<?xml version="1.0" encoding="UTF-8"?><rss version="2.0"
	xmlns:content="http://purl.org/rss/1.0/modules/content/"
	xmlns:dc="http://purl.org/dc/elements/1.1/"
	xmlns:atom="http://www.w3.org/2005/Atom"
	xmlns:sy="http://purl.org/rss/1.0/modules/syndication/"
	xmlns:georss="http://www.georss.org/georss" xmlns:geo="http://www.w3.org/2003/01/geo/wgs84_pos#" xmlns:media="http://search.yahoo.com/mrss/"
		>
<channel>
	<title>Comments on: Top 25 Memes</title>
	<atom:link href="http://mendicantbug.com/2009/02/28/top-25-meme/feed/" rel="self" type="application/rss+xml" />
	<link>http://mendicantbug.com/2009/02/28/top-25-meme/</link>
	<description>Wanderings into computational linguistics, science, social media and life...</description>
	<lastBuildDate>Tue, 17 Jan 2012 18:51:53 +0000</lastBuildDate>
	<sy:updatePeriod>hourly</sy:updatePeriod>
	<sy:updateFrequency>1</sy:updateFrequency>
	<generator>http://wordpress.com/</generator>
	<item>
		<title>By: Jason Adams</title>
		<link>http://mendicantbug.com/2009/02/28/top-25-meme/#comment-1285</link>
		<dc:creator><![CDATA[Jason Adams]]></dc:creator>
		<pubDate>Mon, 02 Mar 2009 00:39:02 +0000</pubDate>
		<guid isPermaLink="false">http://mendicantbug.com/?p=1095#comment-1285</guid>
		<description><![CDATA[And just noticed that wordpress mangled greater than and less than symbols in my first comment.]]></description>
		<content:encoded><![CDATA[<p>And just noticed that wordpress mangled greater than and less than symbols in my first comment.</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Jason Adams</title>
		<link>http://mendicantbug.com/2009/02/28/top-25-meme/#comment-1284</link>
		<dc:creator><![CDATA[Jason Adams]]></dc:creator>
		<pubDate>Mon, 02 Mar 2009 00:37:37 +0000</pubDate>
		<guid isPermaLink="false">http://mendicantbug.com/?p=1095#comment-1284</guid>
		<description><![CDATA[Ahh, ok, I was indeed making it too complicated.  Thanks for clearing that up! :)]]></description>
		<content:encoded><![CDATA[<p>Ahh, ok, I was indeed making it too complicated.  Thanks for clearing that up! :)</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Jon Elsas</title>
		<link>http://mendicantbug.com/2009/02/28/top-25-meme/#comment-1283</link>
		<dc:creator><![CDATA[Jon Elsas]]></dc:creator>
		<pubDate>Sun, 01 Mar 2009 17:19:42 +0000</pubDate>
		<guid isPermaLink="false">http://mendicantbug.com/?p=1095#comment-1283</guid>
		<description><![CDATA[I think you&#039;re making it too complicated.  best case would be  (N - K + K log K), as you pointed out.  If we assume K is a constant, or much smaller than N, then both worst- and best-case are essentially the same, O(N).  If K = O(N), then its basically a full sort algorithm.]]></description>
		<content:encoded><![CDATA[<p>I think you&#8217;re making it too complicated.  best case would be  (N &#8211; K + K log K), as you pointed out.  If we assume K is a constant, or much smaller than N, then both worst- and best-case are essentially the same, O(N).  If K = O(N), then its basically a full sort algorithm.</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Jason Adams</title>
		<link>http://mendicantbug.com/2009/02/28/top-25-meme/#comment-1282</link>
		<dc:creator><![CDATA[Jason Adams]]></dc:creator>
		<pubDate>Sun, 01 Mar 2009 14:30:52 +0000</pubDate>
		<guid isPermaLink="false">http://mendicantbug.com/?p=1095#comment-1282</guid>
		<description><![CDATA[Ah good point, but wouldn&#039;t that just be worst case?  In the best case, it&#039;s in descending order already and you make K log K + N - K comparisons.  I was thinking in the average case you&#039;d have some probability that is decreasing as you progress through the list that the value would be propagated through the tree.  

So building the first 25 items has K log K comparisons.  Then for each remaining item the runtime would be O(1) if  root.  Maybe P(item &gt; root) reduces to some constant value and the runtime is still O(N log K)?]]></description>
		<content:encoded><![CDATA[<p>Ah good point, but wouldn&#8217;t that just be worst case?  In the best case, it&#8217;s in descending order already and you make K log K + N &#8211; K comparisons.  I was thinking in the average case you&#8217;d have some probability that is decreasing as you progress through the list that the value would be propagated through the tree.  </p>
<p>So building the first 25 items has K log K comparisons.  Then for each remaining item the runtime would be O(1) if  root.  Maybe P(item &gt; root) reduces to some constant value and the runtime is still O(N log K)?</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Jon Elsas</title>
		<link>http://mendicantbug.com/2009/02/28/top-25-meme/#comment-1280</link>
		<dc:creator><![CDATA[Jon Elsas]]></dc:creator>
		<pubDate>Sat, 28 Feb 2009 13:42:54 +0000</pubDate>
		<guid isPermaLink="false">http://mendicantbug.com/?p=1095#comment-1280</guid>
		<description><![CDATA[Its O(N log K ) where K is the number of top items (25 in this case).  you need to look at all N data points, but the insertion time is less (log(K)) since the size of the tree is at most K elements.]]></description>
		<content:encoded><![CDATA[<p>Its O(N log K ) where K is the number of top items (25 in this case).  you need to look at all N data points, but the insertion time is less (log(K)) since the size of the tree is at most K elements.</p>
]]></content:encoded>
	</item>
</channel>
</rss>

