Project Euler : problem 265

extra/project-euler/265/265-tests.factor

@@ -0,0 +1,5 @@
+! Copyright (c) 2010 Samuel Tardieu.
+! See http://factorcode.org/license.txt for BSD license.
+USING: project-euler.265 tools.test ;
+
+[ 209110240768 ] [ euler265 ] unit-test

extra/project-euler/265/265.factor

@@ -0,0 +1,63 @@
+! Copyright (c) 2010 Samuel Tardieu.
+! See http://factorcode.org/license.txt for BSD license.
+USING: kernel math math.functions project-euler.common sequences sets ;
+IN: project-euler.265
+
+! http://projecteuler.net/index.php?section=problems&id=265
+
+! 2^(N) binary digits can be placed in a circle so that all the N-digit
+! clockwise subsequences are distinct.
+
+! For N=3, two such circular arrangements are possible, ignoring rotations.
+
+! For the first arrangement, the 3-digit subsequences, in clockwise order, are:
+! 000, 001, 010, 101, 011, 111, 110 and 100.
+
+! Each circular arrangement can be encoded as a number by concatenating
+! the binary digits starting with the subsequence of all zeros as the most
+! significant bits and proceeding clockwise. The two arrangements for N=3 are
+! thus represented as 23 and 29:
+! 00010111 _(2) = 23
+! 00011101 _(2) = 29
+
+! Calling S(N) the sum of the unique numeric representations, we can see that S(3) = 23 + 29 = 52.
+
+! Find S(5).
+
+CONSTANT: N 5
+
+: decompose ( n -- seq )
+    N iota [ drop [ 2/ ] [ 1 bitand ] bi ] map nip reverse ;
+
+: bits ( seq -- n )
+    0 [ [ 2 * ] [ + ] bi* ] reduce ;
+
+: complete ( seq -- seq' )
+    unclip decompose append [ 1 bitand ] map ;
+
+: rotate-bits ( seq -- seq' )
+    dup length iota [ cut prepend bits ] with map ;
+
+: ?register ( acc seq -- )
+    complete rotate-bits
+    dup [ 2 N ^ mod ] map all-unique? [ infimum swap push ] [ 2drop ] if ;
+
+: add-bit ( seen bit -- seen' t/f )
+    over last 2 * + 2 N ^ mod
+    2dup swap member? [ drop f ] [ suffix t ] if ;
+
+: iterate ( acc left seen -- )
+    over 0 = [
+        nip ?register
+    ] [
+        [ 1 - ] dip
+        { 0 1 } [ add-bit [ iterate ] [ 3drop ] if ] with with with each
+    ] if ;
+
+: euler265 ( -- answer )
+    V{ } clone [ 2 N ^ N - { 0 } iterate ] [ sum ] bi ;
+
+! [ euler265 ] time
+! Running time: 0.376389019 seconds
+
+SOLUTION: euler265

extra/project-euler/project-euler.factor

@@ -26,7 +26,7 @@ USING: definitions io io.files io.pathnames kernel math math.parser
     project-euler.134 project-euler.148 project-euler.150 project-euler.151
     project-euler.164 project-euler.169 project-euler.173 project-euler.175
     project-euler.186 project-euler.188 project-euler.190 project-euler.203
-    project-euler.206 project-euler.215 project-euler.255 ;
+    project-euler.206 project-euler.215 project-euler.255 project-euler.265 ;
 IN: project-euler
 
 <PRIVATE