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> <channel><title>Comments on: Programming &#8211; The art of giving percise instructions</title> <atom:link href="http://ripper234.com/p/programming-the-art-of-giving-percise-instructions/feed/" rel="self" type="application/rss+xml" /><link>http://ripper234.com/p/programming-the-art-of-giving-percise-instructions/</link> <description>Stuff Ron Gross Finds Interesting</description> <lastBuildDate>Fri, 27 Jan 2012 03:20:29 +0000</lastBuildDate> <sy:updatePeriod>hourly</sy:updatePeriod> <sy:updateFrequency>1</sy:updateFrequency> <generator>http://wordpress.org/?v=3.3.1</generator> <item><title>By: Friend</title><link>http://ripper234.com/p/programming-the-art-of-giving-percise-instructions/comment-page-1/#comment-1169</link> <dc:creator>Friend</dc:creator> <pubDate>Sat, 31 Jan 2009 09:17:16 +0000</pubDate> <guid
isPermaLink="false">http://ripper234.com/?p=889#comment-1169</guid> <description>This is not the most efficient way, because you don&#039;t need to calculate (X+Y)! at all. Note that the biggest number  will be at most (2X)!/(X!)^{2}, (by setting X=Y). So there is no need to calculate the number (X+Y)! to its bitter end. Probably the most efficient way to calculate this asymptotically, is by using Stirling&#039;s formula. Note that by the nature of this approximation, it is evident that the complexity is way beyond exponential. It is in fact something like X^{X}.</description> <content:encoded><![CDATA[<p>This is not the most efficient way, because you don&#8217;t need to calculate (X+Y)! at all. Note that the biggest number  will be at most (2X)!/(X!)^{2}, (by setting X=Y). So there is no need to calculate the number (X+Y)! to its bitter end. Probably the most efficient way to calculate this asymptotically, is by using Stirling&#8217;s formula. Note that by the nature of this approximation, it is evident that the complexity is way beyond exponential. It is in fact something like X^{X}.</p> ]]></content:encoded> </item> <item><title>By: A Quantum Immortal &#187; Blog Archive &#187; Never rely on tail recursion elimination</title><link>http://ripper234.com/p/programming-the-art-of-giving-percise-instructions/comment-page-1/#comment-1120</link> <dc:creator>A Quantum Immortal &#187; Blog Archive &#187; Never rely on tail recursion elimination</dc:creator> <pubDate>Thu, 29 Jan 2009 12:31:02 +0000</pubDate> <guid
isPermaLink="false">http://ripper234.com/?p=889#comment-1120</guid> <description>[...] previously posted a recursive implementation of Factorial, and asked the readers &#8220;not to bother saying it&#8217;s inefficient because of recursion, [...]</description> <content:encoded><![CDATA[<p>[...] previously posted a recursive implementation of Factorial, and asked the readers &#8220;not to bother saying it&#8217;s inefficient because of recursion, [...]</p> ]]></content:encoded> </item> <item><title>By: ripper234</title><link>http://ripper234.com/p/programming-the-art-of-giving-percise-instructions/comment-page-1/#comment-851</link> <dc:creator>ripper234</dc:creator> <pubDate>Sat, 17 Jan 2009 08:07:23 +0000</pubDate> <guid
isPermaLink="false">http://ripper234.com/?p=889#comment-851</guid> <description>You&#039;re not alone in this confusion. There is one route from the origin to itself - the empty route.
It&#039;s basically a matter of definition, but my definition allows us to have no special cases in the 3rd solution.</description> <content:encoded><![CDATA[<p>You&#8217;re not alone in this confusion. There is one route from the origin to itself &#8211; the empty route.</p><p>It&#8217;s basically a matter of definition, but my definition allows us to have no special cases in the 3rd solution.</p> ]]></content:encoded> </item> <item><title>By: Boaz</title><link>http://ripper234.com/p/programming-the-art-of-giving-percise-instructions/comment-page-1/#comment-850</link> <dc:creator>Boaz</dc:creator> <pubDate>Sat, 17 Jan 2009 07:56:02 +0000</pubDate> <guid
isPermaLink="false">http://ripper234.com/?p=889#comment-850</guid> <description>Yes, I&#039;m stupid.
Though the fact that if x = 0 and y = 0 you return 1 confused me.</description> <content:encoded><![CDATA[<p>Yes, I&#8217;m stupid.<br
/> Though the fact that if x = 0 and y = 0 you return 1 confused me.</p> ]]></content:encoded> </item> <item><title>By: ripper234</title><link>http://ripper234.com/p/programming-the-art-of-giving-percise-instructions/comment-page-1/#comment-849</link> <dc:creator>ripper234</dc:creator> <pubDate>Sat, 17 Jan 2009 07:45:17 +0000</pubDate> <guid
isPermaLink="false">http://ripper234.com/?p=889#comment-849</guid> <description>Thanks Eli.
Boaz, you are incorrect. If x == 0, this means you have no more Left moves left. You have only one route to get to the origin, by going Down all the time. Remember, we are calculating the number of routes to get to the origin, not the length of the route.</description> <content:encoded><![CDATA[<p>Thanks Eli.</p><p>Boaz, you are incorrect. If x == 0, this means you have no more Left moves left. You have only one route to get to the origin, by going Down all the time. Remember, we are calculating the number of routes to get to the origin, not the length of the route.</p> ]]></content:encoded> </item> <item><title>By: Boaz</title><link>http://ripper234.com/p/programming-the-art-of-giving-percise-instructions/comment-page-1/#comment-848</link> <dc:creator>Boaz</dc:creator> <pubDate>Sat, 17 Jan 2009 07:04:26 +0000</pubDate> <guid
isPermaLink="false">http://ripper234.com/?p=889#comment-848</guid> <description>Why &quot;return 1&quot;?
I believe it should be:
if (x == 0) return y;
if (y == 0) return x;</description> <content:encoded><![CDATA[<p>Why &#8220;return 1&#8243;?<br
/> I believe it should be:<br
/> if (x == 0) return y;<br
/> if (y == 0) return x;</p> ]]></content:encoded> </item> <item><title>By: Eli</title><link>http://ripper234.com/p/programming-the-art-of-giving-percise-instructions/comment-page-1/#comment-847</link> <dc:creator>Eli</dc:creator> <pubDate>Sat, 17 Jan 2009 06:25:03 +0000</pubDate> <guid
isPermaLink="false">http://ripper234.com/?p=889#comment-847</guid> <description>Nice post!</description> <content:encoded><![CDATA[<p>Nice post!</p> ]]></content:encoded> </item> </channel> </rss>
