Show that player I can always win a Nim game in which the number of heaps with an odd number of

Show that player I can always win a Nim game in which the...

Key Concepts

Nim Game

Nim is a classic impartial combinatorial game where two players take turns removing objects (or coins) from distinct heaps. The game is characterized by complete information and no chance elements, making it an ideal example for studying winning strategies and the mathematical concept of invariants in combinatorial game theory.

Nim-Sum

The Nim-sum is the bitwise XOR (exclusive or) of the sizes of all heaps. This numerical invariant determines the winning or losing positions in Nim: a position with a zero Nim-sum is losing for the player whose turn it is, while a nonzero Nim-sum indicates a winning position. Analyzing the Nim-sum is crucial to devising a winning strategy.

Parity

Parity, which refers to whether a number is odd or even, plays a key role in certain variations of the Nim game. In problems where the number of heaps with an odd number of coins is critical, understanding parity helps determine how moves affect the overall game state and can be used to guide strategic decisions.

Winning Strategy

A winning strategy in a game like Nim involves making moves that transition the game from a winning position (nonzero Nim-sum) for oneself to a losing position (zero Nim-sum) for the opponent. By carefully adjusting the position and maintaining advantageous invariants such as the Nim-sum and parity conditions, a player can force a win regardless of the opponent’s responses.

Transcript

00:01 We're asked to prove a statement about a variation on the game of nim.

00:07 So recall that in this game, we're given that there are n matches, that each player removes one, two or three matches, and the player to remove the last match loses.

00:20 And we're told that both players use the best strategy.

00:26 We are asked to prove, using strong induction, that if each player uses this best strategy, then the first player wins if n equals three.

00:36 4j, 4j plus 1 or 4j plus 3 for some non -negative j, and the second player wins in the remaining case, which is when n is equal to 4j plus 1 for some non -negative j.

00:53 So first, let's define statement pn, the first player wins.

01:18 If n is congruent to 0, 2, or 3 mod 4, and second wins, this is sort of, implied since somebody has to win in this case if n is congruent to 1 mod 4 for the basis step first we have that p1 is true since player 1 must remove the only match so player 2 wins and we have that 1 is of course congruent to 1 mod 4.

02:57 Next we have that p2 is true.

03:02 First notice that 2 is congruent to 2 mod 4 and we have that player 1 wins.

03:15 This is because player 1 picks one match and then player 2 must take the last match.

03:44 And we have that the statement p3 is also true.

03:50 So notice that p3 is congruent.

03:51 So notice that p3 is congruent to 3 mod 4, we have that player 1 again will win.

04:00 You can see why we have that player 1 removes two matches.

04:11 And so it follows that player 2 must remove the last match.

04:24 And finally we have that statement p4 is true.

04:33 So 4 is congruent to 0 mod 4...