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MathGroup Archive 2007

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Re: Re: Pi upto a Billion Digits

  • To: mathgroup at smc.vnet.net
  • Subject: [mg75727] Re: [mg75672] Re: Pi upto a Billion Digits
  • From: DrMajorBob <drmajorbob at bigfoot.com>
  • Date: Wed, 9 May 2007 04:34:44 -0400 (EDT)
  • References: <f1msfm$rnf$1@smc.vnet.net> <6330655.1178621270967.JavaMail.root@m35>
  • Reply-to: drmajorbob at bigfoot.com

On my machine (WinXP, Athlon 3200+, 2GB RAM, Mathematica 6), I estimate 
 
3.4 hours:

Timing[data =
   Table[Update[];
    Log@{#, First@Timing[N[Pi, #]]} &[2^k 10^5], {k, 1, 8}]]

{231.921, {{Log[200000], -1.06711}, {Log[400000], -0.226901}, {Log[
     800000], 0.669879}, {Log[1600000], 1.50741}, {Log[3200000],
    2.37052}, {Log[6400000], 3.21451}, {Log[12800000],
    4.05123}, {Log[25600000], 4.87675}}}

(I doubled two steps more than Szabolcs did.)

The log-log data is VERY linear:

Needs["LinearRegression`"]
AdjustedRSquared /. Regress[data, x, x]

0.999901

Clear[a, b, z, logSecs, hours]
model = a + b z;
logSecs[z_] = model /. FindFit[data, model, {a, b}, z]
hours[z_] := Exp[logSecs[Log[z]]]/60^2
hours[10^9]

-16.0424 + 1.22792 z

3.37132

Also note, omitting Update[] has NO effect on the timings:

Timing[data =
   Table[Log@{#, First@Timing[N[Pi, #]]} &[2^k 10^5], {k, 1, 8}]]

{231.703, {{Log[200000], -1.07002}, {Log[400000], -0.207024}, {Log[
     800000], 0.628609}, {Log[1600000], 1.52126}, {Log[3200000],
    2.36471}, {Log[6400000], 3.21133}, {Log[12800000],
    4.04989}, {Log[25600000], 4.8771}}}

Bobby

On Tue, 08 May 2007 04:54:55 -0500, Szabolcs <szhorvat at gmail.com> wrote:=


> Raj wrote:
>> hi!
>>
>> Could somebody tell me if they ever tried finding Pi upto a billion
>> digits using the N function:
>> N[Pi,10^9] and how long did it take?
>>
>> Thanks,
>>
>> Raj
>
> I haven't tried it, but you can estimate how long it takes, like this:=

>
> d = Table[Update[]; First@Timing[N[Pi, 2^k 10^5]], {k, 1, 6}]/Second=

>
> {1.437, 3.579, 9.14, 23.375, 58.453, 143.391}
>
> (Results are from Mathematica 5.2, 1.7 GHz processor.)
>
> ListPlot[Log@d]
>
> Now we can extrapolate. Find the slope of Log@d
>
> slope = Mean@ListConvolve[{1, -1}, Log@d]
>
> 0.920604
>
> 10^9 is approximately 2^12 * 10^5, so you can expect
>
> Exp[11*slope + d[[1]] ]
>
> 105202.
>
> This is approximately 30 hours. Of course this result may be an order =
of
> magnitude off, but I suspect that the precise result is less than 30
> hours, not more. With a little patience you can get a much better  =

> estimate.
>
> Or you can leave the calculation running for a day and hope that you
> don't run out of memory (memory usage may be a bigger problem than CPU=

> time). If you do run the calculation, please let us know your findings=


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