Recently I came across the following problem in a contest. We tried the problem for a long time, but could not come up with a correct solution.
Alice and bob are playing a game across $$$ n \leq 10000 $$$, 3×3 Tic-tac-toe boards with both players marking the board with X. Each board already has some X's on it. One player can choose an arbitrary board and put an X on any of the empty cells. Alice always makes the first move. You can not make any further moves on a board after it has three consecutive X's (column, row or diagonal). A game ends when all the boards contain three consecutive X's. The player that made the last move loses the game.
Given a configuration of the game, you have to find out the winner. It is given that none of initial boards contain three consecutive X's.
Sample Input:
5
1
XX.
..X
..X
1
..X
X..
X.X
1
..X
.X.
.X.
1
..X
X..
XX.
1
..X
X..
..X
Sample Output:
Game #1: Alice
Game #2: Alice
Game #3: Bob
Game #4: Bob
Game #5: Bob
Auto comment: topic has been updated by Anachor (previous revision, new revision, compare).
Auto comment: topic has been updated by Anachor (previous revision, new revision, compare).
Isn't this a grundy game?
Each board have a state that can be stored in 2^9 integer (512 possible values).
Compute for each state x, the grundy number g(x). The answer is Alice will win if the xor(g(state of boards)) == 0.
Edit: States are independents, so you can pre-compute the grundy numbers for all 512 states at first.
What would be the output of this case according to your algorithm ?
Since board states are same his algo with say 0, so Alice wins.
Correct.
Then what about this one ?
Now his solution would tell xor is 1, so Bob wins. But here clearly Alice wins.
Edit: I was wrong here. Bob wins in this case actually.
Bob wins and it's correct.
"The player that made the last move loses the game"
Well, I was wrong here. But check the second sample, here we can reach the state
(by making a move in the last row , middle column)
so grundy value of the initial state is going to be non-zero. But still Alice wins here.
https://math.stackexchange.com/questions/375114/nimbers-for-mis%C3%A8re-games
The grundy number for misere game for terminal states isn't 0.
So the grundy number of that original state in the sample is going to be 0.
I have a generated a test file containing all possible games of 2 boards. Also, I have generated the answer for those games, by writing a brute force solution (hopefully the brute force is correct). If you want to test your solution you can find the data here: link
I know that my comment is off-topic. But I'd like to talk about an interesting game.
The numbers from 1 to 9 are written on the board. There're two players. Players take turns. During each turn, the player takes a number (each number can be taken only once). The player who takes 3 numbers such that their sum is 15 first wins. Who will win if both play optimally?
Turns out, the game is equivalent to tic-tac-toe on 3x3 magic square (with row, column and diagonal sum equal to 15). This means that optimal plays will lead to draw.