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Star of David (Posted on 2005-09-20) Difficulty: 2 of 5
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.

See The Solution Submitted by pcbouhid    
Rating: 4.0000 (4 votes)

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Hints/Tips Accomplishing the goal | Comment 1 of 10

I followed a strategy of finding individual combinations of moves that accomplish a small goal, and combining them to solve the whole puzzle.  Then reductions can be made, as what is important is only the parity, odd or even, of any given node's number of flips.

The sequence ACFH (where A represents flipping A, C and D, and C represents flipping A, C, D, B and F; etc.) has the net effect of flipping C and D and leaving the rest of the nodes in their original state.  Thus, any adjacent pair of the inner hexagon can be flipped in this way, and the inner hexagon consists of 3 such pairs.

Since ACFH flips C and F, on net, it follows that ACFHB flips just B, as the B move reflips nodes C and F back to their original state, while flipping B also.  We can use 6 such moves to flip all the outer points.

Thus we have:

ACFH, ADGK, HIJK,
ACFHB, BFILH, HIJKL, LJGEK, KGDAE, EDCBA

Within this group of moves, only parity counts so each individual A or B or C, etc. can be paired with another like it and eliminated.  There are 5 each of A, H and K, and 3 each of each of the other lettered moves. Both 5 and 3 are odd, so each of the lettered moves needs be done once.

The 6 point moves (A, B, H, L, K and E) each flip 3 pieces, totalling 18 flips, and the 6 hexagon-vertex moves each flip 5 points, totalling 30 flips, so this involves a grand total of 48 flips.

I don't think this constitutes proof that this is a minimum, but it does accomplish the reversal of all the pieces.


  Posted by Charlie on 2005-09-20 13:43:24
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