factor/extra/project-euler/175/175.factor

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Factor

! Copyright (c) 2007 Samuel Tardieu.
! See http://factorcode.org/license.txt for BSD license.
USING: combinators kernel math math.parser math.ranges sequences vectors project-euler.common ;
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
! --------
<PRIVATE
: 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 ] }
[ [ 1 mod compute ] 2keep >integer 0 add-bits ]
} cond ;
PRIVATE>
: euler175 ( -- result )
V{ 1 } clone dup 123456789/987654321 compute [ number>string ] map "," join ;
! [ euler175 ] 100 ave-time
! 0 ms ave run time - 0.31 SD (100 trials)
SOLUTION: euler175