From 649fd5643221f33883ba9f4f00ef5f9cf415c9db Mon Sep 17 00:00:00 2001 From: Martin Mares Date: Sat, 5 Apr 2008 22:00:13 +0200 Subject: [PATCH] Fix references. --- macros.tex | 4 ++++ opt.tex | 6 +++--- 2 files changed, 7 insertions(+), 3 deletions(-) diff --git a/macros.tex b/macros.tex index 850c162..19972c8 100644 --- a/macros.tex +++ b/macros.tex @@ -441,6 +441,10 @@ \vfill\supereject\end } +%%% Hyphenation %%% + +\hyphenation{Ra-ma-chan-dran} + %%% The End %%% \catcode`@=12 diff --git a/opt.tex b/opt.tex index 18e5053..8977d34 100644 --- a/opt.tex +++ b/opt.tex @@ -612,7 +612,7 @@ caused by soft heaps, we can easily make the weights unique. If~$C$ is a~subgraph of~$G$, we will refer to the edges of~$R$ whose exactly one endpoint lies in~$C$ by~$R^C$ (i.e., $R^C = R\cap \delta(C)$). -\lemman{Robust contraction}\id{robcont}% +\lemman{Robust contraction, Chazelle \cite{chazelle:almostacker}}\id{robcont}% Let $G$ be a~weighted graph and $C$~its subgraph contractible in~$G\crpt R$ for some set of edges~$R$. Then $\msf(G) \subseteq \msf(C) \cup \msf((G/C) \setminus R^C) \cup R^C$. @@ -669,7 +669,7 @@ We can formulate the exact partitioning algorithm as follows: corrupted edges in the neighborhood of these subgraphs. \endalgo -\thmn{Partitioning to contractible subgraphs, Pettie and Ramachandran \cite{pettie:optimal}}\id{partthm}% +\thmn{Partitioning to contractible subgraphs, Chazelle \cite{chazelle:almostacker}}\id{partthm}% Given a~weighted graph~$G$ and parameters $\varepsilon$ ($0<\varepsilon\le 1/2$) and~$t$, the Partition algorithm \ref{partition} constructs a~collection $\C=\{C_1,\ldots,C_k\}$ of subgraphs and a~set~$R^\C$ of corrupted edges such that: @@ -1080,7 +1080,7 @@ The other inequality is obvious as $D(m,n)$ is an~asymptotic lower bound for the time complexity of every comparison-based algorithm. \qed -\paran{Complexity} +\paran{Complexity of MST} As we have already noted, the exact decision tree complexity $D(m,n)$ of the MST problem is still open and so is therefore the time complexity of the optimal algorithm. However, every time we come up with another comparison-based algorithm, we can use its complexity -- 2.39.2