48

There is an easy way to solve it in the minimum number of moves. The reason this works is:


16

I don't know why this works, but I tried my old strategy from other similar puzzles and it has worked in several cases so far: Then, One final note - since I don't know why it works, I also can't prove that it always works, unfortunately. But in my many years of solving puzzles in this genre, that strategy always seems to eventually land on a solution. ...


16

It is Reasoning:


15

There is a simple solution. Because


13

Edit: Now that I have a bit more time, and my answer is already the accepted one, I'll add a short summary of the theory of this type of game which most of the other answers have touched on. This is a variant of the Lights Out game. There are 9 lights, which can be on (O) or off (X). There are 6 moves available - flipping any column or any row. Applying any ...


13

The winner is The winning strategy To see this,


12

Suppose there are $n\geq 3$ chips. It's easy to see that $n$ moves is possible, making each move exactly once (this flips each chip three times). Since making the same move twice cancels out, any optimal solution also involves making each move at most once. Since each chip needs to be flipped an odd number of times, in any optimal solution each chip is ...


11

Or, create a new blank board, and click on it all the positions of the 'on' switches from the original board. Take this new pattern back to the original board, and click every 'off' position from the new board onto the original board. All lights are on!


11



8

The first variant of Knights Out is solvable. In fact, given any network of lights where toggling a light also toggles its neighbors, it is possible to turn on all the lights! This beautiful fact requires some pretty fancy linear algebra to prove, which I have adapted from page 238 of Algebraic Combinatorics by Richard Stanley. From now on, all addition and ...


8

A less technical solution: We can (try to) make a symmetric solution that makes all numbers equal (where the 6 variables signify how often a cell is chosen): abcba bdedb cefec bdedb abcba For the total we get: T = a+2b = a+b+c+d = 2b+c+e = 2b+d+2e = 2d+c+e+f = 4e+f From which we can extract equalities: (1+3) a = c+e (1+2) b = c+d (3+4) c = d+e (...


7

Every Lights Out game where the lights have two states (on/off) which is reflexive (the pushed light also toggles) and which is symmetric (if light A is a neighbour of B, then B is a neighbour of A) will allow you to go from all-lights on to all-lights off (and vice versa). There are several proofs. You can find two on my Lights Out page. Another ...


7

Yes. Explicit Grid


7

Make these presses: To get these values: You can solve the problem via integer linear programming as follows. Let nonnegative integer decision variable $x_{i,j}$ be the number of times that cell $(i,j)$ is pressed. Let binary decision variable $y_{i,j,v}$ indicate whether cell $(i,j)$ contains value $v$. Let $N_{i,j}$ be the neighborhood of cell $(i,j)$,...


7

You need First, an important thing to know: If S≤L: If S≥L: Time for some math! The setup: The Important Thing: So, what does this tell us?


7

The answer is Reasoning for even $n$: Reasoning for odd $n$:


5

Adding some findings to the answer by Display name. As already noted, Call the coin positions 1 2 3 4 5 6 7 8 9 and label the guard's moves with digits for the flipped coins, e.g. {147} and {2356}. The guard has sequences of moves which can: and sequences which can: By combining these sequences, the guard can solve any initial position where: Though it ...


5

You are basically trying to solve a linear algebra question, but with XOR instead of addition. The good news is that this can be done in the usual way. See this math.SE article. So, let me back up. Draw up a matrix with rows and columns equal to the number of available buttons in the puzzle. In your example, it's 25-3 = 22. So a $22\times 22$ matrix. Each ...


5

This game is a version of Lights Out. Like all these games, they have the following properties: The order of moves does not matter. If you look at the effect of a move sequence on a single square, all that matters to that square is how often it is toggled. It does not matter to that square what order the moves are performed, as all the moves that affect it ...


5

Here is a more straightforward proof than the one given by ffao, plus information about checkerboard patterns This game is a version of Lights Out. Like all these games, they have the following properties: The order of moves does not matter. If you look at the effect of a move sequence on a single square, all that matters to that square is how often it is ...


4

I believe it's where the answer is expressed like this: A B C D E F G H I J K L M N O P Q R S T Background (Step 1): Step 2:


4

I'll show how solve this puzzle in general. First some things to notice To show whether a specific puzzle is solvable An example the Actual puzzle


4

Neater solution, as requested by OP: Most of the credit belongs to @WhatsUp for finding the matrix $A$.


4

Answer to the first question Make these presses To get these values Not sure if this is optimal.


4

I think the least amount of presses is Press the following cells $x$ amount of times yielding


4

PARTIAL ANSWER: NOTE: Credit belongs to WhatsUp for finding the balanced matrix for n = 5 in the original Board with all 2020s problem.


4

Lower bound: Reasoning:


3

Very often it is still possible to apply light chasing techniques to simplify the problem. A B C D E F G H I J K L M N O P Q R S T U V In this case you could start at the corner A, and work your way diagonally down to the opposite corner V. On each diagonal row you do moves that ...


3

Found The Answer. Its called a lights out puzzle


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