2008-02-06 00:53:18 -05:00
|
|
|
|
! Copyright (c) 2008 Aaron Schaefer.
|
|
|
|
|
! See http://factorcode.org/license.txt for BSD license.
|
2009-02-02 07:04:29 -05:00
|
|
|
|
USING: kernel math math.functions math.ranges math.order
|
|
|
|
|
project-euler.common sequences ;
|
2008-02-06 00:53:18 -05:00
|
|
|
|
IN: project-euler.044
|
|
|
|
|
|
|
|
|
|
! http://projecteuler.net/index.php?section=problems&id=44
|
|
|
|
|
|
|
|
|
|
! DESCRIPTION
|
|
|
|
|
! -----------
|
|
|
|
|
|
|
|
|
|
! Pentagonal numbers are generated by the formula, Pn=n(3n−1)/2. The first ten
|
|
|
|
|
! pentagonal numbers are:
|
|
|
|
|
|
|
|
|
|
! 1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...
|
|
|
|
|
|
|
|
|
|
! It can be seen that P4 + P7 = 22 + 70 = 92 = P8. However, their difference,
|
|
|
|
|
! 70 − 22 = 48, is not pentagonal.
|
|
|
|
|
|
|
|
|
|
! Find the pair of pentagonal numbers, Pj and Pk, for which their sum and
|
|
|
|
|
! difference is pentagonal and D = |Pk − Pj| is minimised; what is the value of D?
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
! SOLUTION
|
|
|
|
|
! --------
|
|
|
|
|
|
|
|
|
|
! Brute force using a cartesian product and an arbitrarily chosen limit.
|
|
|
|
|
|
|
|
|
|
<PRIVATE
|
|
|
|
|
|
|
|
|
|
: nth-pentagonal ( n -- seq )
|
|
|
|
|
dup 3 * 1- * 2 / ;
|
|
|
|
|
|
|
|
|
|
: sum-and-diff? ( m n -- ? )
|
2008-11-03 16:20:36 -05:00
|
|
|
|
[ + ] [ - ] 2bi [ pentagonal? ] bi@ and ;
|
2008-02-06 00:53:18 -05:00
|
|
|
|
|
|
|
|
|
PRIVATE>
|
|
|
|
|
|
|
|
|
|
: euler044 ( -- answer )
|
|
|
|
|
2500 [1,b] [ nth-pentagonal ] map dup cartesian-product
|
2009-02-02 07:04:29 -05:00
|
|
|
|
[ first2 sum-and-diff? ] filter [ first2 - abs ] [ min ] map-reduce ;
|
2008-02-06 00:53:18 -05:00
|
|
|
|
|
|
|
|
|
! [ euler044 ] 10 ave-time
|
2008-11-05 01:11:15 -05:00
|
|
|
|
! 4996 ms ave run time - 87.46 SD (10 trials)
|
2008-02-06 00:53:18 -05:00
|
|
|
|
|
|
|
|
|
! TODO: this solution is ugly and not very efficient...find a better algorithm
|
|
|
|
|
|
|
|
|
|
MAIN: euler044
|