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Added solution for problem 35
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Problem35.lua
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82
Problem35.lua
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--ProjectEuler/ProjectEulerLua/Problem35.lua
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--Matthew Ellison
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-- Created: 06-05-21
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--Modified: 06-05-21
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--How many circular primes are there below one million?
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--All of my requires, unless otherwise listed, can be found at https://bitbucket.org/Mattrixwv/luaClasses
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--[[
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Copyright (C) 2021 Matthew Ellison
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This program is free software: you can redistribute it and/or modify
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it under the terms of the GNU Lesser General Public License as published by
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the Free Software Foundation, either version 3 of the License, or
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(at your option) any later version.
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This program is distributed in the hope that it will be useful,
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but WITHOUT ANY WARRANTY; without even the implied warranty of
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MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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GNU Lesser General Public License for more details.
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You should have received a copy of the GNU Lesser General Public License
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along with this program. If not, see <https://www.gnu.org/licenses/>.
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]]
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require "Stopwatch"
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require "Algorithms"
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--Setup the variables
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local timer = Stopwatch:create();
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local MAX_NUM = 999999; --The largest number that we are checking for primes
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local primes = {}; --The primes below MAX_NUM
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local circularPrimes = {}; --The circular primes below MAX_NUM
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--Functions
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--Returns a list of all rotations of a string passed to it
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local function getRotations(str)
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local rotations = {};
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table.insert(rotations, str);
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for cnt = 1, string.len(str) - 1 do
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str = string.sub(str, 2) .. string.sub(str, 1, 1);
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table.insert(rotations, str);
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end
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return rotations;
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end
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--Start the timer
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timer:start();
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--Get all primes under 1,000,000
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primes = getPrimes(MAX_NUM);
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--Go through all primes, get all their rotations, and check if those numbers are also primes
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for cnt = 1, #primes do
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local prime = primes[cnt];
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local allRotationsPrime = true;
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--Get all of the rotations of the prime and see if they are also prime
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local rotations = getRotations(tostring(prime));
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for rotCnt = 1, #rotations do
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local rotation = rotations[rotCnt];
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local p = tonumber(rotation, 10);
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if(not isFound(primes, p)) then
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allRotationsPrime = false;
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break;
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end
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end
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--If all rotations are prime add it to the list of circular primes
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if(allRotationsPrime) then
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table.insert(circularPrimes, prime);
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end
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end
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--Stop the timer
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timer:stop();
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--Print the results
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io.write("The number of all circular prime numbers under " .. MAX_NUM .. " is " .. #circularPrimes .. "\n");
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io.write("It took " .. timer:getString() .. " to run this algorithm\n");
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--[[ Results:
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The number of all circular prime numbers under 999999 is 55
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It took 102.268 seconds to run this algorithm
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]]
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