From 03fa8a4887101563a1cd54f4934d4098c2763008 Mon Sep 17 00:00:00 2001
From: Aaron Schaefer <aaron@elasticdog.com>
Date: Fri, 4 Jan 2008 01:40:01 -0500
Subject: [PATCH] Alternate solution to Project Euler problem 25
---
extra/project-euler/025/025.factor | 33 ++++++++++++++++++++++++++++--
1 file changed, 31 insertions(+), 2 deletions(-)
diff --git a/extra/project-euler/025/025.factor b/extra/project-euler/025/025.factor
index 2da1ee6b57..8d75ef5493 100644
--- a/extra/project-euler/025/025.factor
+++ b/extra/project-euler/025/025.factor
@@ -1,6 +1,7 @@
! Copyright (c) 2007 Aaron Schaefer.
! See http://factorcode.org/license.txt for BSD license.
-USING: kernel math math.parser memoize sequences ;
+USING: alien.syntax kernel math math.functions math.parser math.ranges memoize
+ sequences ;
IN: project-euler.025
! http://projecteuler.net/index.php?section=problems&id=25
@@ -35,6 +36,8 @@ IN: project-euler.025
! SOLUTION
! --------
+! Memoized brute force
+
MEMO: fib ( m -- n )
dup 1 > [ 1 - dup fib swap 1 - fib + ] when ;
@@ -54,4 +57,30 @@ PRIVATE>
! [ euler025 ] 10 ave-time
! 5237 ms run / 72 ms GC ave time - 10 trials
-MAIN: euler025
+
+! 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
+
+FUNCTION: double log10 ( double x ) ;
+
+: phi ( -- phi )
+ 5 sqrt 1+ 2 / ;
+
+: digit-fib* ( n -- term )
+ 1- 5 log10 2 / + phi log10 / ceiling >integer ;
+
+PRIVATE>
+
+: euler025a ( -- answer )
+ 1000 digit-fib* ;
+
+! [ euler025a ] 100 ave-time
+! 0 ms run / 0 ms GC ave time - 100 trials
+
+MAIN: euler025a