12 circular pieces (A-L) white on one side and blue on the other, are arranged in a "Star of David" as shown below, with the white face up.

                          A
                         / \
                        /   \
                 B-----C-----D-----E
                  \   /       \   /
                   \ /         \ /
                    F           G
                   / \         / \
                  /   \       /   \
                 H-----I-----J-----K
                        \   /
                         \ /
                          L

The goal is to turn all the 12 pieces so that they all show its blue side. At any time, you can turn any piece you want, but doing this, all the pieces that are connected to it, must be turned too. That is, if you choose to turn the "A"-piece (so that it becomes blue), the "C"-piece and "D"-piece must be turned too, becoming blue. If next you choose to turn the "E"-piece (so it becomes blue) the "G" and "D"-pieces must be turned too (the "G"-piece becomes blue, but the "D"-piece now becomes white again).

Can you devise a sequence with minimum turns to achieve the goal? Just name the pieces that you should turn; you donīt need to name the pieces that must be turned by the rule.

Iīm not forbidding using a computer (in fact, I canīt) but those who do, please give the others some time before posting a solution obtained by this way.

(In reply to re: Accomplishing the goal by Charlie)

1) Think of an outside pass ABHLKE.  This flips every outside circle once and every inside circle twice.  This makes the outside blue and the inside white.  == 6 flips.

2) An inside pass CFIDGD.  This flips every outside circle twice and every inside circle three times.  Now the Outside stay blue and the inside turn to blue.  === 6 more flips.

12 flips total.

 

Posted by Patrick on 2005-09-20 17:23:49