PLAYING BOARD GAMES ON MULTIDIMENSIONAL SURFACES. BY: STEPHEN GRAVELY & MATTHEW ALLRED MATH MODELS 4910. Three types of surfaces . Torus. Klein Bottle. Mobius Strip. A sample construction of each of the three aforementioned surfaces. Mobius Strip . Klein Bottle (HINT). Torus.
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PLAYING BOARD GAMES ON MULTIDIMENSIONAL SURFACES
BY: STEPHEN GRAVELY & MATTHEW ALLREDMATH MODELS 4910
Torus
Klein Bottle
Mobius Strip
Max ways of winning
Ways of winning
If two players are using winning strategies, why is it always a cat’s game?
If I said these were winning combinations, would it still always be a cat’s game?
Number of Ways to win
(HINT)
New ways to win
Orientation assigning entries corresponding to an Matnxn
Number of ways to win from each entry
Proof that the first player can always win when playing on a Mobius Strip Suppose Matrix A is an 3 x 3 matrix of which the number of entries corresponds to Mat A= a11 =1 a12=2 a13=3 a21=4 a22=5 a23=6 a31=7 a32=8 a33=9
As Player One, Choose a11 =1. Therefore, the winning combinations call them A1j where 2≤j≤9. i.e. A14 = {7,8} are the winning combinations 147 or 148.
Player One will choose either a21=4 or a33=9.
A14 = {7,8} or A19={5,6}
Hence, A14 ∩ A19 = Ø and Player One has four distinct ways of winning.
Player One wins in three moves because Player 2 with only 2 blocks cannot block all distinct ways of winning.
New ways to win on a Klein Bottle (HINT)
Number of ways to win on Klein Bottle
Proof that the first player can always win when playing on a Klein Bottle Suppose Matrix A is an 3 x 3 matrix of which the number of entries corresponds to Mat A= a11 =1 a12=2 a13=3 a21=4 a22=5 a23=6 a31=7 a32=8 a33=9
As Player One, Choose a11 =1. Therefore, the winning combinations call them A1j where 2≤j≤9. i.e. A14 = {7,8} are the winning combinations 147 or 148.
Player One will choose either a21=4 or a33=9.
A14 = {7,8} or A19={5,6}
Hence, A14 ∩ A19 = Ø and Player One has four distinct ways of winning.
Player One wins in three moves because Player 2 with only 2 blocks cannot block all distinct ways of winning.
Therefore, Player One can always win.(HINT)
Connect Four
First move for X if O is blocking
First move for O if O is blocking
Second Move for X if O is blocking
Second move for O if O is blocking
Third move for X if O is blocking
Third move for O if O is blocking(forced)
Fourth move for X if O is blocking
Fourth move for O if O is blocking(forced)
Winning move for X if O is blocking
Alternate fourth move for O
Winning Move after Alternate O Block
Here we are given a 4x4 matrix on a Klein bottle where we are asked to connect four squares in a row... K-nect 4!!! We begin with the matrix... (then put the matrix on there). We will number the boxes as in tic-tac-toe, a11= 1, a22= 2,... a44= 16. In order to prove there is a strategy where the first player always wins we will proceed by cases.
Case 1 (and the most difficult case): Player 2 tries to defend on every move: As with tic-tac-toe, we start with Box 1, giving us A1,j,k, where 2≤ j,k≤ 16. Our second move will be denoted by A l,m,n. Player 2 is now going to choose the spot that blocks the most of our winning combinations. There are a six different choices that block four ways of winning, but the boxes that intersect the most distinct ways of winning are Box 5 and Box 8. There is no difference between the two, so we choose Box 5. This changes player 1's winning combinations and leaves 2 boxes (8, 11) that still give us four distinct winning combinations. Either one is good, so we choose Box 8, giving us A1,8,k, and considering our second move as a different starting point, A8,m,n. Now, for player 2's optimum defense, we see there are 7 combinations between A1,8,k and A8,m,n that can be blocked by choosing Box 11, so that is our choice. For player 1's third move, we don't have one box we can choose that will give us two distinct ways of winning, so we must find a box that will give us as many distinct ways of winning where A1,8,k U A8,m,n ≥2, and A1,8,k (intersect) A8,m,n= 0, while also forcing player 2 to block in a box that affects the least amount of our winning combinations. This box is Box 12, giving us A1,8,12= 1 and A8,12,n ≥2. Since we can win with A1,8,12,16, player two must choose Box 16. Now for this fourth move we must change our k, and find an A1,8,k and A8,12,n, where k= n and the union ≥ 2. This move is Box 15, giving us A1,8,15 and A8,12,15. Here we can win with Box 10 or Box 3, giving A1,8,15,10 or A8,12,15,3. Player 2 can only block one of these moves, therefore player 1 wins.
First move for X if O tries to win
First move for O and second move for X
if O tries to win
Second Move For Oif O tries to win
Third move for X if O tries to win
CASE 2
Case 2: Player two plays offensively, and doesn't defend at all: This case is nearly trivial, because if player two doesn't defend on the first three moves, player 1 easily wins in four moves. We again begin in Box 1. Player 2 will choose a box with the least intersections with player 1's winning combinations in order to have the most ways of winning, so we choose Box 14. Now player 1 uses a strategy of blocking player 2 and producing the most non-intersecting ways of winning. There are a few different options, so we'll go with Box 10, giving us A1,10,k= 4. For player 2's second move, we employ a similar strategy, so we choose Box 13. For player 1's third move, we can go ahead and finish this by moving into Box 15. This gives us A1,10,15= 2, namely A1,10,15,8 and A1,10,15,5. Player 2 can only block one of these moves, so player 1 wins.
First move for X if O tries to block and win
First move for O if O tries to block and win
Second move for X and O if O is trying to block and win
Third move for X if O tries to block and win
CASE 3
Case 3: Player 2 employs a strategy of trying to defend player one as well as trying to win himself. Again we begin in Box 1. As in case 1, the two optimal defensive blocks for player 2 are Box 5 and Box 8, again we'll choose Box 5, giving B5,p,q. For player 1's second move (as in case 1) our optimal choices would be Box 8 and Box 11. This time we'll choose 11, giving us A1,11,k =4. Now player 2 can four of player 1's winning combinations and have B5,p,q= 4 by choosing Box 10, where B5,10,q= 4. This doesn't matter, however, because by choosing either Box 6 or Box 8 player 1 now has two distinct ways to win, so we'll choose Box 8, giving us possible wins of A1,8,11,14 and A1,8,11,15. Player 2 can only block one of these, so again, player 1 wins.
Torus
Klein Bottle