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

http://github.com/abeaumont/factor
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``` 1! Copyright (c) 2008 Aaron Schaefer.
3USING: kernel math math.constants math.functions math.parser memoize
4    project-euler.common sequences ;
5IN: project-euler.025
6
7! http://projecteuler.net/index.php?section=problems&id=25
8
9! DESCRIPTION
10! -----------
11
12! The Fibonacci sequence is defined by the recurrence relation:
13
14!     Fn = Fn-1 + Fn-2, where F1 = 1 and F2 = 1.
15
16! Hence the first 12 terms will be:
17
18!     F1 = 1
19!     F2 = 1
20!     F3 = 2
21!     F4 = 3
22!     F5 = 5
23!     F6 = 8
24!     F7 = 13
25!     F8 = 21
26!     F9 = 34
27!     F10 = 55
28!     F11 = 89
29!     F12 = 144
30
31! The 12th term, F12, is the first term to contain three digits.
32
33! What is the first term in the Fibonacci sequence to contain 1000 digits?
34
35
36! SOLUTION
37! --------
38
39! Memoized brute force
40
41MEMO: fib ( m -- n )
42    dup 1 > [ [ 1 - fib ] [ 2 - fib ] bi + ] when ;
43
44<PRIVATE
45
46: (digit-fib) ( n term -- term )
47    2dup fib number>string length > [ 1 + (digit-fib) ] [ nip ] if ;
48
49: digit-fib ( n -- term )
50    1 (digit-fib) ;
51
52PRIVATE>
53
54: euler025 ( -- answer )
55    1000 digit-fib ;
56
57! [ euler025 ] 10 ave-time
58! 5345 ms ave run time - 105.91 SD (10 trials)
59
60
61! ALTERNATE SOLUTIONS
62! -------------------
63
64! A number containing 1000 digits is the same as saying it's greater than 10**999
65! The nth Fibonacci number is Phi**n / sqrt(5) rounded to the nearest integer
66! Thus we need we need "Phi**n / sqrt(5) > 10**999", and we just solve for n
67
68<PRIVATE
69
70: digit-fib* ( n -- term )
71    1 - 5 log10 2 / + phi log10 / ceiling >integer ;
72
73PRIVATE>
74
75: euler025a ( -- answer )
76    1000 digit-fib* ;
77
78! [ euler025a ] 100 ave-time
79! 0 ms ave run time - 0.17 SD (100 trials)
80
81SOLUTION: euler025a
```