[Haskell-beginners] [Haskell-cafe] For Euler 25; What is the first term in the Fibonacci sequence to contain 1000 digits?; the following seems to work.

[KC]

Project Euler wants the term number not the actual Fibonacci number.

I wonder what the compiled timings would be. :)


On Thu, May 19, 2011 at 11:34 AM, Sebastian Fischer <fischer at nii.ac.jp> wrote:
>
> On Thu, May 19, 2011 at 7:29 PM, KC <kc1956 at gmail.com> wrote:
>>
>> For Euler 25; What is the first term in the Fibonacci sequence to
>> contain 1000 digits?; the following seems to work.
>>
>>
>> -- For number of digits being 5 or more.
>> fibNdigits :: Int -> Int
>> fibNdigits nDigits = floor (((fromIntegral nDigits) - 1.0) / (logBase
>> 10 phi)) + 2
>>  where
>>    sq5 = sqrt 5 :: Double
>>    phi = (1 + sq5) / 2
>
>
> (length . show) is fast enough for me. The following gives the answer in a
> fraction of a second.
> # cabal install fibonacci
> # ghci
> ghci> import Data.Numbers.Fibonacci
> ghci> head . filter ((==1000) . length . show) $ map fib [0..]
> 1070066266382758936764980584457396885083683896632151665013235203375314520604694040621889147582489792657804694888177591957484336466672569959512996030461262748092482186144069433051234774442750273781753087579391666192149259186759553966422837148943113074699503439547001985432609723067290192870526447243726117715821825548491120525013201478612965931381792235559657452039506137551467837543229119602129934048260706175397706847068202895486902666185435124521900369480641357447470911707619766945691070098024393439617474103736912503231365532164773697023167755051595173518460579954919410967778373229665796581646513903488154256310184224190259846088000110186255550245493937113651657039447629584714548523425950428582425306083544435428212611008992863795048006894330309773217834864543113205765659868456288616808718693835297350643986297640660000723562917905207051164077614812491885830945940566688339109350944456576357666151619317753792891661581327159616877487983821820492520348473874384736771934512787029218636250627816
>



-- 
--
Regards,
KC