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85 lines
2.7 KiB
Java
85 lines
2.7 KiB
Java
//ProjectEuler/Java/Problem14.java
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//Matthew Ellison
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// Created: 03-04-19
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//Modified: 03-28-19
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/*
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The following iterative sequence is defined for the set of positive integers:
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n → n/2 (n is even)
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n → 3n + 1 (n is odd)
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Which starting number, under one million, produces the longest chain?
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*/
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//Unless otherwise listed all non-standard includes are my own creation and available from https://bibucket.org/Mattrixwv/JavaClasses
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/*
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Copyright (C) 2019 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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import mattrixwv.Stopwatch;
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import mattrixwv.Algorithms;
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public class Problem14{
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private static final Long MAX_NUM = 1000000L; //This is the top number that you will be checking against the series
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public static void main(String[] argv){
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Stopwatch timer = new Stopwatch(); //This allows the run time of the algorithm to be testd
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Long maxLength = 0L; //This is the length of the longest chain
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Long maxNum = 0L; //This is teh starting number of the longest chain
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//Start the timer
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timer.start();
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//Loop through all numbers less than MAX_NUM and check them against the series
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for(Long currentNum = 1L;currentNum < MAX_NUM;++currentNum){
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Long currentLength = checkSeries(currentNum);
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//If the current number has a longer series than the max then the current becomes the max
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if(currentLength > maxLength){
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maxLength = currentLength;
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maxNum = currentNum;
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}
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}
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//Stop the timer
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timer.stop();
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//Print the results
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System.out.printf("The number %d produced a chain of %d steps\n", maxNum, maxLength);
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System.out.println("It took " + timer.getStr() + " to run this algorithm");
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}
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//This function follows the rules of the sequence and returns its length
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private static Long checkSeries(Long num){
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Long length = 1L; //Start at 1 becuase you need to count the starting number
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//Follow the series, adding 1 for each step you take
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while(num > 1){
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if((num % 2) == 0){
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num /= 2;
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}
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else{
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num = (3 * num) + 1;
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}
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++length;
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}
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//Return the length of the series
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return length;
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}
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}
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/* Results:
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The number 837799 produced a chain of 525 steps
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It took 1.006 seconds to run this algorithm
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*/
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