Author: James D. Skrentny, skrentny@cs.wisc.edu
        copyright 2009-2012 all rights reserved

Our previous Ball2Right solution would fail if there were more cups and/or
more balls. Let's make our code more ROBUST so that it is able to handle
the problem of a single rainbow ball possibly being in the way.

Ball Move Language:
   *The form of our language is
   <object name>.<action>(<arguments>)

   <ball name>.on(<cup number>)  e.g., greenBall.on(2)
   <ball name>.step(<direction>) e.g., greenBall.step("right")
   <ball name>.jump(<direction>) e.g., greenBall.jump("right")


Given: 3 cups in a row numbered 1, 2, 3
       1 green ball   (named greenBall)
       1 rainbow ball (named rainbowBall)
       the balls can start on any one of the cups

Goal:  Move green ball to cup on the right (cup #3).

Note:  Again, if green ball is already on right cup then nothing need be done.


Solution:
---------
if (greenBall.on(2)) {
	if (rainbowBall.on(3)) {
		rainbowBall.jump("left");
	}
	greenBall.step("right");
}
else if (greenBall.on(1)) {
	if (rainbowBall.on(3)) {
		rainbowBall.step("left");
	}
	if (rainbowBall.on(2)) {	
		greenBall.jump("right");
	}
}

Trace the code fragment above to see if it WORKS.

Note: Design a solution that addresses the underlying logic of the problem:
      1. Where's the green ball? (first level decision)
      2. Is the rainbow ball in the way? (second level decision)



Now, let's make the problem more GENERAL by allowing 0 or 1 rainbow balls.

Solution:
---------
Note: added the else part after the last if below.

if (greenBall.on(2)) {
	if (rainbowBall.on(3)) {
		rainbowBall.jump("left");
	}
	greenBall.step("right");
}
else if (greenBall.on(1)) {
	if (rainbowBall.on(3)) {
		rainbowBall.step("left");
	}
	if (rainbowBall.on(2)) {	
		greenBall.jump("right");
	}
	else {
		greenBall.step("right");
		greenBall.step("right");
	}
}

Does this WORK?
TRACE it for the different initial states.