unreality
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it may be notable that the efficiency of hexadecimal over binary is exactly 4 ~~~ anyway, I thought of how it was interesting that at first we were thinking it took a while to converge... anyway, so I made a function g(x) to graph the deviation from the average value (3.321928095) [x] = floor function, ie, take the integer part [log[sub]z[/sub]x]+1 g(x) = ---------------------------- - log[sub]z[/sub]y [log[sub]y[/sub]x]+1 on my calculator for z=2 and y=10, where log(x) is in base 10, it was: (iPart(log(x)/log(2))+1)/(iPart(log(x))+1)-1/log(2) the resulting graph was pretty interesting looking, though not revealing for the smaller numbers. But the calculator has a table feature where you can see the result for large numbers if you want, and it oscillated in a few places, but I went into extremely high numbers and it got smaller and smaller. The deviation gets close to 0, and is within one for almost all numbers. Because the deviation jumps higher at each power of 10 and then gets a little lower and jumps back up, a bit after 10,000 it will always be no more than .5 less than the constant. Anyway just thought that was interesting. I think we've solved this problem in its entirety, but it was a good one while it lasted
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Actually, now that I think about it, instead of the ceiling function, we need to use the floor function and then add 1. The only numbers this makes a difference for are perfect powers of 2 or perfect powers of 10. Ie, in base 2, the number 8 (1000) is 4 digits, not 3. For all other numbers there is no difference Again, I don't expect this to have much impact as x->∞, and the limit should still drop off any ceiling/floor oddities leaving the reduction to log base 2 of 10. So we can say with certainty that the base efficiency for a number approaches 3.322 as x approaches infinity, and then that would be true of the average as well ie, for x=10,000 log210,000 ~= 13.288 log1010,000 = 4 13.288/4 = 3.322 or so So yeah I think this riddle is solved but let me just state it formally from the beginning... * the number of digits of number x in base y is logarithm-base-y of x, floored for the integer part, then added with 1 * to compare the number of digits required in base y and in base z to represent the number x, we just find the digits for each and then divide. To find the efficiency of base 10 over base 2, we do the number of digits it takes in base 2 and divide by the number of digits in base 10 [it's flipped cuz base 2 requires more digits than base 10] * as we get larger, the floor+1 part of the function becomes more and more inconsequential as it's just adding/subtracting tiny parts of a number, between 0 and 1, to both the numerator and denomenator, thus in the end we can drop it out of our final calculations * thus the overall, limit, and average, etc, of the function for the efficiency of base y over base z can be represented by this simple equation: (where z is going to be our smaller base, y the larger one) logzx / logyx recall the change of base formula, where 'b' is ANY base: logac = logbc / logba so we have this: logzx / logyx = ( logxx / logxz ) / ( logxx / logxy ) (a/b) / (c/d) = (a/b)*(d/c) = ad/bc so = (logxx * logxy) / (logxz * logxx) the logxx 's cancel out, leaving: logxy / logxz remember the change of base formula again: logbc / logba = logac where b is ANY base thus logxy / logxz = logzy that's right That's our final result: logzy So the efficiency of base 10 over base 2 is log base 2 of 10 or 3.322 Problem solved, gentlemen ^i've always wanted to say that ;D
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edit2: after reading it, it looks like it's for the defininte integral of a set range of values... we don't have a set range, we want overall average. Although we could figure it out for continually expanding set ranges, turn that into a function, and take the limit. But if we wanted to take the average from, say, 0 to 1 million, we would multiply one millionth by an integral from 0 to million of our equation( Ceiling[Log[2,x]]/Ceiling[Log[10,x]] ) ... but if you think about it, the definite integral sums up the area underenath, and multiplying by the millionth reduces it to the average value from a million values... so this is just a fancy way of saying "take the average of all values from 0 to 1 million" BUT, even though I have no calc training yet, you would just do this then: take the integral from 0 to infinity (I've seen pictures of that, I know it's possible) of our function. Then I don't know if we multiply by 1/inf, ie, infintesimal, or what
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I'm not familiar with OpenOfficeCalc, but why not try Mathematica? Online Integrator this is how to type it in. For the base efficiency of b10 over b2 for a single value of x, it would be Ceiling[Log[2,x]]/Ceiling[Log[10,x]] not sure how exactly you would use integrals (as I said I dont know much calc yet) to find the average for all values of x I'm sure as x -> ∞ in that, the ceilings eventually dont matter so it becomes log2x / log10x = 1 / log102 or log210, both of which equal 3.321928095 but we're more interested in the average of all values of x... hmmm... edit: did a google search, found this, will read it and get back to you
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Yeah I was thinking that, as it stretches out to infinity the convergence rates flattens out until itself is 3.32, the average is 3.32, the average of the average, etc. It just takes longer for each successive average because an infininte infinity more is needed to overcome the more outlying values in the beginning Anyway, I think I sufficiently explained why the value itself approaches log base 2 of 10, and I think we can think informally how the average must approach that same value, but it'd put the final gift wrapping on if we can do it formally with limits or integrals or whatever
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Grayven: the show will not be canceled It has a 6-season story arc that they planned during the very first season. It ends after that. The fifth season starts in 5 days ;D
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They have nothing to do with the Bible I once saw a breakdown of why the writers chose many of the numbers. Some of them they wanted to use because they were already incorporated into the show (Flight 815), others were shout-outs to other great masterpieces (42), others were references to the kinds of intriguing mysteries we love (23), etc. I don't think 4-8-16 is a coincidence either. The Numbers, in other words, don't mean anything in real life. In the LOST world (5 days :o I can't wait! The best show ever, no doubts, no contest ) they mean a lot though 5 days!
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yeah the ones that could be conceivably possible are teleportation (as in Jumper ) and telekinesis, which are probably the best two. Stopping time is up there too, but riddled with physical problems - same with invisibility or shapeshifting.
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I don't know... possibly, if done correctly
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I was actually going to post recently... and there ARE other numbers that share powers, although admittedly they are powers of each other 2^4 = 4^2 works with -2 and -4 as well. In fact many of the lower powers of 2 are very linked with their counterparts... 2^3 = 3^2 - 1, etc. Anyway this is deviating so much from the actual problem I'll stop lol. On topic now: ~~~ if we ignore the brackets around the original equation: Bef(x) in y|z = [logzx] / [logyx] and think of a graph of the function f(x) = log2x / log10x you could rewrite it entirely in log base 10 form (just "log" without a subscript) as: (log(x)/log(2))/log(x) ie, log2x / log10x = 1 / log102 regardless of x this is 3.321928095 which is also equal to log2(10) because that can be written in log base ten as log(10)/log(2) ... but log(10) = 1, so it simplifies to 1 / log(2) ie, log2(10) = 1 / log10(2) = 3.321928095 so if we ignore the brackets, the equation in post #5 will ALWAYS be that exact number. The brackets just add on a bit of decimal (the range of addition is from 0 to 1, so as the numbers get very big, their logarithms get big, and big enough for the ceiling function to become inconsequential on the overall effect) So... the limit of the equation approaches 1/log(2) or log base 2 of 10... so does that mean the average of values up to that point approaches the same limit? Your computer programs have been calculating the AVERAGE (ie, sum of all values up to a result dividing by the total number of values summed) rather than the equation for a single result, right? I'm not that well versed in calculus yet, but it seems to me we could use an integral of some sort to solve this. Or maybe there's a simple formula with the manipulation of limits that can solve it. Is it strange, or commonplace, for the limit of an equation as x approaches infinity to be the same as the average of all the values of x as x goes to infinity?
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Ender's Game Mafia? basically destroying the philosophy of the book ;D
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Why does reading speed matter? Also it'll be fun to discuss, I'll wait til more people finish
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I finished the book earlier today, it's very good Very moving and deep, I thought. I have the sequel too, so I may start reading that, not sure. The book surprised me pretty well in a couple of places, I enjoyed it ;D
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no problem... that was an interesting turn of events Do not fear, Trojans ;D Continue the course!!! *fades away* thanks for hosting, itachi!
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is this a pun because I don't believe in afterdeath/afterlife and thus have no fear of it?
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#1: How often do you find that yous tat in BrainDen longer than you intended 3 #2: How often do you neglect household chores to spend more time on Brainden? 1 #3: How often do you prefer the excitement of BrainDen to face-to-face relations. 1 #4: How often do you form new relationships with fellow BrainDen users? 3 #5: How often do others in your life complain to you about the amount of time you spend on BrainDen? 2 #6: How often do your grades or school work suffer because of the amount of time you spend on Brainden? 1 #7: How often do you check BrainDen before something else that you need to do? 5 #8: How often does your job performance or productivity suffer because of BrainDen? 1 #9: How often do you bcome defensive or secretive when anyone asks you what you do on BrainDen? 4 #10: how often do you block out disturbing thoughts about your life with soothing thoughts of BrainDen? 2 #11: How often do you find yourself anticipating when you will go on BrainDen next? 2 #12: How often do you fear that life without BrainDen would be boring, empty, and pointless? 1 #13: How often do you snap, yell, or act annoyed if someone bothers you while you are on BrainDen? 2 #14: How often do you lose sleep due to late night log-ins? 2 #15: How often do you feel preoccupied with BrainDen when offline, or fantasize about being on BrainDen? 1 #16: How often do you find yourself saying "Just a few more minutes..." when on BrainDen? 3 #17: How often to you try to cut down the amount of time you spend on BrainDen and fail? 2 #18: How often to you choose to spend more time on BrainDen than going out with others? 1 #19: How often do you try to hide how long you've been on BrainDen? 2 #20: How often to you feel depressed, moody, or nervous when you are offline, which goes away once you are back on BrainDen? 1 my result: 40 wow According to this, I'm much less addicted than I thought, lol
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I only got about 20 minutes of reading last night, but I read 5 or so chapters so I still think I'm relatively caught up. I'll read more tonight
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I'm assuming I could be a power that is lol Prof T: yeah we'll be taking out dawh today. I need to go back and look at the day posts, but if SomeGuy is right about Hera having to be in play 100% today then it's even better for the Trojans Right now we think that Athena is dead, but the Achaeans are of course trying to pretend that she/he's still alive Host: itachi 1) A. Person 2) Prof. Templeton 3) Frost -DEAD 4) dawh 5) Cherry Lane -DEAD 6) Twin Pop 7) LIS -DEAD 8) SomeGuy 9) Joe's Student 10) GC 11) PG 12) reaymond 13) impervious -DEAD 14) O'mally 15) unreality - voting for dawh 16) Brandonb [Achaean] 17) Y-San -DEAD [Achaean] ~ Also, I'm not trying to defend the inactives or anything [since usually I am a host, I hate inactivity too ;D] but this game is like Hybrid-hybrid lol, it's not even technically mafia, so you can't expect the days to work the same. It may be more beneficial to not talk and not discuss at all (unless you're organizing your team and whatever). While in normal mafia, it's no excuse to be inactive, in a game like hybrid or this, it's not as important to discuss. What is NOT excusable though is night-action-inactivity
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I read the first 4-5 chapters last night, I'll probably read a bunch more tonight It's pretty good so far, though not super-exceptional yet
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why not Host/creator : soccer11smart 1. unreality 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
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that's very fair If you join us [the Trojans, for anyone who hasn't picked up on who I'm sided with ;D], you'll have made the right choice
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Priam is dead, yes. Paris is cut off and only needs to guess their own actions to get BTSC with us [not sure why it works that way, but it was itachi's decree]. So I was just stressing to Paris that they need to do that tonight
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Dawh seems confident, but the Achaeans may be surprised tonight edit: not only that, but I just realized that dawh is blatantly lying Athena is dead [which hurts the Achaeans a lot] and Hermes CANNOT be recruited by the Achaeans anymore, and has no chance of standing alone independently [so they must join the winners-ie, the Trojans ;D]
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I have a few notes for tonight: * Hermes ~ can you give us a sign that you're alive? Please contact me with your ability tonight * Calchus and Hermes ~ the Trojans are gaining the upper hand, and you can cement that victory. As one of my teammates has so wisely summarized: - We're in majority, especially if Calchus and/or Hermes and/or both were behind us. - You won't be targeted by Achaens, they have more pressing targets. We could protect you every night though [something Achaens can't do so easily], in exchange for your info. - We're in control of Day vote. - You win by being alive at the end of the game. Added bonus of this is that you'll be rewarded by Itachi for the next installment. At the minute, if you stood with us, you would have more chance of staying alive. ~~ Clearly you have an important choice or two up ahead to make - don't make the wrong one To Paris: The more BTSC, the merrier ;D It benefits everyone if you predict your own actions tonight, so as to be added to our BTSC group
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I went to the library and got some books... Slaughterhouse-Five, Hitchhiker's Guide to the Galaxy (hehe), Ender's Game and 'Speaker for the Dead', #2 of the Ender's series (apparently there are 9 total) anyway, I forgot to get GEB, I might next time So I'll start Ender's Game tonight ;D