/*
 * The MIT License (MIT)
 * 
 * Copyright (c) 2015 Camil Staps <info@camilstaps.nl>
 * 
 * Permission is hereby granted, free of charge, to any person obtaining a copy
 * of this software and associated documentation files (the "Software"), to deal
 * in the Software without restriction, including without limitation the rights
 * to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
 * copies of the Software, and to permit persons to whom the Software is
 * furnished to do so, subject to the following conditions:
 * 
 * The above copyright notice and this permission notice shall be included in all
 * copies or substantial portions of the Software.
 * 
 * THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
 * IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
 * FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
 * AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
 * LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
 * OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
 * SOFTWARE.
 */
package com.camilstaps.fibonacci;

/**
 * Calculate a fibonacci number
 * @author Camil Staps s4498062
 */
public class ParFib implements Runnable {
    
    private final int n;
    private int result = 0;
    
    /**
     * Create a new instance
     * @param n the index of the fibonacci number to calculate
     */
    public ParFib(int n) {
        this.n = n;
    }

    @Override
    public void run() {
        calculate(n);
    }
    
    public int getResult() {
        return result;
    }
    
    /**
     * Calculate the nth number of the Fibonacci sequence.
     * 
     * This implementation uses two threads if n > 2: one to calculate the 
     * (n-1)th fibonacci number, and one to calculate the (n-2)th fibonacci
     * number. That is a really silly approach, since many threads will be
     * created and many numbers will be calculated many times. 
     * As an example, to calculate for n=10, we will calculate the 8th number
     * twice: once for n=10 (10-2=8) and once for n=9 (9-1=8). The 7th number
     * will be calculated thrice, and this amount increases for lower numbers.
     * But hey, the assignment asked for a solution with threads.
     * 
     * @param n 
     */
    private void calculate(int n) {
        if (n < 3) {
            result = 1;
            return;
        }
        
        ParFib pf1 = new ParFib(n - 1);
        ParFib pf2 = new ParFib(n - 2);
        Thread t1 = new Thread(pf1);
        Thread t2 = new Thread(pf2);
        t1.start();
        t2.start();
        
        try {
            t1.join();
            t2.join();
        } catch (InterruptedException ex) {
            System.err.println(ex.toString());
        }
        
        if (pf1.getResult() != 0 && pf2.getResult() != 0) {
            result = pf1.getResult() + pf2.getResult();
        }
    }
    
}