Problem Solving - Games

G A M E S

WHAT IS THIS PRESENTATION ALL ABOUT?

B A S I C - T H E O R Y

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LET’S PLAY A GAME FIRST.

is better than

INVESTIGATING THE GAME

Game of two players, A and B. A always moves first but otherwise the rules are the same for A and B.

The game have a set of all position, P and a set of all legal move, M.

POSITIONS AND LEGAL MOVES

First Game

Second Game

Since the players are merely allowed to check 1, 2, or 3 cells, then we have the set of legal moves is .

In this game, a position may be represented by the number of cells remaining.

Therefore, the set of all position in first game is while the set of all position in second game is .

WINNING AND LOSING POSITIONS

The set P can be partitioned intothe set L of losing and the set W of winning positions:

, .

A player finding himself in a position in L will lose provided her opponent plays correctly. A player finding herself in a position in W can force a win whatever his opponent does.

In most games, at his or her turn, a player may have to choose one of several moves. A strategy is a rule or decision making formula that tells the player which

choice to make at each turn.

If this strategy enables the player to win no matter what moves his or her opponent makes, we will call it a

winning strategy for that player.

FINDING A WINNING STRATEGY

Working BackwardI

Frontal AssaultIII Pairing StrategiesIV

Simplifying a GameII

Let’s Find the Winning StrategyThe winning strategy of first and second games that we played before can be investigated by using an approach called working backward.

The winner is a player that occupies position 0 in the end, it means that there is no cell for his opponent to be checked.

No cell to be checked

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Winning Losing Reason

0 Goal has been achieved

1, 2, 3 Opponent can check last cell(s)

4 Opponent must leave 1, 2, or 3 (losing)

5, 6, 7 Opponent may leave 4 (winning)

… … …

20 Opponent must leave 17, 18, or 19 (losing)

21, 22, 23 Opponent may leave 20 (winning)

Finding the Winning and Losing Positions First

and for First Game

Winning Losing Reason

0 Goal has been achieved

1, 2, 3 Opponent can check last cell(s)

4 Opponent must leave 1, 2, or 3 (losing)

5, 6, 7 Opponent may leave 4 (winning)

… … …

25, 26, 27 Opponent may leave 24 (winning)

28 Opponent may leave 20 (winning)

and for Second Game

Therefore, the winning positions of first and second games are 0, 4, 8, … , 20, the multiple of 3 + 1, while

the losing positions are the numbersthat not multiple of 3 + 1.

It implies that, B have the winning strategy for first game, and A have the winning strategy

for second game.

WHAT IF …

The number of cells is n and each player may check 1, 2, …, k – 1, or k cells at each turn.

The winning positions are all multiples of k + 1 and the losing positions are all other numbers.

Thus, the player who want to be a winner should always leave multiples of k + 1 cells.

M SL EBR OP

LATTICE SQUARE

Two players A and B alternately color lattice squares of a 19 × 94 square. Who has a winning strategy?A lattice square is any square of the board whose vertices are lattice points of the 19 × 94 board.

– MMO 1994

Coloring lattice squares can be like this!

CONSIDER THE BOARD

Consider a board with a small odd height and a large even width below!

What do you think about that board?Yes, the board is symmetric with respect to the line s

HOW WE CAN SOLVE IT?

Can we use pairing strategy to solve itwith ?

So, what is the pair of then?

THE SOLUTION IS …

Anyway, we can use the pairing strategy after we color the square (a square with maximum width and symmetric to s).

And finally we have . And according to , we can conclude that A have the winning strategy.

For 19 × 94 square have a similar solution with the illustration before. The first move should

be to color the square . Now the board is split into two parts which are symmetric with respect to the line s. B is forced to color a square on one side of s. A responds by coloring the square which is symmetric to B’s choice

with respect to s.

KNIGHT’S TOURA and B alternately move a knight on a 1994 × 1994 chessboard. A makes only horizontal moves , B makes only vertical moves . A starts by choosing a square and making a move. Visiting a square for a second time is not permitted. The loser is the one who cannot move.Prove that A has a winning strategy.

–ARO 1994

&SIMPLIFY, TRY, TRY, AND TRY

Winning strategy I Winning strategy II

Winning strategy III

From examples, A have several winning strategies for 4 × 4 chessboard knight’s tour game.

Also, A can lose the game if he don’t have a right strategy.

Wrong strategy

EXTEND THE WINNING STRATEGY

We can extend 4 × 4 first chessboard’s winning strategy to become 8 × 8 chessboard’s winning strategy.

First, imagine that 8 × 8 chessboard can be made from 4 connected 4 × 4 chessboard (look the figure beside). But there are the possible position that made by B that don’t have the “connector” arrow.

FIND THE CONNECTOR ARROW

The winning strategy can be like this.

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Assume that 4k × 4k have a winning strategy like figure below.

Then 4(k + 1) × 4(k + 1) also have similar winning strategy.

BONUSBeside the previous winning strategy, A also have another winning strategies for this game.These are another winning strategies of A.

The value of mathematics in any science lies more in disciplined analysis and abstract thinking than in particular theories and techniques. —Alan Tucker, 1982

THANK YOU.