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MathGroup Archive 2004

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Re: Re: Fibonacci[1,000,000,000] contains 208,987,640 decimal digits (was: Fibonachi[5,000,000] contains 1044938 decimal digits)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg49703] Re: [mg49668] Re: Fibonacci[1,000,000,000] contains 208,987,640 decimal digits (was: Fibonachi[5,000,000] contains 1044938 decimal digits)
  • From: Daniel Lichtblau <danl at wolfram.com>
  • Date: Thu, 29 Jul 2004 07:43:15 -0400 (EDT)
  • References: <7f4ffu$6dj$1@nnrp1.dejanews.com>, <Mo2OZGACWlF3Ewez@raos.demon.co.uk>, <7g0qsd$dr9@smc.vnet.net> <cdvm0e$7hv$1@smc.vnet.net> <ce2ftd$8rh$1@smc.vnet.net> <200407271100.HAA11140@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Alex Vinokur wrote:
> "Michael Taktikos" <michael.taktikos at hanse.net> wrote in message news:ce2ftd$8rh$1 at smc.vnet.net...
> 
>>Alex Vinokur <alexvn at bigfoot.com> wrote:
>>
>>
>>>>>>Several large Fibonacci numbers were calculated using only
>>>>>>the well-known explicit formula:
>>>>>>       Fib (0) = 0, Fib (1) = 1,
>>>>>>       Fib (n) = Fib (n-1) + Fib (n-2), n >= 2
>>>>>>All the (decimal) digits of these numbers were obtained.
>>>>>
>>[...]
>>
>>>>>>But to get ALL digits of the large Fibonacci number is very
>>>>>
>>>advisable one.
>>>[C++ code, snip]
>>
>>Thanks for your code.
> 
> 
> Impoved version of that algorithm can be seen at
> http://groups.google.com/groups?selm=2mk62rFo1mv0U1%40uni-berlin.de
> 
> 
>>Until now, the fastest way to get with "Mathematica-alone" all digits of
>>large Fibonacci numbers seems to be that of Roman Maeder:
>>
>>********************************************************
>>fibmaeder[n_] :=
>> Module[{r11 = 1, r12 = 0, r22 = 1, digits = IntegerDigits[n-1, 2], i, t},
>>  Do[ If[ digits[[i]] == 1,
>>          {r11, r22} = {r11(r11 + 2r12), r12(r11 + r22)};
>>          r12 = r11 - r22
>>        , t = r12(r11 + r22);
>>          {r11, r12} = {r11(r11 + 2r12) - t, t};
>>          r22 = r11 - r12
>>        ],
>>     {i, Length[digits]-1}
>>    ];
>>  If[ digits[[-1]] == 1,
>>      r11(r11 + 2r12),
>>      r11(r11 + r22) - (-1)^((n-1)/2)
>>  ]
>> ]
>>
> 
> [snip]
> 
> I would like to measure the comparative performance of my C++ algorithm and "Mathematica-alone" code above.
> However I have never worked with "Mathematica-alone".
> How can I compare those algorithms?

You could run them both on identical machines/inputs and check the 
timings. You might also want to read what Mark Sofroniou said about this 
five years ago (in response to your similar post back then).

http://forums.wolfram.com/mathgroup/archive/1999/Apr/msg00349.html

His point remains unchanged. Not knowing then about Mark's posts, I also 
wrote a bit about this in 2001:

http://forums.wolfram.com/mathgroup/archive/2001/Jan/msg00234.html

But what Mark said is most relevant here. In essence, the method in 
Maeder's code is a good divide-and-conquer, based on doubling 
recurrences in Fibonacci numbers. This is much faster than anything 
based on the linear recurrences, as it takes advantage of asymptotically 
fast multiplication.


Daniel Lichtblau
Wolfram Research


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