57 lines
1.6 KiB
Ruby
57 lines
1.6 KiB
Ruby
|
problem=%{
|
||
|
-- Starting with 1 and spiralling anticlockwise in the following way, a square spiral 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 spiral above, a square spiral with side length 9 will be formed. If this process is continued, what is the side length of the square spiral for which the ratio of primes along both diagonals first falls below 10%?
|
||
|
--
|
||
|
-- Reflexion:
|
||
|
-- if n is the side length of the square spiral. The numbers at each 'coin' are
|
||
|
--
|
||
|
-- 1 -> 1
|
||
|
-- 2 -> 3 5 7 9
|
||
|
-- 3 -> 13 17 21 25
|
||
|
-- ...
|
||
|
--
|
||
|
-- The rule is
|
||
|
-- (last_max_number + n.(size-1) | n in [1..4])
|
||
|
-- examples:
|
||
|
-- last_max_number=1
|
||
|
-- size = 3
|
||
|
-- 1+2 = 3, 1+2*2=5, 1+3*2=7, 1+4*2=9
|
||
|
--
|
||
|
-- last_max_number=9
|
||
|
-- size=5
|
||
|
-- 9+4=13, 9+2*4=17, 9+3*4=21, 9+4*4=25
|
||
|
}
|
||
|
require 'primes'
|
||
|
$po = Primes.new(100)
|
||
|
|
||
|
size=1
|
||
|
ratio=1
|
||
|
last_max_number=1
|
||
|
nb_prime=0.0
|
||
|
nb_numbers_on_diag=1
|
||
|
|
||
|
while ratio > 0.1 do
|
||
|
size +=2
|
||
|
(1..4).each do |n|
|
||
|
last_max_number += (size-1)
|
||
|
puts last_max_number
|
||
|
nb_prime += 1 if $po.is_prime(last_max_number)
|
||
|
nb_numbers_on_diag += 1
|
||
|
end
|
||
|
ratio= nb_prime / nb_numbers_on_diag
|
||
|
puts ratio
|
||
|
end
|
||
|
|
||
|
puts "size = #{size}"
|