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He has created a software implementation of this micr..." 13:04:32 [[Gb gates RISC]] https://esolangs.org/w/index.php?diff=53641&oldid=53640 * B jonas * (+90) 13:05:12 [[Language list]] https://esolangs.org/w/index.php?diff=53642&oldid=53628 * B jonas * (+20) 13:05:21 mroman: H should be inlined. And in any case the proofs rely on interpreters, because you cannot pass programs around, only descriptions of programs. 13:05:54 but then this means that P knows what H is 13:05:57 otherwise it can't inline it. 13:06:02 `recipe 13:06:02 2 tablespoon of the whipped cream into the pan in lightly floured \ get and deep flour and including for 1 hour. Top with side of \ the sugar, chopped fresh chop more. \ \ Recipe By : Low-Bobbie Cooking \ \ MMMMM \ \ MMMMM----- Recipe via Meal-Master (tm) v8.05 \ \ Title: ICE ENCION COOKIES \ Categories: Desserts, Germanaus \ Yield: 4 13:06:23 chop freshed chop more! 13:06:31 so the program would contain its own "halting function" 13:07:59 mroman: So what you really do is write a program P'(h,p) = if H(eval(p(h,p))) then endless_loop else halt; And then look at P'([H],[P']), where [_] is the description of the program inside the brackets. 13:08:23 and eval is the interpreter, and the p(h,p) inside there is actually a kind of substitution. 13:08:40 yeah but where is H 13:08:54 H is assumed to exist, so you know its description. 13:09:12 so you pass the halting program to P' 13:09:36 or what's small h 13:09:40 P' knows what H is. P' doesn't know what P is, but it will be told what P' is and can then reconstruct P from P' and H. 13:09:45 and then run it. 13:10:06 Small h is a parameter that will be replaced by [H]. 13:10:14 wNom nom nom 13:10:16 -w 13:10:39 but this again assumes that you can actually call H within P. 13:10:40 But actually the h [H] part isn't needed 13:10:49 or P' 13:11:09 I could define P'(p) = if H(eval(p(p))) then loop else halt. 13:11:20 The trick is to then consider P'([P']). 13:12:58 that makes no sense either 13:13:04 you're missing a parameter then. 13:13:29 (the p(p) inside is still cheating with notation, it's an operation that takes p to be a description of a program P(x) with a single parameter, and replaces x by a description of the value p, syntactically.) 13:14:08 so the result is a description of a program that behaves like P(p). 13:14:50 ... which will end up being P'([P']), I'm afraid I'm not explaining this very well. 13:15:24 I assume that there's a program H that can determine whether P halts or not but P doesn't know H. 13:15:27 But anyway, in rthe if H(P) then loop else halt, H is never the problem. 13:15:37 and P can't communicate with H (H being a seperate program) 13:15:45 P is essentially I/O-free. 13:16:06 But anyway, in "if H(P) then loop else halt", H is never the problem. It's the reference of P to itself which you need to escape. 13:16:13 H takes the source code of P (or some representation of P) 13:16:31 but for practical reason I'll refer to that as source code of P :D 13:16:54 so P can't call H unless P contains H 13:17:07 mroman: btw this isn't anything unusual. The trick here is essentially the same as that for writing quines. 13:17:09 which means P would have to know what Hlooks like. 13:17:37 Rather than having P store its full source code, it stores about half of it, and reconstructs the full source code from that. 13:17:56 well it's certainly possible to feed yourself to print 13:18:00 The half that is being stored is [P']. 13:18:02 == that's a quine 13:18:14 but you can only feed yourself to H if you know what H is 13:18:17 because you have to include H 13:18:22 Yes, you do include H 13:18:24 P isn' 13:18:29 t uniform. 13:18:43 (in H) 13:19:01 I'm not sure why you think that including H is problematic. 13:19:23 because this makes the assumption that H actually exists. 13:19:32 Yes. 13:19:45 You assume that H exists, then you make a program that H cannot decide the halting problem for. 13:20:16 Yes, but I assume that an external H exists 13:20:22 So the assumption is wrong. End of story. 13:20:24 but I don't know whether P can actually construt H itself. 13:20:53 maybe H is infinitely large 13:20:57 but will terminate in a finite number of steps 13:21:26 which would make it really hard for P to construct H 13:21:54 because if P is infinitely large you can't really include it in P 13:22:02 unless you can reduce it to a finitely large representation 13:22:39 so to be able to use this as a proof 13:22:44 I'd have to proof first that P can include H. 13:22:54 because if that is impossible to begin with 13:23:06 I can't make arguments based on the assumption that P can include H and then call H 13:23:22 But you /can/ make a description of H a parameter of P, which I mostly did when I wrote (almost) "P'(h,p) = if eval(h(p(h,p))) then loop else halt" By partial substitution you can make P(h) = if eval(h(P'(h,[P'])) then halt alse loop", so that's a program for which, H cannot decide termination of P([H]). 13:23:43 But that's really proving more than the halting problem entails. 13:24:08 I'm also assuming P is parameterless 13:24:09 if "H is infinitely large" then it's not a Turing machine. 13:24:12 because paremeters are basically I/O 13:24:16 You cannot disprove the existence of oracles. 13:24:17 *parameters 13:24:26 but turing machines have infinite memory? 13:24:38 unbounded. 13:24:59 so I could fit an infinitely large program in there and then my turing machine interprets that program 13:24:59 every TM configuration is finite 13:25:05 oh? 13:25:06 hm. 13:25:09 I did not know that. 13:26:16 so the initial state of a TM is finite 13:26:35 or at least has a finite representation 13:26:46 (arguably in BF the inital state is an infinite amount of zeroes) 13:26:52 (that's not really a finite initial state then) 13:29:25 and quines aren't exactly turing machines either? 13:29:52 the proof is essentially that for a turing machine T(x) = y an input i exists such that T(i) = i 13:30:03 meaning that T has a fix point. 13:30:16 *fixed point 13:30:21 or specifically that it has at least one fixed point. 13:30:47 yeah the common trick here is called "diagonalization". 13:31:10 so if P contains both itself and H 13:31:15 it must be a fixed point as well? 13:31:51 it must be a program that contains its own halting program 13:32:50 but it can only do that if H exists. 13:32:55 if H doesn't exist then it can't contain H 13:33:01 and that's what I struggle with 13:33:07 It sounds like an impossible circular argument. 13:34:16 The program doesn't contain itself, just a description of something that behaves like itself. 13:35:35 -!- Guest45071 has changed nick to ^arcade_droid. 13:35:41 -!- ^arcade_droid has quit (Changing host). 13:35:41 -!- ^arcade_droid has joined. 13:36:04 > text$ap(++)show"> text$ap(++)show" 13:36:06 > text$ap(++)show"> text$ap(++)show" 13:37:33 Assume a halting program H exists that can answer whether P halts or not (P being a SPECIFIC program). 13:37:59 For the argument `P = if H(P) then loop() else halt` to work P needs to contain itself and it needs to contain its halting program (in whatever form) 13:38:01 (the "> text$ap(++)show" isn't even the program itself; it's missing a constant. the ap(++)show does the substitution (degenerated; it appends the string to the partial program); that is the description of the program itself. And for the halting problem, you'd pass something like that to H.) 13:39:09 the output of H for every other program other than P is nonsense 13:39:23 (or undetermined) 13:39:31 essentially, H isn't defined on programs other than P 13:39:42 (I think in english this is called the domain of H) 13:40:02 For any specific program P, one of the two TMs that always reject or always accept will decide the halting problem for P. 13:40:17 * int-e shrugs 13:40:56 I'm trying to understand if for every P an H exists that answers whether P halts or not 13:41:25 The halting problem is the problem of finding a program H such that for any program P, H accepts [P] if and only if P halts. 13:42:44 mroman: I just answered that question. Yes, but not in any interesting way, because unless we already know whether P terminates of not, we don't know which of the two candidates for H is the right one. 13:43:50 but how would P know then which H to include? 13:44:19 However, I was under the impression that we were discussing the halting problem. Now that I've figured out that we're not doing that I'm no longer interested. 13:44:25 :) 13:44:47 [[User:B jonas]] https://esolangs.org/w/index.php?diff=53643&oldid=53639 * B jonas * (-26) 13:45:06 I don't even understand what the question is anymore. 13:45:27 I accept the proof for the halting problem based on the assumption that P can communicate with H over some form of I/O. 13:45:40 I don't think the proof holds if P can't do that. 13:46:36 because this way P doesn't need to know what H actually is. 13:46:39 P can treat H as a black box. 13:46:42 Well you're wrong. And I think you have a wrong statement of the halting problem in mind. I don't know what that statement is. 13:48:19 if P can't communicate with H it could only do that if P contains a copy of H but if P contains a copy of H it inherently needs to know what H actually is. 13:48:44 but since your proof essentially says that H can't exist 13:48:57 and that's what I take is what the proof is saying 13:49:01 It's a proof by contradiction. 13:49:12 P can't possible ever embed H to begin with 13:49:22 making the argument based on the assumption that P can actually embed H invalid. 13:49:41 (Actually it's not, it's negation introduction, so even intuitionistically you should not have any complaints.) 13:49:42 H doesn't exist if and only if P can embed H. 13:50:02 but if P can embed H, then H must exist. 13:50:09 -!- variable has joined. 13:50:25 Suppose H exists. Then there would be a program P such that H gets termination of P wrong. Hence H doesn't exist. 13:51:39 Essentially I'm saying the program you construct to proof the non-existence can itself not exist. 13:52:10 You're given a program P. 13:52:22 All the programs "P" in that proof are necessarily hypothetical, though they can be obtained by making a description of H a parameter of P. But whatever you ultimately pass to H in order to obtain a contradiction will contain H or a description of H. 13:52:51 for the P that was given to you you write a program that answers whether P halts or not (without prior knowledge about whether P halts or not) 13:52:52 mroman: If you fix P you're discussing something entirely different from the halting problem. 13:53:07 int-e: apparentely :) 13:53:11 Hm 13:53:24 * APic just thought of a nice Project 13:53:33 A figlet/toilet Reverser 13:53:40 That decomposes Text on IRC or similar 13:53:53 banner is traditional :P 13:54:01 With Font-Auto-Detection 13:54:04 [[BuzzFizz]] https://esolangs.org/w/index.php?diff=53644&oldid=53634 * B jonas * (+1) /* External resources */ 13:54:04 Yes, banner too 13:54:40 One could hack some OCR-Software to convert Letters back to Pixels 13:55:00 Hm 13:55:30 Or just render the Font and then use a stock OCR 13:55:51 So quite simple, not _that_ interesting Project 😉 13:56:04 int-e: yeah I guess I need to reformulate that properly 13:56:15 so knowledged people actually know what I'm talking about :D 13:56:22 APic: or just start feeding parameters and strings of increasing length to $program 13:56:58 (you'll need some lenience because of kerning and the like, but it seems to me that this stupid approach should be good enough anyway) 13:57:17 int-e: lol, good Brute-Force-Idea 13:58:36 in fact guessing the right font may well be the hardest part for this reversal 14:02:51 int-e: http://codepad.org/uKMXVOaJ 14:02:56 those are the two cases I was talking about. 14:04:02 how can Q exist? 14:04:44 mroman: it doesn't. Q is entirely hypothetical, based on the assumption that H exists. 14:05:19 but if Q doesn't exist 14:05:25 how can you construct it to proof that H doesn't exist? 14:06:10 mroman: This is like the largest prime. If 2 = p_1 < p_2 < ... < p_n is a complete list of primes, then p_1*...*p_n+1 is not divisible by any (other) prime, hence must be prime as well, a contradiction. So there is no such complete list of primes. 14:06:26 yes, that's a traditional proof by contradiction 14:06:32 like sqrt(2) being not a rational number. 14:06:44 but those are entirely different. 14:06:47 And then it turns out that in most cases, p_1*...*p_n+1 for a prefix of the list of primes is not actually prime but divisible by some other prime... 14:07:47 e.g., 2*3*5*7*11*13+1 = 59*509 14:08:55 despite the fact that under the hypothesis that 2,3,5,7,11,13 contains all primes, it would be prime. Ex falso quod libet. 14:09:15 hmm, single word? 14:09:19 * int-e should stay away from latin. 14:09:35 http://www.oxfordreference.com/view/10.1093/oi/authority.20110803095804354 14:12:01 mroman: and for part 1), the trouble is that by the law of excluded middle, P either terminates (and then H exists), or P doesn't terminate (and H exists). So classically, H exists. Constructively, you're back in halting problem territory, because without further information on P, you need a /uniform/ construction for H that works for all possible P. 14:13:59 (Such a construction can be performed by a Turing Machine, and the result can be evaluated using a universal TM (vulgo interpreter)) 14:15:08 I'll readily admit that this is all rather subtle and full of pitfalls. I can't even promise that I'm avoiding all of them successfully, but I'm trying. :) 14:32:07 -!- idris-bot has quit (Ping timeout: 268 seconds). 14:32:16 `coins 14:32:17 -!- Melvar has quit (Ping timeout: 256 seconds). 14:32:19 ​nuntcoin blintercoin obstocoin coborcoin micrcoin lndcoin crabicoin sholdcoin eldcoin brycoin mirecoin redchcoin trainccoin omercoin codecoin mdccoin djousioncoin q-balcoin surfcoin osseacoin 14:35:17 -!- doesthiswork has joined. 14:37:52 [05:54:40] One could hack some OCR-Software to convert Letters back to Pixels 14:38:21 https://twitter.com/reverseocr 14:38:35 pfft 14:38:48 int-e: ? 14:39:27 variable: maybe I shouldn't laugh. undoing this kind of OCR could be quite tricky: http://int-e.eu/~bf3/tmp/ocr.png 14:39:30 int-e: what happens if I take your derivative. Do you become 'e'? or int-int-e? or something else ? 14:39:36 (original above, OCR below) 14:39:46 lol true 14:40:02 variable: Sorry, but I'm discrete. 14:41:41 variable: nice feed, though not quite as tautological as I expected. 14:42:15 `? fonts 14:42:16 ​#esoteric bitmap fonts include: \oren\'s font http://www.orenwatson.be/fontdemo.htm , lifthrasiir's font https://github.com/lifthrasiir/unison/ https://lifthrasiir.github.io/unison/sample.png , b_jonas's font http://www.math.bme.hu/~ambrus/pu/fecupboard20-c.pcf.gz , fizzie's font https://github.com/fis/rfk86/tree/master/web/font 14:42:36 my IRC client displays images inline 14:42:41 that was um.... fun to see 14:47:47 -!- Melvar has joined. 15:04:48 -!- variable has changed nick to function. 15:07:18 -!- function has changed nick to constant. 15:08:19 -!- laerling has joined. 15:10:53 -!- friendlyGoat has joined. 15:27:58 -!- friendlyGoat has left. 15:30:55 -!- constant has quit (Quit: Found 1 in /dev/zero). 15:33:17 -!- sebbu2 has joined. 15:34:38 -!- sebbu has quit (Ping timeout: 272 seconds). 15:34:41 -!- sebbu2 has changed nick to sebbu. 15:37:59 -!- mroman has quit (Ping timeout: 260 seconds). 16:08:24 -!- xkapastel has joined. 16:28:53 -!- LKoen has quit (Remote host closed the connection). 16:35:00 -!- ais523 has joined. 16:45:19 -!- ais523 has quit (Ping timeout: 248 seconds). 16:45:21 -!- callforjudgement has joined. 16:48:21 -!- callforjudgement has changed nick to ais253. 16:48:24 -!- ais253 has changed nick to ais523. 16:57:42 Would it possible to improve the OCR by cropping pictures of each individual letter and then specifying what letters they are, and see if that improve them? 17:00:34 What? 17:00:43 If you know what letter they are why do you need to OCR them 17:01:10 To avoid having to retype the entire document, so that you only have to retype a part of it. 17:06:49 -!- sleffy has joined. 17:19:57 -!- LKoen has joined. 17:59:57 -!- Phantom_Hoover has joined. 18:00:02 -!- Phantom_Hoover has quit (Changing host). 18:00:02 -!- Phantom_Hoover has joined. 18:20:59 -!- jaboja has joined. 18:33:48 -!- LKoen has quit (Remote host closed the connection). 18:55:05 -!- Hoolootwo has changed nick to Hooloovo0. 19:17:35 -!- LKoen has joined. 19:30:19 -!- LKoen has quit (Remote host closed the connection). 19:40:32 -!- zseri has joined. 20:04:22 -!- sprocklem has joined. 20:33:32 -!- oerjan has joined. 20:38:27 -!- moony has joined. 20:46:23 -!- zseri has quit (Quit: Leaving). 20:46:35 -!- sprocklem has quit (Ping timeout: 252 seconds). 20:52:05 -!- moony has quit (Ping timeout: 252 seconds). 21:05:09 -!- ais523 has quit (Quit: sorry for my connection). 21:05:23 -!- ais523 has joined. 21:35:16 . o O ( that halting problem discussion was painful to read ) 21:42:50 -!- LKoen has joined. 21:42:50 -!- LKoen has quit (Client Quit). 22:14:37 o 22:28:44 -!- laerling has quit (Quit: Leaving). 22:36:02 -!- sleffy has quit (Ping timeout: 252 seconds). 22:37:14 -!- sprocklem has joined. 22:47:36 -!- newsham has quit (Read error: Connection reset by peer). 22:47:52 -!- newsham has joined. 22:48:10 -!- augur has joined. 23:03:51 -!- sleffy has joined. 23:12:26 oerjan: did I get it horribly wrong? 23:32:19 -!- vertrex has quit (Ping timeout: 255 seconds). 23:33:42 -!- vertrex has joined. 23:45:31 -!- variable has joined. 23:45:52 int-e: well i don't think your explanation was precise enough to get past the horrible miscommunication barrier at a reasonable speed. 23:46:10 it may not have been fundamentally wrong. 23:48:25 i suppose this may have some of the same flavor as those arguments people have over cantor's theorem for real numbers... 23:56:39 -!- newsham_ has joined. 23:57:54 -!- newsham has quit (Read error: Connection reset by peer).