Problem Description

Starting with 1 and spiralling counterclockwise 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%?

Analysis

Forget about the geometry and determine the series: (3, 5, 7, 9), (13, 17, 21, 25), (31, 37, 43, 49), … This represents the “corners” for every square layer.
Determine and count the primes, p_n, in the series ignoring every 4th one since it will always be a square and therefore composite. As soon as we reach a ratio of primes to series length (n), p_n/n < 10% we can calculate a side length as n/2.

It’s important to have a decent is_prime() function to achieve a better run time.

Solution

Runs < 8 seconds in Python.

from Euler import is_prime
n_prime, d, avg, n = 0, 1, 1, 2;
 
while avg >= 0.10:
  n_prime += is_prime(d + n) + is_prime(d + n*2) + is_prime(d + n*3)
  d += n*4
  n += 2
  avg = float(n_prime) / (2 * n)
 
print "Answer to PE58 = ",n-1

Comments

• More information on the Euler module can be found on the tools page.
• Note how we calculate the answer as n-1. This is because we count only even n and the sides of our layer has to be odd.
• See problem 28.