--- /dev/null
+! Copyright (c) 2007 Samuel Tardieu.
+! See http://factorcode.org/license.txt for BSD license.
+USING: combinators kernel math math.parser math.ranges sequences vectors ;
+IN: project-euler.175
+
+! http://projecteuler.net/index.php?section=problems&id=175
+
+! DESCRIPTION
+! -----------
+
+! Define f(0)=1 and f(n) to be the number of ways to write n as a sum of
+! powers of 2 where no power occurs more than twice.
+
+! For example, f(10)=5 since there are five different ways to express
+! 10: 10 = 8+2 = 8+1+1 = 4+4+2 = 4+2+2+1+1 = 4+4+1+1
+
+! It can be shown that for every fraction p/q (p0, q0) there exists at
+! least one integer n such that f(n)/f(n-1)=p/q.
+
+! For instance, the smallest n for which f(n)/f(n-1)=13/17 is 241. The
+! binary expansion of 241 is 11110001. Reading this binary number from
+! the most significant bit to the least significant bit there are 4
+! one's, 3 zeroes and 1 one. We shall call the string 4,3,1 the
+! Shortened Binary Expansion of 241.
+
+! Find the Shortened Binary Expansion of the smallest n for which
+! f(n)/f(n-1)=123456789/987654321.
+
+! Give your answer as comma separated integers, without any whitespaces.
+
+! SOLUTION
+! --------
+
+: add-bits ( vec n b -- )
+ over zero? [
+ 3drop
+ ] [
+ pick length 1 bitand = [ over pop + ] when swap push
+ ] if ;
+
+: compute ( vec ratio -- )
+ {
+ { [ dup integer? ] [ 1- 0 add-bits ] }
+ { [ dup 1 < ] [ 1 over - / dupd compute 1 1 add-bits ] }
+ { [ t ] [ [ 1 mod compute ] 2keep >integer 0 add-bits ] }
+ } cond ;
+
+: euler175 ( -- result )
+ V{ 1 } clone dup 123456789/987654321 compute [ number>string ] map "," join ;
+
+! [ euler175 ] 100 ave-time
+! 0 ms run / 0 ms GC ave time - 100 trials
+
+MAIN: euler175