←2019-06-18 2019-06-19 2019-06-20→ ↑2019 ↑all
00:10:47 <fizzie> Internet says the 9th and 10th digit of a UPS "1Z..." format tracking number denotes the service type, but this one has a code that doesn't appear on any of the lists of service codes.
00:46:59 <int-e> . o O ( Re-routed through Fort Meade. )
00:56:21 <shachaf> hi int-e
00:56:30 <shachaf> imo where's the application form to join the int-e fan club
00:58:49 <int-e> what did I do now?
00:59:59 <shachaf> maybe it'll say it on the form
01:00:47 <int-e> fungot: do you have the forms?
01:00:47 <fungot> int-e: you probably do not know whether this might not have been altogether fnord but it drove out a cant far more offensive.
01:00:58 <shachaf> ^style
01:00:59 <fungot> Available: agora alice c64 ct darwin discworld enron europarl ff7 fisher fungot homestuck ic irc iwcs jargon lovecraft nethack oots pa qwantz sms speeches* ss wp ukparl youtube
01:01:02 <int-e> for once I approve of the fnord.
01:01:07 <shachaf> ^style europarl
01:01:07 <fungot> Selected style: europarl (European Parliament speeches during approx. 1996-2006)
01:30:37 -!- rodgort has quit (Quit: Leaving).
01:39:49 -!- rodgort has joined.
02:05:53 -!- xkapastel has quit (Quit: Connection closed for inactivity).
05:16:20 -!- Lord_of_Life has quit (Ping timeout: 272 seconds).
05:22:15 -!- Lord_of_Life has joined.
06:16:12 -!- adu has quit (Quit: adu).
06:52:43 -!- imode has quit (Ping timeout: 248 seconds).
06:53:23 -!- AnotherTest has joined.
08:30:24 -!- FreeFull has quit (Ping timeout: 248 seconds).
09:35:48 -!- jix has quit (Ping timeout: 244 seconds).
09:36:16 -!- jix has joined.
09:45:37 -!- user24 has joined.
10:11:54 -!- arseniiv has joined.
10:55:38 -!- xkapastel has joined.
13:25:08 -!- xkapastel has quit (Quit: Connection closed for inactivity).
13:38:14 -!- AnotherTest has quit (Ping timeout: 252 seconds).
13:52:45 -!- john_metcalf has joined.
13:52:48 -!- user24 has quit (Quit: Leaving).
13:55:02 -!- xkapastel has joined.
13:59:58 -!- Melvar has quit (Quit: rebooting).
14:10:34 -!- john_metcalf has quit (Ping timeout: 246 seconds).
14:14:28 -!- Melvar has joined.
14:25:49 -!- mniip has quit (Ping timeout: 633 seconds).
14:27:41 -!- imode has joined.
14:40:53 -!- AnotherTest has joined.
15:39:23 -!- atslash has quit (Quit: This computer has gone to sleep).
15:42:16 -!- tuxcrafting has joined.
16:48:15 -!- b_jonas has joined.
16:52:41 <b_jonas> `? fundamental theorem of taneb
16:52:42 <HackEso> The Fundamental Theorem of Taneb states that for all strings S, if S describes a thing not involving sex, then it is provable that Taneb invented the thing described by S; and, furthermore, that it is provable that there exists a string T that describes a thing not involving sex that Taneb did not invent.
17:18:22 -!- Lord_of_Life has quit (Ping timeout: 245 seconds).
17:19:08 -!- Lord_of_Life has joined.
17:21:14 -!- TryingGeorge has joined.
17:25:25 -!- GeekDude has joined.
17:31:09 <shachaf> I don't understand that theorem.
17:31:10 -!- LKoen has joined.
17:33:20 -!- TryingGeorge has quit (Ping timeout: 248 seconds).
17:35:56 -!- LKoen has quit (Remote host closed the connection).
17:38:08 -!- LKoen has joined.
17:51:09 -!- FreeFull has joined.
17:56:17 -!- tuxcrafting has quit (Ping timeout: 245 seconds).
18:13:38 -!- LKoen has quit (Remote host closed the connection).
18:14:12 -!- adu has joined.
18:14:22 -!- arseniiv_ has joined.
18:15:53 -!- arseniiv has quit (Ping timeout: 245 seconds).
18:36:11 <int-e> `cwlprits fundamental theorem of taneb
18:36:15 <HackEso> oerjän oerjän tsweẗt
18:36:24 <int-e> not quite what I expected
18:39:14 <int-e> https://esolangs.org/logs/2015-07-22.html#l0d <-- I guess it's "you can't just claim something; you have to prove it"
18:40:37 <int-e> shachaf: someone complained about it before... leading to an... improvement? https://esolangs.org/logs/2017-05-27.html#lZ
19:01:50 -!- atslash has joined.
19:13:28 <arseniiv_> `? ℵrjan
19:13:29 <HackEso> ​ℵrjan? ¯\(°​_o)/¯
19:13:48 <arseniiv_> hm it should involve an actual Hebrew aleph then
19:13:53 -!- arseniiv_ has changed nick to arseniiv.
19:14:48 <arseniiv> BTW I was thinking about “all values are Xs” esolangs and I think I stumbled on a unused X
19:16:02 <arseniiv> namely, zippers and “half zippers” (terms with one hole, or maybe even several holes, homogeneous or heterogeneous) of various kinds
19:16:28 <arseniiv> you can combine terms with holes in a straightforward way
19:17:20 <arseniiv> though zipper operations—like go left, go right, go up for trees—are harder to realize if all we have is half-zippers
19:18:05 -!- sebbu has quit (Ping timeout: 258 seconds).
19:21:06 <kmc> hm interesting
19:21:22 <kmc> is the code a zipper too
19:25:04 <arseniiv> (cont.) but I seemed to see an easy fix. E. g. take 1-holed binary trees T made from 2: T² → T, 0: T and _: T (this is a hole),
19:25:04 <arseniiv> let A = …2(InnerA, _)…, and let there be B and C; we can make A′ = …2(_, C)… and B′ = B[_ ↦ InnerA] or something. I’ve already forgot what it was
19:25:04 <arseniiv> kmc: in a language where it’s representable as a value, why not! It could even do something continuation-like maybe. I hadn’t think about code yet
19:26:33 <arseniiv> so I was more interested in a holed terms, not actual zippers, but I think zippers are OK too
19:26:52 <kmc> mm
19:28:03 <arseniiv> also for all operations to conserve count of _’s is fairly esoteric in itself I think
19:29:09 <arseniiv> hopefully it will be useful for someone! I have no ideas yet
20:05:10 -!- sebbu has joined.
20:18:26 -!- Phantom_Hoover has joined.
20:33:07 -!- sebbu has quit (Ping timeout: 246 seconds).
20:35:01 -!- TryingGeorge has joined.
20:36:43 -!- TryingGeorge has quit (Client Quit).
20:37:54 -!- sebbu has joined.
21:10:36 -!- moony has quit (Quit: Bye!).
21:12:02 -!- Bowserinator has quit (Ping timeout: 272 seconds).
21:14:03 -!- Bowserinator has joined.
21:14:37 -!- moony has joined.
21:16:16 -!- ais523 has joined.
21:19:42 -!- Sgeo has quit (Read error: Connection reset by peer).
21:20:06 -!- Sgeo has joined.
21:22:30 <esowiki> [[RarVM]] M https://esolangs.org/w/index.php?diff=63579&oldid=63438 * Void * (+52) /* See also */
21:23:02 <b_jonas> In the same vein as finding universal Turing-machines with a small number of symbols and control states, has there been attempts to improve on that if you're allowed to have two stacks instead of a tape, such as the one you can emulate in Underload?
21:23:39 -!- MDead has joined.
21:23:51 <kmc> isn't the correspondance pretty direct there
21:23:53 <ais523> b_jonas: I've mostly been trying that for Turing machines, The Waterfall Model, and Jelly, but not two-stack machines
21:24:01 <b_jonas> kmc: in one direction yes
21:24:08 <kmc> hm ok
21:24:09 <ais523> kmc: having two stacks is more flexible than having a tape because it allows you to insert and delete symbols
21:24:11 <kmc> true
21:25:12 <ais523> that said, I'm not sure how much the insert and delete operations help, normally instead of an AAAABBBBBCCCC type encoding you can use ABCABCABCABC-B-
21:25:16 <b_jonas> I'm reminded of the minimization partly because I thought of the amazing brainfuck two bracket universality, and then I was wondering on Consumer society, and realized that in the restricted variant of Consumer Society where you can't nest brackets, you can still emulate two-stack machines
21:25:35 <ais523> so it only seems to help if you're trying to simulate unboundedly many counters/stacks, rather than any specific fixed number
21:25:47 <b_jonas> ais523: hmm
21:25:56 <b_jonas> yes, that might help
21:26:09 <b_jonas> I didn't think of that
21:26:18 -!- MDude has quit (Ping timeout: 245 seconds).
21:26:24 -!- MDead has changed nick to MDude.
21:27:17 <b_jonas> I have to admit that I was also thinking of simulating multiple (but a fixed number of) tapes
21:28:22 <ais523> if you can have arbitrarily many tapes you can use them as counters by just marking a single 1 on each of them and using the position of the tape head as the counter value
21:28:22 <b_jonas> I guess Underload and Brainfuck makes that sort of thing easy, because they let you skip over cells without having to enumerate each symbol, unlike in say Thue
21:29:18 <ais523> that said, actually editing the tapes at runtime probably makes things more efficient
21:29:27 <b_jonas> I didn't want arbitrarily many tapes, but a finite number of tapes is useful for things like simulating a machine that has both code and data
21:29:56 <ais523> b_jonas: heh, I actually see programming languages the other way round to some extent: having separate code and data is hard to simulate, thus a language is better if it doesn't
21:30:16 <ais523> (but memory-mapping the code doesn't really help much because you still have to access two separate points in it at once)
21:30:22 <b_jonas> ais523: maybe, but I don't like the restriction of too few tapes
21:30:54 <ais523> what would be really nice from a tarpit point of view would be a simple ZISC that doesn't rely on random access
21:31:16 <ais523> which is, I guess, what a Turing machine is
21:31:54 <b_jonas> does "doesn't rely on random access" allow a machine with twelve stacks?
21:32:10 <b_jonas> twelve stacks of bits
21:32:19 <ais523> technically, but I'm aiming to have just a single data structure (stack is impossible, but queue/tape are reasonable)
21:33:20 -!- AnotherTest has quit (Ping timeout: 248 seconds).
21:34:32 <b_jonas> what I'd like to know is, is there a natural number n such that for any natural number N, you can simulate any N-stack finite control machine with an n-stack finite control machine without more than polylog factor slowdown?
21:34:40 <b_jonas> with some reasonably limited translation
21:35:05 <b_jonas> polylog in the runtime but only for fixed N, exponent may grow in N
21:35:10 <b_jonas> s/exponent/degree/
21:35:19 <ais523> which one is polylog again?
21:35:27 <b_jonas> polynomial of logarithm
21:35:35 <ais523> times n, I assume
21:35:59 <b_jonas> so simulating t timesteps in O(t*log**10(t)) is fine, but simulating t timesteps in O(t**1.01) timesteps is not fine
21:36:07 <b_jonas> yes, factor slowdown
21:36:20 <b_jonas> as in how many time more step it takes
21:36:26 <b_jonas> yeah, I know my phrasing is weird
21:36:45 <b_jonas> but basically you need O(n**2) time to simulate a two-tape turing-machine on a one-tape turing-machine
21:37:01 <b_jonas> and that's actually Theta(n**2) or something like that I think
21:37:11 <b_jonas> because it can't move data to distances faster
21:37:43 <ais523> my guess is that you can't because once you're storing two of the N stacks on the same one of the n stack, you can't change from an operation on one of those stacks to an operation on another in less than time proportional to the difference in their sizes
21:37:47 -!- LKoen has joined.
21:37:57 <b_jonas> but my question is like, are five tapes enough? is a fixed number of tapes enough?
21:38:12 <ais523> so any polylog-speed interpreter would have to be optimising somehow, and it seems implausible that you could optimise /all/ programs
21:38:22 <b_jonas> `? oerjan
21:38:23 <HackEso> Your omnidryad saddle principal ideal golfing toe-obsessed "Darth Ook" oerjan the shifty eldrazi grinch is a punctual expert in minor compaction. Also a Groadep who minces Roald Dahl. He could never remember the word "amortized" so he put it here for convenience. His arkup-nemesis is mediawiki's default diff. He twice punned without noticing it.
21:38:32 <b_jonas> ais523: amortized is a relevant word
21:39:01 <ais523> I don't think you could amortize a random pattern of jumping around between the stacks
21:39:14 <b_jonas> at some point it did seem plausible to me that there may be a general scheme to optimize compressing multiple tapes that way
21:39:24 <b_jonas> I'm not so sure anymore
21:39:24 <ais523> maybe you could, though? you'd need some sort of cache
21:41:43 <b_jonas> like, when one stack is consumed faster than another, you could spend some of that time swapping around their elements on the tape that stores both, so the top of both are stored close to your head
21:42:08 <b_jonas> you don't rearrange the whole tape of course,
21:42:16 <b_jonas> just a part proportional with how much time you spent
21:43:14 <b_jonas> you need a second tape for this of course, otherwise even exchanging k elements with another k elements would take O(k) time, but you can have a second tape
21:43:33 <b_jonas> s/O(k)/O(k**2)/
21:44:14 <b_jonas> but you do have a second tape, so you can collect amortized credit for k time, then spend another k time to swap k elements of one stack with k elements of another stack, as represented on the tape, in O(k) time with the help of another tape
21:44:51 <b_jonas> I couldn't get this to work, but neither do I see why it must be impossible
21:47:05 <ais523> the problem case is the one where tape n contains nk elements, for some constant k
21:47:12 <ais523> and all the tapes are growining
21:47:14 <ais523> err, s/tape/stack
21:47:22 <ais523> (but not forever, eventually you start reading the values again)
21:48:17 <ais523> admittedly, you can deal with /that/ case by changing the interleave pattern, but it's hard to come up with a general rule for doing that
21:49:14 <b_jonas> yes, there can't be a fixed interleave pattern, because you can't predict (on the thread of halting problem) how the stacks will be accessed in the future. you need some sort of tricky dynamic structure.
21:49:41 <b_jonas> with metadata stored too
21:49:54 <b_jonas> at runtime
21:55:14 <b_jonas> the polylog of runtime allowance means you can even store each original stack element repeatedly log time times, and tag each occurance with the depth of that element from the bottom of the stack in binary
21:55:27 <ais523> I'm not sure about that "you can't predict"; it's /probably/ correct, but if the program you're predicting produces its output quickly it'll be quite predictable, and if it's slow you'll be able to use the time it's taking as part of the time you have
21:55:57 <int-e> Can we do something adverserial? It seems that there's a k+1 tape universal machine for k-tape TMs... so we can actually run the assumed k-tape version of our (say) 3k-tape machine and act upon what it writes and reads on a tape...
21:56:00 <ais523> so things like the halting theorem don't prove that it's impossible
21:56:36 <int-e> *adversarial
21:56:56 <b_jonas> ais523: maybe
21:57:30 <int-e> (The problem with that adversarial approach is that we can't know what the data written *means*... halting problem and all that; the transformation to the k-tape machine can be all sorts of magical. :-( )
21:58:48 <b_jonas> unrelated question. ais523: you said you were writing a library for balanced trees, copy-on-write. will that library allow me to implement a tree-in-a-tree, where I store the nodes of a tree as values inside a tree and the address of those nodes as keys, to be able to store all previous versions of the tree in memory together and any number of iterators to them in such a way that the iterators never go
21:58:54 <b_jonas> stale as long as the node exists?
21:59:29 <ais523> I'm not sure
21:59:57 <b_jonas> ok, that last part was phrased a bit confusingly
22:00:32 <ais523> I haven't really got anywhere with personal programming projects in the last several weeks
22:00:41 <ais523> it's all been my day job and non-programming things
22:00:50 <b_jonas> yeah, I know
22:00:54 <b_jonas> I've been like that for years now
22:01:02 <b_jonas> :-(
22:02:01 <b_jonas> and there's of course so much that I'd _like_ to write
22:02:14 <b_jonas> but there's all sort of other non-job projects that I'd also like to work on
22:02:17 <b_jonas> not only programming
22:02:22 <b_jonas> life is hard
22:02:34 <b_jonas> or being an adult is hard or something
22:40:06 <arseniiv> life is more specifically too short :(
22:42:41 -!- adu has quit (Quit: adu).
22:49:16 -!- LKoen has quit (Remote host closed the connection).
23:03:55 -!- LKoen has joined.
23:50:53 -!- Phantom_Hoover has quit (Read error: Connection reset by peer).
23:53:36 -!- arseniiv has quit (Ping timeout: 248 seconds).
←2019-06-18 2019-06-19 2019-06-20→ ↑2019 ↑all