer edge weights can be found in time $\O(m)$ on the Word-RAM.
523 We will combine the Iterated Jarn\'\i{}k's algorithm with the Q-heaps from section \ref{qheaps}.
524 We modify the first pass of the algorithm to choose $t=\log n$ and use the Q-heap tree instead
525 of the Fibonacci heap. From Theorem \ref{qh} and Remark \ref{qhtreerem} we know that the
526 operations on the Q-heap tree run in constant time, so the modified first phase takes time~$\O(m)$.
527 Following the analysis of the original algorithm in the proof of Theorem \ref{itjarthm} we obtain
528 $t_2\ge 2^{t_1} = 2^{\log n} = n$, so the algorithm stops after the second phase.\foot{%
529 Alternatively, we can use the Q-heaps directly with $k=\log^{1/4}n$ and then stop
530 after the third phase.}
534 Gabow et al.~\cite{gabow:mst} have shown how to speed up the Iterated Jarn\'\i{}k's algorithm to~$\O(m\log\beta(m,n))$.
535 They split the adjacency lists of the vertices to small buckets, keep each bucket
536 sorted and consider only the lightest edge in each bucket until it is removed.
537 The mechanics of the algorithm is complex and there is a~lot of technical details
538 which need careful handling, so we omit the description of this algorithm.
540 \FIXME{Reference to Chazelle.}
542 %--------------------------------------------------------------------------------
544 \section{Verification of minimality}
546 Now we will turn our attention to a~slightly different problem: given a~spanning
547 tree, how to verify that it is minimum? We will show that this can be achieved
548 in linear time and it will serve as a~basis for the randomized linear-time
549 MST algorithm in the next section.
551 MST verification has been studied by Koml\'os \cite{komlos:verify}, who has
552 proven that $\O(m)$ edge comparisons are sufficient, but his algorithm needed
553 superlinear time to find the edges to compare. Dixon, Rauch and Tarjan
554 have later shown in \cite{dixon:verify} that the overhead can be reduced
555 to linear time on the RAM using preprocessing and table lookup on small
556 subtrees. This algorithm was then simplified by King in \cite{king:verifytwo}.
557 We will follow the results of Koml\'os and King, but as we have the power of the
558 RAM data structures from Section~\ref{bitsect} at our command, we will not have
559 to worry about much technical details.
561 To verify that a~spanning~$T$ is minimum, it is sufficient to check that all
562 edges outside~$T$ are $T$-heavy (by Theorem \ref{mstthm}). In fact, we will be
563 able to find all $T$-light edges efficiently. For each edge $uv\in E\setminus T$,
564 we will find the heaviest edge of the tree path $T[u,v]$ and compare its weight
565 to $w(uv)$. It is therefore sufficient to solve this problem:
568 Given a~weighted tree~$T$ and a~set of \df{query paths} $Q \subseteq \{ T[u,v] ; u,v\in V(T) \}$
569 specified by their endpoints, find the heaviest edge for every path in~$Q$.
572 Finding the heaviest edges can be burdensome if the tree~$T$ is degenerated,
573 so we will first reduce it to the same problem on a~balanced tree. We run
574 the Bor\o{u}vka's algorithm on~$T$, which certainly produces $T$ itself, and we
575 record the order in which the subtrees have been merged in another tree~$B(T)$.
576 The queries on~$T$ can be then easily translated to queries on~$B(T)$.
579 For a~weighted tree~$T$ we define its \df{Bor\o{u}vka tree} $B(T)$ as a~rooted tree which records
580 the execution of the Bor\o{u}vka's algorithm run on~$T$. The leaves of $B(T)$
581 are exactly the vertices of~$T$, every internal vertex~$v$ at level~$i$ from the bottom
582 corresponds to a~component tree~$C(v)$ formed in the $i$-th phase of the algorithm. When
583 a~tree $C(v)$ selects an adjacent edge~$xy$ and gets merged with some other trees to form
584 a~component $C(u)$, we add an~edge $uv$ to~$B(T)$ and set its weight to $w(xy)$.
586 \figure{bortree.eps}{\epsfxsize}{An octipede and its Bor\o{u}vka tree}
589 As the algorithm finishes with a~single component after the last phase, the Bor\o{u}vka tree
590 is really a~tree. All its vertices are on the same level and each internal vertex has
591 at least two sons. Such trees will be called \df{complete branching trees.}
594 For every tree~$T$ and every pair of its vertices $x,y\in V(T)$, the heaviest edge
595 of the path $T[x,y]$ has the same weight as the heaviest edge of~the path
599 Let us denote the path $T[x,y]$ by~$P$ and its heaviest edge by~$h=ab$. Similarly,
600 let us use $P'$ for $B(T)[x,y]$ and $h'$ for the heaviest edge of~$P'$.
602 We will first prove that~$h$ has its counterpart of the same weight in~$P'$,
603 so $w(h') \ge w(h)$. Consider the lowest vertex $u$ of~$T$ such that the
604 component $C(u)$ contains both $a$ and~$b$, and consider the sons $v_a$ and $v_b$ of~$u$
605 for which $a\in C(v_a)$ and $b\in C(v_b)$. As the edge~$h$ must have been
606 selected by at least one of these components, assume without loss of generality that
607 it was $C(v_a)$, and hence we have $w(uv_a)=w(h)$. We will show that the
608 edge~$uv_a$ lies in~$P'$, because exactly one of the endpoints of~$h$ lies
609 in~$C(v_a)$. Both endpoints cannot lie there, since it would imply that $C(v_a)$,
610 being connected, contains the whole path~$P$, including~$h$. On the other hand,
611 if $C(v_a)$ contained neither~$x$ nor~$y$, it would have to be incident with
612 another edge of~$P$ different from~$h$, so this lighter edge would be selected
615 For the other direction: for any edge~$uv\in P'$, the tree~$C(v)$ is incident
616 with at least one edge of~$P$, so the selected edge must be lighter or equal
617 to this edge and hence also than~$h$.
621 We will make one more simplification: For an~arbitrary tree~$T$, we split each
622 query path $T[x,y]$ to two half-paths $T[x,a]$ and $T[a,y]$ where~$a$ is the
623 \df{lowest common ancestor} of~$x$ and~$y$ in~$T$. It is therefore sufficient to
624 consider only paths which connect a~vertex with one of its ancestors.
626 Finding the common ancestors is not trivial, but Harel and Tarjan have shown
627 in \cite{harel:nca} that linear time is sufficient on the RAM. Several more
628 accessible algorithms have been developed since then (see the Alstrup's survey
629 paper \cite{alstrup:nca} and a~particularly elegant algorithm shown by Bender
630 and Falach-Colton in \cite{bender:lca}). Any of them implies the following
633 \thmn{Lowest common ancestors}\id{lcathm}%
634 On the RAM, it is possible to preprocess a~tree~$T$ in time $\O(n)$ and then
635 answer lowest common ancestor queries presented online in constant time.
638 See for example Bender and Falach-Colton \cite{bender:lca}.
642 We summarize the reductions in the following lemma, also showing that they can
643 be performed in linear time:
645 \lemma\id{verbranch}%
646 For each tree~$T$ and a~set of query paths~$Q$ on~$T$, it is possible to find
647 a~complete branching tree~$T'$ in linear time together with a~set~$Q'$ of query
648 paths on~$T'$, such that the weights of the heaviest edges of the new paths can
649 be transformed to the weights of the heaviest edges of the paths in~$Q$ in
651 The tree $T$ has at most $2n(T)$ vertices and $\O(\log n(T))$ levels. The set~$Q'$ contains
652 at most~$2\vert Q\vert$ paths and each of them connects a~vertex of~$T'$ with one
656 The tree~$T'$ will be the Bor\o{u}vka tree for~$T$. We run the contractive version
657 of the Bor\o{u}vka's algorithm (Algorithm \ref{contbor}) on~$T$. It runs in linear time,
658 for example because trees are planar (Theorem \ref{planarbor}). The construction of~$T'$
659 itself adds only a~constant overhead to every step of the algorithm. As~$T'$ has~$m(T)=n(T)-1$
660 leaves and it is a~complete branching tree, it has at most~$m(T)$ internal vertices,
661 so~$n(T')\le 2n(T)$ as promised. Since all internal vertices have at least two sons,
662 the depth must be logarithmic.
664 For each query path $T[x,y]$ we find the lowest common ancestor of~$x$ and~$y$
665 using Theorem \ref{lcathm} and replace the path by the two half-paths. This
666 produces a~set~$Q'$ of at most~$2\vert Q\vert$ paths. If we remember the origin
667 of each of the new paths, the reconstruction of answers to the original queries
672 We will now describe a~simple variant of depth-first search which finds the
673 maximum-weight edges for all query paths of the transformed problem. For the
674 time being, we will not care about the time complexity of the algorithm (as long
675 as it is polynomial) and we will minimize only the number of edge weight comparisons
678 For every edge~$e=uv$, we consider the set $Q_e$ of all query paths containing~$e$.
679 The vertex of a~path, which is closer to the root, will be called its \df{top,}
680 the other vertex its \df{bottom.}
681 We define arrays $T_e$ and~$H_e$ as follows: $T_e$ contains
682 the tops of the paths in~$Q_e$ in order of their increasing depth (we
683 will call them \df{active tops} and each of them will be stored exactly once). For
684 each active top~$t=T_e[i]$, we define $H_e[i]$ as the heaviest edge of the path $T[v,t]$.
685 As for every~$i$ the path $T[v,T_e[i+1]]$ is contained within $T[v,T_e[i]]$,
686 the edges in~$H_e$ must have non-increasing weights, that is $w(H_e[i+1]) \le
689 \alg $\<FindHeavy>(u,p,T_p,H_p)$ --- process all queries in the subtree rooted
690 at~$u$ entered from its parent via an~edge~$p$.
694 \:Process all query paths whose bottom is~$u$. For each of them, find the top of
695 the path in~$T_p$ and record the heaviest edge remembered for it in~$H_p$.
697 \:First find the tops~$T$ which will be shared by all edges going from~$u$ downwards.
698 These are the tops from~$T_p$ except for the ones which have ceased to be active,
699 because all query paths which were referring to them have~$u$ as their bottom.
700 Select the corresponding array of the heaviest edges~$H$ from~$H_p$.
702 \:For every son~$v$ of~$u$, do:
704 \::Construct the array of tops~$T_e$ for the edge $e=uv$. It contains all tops
705 from~$T$ and possibly also the vertex~$u$ itself if there is a~query path
706 which has~$u$ as its top and which has bottom somewhere in the subtree
709 \::Construct the array of the heaviest edges~$H_e$. For all tops from~$T$, just compare
710 the weight of the edge recorded in~$H$ with $w(e)$ and if it was smaller,
711 replace the remembered edge by~$e$. Since $H_p$ was sorted, so is~$H$
712 and we can use binary search to locate the boundary between lighter and
713 heavier edges in~$H$.
714 For the new top~$u$, the edge~$e$ itself is obviously the heaviest.
716 \::Recurse on~$v$: call $\<FindHeavy>(v,e,T_e,H_e)$.
719 \>As we need a~parent edge to start the recursion, we add an~imaginary parent
720 edge~$p_0$ of the root vertex~$r$, for which no queries are defined. We can
721 therefore start with $\<FindHeavy>(r,p_0,\emptyset,\emptyset)$.
723 Let us account for the comparisons:
725 \lemma\id{vercompares}%
726 When the procedure \<FindHeavy> is called on the transformed problem, it
727 performs $\O(n+m)$ comparisons, where $n$ is the size of the tree and
728 $m$ is the number of query paths.
731 We will calculate the number of comparisons~$c_i$ performed when processing the edges
732 going from the $(i+1)$-th to the $i$-th level of the tree.
733 The levels are numbered from the bottom, so leaves are at level~0, the root
734 is at level $\ell\le \lceil \log_2 n\rceil$ and there are $n_i\le n/2^i$ vertices
735 at the $i$-th level, so we consider exactly $n_i$ edges.
736 \def\eqalign#1{\null\,\vcenter{\openup\jot
737 \ialign{\strut\hfil$\displaystyle{##}$&$\displaystyle{{}##}$\hfil
739 $$\vcenter{\openup\jot\halign{\strut\hfil $\displaystyle{#}$&$\displaystyle{{}#}$\hfil&\quad#\hfil\cr
740 c_i &\le \sum_e \left( 1 + \log \vert T_e\vert \right)&(Total cost of the binary search.)\cr
741 &\le n_i + \sum_e \log\vert T_e\vert&(We sum over $n_i$ edges.)\cr
742 &\le n_i + n_i \cdot {\sum_e \log\vert T_e\vert \over n_i}&(Consider the average of logarithms.) \cr
743 &\le n_i + n_i \cdot \log{\sum_e \vert T_e\vert \over n_i}&(Logarithm is concave.) \cr
744 &\le n_i + n_i \cdot \log{m\over n_i}&(Bound the number of tops by queries.) \cr
745 &= n_i \cdot \left( 1 + \log\left({m\over n}\cdot{n\over n_i}\right) \right)\cr
746 &= n_i + n_i\log{m\over n} + n_i\log{n\over n_i}.\cr
748 Summing over all levels, we estimate the total number of comparisons as:
750 c = \sum_i c_i = \left( \sum_i n_i \right) + \left( \sum_i n_i \log{m\over n}\right) + \left( \sum_i n_i \log{n\over n_i}\right).
752 The first part is equal to~$n$, the second one to $n\log(m/n)\le m$. For the third
753 one, we would like to plug in $n_i \le n/2^i$, but we unfortunately have one~$n_i$
754 in the denominator. We save the situation by observing that the function $f(x)=x\log(n/x)$
755 is decreasing\foot{We can easily check the derivative: $f(x)=(x\ln n-x\ln x)/\ln 2$, so $f'(x)\cdot \ln2 =
756 \ln n - \ln x - 1$. We want $f'(x)<0$ and therefore $\ln x > \ln n - 1$, i.e., $x > n/e$.}
757 for $x > n/e$, so except for $i=0$ and $i=1$ it holds that:
759 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.
761 We can therefore rewrite the third part as:
763 \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
764 &\le n\log1 + n_1\cdot {n\over n_1} + n\cdot\O(1) = \O(n).\cr
766 Putting all three parts together, we conclude that:
768 c \le n + m + \O(n) = \O(n+m).
773 When we combine this lemma with the above reduction, we get the following theorem:
775 \thmn{Verification of the MST, Koml\'os \cite{komlos:verify}}
776 For every weighted graph~$G$ and its spanning tree~$T$, it is sufficient to
777 perform $\O(m)$ comparisons of edge weights to determine whether~$T$ is minimum
778 and to find all $T$-light edges in~$G$.
781 We first transform the problem to finding the heaviest edges for a~set
782 of query paths in~$T$ (these are exactly the paths covered by the edges
783 of $G\setminus T$). We use the reduction from Lemma \ref{verbranch} to get
784 an~equivalent problem with a~full branching tree and a~set of parent-descendant
785 paths. Then we run the \<FindHeavy> procedure (\ref{findheavy}) to find
786 the heaviest edges and we employ Lemma \ref{vercompares} to bound the number
791 We will now show an~efficient implementation of \<FindHeavy>, which will
792 run in linear time on the RAM.
801 %--------------------------------------------------------------------------------
803 \section{Special cases and related problems}
805 Finally, we will focus our attention on various special cases of the minimum
806 spanning tree problem which frequently arise in practice.
808 \examplen{Graphs with sorted edges}
809 When the edges are already sorted by their weights, we can use the Kruskal's
810 algorithm to find the MST in time $\O(m\timesalpha(n))$ (Theorem \ref{kruskal}).
811 We however can do better: As the minimality of a~spanning tree depends only on the
812 order of weights and not on the actual values (Theorem \ref{mstthm}), we can
813 renumber the weights to $1, \ldots, m$ and find the MST using the Fredman-Willard
814 algorithm for integer weights. According to Theorem \ref{intmst} it runs in
815 time $\O(m)$ on the Word-RAM.
817 \examplen{Graphs with a~small number of distinct weights}
818 When the weights of edges are drawn from a~set of a~fixed size~$U$, we can
819 sort them in linear time and so reduce the problem to the previous case.
820 A~more practical way is to use the Jarn\'\i{}k's algorithm (\ref{jarnimpl}),
821 but replace the heap by an~array of $U$~buckets. As the number of buckets
822 is constant, we can find the minimum in constant time and hence the whole
823 algorithm runs in time $\O(m)$, even on the Pointer Machine. For large
824 values of~$U,$ we can build a~binary search tree or the van Emde-Boas
825 tree (see Section \ref{ramdssect} and \cite{boas:vebt}) on the top of the buckets to bring the complexity
826 of finding the minimum down to $\O(\log U)$ or $\O(\log\log U)$ respectively.
828 \examplen{Graphs with floating-point weights}
829 A~common case of non-integer weights are rational numbers in floating-point (FP)
830 representation. Even in this case we will be able to find the MST in linear time.
831 The most common representation of binary FP numbers specified by the IEEE
832 standard 754-1985 \cite{ieee:binfp} has a~useful property: When the
833 bit strings encoding non-negative FP numbers are read as ordinary integers,
834 the order of these integers is the same as of the original FP numbers. We can
835 therefore once again replace the edge weights by integers and use the linear-time
836 integer algorithm. While the other FP representations (see \cite{dgoldberg:fp} for
837 an~overview) need not have this property, the corresponding integers can be adjusted
838 in $\O(1)$ time to the format we need. (More advanced tricks of this type have been
839 employed by Thorup in \cite{thorup:floatint} to extend his linear-time algorithm
840 for single-source shortest paths to FP edge lengths.)
842 \examplen{Graphs with bounded degrees}
843 For graphs with vertex degrees bounded by a~constant~$\Delta$, the problem is either
844 trivial (if $\Delta<3$) or as hard as for arbitrary graphs. There is a~simple linear-time
845 transform of arbitrary graphs to graphs with maximum degree~3 which preserves the MST:
847 \lemman{Degree reduction}\id{degred}%
848 For every graph~$G$ there exists a~graph~$G'$ with maximum degree at most~3 and
849 a~function $\pi: E(G)\rightarrow E(G')$ such that $\mst(G) = \pi^{-1}(\mst(G'))$.
850 The graph $G'$ and the embedding~$\pi$ can be constructed in time $\O(m)$.
852 \figure{french.eps}{\epsfxsize}{Degree reduction in Lemma~\ref{degred}}
855 We show how to eliminate a~single vertex~$v$ of degree $d>3$ and then apply
858 Assume that $v$~has neighbors $w_1,\ldots,w_d$. We replace~$v$ and the edges~$vw_i$
859 by $d$~new vertices $v_1,\ldots,v_d$, joined by a~path $v_1v_2\ldots v_d$, and
860 edges~$v_iw_i$. Each edge of the path will receive a~weight smaller than all
861 original weights, the other edges will inherit the weights of the edges $vw_i$
862 they replace. The edges of the path will therefore lie in the MST (this is
863 obvious from the Kruskal's algorithm) and as~$G$ can be obtained from the
864 new~$G'$ by contracting the path, the rest follows from the Contraction lemma
867 This step can be carried out in time $\O(d)$. As it replaces a high-degree
868 vertex by vertices of degree~3, the whole procedure stops in at most~$n$ such
869 steps, so it takes time $\O(\sum_{v\in V}\deg(v)) = \O(m)$ including the
870 time needed to find the high-degree vertices at the beginning.
873 \examplen{Euclidean MST}
874 The MST also has its counterparts in the realm of geometric algorithms. Suppose
875 that we have $n$~points $x_1,\ldots,x_n$ in the plane and we want to find the
876 shortest system of segments connecting these points. If we want the segments to
877 touch only in the given points, this is equivalent to finding the MST of the
878 complete graph on the vertices $V=\{x_1,\ldots,x_n\}$ with edge weights
879 defined as the Euclidean distances of the points. Since the graph is dense, many of the MST
880 algorithms discussed run in linear time with the size of the graph, hence
883 There is a~more efficient method based on the observation that the MST
884 is always a~subgraph of the Delaunay's tesselation for the given points
885 (this was first noted by Shamos and Hoey in~\cite{shamos:closest}). The
886 tesselation is a~planar graph, which guarantees that it has $\O(n)$ edges,
887 and it is a~dual graph of the Voronoi diagram of the given points, which can
888 be constructed in time $\O(n\log n)$ using for example the Fortune's
889 algorithm \cite{fortune:voronoi}. We can therefore reduce the problem
890 to finding the MST of the tesselation for which $\O(n\log n)$ time
891 is more than sufficient.
893 This approach fails for non-Euclidean metrics, but in some cases
894 (in particular for the rectilinear metric) the $\O(n\log n)$ time bound
895 is also achievable by the algorithm of Zhou et al.~\cite{zhou:nodel}
896 based on the sweep-line technique and the Red rule. For other
897 variations on the geometric MST, see Eppstein's survey paper
898 \cite{eppstein:spanning}.
900 \examplen{Steiner trees}
901 The constraint that the segments in the previous example are allowed to touch
902 each other only in the given points looks artificial and it is indeed uncommon in
903 practical applications (including the problem of designing electrical transmission
904 lines originally studied by Bor\o{u}vka). If we lift this restriction, we get
905 the problem known by the name Steiner tree.\foot{It is named after the Swiss mathematician
906 Jacob Steiner who studied a~special case of this problem in the 19th century.}
907 We can also define it in terms of graphs:
909 \defn A~\df{Steiner tree} of a~weighted graph~$(G,w)$ with a~set~$M\subseteq V$
910 of \df{mandatory notes} is a~tree~$T\subseteq G$ which contains all the mandatory
911 vertices and its weight is minimum possible.
913 For $M=V$ the Steiner tree is identical to the MST, but if we allow an~arbitrary
914 choice of the mandatory vertices, it is NP-hard. This has been proven by Garey and Johnson
915 \cite{garey:steiner,garey:rectisteiner} for not only the graph version with
916 weights $\{1,2\}$, but also for the planar version with Euclidean or rectilinear
917 metric. There is a~polynomial approximation algorithm with ratio $5/3$ for
918 graphs due to Pr\"omel and Steger \cite{proemel:steiner} and a~polynomial-time
919 approximation scheme for the Euclidean Steiner tree in an~arbitrary dimension
920 by Arora \cite{arora:tspapx}.
922 \examplen{Approximating the weight of the MST}
923 Sometimes we are not interested in the actual edges forming the MST and only
924 the weight matters. If we are willing to put up with a~randomized approximation,
925 we can even achieve sub-linear complexity. Chazelle et al.~\cite{chazelle:mstapprox}
926 have shown an~algorithm which, given $0 < \varepsilon < 1/2$, approximates
927 the weight of the MST of a~graph with average degree~$d$ and edge weights from the set
928 $\{1,\ldots,w\}$ in time $\O(dw\varepsilon^{-2}\cdot\log(dw/\varepsilon))$,
929 producing a~weight which has relative error at most~$\varepsilon$ with probability
930 at least $3/4$. They have also proven an~almost matching lower bound $\Omega(dw\varepsilon^{-2})$.
932 For the $d$-dimensional Euclidean case, there is a~randomized approximation
933 algorithm by Czumaj et al.~\cite{czumaj:euclidean} which with high probability
934 produces a~spanning tree within relative error~$\varepsilon$ in $\widetilde\O(\sqrt{n}\cdot \poly(1/\varepsilon))$\foot{%
935 $\widetilde\O(f) = \O(f\cdot\log^{\O(1)} f)$ and $\poly(n)=n^{\O(1)}$.}
936 queries to a~data structure containing the points. The data structure is expected
937 to answer orthogonal range queries and cone approximate nearest neighbor queries.
938 There is also a~$\widetilde\O(n\cdot \poly(1/\varepsilon))$ time approximation
939 algorithm for the MST weight in arbitrary metric spaces by Czumaj and Sohler \cite{czumaj:metric}.
940 (This is still sub-linear since the corresponding graph has roughly $n^2$ edges.)