Birthday dates using pi continued fractions
Birthday dates using pi continued fractions
Author: Bill Gosper
For π Day. 2015, WRI stirred up publicity with
https://blog.stephenwolfram.com/2015/03/pi-or-pie-celebrating-pi-day-of-the-centuryand-how-to-get-your-very-own-piece-of-pi/
Off-list I grumbled that no self-respecting Deity would bother sending clues to worshipers dumb enough to use decimal instead of continued fractions. However, in 2019 we’re barely able to afford a full CF version of Wolfram’s birthday games.
Assuming Gauss–Kuzmin distribution, define
https://blog.stephenwolfram.com/2015/03/pi-or-pie-celebrating-pi-day-of-the-centuryand-how-to-get-your-very-own-piece-of-pi/
Off-list I grumbled that no self-respecting Deity would bother sending clues to worshipers dumb enough to use decimal instead of continued fractions. However, in 2019 we’re barely able to afford a full CF version of Wolfram’s birthday games.
Assuming Gauss–Kuzmin distribution, define
cfprob[L_List]:=Abs@Log[2,(1+1/FromContinuedFraction@L)/(1+1/FromContinuedFraction@MapAt[#+1&,L,-1])]
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cfprob[r:(Integer_|Rational_)]:=cfprob@ContinuedFraction@r
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Then
cfprob/@Range@3
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Log
4
3
Log[2]
Log
9
8
Log[2]
Log
16
15
Log[2]
Out[]=
N@%
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{0.415037,0.169925,0.0931094}
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I.e, 41.5% of terms should be 1, 17% should be 2, etc.
But cfprob also gives us the probabilities of term sequences:
But cfprob also gives us the probabilities of term sequences:
cfprob/@{{1,2,3},1+1/(2+1/3),{3,2,1},{2,1,3}}
{Log[221/220]/Log[2],Log[221/220]/Log[2],Log[221/220]/Log[2],Log[210/209]/Log[2]}
N@%
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{0.415037,0.169925,0.0931094}
Out[]=
(Invariant under reversal but not shuffling.)
This says to expect about six 1,2,3’s in every burst of 1000 terms:
SequencePosition[ContinuedFraction[π,10^3],{1,2,3}]//tim
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0.00044,6
{{47,49},{293,295},{512,514},{542,544},{841,843},{987,989}}
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Try a million :
SequencePosition[ContinuedFraction[π,10^6],{1,2,3}]//tim
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0.440412,6561
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(0.5 seconds for a million terms. Have I actually lived to see this?)
3^8
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6561
Out[]=
SequencePosition[ContinuedFraction[π,10^6],{3,2,1}]//tim;0;
In[]:=
0.448478,6477
Continued fractions accommodate fancier date formats:
cfprob@{3,14,15}
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Log
593569
593568
Log[2]
Out[]=
N@%
In[]:=
2.43055×10
-6
Out[]=
SequencePosition[ContinuedFraction[π,10^6],{3,14,15}]//tim
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0.443411,2
{{415314,415316},{607114,607116}}
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Sure enough, there were two of them.
But for really fancy dates,
But for really fancy dates,
cfprob@{14,3,2015}//N
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1.79491×10
-10
Out[]=
Utility functions
Utility functions
These are the definitions of the utility function tim you will need for the above evaluations:
Clear[tim];tim[xp_]:=(Print[#[[1]],",",Length[#[[2]]]];If[#[[1]]>69,Speak[#[[1]]]];#[[2]])&@AbsoluteTiming[xp]SetAttributes[tim,HoldAll]
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