←2010-02-16 2010-02-17 2010-02-18→ ↑2010 ↑all
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00:58:06 <MissPiggy> yeah
00:58:27 <MissPiggy> I still feel defeated wrt. that robozzle thing
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05:39:41 <Gregor> Heh, http://xkcd.com/703/ is actually pretty good :P
05:40:44 <lament> it is.
05:43:10 <pikhq> Alas.
05:43:31 <pikhq> It's like an xkcd from the time span where it got really popular because it was funny.
05:50:40 <bsmntbombdood> bah
05:50:43 <bsmntbombdood> still no questionable content
05:57:02 <augur_> http://www.manjulaskitchen.com/
05:57:04 <augur_> indian food :
05:57:05 <augur_> :D
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06:12:33 <oerjan> grmph xkcd loads just a ... blank page?
06:13:15 <Gregor> XKCD don't need you Nordic types.
06:13:40 <oerjan> hmph the page source is definitely not blank
06:15:51 <oerjan> old comics too...
06:17:23 <Gregor> Works fine for me.
06:17:30 <Gregor> XKCD just don't need you Nordic types.
06:17:51 <oerjan> oh well
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06:32:09 <Sgeo> Forgot about XKCD tonight
06:32:59 <coppro> it's awesome
06:33:31 <augur> smbc is more regularly humorous
06:35:03 <dbc> That "most dangerous game" one was cute.
06:35:38 <pikhq> augur: Yes, yes it is.
06:35:49 <pikhq> It is, in fact, *usually* humorous.
06:37:21 <augur> oh god this woman has a tamarind chutney recipe
06:37:27 <augur> god this is the best recipe website ever
06:38:50 <Sgeo> If 1,000,000 people join, wouldn't the group have 1,000,001, including the creator?
06:38:59 <Sgeo> Or what if a bunch of people join, then leave?
06:39:13 <augur> D:
06:39:33 <pikhq> augur: Omnomnom.
06:39:41 <augur> i know :(
06:39:45 <augur> tamarind chutney is the best
06:39:46 <augur> BEST
06:39:49 <augur> fucking chutney
06:39:51 <augur> ever
06:43:24 <Sgeo> Night all
06:43:29 * Sgeo goes to cry himself to sleep
06:43:49 <oerjan> O_O
06:44:14 <Sgeo> http://raglanshire.com/2010/02/in-memory-of-our-friend-stillpink/
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06:51:00 <oerjan> *BOOOOOM*
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07:04:36 <bsmntbombdood> i like that show
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11:27:01 <cheater> hi
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19:23:52 <cpressey> FWIW, I just added 3 entries to my "unfinished esolang designs" list ( http://catseye.tc/cpressey/louie.html ) -- Faradaisical, Kig, and Milab.
19:29:18 <Wareya> Is that supposed to be an all-inclusive list? :D
19:29:34 <cpressey> Ha.
19:35:01 <cpressey> In case it's not clear from the page, those are the ideas I have basically given up on, but which I consider too interesting to be simply discarded.
19:35:56 <Wareya> k
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21:04:06 <oklogon> faradaisical has a great name
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21:12:36 <Ilari> Hmm... Are there groups that have multiple-element minimal generators that don't decompose into multiple groups (w.r.t. cartesian product). If there are, wonder if they would be useful for constructing esolang...
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21:15:18 <Ilari> Ah, there are, at least D2 (Klein-Four group). That has minimal generator size of 2 but doesn't decompose.
21:17:15 <Ilari> But are there groups that have all elements in generator set have sizable order?
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22:50:52 <AnMaster> Ilari, that went completely over my head XD
22:51:10 <AnMaster> is it about topology again?
22:56:34 <Ilari> AnMaster: No. But I think I have solved most problems with wormholes except for how various operations affecting IP delta would behave.
22:56:57 <coppro> is this in Funge?
22:57:03 <AnMaster> Ilari, hm?
22:57:38 <AnMaster> Ilari, you mean like the MODE fingerprint which changes <>^v and such to affect position not delta (or various other variants)?
22:58:05 <AnMaster> http://catseye.tc/projects/funge98/library/MODE.html
22:58:13 <Ilari> AnMaster: Nope, just basic befunge operations that set IP delta.
22:59:46 <AnMaster> Ilari, well, those are ^v<>x for unconditional. A few more in trefunge
22:59:56 <AnMaster> plus some conditional ones
22:59:59 <AnMaster> like w_|
23:00:18 <AnMaster> oh and the random cardinal direction ? of course
23:00:40 <coppro> MODE doesn't work on h or l?
23:00:43 <AnMaster> oh and I forgot [ and ] that turns 90 degrees (counterclockwise/clockwise
23:00:44 <AnMaster> )
23:00:47 <AnMaster> coppro, no clue
23:01:10 <AnMaster> coppro, it doesn't work on w either it seems
23:01:13 <AnMaster> but then
23:01:18 <AnMaster> w is relative current always
23:01:23 <AnMaster> it is turn left/right
23:01:29 <AnMaster> or go straight ahead
23:01:30 <Ilari> Well, if there couldn't be operators inside wormhole, that would mean no IP delta changes there either...
23:01:32 <AnMaster> depending on compare
23:01:42 <AnMaster> Ilari, why couldn't there be?
23:02:24 <AnMaster> Ilari, and of course there should be possible to have wormholes between wormholes and normal space
23:03:04 <Ilari> AnMaster: Because normal space have two equivalent dimensions, but wormholes have longitudial direction and transverse direction, and those most definitely aren't equivalent. In fact, what direction is longitudial depends on direction wormhole was hit.
23:03:42 <coppro> Ilari: Do you have docs on how wormholes work as of now?
23:03:44 <AnMaster> Ilari, so there could be no code inside a wormhole?
23:03:48 <coppro> or do I need to look in the logs?
23:03:58 <AnMaster> coppro, they aren't implemented anywhere yet
23:04:02 <AnMaster> it is theorising about it
23:04:07 <coppro> yeah
23:04:14 <coppro> but one can have docs without an implementation
23:04:27 <AnMaster> well, logs the last week or so I guess
23:04:30 <AnMaster> grep for wormhole
23:04:38 <AnMaster> can't have too many irrelevant hits
23:05:36 <Ilari> Pretty much the only basic operations that are problematic are those that set absolute IP delta... '[' and ']' wouldn't be...
23:06:26 <AnMaster> Ilari, well, if you hit a cell with ^ in, you go the direction where ^ is written
23:06:29 <Ilari> Wormholes within wormholes would really mess up tracking about what direction wormhole was entered from.
23:06:37 <AnMaster> Ilari, one issue would be *how to write* inside the wormhole
23:06:45 <AnMaster> since what is the x/y coordinate
23:07:02 <Ilari> Wormhole endpoint has x/y coordinate, but that's a gap in space.
23:07:20 <AnMaster> Ilari, you need to be able to give current x/y coord, the y command (sysinfo) includes that amongst many other things
23:08:25 <AnMaster> and even you made y inside a wormhole invalid, you would still need some coordinate system for use by the interpreter itself
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23:10:42 <coppro> Ilari: how do wormholes work?
23:11:31 <Ilari> The standard befunge coordinates form a group. Wormholes would break that. WORM would be one heck of a feral fingerprint (not as bad as TRDS, but quite bad).
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23:12:18 <AnMaster> Ilari, agreed. I doubt I would implement it
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23:14:55 <coppro> Ilari: Explain?
23:15:21 <Ilari> coppro: Coordinate addition is no longer associative in presence of wormholes.
23:15:36 <coppro> O_o
23:15:38 <MissPiggy> huh
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23:16:28 <oklopol> cpressey: why couldn't a ring have absorptino?
23:16:31 <oklopol> *absorption
23:16:48 <Ilari> There are also proerties like: If at t = 0, two IPs are at the same point, have different IP deltas that will not change, in standard funge, those IPs will never be at same point again, whereas with wormholes, they could.
23:17:16 <oklopol> "Ilari: But are there groups that have all elements in generator set have sizable order?" <<< what's sizable order?
23:17:26 <coppro> why would they never return to the same place?
23:17:34 <coppro> in fact, they would have to
23:17:48 <Ilari> oklopol: There prsumably means much larger than 2.
23:17:50 <AnMaster> <Ilari> There are also proerties like: If at t = 0, two IPs are at the same point, have different IP deltas that will not change, in standard funge, those IPs will never be at same point again, whereas with wormholes, they could. <-- wraparound
23:18:00 <Ilari> Ah yeah.
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23:18:19 <cpressey> oklopol: The only x where x * x = x in a ring is x = 1. And the only y where y + y = y is y = 0.
23:18:36 <AnMaster> Ilari, one goes straight up one goes straight left, they will hit the same place again assuming width/height ever match up
23:19:09 <oklopol> Ilari: the free group generated by {x, y, z} has pretty sizable order for the generators.
23:19:31 <Ilari> oklopol: But doesn't it also decompose w.r.t. cartesian product?
23:19:45 <oklopol> cpressey: why?
23:19:55 <oklopol> the latter is obvious, i don't see why the first one is
23:20:06 <coppro> Ilari: I'm still really not clear what this is supposed to be doing :/
23:20:38 <cpressey> oklopol: My ring theory is rusty.
23:20:41 <oklopol> + forms an abelian group, but . just needs to distribute over + and possibly have 1.
23:21:08 <AnMaster> cpressey, some questions about TURT btw.
23:21:19 <oklopol> y + y = y => y = 0, by adding -y to both sides
23:21:44 <oklopol> but x^2 = x <=> xx - x = 0 <=> x(x - 1) = 0
23:21:48 <MissPiggy> you can have
23:21:48 <Ilari> oklopol: I think free group from {x, y, z} is the same as Z x Z x Z.
23:21:56 <MissPiggy> (H + x) + (H + y) = H + (x + y) if H is a coset
23:21:57 <AnMaster> cpressey, mainly, is U supposed to return the same value all the time (max bounds supported)? Or is it current bounds of the drawing?
23:22:10 <oklopol> Ilari: what would xyx be in Z^3?
23:22:15 <AnMaster> cpressey, if the former I have an issue, bignum again :/
23:22:30 <oklopol> cpressey: and if it's not a domain, i don't see why x would have to be 1 or 0 if x(x - 1) = 0
23:22:45 <Ilari> Ah, free groups are noncommutative.
23:22:55 <AnMaster> cpressey, another thing, how is the coordinate supposed to be handled. Round to nearest integer? You can easily get non-integers when you move an integer distance along an integer heading
23:23:04 <AnMaster> non-integers for x and y that is
23:23:44 <cpressey> AnMaster: Your guess is as good as mine.
23:23:54 <AnMaster> cpressey, you wrote TURT though
23:24:08 <oklopol> Ilari: for abelian groups, we can in fact split all groups up into pieces sort of like you guessed, but groups in general can do pretty much anything, at least afaik.
23:24:10 <cpressey> Yes, more than a decade ago.
23:24:18 <AnMaster> cpressey, since I render to svg in cfunge (and will do same in efunge) I can use floating point coordinates
23:24:42 <AnMaster> but you can query coordinates in TURT, but that should return integer
23:25:09 <Ilari> x(x - 1) = 0 could have other solutions than just x = 0 and x = 1 if operation is done in ring that's not integral domain.
23:25:45 <MissPiggy> 5*4=0
23:25:50 <MissPiggy> mod 20 :P
23:25:55 <cpressey> AnMaster: I'm sure you won't break many existing TURT-using Befunge-98 programs if you just pick a behaviour that sounds reasonable :)
23:26:16 <oklopol> 2*3 = 0 in Z_6 if you want an example
23:26:18 <Ilari> And indeed 5^2 = 25 = 5 (mod 20).
23:26:29 <oklopol> oh
23:26:34 <AnMaster> Ilari, that's x^2-x=0 ? Which gives x= 1/2 +/- sqrt(2^2/4) ?
23:26:47 <AnMaster> err wait
23:26:49 <AnMaster> typoed that
23:26:55 <AnMaster> x= 1/2 +/- sqrt(1^2/4) ?
23:26:59 <AnMaster> of course
23:27:40 <AnMaster> modulo style thingies are confusing ;P
23:27:59 <AnMaster> at least when using it that way
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23:29:35 <oklopol> x^2 - x = 0 gives x(x - 1) = 0 which gives x = 0 or x = 1 on R, silly AnMaster.
23:29:58 <AnMaster> oklopol, well yes
23:30:25 <AnMaster> oklopol, I was just checking if there was any solutions in C by using the general formula
23:30:32 <AnMaster> and of course there wasn't
23:30:35 <AnMaster> I'm too sleepy
23:30:46 <oklopol> modulo isn't confusing, Z_n is an abelian ring, and abelian rings are nice and simple things.
23:31:12 <AnMaster> oklopol, only if you are used to them
23:31:14 <oklopol> yeah C is an integral domain, you can't have ab = 0 unless a = 0 or b = 0
23:31:30 <oklopol> oh well also you need to know it's a unique factorization domain
23:31:40 <oklopol> or wait
23:31:49 <oklopol> nooo you don't
23:31:54 <oklopol> silly oklopol
23:32:12 <AnMaster> who said anything about ab=0?
23:32:48 <oklopol> x^2 - x = 0 <=> x(x - 1) = 0
23:33:37 <oklopol> in case your pattern matching skills are sleepy too, a = x, b = x - 1
23:33:38 <cpressey> oklopol: Say a * a = a and b * b = b. But a * e = a and b * e = b. Therefore a = b.
23:34:05 <cpressey> Wait, is that complete bullshit? Hm.
23:34:11 <oklopol> well
23:34:16 <oklopol> your therefore isn't really based on anything
23:34:26 <oklopol> or it's based on division
23:34:38 <oklopol> oh umm
23:35:34 <cpressey> b * b = b, b * e = b, therefore e = b. a * a = a, a * e = a, therefore e = a. Therefore a = b.
23:35:40 <oklopol> yeah i don't see where you get an equality
23:35:56 <oklopol> b * b = b, b * e = b, therefore e = b <<< why?
23:36:02 <cpressey> Substitution?
23:36:17 <AnMaster> <oklopol> in case your pattern matching skills are sleepy too, a = x, b = x - 1 <-- yeah thanks
23:36:22 <cpressey> I can substitute e for b and b for e, therefore e = b, no?
23:36:56 <AnMaster> oklopol, any solutions that are quaternions? If not, what would be needed to get a polynomial with quaternionious solutions?
23:36:58 <oklopol> you can substitute e for b and b for e if they are then multiplied from the left by b, yes.
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23:37:23 <oklopol> AnMaster: no. there are just those two solutions, because also quaternions are an integral domain (i should hope)
23:37:29 <cpressey> Well, just add the mirror-image to get multiplied from the right. e is both left and right identity.
23:37:31 <AnMaster> oklopol, well yes, for that given one
23:37:42 <oklopol> what's a mirror image?
23:37:44 <oklopol> a^-1?
23:37:45 <AnMaster> oklopol, but what sort of polynomial would give a quaternion as a solution?
23:38:01 <MissPiggy> AnMaster, well a polynomial in H would!
23:38:07 <cpressey> oklopol: b * b = b, e * b = b, therefore e = b
23:38:14 <MissPiggy> for example x - q = 0 (with q in H)
23:38:19 <oklopol> cpressey: my guess is you're confusing rings and fields
23:38:27 <AnMaster> MissPiggy, oh good point
23:38:36 <cpressey> oklopol: what axiom from field theory am I using?
23:38:41 <MissPiggy> another thing is, since C is a subset of H
23:38:50 <MissPiggy> then any polynomial with roots in C has roots in H
23:38:59 <oklopol> i mean you're basically getting b*b = e*b ==> b = e, right?
23:39:17 <AnMaster> MissPiggy, well, I meant ones that doesn't fit into "simpler" "subsets"
23:39:30 <cpressey> oklopol: Yes, I believe I stated that
23:39:35 <oklopol> if b has a right inverse, then that's true
23:39:38 <oklopol> but what if it doesn't?
23:39:51 <MissPiggy> AnMaster, I wonder if H is algebraically closed (like C is)
23:40:58 <oklopol> i mean by your logic b*0 = 0, e*0 = 0 => b = e
23:41:05 <oklopol> because you can substitute them for each other
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23:42:05 <cpressey> oklopol: b*0 = 0, e*0 = 0 => b = e = 0
23:42:11 <cpressey> substitute all occurrences of b
23:42:19 <oklopol> are you sure about that?
23:42:21 <oklopol> i mean
23:42:30 <oklopol> you're basically saying all rings only have the one element 0
23:42:35 <oklopol> and i'm not sure that's true
23:42:52 <oklopol> a*0 = 0 for all a
23:42:53 <cpressey> oklopol: No, I'm not saying that.
23:43:17 <oklopol> so take any two elements a and b, and you'll have a*0 = 0, b*0 = 0 => a=b=0
23:43:21 <MissPiggy> 23:40 < oklopol> i mean by your logic b*0 = 0, e*0 = 0 => b = e
23:43:22 <MissPiggy> haha
23:43:43 <MissPiggy> nice misinterpretation of x and y are equal if P(x) <=> P(y) ==> x = y
23:43:57 <MissPiggy> it has to be true for ALL P, not just one P
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23:44:25 <oklopol> i'm not sure that's what's being misinterpreted
23:44:36 <oklopol> i think he's just confusing rings and fields
23:44:43 <MissPiggy> oh don't get algebraic about this
23:44:48 <MissPiggy> for goodness sake
23:44:50 <oklopol> :P
23:45:39 <AnMaster> MissPiggy, no idea
23:46:55 <cpressey> Still not sure what field axiom I'm relying on. If x * x = x then x has a right inverse (it's x.)
23:47:03 <oklopol> cpressey: anyway in Z_6, 4*4 = 16 = 4, so 4 is an idempotent element
23:47:48 <oklopol> i just want to know if you're really not seeing you're doing a division there
23:47:56 <oklopol> hmm
23:48:07 <oklopol> if x*x = x, then x has a right inverse, huh?
23:48:17 <oklopol> that would be x*x = 1
23:49:59 <cpressey> oklopol: I may well be wrong.
23:50:05 <oklopol> cpressey: i can't say what field axiom you're relying on, because i just see a gap in logic. all i can say is that from the existence of inverses, e=b follows trivially from eb = bb
23:50:09 <cpressey> But I don't see how I'm doing division.
23:50:32 <oklopol> that's because you just have a gap in logic, you take what you have, and you say "therefore b=e".
23:50:51 <cpressey> Well, you agree that eb = bb -> e = b?
23:51:08 <oklopol> of course not, i just showed you a counterexample
23:51:21 <cpressey> Then what was your "all i can say" statement about?
23:51:52 <oklopol> that i don't know what field axioms you're relying on. because you aren't relying on anything.
23:52:20 <cpressey> I mean, it sounded like you just said, "All I can say is that from the existence of inverses, eb = bb implies e = b" -- is that not what you meant?
23:52:38 <oklopol> yeah, *if we had inverses*, then that'd trivially be true
23:52:49 <oklopol> so given that we don't, why is it true?
23:54:13 <cpressey> Would you agree it would be true if we knew b had an inverse (regardless of any other element having one)?
23:54:20 <oklopol> yeah
23:54:36 <oklopol> we could then multiply by that inverse.
23:56:08 <cpressey> Well, I wish I could show it had an inverse, then.
23:56:37 <oklopol> yeah good luck showing there are no rings that aren't fields.
23:57:11 <cpressey> I'm not saying EVERY element in the ring has to have an inverse.
23:57:11 <oklopol> (Z_4)
23:57:28 <cpressey> That's where your angle is confusing me.
23:57:34 <oklopol> well right, the idempotent element does
23:58:14 <cpressey> wait
23:58:22 <oklopol> if you don't believe in Z_6, then hmm... if you know linear algebra, you'll probably know a matrix can be idempotent without being identity?
23:58:27 <cpressey> What claim of mine are you trying to refute, anyway?
23:58:38 <oklopol> that you can't have idempotent elements
23:58:41 <cpressey> Is it the thing about Potro on my LoUIE page?
23:58:46 <oklopol> oh wait
23:58:55 <oklopol> i'm trying to refure a*b = a isn't possible
23:59:23 <oklopol> but if a*a = a is possible (and it is), then a*b = a is possible, so doesn't matter
23:59:47 <oklopol> yeah potro
23:59:51 <AnMaster> night →
23:59:56 <oklopol> night
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