From 9a870b7760e0d8e042a0aba678844fe0abdcf75a Mon Sep 17 00:00:00 2001
From: Aaron Schaefer <aaron@elasticdog.com>
Date: Thu, 20 Nov 2008 01:48:43 -0500
Subject: [PATCH] Solution to Project Euler problem 50

---
 extra/project-euler/050/050-tests.factor |  6 ++
 extra/project-euler/050/050.factor       | 90 ++++++++++++++++++++++++
 2 files changed, 96 insertions(+)
 create mode 100644 extra/project-euler/050/050-tests.factor
 create mode 100644 extra/project-euler/050/050.factor

diff --git a/extra/project-euler/050/050-tests.factor b/extra/project-euler/050/050-tests.factor
new file mode 100644
index 0000000000..2bd5482f7e
--- /dev/null
+++ b/extra/project-euler/050/050-tests.factor
@@ -0,0 +1,6 @@
+USING: project-euler.050 project-euler.050.private tools.test ;
+IN: project-euler.050.tests
+
+[ 41 ] [ 100 solve ] unit-test
+[ 953 ] [ 1000 solve ] unit-test
+[ 997651 ] [ euler050 ] unit-test
diff --git a/extra/project-euler/050/050.factor b/extra/project-euler/050/050.factor
new file mode 100644
index 0000000000..f8ce68d173
--- /dev/null
+++ b/extra/project-euler/050/050.factor
@@ -0,0 +1,90 @@
+! Copyright (c) 2008 Aaron Schaefer.
+! See http://factorcode.org/license.txt for BSD license.
+USING: arrays kernel locals math math.primes sequences ;
+IN: project-euler.050
+
+! http://projecteuler.net/index.php?section=problems&id=50
+
+! DESCRIPTION
+! -----------
+
+! The prime 41, can be written as the sum of six consecutive primes:
+
+!     41 = 2 + 3 + 5 + 7 + 11 + 13
+
+! This is the longest sum of consecutive primes that adds to a prime below
+! one-hundred.
+
+! The longest sum of consecutive primes below one-thousand that adds to a
+! prime, contains 21 terms, and is equal to 953.
+
+! Which prime, below one-million, can be written as the sum of the most
+! consecutive primes?
+
+
+! SOLUTION
+! --------
+
+! 1) Create an sequence of all primes under 1000000.
+! 2) Start summing elements in the sequence until the next number would put you
+!    over 1000000.
+! 3) Check if that sum is prime, if not, subtract the last number added.
+! 4) Repeat step 3 until you get a prime number, and store it along with the
+!    how many consecutive numbers from the original sequence it took to get there.
+! 5) Drop the first number from the sequence of primes, and do steps 2-4 again
+! 6) Compare the longest chain from the first run with the second run, and store
+!    the longer of the two.
+! 7) If the sequence of primes is still longer than the longest chain, then
+!    repeat steps 5-7...otherwise, you've found the longest sum of consecutive
+!    primes!
+
+<PRIVATE
+
+:: sum-upto ( seq limit -- length sum )
+    0 seq [ + dup limit > ] find
+    [ swapd - ] [ drop seq length swap ] if* ;
+
+: pop-until-prime ( seq sum -- seq prime )
+    over length 0 > [
+        [ unclip-last-slice ] dip swap -
+        dup prime? [ pop-until-prime ] unless
+    ] [
+        2drop { } 0
+    ] if ;
+
+! a pair is { length of chain, prime the chain sums to }
+
+: longest-prime ( seq limit -- pair )
+    dupd sum-upto dup prime? [
+        2array nip
+    ] [
+        [ head-slice ] dip pop-until-prime
+        [ length ] dip 2array
+    ] if ;
+
+: longest ( pair pair -- longest )
+    2dup [ first ] bi@ > [ drop ] [ nip ] if ;
+
+: continue? ( pair seq -- ? )
+    [ first ] [ length 1- ] bi* < ;
+
+: (find-longest) ( best seq limit -- best )
+    [ longest-prime longest ] 2keep 2over continue? [
+        [ rest-slice ] dip (find-longest)
+    ] [ 2drop ] if ;
+
+: find-longest ( seq limit -- best )
+    { 1 2 } -rot (find-longest) ;
+
+: solve ( n -- answer )
+    [ primes-upto ] keep find-longest second ;
+
+PRIVATE>
+
+: euler050 ( -- answer )
+    1000000 solve ;
+
+! [ euler050 ] 100 ave-time
+! 291 ms run / 20.6 ms GC ave time - 100 trials
+
+MAIN: euler050