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	<title>Comments on: A little (combinatorial) graph problem</title>
	<atom:link href="http://www.tangentspace.net/cz/archives/2008/07/a-little-combinatorial-graph-problem/feed/" rel="self" type="application/rss+xml" />
	<link>http://www.tangentspace.net/cz/archives/2008/07/a-little-combinatorial-graph-problem/</link>
	<description>somewhere near the beginning.</description>
	<pubDate>Fri, 21 Nov 2008 20:52:54 +0000</pubDate>
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		<title>By: Alex</title>
		<link>http://www.tangentspace.net/cz/archives/2008/07/a-little-combinatorial-graph-problem/#comment-392162</link>
		<dc:creator>Alex</dc:creator>
		<pubDate>Tue, 15 Jul 2008 00:28:19 +0000</pubDate>
		<guid isPermaLink="false">http://www.tangentspace.net/cz/?p=940#comment-392162</guid>
		<description>Oh. Thanks :P</description>
		<content:encoded><![CDATA[<p>Oh. Thanks <img src='http://www.tangentspace.net/cz/wp-includes/images/smilies/icon_razz.gif' alt=':P' class='wp-smiley' /></p>
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		<title>By: Pete</title>
		<link>http://www.tangentspace.net/cz/archives/2008/07/a-little-combinatorial-graph-problem/#comment-391987</link>
		<dc:creator>Pete</dc:creator>
		<pubDate>Mon, 14 Jul 2008 21:01:44 +0000</pubDate>
		<guid isPermaLink="false">http://www.tangentspace.net/cz/?p=940#comment-391987</guid>
		<description>As stated, the LHS is a vector, which you're saying is equal to the number k.

nk is the total number of vertex-edge pairs, which must be even since each edge touches 2 vertices. Or another way to look at it, it's the number of nonzeros in the incidence matrix, which must be even since it's symmetric hollow.</description>
		<content:encoded><![CDATA[<p>As stated, the LHS is a vector, which you&#8217;re saying is equal to the number k.</p>
<p>nk is the total number of vertex-edge pairs, which must be even since each edge touches 2 vertices. Or another way to look at it, it&#8217;s the number of nonzeros in the incidence matrix, which must be even since it&#8217;s symmetric hollow.</p>
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	<item>
		<title>By: Alex</title>
		<link>http://www.tangentspace.net/cz/archives/2008/07/a-little-combinatorial-graph-problem/#comment-391974</link>
		<dc:creator>Alex</dc:creator>
		<pubDate>Mon, 14 Jul 2008 20:47:11 +0000</pubDate>
		<guid isPermaLink="false">http://www.tangentspace.net/cz/?p=940#comment-391974</guid>
		<description>heh, circulant matrices work, true. The LHS isn't a sum. So, why won't it work if nk is odd?</description>
		<content:encoded><![CDATA[<p>heh, circulant matrices work, true. The LHS isn&#8217;t a sum. So, why won&#8217;t it work if nk is odd?</p>
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		<title>By: Pete</title>
		<link>http://www.tangentspace.net/cz/archives/2008/07/a-little-combinatorial-graph-problem/#comment-391908</link>
		<dc:creator>Pete</dc:creator>
		<pubDate>Mon, 14 Jul 2008 19:12:10 +0000</pubDate>
		<guid isPermaLink="false">http://www.tangentspace.net/cz/?p=940#comment-391908</guid>
		<description>Nice warm up to start the day. At noon.

The answer is any n greater than k, provided nk is even. It's pretty easy to construct example graphs with circulant connectivity matrices.

BTW, you forgot to sum the LHS of your equation. =D</description>
		<content:encoded><![CDATA[<p>Nice warm up to start the day. At noon.</p>
<p>The answer is any n greater than k, provided nk is even. It&#8217;s pretty easy to construct example graphs with circulant connectivity matrices.</p>
<p>BTW, you forgot to sum the LHS of your equation. =D</p>
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