<rss version="2.0" xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:atom="http://www.w3.org/2005/Atom"><channel><title>Hacker News: ezwoodland</title><link>https://news.ycombinator.com/user?id=ezwoodland</link><description>Hacker News RSS</description><docs>https://hnrss.org/</docs><generator>hnrss v2.1.1</generator><lastBuildDate>Mon, 04 May 2026 15:52:40 +0000</lastBuildDate><atom:link href="https://hnrss.org/user?id=ezwoodland" rel="self" type="application/rss+xml"></atom:link><item><title><![CDATA[New comment by ezwoodland in "BYOMesh – New LoRa mesh radio offers 100x the bandwidth"]]></title><description><![CDATA[
<p>What makes OFDM inherently worse at long range? Don't you just lengthen your symbols and use the extra frequency bins until you have tolerable losses?</p>
]]></description><pubDate>Mon, 04 May 2026 04:56:54 +0000</pubDate><link>https://news.ycombinator.com/item?id=48004800</link><dc:creator>ezwoodland</dc:creator><comments>https://news.ycombinator.com/item?id=48004800</comments><guid isPermaLink="false">https://news.ycombinator.com/item?id=48004800</guid></item><item><title><![CDATA[New comment by ezwoodland in "The Wolfram S Combinator Challenge"]]></title><description><![CDATA[
<p>No? Why would it?<p>In the negative case, it would say the idea doesn't pan out.<p>In the positive case, it would mean that you can use just S instead of S and K when doing combinator reduction, but doesn't change that this kind of reduction is not super efficient practically speaking.</p>
]]></description><pubDate>Thu, 26 Feb 2026 21:39:27 +0000</pubDate><link>https://news.ycombinator.com/item?id=47172335</link><dc:creator>ezwoodland</dc:creator><comments>https://news.ycombinator.com/item?id=47172335</comments><guid isPermaLink="false">https://news.ycombinator.com/item?id=47172335</guid></item><item><title><![CDATA[New comment by ezwoodland in "The Wolfram S Combinator Challenge"]]></title><description><![CDATA[
<p>That doesn't matter.
You could imagine a system that accumulates two terms: (actual result, junk). Instead of deleting something it just adds it to the junk part of the pair.
Maybe the junk part itself has computation which never ends, but it doesn't matter because you just extract your result from the left part of the pair.</p>
]]></description><pubDate>Thu, 26 Feb 2026 20:40:30 +0000</pubDate><link>https://news.ycombinator.com/item?id=47171731</link><dc:creator>ezwoodland</dc:creator><comments>https://news.ycombinator.com/item?id=47171731</comments><guid isPermaLink="false">https://news.ycombinator.com/item?id=47171731</guid></item><item><title><![CDATA[New comment by ezwoodland in "Show HN: If you lose your memory, how to regain access to your computer?"]]></title><description><![CDATA[
<p>Something along the lines of reed-solomon codes could work for you:<p>If you want to share your password with M family members such that you only need N to agree to recover the original:<p>Split your password into ordered chunks.<p>Make a polynomial p, of power N where the p(1) = chunk1, p(2) = chunk2, ...<p>Evaluate the polynomial at M other points: p(N+1),p(N+2)...<p>Gives those M new points to your family along with their index (+1,+2,...).<p>If less than N family members get together, they will not be able to figure out the password much better than guessing. If N get together, they can interpolate their points to form the unique polynomial which will match p. Then evaluate p at p(1),p(2),... to get your original password.<p>If you put the whole password into 1 chunk, and pad the polynomial with random extra coefficients or points to make the polynomial of sufficient degree, then they get literally no information on the password without having at least N cooperate. If you make multiple chunks then they can do a little correlation between the chunks without knowing the whole thing.<p>This is sufficiently simple you can even work this out by hand without a computer, though it would be somewhat tedious.</p>
]]></description><pubDate>Fri, 06 Feb 2026 22:52:50 +0000</pubDate><link>https://news.ycombinator.com/item?id=46919289</link><dc:creator>ezwoodland</dc:creator><comments>https://news.ycombinator.com/item?id=46919289</comments><guid isPermaLink="false">https://news.ycombinator.com/item?id=46919289</guid></item><item><title><![CDATA[New comment by ezwoodland in "Zig's comptime is bonkers good"]]></title><description><![CDATA[
<p>I think the parsed c program is the DSL. You have to write a parser and compiler for the c-like source to the actual c code.</p>
]]></description><pubDate>Tue, 07 Jan 2025 19:11:46 +0000</pubDate><link>https://news.ycombinator.com/item?id=42626107</link><dc:creator>ezwoodland</dc:creator><comments>https://news.ycombinator.com/item?id=42626107</comments><guid isPermaLink="false">https://news.ycombinator.com/item?id=42626107</guid></item></channel></rss>