Two players P1 and P2 play the following game n cards are put on the table face down in a circle Originally everv card overlaps with two neighboring cards as shown on the picture belowLet us continue

Let's call a group of three neighboring cards a "triad". We will first show that if there are only triads left on the table, then the player who moves last wins.

Suppose that there are only triads left on the table. Since each triad overlaps with its neighbors, removing a triad will result in two separate groups of cards. Therefore, each move by a player will split the remaining cards into two groups, one with an odd number of triads and one with an even number. Since the player who moves last will make the final split, they can ensure that they are the one with an odd number of triads, guaranteeing their win.

Now, let's consider the case where there are non-triad groups of cards on the table.

If there is only one non-triad group of cards, then the optimal strategy is to move second and mirror the opponent's moves. This way, you will always be the one to take the last card in the non-triad group, leaving only triads on the table. By the above argument, you will win.

If there are multiple non-triad groups of cards, then the optimal strategy is to move first and split the table into two groups, one with an odd number of non-triad groups and one with an even number. You can do this by either removing a non-triad group or by removing enough triads to split the non-triad groups into two separate groups. Then, you mirror your opponent's moves, ensuring that you are the one with an odd number of non-triad groups when it is their turn to move last. This way, you can force them to take the last card in a non-triad group, leaving only triads on the table. By the above argument, you will win