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