<instructions>Your goal is to analyze the following comment and estimate how highly it will be upvoted by the Hacker News community.</instructions><comment><author>gowld</author><text>This description makes it sounds like large areas of computer science are just goofy, meaningless, games.&lt;p&gt;But what&amp;#x27;s really happening is that &amp;quot;infinity&amp;quot; is standing in for &amp;quot;approximate, eventual, steady state behavior for sufficiently large N,  larger than any specific one-off gimmick you might think of&amp;quot;.&lt;p&gt;In the real world, though, those gimmicks are important, and the constants and low-order terms ignored in a Big-O comparison are important to real world performance.&lt;p&gt;There is constant tension between &amp;quot;big enough problem that the contant factors don&amp;#x27;t matter&amp;quot;, and &amp;quot;small enough problem that it conforms to the (often implicit)  of what &amp;#x27;constant&amp;#x27; means (example: 32bit ints masquerading as integers)&amp;quot;</text><parent_chain><item><author>Xcelerate</author><text>It can be sort of unintuitive how the concept of computability necessarily involves infinity.&lt;p&gt;For example: does there exist an algorithm that computes the Kolmogorov complexity, K(s), of string s for arbitrary s? It is well-known that the answer is &amp;quot;no&amp;quot; — there is no Turing machine that takes as input a string of arbitrary length and computes K(s). The proof is quite brief and involves the halting problem.&lt;p&gt;But if we ask a similar question: does there exist an algorithm that computes K(s) of string s for arbitrary string s with length &amp;lt; n? The answer is yes! And there exists such an algorithm for any value of n.&lt;p&gt;How is that possible? Think about it for a second, because the answer is going to disappoint you: simply create a Turing machine that consists of a giant lookup table for all 2^n possible strings that prints the value of K(s) for each one.&lt;p&gt;But wait, that&amp;#x27;s cheating! Maybe so, but any specific implementation of the algorithm has a finite description. And by definition, K(s) is also finite for all s. While it&amp;#x27;s true that I haven&amp;#x27;t provided any particular method for determining the value of K(s) for all 2^n strings in order to actually create the lookup table, that doesn&amp;#x27;t matter. Such an algorithm nevertheless exists, regardless of whether you can find it or prove that it does what you want it to.&lt;p&gt;So in a sense, finite questions about a finite number of things are sort of uninteresting from the perspective of computability, because you can always write a program that just prints the answer for all of those things (how quickly it does this is another matter). But when you extend the question to an infinite number of things, computability becomes much more interesting, because you don&amp;#x27;t know whether something finite can provide answers to questions about an infinite number of things.</text></item></parent_chain></comment><story><title>The zombie misconception of theoretical computer science</title><url>https://scottaaronson.blog/?p=8106</url></story>
<instructions>Your goal is to analyze the following comment and estimate how highly it will be upvoted by the Hacker News community.</instructions><comment><author>aidenn0</author><text>Similar to how all real-world computers have a finite number of states and are thus not Turing machines, but rather finite state machines.</text><parent_chain><item><author>Xcelerate</author><text>It can be sort of unintuitive how the concept of computability necessarily involves infinity.&lt;p&gt;For example: does there exist an algorithm that computes the Kolmogorov complexity, K(s), of string s for arbitrary s? It is well-known that the answer is &amp;quot;no&amp;quot; — there is no Turing machine that takes as input a string of arbitrary length and computes K(s). The proof is quite brief and involves the halting problem.&lt;p&gt;But if we ask a similar question: does there exist an algorithm that computes K(s) of string s for arbitrary string s with length &amp;lt; n? The answer is yes! And there exists such an algorithm for any value of n.&lt;p&gt;How is that possible? Think about it for a second, because the answer is going to disappoint you: simply create a Turing machine that consists of a giant lookup table for all 2^n possible strings that prints the value of K(s) for each one.&lt;p&gt;But wait, that&amp;#x27;s cheating! Maybe so, but any specific implementation of the algorithm has a finite description. And by definition, K(s) is also finite for all s. While it&amp;#x27;s true that I haven&amp;#x27;t provided any particular method for determining the value of K(s) for all 2^n strings in order to actually create the lookup table, that doesn&amp;#x27;t matter. Such an algorithm nevertheless exists, regardless of whether you can find it or prove that it does what you want it to.&lt;p&gt;So in a sense, finite questions about a finite number of things are sort of uninteresting from the perspective of computability, because you can always write a program that just prints the answer for all of those things (how quickly it does this is another matter). But when you extend the question to an infinite number of things, computability becomes much more interesting, because you don&amp;#x27;t know whether something finite can provide answers to questions about an infinite number of things.</text></item></parent_chain></comment><story><title>The zombie misconception of theoretical computer science</title><url>https://scottaaronson.blog/?p=8106</url></story>