You could consider the problem as a state machine, with an infinite amount of nodes. From each node, there are only a finite number of transitions: fill X from supply, fill Y from supply, pour X in Y, pour X in Z, pour Y in X, pour Y in Z, pour Z in X, pour Z in Y, empty X and empty Y.

So, I'd apply Dijkstra's algorithm. Start with a queue containing a single state (X = Y = Z = 0). Now, loop and do the following:

• Pop a state from the queue.
• For each transition, apply the transition to the popped state, creating a new state. For each such state:
  • If the new state satisfies the condition, you're done.
  • Else, if we've seen this state before, try next transition.
  • Otherwise, put the state at the back of the queue.

This will garantee you find a solution with the minimal number of steps needed.

The drawback is that the algorithm will not terminate if there is no solution. Nor is it necessarely efficient.

You can easily change the algorithm to find the solution using the smallest amount of water to keep track of the amount of water used, and keeping the queue partially ordered (a heap will do) on water usage.