00:01:01 what do you call it when a piece of music starts on a beat other than #1, and how do you get Rosegarden to do this? 00:03:48 -!- Ilari has joined. 00:04:45 my mind is blanking 00:09:50 -!- scshunt has quit (Quit: I am leaving. You are about to explode.). 00:15:12 -!- Ilari_antrcomp has joined. 00:15:58 -!- Oranjer has left (?). 00:18:04 -!- Ilari has quit (Read error: Operation timed out). 00:18:04 -!- Ilari_antrcomp has quit (Read error: Operation timed out). 00:21:48 -!- SevenInchBread has joined. 00:22:33 -!- Ilari_antrcomp has joined. 00:23:32 -!- CakeProphet has quit (Ping timeout: 245 seconds). 00:24:32 -!- Ilari has joined. 00:30:31 -!- relet has quit (Quit: Leaving.). 00:35:03 -!- mycroftiv has quit (Ping timeout: 260 seconds). 00:36:35 -!- mycroftiv has joined. 00:49:10 -!- oklopol has quit (Ping timeout: 240 seconds). 01:07:43 -!- FireFly has quit (Quit: swatted to death). 01:08:46 -!- Ilari_antrcomp2 has joined. 01:09:15 -!- Ilari_antrcomp has quit (Quit: leaving). 01:09:27 -!- Ilari_antrcomp2 has changed nick to Ilari_antrcomp. 01:48:22 -!- SevenInchBread has changed nick to CakeProphet. 01:48:29 -!- CakeProphet has quit (Changing host). 01:48:29 -!- CakeProphet has joined. 01:48:42 :o 01:48:56 I am a legend. 01:50:24 -!- augur has quit (Remote host closed the connection). 01:50:32 how did that happen? 01:51:56 So, I made my first good CA today. 01:55:04 Maybe you can replace the one south of Oregon with your good one. Or the one north of the US. 01:55:17 *hyuk hyuk* 01:55:29 lol 01:55:42 I meant cellular automaton. :P 01:58:49 -!- BeholdMyGlory has quit (Ping timeout: 258 seconds). 02:01:08 I realize that :P 02:08:57 -!- BeholdMyGlory has joined. 02:10:08 CakeProphet, oh? 02:10:49 leBMD, link? 02:11:13 http://conwaylife.com/forums/viewtopic.php?f=11&t=434 02:18:37 -!- Mathnerd314 has quit (Quit: ChatZilla 0.9.86-rdmsoft [XULRunner 1.9.2.3/20100401080539]). 02:22:11 -!- SevenInchBread has joined. 02:22:48 -!- CakeProphet has quit (Disconnected by services). 02:22:52 -!- SevenInchBread has changed nick to CakeProphet. 02:22:59 -!- CakeProphet has quit (Changing host). 02:23:00 -!- CakeProphet has joined. 02:23:02 -!- Mathnerd314 has joined. 02:23:50 AnMaster, what do you think? 02:28:50 -!- BeholdMyGlory has quit (Read error: Connection reset by peer). 02:38:49 leBMD, sorry was busy 02:39:00 it's ok 02:39:19 leBMD, um any images? 02:39:23 I don't have golly locally 02:39:44 er, not right now 02:39:54 leBMD, will comment on it when that is up 02:40:00 ok 02:40:59 -!- leBMD has quit (Quit: stuff). 02:44:39 Have there been any grammatical case based languages? 02:47:10 night 02:47:17 (no not an answer to the question) 02:47:21 (just, good night) 02:57:57 wareya: Perl versions < 6.0 have features inspired by case in natural languages. 02:58:04 scalary and array context. 02:58:06 -y 02:58:50 nothing too revolutionary 03:03:39 -!- CakeProphet has quit (Quit: leaving). 03:04:50 hello sweeties 03:05:09 * cheater99 is burning through haskell, so far it's a nice language with no shittiness 03:07:05 -!- pikhq has joined. 03:17:06 FUTURAMAFUTURAMAFUTURAMFAUFMA 03:18:03 HOLY FEKK THAT'S RIGHT 03:47:39 -!- Oranjer has joined. 04:09:40 -!- Ilari has quit (Quit: leaving). 04:09:40 -!- Ilari_antrcomp has changed nick to Ilari. 04:17:13 -!- Mathnerd314 has quit (Ping timeout: 264 seconds). 04:49:40 Anyone know a web server that responds on every port? 05:02:46 -!- MizardX has quit (Ping timeout: 260 seconds). 06:10:00 -!- zzo38 has joined. 06:11:08 I have some crazy idea, that if you have /dev/webcam is a block device to access picture from webcam, then you can mount it as a filesystem by putting some paper with barcodes in front of the camera and then use the mount command....... 06:15:53 Holy SHIT 06:15:59 Perry Bible Fellowship updated 06:16:52 This crazy idea is one of the most crazy idea probably 06:18:16 Probably. 06:18:29 Certainly possible, but not very *useful*. 06:21:22 -!- wareya has quit (Ping timeout: 276 seconds). 06:21:45 It probably isn't very useful. But it is useful to have /dev/webcam to be a block device to access the raw picture. It is less useful to mount it as a filesystem. 06:36:10 Hrm. That's stupid. There's a "list of the best 10 Pixar films". 06:36:17 There have only *been* 10 Pixar films. 06:37:54 Then, the list is all of them, it should be? Until, they make more films, and then they will update the list if any of the new ones are better. (If they make bad films then they don't have to update the list) 06:38:09 Yeah... 06:38:23 I'm not sure Pixar is (currently) capable of making bad films. 06:38:44 Their worst one so far was merely "pretty good". 06:46:44 And that is? 06:47:24 Cars. 06:47:38 -!- oklopol has joined. 06:50:12 -!- Oranjer has left (?). 06:51:49 pikhq: IMDB claims 55 :P 06:52:18 -!- oklopol has quit (Ping timeout: 272 seconds). 06:53:12 Oh, that list includes shorts and TV episodes. Foo. 06:53:22 Also, specials and other irrelevant drivel. 06:53:26 Pixar's done quite a lot of shorts. 06:53:40 That list also includes movies in production. 07:00:19 -!- pikhq has quit (Remote host closed the connection). 07:14:55 -!- scshunt has joined. 07:28:33 -!- zzo38 has quit (Remote host closed the connection). 07:59:59 -!- clog has quit (ended). 08:00:00 -!- clog has joined. 08:05:07 -!- scshunt has changed nick to coppro. 08:05:30 -!- myndzi has joined. 08:09:27 -!- myndzi\ has quit (Ping timeout: 260 seconds). 08:15:39 AAARGH 08:15:55 The compression of my home folder failed. 08:19:32 -!- coppro has quit (Remote host closed the connection). 08:26:18 -!- MigoMipo has joined. 08:33:06 -!- ais523 has quit (Remote host closed the connection). 08:39:23 -!- CakeProphet has joined. 08:39:27 I just got an idea 08:40:05 perhaps we could solve N=NP by using the process calculii to create constructions of a UTM that can solve nondeterministic problems at O(n) 09:24:23 -!- FireFly has joined. 09:24:24 -!- FireFly has quit (Changing host). 09:24:24 -!- FireFly has joined. 09:25:08 CakeProphet, P=NP surely? 09:36:11 -!- relet has joined. 09:45:14 Hmm, will the compression of my home directory work if I log out and switch user? 09:45:21 So that nothing is modified? 09:50:35 -!- Phantom_Hoover has left (?). 09:55:00 -!- oerjan has joined. 10:07:31 -!- tombom has joined. 10:08:01 although I don't think there's anything in Formula that fails if you implement it by calculating decimal expansions lazily 10:08:08 -!- oklopol has joined. 10:09:23 the parts which need to be tested for exact equality with an integer might fail if you hit upon a formula which is unexpectedly an integer (like some n*e*pi things) 10:10:02 (i.e. it's an unsolved problem whether e*pi is rational) 10:11:53 that might be hard even with something more complicated than lazy evaluation, too 10:14:55 wareya: Perl versions < 6.0 have features inspired by case in natural languages. 10:14:59 also Perligata 10:19:17 perhaps we could solve N=NP by using the process calculii to create constructions of a UTM that can solve nondeterministic problems at O(n) 10:20:08 N? 10:20:23 not without doing something insanely clever. an ordinary UTM needs something like time proportional to the number of processes to simulate them 10:20:49 oklopol: obviously a typo for P 10:20:56 i just learned that windows has a log viewer, and i used it to find out why it shut down during the night, god i'm cool 10:20:58 oh 10:21:49 i have no idea what process calculii is so i assumed N is a complexity class that has to do with it 10:21:57 and redefining the P=NP problem to use a different kind of underlying TM doesn't solve the _original_ problem. 10:22:08 oklopol: i assumed he meant something like pi calculus 10:23:10 actually my guess was N means some class to do with pi calculus 10:23:49 if N was a weird name for its polynomial time stuff-it-can-solve problem, then N=NP might be open mightn't it 10:24:20 and CakeProphet would definitely know this because he's a complexity theory guru 10:24:41 in fact it's a well known theorem that if you try to extend your underlying model with _oracles_ (essentially a subroutine for solving some specific problem), then you can choose the oracle either such that P=NP or such that P != NP in the "relativized" model 10:24:53 yeah 10:25:18 is it really well known 10:25:27 i learned it like a month ago 10:26:09 oklopol: i don't know what N alone means in complexity theory (probably something), but if he typoed an L instead then it _would_ have been an open problem 10:26:17 btw isn't P=NP with an oracle sort of trivial if you just take some ridiculously good oracle? 10:26:25 oklopol: it's well known among complexity theorists :D 10:26:31 yeah, L!=PSPACE but otherwise all open 10:26:39 right 10:27:01 -!- Gregor has quit (Ping timeout: 252 seconds). 10:27:17 oklopol: because it makes a _lot_ of the methods they use unusable for deciding P=NP 10:27:32 (most of the methods "relativize") 10:27:42 yeah okay i guess it might be one of the first things you hear about complexity theory because it has to do with the only problem people are interested in, the famous P=NP problem that asks whether polynomial time is the same as not-polynomial time, otherwise known as the quantum computation model. 10:28:06 yeah 10:28:11 i know that too! 10:28:13 * oerjan swats oklopol -----### 10:28:34 yeah i know you were joking, but still... 10:28:51 i don't know the specifics yet tho, but soon i will know more than you... wait how much more do you know? 10:28:56 do you know left sets 10:29:08 what the heck is a left set 10:29:14 oh wait also we have a course about complexity theory next fall 10:31:30 er i forgot a detail about the definition :D 10:32:06 hm wait is it that thing about context-free grammars? 10:33:17 where you construct a _regular_ language that describes how a left part of a string in the original language can be constructed 10:34:02 probably not though 10:34:06 okay umm lol i can't remember exactly under pressure, but i think for a problem in NP, its left set is (x, w) where x is an instance in the language, and w is a string such that there's a witness y with w<=y 10:34:28 and the idea is this problem is in NP too because you can guess the witness 10:34:40 so then we can do sort of interval pruning to find the maximal witness 10:35:11 this is used to show P=NP if(f) there are sparse sets that are NP-hard (with bounded truth-table reduction) 10:35:12 s 10:35:27 oh 10:35:47 the complexity theory book i'm reading starts with two trivial algorithms and then this theorem, the proof of which is something like 8 pages 10:36:05 ouch 10:36:12 complexity theory companion 10:38:28 i haven't actually read a whole book, i've read _some_ of papadimitriou 10:40:16 so i know like why QBF is complete for PSPACE, that was sort of the most complicated thing i learned there i think 10:40:32 P=NP if co-NP has a sparse hard set is pretty easy btw, lets say you can reduce to that sparse set from sat with f, then basically you just enumerate partially solved sat instances, and the amount of them you need to keep track of is polynomial, because if you have two partially solved instances a and b and f(a) = f(b) then one is true if the other is, and if there are more than p(n) f(x) values where p is the growth rate polynomial 10:40:36 i'm wordy 10:40:43 i'll repeat half for fun 10:40:44 because if you have two partially solved instances a and b and f(a) = f(b) then one is true if the other is, and if there are more than p(n) f(x) values where p is the growth rate polynomial of the sparse set, then one of the instances must be satisfiable 10:42:32 my brain is too rusty for that today 10:42:51 because the set contains at most p(n) strings of length n, where n is your sat instance length, so if you map to p(n)+1 different strings of length n, one of them can't be in the set 10:43:22 okay, well i realized there are a few details you need to understand before you can appreciate the easiness of the underlying idea 10:43:27 :P 10:43:36 ok 10:43:52 for the left set proof i still have no idea what the underlying idea is, and i've read it like 3 times 10:44:33 o 10:44:33 o 10:44:33 o 10:44:33 o 10:44:42 lately, the last complexity proof i've read is a proof that you can do integer division and arithmetic stuff in logspace. hm i probably mentioned it. 10:44:51 yep 10:45:00 division is surprisingly hard 10:45:02 i hear 10:45:05 -!- relet has quit (Quit: Leaving.). 10:45:10 (it was actually somewhat stronger than logspace, but that's the result i was interested in) 10:46:48 okay so if there's a tally set, subset of 1* that is, that's np-hard, then also p=np, we reduce sat to it, then again you only need a polynomial amt of instances, because if one of them has f-image that's not of 1* form, then it's unsatisfiable, and if two have the same image, again you can remove one of them because one is satisfiable if the other one is 10:46:50 is that easier? 10:47:04 i _tried_ starting to read the result that SL = L but that had a lot of unknown graph theory at the bottom so just too much work 10:47:14 SL= 10:47:15 ? 10:47:20 superlog 10:47:24 symmetric 10:47:38 okay that tells me everything 10:47:59 it actually tells me nothing 10:48:06 essentially you have a nondeterministic turing machine where if you take a step one way you can always take it in reverse 10:48:07 i'm just screwing you 10:48:16 oh 10:48:35 is L with a nondet tm or det? 10:48:38 or does it even matter 10:48:51 there might be technical reasons why that is too simple explanation but that's the gist 10:49:02 L is det, NL is nondet. N = NL is still open 10:49:11 :D 10:49:45 okay 10:49:58 oklopol: the other simpler way to say it is that SL is the set of problems for which "undirected graph reachability" is complete (under log-reductions) 10:50:09 for NL drop the undirected 10:50:41 8\ 10:51:06 so it's open whether dijkstra's algorithm can solve all problems in NP? 10:51:10 so restricting to undirected graph makes it simpler 10:51:17 oklopol: um what is that? 10:51:35 isn't it graph reachability 10:52:00 i thought it's one of the three algorithms people know by name 10:52:40 graph reachability is complete for NL. that is an important theorem as it also can be used to deduce other space things 10:52:40 my point was just to show how badly i must've misunderstood the sense in which reachability is complete 10:53:08 do you know why L <= P, is that obvious? 10:53:12 oklopol: NL, not NP 10:53:19 yes that's obvious 10:53:41 if you have only logarithmic space then if you run for more than P time you must cycle 10:53:46 oh lol true 10:54:12 i mean intuitively, i'm not sure i could fill in the details 10:54:40 SPACE(f(n)) <= TIME(exp(f(n)) 10:55:04 oh okay 10:55:51 that is sort of obvious 10:56:05 dijkstra's algorithm is that just marking every vertex you can reach from those you already have, until you've either got to the one you want or have none left to try? 10:56:27 (with some book keeping to avoid duplicating work) 10:57:01 because there's another important algorithm for graph reachability which uses more time but less space 10:57:07 yeah but you also find the shortest path so you need to be greedy in that aspect 10:57:28 specifically it does reachability in O(log(n)^2) space 10:57:36 go where you've gone least distance sofar 10:58:07 and that with the completeness result gives the important theorem NSPACE(f(n)) <= SPACE(f(n)^2) as a corollary 10:58:23 so n would be number of edges or what? 10:58:30 vertices + edges maybe 10:58:31 lol 10:58:40 and then log(n)^2 means.... 10:58:46 well it doesn't matter when you log 10:59:07 log(n) means that you can essentially store an index to a vertex 10:59:26 linear amount of indices and linear additional shit 10:59:45 yeah okay i think i have a grasp of what logarithms mean 10:59:57 log(n)^2 means you can essentially store a stack of indexes while recursing in a divide and conquer manner. at least that's what the algorithm does :D 10:59:59 i have a hard time getting any sort of concrete view of what logspace actually means 11:00:10 i mean for the implementer 11:00:15 of an algo 11:00:43 oklopol: as wikipedia said, a finite set of pointers + a logarithmic set of bit values 11:01:29 okay i don't see how you can do the d&q 11:02:28 oklopol: to check if there is a path from v1 to v2 of length <= n, iterate through all vertices v checking if there's a path from v1 to v and from v to v2 of length <= n/2 11:02:28 how can you know when you're in a loop with finite pointers 11:02:35 hmm 11:02:54 okay that's the same idea they use for PSPACE searches 11:02:57 of course with some base case 11:03:11 oklopol: yep, that's essentially the same thing 11:03:44 the QBF proof used a lot of that reachability stuff 11:03:58 or well the proof i have internalized, anyhow 11:04:04 oh lol iterate through all vertices, ofc you can do THAT, for some reason i thought you meant iterate through all vertices ON THE PATH BETWEEN V1 AND V2, which i found hard to do for some reason....... :D 11:04:18 yeah 11:04:53 oh and of course there is the even weirder proof that reachability _is_ in NL 11:05:14 quantum brainfuck is... quantified... okay what 11:05:18 quantum bit fields 11:05:25 i should probably know this 11:05:26 :D 11:05:35 boolean formulas? 11:05:47 not all the "quantum" languages on the wiki may be properly quantum 11:06:01 oh there was a quantum brianfuck? 11:06:04 poor brian 11:06:37 * oerjan checks 11:07:51 hm not sure if that's a proper one 11:08:04 it has qubits at least 11:08:15 so okay qbf is qsat 11:08:35 uorygl usually knows what he's talking about, or at least knows if he doesn't 11:08:36 uorygl 11:08:38 is it proper 11:09:15 oh and so 11:09:23 then i've read the proof that it's complete too 11:09:29 i think 11:10:11 no wait actually you need to implement a tm in the circuitry right? 11:10:40 i think i've just read a rough description of how doing it is similar to proving sat is complete for np 11:10:53 or maybe i haven't really read anything 11:11:25 oh wait 11:11:44 you're talking about the qbf i mentioned, not some quantum stuff? 11:11:49 the sat completeness proof i read from the 50's or something implemented some sort of BASIC derivative with guessing 11:11:51 yes :D 11:12:13 sorry quantum brainfuck was just the acronym that got expanded by accident 11:12:17 i got confused there and thought i was heavily out-maneuvered :D 11:12:23 hehe 11:12:48 i don't know anything about quantum computation, please teach me everything about it 11:12:58 um i know not very much more 11:13:08 we have a few researchers doing exactly that here, but they don't teach 11:13:47 or well they do teach engineers how to copypaste laplace transforms from their cheatsheets 11:14:04 or actually they have this massive book of them 11:14:10 for qbf you can essentially encode the complete state graph of a turing machine using boolean logical predicates 11:14:47 what exactly do you mean by complete state graph 11:14:51 and then you formulate that O(log(n)^2) search in logic, and get a qbf formula 11:15:27 a graph where a vertex is a possible state of the whole TM, including the tape (polynomially bounded though) 11:15:46 and an edge if there is a transition step 11:16:44 note that the size of this graph is exponential, but when you do the search you take the log again so it all works out 11:17:21 yeah i get that, but what i don't get, i think, is how exactly that shows completeness 11:18:24 oh hmm 11:18:27 oh well you search for a path between the initial state of the TM and any accepting state 11:18:37 now that i recalled the definition of PSPACE, i think i'm starting to see it... 11:18:45 who knew that might be important 11:19:30 the qbf formula you end up with is of polynomial size, naturally 11:19:48 where was this proof? 11:19:54 papa? 11:20:05 i assume that's where i read it 11:20:32 and i'm not sure if i really read it that way or just distilled out the essence of what it said 11:20:40 there's probably a mention of it in CTC, but at least the proof of completeness of sat they just have as a note in the back of the book 11:21:02 so probably not very detailed 11:21:12 :D 11:21:23 i like the proof because it combines so many of the other methods used for L,NL,P,NP stuff 11:21:53 like using the evolution state graph of a TM, encoding that using boolean logic... 11:22:12 where else are evolution graphs used 11:22:30 NL completeness 11:22:40 makes sense 11:22:49 and in a sense for proving SAT is NP-complete too 11:23:11 except you split up each vertex into many boolean variables again 11:24:35 err umm. 11:24:48 but that's not a very convincing example 11:25:07 but the SAT proof includes the time evolution of the TM, anyhow 11:25:54 mm yes 11:26:38 while the NL and PSPACE cases actually use graph algorithms on the resulting time steps 11:27:59 if you have a prob P in np solved by tm T you, given a string s, to check whether it's in P, make a sat instance that is true iff T accepts s, and this is easy because you have a polynomial amt of steps you need at most so you just need to encode, ah, a time evolution of T 11:28:27 yep 11:29:52 basically if you need p(n) steps at most, you make a p(n) x p(n) square, and assert for each line that you get it from the previous line by using the rewrite rules of the tm (each line has an ID of the tm) 11:30:03 ofc you know all this, i'm just saying it for me 11:30:35 yeah 11:31:55 and then the sat is true for a configuration if there's a tape head in accepting state in one of the squares, so it's satisfiable iff there's a run of the tm, starting from the config induced by s, that accepts 11:32:11 cool, i could probably fill in the exact proof 11:32:16 i suppose it's sort of trivial 11:32:54 (configuration being a consistent placement of values in the rest of the squares) 11:34:01 it's a simple but powerful idea 11:34:09 i need to do some work now i think, i feel all dirty if i don't work during my weekends 11:34:12 yes 11:34:15 and sexy 11:35:04 do you know "shifts dynamical systems"? 11:35:09 *shift 11:35:17 certainly 11:35:28 do you know why they have such a weird name 11:36:00 they are dynamical systems where the transformation is a shift of a sequence 11:36:17 you know all finite groups can be embedded in the automorphism group of the full 1-dimensional shift? 11:36:40 sure, but why not say dynamical shift systems or something, i don't see how shift dynamical system is even english 11:36:41 i cannot say i recall that result :D 11:37:18 "dynamical system" is a unit there 11:38:08 oh well 11:39:11 you can embed a finite permutation group: given permutation p you make an automorphism by having u*string*v always be transformed to u*p(string)*v, where u and v are chosen so that these can't overlap 11:39:53 well i know all groups are isomorphic to permutation groups 11:41:47 -!- augur has joined. 11:42:00 now clearly the map from a permutation group to those is injective (just give them something to permute and they'll permute differently), and t(p1)t(p2) = t(p1p2) because clearly doing two transformations after each other does u*s*v -> u*p2(s)*v -> u*p1(p2(s))*v 11:42:12 the transformations always leave everything outside u*s*v patterns unchanged 11:42:34 -!- augur has quit (Remote host closed the connection). 11:42:38 ok, the only thing i don't see is why you can choose u and v like that 11:42:49 -!- augur has joined. 11:43:00 1^(2n) 0 s 0 (10)^n 11:43:08 where |s| = n 11:43:20 IIRC, maybe you need to require something of s too 11:43:25 ...no 11:43:52 1^(2n) can't be anywhere but in the beginning 11:43:56 so no overlap 11:44:24 right you divide the whole sequence at 1^(2n) 0 points i guess 11:44:24 and u = 1^(2n) 0, v = 0 (10)^n of course 11:44:41 what do you mean? 11:45:08 if a transformation given by permutation p sees such a pattern, it'll rewrite what comes after 11:45:29 by permuting it 11:45:50 oh and, it should be easy to see that this is a block map, and that it's invertible 11:46:26 i'm not sure you actually need (10)^n at the end... 11:46:58 otherwise what if there's ... 1^(2n) 0 1^(2n) 0 ... inside a string 11:47:27 the strings have length n you said 11:47:29 how do you rewrite (1^(2n) 0)^ww (w being omega) 11:47:32 err 11:47:49 the transformations we're compiling permutations to are maps from the full shift to itself 11:48:27 i know 11:48:31 so the transformations need to behave like the group on ANY bi-infinite string in the full shift 11:49:22 ... 1^(2n) 0 1^(2n) 0 1^(2n) 0 ... has 1^(2n) 0 1^n inside it in two ways, what does a transformation t permutation p compiles to do on such input? 11:49:56 oklopol: i said to drop (10)^n, not 0 (10)^n 11:50:03 the 0 alone is enough, i think 11:50:05 okay sorry 11:50:22 i forgot it was there so i thought you were somehow fundamentally lost 11:50:30 yeah should be enough 11:50:47 i actually used that when proving to myself that the pattern can't overlap itself 11:50:53 and couldn't you also reduce 2n to n+1 11:51:02 yes clearlyy 11:51:04 *clearly 11:51:21 ok then 11:51:48 but i haven't really thought about that, because i think it's intuitively clear u and v can be chosen SOMEHOW, perhaps because it's rather common to do that in combinatorics on words 11:52:23 anyway the paper this is from is not the best thing ever written 11:52:35 scandalous! 11:53:03 -!- sebbu2 has joined. 11:53:12 it's the same one that proved stuff like "ca rules that look only to the right can only rotate a sequence to the left" with a rather complicated formal proof 11:53:41 -!- sebbu has quit (Ping timeout: 240 seconds). 11:53:42 -!- sebbu2 has changed nick to sebbu. 11:54:38 ...right 11:54:48 it's an old classic, a collection of stuff proved sofar when the field was young, by hedlund, sort of starting the thing 11:54:55 that _does_ sound obvious (that you cannot rotate right) 11:56:10 even if it needs proof, the proof could be as follows: ...1:000... and ...0:000... are rewritten in the same way 11:56:16 err the 000... part is 11:56:34 so you can't get 1 to pop up in just one 11:57:07 mhm 11:57:11 well i dunno maybe the details aren't completely trivial 11:57:20 shut up 11:57:47 wait what, mhm is my way of saying i have nothing to say, without looking absent 11:57:53 :P 11:58:07 -!- Gregor-P has quit (Ping timeout: 260 seconds). 11:58:38 i know that on an intellectual level, but i... wait actually i just recalled i don't admit i have anything else 11:58:46 -!- sebbu2 has joined. 11:58:55 so i have no idea what just happened 11:59:01 aaaaaaaaanyway about that stuff i have to do 11:59:03 i should do it 11:59:07 soon i will 11:59:22 maybe one more proof first 11:59:38 or maybe not 11:59:43 human communication should be more like this 11:59:52 "hey dude listen to this result i just read" 12:00:16 you don't say 12:00:31 -!- sebbu has quit (Ping timeout: 252 seconds). 12:00:31 -!- sebbu2 has changed nick to sebbu. 12:00:55 seriously! ---> 12:00:58 cu 12:01:26 -!- oerjan has quit (Quit: Me four, er, two, wait what). 12:04:00 -!- sebbu2 has joined. 12:05:49 -!- sebbu has quit (Ping timeout: 264 seconds). 12:05:49 -!- sebbu2 has changed nick to sebbu. 12:28:33 -!- tombom has quit (Quit: Leaving). 12:40:37 -!- waga has joined. 12:40:39 hi 12:43:00 bye 12:43:02 -!- waga has quit (Client Quit). 12:52:16 -!- BeholdMyGlory has joined. 13:26:10 -!- Phantom_Hoover has joined. 13:48:48 -!- alise has joined. 13:55:01 -!- MizardX has joined. 14:03:28 -!- Gracenotes has quit (Ping timeout: 240 seconds). 14:05:50 -!- Phantom_Hoover has quit (Ping timeout: 265 seconds). 14:08:18 -!- MigoMipo has quit (Remote host closed the connection). 14:12:52 -!- MizardX has quit (Read error: Connection reset by peer). 14:16:41 -!- MizardX has joined. 14:19:32 I just realised that if you watch "Caretaker", then "Endgame", you miss nothing except that Seven of Nine suddenly appears. 14:19:37 Such is the power of the HMS Reset Button. 14:26:45 -!- Gracenotes has joined. 14:40:45 -!- zzo38 has joined. 14:41:22 Now I added a way to add IRC commands by script 14:42:31 http://sprunge.us/HKGi 14:44:10 I tried to make it as simple as reasonable possible so as to not overload the server 14:48:41 Wow, someone actually bothered to clean up the video on the DVD release of The Next Generation and then encoded it all with x264 on super-high settings, yielding a 550 MiB-per-episode average. 14:48:59 The only issue is that it's over 10 gibibytes per season. 14:50:32 that fits on 3 discs per season, doesn't it? 14:50:51 Define disc... 14:51:17 pineapple: No player that takes discs could play H.264, I don't think, and it'd be slow anyway. This 100 gibibyte rip is obviously designed for hard disk storage. 14:51:26 4.3ish gb dvd, single layer single sided 14:51:38 Well, play H.264 in Matroska, anyway. 14:52:02 double sided is annoying, double layer doesn't work on some hardware 14:52:12 You could fit the whole thing on two dual-layer Blu-Ray discs, or three or four one-sided discs. 14:52:15 and... well, yeah, but for distribution? 14:52:24 pineapple: That would be Illegal. :P 14:52:29 true 14:52:43 pineapple: Although admittedly faster than downloading it. 14:53:03 31 KiB/s download right now, but tons of peers; why?! It will take 4 days, apparently. I think I'll cancel it. And that's just the first season. 14:53:03 which would also be Illegal :-P 14:53:12 Yes, but less detectably so. 14:53:17 true 14:53:34 14:19:08 I just realised that if you watch "Caretaker", then "Endgame", you miss nothing except that Seven of Nine suddenly appears. 14:53:46 also, that Janeway gave up coffee 14:53:49 not being a trekkie... explain? 14:54:06 i know it's voyager, and who the characters are 14:54:15 but not the relevance of skipping, nor what's inbetween 14:54:16 Caretaker and Endgame are the first and last episodes of Star Trek: Voyager, whose titular ship is also known as the "HMS Reset Button". 14:54:33 1.1 and 7.22 ? 14:54:43 What's in-between is a lot of attempts to get home that inevitably fail because of the whole premise of the series, and also a whole lot of Captain Janeway being a bipolar sociopath. 14:54:58 the former i'd noted... the latter? 14:55:12 (Switching between righteously and suicidally upholding Universal Principles, and sometimes just breaking every rule to an ultimately futile end.) 14:55:13 "HMS Reset Button"? 14:55:14 -!- BeholdMyGlory has quit (Remote host closed the connection). 14:55:23 (But always making the most insane decision possible.) 14:55:36 pineapple: The attempts to get home always fail, and the characters are never developed. 14:55:43 The status quo is resumed at the end of every episode. 14:55:59 One could very easily watch just the first and last episode and not be confused at all. 14:56:00 sounds like the writers were applying discworld logic 14:57:27 Well, they pretty much set themselves up for this with the premise. By definition, them getting home is the end of the series. 14:57:34 By definition, their main, overriding goal is to get home. 14:57:44 Therefore, every episode consists of them getting some hope that might let them get home, then being let down. 14:58:21 did the characters not develop, though? 14:58:37 i mean, there was the relationship 14:58:46 and the doctor 14:59:03