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(2): Parity rocks! (no computer) by Steve Herman)

Steve, Iīm trying to remember, and Iīll explain.

This kind of problem, as far as I know (not the idea, their drawings) are original (and I got others), and when I started with a six-pointed star and found that every piece must be turned once, and that the same applied to 7, 8, etc...I challenged a friend with a 5-pointed star, believing that he never would found this "easy" way. But, to my surprise, he achieved the solution in few turns, and start laughing at me, saying: "ridiculous...donīt you have anything more difficult to me?". But, now, Iīm not seeing how could be done (I tried only a few minutes).

Maybe, when solving (the 5-), something was made wrong and I missed that.

With all the analysis, thatīs the explanation that now holds for me, until somebody show me that the 5- could be turned in fewer than 10 turns.

Tk you for your comments.               

Posted by pcbouhid on 2005-09-22 12:23:26