tag:blogger.com,1999:blog-5204863782883637837.post4737295808540051230..comments2024-03-27T20:51:11.303-04:00Comments on Accepting Ignorance: Empirical Disproof of Turing's Halting ProblemAlrenoushttp://www.blogger.com/profile/11119846531341190283noreply@blogger.comBlogger10125tag:blogger.com,1999:blog-5204863782883637837.post-28489114163269344292022-02-18T16:04:51.019-05:002022-02-18T16:04:51.019-05:00You're explaining something everyone else here...You're explaining something everyone else here already knows. You're the one who doesn't get it. You're probably crazy and should be in an asylum. Alrenoushttps://www.blogger.com/profile/11119846531341190283noreply@blogger.comtag:blogger.com,1999:blog-5204863782883637837.post-61996334351509626782022-02-18T12:56:11.454-05:002022-02-18T12:56:11.454-05:00This comment has been removed by a blog administrator.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5204863782883637837.post-59195989448650028352022-02-14T20:31:08.438-05:002022-02-14T20:31:08.438-05:00Possibly. You weren't entirely unambiguous.
T...Possibly. You weren't entirely unambiguous.<br /><br />That said yes that's the keystone and I myself understand better what I'm saying now you've highlighted it.<br /><br />For the proof to work, program g has to actually exist. Ideally you would be able to physically write program g in C++ or python. <br />Program g is a function of f, which is a function of g, which is a function of f...<br />This function does not converge. You definitely absolutely cannot write this in C++. Even if you somehow pseudocoded it properly, it would fail to compile. <br /><br />Turing has instead proven the opposite of what he's alleged to prove. "Halting is undecidable if program g exists. Program g does not exist. Modus tollens: halting is decidable."<br /><br />Since Godel's first incompleteness and Turing's halting are isomorphic, Turing has likewise proven that formal systems are complete. Alrenoushttps://www.blogger.com/profile/11119846531341190283noreply@blogger.comtag:blogger.com,1999:blog-5204863782883637837.post-82354058008632353082022-02-14T06:48:20.984-05:002022-02-14T06:48:20.984-05:00Okay, I think I finally understood the flaw in the...Okay, I think I finally understood the flaw in the proof of the lack of any solutions to halting problem.<br /><br /><br />From wikipedia:<br /><br />"For any program f that might determine if programs halt, a "pathological" program g, called with some input, can pass its own source and its input to f and then specifically do the opposite of what f predicts g will do. No f can exist that handles this case."<br /><br />If f cannot be represented by any program, then g is not a program. f, which takes as input only programs, cannot be expected to tackle the case of g.<br /><br />Is this the case you are making?BSRK Adityahttps://www.blogger.com/profile/17264227679725849601noreply@blogger.comtag:blogger.com,1999:blog-5204863782883637837.post-11379616790651079532022-01-06T20:45:45.888-05:002022-01-06T20:45:45.888-05:00I don't. I investigate each while individually...I don't. I investigate each while individually. Then I investigate the [nesting] relationship. The program has three components, and I need to understand three components. If I don't know the components then I don't use them. <br /><br />However, I don't have to investigate every instance of nesting as if it were some unique species. I can investigate nesting per se, once, and then understand it for all programs. <br /><br />It turns out nesting doesn't affect the issue unless it breaks certain rules of variable ownership, in which case the problem isn't [nesting] it's spaghetti code. I consider the latter a distinct kind of relationship. <br /><br />If the while inside halts, and the while outside halts, then the program halts. No amount of nesting finite programs will get you an infinite program. Alrenoushttps://www.blogger.com/profile/11119846531341190283noreply@blogger.comtag:blogger.com,1999:blog-5204863782883637837.post-69067033843204163102022-01-05T22:05:00.054-05:002022-01-05T22:05:00.054-05:00How will you predict the behaviour of two nested w...How will you predict the behaviour of two nested whiles using just the behaviour of one while? I.e.<br /><br /><br />While (test1) {<br /> While (test2) {<br /> ...<br /> }<br /> ...<br />}<br /><br /><br /><br />Using just<br /><br />While(test) {<br /> ...<br />}<br /><br />& ...?<br /><br /><br />(... represents non looping code)<br />BSRK Adityahttps://www.blogger.com/profile/17264227679725849601noreply@blogger.comtag:blogger.com,1999:blog-5204863782883637837.post-18718322005620766032022-01-05T20:57:29.225-05:002022-01-05T20:57:29.225-05:00It's not a hidden hypothesis, I said it straig...It's not a hidden hypothesis, I said it straight out: "Some of their relationships result in non-halting, and you can just list these relationships if you want, because there's like six simple machines and thus like 15 binary relationships between machines"<br /><br />Don't use the non-halting relationships, don't get infinite loops. You can pack as many halting algorithms together as you want and they'll still halt. If you're not sure, then check. Alrenoushttps://www.blogger.com/profile/11119846531341190283noreply@blogger.comtag:blogger.com,1999:blog-5204863782883637837.post-85378356561088614512022-01-05T20:52:15.654-05:002022-01-05T20:52:15.654-05:00> A program is nothing more than a series of al...> A program is nothing more than a series of algorithms. If all parts of a construct are clay, it is a clay construct.<br /><br />Okay. Are all algorithms finite in length?<br /><br />> If the trinary relationships matter, I think there's 90 of them.<br /><br />I can see how you can enumerate all programs with a program. But how do you determine whether a program halts or not based on this?<br /><br />There seems to be a hidden hypothesis here - namely that it's possible to determine if a program of size N halts based on the behaviour of smaller programs.<br /><br /><br />> Here's my algorithm to determine if a program halts: run it and see what happens.<br /><br />This will say that the program halts for all halting programs. However, for non halting programs it will not halt. It's supposed to output "it will not halt".<br />BSRK Adityahttps://www.blogger.com/profile/17264227679725849601noreply@blogger.comtag:blogger.com,1999:blog-5204863782883637837.post-22997648425781541652022-01-05T15:18:25.600-05:002022-01-05T15:18:25.600-05:00A program is nothing more than a series of algorit...A program is nothing more than a series of algorithms. If all parts of a construct are clay, it is a clay construct. Alrenoushttps://www.blogger.com/profile/11119846531341190283noreply@blogger.comtag:blogger.com,1999:blog-5204863782883637837.post-26908280297181549632022-01-05T08:28:38.888-05:002022-01-05T08:28:38.888-05:00> How can there be no general algorithm to dete...> How can there be no general algorithm to determine if a program halts if every particular program has a particular proof? <br /><br />Do you define an algorithm to be the same thing as a program?BSRK Adityahttps://www.blogger.com/profile/17264227679725849601noreply@blogger.com