Either $e$~is $F$-heavy and so the forest~$F$ is also the MSF of the new graph. Or it is $F$-light
and we have to modify~$F$ by exchanging the heaviest edge~$f$ of the path $F[e]$ with~$e$.
-Correctness of the former case follows immediately from the Theorem on Minimality by order
-(\ref{mstthm}), because any $F'$-light would be also $F$-light, which is impossible as $F$~was
+Correctness of the former case follows immediately from the Minimality Theorem (\ref{mstthm}),
+because any $F'$-light would be also $F$-light, which is impossible as $F$~was
minimum. In the latter case, the edge~$f$ is not contained in~$F'$ because it is the heaviest
on the cycle $F[e]+e$ (by the Red lemma, \ref{redlemma}). We can now use the Blue lemma
(\ref{bluelemma}) to prove that it should be replaced with~$e$. Consider the tree~$T$
If there was only a~single occurrence of~$v$, then $v$~was a~leaf and thus the sequence
transforms from $AuvuC$ to $AuC$ and $v$~alone.
-\em{Changing the root} of the tree~$T$ from~$v$ to~$w$ changes the sequence from $vAwBwCv$ to $wBwCvAw$.
+\em{Changing the root} of the tree~$T$ from~$v$ to~$w$ changes its ET-sequence from $vAwBwCv$ to $wBwCvAw$.
If $w$~was a~leaf, the sequence changes from $vAwCv$ to $wCvAw$. If $vw$ was the only edge of~$T$,
the sequence $vw$ becomes $wv$. Note that this works regardless of the possible presence of~$w$ inside~$B$.
}
At the beginning, the graph contains no edges, so both invariants are trivially
-satistifed. Newly inserted edges can enter level~0, which cannot break I1 nor~I2.
+satisfied. Newly inserted edges can enter level~0, which cannot break I1 nor~I2.
When we delete a~tree edge at level~$\ell$, we split a~tree~$T$ of~$F_\ell$ to two
trees $T_1$ and~$T_2$. Without loss of generality, let us assume that $T_1$ is the
\::Insert $f$ to~$F_0,\ldots,F_{\ell(f)}$.
\endalgo
-\algn{$\<Replace>(uv,i)$ -- Search for an~replacement edge for~$uv$ at level~$i$}
+\algn{$\<Replace>(uv,i)$ -- Search for replacement for edge~$uv$ at level~$i$}
\algo
\algin An~edge~$uv$ to replace and a~level~$i$ such that there is no replacement
at levels greater than~$i$.
\:Enumerate non-tree edges incident with vertices of~$T_1$ and stored in ${\cal E}_i$.
For each edge~$xy$, $x\in T_1$, do:
\::If $y\in T_2$, remove~$xy$ from~${\cal E}_i$ and return it to the caller.
-\::Otherwise increase $\ell(xy)$ by one. This includes deleting~$xy$ from~${\cal E}_i$
- and inserting it to~${\cal E}_{i+1}$.
+\::Otherwise increase $\ell(xy)$ by one.
+ \hfil\break
+ This includes deleting~$xy$ from~${\cal E}_i$ and inserting it to~${\cal E}_{i+1}$.
\:If $i>0$, call $\<Replace>(xy,i-1)$.
\:Otherwise return \<null>.
\algout The replacement edge.
\<Replace> costs $\O(\log n)$ per call, the rest is paid for by the edge level
increases.
-To bring the complexity of \<Connected> from $\O(\log n)$ down to $\O(\log n/\log\log n)$,
+To bring the complexity of the operation \<Connected> from $\O(\log n)$ down to $\O(\log n/\log\log n)$,
we apply the trick from Example \ref{accel} and store~$F_0$ in a~ET-tree with $a=\max(\lfloor\log n\rfloor,2)$.
This does not hurt the complexity of insertions and deletions, but allows for faster queries.
\qed
+\rem
+An~$\Omega(\log n/\log\log n)$ lower bound for the amortized complexity of the dynamic connectivity
+problem has been proven by Henzinger and Fredman \cite{henzinger:lowerbounds} in the cell
+probe model with $\O(\log n)$-bit words. Thorup has answered by a~faster algorithm
+\cite{thorup:nearopt} that achieves $\O(\log n\log^3\log n)$ time per update and
+$\O(\log n/\log^{(3)} n)$ per query on a~RAM with $\O(\log n)$-bit words. (He claims
+that the algorithm runs on a~Pointer Machine, but it uses arithmetic operations,
+so it does not fit the definition of the PM we use. The algorithm only does not
+need direct indexing of arrays.) So far, it is not known how to extend this algorithm
+to fit our needs, so we omit the details.
+
%--------------------------------------------------------------------------------
-\section{Dynamic MSF}
+\section{Dynamic spanning forests}\id{dynmstsect}%
Let us turn our attention back to the dynamic MSF now.
Most of the early algorithms for dynamic connectivity also imply $\O(n^\varepsilon)$
algorithms for dynamic maintenance of the MSF. Henzinger and King \cite{henzinger:twoec,henzinger:randdyn}
have generalized their randomized connectivity algorithm to maintain the MSF in $\O(\log^5 n)$ time per
-operation, or $\O(k\log^3 n)$ if only~$k$ different values of edge weights are allowed. They have solved
+operation, or $\O(k\log^3 n)$ if only $k$ different values of edge weights are allowed. They have solved
the decremental version of the problem first (which starts with a~given graph and only edge deletions
are allowed) and then presented a~general reduction from the fully dynamic MSF to its decremental version.
We will describe the algorithm of Holm, de Lichtenberg and Thorup \cite{holm:polylog}, who have followed
\proof
Let us consider any two edges $f_1$ and~$f_2$ replacing~$e$. By minimality of~$F$ and the Cycle
-rule (\ref{cyclerule}), each $f_i$ is the heaviest edge on the cycle~$C_i = F[f_i] + f_i$.
+rule (Lemma \ref{redlemma}), each $f_i$ is the heaviest edge on the cycle~$C_i = F[f_i] + f_i$.
In a~moment, we will show that the symmetric difference~$C$ of these two cycles is again a~cycle.
This implies that if $f_1$ is heavier than~$f_2$, then $f_1$~is the heaviest edge on~$C$, so
$\ell(f_1) \le \ell(f_2)$ by I3. Therefore the lightest of all replacement edges must have
are at level~0. Sole deletions of edges (both tree and non-tree) cannot violate I3, so we need
to check only the replaces, in particular the place when an~edge~$e$ gets its level increased.
-For the violation to happen, $e$~must be the heaviest on some cycle~$C$, so by the Cycle
-rule, $e$~must be non-tree. The increase of $\ell(e)$ must therefore take place when~$e$ is
-considered as a~replacement edge incident with some tree~$T_1$ at level $\ell=\ell(e)$.
-We will pause the computation just before this increase and we will prove that
-all other edges of~$C$ already are at levels greater than~$\ell$.
+For the violation to happen for the first time, $e$~must be the heaviest on
+some cycle~$C$, so by the Cycle rule, $e$~must be non-tree. The increase of
+$\ell(e)$ must therefore take place when~$e$ is considered as a~replacement
+edge incident with some tree~$T_1$ at level $\ell=\ell(e)$. We will pause the
+computation just before this increase and we will prove that all other edges
+of~$C$ already are at levels greater than~$\ell$, so the violation cannot occur.
-Let us first show that for edges of~$C$ incident with~$T_1$. All edges of~$T_1$ itself
+Let us first show this for edges of~$C$ incident with~$T_1$. All edges of~$T_1$ itself
already are at the higher levels as they were moved there at the very beginning of the
-search for the replacement edge. As the algorithm always picks the lightest candidate
-for the replacement edge available, all non-tree edges incident with~$T_1$ that are lighter
-than~$e$ were already considered and thus also moved one level up. This includes all
-other edges of~$C$ that are incident with~$T_1$.
+search for the replacement edge. The other tree edges incident with~$T_1$ would have
+lower levels, which is impossible since the invariant would be already violated.
+Non-tree edges of~$C$ incident with~$T_1$ are lighter than~$e$, so they were already considered
+as~candidates for the replacement edge, because the algorithm always picks the lightest
+candidate first. Such edges therefore have been already moved a~level up.
The case of edges of~$C$ that do not touch~$T_1$ is easy to handle: Such edges do not exist.
-If they did, at least two edges of~$C$ would have to be non-tree edges connecting~$T_1$
-with the other trees at level~$\ell$, so one of them that is lighter than~$e$ would be selected as the
-replacement edge before~$e$ could be considered.
+If they did, at least one more edge of~$C$ besides~$e$ would have to connect~$T_1$ with the other
+trees of level~$\ell$. We already know that this could not be a~tree edge. If it were a~non-tree
+edge, it could not have level greater than~$\ell$ by~I1 nor smaller than~$\ell$ by~I3. Therefore
+it would be a~level~$\ell$ edge lighter than~$e$, and as such it would have been selected as the
+replacement edge before $e$~was.
\qed
We can conclude:
\thmn{Decremental MSF, Holm et al.~\cite{holm:polylog}}
When we start with a~graph on $n$~vertices with~$m$ edges and we perform a~sequence of
edge deletions, the MSF can be initialized in time $\O((m+n)\cdot\log^2 n)$ and then
-updated in time $\O(log^2 n)$ amortized per operation.
+updated in time $\O(\log^2 n)$ amortized per operation.
\paran{Fully dynamic MSF}%
The decremental MSF algorithm can be turned to a~fully dynamic one by a~blackbox
reduction whose properties are summarized in the following theorem:
-\thmn{MSF dynamization, Henzinger et al.~\cite{henzinger:twoec}, Holm et al.~\cite{holm:polylog}}
+\thmn{MSF dynamization, Holm et al.~\cite{holm:polylog}}
Suppose that we have a~decremental MSF algorithm with the following properties:
\numlist\ndotted
\:For any $a$,~$b$, it can be initialized on a~graph with~$a$ vertices and~$b$ edges.
sought among the replacement edges found in these structures. The unused replacement edges then have
to be reinserted back to the structure.
-The original reduction of Henzinger et al.~handles these reinserts by a~mechanism of batch insertions
+The original reduction of Henzinger et al.~\cite{henzinger:twoec} handles these reinserts by a~mechanism of batch insertions
supported by their decremental structure, which is not available in our case. Holm et al.~have
replaced it by a~system of auxiliary edges inserted at various places in the structure.
We refer to the article \cite{holm:polylog} for details.
to only edge deletions seems to be unavoidable. The invariant I3 could be easily
broken for many cycles at once whenever a~very light non-tree edge is inserted.
We could try increasing the level of the newly inserted edge, but we would quite
-possibly hit I1 before we skipped the levels of all the heaviest edges on the
+likely hit I1 before we managed to skip the levels of all the heaviest edges on the
particular cycles.
-On the other hand, if we decided to drop I3, we would encounter different problems.
-We have enough time to scan all non-tree edges incident to the current tree~$T_1$
---- we can charge it on the level increases of its tree edges and if we use the
-degree reduction from Lemma \ref{degred}, there are at most two non-tree edges
-per vertex. (The reduction can be used dynamically as it always translates a~single
-change of the original graph to $\O(1)$ changes of the reduced graph.) The lightest
-replacement edge however could also be located in the super-trees of~$T_1$ at the
-lower levels, which are too large to scan and both I1 and I2 prevent us from
-charging the time on increasing levels there.
+On the other hand, if we decided to drop I3, we would encounter different problems. The ET-trees can
+bring the lightest non-tree incident with the current tree~$T_1$, but the lightest replacement edge
+could also be located in the super-trees of~$T_1$ at the lower levels, which are too large to scan
+and both I1 and I2 prevent us from charging the time on increasing levels there.
An~interesting special case in which insertions are possible is when all non-tree
edges have the same weight. This leads to the following algorithm for dynamic MSF
\paran{Dynamic MSF with limited edge weights}%
Let us assume for a~while that our graph has edges of only two different weights (let us say
-1~and~2). We will forget our rule that all edge weights are distinct for a~moment and we recall that
-the basic structural properties of the MST's from Section \ref{mstbasics} still hold.
+1~and~2). We will forget our rule that all edge weights are distinct for a~moment and we recall
+the observation in \ref{multiweight} that the basic structural properties of
+the MST's from Section \ref{mstbasics} still hold.
We split the graph~$G$ to two subgraphs~$G_1$ and~$G_2$ according to the edge
weights. We use one instance~$\C_1$ of the dynamic connectivity algorithm maintaining
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