91 lines
2.6 KiB
Factor
91 lines
2.6 KiB
Factor
! Copyright (c) 2008 Aaron Schaefer.
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! See http://factorcode.org/license.txt for BSD license.
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USING: arrays kernel locals math math.primes sequences project-euler.common ;
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IN: project-euler.050
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! http://projecteuler.net/index.php?section=problems&id=50
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! DESCRIPTION
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! -----------
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! The prime 41, can be written as the sum of six consecutive primes:
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! 41 = 2 + 3 + 5 + 7 + 11 + 13
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! This is the longest sum of consecutive primes that adds to a prime below
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! one-hundred.
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! The longest sum of consecutive primes below one-thousand that adds to a
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! prime, contains 21 terms, and is equal to 953.
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! Which prime, below one-million, can be written as the sum of the most
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! consecutive primes?
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! SOLUTION
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! --------
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! 1) Create an sequence of all primes under 1000000.
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! 2) Start summing elements in the sequence until the next number would put you
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! over 1000000.
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! 3) Check if that sum is prime, if not, subtract the last number added.
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! 4) Repeat step 3 until you get a prime number, and store it along with the
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! how many consecutive numbers from the original sequence it took to get there.
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! 5) Drop the first number from the sequence of primes, and do steps 2-4 again
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! 6) Compare the longest chain from the first run with the second run, and store
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! the longer of the two.
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! 7) If the sequence of primes is still longer than the longest chain, then
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! repeat steps 5-7...otherwise, you've found the longest sum of consecutive
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! primes!
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<PRIVATE
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:: sum-upto ( seq limit -- length sum )
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0 seq [ + dup limit > ] find
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[ swapd - ] [ drop seq length swap ] if* ;
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: pop-until-prime ( seq sum -- seq prime )
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over length 0 > [
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[ unclip-last-slice ] dip swap -
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dup prime? [ pop-until-prime ] unless
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] [
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2drop { } 0
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] if ;
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! a pair is { length of chain, prime the chain sums to }
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: longest-prime ( seq limit -- pair )
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dupd sum-upto dup prime? [
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2array nip
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] [
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[ head-slice ] dip pop-until-prime
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[ length ] dip 2array
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] if ;
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: longest ( pair pair -- longest )
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2dup [ first ] bi@ > [ drop ] [ nip ] if ;
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: continue? ( pair seq -- ? )
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[ first ] [ length 1 - ] bi* < ;
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: (find-longest) ( best seq limit -- best )
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[ longest-prime longest ] 2keep 2over continue? [
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[ rest-slice ] dip (find-longest)
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] [ 2drop ] if ;
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: find-longest ( seq limit -- best )
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{ 1 2 } -rot (find-longest) ;
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: solve ( n -- answer )
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[ primes-upto ] keep find-longest second ;
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PRIVATE>
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: euler050 ( -- answer )
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1000000 solve ;
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! [ euler050 ] 100 ave-time
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! 291 ms run / 20.6 ms GC ave time - 100 trials
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SOLUTION: euler050
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