From: Martin Mares <mj@ucw.cz>
Date: Mon, 21 Jan 2008 18:06:33 +0000 (+0100)
Subject: A remark.
X-Git-Tag: printed~280
X-Git-Url: http://mj.ucw.cz/gitweb/?a=commitdiff_plain;h=0a7de2c6c259fba4086479e3993318b5ac276f99;p=saga.git

A remark.
---

diff --git a/mst.tex b/mst.tex
index 58a8d5f..b1814c3 100644
--- a/mst.tex
+++ b/mst.tex
@@ -812,11 +812,21 @@ 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 $\sum_v d(v) < 3n$,
-hence by the same argument as in the previous proof, for at least $n/2$
+hence 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
+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 increases 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.
 
 
 \section{Using Fibonacci heaps}