1 ! Copyright (c) 2009 Jon Harper.
2 ! See https://factorcode.org/license.txt for BSD license.
3 USING: kernel math math.functions project-euler.common ;
6 ! https://projecteuler.net/problem=255
11 ! We define the rounded-square-root of a positive integer n as
12 ! the square root of n rounded to the nearest integer.
14 ! The following procedure (essentially Heron's method adapted to
15 ! integer arithmetic) finds the rounded-square-root of n:
17 ! Let d be the number of digits of the number n.
18 ! If d is odd, set x_(0) = 2×10^((d-1)⁄2).
19 ! If d is even, set x_(0) = 7×10^((d-2)⁄2).
21 ! Repeat: [see URL for figure ]
23 ! until x_(k+1) = x_(k).
25 ! As an example, let us find the rounded-square-root of n =
26 ! 4321. n has 4 digits, so x_(0) = 7×10^((4-2)⁄2) = 70.
28 ! [ see URL for figure ]
30 ! Since x_(2) = x_(1), we stop here.
32 ! So, after just two iterations, we have found that the
33 ! rounded-square-root of 4321 is 66 (the actual square root is
36 ! The number of iterations required when using this method is
37 ! surprisingly low. For example, we can find the
38 ! rounded-square-root of a 5-digit integer (10,000 ≤ n ≤ 99,999)
39 ! with an average of 3.2102888889 iterations (the average value
40 ! was rounded to 10 decimal places).
42 ! Using the procedure described above, what is the average
43 ! number of iterations required to find the rounded-square-root
44 ! of a 14-digit number (10^(13) ≤ n < 10^(14))? Give your answer
45 ! rounded to 10 decimal places.
47 ! Note: The symbols ⌊x⌋ and ⌈x⌉ represent the floor function and
48 ! ceiling function respectively.
55 ! same as produce, but outputs the sum instead of the sequence of results
56 : produce-sum ( id pred quot -- sum )
57 [ 0 ] 2dip [ [ dip swap ] curry ] [ [ dip + ] curry ] bi* while ; inline
60 number-length dup even?
61 [ 2 - 2 / 10 swap ^ 7 * ]
62 [ 1 - 2 / 10 swap ^ 2 * ] if ;
64 : ⌈a/b⌉ ( a b -- ⌈a/b⌉ )
67 : xk+1 ( n xk -- xk+1 )
68 [ ⌈a/b⌉ ] keep + 2 /i ;
70 : next-multiple ( a multiple -- next )
71 [ [ 1 - ] dip /i 1 + ] keep * ;
74 ! Gives the number of iterations when xk+1 has the same value for all a<=i<=n
75 :: (iteration#) ( i xi a b -- # )
77 [ drop i b a - 1 + * ]
78 [ i 1 + swap a b iteration# ] if ;
80 ! Gives the number of iterations in the general case by breaking into intervals
81 ! in which xk+1 is the same.
82 :: iteration# ( i xi a b -- # )
87 ! set up the values for the next iteration
88 [ nip [ 1 + ] [ xi + ] bi ] 2keep
89 ! set up the arguments for (iteration#)
90 [ i xi ] 2dip (iteration#)
92 ! deal with the last numbers
93 [ drop b [ i xi ] 2dip (iteration#) ] dip
96 : (euler255) ( a b -- answer )
98 [ [ drop x0 1 swap ] 2keep iteration# ] 2keep
103 : euler255 ( -- answer )
104 13 14 (euler255) 10 nth-place ;
106 ! [ euler255 ] gc time
107 ! Running time: 37.468911341 seconds