Four glasses are placed on the corners of a square Lazy Susan (turntable). Some of the glasses are upright (up) and some upside-down (down).
A blindfolded person is seated next to the Lazy Susan and is required to re-arrange the glasses so that they are all up or all down, either arrangement being acceptable.
The glasses may be re-arranged in turns subject to the following rules.
- Any two glasses may be inspected in one turn and after feeling their orientation the person may reverse the orientation of either, neither or both glasses.
- After each turn the Lazy Susan is rotated through a random angle.
- At any point of time if all four glasses are of the same orientation, a bell will ring.
Can you devise an algorithm which allows the blindfolded person to ensure that all glasses have the same orientation (either up or down) in a finite number of turns? The algorithm must not depend on luck.
Riddle Answer