Actual responses to actual questions from students of Math 344, Introduction to Combinatorics (Spring 2004).
Can you check out my solution (see the attached file)? In this representation, I have noted at each vertex, the state of the two hour-glasses and the total time passed as a triple (time left in the 4 min, time left in the 7 min, total time elapsed).
In this 3-ary tree, travel down the left branch indicates that the 4-min hour glass gets flipped, travel down the right branch indicates that the 7-min hour glass gets flipped, and travel down the centre branch indicates that both hour-glasses have been flipped.
While this could have been done with a simple pair, (time remaining, time elapsed) it was easier to determine which hour-glass had time remaining by using a triple instead.
By my reading of the question, there are two possible ways of measuring out a 9-minute period of time, but only one of those two ways results in both hour-glasses being empty at the 9 minute mark. (start both; flip the 4 when it runs out - 4 minutes, flip the 7 when it runs out - 3 more minutes, flip the 7 again when the 4 runs out - one more minute.)
Thanks to John Calvin!
