factor: trim using lists

[factor.git] / extra / project-euler / 134 / 134.factor

diff --git a/extra/project-euler/134/134.factor b/extra/project-euler/134/134.factor
index 90d8404760e0fb759523d0d188d3060b8c1079ad..36e45b989ceff2517041704d5237d5678785d6e4 100644 (file)
--- a/extra/project-euler/134/134.factor
+++ b/extra/project-euler/134/134.factor
@@ -1,7 +1,7 @@
 ! Copyright (c) 2007 Samuel Tardieu.
 ! See http://factorcode.org/license.txt for BSD license.
-USING: arrays kernel lazy-lists math.algebra math math.functions math.primes
-       math.ranges sequences ;
+USING: arrays kernel lists lists.lazy math math.algebra
+math.functions math.primes.lists project-euler.common ;
 IN: project-euler.134
 
 ! http://projecteuler.net/index.php?section=problems&id=134
@@ -9,34 +9,40 @@ IN: project-euler.134
 ! DESCRIPTION
 ! -----------
 
-! Consider the consecutive primes p1 = 19 and p2 = 23. It can be
-! verified that 1219 is the smallest number such that the last digits
-! are formed by p1 whilst also being divisible by p2.
+! Consider the consecutive primes p1 = 19 and p2 = 23. It can be verified that
+! 1219 is the smallest number such that the last digits are formed by p1 whilst
+! also being divisible by p2.
 
 ! In fact, with the exception of p1 = 3 and p2 = 5, for every pair of
-! consecutive primes, p2 p1, there exist values of n for which the last
-! digits are formed by p1 and n is divisible by p2. Let S be the
-! smallest of these values of n.
+! consecutive primes, p2 p1, there exist values of n for which the last digits
+! are formed by p1 and n is divisible by p2. Let S be the smallest of these
+! values of n.
 
 ! Find S for every pair of consecutive primes with 5 p1 1000000.
 
+
 ! SOLUTION
 ! --------
 
-! Compute the smallest power of 10 greater than m or equal to it
+! Compute the smallest power of 10 greater than or equal to m
 : next-power-of-10 ( m -- n )
-  10 swap log 10 log / ceiling >integer ^ ; foldable
+    10 swap log10 ceiling >integer ^ ; foldable
+
+<PRIVATE
 
 ! Compute S for a given pair (p1, p2) -- that is the smallest positive
 ! number such that X = p1 [npt] and X = 0 [p2] (npt being the smallest
 ! power of 10 above p1)
 : s ( p1 p2 -- s )
-  over 0 2array rot next-power-of-10 rot 2array chinese-remainder ;
+    over 0 2array rot next-power-of-10 rot 2array chinese-remainder ;
+
+PRIVATE>
 
 : euler134 ( -- answer )
-  5 lprimes-from [ 1000000 > ] luntil [ [ s + ] keep ] leach drop ;
+    0 5 lprimes-from uncons [ 1000000 > ] luntil
+    [ [ s + ] keep ] leach drop ;
 
 ! [ euler134 ] 10 ave-time
-! 6743 ms run / 79 ms GC ave time - 10 trials
+! 933 ms ave run timen - 19.58 SD (10 trials)
 
-MAIN: euler134
+SOLUTION: euler134