Talk:Lights Out (game)

message: Let \mathcal{L} be the set of all 5×5 matrices over the two-element field {0,1}. Then any board setup in Lights Out can be represented by a matrix in \mathcal{L}, where 1 corresponds to a light that is on and 0 to one that is off.
However, if we regard the elements of \mathcal{L} as simply matrices over {0,1}, then we can add them according to the rules of matrix addition. And we know from matrix theory that \mathcal{L} is an Abelian group under matrix addition.
If m,n\in\{1,2,3,4,5\}, let P(m,n) denote the matrix whose (i,j)th entry is 1 if (i,j) = (m,n) or (m±1,n) or (m,n±1) , and 0 otherwise. Then for a board setup L, pressing the (m,n)th light corresponds to adding the matrix P(m,n) to L.
This analysis proves the two statements stated in the main article. The first statement is proved by the fact that matrix addition is commutative and associative, while the second statement stems from the fact that \forall\,L\in\mathcal{L}, L+L=\mathbf{0}, the zero matrix (board with all lights off).
Jane Fairfax 13:38, 12 May 2007 (UTC)
thread_title:  Mathematical proof by matrix analysis 

message: using the strategy, what happens when you're left with a light at D or E (single light at the last row in pos E for example)? gujamin 16:52, 24 August 2007 (UTC)
replies: text: If that were to happen, it would mean that the position you started with was insoluble. Spacepotato 21:47, 18 October 2007 (UTC)
thread_title:  strategy missing something 

message: In the illustration, the upper-left square of the board appears to change erroneously in the last step.  Am I missing something?  —Preceding unsigned comment added by BHSPitMonkey (talk • contribs) 01:26, 16 January 2009 (UTC)
replies: text: No, it was a mistake.  It's been fixed now. Spacepotato (talk) 01:44, 16 January 2009 (UTC)
thread_title:  Error in illustration? 

message: The table for 3 x 3 grid is:
 3x3
 ===
 A -> -BC
 B -> ABC
 C -> AB-
So, the strategy becomes:
# If A is the first light that is on in the bottom row, then press B in the top row and bring down the lights to the bottom row.
# If B is the first light that is on in the bottom row, then press A in the top row and bring down the lights to the bottom row.
Now, if you play with this starting configuration where 1 indicates ON and 0 indicates OFF, then this strategy doesn't lead to a solution.
 ABC
 ===
 101
 101
 011
Next, I do as the article says: "The most common method to solve this puzzle is to start by wiping all the lights except in the bottom, or last, row. This is done by pressing lights that are directly below lights that are turned on to cancel them out until only lights in the last row remain." and I get this:
 ABC
 ===
 000
 000
 110
Now, since the first light that is on in the bottom row is A, I press B on the top row and start wiping again. I get this:
 ABC
 ===
 000
 000
 001
Now, how do we proceed? The first light that is on in the bottom row is C, but we do not have a rule for C. I think something is wrong with the strategy section or it requires more clarification of the strategy.
----
It looks to me that this question is answered by the article (at least it was at some point in time, by the deleted strategy section):
"Some calculation must be done to figure out what combination of top row lights must be pressed in order to blank out the bottom row. (This is an example of the decomposition in linear algebra of a vector into basis vectors.)"
So, to blank out 001, you would press B & C on the top row, then wipe all the way down. This is because 
   B -> ABC
 + C -> AB-
 ----------
  BC -> --C (mod 2)
Of course, you could have avoided this step by noticing from the beginning that pressing C results in AB-, which is exactly the original pattern you had after the first wipe to the bottom. I believe you were misled by the simplification which applies only to the 5x5 case. I suppose an edit would make this more obvious, but I'm not about to jump into the middle of the edit war mentioned below.  Cxseven (talk) 07:49, 28 November 2010 (UTC)
replies: text: Well, I clarified the strategy section and added several references as you can see here, but User:Sottolacqua promptly deleted the entire section again. This should be considered vandalism. Cxseven (talk) 07:49, 28 November 2010 (UTC)
thread_title:  Strategy section seems incorrect - w/ Response 

message: I could take and upload a photograph of a Lights Out game if potential copyright that I do not know about does not forbid it? Zeyra (talk) 16:02, 1 March 2010 (UTC)
thread_title:  Photography of game device 

message: Surely, the most obvious question that can be asked (and, incidentally, the information I came here especially hoping to find) is: are all starting positions solvable? Perhaps this was answered in the deleted sections mentioned just above. If so, their removal is a sad loss, because surely the answer must be known and a good source could be found.
Currently, the article states "'Light chasing'" is a method similar to Gaussian elimination which always solves the puzzle", implying that a solution always exists, but further down, "If the 5x5 puzzle is unsolvable under legal game creation", implying that unsolvable positions exist. 2.25.120.117 (talk) 22:47, 7 December 2012 (UTC)
thread_title:  Existence of solution? 

message: There is a 1985 computer game by MECC for the Apple II called Multiplication Puzzles which includes a mode called "Lights Out" which is very much a variation of this game (mixed with an arithmetic quiz).
Reference:
* video of the game
* archive.org emulation
- Rainwarrior (talk) 08:13, 7 April 2018 (UTC)
thread_title:  Multiplication Puzzles for Apple II, 1985 

message: How can it be that "There exists a solution for every N×N case"?  The article gave two 5x5 cases which are insoluble.
Jmichael ll (talk) 21:29, 27 February 2022 (UTC)
thread_title:  There exists a solution for every N×N case?