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Comments:
<0> He likes smacks.
<1> @slap Olathe
<2> http://www.haskell.org/hawiki/HaskellUserLocations
<0> :(
<1> hehe
<3> ?
<1> apparently mbot doesn't particiate in slapfests
<1> (s//p/)
<3> ?
<0> pappppaprpepnptplpyp pmpbpoptp ...?
<1> could be
<3> % #%"Bad day to you!"& /@ {0,10,20}
<2> ToExpression::notstrbox: #% & /@ {0,10,20} Bad day to you! is not a string or a box. ToExpression can only interpret strings or boxes as Mathematica input.
<2> $Failed
<3> i failed.
<3> % 3%4
<2> pyenos: 3*"Good day to you!"
<3> ok
<3> % 4%3
<2> pyenos: {4*OutputStream["stderr", 2]}
<3> % 5%4
<2> pyenos: 5*"Good day to you!"
<3> mbot is random
<3> % 6%5
<2> pyenos: 6*"Good day to you!"
<3> % 4%3
<2> pyenos: {4*OutputStream["stderr", 2]}
<3> why?
<3> % 4%4
<1> % %
<2> pyenos: 4*"Good day to you!"
<2> ski: "No state is preserved between computations, so % doesn't work."
<1> mrf
<3> % 3%3
<2> pyenos: {3*OutputStream["stderr", 2]}
<0> % Explode[]
<2> Olathe: Explode[]
<1> % %4
<2> ski: "Good day to you!"
<1> % %5
<2> ski: "Good day to you!"
<1> % %2
<2> ski: {Null, Null, Null, Null, Null}
<1> % %1
<2> ski: {BinaryFormat -> False, FormatType -> OutputForm, PageWidth -> 300, PageHeight -> 22, TotalWidth -> Infinity, TotalHeight -> Infinity, CharacterEncoding :> $CharacterEncoding, NumberMarks :> $
<2> NumberMarks}
<1> % %0
<2> ski: Out[0]
<3> ?
<1> strange :)
<0> % 5
<2> Olathe: 5
<0> mbot isn't responding in private.
<3> ski: what does % %2 mean?
<1> pyenos : who knows ? :)
<4> ski: The square of a summation is called a "change in variable"?
<1> nono
<4> Oh.
<4> It seemed awkward. ;P
<1> the step from '(sum of k+1 from k=0 to n)^2' to '(sum of k from k=1 to n+1)^2' was a change in variable
<4> Ah, I see.
<3> well, ski answered Capso's question earlier:
<5> %n is the result of the nth output in the current session
<4> pyenos: Yeah, I got that.
<3> ((n+1)(n+2)/2)^2 is not equal to (n+1)(n+2)(2n+1)/3.
<3> cool....
<5> mbot starts a new session each time, but there's a few initialisation lines
<4> ski: How would those equivalencies work out for summations from k=0 to infinity? Would it be the same, simply with limits as n approaches infinity?
<4> Cale: Hey.
<5> hi
<3> you can you limit
<3> i think
<3> just a sec
<3> Capso: how about this?
<3> i'll simplify the problem first
<3> let's say
<3> sum_(k=0)^(inf)(k+1) = lim_(c->inf)sum_(k=0)^(c)(K+1)
<3> =lim_(c->inf)sum_(k=1)^(c+1)(k)
<3> =lim_(c->inf)((c+1)(c+2)/2)
<3> ->inf
<3> i meant ->+inf
<3> so the sum is divergent.
<3> what do you think?
<3> Capso: hello?
<3> :( ok....
<3> anyways therefore, (sum_(k=0)^(inf)(k+1))^2 -> +inf
<1> um .. i think that's right
<3> also, i can guess that sum_(k=0)^(n)((k+1)^2) -> +inf also.
<3> oops
<3> substitue n with inf
<3> so you can't compare the two.. since they all approach +inf
<3> hello??????????
<0> Ahoy hoy !
<3> Olathe: hello, are you dutch?
<0> No.
<3> Olathe: ahoy is dutch?
<0> It is ?
<3> Olathe: i'm not sure
<0> I don't know. I just heard people saying ahoy hoy.
<3> ok
<3> anyways....................... i guess i can find solace in the fact that i have a place in the internet where there are people who agree that we can all learn mathematics by helping each other. (unfortunate things happened to my life and i'm not proud of myself) we can all learn anything by helping each other. (please excuse me for irrelevent rant). i think i can learn any topic of mathematics whether other people believe on me or not
<3> by thinking clearly as i can. (sorry that's the end).
<6> guys, take a look at this blurring technique, used to cencor ip addresses: http://lighttpd.net/screenshots/lighttpd-status-large.png
<6> isn't it possible to "uncencor" it and get the original ip addresses?
<6> since we can just run through all the ip address string images, and only a single one will give the exact blurred result
<6> right?
<3> ?
<6> eh?
<3> ?
<6> sorry i don't understand
<1> possibly the blurring function is not injective
<3> xic: me too :(
<6> ski: what do you mean?
<3> ?
<6> the blurring function most likely just takes the average color of all of the pixels in it's square region
<1> xic : it could be that two different ip adresses give the same blurred pattern
<6> it could be, but not very likely i don't think
<3> :-O
<6> each digit falls into about 2 squares, there are 10 digits, and 2^16 2-square combinations
<3> what is 2-square combination?
<7> hi #maths !
<6> 2 adjacent blurred squares. each square is a gray color value ranging 0-255
<7> good mornin' all !
<8> hi
<6> ski: what do you think?
<9> censor
<3> xic: each digit is 2^16 combinations because each square is 2^8 so 2^8.2^8, right?
<6> pyenos: yes
<3> so what is your exact question again?
<6> pyenos: but it's just an approximation, since each digit doesn't exactly fall into 2 squares. the point is that i think that there's a good chance that only a single ip address string will give the exact blurred representation
<3> wow
<3> that's awesome
<3> why do you think that xic?
<6> hm.... sorry but i'm not sure that i can explain it any better then i did :|
<3> xic: but you must know the reason why you think that there's a good chance that an unique ip and an unique blurring.
<3> i want to know why you thin so.
<6> because there is only a small chance that 2 digits, out of a set of 10, will both blur into the same blur pattern, out of more then 60,000 blur patterns
<3> where did you get the lower estimate of 60,000 from?
<6> 2^16, it's just a rough estimate
<3> i'm not convinced of your thinkining
<3> because practically speaking, it is advantageous of blurring function to have more than one digit key matchnig with the same color value.
<6> by looking at the blurred squares that are in between the lines, it appears that no randomness is being used
<3> xic: why can't the blurring function use for every set of key S, map them to the color C.
<6> it can, but this particular blurring function just takes the average color value of all of the 100 pixels in it's 10x10 pixel square region. i've just confirmed this with the gimp
<3> xic: wow......... you are very insightful...
<3> xic: it is certainly extremely interesting... i just wish i had the knowledge and insight at the moment. (but i'll keep this problem in mind)
<6> i'm almost positive now that the original ips can be discovered
<3> xic: i don't have the math background to help you... what a shame at the moment...
<6> the math is actually very simple, just basic arithmetic
<3> xic: i wish i had more.... skills...
<10> what kinda skill
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