$T^*$-light, i.e., that it is heavier than all edges on $T^*[f]$. The path $T^*[f]$ is
either identical to the original path $T[f]$ (if $e\not\in T[f]$) or to $T[f] \symdiff C$,
where $C$ is the cycle $T[e']+e'$. The former case is trivial, in the latter we have
-$w(f)\ge w(e')$ due to the choice of $e'$ and all other edges on~$C$ are lighter
+$w(f)\ge w(e')$ due to the choice of~$e'$ and all other edges on~$C$ are lighter
than~$e'$ as $e'$ was not $T$-light.
\qed
these edge exchanges must be in fact strictly increasing. On the other hand,
we know that $w(T_1)=w(T_2)$, so the exchange sequence must be empty and indeed
$T_1$ and $T_2$ must be identical.
+\looseness=1 %%HACK%%
\qed
\nota\id{mstnota}%
we do not need the Red rule to explicitly exclude edges.
It remains to show that adding the edges simultaneously does not
-produce a cycle. Consider the first iteration of the algorithm where $T$ contains a~cycle~$C$. Without
+produce a~cycle. Consider the first iteration of the algorithm where $T$ contains a~cycle~$C$. Without
loss of generality we can assume that:
$$C=T_1[u_1,v_1]\,v_1u_2\,T_2[u_2,v_2]\,v_2u_3\,T_3[u_3,v_3]\, \ldots \,T_k[u_k,v_k]\,v_ku_1.$$
Each component $T_i$ has chosen its lightest incident edge~$e_i$ as either the edge $v_iu_{i+1}$
or $v_{i-1}u_i$ (indexing cyclically). Suppose that $e_1=v_1u_2$ (otherwise we reverse the orientation
of the cycle). Then $e_2=v_2u_3$ and $w(e_2)<w(e_1)$ and we can continue in the same way,
-getting $w(e_1)>w(e_2)>\ldots>w(e_k)>w(e_1)$, which is a contradiction.
+getting $w(e_1)>w(e_2)>\ldots>w(e_k)>w(e_1)$, which is a~contradiction.
(Note that distinctness of edge weights was crucial here.)
\qed
\lemman{Contraction of MST edges}\id{contlemma}%
Let $G$ be a weighted graph, $e$~an arbitrary edge of~$\mst(G)$, $G/e$ the multigraph
produced by contracting~$e$ in~$G$, and $\pi$ the bijection between edges of~$G-e$ and
-their counterparts in~$G/e$. Then: $$\mst(G) = \pi^{-1}[\mst(G/e)] + e.$$
+their counterparts in~$G/e$. Then $\mst(G) = \pi^{-1}[\mst(G/e)] + e.$
\proof
% We seem not to need this lemma for multigraphs...
\figure{hedgehog.eps}{\epsfxsize}{A~hedgehog $H_{5,2}$ (quills bent to fit in the picture)}
\lemma
-A~single iteration of the contractive algorithm reduces~$H_{a,k}$ to a graph isomorphic with $H_{a,k-1}$.
+A~single iteration of the contractive algorithm reduces~$H_{a,k}$ to a~graph isomorphic with $H_{a,k-1}$.
\proof
Each vertex is incident with an edge of some distractor, so the algorithm does not select
The first linear-time algorithm that partitions all subtrees to isomorphism equivalence
classes is probably due to Zemlayachenko \cite{zemlay:treeiso}, but it lacks many
details. Dinitz et al.~\cite{dinitz:treeiso} have recast this algorithm in modern
-terminology and filled the gaps. Our algorithm is easier to formulate,
+terminology and filled the gaps. Our algorithm is easier to formulate than those,
because it replaces the need for auxiliary data structures by more elaborate bucket
sorting.
of possible values.
As in the case of tree decompositions, we would like to identify the equivalent subgraphs
-and process only a~single instance from each equivalence class. The obstacle is that
+and process only a~single instance from each equivalence class. We need to be careful
+with the definition of the equivalence classes, because
graph isomorphism is known to be computationally hard (it is one of the few
problems that are neither known to lie in~$\rm P$ nor to be $\rm NP$-complete;
see Arvind and Kurur \cite{arvind:isomorph} for recent results on its complexity).
the output vertices that refer to an~input vertex with the corresponding places
in the encoding of the input graph. This way, the whole output can be generated in time
$\O(\Vert\C\Vert + \Vert{\cal G}\Vert)$.
+\looseness=1 %%HACK%%
\qed
\rem
\defn
The \df{bitwise encoding} of a~vector ${\bf x}=(x_0,\ldots,x_{d-1})$ of~$b$-bit numbers
-is an~integer~$x$ such that $\(x)=\(x_{d-1})_b\0\(x_{d-2})_b\0\ldots\0\(x_0)_b$, i.e.,
-$x = \sum_i 2^{(b+1)i}\cdot x_i$. (We have interspersed the elements with \df{separator bits.})
+is an~integer~$x$ such that $\(x)=\(x_{d-1})_b\0\(x_{d-2})_b\0\ldots\0\(x_0)_b$. In other
+words, $x = \sum_i 2^{(b+1)i}\cdot x_i$. (We have interspersed the elements with \df{separator bits.})
\notan{Vectors}\id{vecnota}%
We will use boldface letters for vectors and the same letters in normal type
\[r_{d-1}] \dd \[r_2] \[r_1] \[s_d] \dd \[s_3] \[s_2] \[s_1] \cr
}
-This way, we also get the vector of all partial sums:
+This way, we also get all partial sums:
$s_k=\sum_{i=0}^{k-1}x_i$, $r_k=\sum_{i=k}^{d-1}x_i$.
\:$\<Cmp>(x,y)$ --- Compare vectors ${\bf x}$ and~${\bf y}$ element-wise,
can use the previous algorithm. The same trick of course applies to for finding the
\<MSB>, too.
-The following algorithm shows the details.
+The following algorithm shows the details:
\algn{LSB in linear workspace}
\paran{Combining Q-heaps}%
We can also use the Q-heaps as building blocks of more complex structures
like Atomic heaps and AF-heaps (see once again \cite{fw:transdich}). We will
-show a~simpler, but often sufficient construction, sometimes called the \df{Q-heap tree.}
+show a~simpler, but often sufficient construction, sometimes called the \df{\hbox{Q-heap} tree.}
Suppose we have a~Q-heap of capacity~$k$ and a~parameter $d\in{\bb N}^+$. We
can build a~balanced $k$-ary tree of depth~$d$ such that its leaves contain
a~given set and every internal vertex keeps the minimum value in the subtree
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