--ProjectEuler/ProjectEulerLua/Problem37.lua --Matthew Ellison -- Created: 07-01-21 --Modified: 07-01-21 --Find the sum of the only eleven primes that are both truncatable from left to right and right to left (2, 3, 5, and 7 are not counted). --All of my requires, unless otherwise listed, can be found at https://bitbucket.org/Mattrixwv/luaClasses --[[ Copyright (C) 2021 Matthew Ellison This program is free software: you can redistribute it and/or modify it under the terms of the GNU Lesser General Public License as published by the Free Software Foundation, either version 3 of the License, or (at your option) any later version. This program is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU Lesser General Public License for more details. You should have received a copy of the GNU Lesser General Public License along with this program. If not, see . ]] require "Stopwatch" require "Algorithms" require "SieveOfEratosthenes" --Setup the variables local timer = Stopwatch:create(); local LAST_PRIME_BEFORE_CHECK = 7; --The last prime before 11 since single digit primes aren't checked local truncPrimes = {}; --All numbers that are truncatable primes local sum = 0; --The sum of all elements in truncPrimes --Start the timer timer:start(); --Create the sieve and get the first prime number local sieve = SieveOfEratosthenes:create(); local currentPrime = sieve:next(); --Loop through the sieve until you get to LAS_PRIME_BEFORE_CHECK while(currentPrime < LAST_PRIME_BEFORE_CHECK) do currentPrime = sieve:next(); end --Loop until truncPrimes contains 11 elements while(#truncPrimes < 11)do local isTruncPrime = true; --Get the next rpime currentPrime = sieve:next(); --Convert the prime to a string local primeString = tostring(currentPrime); --If the string contains an even digit move to the next prime local strLoc = 1; while((strLoc <= #primeString) and (isTruncPrime)) do local char = string.sub(primeString, strLoc, strLoc); --Allow 2 to be the first digit if((strLoc == 1) and (char == "2")) then else if((char == "0") or (char == "2") or (char == "4") or (char == "6") or (char == "8")) then isTruncPrime = false; end end strLoc = strLoc + 1; end --Start removing digits from the left and see if the number stays prime if(isTruncPrime) then for truncLoc = 2, #primeString do --Create a substring of the prime, removing the needed digits from the left local primeSubstring = string.sub(primeString, truncLoc); --Convert the string to an int and see if the number is still prime local newPrime = tonumber(primeSubstring); if(not isPrime(newPrime)) then isTruncPrime = false; break; end end end --Start removing digits from the right and see if the number stays prime if(isTruncPrime) then for truncLoc = 1, #primeString - 1 do --Create a substring of the prime, removing the needed digits from the right local primeSubstring = string.sub(primeString, 1, #primeString - truncLoc); --Convert the string to an int and see if the number is still a prime local newPrime = tonumber(primeSubstring); if(not isPrime(newPrime)) then isTruncPrime = false; break; end end end --If the number remained prime through all operations add it to the table if(isTruncPrime) then table.insert(truncPrimes, currentPrime); end end --Get the sum of all elements in the truncPrimes vector sum = getSum(truncPrimes); --Stop the timer timer:stop(); --Print the results io.write("The sum of all left and right truncatable primes is " .. sum .. "\n"); io.write("It took " .. timer:getString() .. " to run this algorithm\n"); --[[ Results: The sum of all left and right truncatable primes is 748317 It took 491.000 milliseconds to run this algorithm ]]