50 lines
1.2 KiB
Julia
50 lines
1.2 KiB
Julia
#=
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Created on 14 Sep 2021
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@author: David Doblas Jiménez
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@email: daviddoji@pm.me
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Solution for Problem 44 of Project Euler
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https://projecteuler.net/problem=44 =#
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using BenchmarkTools
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using Combinatorics
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function pentagonal(n)
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return Int(n * (3 * n - 1) / 2)
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end
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function Problem44()
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#=
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Pentagonal numbers are generated by the formula, Pn=n(3n−1)/2.
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The first ten pentagonal numbers are:
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1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...
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It can be seen that P4 + P7 = 22 + 70 = 92 = P8. However, their
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difference, 70 − 22 = 48, is not pentagonal.
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Find the pair of pentagonal numbers, Pj and Pk, for which their
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sum and difference are pentagonal and D = |Pk − Pj| is minimised.
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What is the value of D? =#
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dif = 0
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pentagonal_list = [pentagonal(n) for n in 1:2500]
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pairs = combinations(pentagonal_list, 2)
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for p in pairs
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if reduce(+, p) in pentagonal_list && abs(reduce(-, p)) in pentagonal_list
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dif = (abs(reduce(-, p)))
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# the first one found would be the smallest
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break
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end
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end
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return dif
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end
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println("Time to evaluate Problem $(lpad(44, 3, "0")):")
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@btime Problem44()
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println("")
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println("Result for Problem $(lpad(44, 3, "0")): ", Problem44())
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