clean up PE solution 255

extra/project-euler/255/255.factor

@@ -1,40 +1,54 @@
-! Copyright (C) 2009 Jon Harper.
+! Copyright (c) 2009 Jon Harper.
 ! See http://factorcode.org/license.txt for BSD license.
-USING: project-euler.common math kernel sequences math.functions math.ranges prettyprint io threads math.parser locals arrays namespaces ;
+USING: arrays io kernel locals math math.functions math.parser math.ranges
+    namespaces prettyprint project-euler.common sequences threads ;
 IN: project-euler.255
 
 ! http://projecteuler.net/index.php?section=problems&id=255
 
 ! DESCRIPTION
 ! -----------
-! We define the rounded-square-root of a positive integer n as the square root of n rounded to the nearest integer.
+
-! 
+! We define the rounded-square-root of a positive integer n as the square root
-! The following procedure (essentially Heron's method adapted to integer arithmetic) finds the rounded-square-root of n:
+! of n rounded to the nearest integer.
-! 
+
+! The following procedure (essentially Heron's method adapted to integer
+! arithmetic) finds the rounded-square-root of n:
+
 !     Let d be the number of digits of the number n.
 !     If d is odd, set x_(0) = 2×10^((d-1)⁄2).
 !     If d is even, set x_(0) = 7×10^((d-2)⁄2).
-! Repeat:
+
-! 
+!     Repeat: [see URL for figure ]
+
 !     until x_(k+1) = x_(k).
-! 
+
 ! As an example, let us find the rounded-square-root of n = 4321.
 ! n has 4 digits, so x_(0) = 7×10^((4-2)⁄2) = 70.
-! 
-! Since x_(2) = x_(1), we stop here.
-! So, after just two iterations, we have found that the rounded-square-root of 4321 is 66 (the actual square root is 65.7343137…).
-! 
-! The number of iterations required when using this method is surprisingly low.
-! For example, we can find the rounded-square-root of a 5-digit integer (10,000 ≤ n ≤ 99,999) with an average of 3.2102888889 iterations (the average value was rounded to 10 decimal places).
-! 
-! Using the procedure described above, what is the average number of iterations required to find the rounded-square-root of a 14-digit number (10^(13) ≤ n < 10^(14))?
-! Give your answer rounded to 10 decimal places.
-! 
-! Note: The symbols ⌊x⌋ and ⌈x⌉ represent the floor function and ceiling function respectively.
-! 
-<PRIVATE
 
-: round-to-10-decimals ( a -- b ) 1.0e10 * round 1.0e10 / ;
+!     [ see URL for figure ]
+
+! Since x_(2) = x_(1), we stop here.
+
+! So, after just two iterations, we have found that the rounded-square-root of
+! 4321 is 66 (the actual square root is 65.7343137…).
+
+! The number of iterations required when using this method is surprisingly low.
+! For example, we can find the rounded-square-root of a 5-digit integer
+! (10,000 ≤ n ≤ 99,999) with an average of 3.2102888889 iterations (the average
+! value was rounded to 10 decimal places).
+
+! Using the procedure described above, what is the average number of iterations
+! required to find the rounded-square-root of a 14-digit number
+! (10^(13) ≤ n < 10^(14))? Give your answer rounded to 10 decimal places.
+
+! Note: The symbols ⌊x⌋ and ⌈x⌉ represent the floor function and ceiling
+! function respectively.
+
+! SOLUTION
+! --------
+
+<PRIVATE
 
 ! same as produce, but outputs the sum instead of the sequence of results
 : produce-sum ( id pred quot -- sum )
@@ -44,6 +58,7 @@ IN: project-euler.255
     number-length dup even?
     [ 2 - 2 / 10 swap ^ 7 * ]
     [ 1 - 2 / 10 swap ^ 2 * ] if ;
+
 : ⌈a/b⌉  ( a b -- ⌈a/b⌉ )
     [ 1 - + ] keep /i ;
 
@@ -76,18 +91,17 @@ DEFER: iteration#
     [ drop b [ i xi ] 2dip (iteration#) ] dip
     + ;
 
-: 10^ ( a -- 10^a ) 10 swap ^ ; inline
-
 : (euler255) ( a b -- answer )
     [ 10^ ] bi@ 1 -
     [ [ drop x0 1 swap ] 2keep iteration# ] 2keep
     swap - 1 + /f ;
 
-
 PRIVATE>
 
 : euler255 ( -- answer )
-    13 14 (euler255) round-to-10-decimals ;
+    13 14 (euler255) 10 nth-place ;
+
+! [ euler255 ] gc time
+! Running time: 37.468911341 seconds
 
 SOLUTION: euler255
-

extra/project-euler/common/common.factor

@@ -92,6 +92,9 @@ PRIVATE>
         [ [ 10 * ] [ 1 + ] bi* ] while 2nip
     ] if-zero ;
 
+: nth-place ( x n -- y )
+    10^ [ * round >integer ] keep /f ;
+
 : nth-prime ( n -- n )
     1 - lprimes lnth ;