Deemo II Candle Puzzle

Question

I have come across an interesting candle puzzle when playing Deemo II. The puzzle has 8 candles with 2 states: on(○) and off(●).

Initial State

All candles are off:

● ● ● ● ● ● ● ●

Allowed Operation

You can switch the state of one of these candles; however, doing so will also switch the states of the two adjacent candles. Note that the 1st candle is adjacent to the 2nd and 8th candles and, similarly, the 8th candle is adjacent to the 7th and 1st candles.

For example, if I switch the 1st candle from its initial state, the scenario will become:

○ ○ ● ● ● ● ● ○

And then if I switch the 2nd candle, it will become:

● ● ○ ● ● ● ● ○

Question

If I want to light all candles from their initial state, what would be the operation order?

In general, given a random candle state 'A', is there an operation order to transfer to another state 'B'?

Answer 1

This puzzle is a version of lights-out.

There is a large amount of theory behind these games, but this is quite a simple variant. Two observations help enormously:

• Doing a move twice has no effect.
• Two moves can be done in either order, and will have the same effect.

From this it follows that the order does not matter, and no move ever needs to be done twice (because they can be reordered to be adjacent and then cancel out).

To switch all the candles, you simply:

Do all 8 possible moves in any order. Every candle will be affected three times, by its own move and by the moves on the adjacent candles.

To solve any position, it is easy to solve it until at most one candle incorrect as follows:

Solve from left to right, and for each incorrect candle you encounter switch it by doing a move on the candle to its right. Repeat until no more than one wrong candle is left.

To switch any final incorrect candle, do the following five moves:

The candle itself, and the candles at a distance of two and three to the left and to the right of it.

Answer 2

Touching a candle twice will negate its effect, so each can be touched once or twice. Let the numbers of times they were touched be $n_1,...,n_8$. Then:

$n_1+n_2+n_3=2k+1$
$n_2+n_3+n_4=2k+1$
...
$n_7+n_8+n_1=2k+1$
$n_8+n_1+n_2=2k+1$

This leads to the conclusion that $n_1=n_4=n_7=n_2=...$, so all $n$s must be equal, and they all must be 1.

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PS. Note that if the number of candles is divisible by 3, $n_1$ will wrap over to itself as soon as we're on the last line anyway. For example, 9 candles will allow $n_1=n_4=n_7$ to wrap onto itself, which also applies to $n_2=n_5=n_8$ and $n_3=n_6=n_9$.

Besides, we can prove all $k$s are equal in this particular case. If some of the sums were 3 and the others 1, they'd meet somewhere with a sum being followed by a different one, but that can't happen because the difference between two consecutive sums must amount to the difference between two particular ns, which can be either 0 or 1. Not 2. So all the sums must be equal.