Solution to Project Euler problem 47
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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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USING: arrays combinators.lib kernel math math.primes math.primes.factors math.ranges namespaces sequences ;
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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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pick pick = [
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swap - nip
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] [
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dup prime? [ [ drop 0 ] dipd ] [
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2dup unique-factors length = [ [ 1+ ] dipd ] [ [ drop 0 ] dipd ] 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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! 542708 ms run / 60548 ms GC time
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! ALTERNATE SOLUTIONS
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! -------------------
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! Use a sieve to generate prime factor counts up to a limit, then look for a
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! 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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sieve get nth zero? ;
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: multiples ( n -- seq )
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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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prime-tau-upto [ dup <repetition> ] dip start ;
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PRIVATE>
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: euler047a ( -- answer )
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4 1000000 consecutive-under ;
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! [ euler047a ] 100 ave-time
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! 2589 ms run / 45 ms GC ave time - 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 will also short-circuit when it finds the answer?
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MAIN: euler047a
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@ -13,10 +13,10 @@ USING: definitions io io.files kernel math math.parser project-euler.ave-time
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project-euler.033 project-euler.034 project-euler.035 project-euler.036
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project-euler.037 project-euler.038 project-euler.039 project-euler.040
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project-euler.041 project-euler.042 project-euler.043 project-euler.044
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project-euler.045 project-euler.046 project-euler.048 project-euler.052
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project-euler.053 project-euler.056 project-euler.067 project-euler.075
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project-euler.079 project-euler.092 project-euler.097 project-euler.134
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project-euler.169 project-euler.173 project-euler.175 ;
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project-euler.045 project-euler.046 project-euler.047 project-euler.048
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project-euler.052 project-euler.053 project-euler.056 project-euler.067
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project-euler.075 project-euler.079 project-euler.092 project-euler.097
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project-euler.134 project-euler.169 project-euler.173 project-euler.175 ;
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IN: project-euler
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<PRIVATE
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