USING: math math.functions math.primes
project-euler.common sequences sets ;

IN: project-euler.087

! https://projecteuler.net/index.php?section=problems&id=87

! DESCRIPTION
! -----------

! The smallest number expressible as the sum of a prime square,
! prime cube, and prime fourth power is 28. In fact, there are
! exactly four numbers below fifty that can be expressed in such
! a way:

! 28 = 2^2 + 2^3 + 2^4
! 33 = 3^2 + 2^3 + 2^4
! 49 = 5^2 + 2^3 + 2^4
! 47 = 2^2 + 3^3 + 2^4

! How many numbers below fifty million can be expressed as the
! sum of a prime square, prime cube, and prime fourth power?

<PRIVATE

:: prime-powers-less-than ( primes pow n -- prime-powers )
    primes [ pow ^ ] map [ n <= ] filter ;

! You may think to make a set of all possible sums of a prime
! square and cube and then subtract prime fourths from numbers
! ranging from 1 to 'n' to find this. As n grows large, this is
! actually more inefficient!
!
! Prime numbers grow ~ n / log n
!
! Thus there are (n / log n)^(1/2) prime squares <= n,
!                (n / log n)^(1/3) prime cubes   <= n,
!            and (n / log n)^(1/4) prime fourths <= n.
!
! If we compute the cartesian product of these, this takes
! O((n / log n)^(13/12)).
!
! If we instead precompute sums of squares and cubes, and
! iterate up to n, checking each fourth power against it, this
! takes:
!
! O(n * (n / log n)^(1/4)) = O(n^(5/4)/(log n)^(1/4)) >> O((n / log n)^(13/12))
!
! When n = 50,000,000, the first equation is approximately 10
! million and the second is approximately 2 billion.

:: prime-triples ( n -- answer )
    n sqrt primes-upto                :> primes
    primes 2 n prime-powers-less-than :> primes^2
    primes 3 n prime-powers-less-than :> primes^3
    primes 4 n prime-powers-less-than :> primes^4
    primes^2 primes^3 [ + ] cartesian-map concat
             primes^4 [ + ] cartesian-map concat
    [ n <= ] filter members length ;

PRIVATE>

:: euler087 ( -- answer )
    50,000,000 prime-triples ;

SOLUTION: euler087