factor/extra/project-euler/255/255.factor

! Copyright (c) 2009 Jon Harper.
! See http://factorcode.org/license.txt for BSD license.
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.

! 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: [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.

!     [ 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 )
    [ 0 ] 2dip [ [ dip swap ] curry ] [ [ dip + ] curry ] bi* while ; inline

: x0 ( i -- x0 )
    number-length dup even?
    [ 2 - 2 / 10 swap ^ 7 * ]
    [ 1 - 2 / 10 swap ^ 2 * ] if ;

: ⌈a/b⌉  ( a b -- ⌈a/b⌉ )
    [ 1 - + ] keep /i ;

: xk+1 ( n xk -- xk+1 )
    [ ⌈a/b⌉ ] keep + 2 /i ;

: next-multiple ( a multiple -- next )
    [ [ 1 - ] dip /i 1 + ] keep * ;

DEFER: iteration#
! Gives the number of iterations when xk+1 has the same value for all a<=i<=n
:: (iteration#) ( i xi a b -- # )
    a xi xk+1 dup xi =
    [ drop i b a - 1 + * ]
    [ i 1 + swap a b iteration# ] if ;

! Gives the number of iterations in the general case by breaking into intervals
! in which xk+1 is the same.
:: iteration# ( i xi a b -- # )
    a
    a xi next-multiple
    [ dup b < ]
    [
        ! set up the values for the next iteration
        [ nip [ 1 + ] [ xi + ] bi ] 2keep
        ! set up the arguments for (iteration#)
        [ i xi ] 2dip (iteration#)
    ] produce-sum
    ! deal with the last numbers
    [ drop b [ i xi ] 2dip (iteration#) ] dip
    + ;

: (euler255) ( a b -- answer )
    [ 10^ ] bi@ 1 -
    [ [ drop x0 1 swap ] 2keep iteration# ] 2keep
    swap - 1 + /f ;

PRIVATE>

: euler255 ( -- answer )
    13 14 (euler255) 10 nth-place ;

! [ euler255 ] gc time
! Running time: 37.468911341 seconds

SOLUTION: euler255