00:04:54 <ehird`> both y and z then apply, right?
00:06:09 <olsner> no, if the first x-rule fails, the second one is tried
00:06:34 <olsner> so that's x :- y ; z. (y or z)
00:07:09 <ehird`> olsner: so if i can't overload :-
00:07:16 <ehird`> what would you reccomend as the operator for that? :P
00:07:30 <ehird`> i've been using == but that doesn't make much sense for multiple cases
00:07:38 <ehird`> or <=, but that's really evil :-)
00:12:20 <ehird`> olsner: if you don't say anything i'll override the goddarn less-than-or-equal-to op
00:12:50 <olsner> I don't understand, are you doing this in prolog?
00:13:18 <olsner> obviously, if you've changed the meaning of :-, my previous answer was wrong
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00:33:55 <ehird`> writing a prolog interp as a DSL in Ruby
00:44:46 <olsner> aha, so you're looking for ruby operators to overload into their prolog meaning?
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00:51:43 <ehird`> prolog would be really suited for something i want to code up
00:52:09 <ehird`> take a few numbers, and a target number, then using some basic maths operators get from the numbers to the target
00:52:16 <ehird`> probably np-complete or something
00:53:06 <ehird`> i'd probably have to do my own math ops for it...
00:53:17 <ehird`> solve([target,...],dest,X).
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03:17:34 * pikhq would like to take this opportunity to curse at calculus
03:17:49 <pikhq> \int{{x^2}\over{x^2+1}}dx
03:21:39 <RodgerTheGreat> to preface this, I understand coding theory and the entropy of data. I understand that it is impossible to create a "magic" compression algo that would work for *any* input, and I'm asking about this idea merely to discover wether people think it's sufficiently plausible for Sci-fi, not real implementation
03:22:29 <RodgerTheGreat> There are many operations you can do to pairs of operands that generate infinite, nonrepeating sequences
03:22:41 <RodgerTheGreat> Pi would be an example of one of these nonrepeating sequences
03:23:53 <RodgerTheGreat> it is reasonable to assume that you can find an arbitrary pattern (arbitrary data) in Pi, and encode data as merely an index and run length into the digits of Pi, but you'd probably have such huge numbers that you wouldn't actually save any space, and it'd take forever to calculate
03:24:29 <RodgerTheGreat> however, there are many possible numbers you could index into- this would improve your odds of finding a "close" match dramatically
03:25:07 <RodgerTheGreat> Also, in many kinds of digital data, the actual arrangement of information is largely arbitrary- take a look at a .TAR file.
03:25:30 <RodgerTheGreat> there are a massive number of arrangements of data within a .TAR that would reconstruct exactly the same data
03:26:55 <RodgerTheGreat> so, if you had *MASSIVE* computing power to try every combination of rearranging your data, breaking it up into pieces and adding noncoding padding, in combination with a large pool of sequences to search, you could theoretically compress enormous amounts of data into something that could be written on a piece of paper.
03:27:37 <RodgerTheGreat> And there's always the possibility that you simply won't be able to find the sequence you want, so there would be a certain probability that specific data would be just plain impossible to compress
03:28:20 <RodgerTheGreat> but in essence, what do you guys think? Complete technobabble, or vaguely plausible?
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03:36:45 <calamari> pikhq: I just put that into my hp48gx calculator..
03:37:24 <calamari> pikhq: and I must have done something wrong because that's not the answer :P
03:38:06 <GregorR> RodgerTheGreat: Once you have the index, you have to find the location of the index within pi. Statistically speaking, since the index is smaller than the original data, it is more likely to appear, sooner. So you can usually reduce it further. Of course, this will break down at an unpredictable point, but you could probably get it down to a few billion gigs of data.
03:38:20 <calamari> RodgerTheGreat: that was latex?
03:39:49 <RodgerTheGreat> GregorR: that's the value of being able to search many, many repeating sequences, choosing from a large selection of binary "targets" that can all decode identically
03:41:03 <RodgerTheGreat> the more variables you can play with, the higher your odds become to find a usable index that's some kind of sane distance in
03:42:03 * pikhq is just cursing at partial fractions
03:42:16 <pikhq> Screw it: I've spent 40 minutes at it and have no answer. I'm not going to find anything.
03:42:30 <RodgerTheGreat> your odds would also improve significantly for smaller filesizes, too
03:42:41 <pikhq> Unless someone can tell me where to start.
03:43:13 <calamari> pikhq: sorry, my calculator isn't smart enough for that one
03:43:29 <pikhq> RodgerTheGreat: Damned clever.
03:43:34 <calamari> seems like you could use substitution tho
03:43:48 <pikhq> calamari: u=x^2. du=2xdx. How does that help?
03:44:41 <RodgerTheGreat> I think if you're willing to accept the caveats that it will take ludicrous amounts of time to compress data, your mileage may vary, and some sequences are incompressible, it sounds semi-realistic
03:44:50 <calamari> pikhq: is that a real question, because Itook calc 2 a number of years ago and I don't remember much :P
03:45:19 <pikhq> calamari: I'm in it now, and that stupid problem is going to give me nightmares.
03:46:04 * pikhq moves on to the next problem set. . .
03:46:10 <pikhq> Where I can use a table of integrals. Yay!
03:46:15 <calamari> arctan u = u'/(1+u^2) does that help?
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04:00:54 <Sgeo> No immibis (no=hi)
04:11:09 <Sgeo> Does anyone know what can cause a blocked port 80?
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04:22:52 <calamari> hrm, I did something wrong because that doesn't work :)
04:35:45 <calamari> argh.. I think I've completely forgotten how to do integration by parts
04:35:50 <immibis> er, 1-1/1+ is surplus isn't it?
04:53:30 <calamari> (x^2 * arctan(x)) - integrate(x * arctan(x) dx)
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11:09:28 <oerjan> x^2/(1+x^2) = 1-1/(1+x^2), so calamari had it right
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11:14:17 <oerjan> (long division of polynomials)
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21:58:04 <Sgeo> Is ANYONE going to play with PSOX?
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