//ProjectEuler/Java/Problem14.java
//Matthew Ellison
// Created: 03-04-19
//Modified: 03-28-19
/*
The following iterative sequence is defined for the set of positive integers:
n → n/2 (n is even)
n → 3n + 1 (n is odd)
Which starting number, under one million, produces the longest chain?
*/
//Unless otherwise listed all non-standard includes are my own creation and available from https://bibucket.org/Mattrixwv/JavaClasses
/*
	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/>.
*/


import mattrixwv.Stopwatch;
import mattrixwv.Algorithms;


public class Problem14{
	private static final Long MAX_NUM = 1000000L;	//This is the top number that you will be checking against the series
	public static void main(String[] argv){
		Stopwatch timer = new Stopwatch();	//This allows the run time of the algorithm to be testd
		Long maxLength = 0L;	//This is the length of the longest chain
		Long maxNum = 0L;	//This is teh starting number of the longest chain

		//Start the timer
		timer.start();

		//Loop through all numbers less than MAX_NUM and check them against the series
		for(Long currentNum = 1L;currentNum < MAX_NUM;++currentNum){
			Long currentLength = checkSeries(currentNum);
			//If the current number has a longer series than the max then the current becomes the max
			if(currentLength > maxLength){
				maxLength = currentLength;
				maxNum = currentNum;
			}
		}

		//Stop the timer
		timer.stop();

		//Print the results
		System.out.printf("The number %d produced a chain of %d steps\n", maxNum, maxLength);
		System.out.println("It took " + timer.getStr() + " to run this algorithm");
	}
	//This function follows the rules of the sequence and returns its length
	private static Long checkSeries(Long num){
		Long length = 1L;	//Start at 1 becuase you need to count the starting number

		//Follow the series, adding 1 for each step you take
		while(num > 1){
			if((num % 2) == 0){
				num /= 2;
			}
			else{
				num = (3 * num) + 1;
			}
			++length;
		}

		//Return the length of the series
		return length;
	}
}

/* Results:
The number 837799 produced a chain of 525 steps
It took 1.006 seconds to run this algorithm
*/