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Project Euler Solutions

Project Euler 58 Solution

Project Euler 58 Solution

Project Euler 58: Counting primes that lie on the diagonals of the spiral grid


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 instead: (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, pn, 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), pn/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.

Project Euler 58 Solution

Runs < 0.500 seconds in Python 2.7.

download arrowUse this link to get the Project Euler 58 Solution Python 2.7 source.

Afterthoughts

Project Euler 58 Solution last updated

Discussion

2 Responses to “Project Euler 58 Solution”

  1. Could someone tell me what is the final ratio at the answer length? I feel like I am not getting the right answer because of floating point error. I don’t know what to do about it. I am using C++

    Posted by amir rasouli | September 4, 2017, 11:06 PM
  2. It is fun to note that this solution provides incorrect answers around bound<=0.0895. This is because the floating point error on the ratio is enough by then to push us just under the limit one iteration too soon. If we re-compute n from the precise integrals we seem to last longer, but it is still a matter of time before that result differs from a result using exact rational values.

    Posted by Thomas DuBuisson | December 9, 2015, 9:40 PM

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