Added solution for problem 35

Problem34.m

@@ -1,7 +1,7 @@
 function [] = Problem34()
 %ProjectEuler/ProjectEulerOctave/Problem34.lua
 %Matthew Ellison
-% Createe: 06-01-21
+% Created: 06-01-21
 %Modified: 06-01-21
 %Find the sum of all numbers which are equal to the sum of the factorial of their digits
 %{
@@ -22,13 +22,13 @@ function [] = Problem34()
 %}
 
 %Setup the variables
-
-%Start the timer
-startTime = clock();
 MAX_NUM = 1499999;	%The largest num that can be the sum of its own digits
 numberFactorials = [];	%Holds the pre-computed factorials of the numbers 0-9
 totalSum = 0;	%Holds the sum of all numbers equal to the sum of their digit's factorials
 
+%Start the timer
+startTime = clock();
+
 %Pre-compute the possible factorials from 0! to 9!
 for cnt = 1 : 9
 	numberFactorials(cnt) = factorial(cnt);

Problem35.m

@@ -0,0 +1,89 @@
+function [] = Problem35()
+%ProjectEuler/ProjectEulerOctave/Problem35.lua
+%Matthew Ellison
+% Created: 06-05-21
+%Modified: 06-05-21
+%Find the sum of all numbers which are equal to the sum of the factorial of their digits
+%{
+	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 <https://www.gnu.org/licenses/>.
+%}
+
+%Setup the variables
+MAX_NUM = 999999;
+%MAX_NUM = 100;
+circularPrimes = [];
+
+%Start the timer
+startTime = clock();
+
+%Get all primes under 1,000,000
+primeList = primes(MAX_NUM);
+%Go through all primes, get all their rotations, and check if those numbers are also prime
+for cnt = 1 : size(primeList)(2)
+	prime = primeList(cnt)
+	allRotationsPrime = true;
+	%Get all of the rotations of the pirme and see if they are also prime
+	rotations = getRotations(num2str(prime));
+	for rotCnt = 1 : size(rotations)(2)
+		rotation = rotations(rotCnt);
+		p = rotation;
+		if(~isprime(p))
+			allRotationsPrime = false;
+			break;
+		end
+	end
+	%If all rotations are prime add it to the list of circular primes
+	if(allRotationsPrime)
+		circularPrimes(end + 1) = prime;
+	end
+end
+
+%Stop the timer
+endTime = clock();
+
+%Print the results
+printf("The number of all circular prime numbers under %d is %d\n", MAX_NUM, size(circularPrimes)(2));
+printf("It took %f seconds to run this algorithm\n", etime(endTime, startTime))
+
+end
+
+%Returns a list of all rotations of a string passed to it
+function [rotations] = getRotations(str)
+	rotations = [];
+	rotations(end + 1) = str2num(str);
+	for cnt = 2 : size(str)(2)
+		temp = [substr(str, 2, size(str)(2) - 1) str(1)(1)];
+		num = str2num(temp);
+		rotations(end + 1) = num;
+		str = temp;
+	end
+end
+
+function [found] = isFound(array, key)
+	found = false;
+	for location = 1 : size(array)(2)
+		if(key == array(location))
+			found = true;
+			return;
+		end
+	end
+end
+
+%{
+Results:
+The number of all circular prime numbers under 999999 is 55
+It took 712.060379 seconds to run this algorithm
+%}