function [] = Problem28()
%ProjectEuler/Octave/Problem28.m
%Matthew Ellison
% Created: 09-29-19
%Modified: 10-06-19
%Find the product of the coefficients, |a| < 1000 and |b| <= 1000, for the quadratic expression that produces the maximum number of primes for consecutive values of n, starting with n=0.
%{
	Copyright (C) 2019  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
finalLocation = false;	%A flag to indicate if the final location to be filled has been reached
currentNum = 1;	%Set the number that is going to be put at each location
%Make a 1001x1001 grid full of 0's
square = zeros(1001, 1001);

%Start the timer
startTime = clock();

%Start with the middle location and set it correctly and advance the tracker to the next number
xLocation = 501;
yLocation = 501;
square(yLocation, xLocation) = currentNum++;
%Move right the first time
++xLocation;
%Move in a circular pattern until you reach the final location
while(~finalLocation)
	%Move down until you reach a blank location on the left
	while(square(yLocation, xLocation - 1) ~= 0)
		square(yLocation, xLocation) = currentNum++;
		++yLocation;
	end
	%Move left until you reach a blank location above
	while(square(yLocation - 1, xLocation) ~= 0)
		square(yLocation, xLocation) = currentNum++;
		--xLocation;
	end
	%Move up until you reach a blank location to the right
	while(square(yLocation, xLocation + 1) ~= 0)
		square(yLocation, xLocation) = currentNum++;
		--yLocation;
	end
	%Move right until you reach a blank location below
	while(square(yLocation + 1, xLocation) ~= 0)
		square(yLocation, xLocation) = currentNum++;
		++xLocation;
		%Check if you are at the final location and break the loop if you are
		if(xLocation > size(square)(2))
			finalLocation = true;
			break;
		end
	end
end

%Get the sum of the diagonals
sumOfDiag = 0;
leftSide = 1;
rightSide = size(square)(2);
row = 1;
while(row <= size(square)(2))
	%This ensure the middle location is only counted one
	if(leftSide == rightSide)
		sumOfDiag += square(row, leftSide);
	else
		sumOfDiag += square(row, leftSide);
		sumOfDiag += square(row, rightSide);
	end
	++row;
	++leftSide;
	--rightSide;
end

%Stop the timer
endTime = clock();

%Print the results
printf("The sum of the diagonals in the given grid is %d\n", sumOfDiag);
printf("It took %f to run this algorithm\n", etime(endTime, startTime));

end

%{
Results:
The sum of the diagonals in the given grid is 669171001
It took 8.751038 to run this algorithm
%}