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Added solution to problem 37
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--ProjectEuler/ProjectEuler/Problem36.luaClasses
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--ProjectEuler/ProjectEulerLua/Problem36.lua
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--Matthew Ellison
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-- Created: 06-29-21
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--Modified: 06-29-21
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111
Problem37.lua
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111
Problem37.lua
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--ProjectEuler/ProjectEulerLua/Problem37.lua
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--Matthew Ellison
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-- Created: 07-01-21
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--Modified: 07-01-21
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--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).
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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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require "SieveOfEratosthenes"
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--Setup the variables
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local timer = Stopwatch:create();
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local LAST_PRIME_BEFORE_CHECK = 7; --The last prime before 11 since single digit primes aren't checked
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local truncPrimes = {}; --All numbers that are truncatable primes
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local sum = 0; --The sum of all elements in truncPrimes
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--Start the timer
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timer:start();
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--Create the sieve and get the first prime number
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local sieve = SieveOfEratosthenes:create();
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local currentPrime = sieve:next();
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--Loop through the sieve until you get to LAS_PRIME_BEFORE_CHECK
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while(currentPrime < LAST_PRIME_BEFORE_CHECK) do
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currentPrime = sieve:next();
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end
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--Loop until truncPrimes contains 11 elements
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while(#truncPrimes < 11)do
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local isTruncPrime = true;
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--Get the next rpime
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currentPrime = sieve:next();
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--Convert the prime to a string
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local primeString = tostring(currentPrime);
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--If the string contains an even digit move to the next prime
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local strLoc = 1;
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while((strLoc <= #primeString) and (isTruncPrime)) do
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local char = string.sub(primeString, strLoc, strLoc);
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--Allow 2 to be the first digit
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if((strLoc == 1) and (char == "2")) then
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else
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if((char == "0") or (char == "2") or (char == "4") or (char == "6") or (char == "8")) then
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isTruncPrime = false;
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end
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end
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strLoc = strLoc + 1;
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end
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--Start removing digits from the left and see if the number stays prime
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if(isTruncPrime) then
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for truncLoc = 2, #primeString do
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--Create a substring of the prime, removing the needed digits from the left
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local primeSubstring = string.sub(primeString, truncLoc);
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--Convert the string to an int and see if the number is still prime
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local newPrime = tonumber(primeSubstring);
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if(not isPrime(newPrime)) then
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isTruncPrime = false;
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break;
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end
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end
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end
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--Start removing digits from the right and see if the number stays prime
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if(isTruncPrime) then
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for truncLoc = 1, #primeString - 1 do
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--Create a substring of the prime, removing the needed digits from the right
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local primeSubstring = string.sub(primeString, 1, #primeString - truncLoc);
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--Convert the string to an int and see if the number is still a prime
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local newPrime = tonumber(primeSubstring);
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if(not isPrime(newPrime)) then
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isTruncPrime = false;
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break;
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end
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end
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end
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--If the number remained prime through all operations add it to the table
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if(isTruncPrime) then
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table.insert(truncPrimes, currentPrime);
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end
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end
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--Get the sum of all elements in the truncPrimes vector
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sum = getSum(truncPrimes);
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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 sum of all left and right truncatable primes is " .. sum .. "\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 sum of all left and right truncatable primes is 748317
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It took 491.000 milliseconds to run this algorithm
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]]
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