2008-02-12 23:02:33 -05:00
										 
									 
								 
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								! Copyright (c) 2008 Aaron Schaefer.
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								! See http://factorcode.org/license.txt for BSD license.
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											2008-10-02 18:40:49 -04:00
										 
									 
								 
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								USING: arrays kernel math math.primes math.primes.factors
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											2009-03-19 00:05:32 -04:00
										 
									 
								 
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								    math.ranges namespaces sequences project-euler.common ;
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											2008-02-12 23:02:33 -05:00
										 
									 
								 
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								IN: project-euler.047
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								! http://projecteuler.net/index.php?section=problems&id=47
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								! DESCRIPTION
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								! -----------
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								! The first two consecutive numbers to have two distinct prime factors are:
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								!     14 = 2 * 7
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								!     15 = 3 * 5
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								! The first three consecutive numbers to have three distinct prime factors are:
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								!     644 = 2² * 7 * 23
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								!     645 = 3 * 5 * 43
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								!     646 = 2 * 17 * 19.
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								! Find the first four consecutive integers to have four distinct primes
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								! factors. What is the first of these numbers?
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								! SOLUTION
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								! --------
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								! Brute force, not sure why it's incredibly slow compared to other languages
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								<PRIVATE
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								: (consecutive) ( count goal test -- n )
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											2008-11-17 18:41:21 -05:00
										 
									 
								 
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								    2over = [
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								        swap - nip
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								    ] [
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											2008-05-23 23:48:58 -04:00
										 
									 
								 
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								        dup prime? [ [ drop 0 ] 2dip ] [
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											2009-08-13 20:21:44 -04:00
										 
									 
								 
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								            2dup unique-factors length = [ [ 1 + ] 2dip ] [ [ drop 0 ] 2dip ] if
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								        ] if 1 + (consecutive)
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								    ] if ;
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								: consecutive ( goal test -- n )
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								    0 -rot (consecutive) ;
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								PRIVATE>
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								: euler047 ( -- answer )
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								    4 646 consecutive ;
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								! [ euler047 ] time
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											2008-11-05 01:11:15 -05:00
										 
									 
								 
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								! 344688 ms run / 20727 ms GC time
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								! ALTERNATE SOLUTIONS
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								! -------------------
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											2008-02-12 23:31:10 -05:00
										 
									 
								 
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								! Use a sieve to generate prime factor counts up to an arbitrary limit, then
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								! look for a repetition of the specified number of factors.
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								<PRIVATE
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								SYMBOL: sieve
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								: initialize-sieve ( n -- )
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								    0 <repetition> >array sieve set ;
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								: is-prime? ( index -- ? )
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											2008-11-09 22:10:42 -05:00
										 
									 
								 
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								    sieve get nth 0 = ;
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								: multiples ( n -- seq )
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											2009-08-13 20:21:44 -04:00
										 
									 
								 
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								    sieve get length 1 - over <range> ;
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								: increment-counts ( n -- )
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								     multiples [ sieve get [ 1 + ] change-nth ] each ;
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								: prime-tau-upto ( limit -- seq )
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								    dup initialize-sieve 2 swap [a,b) [
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								        dup is-prime? [ increment-counts ] [ drop ] if
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								    ] each sieve get ;
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								: consecutive-under ( m limit -- n/f )
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											2017-06-01 15:45:54 -04:00
										 
									 
								 
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								    prime-tau-upto [ dup <repetition> ] dip subseq-start ;
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								PRIVATE>
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								: euler047a ( -- answer )
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											2008-02-12 23:31:10 -05:00
										 
									 
								 
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								    4 200000 consecutive-under ;
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								! [ euler047a ] 100 ave-time
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											2008-11-05 01:11:15 -05:00
										 
									 
								 
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								! 331 ms ave run time - 19.14 SD (100 trials)
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								! TODO: I don't like that you have to specify the upper bound, maybe try making
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								! this lazy so it could also short-circuit when it finds the answer?
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											2009-03-19 00:05:32 -04:00
										 
									 
								 
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								SOLUTION: euler047a
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