Lights out linear algebra
This makes things a little more complicated.
Lights out instructions
Some general definitions. Thus you must have an odd number of buttons in regular solution, all unlit, or an even number, all lit. In terms of the game itself, this means that if pressing one light changes a second light, then pressing that second light will also change the first. Proof of Theorem 4: Click to hide proof. Lemma 3a: Suppose you are pressing buttons in order to change position A to position B in an ordinary symmetric reflexive game. This is a proper integer only if the sum is even, which can only mean that not all mi can be odd. For example, going from lights all on to all off in the case, there are four possible solutions to the all-lights pattern, illustrated above. The above strategy is not a rigorous proof of the converse of theorem 5. Suppose you have a sequence of button presses that changes position A to position B. The change in state of these neighbours will therefore not affect conditions 1 and 2, nor the parity of the number of neighbouring pairs. Note that [EES] contains a supposed counterexample, but that is false. I prefer this proof, despite the fact that it is much longer. It is possible to generalise the theorem a bit. Find out the buttons in the regular solution to the position, and then solve the toggle game as far as possible, pressing only the buttons in that regular solution.
So far we have used the standard Lights Out game as a basis. To prove the theorem, I will need to prove a simpler result first.
Lights out linear algebra
There are four neighbouring pairs the centre is adjacent to each of the four others , and there are four unlit buttons. Eventually you reach a position in which all remaining buttons of the solution are unlit. Conlon, M. Proof of Theorem 6: Click to hide proof. In a symmetric game there is an easy way to test if a pattern of lights can be solved. In that case let k be any light that is affected by m but not by n, and replace the moves mn in the sequence by the moves kmnk. So we may assume that each occurs at most once. The button b between a and n and the button c between a and f are neighbours. Therefore there can be no unsolvable reflexive symmetric Lights Out game. Eriksson, J. I found this out when I tried to prove it and failed. We will generalise the theory a little, and must be careful not to implicitly assume that all types of Lights Out game have the same properties. Pressing both b and c would then leave all their neighbours and themselves unaffected. Whitman College Department of Mathematics.
It also applies to Merlin's Magic Square, or in fact to any reflexive game on a rectangular grid where each button affects only some or all of the four neighbouring lights, due to the fact that the grid can be given a checkerboard colouring. The move sequence nbn aca b a fcf is a lit-only equivalent to pressing a.
The buttons of the solution consists of one or more regions of connected buttons, which satisfy the following two conditions: Each region has at least one lit button.
Lights out 90s game
Finish with a2a3a2, a3a4a3, Corollary: Every quiet pattern in a symmetric reflexive game has an even number of buttons. One of the following cases must then occur: There is a button b that is adjacent to a, n, and f. Now we press a lit button that is not yet one of the solution buttons. Eriksson, K. Note however that all of these have a surplus of four. Theorem 2: A standard Lights Out game on a rectangular grid which starts completely lit can be switched off by pressing only lit buttons. We will generalise the theory a little, and must be careful not to implicitly assume that all types of Lights Out game have the same properties.
Now we press a lit button that is not yet one of the solution buttons. Then q is clearly also a permutation, as it is simply the transposition x y composed with p, and furthermore q will be of opposite parity to p.
Lights out algorithm
It turns out that the bulk of the proof above can be bypassed. Below is a second, much shorter proof found by Robert Torrence. Additionally, all four horizontally and vertically adjacent lights - forming a cross - get inverted as well when a single light gets pressed. There are now two cases to consider - n is a neighbour of only one of b or c, or n is a neighbour of both b and c. Removing solutions that are equivalent by rotation or reflection gives the distinct solutions illustrated above, of which there are 1, 1, 1, 5, 1, 1, 1, 1, 43, 1, 10, 1, 1, 5, 1, Another property of the standard Lights Out matrix is that it is reflexive i. The shortest way to prove it is probably as follows: Proof 1 of theorem 1: Click to hide proof. For example, in the pattern shown above, it is impossible to turn off all the lights. The determinant is therefore the same apart from the changed sign. This is a proper integer only if the sum is even, which can only mean that not all mi can be odd. If n were even, this would mean that every light has changed state which would make our game solvable.
Before we do the second case, consider what would happen if b and c had exactly the same neighbours. Do this for all n! As shown by Sutnergoing from all lights on to all lights off is always possible for any size square lattice.
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