factor/extra/project-euler/058/058.factor

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1.8 KiB
Factor

! Copyright (c) 2009 Aaron Schaefer.
! See http://factorcode.org/license.txt for BSD license.
USING: fry kernel math math.primes math.ranges project-euler.common sequences ;
IN: project-euler.058
! http://projecteuler.net/index.php?section=problems&id=58
! DESCRIPTION
! -----------
! Starting with 1 and solveling anticlockwise in the following way, a square
! solve with side length 7 is formed.
! 37 36 35 34 33 32 31
! 38 17 16 15 14 13 30
! 39 18 5 4 3 12 29
! 40 19 6 1 2 11 28
! 41 20 7 8 9 10 27
! 42 21 22 23 24 25 26
! 43 44 45 46 47 48 49
! It is interesting to note that the odd squares lie along the bottom right
! diagonal, but what is more interesting is that 8 out of the 13 numbers lying
! along both diagonals are prime; that is, a ratio of 8/13 ≈ 62%.
! If one complete new layer is wrapped around the solve above, a square solve
! with side length 9 will be formed. If this process is continued, what is the
! side length of the square solve for which the ratio of primes along both
! diagonals first falls below 10%?
! SOLUTION
! --------
<PRIVATE
CONSTANT: PERCENT_PRIME 0.1
! The corners of a square of side length n are:
! (n-2)² + 1(n-1)
! (n-2)² + 2(n-1)
! (n-2)² + 3(n-1)
! (n-2)² + 4(n-1) = odd squares, no need to calculate
: prime-corners ( n -- m )
3 [1,b] swap '[ _ [ 1 - * ] keep 2 - sq + prime? ] count ;
: total-corners ( n -- m )
1 - 2 * ; foldable
: ratio-below? ( count length -- ? )
total-corners 1 + / PERCENT_PRIME < ;
: next-layer ( count length -- count' length' )
2 + [ prime-corners + ] keep ;
: solve ( count length -- length )
2dup ratio-below? [ nip ] [ next-layer solve ] if ;
PRIVATE>
: euler058 ( -- answer )
8 7 solve ;
! [ euler058 ] 10 ave-time
! 12974 ms ave run time - 284.46 SD (10 trials)
SOLUTION: euler058