/**
 * Matt Weiss
 * CSE 2231
 * 3:00 PM
 */
public class Homework4 {

	/*
	 	Based on Integer artihmetic in the Java language, there are automatically no decimals when dealing 
	 	with the division of two integers. This being the case, integers are limited by INTEGER_MAX_VALUE,
	 	a value around 2.147 billion. The average of any number of these max values will always result back
	 	to INTEGER_MAX_VALUE or lower, as can be seen with basic mathematics. The same applies for any average
	 	of an arbitrary number of INTEGER_MIN_VALUE's.

	 */

	/**
	 * Returns the integer average of two given {@code int}s.
	 *
	 * @param j
	 *            the first of two integers to average
	 * @param k
	 *            the second of two integers to average
	 * @return the integer average of j and k
	 * @ensures <pre>
	 * {@code average = (j+k)/2}
	 * </pre>
	 */
	public static int average(int j, int k) {
		int signTwo = (int) Math.signum(k);
		int value = (j+k) / 2;
		int reductions = 1;
		
		//If bothOdd is true, then we need to add or subtract one at the end, 
		//as there will be two cases where 0.5 was chopped off in the end due to integer arithmetic.
		boolean bothOdd = j % 2 != 0 && k % 2 != 0;
		
		//If 2nd val pos, need to check for positive overflow (can't add into a neg overflow)
		if (signTwo > 0) {
			while ((value = (j + k) / 2) <= 0) {
				j /= 2;
				k /= 2;
				reductions++;
			}
			if (bothOdd)
				value++;
		}
		//If 2nd val neg, need to check for negative overflow (can't subtract into pos overflow)
		else if (signTwo < 0) {
			while ((value = (j + k) / 2) >= 0) {
				j /= 2;
				k /= 2;
				reductions++;
			}
			if (bothOdd)
				value--;
		}

		/*
		 * Mathematically speaking, the average of any two values, i and j, and be represented either:
		 * (i + j) / 2
		 * or...
		 * (i / 2) + (j / 2)
		 * by the Distributive Property.
		 * 
		 * This being the case, we can simply reduce the values 
		 */

		System.out.println("Adjusting the final value of " + value + " by a factor of " + reductions);
		return value * reductions;
	}



}