factor/extra/project-euler/025/025.factor

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1.7 KiB
Factor

! Copyright (c) 2008 Aaron Schaefer.
! See http://factorcode.org/license.txt for BSD license.
USING: kernel math math.constants math.functions math.parser memoize
project-euler.common sequences ;
IN: project-euler.025
! http://projecteuler.net/index.php?section=problems&id=25
! DESCRIPTION
! -----------
! The Fibonacci sequence is defined by the recurrence relation:
! Fn = Fn-1 + Fn-2, where F1 = 1 and F2 = 1.
! Hence the first 12 terms will be:
! F1 = 1
! F2 = 1
! F3 = 2
! F4 = 3
! F5 = 5
! F6 = 8
! F7 = 13
! F8 = 21
! F9 = 34
! F10 = 55
! F11 = 89
! F12 = 144
! The 12th term, F12, is the first term to contain three digits.
! What is the first term in the Fibonacci sequence to contain 1000 digits?
! SOLUTION
! --------
! Memoized brute force
MEMO: fib ( m -- n )
dup 1 > [ [ 1 - fib ] [ 2 - fib ] bi + ] when ;
<PRIVATE
: (digit-fib) ( n term -- term )
2dup fib number>string length > [ 1 + (digit-fib) ] [ nip ] if ;
: digit-fib ( n -- term )
1 (digit-fib) ;
PRIVATE>
: euler025 ( -- answer )
1000 digit-fib ;
! [ euler025 ] 10 ave-time
! 5345 ms ave run time - 105.91 SD (10 trials)
! ALTERNATE SOLUTIONS
! -------------------
! A number containing 1000 digits is the same as saying it's greater than 10**999
! The nth Fibonacci number is Phi**n / sqrt(5) rounded to the nearest integer
! Thus we need we need "Phi**n / sqrt(5) > 10**999", and we just solve for n
<PRIVATE
: digit-fib* ( n -- term )
1 - 5 log10 2 / + phi log10 / ceiling >integer ;
PRIVATE>
: euler025a ( -- answer )
1000 digit-fib* ;
! [ euler025a ] 100 ave-time
! 0 ms ave run time - 0.17 SD (100 trials)
SOLUTION: euler025a