There are $2^n$ lamps on a roulette wheel. Each time player A decides
on a list of lamps to flip (so on becomes off and off becomes on).
Player B sees player A's list, and rotates the roulette wheel to a
different location. Player A still flip the lights as if they were on
the original location. For example, say there are four lights
$1,2,3,4$. Player A decides to flip the switch of lamp $1$. Player B
rotates the wheel to $3,4,1,2$. Then after this round, lamp $3$ is
flipped.At the beginning, some lamps are on and some are off. Prove that regardless of the initial state, for no more than
$2^{2n}$ rounds, player A can always turn on all the lights.
There is a hint to this problem. Prove by mathematical induction. Suppose there is a way for $2^{k}$ lamps. Consider the case when $2^{k+1}$. Pair up the lamps that are opposite of each other…
I am thinking after we pair up the lamps that are opposite of each other, We define a new state for this $2^{k}$ pairs. If the pair has different states, we treat them as an "off" lamp. Otherwise we treat them as an "on" light. But there is no way to turn on this "off" light in this new, paired up roulette. My idea could be completely off.
Certainly, a solution without using this hint is welcome too.
Best Answer
It looks to me that you've made significant progress. Before reading further, I encourage you to push through from what you have, with the knowledge that your path does indeed lead to a solution. In fact, my solution is just what you have and what's in the comments.
If you want to read further, scan through the hints as a suggestion, before reading in more detail.
I also provided some motivational steps that hint we're proceeding in the right direction. These obviously could be skipped when writing up an actual solution.
Hint 1: Consider the last step: What possible state could A be in which guarantees no matter what B choses, A will win?
Hint 2: Consider the second last step: How can A get to the last step.
So, this line of reasoning indicates that we likely really want to get to "diametrically opposite lamps are both on or both off".
Hint 3: Setting up the induction.
Hint 4: Run the induction, see what we get.
Hint 5: Setup the induction again.
Hint 6: Run the induction again.
Hence, we are done.
Bonus: Note that in the induction, if $2^k$ lamps takes $f(k)$ steps, then $2^{k+1}$ lamps takes $2f(k)$ steps. Since we can easily verify that $2^0$ lamps takes (at most) $1$ step, we can conclude that $f(k) = 2^k$ steps, which is much better than the $2^{2k}$ steps that they gave us.