To determine which player has a winning strategy, we can analyze the game using game theory, specifically focusing on the concepts of winning and losing positions.

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Key Concepts

1. Winning Position: A position where a player can guarantee a win regardless of the opponent's moves.
2. Losing Position: A position where any move leaves the opponent in a winning position.

The game starts with 117 squares. A player can:

• Remove 1 square.
• Remove 2 consecutive squares.

This creates smaller sub-games for the opponent.

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Step-by-Step Analysis

Base Cases:

1. $n = 1$: Only 1 square exists.
   • The first player removes it and wins. This is a winning position.
2. $n = 2$: Two squares exist.
   • The first player removes both (taking 2 consecutive squares) and wins. This is a winning position.
3. $n = 3$: Three squares exist.
   • The first player can:
     • Remove 1 square, leaving 2 squares (a winning position for the opponent).
     • Remove 2 squares, leaving 1 square (a winning position for the opponent).
   • Thus, $n = 3$ is a losing position.

Generalizing the Pattern:

From the base cases, observe:

• If the opponent is left with $n = 3$, they are in a losing position.
• Players aim to leave their opponent in a losing position.

Recursive Strategy:

• For $n > 3$, consider the current player’s moves:
  • Remove 1 square: Leaves $n - 1$.
  • Remove 2 squares: Leaves $n - 2$.

If either $n - 1$ or $n - 2$ is a losing position, the current position $n$ is a winning position.

Winning and Losing Positions:

Using the above logic, compute the sequence of positions:

• $n = 1$: Winning position.
• $n = 2$: Winning position.
• $n = 3$: Losing position.
• $n = 4$: Winning position (leave $n = 3$ for the opponent).
• $n = 5$: Winning position (leave $n = 3$ for the opponent).
• $n = 6$: Losing position (any move leaves $n = 4$ or $n = 5$, both winning for the opponent).
• Continuing this pattern, $n = 3, 6, 9, 12, \dots$ (multiples of 3) are losing positions.

Starting Position ($n = 117$):

• $117$ is a multiple of 3 ($117 = 3 \times 39$).
• Thus, $n = 117$ is a losing position.

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Conclusion

• Player 1 starts in a losing position ($n = 117$).
• Player 2 has a winning strategy by always leaving a multiple of 3 for Player 1 after their turn.

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Would you like a detailed explanation of the optimal moves for Player 2 or further questions on game theory concepts? Here are 5 related questions:

1. How would the strategy change if the board had a different number of squares?
2. What if players could remove 1, 2, or 3 consecutive squares?
3. How can this concept be extended to boards with multiple rows?
4. How are winning and losing positions related to the concept of "Nimbers" in combinatorial game theory?
5. What algorithms can efficiently compute winning positions for larger games?

Tip: Understanding patterns in simple cases is key to solving combinatorial games.