←2017-12-20 2017-12-21 2017-12-22→ ↑2017 ↑all
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00:03:18 <\oren\_> oerjan: he's a famous pop singer
00:09:42 <boily> ヘ\\オレン\_!
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00:18:27 <oerjan> \oren\_: was hth
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01:55:40 <esowiki> [[User:B jonas]] https://esolangs.org/w/index.php?diff=53638&oldid=53333 * B jonas * (+26)
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08:08:43 <ais523> @messages?
08:08:43 <lambdabot> Sorry, no messages today.
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12:06:38 <esowiki> [[User:B jonas]] https://esolangs.org/w/index.php?diff=53639&oldid=53638 * B jonas * (+30)
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13:01:39 <mroman> Are there proofs for the halting problem that don't rely on I/O?
13:02:23 <mroman> `if H(P) then endless_loop(); else halt();` usually assumes that H is an external "machine/program" you invoke
13:02:24 <HackEgo> bash: -c: line 0: syntax error near unexpected token `P' \ bash: -c: line 0: `if H(P) then endless_loop(); else halt();` usually assumes that H is an external "machine/program" you invoke'
13:02:30 <mroman> but that's I/O.
13:02:44 <mroman> if you disallow I/O then P would have to contain H
13:03:20 <esowiki> [[Gb gates RISC]] N https://esolangs.org/w/index.php?oldid=53640 * B jonas * (+4593) Created page with "{{DISPLAYTITLE: Gb_gates RISC}} '''Gb_gates RISC''' is a very small hypothetical microprocessor created by Donald Knuth. He has created a software implementation of this micr..."
13:04:32 <esowiki> [[Gb gates RISC]] https://esolangs.org/w/index.php?diff=53641&oldid=53640 * B jonas * (+90)
13:05:12 <esowiki> [[Language list]] https://esolangs.org/w/index.php?diff=53642&oldid=53628 * B jonas * (+20)
13:05:21 <int-e> 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 <mroman> but then this means that P knows what H is
13:05:57 <mroman> otherwise it can't inline it.
13:06:02 <b_jonas> `recipe
13:06:02 <HackEgo> 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 <b_jonas> chop freshed chop more!
13:06:31 <mroman> so the program would contain its own "halting function"
13:07:59 <int-e> 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 <int-e> and eval is the interpreter, and the p(h,p) inside there is actually a kind of substitution.
13:08:40 <mroman> yeah but where is H
13:08:54 <int-e> H is assumed to exist, so you know its description.
13:09:12 <mroman> so you pass the halting program to P'
13:09:36 <mroman> or what's small h
13:09:40 <int-e> 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 <int-e> and then run it.
13:10:06 <int-e> Small h is a parameter that will be replaced by [H].
13:10:14 <APic> wNom nom nom
13:10:16 <APic> -w
13:10:39 <mroman> but this again assumes that you can actually call H within P.
13:10:40 <int-e> But actually the h [H] part isn't needed
13:10:49 <mroman> or P'
13:11:09 <int-e> I could define P'(p) = if H(eval(p(p))) then loop else halt.
13:11:20 <int-e> The trick is to then consider P'([P']).
13:12:58 <mroman> that makes no sense either
13:13:04 <mroman> you're missing a parameter then.
13:13:29 <int-e> (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 <int-e> so the result is a description of a program that behaves like P(p).
13:14:50 <int-e> ... which will end up being P'([P']), I'm afraid I'm not explaining this very well.
13:15:24 <mroman> 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 <int-e> But anyway, in rthe if H(P) then loop else halt, H is never the problem.
13:15:37 <mroman> and P can't communicate with H (H being a seperate program)
13:15:45 <mroman> P is essentially I/O-free.
13:16:06 <int-e> 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 <mroman> H takes the source code of P (or some representation of P)
13:16:31 <mroman> but for practical reason I'll refer to that as source code of P :D
13:16:54 <mroman> so P can't call H unless P contains H
13:17:07 <int-e> mroman: btw this isn't anything unusual. The trick here is essentially the same as that for writing quines.
13:17:09 <mroman> which means P would have to know what Hlooks like.
13:17:37 <int-e> 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 <mroman> well it's certainly possible to feed yourself to print
13:18:00 <int-e> The half that is being stored is [P'].
13:18:02 <mroman> == that's a quine
13:18:14 <mroman> but you can only feed yourself to H if you know what H is
13:18:17 <mroman> because you have to include H
13:18:22 <int-e> Yes, you do include H
13:18:24 <int-e> P isn'
13:18:29 <int-e> t uniform.
13:18:43 <int-e> (in H)
13:19:01 <int-e> I'm not sure why you think that including H is problematic.
13:19:23 <mroman> because this makes the assumption that H actually exists.
13:19:32 <int-e> Yes.
13:19:45 <int-e> You assume that H exists, then you make a program that H cannot decide the halting problem for.
13:20:16 <mroman> Yes, but I assume that an external H exists
13:20:22 <int-e> So the assumption is wrong. End of story.
13:20:24 <mroman> but I don't know whether P can actually construt H itself.
13:20:53 <mroman> maybe H is infinitely large
13:20:57 <mroman> but will terminate in a finite number of steps
13:21:26 <mroman> which would make it really hard for P to construct H
13:21:54 <mroman> because if P is infinitely large you can't really include it in P
13:22:02 <mroman> unless you can reduce it to a finitely large representation
13:22:39 <mroman> so to be able to use this as a proof
13:22:44 <mroman> I'd have to proof first that P can include H.
13:22:54 <mroman> because if that is impossible to begin with
13:23:06 <mroman> I can't make arguments based on the assumption that P can include H and then call H
13:23:22 <int-e> 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 <int-e> But that's really proving more than the halting problem entails.
13:24:08 <mroman> I'm also assuming P is parameterless
13:24:09 <int-e> if "H is infinitely large" then it's not a Turing machine.
13:24:12 <mroman> because paremeters are basically I/O
13:24:16 <int-e> You cannot disprove the existence of oracles.
13:24:17 <mroman> *parameters
13:24:26 <mroman> but turing machines have infinite memory?
13:24:38 <int-e> unbounded.
13:24:59 <mroman> so I could fit an infinitely large program in there and then my turing machine interprets that program
13:24:59 <int-e> every TM configuration is finite
13:25:05 <mroman> oh?
13:25:06 <mroman> hm.
13:25:09 <mroman> I did not know that.
13:26:16 <mroman> so the initial state of a TM is finite
13:26:35 <mroman> or at least has a finite representation
13:26:46 <mroman> (arguably in BF the inital state is an infinite amount of zeroes)
13:26:52 <mroman> (that's not really a finite initial state then)
13:29:25 <mroman> and quines aren't exactly turing machines either?
13:29:52 <mroman> the proof is essentially that for a turing machine T(x) = y an input i exists such that T(i) = i
13:30:03 <mroman> meaning that T has a fix point.
13:30:16 <mroman> *fixed point
13:30:21 <mroman> or specifically that it has at least one fixed point.
13:30:47 <int-e> yeah the common trick here is called "diagonalization".
13:31:10 <mroman> so if P contains both itself and H
13:31:15 <mroman> it must be a fixed point as well?
13:31:51 <mroman> it must be a program that contains its own halting program
13:32:50 <mroman> but it can only do that if H exists.
13:32:55 <mroman> if H doesn't exist then it can't contain H
13:33:01 <mroman> and that's what I struggle with
13:33:07 <mroman> It sounds like an impossible circular argument.
13:34:16 <int-e> The program doesn't contain itself, just a description of something that behaves like itself.
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13:36:04 <int-e> > text$ap(++)show"> text$ap(++)show"
13:36:06 <lambdabot> > text$ap(++)show"> text$ap(++)show"
13:37:33 <mroman> Assume a halting program H exists that can answer whether P halts or not (P being a SPECIFIC program).
13:37:59 <mroman> 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 <int-e> (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 <mroman> the output of H for every other program other than P is nonsense
13:39:23 <mroman> (or undetermined)
13:39:31 <mroman> essentially, H isn't defined on programs other than P
13:39:42 <mroman> (I think in english this is called the domain of H)
13:40:02 <int-e> 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 <mroman> I'm trying to understand if for every P an H exists that answers whether P halts or not
13:41:25 <int-e> 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 <int-e> 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 <mroman> but how would P know then which H to include?
13:44:19 <int-e> 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 <mroman> :)
13:44:47 <esowiki> [[User:B jonas]] https://esolangs.org/w/index.php?diff=53643&oldid=53639 * B jonas * (-26)
13:45:06 <int-e> I don't even understand what the question is anymore.
13:45:27 <mroman> 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 <mroman> I don't think the proof holds if P can't do that.
13:46:36 <mroman> because this way P doesn't need to know what H actually is.
13:46:39 <mroman> P can treat H as a black box.
13:46:42 <int-e> 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 <mroman> 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 <mroman> but since your proof essentially says that H can't exist
13:48:57 <mroman> and that's what I take is what the proof is saying
13:49:01 <int-e> It's a proof by contradiction.
13:49:12 <mroman> P can't possible ever embed H to begin with
13:49:22 <mroman> making the argument based on the assumption that P can actually embed H invalid.
13:49:41 <int-e> (Actually it's not, it's negation introduction, so even intuitionistically you should not have any complaints.)
13:49:42 <mroman> H doesn't exist if and only if P can embed H.
13:50:02 <mroman> but if P can embed H, then H must exist.
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13:50:25 <int-e> 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 <mroman> Essentially I'm saying the program you construct to proof the non-existence can itself not exist.
13:52:10 <mroman> You're given a program P.
13:52:22 <int-e> 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 <mroman> 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 <int-e> mroman: If you fix P you're discussing something entirely different from the halting problem.
13:53:07 <mroman> int-e: apparentely :)
13:53:11 <APic> Hm
13:53:24 * APic just thought of a nice Project
13:53:33 <APic> A figlet/toilet Reverser
13:53:40 <APic> That decomposes Text on IRC or similar
13:53:53 <int-e> banner is traditional :P
13:54:01 <APic> With Font-Auto-Detection
13:54:04 <esowiki> [[BuzzFizz]] https://esolangs.org/w/index.php?diff=53644&oldid=53634 * B jonas * (+1) /* External resources */
13:54:04 <APic> Yes, banner too
13:54:40 <APic> One could hack some OCR-Software to convert Letters back to Pixels
13:55:00 <APic> Hm
13:55:30 <APic> Or just render the Font and then use a stock OCR
13:55:51 <APic> So quite simple, not _that_ interesting Project 😉
13:56:04 <mroman> int-e: yeah I guess I need to reformulate that properly
13:56:15 <mroman> so knowledged people actually know what I'm talking about :D
13:56:22 <int-e> APic: or just start feeding parameters and strings of increasing length to $program
13:56:58 <int-e> (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 <APic> int-e: lol, good Brute-Force-Idea
13:58:36 <int-e> in fact guessing the right font may well be the hardest part for this reversal
14:02:51 <mroman> int-e: http://codepad.org/uKMXVOaJ
14:02:56 <mroman> those are the two cases I was talking about.
14:04:02 <mroman> how can Q exist?
14:04:44 <int-e> mroman: it doesn't. Q is entirely hypothetical, based on the assumption that H exists.
14:05:19 <mroman> but if Q doesn't exist
14:05:25 <mroman> how can you construct it to proof that H doesn't exist?
14:06:10 <int-e> 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 <mroman> yes, that's a traditional proof by contradiction
14:06:32 <mroman> like sqrt(2) being not a rational number.
14:06:44 <mroman> but those are entirely different.
14:06:47 <int-e> 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 <int-e> e.g., 2*3*5*7*11*13+1 = 59*509
14:08:55 <int-e> 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 <int-e> hmm, single word?
14:09:19 * int-e should stay away from latin.
14:09:35 <int-e> http://www.oxfordreference.com/view/10.1093/oi/authority.20110803095804354
14:12:01 <int-e> 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 <int-e> (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 <int-e> 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. :)
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14:32:16 <b_jonas> `coins
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14:32:19 <HackEgo> nuntcoin blintercoin obstocoin coborcoin micrcoin lndcoin crabicoin sholdcoin eldcoin brycoin mirecoin redchcoin trainccoin omercoin codecoin mdccoin djousioncoin q-balcoin surfcoin osseacoin
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14:37:52 <variable> [05:54:40] <APic>One could hack some OCR-Software to convert Letters back to Pixels
14:38:21 <variable> https://twitter.com/reverseocr
14:38:35 <int-e> pfft
14:38:48 <variable> int-e: ?
14:39:27 <int-e> 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 <variable> int-e: what happens if I take your derivative. Do you become 'e'? or int-int-e? or something else ?
14:39:36 <int-e> (original above, OCR below)
14:39:46 <variable> lol true
14:40:02 <int-e> variable: Sorry, but I'm discrete.
14:41:41 <int-e> variable: nice feed, though not quite as tautological as I expected.
14:42:15 <b_jonas> `? fonts
14:42:16 <HackEgo> ​#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 <variable> my IRC client displays images inline
14:42:41 <variable> that was um.... fun to see
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16:57:42 <zzo38> 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 <Slereah_> What?
17:00:43 <Slereah_> If you know what letter they are why do you need to OCR them
17:01:10 <zzo38> To avoid having to retype the entire document, so that you only have to retype a part of it.
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21:35:16 <oerjan> . o O ( that halting problem discussion was painful to read )
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22:14:37 <quintopia> o
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23:12:26 <int-e> oerjan: did I get it horribly wrong?
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23:45:52 <oerjan> 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 <oerjan> it may not have been fundamentally wrong.
23:48:25 <oerjan> i suppose this may have some of the same flavor as those arguments people have over cantor's theorem for real numbers...
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