Project Euler 351 - Hexagonal Orchards

Official link: https://projecteuler.net/problem=351

Thought Process

Clearly we only need to consider one sixth of the hexagon and then we can multiply the final answer by 6.

Lets focus on the top right triangle

I include a diagram to help readers understand.

Lets denote the angle between P0-P1 and P0-P2 as Angle 1. Then the angle between P0-P1 and P0-P3 is clearly 1/2 of Angle 1, lets call it Angle 1/2. Now clearly the angle between P0-P1 and P0-P4 is also Angle 1, we've already seen this angle before, hence the point will be hidden. Similarly we can see that between P0-P1 and P0-P5 we have Angle 1/3. In short each point, x, on the n-th layer represents Angle x/n.

For example on the 4-th layer we have the 3rd point, P6, will have Angle 2/4 = Angle 1/2

Now the problem is simple.

We just need to find all of the points which have an angle x/n such that gcd(x, n) > 1. We know that on the n-th layer there are gonna be φ(n), where φ(n) is Euler's Totient Function, numbers, x, such that gcd(x, n) = 1, therefore there are n - φ(n) numbers such that gcd(x, n) > 1.

Then we have a nice formula for H(n). Take note of the Totient Summatory Function

Originally just using the wiki page and my mobius sieve I could get the answer in ~85 seconds.

After solving the problem you have access to the PDF overview of the problem, which shows you how to implement the Totient Summatory Function efficiently. I implemented algorithm 6 detailed on the PDF, then runtime has now dropped to ~1 second!

This new algorithm has been added to my python package mathslib

Interactive Code

Input an Integer (yourinput)

Code will output H(n)

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