Title: CAT 2024 Slot 1 Question: Indian Game Strategy Analysis
Content: English Solution for a Sample Game Theory Problem
Problem Statement (Hypothetical Example):
In a game called "Stone Piles," two players take turns removing stones from three piles. A player can remove any number of stones (≥1) from a single pile. The player who takes the last stone wins. If the initial piles have (5, 3, 2) stones, does Player 1 (the first player) have a winning strategy? Explain using game theory principles.
Solution:
This problem is a classic example of the Nim Game, a combinatorial game studied in game theory. The key to solving such problems lies in calculating the Nim-sum (binary XOR) of the pile sizes.
Step 1: Calculate the Nim-sum
Convert each pile size to binary and compute the XOR:
5 = 101
3 = 011
2 = 010
Nim-sum = 101 XOR 011 XOR 010
101 XOR 011 = 110
110 XOR 010 = 100 (Decimal: 4)
Step 2: Determine the Winning Condition
If Nim-sum ≠ 0: Player 1 can force a win.
If Nim-sum = 0: Player 2 can force a win.
Here, Nim-sum = 4 ≠ 0, so Player 1 has a winning strategy.
Step 3: Construct the Winning Move
To win, Player 1 must make a move that results in a Nim-sum of 0.
Identify the pile with the highest binary digit in the Nim-sum:
Nim-sum = 100 (Binary), so focus on the leftmost 1 (third position).
Modify the largest pile (5 stones) to reduce its binary representation:
Current binary: 101
Desired binary: 100 (Decimal: 4)
Remove 1 stone from the first pile.
New pile sizes: (4, 3, 2).
Step 4: Verify the New Nim-sum
4 = 100
3 = 011
2 = 010
Nim-sum = 100 XOR 011 XOR 010 = 001 (Decimal: 1).
Wait—this calculation is incorrect. Let’s correct it:
100 XOR 011 = 111
111 XOR 010 = 101 (Decimal: 5).
Error Note: A common mistake is miscalculating the Nim-sum after the move. Let’s revise the strategy.
Corrected Step 3:
To achieve a Nim-sum of 0, Player 1 should:
Find the pile where XORing its size with the Nim-sum reduces it:
For each pile: size XOR Nim-sum.
5 XOR 4 = 1 → Remove 5 – 1 = 4 stones from the first pile.
New pile sizes: (1, 3, 2).
New Nim-sum:
1 XOR 3 XOR 2 = 001 XOR 011 XOR 010 = 000 (Decimal: 0).
Step 5: Strategy Execution
Player 1 removes 4 stones from the first pile, leaving (1, 3, 2). Now, no matter what Player 2 does, Player 1 can mirror moves to maintain a Nim-sum of 0, eventually winning.
Final Answer:
Player 1 has a winning strategy. They should remove 4 stones from the first pile, reducing it to (1, 3, 2). This forces Player 2 into a losing position.
Key Takeaways:
Nim-sum determines the winner.
Always leave a Nim-sum of 0 for the opponent.
Practical applications include scheduling, resource allocation, and competitive programming.
Formula:
Nim-sum = ( a \oplus b \oplus c ) (where ( \oplus ) denotes XOR).
Note: Replace the example numbers with actual CAT 2024 question values if provided. Let me know if you need further clarification!
This solution aligns with CAT exam patterns, emphasizing logical reasoning and mathematical modeling. Let me know if you need help with other similar problems! 🎯
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