Solution to problem 58 in julia
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src/Julia/Problem058.jl
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src/Julia/Problem058.jl
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Created on 11 Oct 2021
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@author: David Doblas Jiménez
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@email: daviddoji@pm.me
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Solution for Problem 58 of Project Euler
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https://projecteuler.net/problem=58
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=#
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using BenchmarkTools
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using Primes
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function Problem58()
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Starting with 1 and spiralling anticlockwise in the following way,
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a square spiral with side length 7 is formed.
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37 36 35 34 33 32 31
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38 17 16 15 14 13 30
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39 18 5 4 3 12 29
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40 19 6 1 2 11 28
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41 20 7 8 9 10 27
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42 21 22 23 24 25 26
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43 44 45 46 47 48 49
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It is interesting to note that the odd squares lie along the bottom right
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diagonal, but what is more interesting is that 8 out of the 13 numbers
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lying along both diagonals are prime; that is, a ratio of 8/13 ≈ 62%.
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If one complete new layer is wrapped around the spiral above, a square
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spiral with side length 9 will be formed. If this process is continued,
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what is the side length of the square spiral for which the ratio of primes
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along both diagonals first falls below 10%?
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=#
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ratio = 1
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corners = []
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side = 0
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num = 1
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while ratio > 0.1
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side += 2
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for _ in 1:4
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num += side
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push!(corners, isprime(num))
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end
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ratio = sum(corners) / length(corners)
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end
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return side + 1
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end
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println("Time to evaluate Problem 58:")
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@btime Problem58()
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println("")
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println("Result for Problem 58: ", Problem58())
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