OctaveFunctions/ProjectEuler/Problem4.m

%ProjectEuler/Problem4.m
%This is a script to answer Problem 4 for Project Euler
%Find the largest palindrome made from the product of two 3-digit numbers

%Make your variables
answer = 0;	%For the product of the two numbers
numbers = [100:999];	%Create an array with a list of all 3 digit numbers
palindromes = [];	%Holds all the numbers that are palindromes
%Create 2 counters for an inner loop and an outer loop
%This allows you to multiply 2 numbers from the same array
outerCounter = 1;
innerCounter = 1;

startTime = clock();	%This is for timing purposes
while(outerCounter < size(numbers)(2))
	innerCounter = outerCounter;	%Once you have multiplied 2 numbers there is no need to multiply them again, so skip what has already been done
	while(innerCounter < size(numbers)(2))
		%Multiply the two numbers
		answer = numbers(outerCounter) * numbers(innerCounter);

		%See if the number is a palindromes
		%%WARNING - Ocatave does not have a Reverse function. I had to create one that reversed strings
		if(num2str(answer) == Reverse(num2str(answer)))
			%Add it to the palindromes list
			palindromes(end + 1) = answer;
		end
		++innerCounter;	%Increment
	end
	++outerCounter;	%Increment
end
endTime = clock();	%This is for timing purposes
timeTaken = etime(endTime - startTime)	%This is for timing purposes

max(palindromes)

%Cleanup your variables
clear outerCounter;
clear innerCounter;
clear answer;
clear numbers;
clear palindromes;
clear startTime;
clear endTime;
clear timeTaken;


%This way is slow. I would like to find a faster way
%{
The palindrome can be written as: abccba Which then simpifies to: 100000a + 10000b + 1000c + 100c + 10b + a And then: 100001a + 10010b + 1100c Factoring out 11, you get: 11(9091a + 910b + 100c) So the palindrome must be divisible by 11. Seeing as 11 is prime, at least one of the numbers must be divisible by 11
%}