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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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