Nim Game: Find Proof for D > A, E > B, F > C

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In summary, the conversation is discussing the concept of a nim-sum and whether there is a case where D > A, E > B, and F > C. The proof of this is unknown and may depend on the definition of a nim-sum.
  • #1
unscientific
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Hi all, ok there are 3 piles containing objects; Pile A, Pile B and Pile C.

Let the Nim-sum of all 3 piles be denoted as X.

Let Pile A and Pile B and Pile C contain objects.
( In Nim-Sum )

A + X = D

B + X = E

C + X = F

As we are currently in research of a math project, we are wondering if there is a case whereby D > A, E > B, and F > C. If not, what is the proof?
 
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  • #2
Perhaps I am not understanding the question. The number of objects in each pile are A, B, C respectively. What do you mean by "Nim-sum"? Certainly if X> 0 then it must be the case that D> A, E> B, and F> C.
 
  • #3
unscientific said:
Hi all, ok there are 3 piles containing objects; Pile A, Pile B and Pile C.

Let the Nim-sum of all 3 piles be denoted as X.

Let Pile A and Pile B and Pile C contain objects.
( In Nim-Sum )

A + X = D

B + X = E

C + X = F

As we are currently in research of a math project, we are wondering if there is a case whereby D > A, E > B, and F > C. If not, what is the proof?

I don't know what a Nim-Sum is, but from what you have
[tex]D>A\Leftrightarrow E>B\Leftrightarrow F>C\Leftrightarrow X>0[/tex]

so you're just asking if the Nim-Sum is always negative. Well, define it for us and we'll see!
 
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  • #4
According to wikipedia, a nim-sum is an operation used in game theory.
 

1. What is the Nim Game?

The Nim Game is a mathematical strategy game where two players take turns removing objects from a pile. The player who removes the last object wins.

2. What does "D > A, E > B, F > C" mean in the Nim Game?

This notation means that for a game with four piles (A, B, C, D), the player who has the first turn and follows the strategy of always removing more objects from pile D than their opponent removes from pile A will win. Similarly, for a game with six piles (A, B, C, D, E, F), the player who follows the strategy of removing more from pile E than their opponent removes from pile B and more from pile F than their opponent removes from pile C will win.

3. What is the proof for D > A, E > B, F > C in the Nim Game?

The proof for this strategy in the Nim Game is based on the concept of "nim sums." A nim sum is the bitwise exclusive OR (XOR) of the number of objects in each pile. By following the strategy of always maintaining a nim sum of 0, the player can guarantee a win. This is because a nim sum of 0 makes it impossible for the opponent to make a move that keeps the nim sum at 0, forcing them into a losing position.

4. Can the strategy of D > A, E > B, F > C be applied to games with more than six piles?

Yes, this strategy can be applied to games with any number of piles. The key is to always maintain a nim sum of 0 by following the same principles of removing more objects from certain piles than the opponent removes from others. However, the strategy becomes increasingly complex as the number of piles increases, making it more difficult to determine the optimal moves.

5. Are there any other winning strategies for the Nim Game?

Yes, there are several winning strategies for the Nim Game, including the "Fibonacci strategy" and the "Wythoff's Game strategy." These strategies involve more advanced mathematical concepts and can be applied to games with any number of piles. However, the D > A, E > B, F > C strategy is commonly used and is often the simplest solution for winning the game.

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