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.

As has been pointed out already, the pieces can be flipped in any sequence, and flipping something twice has no net effect, so a minimum solution has every piece flipped either 0 or 1 times.

From the point of view of B: B and C and F must be flipped an odd number of times, in total.

From the point of view of F: B, C, F, H and I must be flipped an odd number of times.

Therefore, H and I must be flipped an even number of times.

By a similar argument, H and F must be flipped an even number of times. 

Therefore, I and F must be flipped the same number of times.

Therefore, all interior points must be flipped the same number of times (because the same reasoning applies to all pairs of adjacent interior points).

Therefore, all exterior points must be flipped an odd number of times. (In order to have an odd number of flips from the point of view of the outer point).

Therefore, all interior points must be flipped an odd number of times also, (In order to have an odd number of flips from the point of view of the interior point).

Therefore, flipping every point once is the minimum and (sequence aside) only solution.

Also, this same logic applies to a 5 pointed star, with the same solution.

Posted by Steve Herman on 2005-09-21 18:14:59