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Oct-18-17
 | | beatgiant: <alexmagnus>
Funny you should mention e and pi, I was thinking given the factorial function, we might try Stirling's formula.n! =approx (2*pi*n)^.5 * (n/e)^n
Solving for pi, we have
((n!*(e/n)^n)^2)/2n = pi
I tried this for n=29, the highest n that shows as an exact value on my calculator. ((29!*(2.71828/29)^29)^2)/(2*29) = 3.15...
It must better with higher n, but my calculator itself is probably using some version of Stirling's formula to estimate the factorials. Leading to... you guessed it... a circularity. |
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Oct-18-17
 | | SwitchingQuylthulg: (97+9/22+11/88062328.3)^(1/4) has an efficiency of 22 - 18 = +4. Of course, there's a trivial way to beat that - recite pi up to the first spot where five consecutive zeros appear, and stop. |
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Oct-18-17
 | | al wazir: 5*7 + 5! = 155. Try 3*5! - 5 instead.
So (3*5! - 5)/((5! - 7) = 355/(5! - 7) = 71/(4! - 7/5) = 3.1415929... is a co-leader (efficiency = 7-5 = 2). |
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Oct-18-17
| | beatgiant: With our factorial capability, we can generate arbitrarily large numbers with a single digit, like 9!, 9!!, 9!!!, etc. and if Stirling's formula converges, and if one has an arbitrary precision math package and a lot of compute power, the method should become very <al wazir>-efficient. That's the power of composing with looping constructs. <SwitchingQuylthulg>
<recite pi up to the first spot where five consecutive zeros appear>
And that's a big weakness in our definition of efficiency. But is it definitely known that there is a spot where five consecutive zeros appears? |
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Oct-18-17
| | beatgiant: <SwitchingQuylthulg>
Checking on MathOverflow, I learned <There is a sequence of 12 zeroes starting at position 1755524129973> citing Fabrice Bellard https://bellard.org/pi/pi2700e9/pid... So there's our new champion: The first 1755524129972 digits of pi yeild 1755524129972 + 12 digits of accuracy, for efficiency of 12. |
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Oct-18-17
| | al wazir: <beatgiant: But is it definitely known that there is a spot where five consecutive zeros appears?> Although the digits in the decimal expansion of pi are self-evidently not random, they satisfy every test of randomness. So there must be a place where this occurs. Since the chances of this happening as a result of a truly random process are 1 in 10^5, we would expect it to occur within the first 10^5 or so places. That said, the first occurrence of six consecutive 9s starts in the 762nd place: https://en.wikipedia.org/wiki/Six_n... |
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Oct-18-17
 | | alexmagnus: <That said, the first occurrence of six consecutive 9s starts in the 762nd place: >
What's more interesting, the <next> occurence of six identical digits is again that of 9s. |
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Oct-18-17
| | alexmagnus: We all surely remember: last year AlphaGo beat one of the best Go players in the world 4:1. Well, now <that> version of AlphaGo lost to a newer version, called AlphaGo Zero, by a score of <0 to 100>. With taking up less resources than the original. https://gizmodo.com/stunning-ai-bre... Imagine. It's like going from Deep Blue to something stronger than Stockfish within a year. After AphaGo itself took a leap from something comparable to the chess engine Kaissa within a year. |
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Oct-18-17
| | beatgiant: <al wazir>
<there must be a place where this occurs>... is actually not <definitely known> but is an open conjecture, unless you've got a major new result up your sleeve. |
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Oct-18-17
 | | WannaBe: I hope the contestant who wrote down Tibet, realize that it's a land locked country... Dumbazz http://www.sfgate.com/entertainment... And a naval officer does not know ocean geography. Ouch. |
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Oct-18-17
| | SwitchingQuylthulg: (7.85 + 738/357823923.23114723657983449515923235473) ^ (5/9) has an efficiency of 52 - 46 = 6, unless that giant decimal is huge enough to throw off multiple online calculators. With a factorial, it gets an efficiency of 7:
(7.85 + 738 / (8!*8874.6+51.23114723657983449515923235473)) ^ (5/9) |
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Oct-18-17
| | al wazir: <SwitchingQuylthulg: (7.85 + 738/357823923.23114723657983449515923235473) ^ (5/9) has an efficiency of 52 - 46 = 6, unless that giant decimal is huge enough to throw off multiple online calculators.> I'm a little skeptical about those online calculators. For example, the one at http://web2.0calc.com/ evaluates (97+9/22+11/88062328.3)^(1/4) as differing from pi in the 16th decimal place. Which calculators did you use? |
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Oct-18-17
| | SwitchingQuylthulg: <al wazir> WolframAlpha's calculator, https://apfloat.appspot.com/ (with lots of extra 0s after the decimal points to encourage precision) and http://keisan.casio.com/calculator. The first two gave the same result, while the last only displayed 50 digits, all identical with pi. |
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Oct-18-17
| | al wazir: <beatgiant: <there must be a place where this occurs>... is actually not <definitely known> but is an open conjecture, unless you've got a major new result up your sleeve.> No, it's not a new result. The first 0 occurs only in the 40th place. If no zeros ever occurred, the sequence would fail the tests for "randomness." The same thing is true if 00, 000, 0000, 00000, or any other finite sequence you care to choose never occurs. It's just a little less blatant. |
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Oct-18-17
| | beatgiant: <al wazir>
We can only ever have finite subsets of the decimal expansion for pi to test for randomness. That is good evidence, but not proof.It's widely believed, but not as far as I know proven, that <any finite sequence you care to choose> does eventually occur. |
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Oct-18-17
| | al wazir: <beatgiant: We can only ever have finite subsets of the decimal expansion for pi to test for randomness. That is good evidence, but not proof.> You're right. Pi has been calculated out to one trillion digits or so. But that means we can't know anything about the frequency of occurrence of sequences two trillion digits long. However, I'm sure that the frequency with which short sequences like 00000 occur has been checked -- though not beyond the trillionth digit. |
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Oct-19-17
| | beatgiant: Updated leaderboard with the contributions of our world-beater <SwitchingQuylthulg>: 3 + (the next 1755524129972 digits of pi)
Efficiency: 12
Method: Look for a long string of 0's in the digits of pi.
This method will scale to arbitrarily high efficiency if the conjecture about the normality of pi is true. That would kill the whole contest or motivate a new definition of efficiency. Runner-up:
(7.85 + 738 / (8!*8874.6+51.23114723657983449515923235473)) ^ (5/9)
Efficiency: 7
Method: fine-tune a previous efficient solution. (97+9/22+11/88062328.3)^(1/4)
Efficiency: 4
Method: fine-tune a previous efficient solution. And one up-and-comer:
((n!*(e/n)^n)^2)/2n = pi
Method: Stirling's formula. The initial test with n=29 and using 6 digits of e was not very efficient, but it might scale with very large values of n, which we can generate with a single digit via repeated factorials. But I lack the resources to test the idea. |
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| Oct-19-17 | | Count Wedgemore: <al wazir: The first 0 occurs only in the 40th place> Actually, the 32nd..
<beatgiant: It's widely believed, but not as far as I know proven, that <any finite sequence you care to choose> does eventually occur.> While it is fascinating to ponder the distinct possibilty that pi may contain the answer to every question one could possibly want to ask, it is also not of any practical use, is it? Ater all, a string that contains everything contains nothing. |
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| Oct-19-17 | | john barleycorn: < Count Wedgemore: ...
Ater all, a string that contains everything contains nothing.> Amen to that. However, the irrationality of pi may be contagious to the mind of its researchers. Anyway, I never gave up upon a search for next week's lotto numbers under the known digits of pi. :-) |
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Oct-19-17
| | al wazir: <Count Wedgemore: <al wazir: The first 0 occurs only in the 40th place<<>>> Actually, the 32nd.> I slipped and wrote that in octal there. Sorry about that. |
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Oct-19-17
| | beatgiant: Here's a new spinoff stumper based on <al wazir>'s stumper. As we saw, the <al wazir>-efficiency of representations of pi is increasing without bound but apparently very slowly. The new stumper is:
Define AW(b) as the best attainable al wazir efficiency for a given budget of b digits. The rules are same as before - we can use +, -, x, /, ^, !, ., (, ), and we have an nth-root oracle. I show the first few values below. The new stumper is, what's the smallest number b such that AW(b) > 2? AW(1) = 0, solution 3
AW(2) = 0, solutions 3.1, 3 + .1
AW(3) = 0, solutions 3.14, 3 + .14, 22/7, (4!-2)/7 AW(4) = 1, solution 2 + 4!^(1/4!)
AW(5) = 2, solutions 7.85^(5/9), 355/(4! - 7) |
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Oct-20-17
| | WannaBe: Okay, any predictions on when SkyNet will come online? 2030? 2040? http://www.foxnews.com/tech/2017/10... |
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| Oct-20-17 | | diceman: <WannaBe:
Okay, any predictions on when SkyNet will come online? 2030? 2040?> 2035
August 13 @ 8:38AM EST.
(all Fibonacci numbers) |
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Oct-25-17
| | WannaBe: Garbage in, garbage out... Guess programmers still have not yet learned. http://mashable.com/2017/10/25/goog... |
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| Oct-25-17 | | Marmot PFL: Bet cat owners and atheists scored even worse but I am not about to give them my CC number just to find out. |
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