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Problem 1.1 from KP Consider a game of Nim with four piles, of sizes 9, 10, 11, 12. (a) Is this position a win for the next player or the previous player (assuming optimal play)? Describe the winning first move. (b) Consider the same initial position, but suppose that each player is allowed to remove at most 9 chips in a single move (other rules of Nim remain in force). Is this an N- or P-position? Solution (a) We can use Bouton's theorem for this part. Just compute the Nim-sum of the components. We see that: 9 = (1001) 2 10 = (1010) 2 11 = (1011) 2 12 = (1100) ₂ The Nim-sum, (1001) 2 (1010) ₂ (1011) ₂ (1100)₂ = 0, therefore the position is an N-position. A winning move would be to take the pile of 12 chips to 8 chips. (b) If there is a restriction at most 9 chips can be removed in a single move, this game is now a sum of 4 subtraction games, all with subtraction sets {1,2,3,..., 9}. We will use the Sprague- Grundy theorem to see if the position is in P or in N. The Sprague-Grundy value of this game is g,(x) = x(mod 10) or the remainder of x when divided by 10. The SG value of the position (9, 10, 11, 12) is given by the Nim-sum of the SG values of the component positions. We can see that the SG value of (9, 10, 11, 12) is not zero, and so it is not a P position, but in N. There are at least 2 winning moves: 128 still works, or we could take 9 →3. Note that on the exam, you would have to explicitly show your reasoning.

Problem 1.1 from KP Consider a game of Nim with four piles, of sizes 9, 10, 11, 12. (a) Is this position a win for the next player or the previous player (assuming optimal play)? Describe the winning first move. (b) Consider the same initial position, but suppose that each player is allowed to remove at most 9 chips in a single move (other rules of Nim remain in force). Is this an N- or P-position? Solution (a) We can use Bouton's theorem for this part. Just compute the Nim-sum of the components. We see that: 9 = (1001) 2 10 = (1010) 2 11 = (1011) 2 12 = (1100) ₂ The Nim-sum, (1001) 2 (1010) ₂ (1011) ₂ (1100)₂ = 0, therefore the position is an N-position. A winning move would be to take the pile of 12 chips to 8 chips. (b) If there is a restriction at most 9 chips can be removed in a single move, this game is now a sum of 4 subtraction games, all with subtraction sets {1,2,3,..., 9}. We will use the Sprague- Grundy theorem to see if the position is in P or in N. The Sprague-Grundy value of this game is g,(x) = x(mod 10) or the remainder of x when divided by 10. The SG value of the position (9, 10, 11, 12) is given by the Nim-sum of the SG values of the component positions. We can see that the SG value of (9, 10, 11, 12) is not zero, and so it is not a P position, but in N. There are at least 2 winning moves: 128 still works, or we could take 9 →3. Note that on the exam, you would have to explicitly show your reasoning.

Database System Concepts
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
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Problem 1.1 from KP Consider a game of Nim with four piles, of sizes 9, 10, 11, 12. (a) Is this position a win for the next player or the previous player (assuming optimal play)? Describe the winning first move. (b) Consider the same initial position, but suppose that each player is allowed to remove at most 9 chips in a single move (other rules of Nim remain in force). Is this an N- or P-position? Solution (a) We can use Bouton's theorem for this part. Just compute the Nim-sum of the components. We see that: 9 = (1001) 2 10 = (1010) 2 11 = (1011) 2 12 = (1100) 2 The Nim-sum, (1001) 2 (1010)₂ (1011) ₂ (1100)₂ = 0, therefore the position is an A winning move would be to take the pile of 12 chips to 8 chips. N-position. (b) If there is a restriction at most 9 chips can be removed in a single move, this game is now {1,2,3,...,9}. We will use the Sprague- a sum of 4 subtraction games, all with subtraction sets Grundy theorem to see if the position is in P or in N. The Sprague-Grundy value of this game is g(x)= x(mod 10) or the remainder of x when divided by 10. The SG value of the position (9, 10, 11, 12) is given by the Nim-sum of the SG values of the component positions. We can see that the SG value of (9, 10, 11, 12) is not zero, and so it is not a P position, but in N. There are at least 2 winning moves: 128 still works, or we could take 9 → 3. Note that on the exam, you would have to explicitly show your reasoning.
Transcribed Image Text:Problem 1.1 from KP Consider a game of Nim with four piles, of sizes 9, 10, 11, 12. (a) Is this position a win for the next player or the previous player (assuming optimal play)? Describe the winning first move. (b) Consider the same initial position, but suppose that each player is allowed to remove at most 9 chips in a single move (other rules of Nim remain in force). Is this an N- or P-position? Solution (a) We can use Bouton's theorem for this part. Just compute the Nim-sum of the components. We see that: 9 = (1001) 2 10 = (1010) 2 11 = (1011) 2 12 = (1100) 2 The Nim-sum, (1001) 2 (1010)₂ (1011) ₂ (1100)₂ = 0, therefore the position is an A winning move would be to take the pile of 12 chips to 8 chips. N-position. (b) If there is a restriction at most 9 chips can be removed in a single move, this game is now {1,2,3,...,9}. We will use the Sprague- a sum of 4 subtraction games, all with subtraction sets Grundy theorem to see if the position is in P or in N. The Sprague-Grundy value of this game is g(x)= x(mod 10) or the remainder of x when divided by 10. The SG value of the position (9, 10, 11, 12) is given by the Nim-sum of the SG values of the component positions. We can see that the SG value of (9, 10, 11, 12) is not zero, and so it is not a P position, but in N. There are at least 2 winning moves: 128 still works, or we could take 9 → 3. Note that on the exam, you would have to explicitly show your reasoning.
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