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"Double Lights Out" by bpotetz
|Comments (turn spoilers off)|
|20282||Double Lights Out||noname (559)||Thu 26 Jul 2018 11:25||SPOILER|
|Added speedrun: 322 moves (old: 324).
|14005||Double Lights Out||radiant (1469)||Sun 28 Mar 2010 22:29||SPOILER|
|Added speedrun: 324 moves (old: 326).|
|5599||Double Lights Out||Tom 7 (1)||Wed 15 Mar 2006 10:11||SPOILER|
|I'm impressed that you guys are daring enough to compose such long messages in the comment box, which has no cut/paste and can easily lose your comment for good!|
|5598||Double Lights Out||bpotetz (144)||Wed 15 Mar 2006 04:24||SPOILER|
|Thanks for posting your perl script, bje. I'm happy that you were interested enough in the level to write a brute force solver. To encourage more analysis & thinking, I will post some basic deductions about the game that I used in selecting the start-state & other parameters.
ANOTHER EXTREME SPOILER:
First, John is right - pressing a button & its 4 nearest neighbors will toggle the central button. That is enough to completely solve the simpler level (1080). Similar theorems exist for all lights-out puzzles played on a regular closed surface (torus, klein bottle, of any size). On the other hand, if the game has hard boundaries, you can solve it easily using a technique that I recently discovered already has a name: "light chasing" (google for more info). Maybe if the game was played on a less regular graph, harder techniques would be needed.
Anyway, this means that you can determine the set of buttons that must be pressed by just summing up all the red panels & their 4 nearest neighbors in mod 2 arithmetic. For this level, that set should contain 9 buttons (I won't list them here - that is *too* spoiling). Unfortunately, because the transports also toggle on & off, there is no way to press those buttons & only those buttons, which we can prove.
Note that pressing a button toggles the same set of red/green lights as it does the transports. I considered using a different pattern for the transports, but I did not like how most other simple patterns lie on a checkerboard - that makes it too easy to construct parity arguments to solve the puzzle.
Anyway, using the same pattern for lights & transports means that at any given time, the state of the lights & the state of the transports only differs by a constant set. In this case, that is just 4 squares: the three that are already out (green) in the start state, plus the one that has a transport in the start state. Since the game is finished when all the lights are green, the end game must have all but those 4 transports "open". Note that in your final move, you must walk into an open transporter, which will turn that transporter off. So the final move must be made at one of those 4 squares that has no transporter in the final state. Only one of those 4 squares is in the set of buttons that MUST get pressed. And that is the one that MUST be pressed in your first move (it is like a joke at your expense). So at least one of those 4 differing squares must be pressed twice.
Knowing all this makes solving the puzzle by hand much easier.
|5596||Double Lights Out||bje (262)||Wed 15 Mar 2006 01:06||SPOILER|
For those interested, the perl script I used to do all the thinking for me can be downloaded from:
I numbered the teleporters from left to right, top to bottom, 1 to 16. The final output is the state (0 = all green), a bitmask for which teleporters are visible, and then a list of teleporters to walk through in order.
It's not a fastest solution finder, though the final output is in the set of "fewest teleports" solutions.
|5533||Double Lights Out||John Lewis (411)||Sat 11 Mar 2006 18:10||SPOILER|
|Finally solved this. Even though it's tough to do it on this level, it's helpful to remember that to turn a single red light to green, "push" it and all other reds around it.|
|5510||Double Lights Out||bpotetz (144)||Fri 10 Mar 2006 17:13|
|'Double Lights Out' uploaded by bpotetz:
This version is supposed to be harder, but it's still not too difficult.