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PLAYING BOARD GAMES ON MULTIDIMENSIONAL SURFACES

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PLAYING BOARD GAMES ON MULTIDIMENSIONAL SURFACES

BY: STEPHEN GRAVELY & MATTHEW ALLREDMATH MODELS 4910

Torus

Klein Bottle

Mobius Strip

- Mobius Strip

- Klein Bottle (HINT)

- Torus

Max ways of winning

Ways of winning

If two players are using winning strategies, why is it always a catâ€™s game?

If I said these were winning combinations, would it still always be a catâ€™s game?

Number of Ways to win

(HINT)

New ways to win

- A torus is product of two circles
- Atorus is a surface of revolution generated by revolving a circle in three dimensional space about an axis coplanar with the circle
- Max ways of winning is the same
- The board shifts

Orientation assigning entries corresponding to an Matnxn

- A Mobius Strip is a surface with only one side
- A Mobius Strip can be easily created by a taking a rectangular piece of paper giving it a half twist and then connecting the two ends

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 2â‰¤jâ‰¤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 âˆ© A19 = Ã˜ 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

- A Klein Bottle is two Mobius Strips sewn together to make a single sided bottle with no boundary
- A true Klein Bottle only lives in four dimensions
- A Klein Bottle has one hole and one handle
- Its inside is its outside
- It contains itself

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 2â‰¤jâ‰¤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 âˆ© A19 = Ã˜ 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!!!Â We begin with the matrix... (then put the matrix on there).Â We will number the boxes as in tic-tac-toe, a11= 1, a22= 2,... a44= 16.Â 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 2â‰¤ j,kâ‰¤ 16.Â Our second move will be denoted by A l,m,n.Â Player 2 is now going to choose the spot that blocks the most of our winning combinations.Â 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.Â There is no difference between the two, so we choose Box 5.Â This changes player 1's winning combinations and leaves 2 boxes (8, 11) that still give us four distinct winning combinations.Â 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.Â 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.Â 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 â‰¥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.Â This box is Box 12, giving us A1,8,12= 1 and A8,12,n â‰¥2.Â Since we can win with A1,8,12,16,Â player two must choose Box 16.Â 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 â‰¥ 2.Â This move is Box 15, giving usÂ Â 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.Â 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.Â We again begin in Box 1.Â 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.Â Now player 1 uses a strategy of blocking player 2 and producing the most non-intersecting ways of winning.Â There are a few different options, so we'll go with Box 10, giving us A1,10,k= 4.Â For player 2's second move, we employ a similar strategy, so we choose Box 13.Â For player 1's third move, we can go ahead and finish this by moving into Box 15.Â This gives us A1,10,15= 2, namely A1,10,15,8 and A1,10,15,5.Â 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.Â Again we begin in Box 1.Â 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.Â For player 1's second move (as in case 1) our optimal choices would be Box 8 and Box 11.Â This time we'll choose 11, giving us A1,11,k =4.Â 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.Â 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.Â Player 2 can only block one of these, so again, player 1 wins.

Torus

Klein Bottle