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.
\ / \ /
\ / \ /
/ \ / \
/ \ / \
LThe 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): Accomplishing the goal - WRONG
by Old Original Oskar!)
When I said 48 flips, that was counting those turned by the rule as well as those turned by the nominal name of the move. That is, the solution was to make moves ABCDEFGHIJKL--yes, 12 moves. But A, B, H, L, K and E each require 3 flips (for example, A requires that pieces A, C and D be flipped), while C, D, F, G and I each require 5 flips (for example, C requires that pieces A, D, B, F and C be flipped). A flip is not a turn; it's only a subset of a turn. Any turn requires either 3 or 5 flips.
While the puzzle says you need not name the pieces that must be turned by the rule, I thought at least I'd count the number of flips, as indeed by following the rules, each individual piece cannot be flipped merely once. So I was giving more information than necessary, but not flipping more pieces than necessary.
As I said originally:
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.
Edited on September 20, 2005, 3:38 pm
Posted by Charlie
on 2005-09-20 15:33:23