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MATH 310

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MATH 310 GAME THEORY Fall 2008 Instructor: CHENG Shiu Yuen e-mail: macheng_at_ust.hk, Tel #: x-7267 Office hour: M W 9:15-10:15 am at Rm 6520 or by appointment. – PowerPoint PPT presentation

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Title: MATH 310


1
  • MATH 310
  • GAME THEORY
  • Fall 2008
  • Instructor CHENG Shiu Yuen e-mail
    macheng_at_ust.hk, Tel x-7267
  • Office hour M W 915-1015 am at Rm 6520 or by
    appointment.
  • Course Description This course will follow
    closely Prof. Thomas Fergusons MATH 167 at UCLA.
    We will add a few more topics if time permits. We
    will use the electronic text by Ferguson
    (http//www.math.ucla.edu/tom/math167.html) as
    our text. Our assignment strictly contains that
    of MATH 167. From time to time, I will also
    assign additional problems outside that of MATH
    167.

2
  • Learning Outcomes
  • Develop an understanding of the core ideas and
    concepts of Game Theory.
  • Be able to recognize the power of abstraction and
    generalization, and to carry out investigative
    mathematical work with independent judgment.
  • Be able to apply rigorous, analytic, highly
    numerate approach to analyze and solve problems.
  • Be able to communicate problem solutions using
    correct mathematical terminology and good
    English.

3
  • Assessment Various assessment tasks are
    intended for the enhancement of student learning.
  • Midterm Examination (30) Midterm examination
    will be scheduled on evening of 25 March 2010.
  • Final Examination (50)
  • Project Presentation (20)Students should form
    groups of 3 (exactly 3 unless with prior
    approval) to submit a written report and make
    presentation on a research article in game
    theory. The list of articles and the format of
    the written report will be announced in March.
    The written report is due on 5 May and the
    presentation is scheduled on Sunday, 9 May 2010.

4
  • It is important to do homework assignments even
    though students are not required to submit the
    solutions. At least 50 of problems in the
    midterm and final examinations will be based on
    the problems in the homework assignments.

5
  • Grading Policy The course grade (G) will be
    based on the midterm examination (M), the final
    examination (F), the project presentation (P),
    and challenging problems (?).
  • G M F P ?
  • M30G, F50G, P20G, 0 ? 5 G
  • Only students in passing status can receive ?.
    It is the grade for doing assigned challenging
    problems. It is intended as a bonus and not
    intended for all students. All works on the
    assigned challenging exercise problems must be
    submitted within one week.

6
  • This course will be graded on an absolute scale
    in the following.
  • F 0G54,
  • D 55G59,
  • C range 60G69,
  • B range 70G84,
  • A range 85G100

7
INTRODUCTION
  • Game theory is a fascinating subject.

Game Theory studies the competition or
cooperation between rational and intelligent
decision makers.
It has its origin in the entertaining games that
people play, such as chess, poker, tic-tac-toe,
bridge the list is quite varied and almost
endless.
8
  • In addition, there is a vast area of economic
    games, as discussed in Myerson (1991) and Kreps
    (1990), and the related political games,
    Ordeshook (1986), Shubik (1982), and Taylor
    (1995). The competition between firms, the
    conflict between management and labor, the fight
    to get bills through congress, the power of the
    judiciary, war and peace negotiations between
    countries, and so on, all provide examples of
    games in action.

9
  • Games are characterized by a number of players
    or decision makers who interact, possibly
    threaten each other and form coalitions, take
    actions under uncertain conditions, and finally
    receive some benefit or reward or possibly some
    punishment or monetary loss.

We will study various mathematical models of
games and create a theory or a structure of the
phenomena that arise. In some cases, we will be
able to suggest what courses of action should be
taken by the players.
10
  • The number of players will be denoted by n. Let
    us label the players with the integers 1 to n,
    and denote the set of players by
  • N 1, 2, . . . , n.

There are three main mathematical models or
forms used in the study of games, the extensive
form, the strategic form and the coalitional form.
These 3 differ in the amount of detail on the
play of the game built into the model.
11
  • 1.2 What is a Combinatorial Game?

(1) There are two players.
(2) There is a set, usually finite, of possible
positions of the game.
(3) The rules of the game specify for both
players and each position which moves to other
positions are legal moves. If the rules make
no distinction between the players, that is if
both players have the same options of moving
from each position, the game is called
impartial otherwise, the game is called
partizan.
12
  • (4) The players alternate moving.

(5) The game ends when a position is reached
from which no moves are possible for the player
whose turn it is to move. Under the normal play
rule, the last player to move wins (If you cant
move, you lose.) Under the misère play rule the
last player to move loses.
(6) The game ends in a finite number of moves
no matter how it is played.
13
  • 1.1 A Simple Take-Away Game.

(1) There are two players. We label them I
and II.
(2) There is a pile of 21 chips in the center of
a table.
(3) A move consists of removing one, two, or
three chips from the pile. At least one chip
must be removed, but no more than three may
be removed.
(4) Players alternate moves with Player I
starting.
(5) The player that removes the last chip wins.
(The last player to move wins. If you cant
move, you lose.)
14
  • How can we analyze this game?

Can one of the players force a win in this game?
Which player would you rather be, the player who
starts or the player who goes second?
What is a good strategy?
15
  • We analyze this game from the end back to the
    beginning. This method is sometimes called
    backward induction.

If there are just one, two, or three chips left,
the player who moves next wins simply by taking
all the chips.
Suppose there are four chips left. Then the
player who moves next must leave either one, two
or three chips in the pile and his opponent will
be able to win. So four chips left is a loss for
the next player to move and a win for the
previous player, i.e. the one who just moved.
16
  • With 5, 6, or 7 chips left, the player who moves
    next can win by moving to the position with four
    chips left.

With 8 chips left, the next player to move must
leave 5, 6, or 7 chips, and so the previous
player can win.
We see that positions with 0, 4, 8, 12, 16, . .
. chips are target positions we would like to
move into them. We may now analyze the game with
21 chips.
Since 21 is not divisible by 4, the first player
to move can win. The unique optimal move is to
take one chip and leave 20 chips which is a
target position.
17
  • 1.3 P-positions, N-positions.

We see that 0, 4, 8, 12, 16, . . . are positions
that are winning for the Previous player (the
player who just moved) and that
1, 2, 3, 5, 6, 7, 9, 10, 11, . . . are winning
for the Next player to move.
The former are called P-positions, and the
latter are called N-positions. We say a
position in a game is a terminal position, if no
moves from it are possible.
18
  • Labeling Algorithm

Step 1 Label every terminal position as a
P-position.
Step 2 Label every position that can reach a
labeled P-position in one move as an N-position.
Step 3 Find those positions whose only moves
are to labeled N-positions label such positions
as P-positions.
Step 4 If no new P-positions were found in step
3, stop otherwise return to step 2.
19
  • It is easy to see that the strategy of moving to
    P-positions wins.
  • From a P-position, your opponent can move only
    to an N-position (Step 3).


Then you may move back to a P-position (Step 2).
Eventually the game ends at a terminal position
and since this is a P-position, you win (Step 1).
20
  • Characteristic Property. P-positions and
    N-positions are defined recursively by the
    following three statements.

(1) All terminal positions are P-positions.
(2) From every N-position, there is at least
one move to a P-position.
(3) From every P-position, every move is to an
N- position.
21
  • 1.4 Subtraction Games.

Let S be a set of positive integers. The
subtraction game with subtraction set S is played
as follows.
From a pile with a large number, say n, of
chips, two players alternate moves.
A move consists of removing s chips from the
pile where s ? S.
Last player to move wins
22
  • For illustration, let us analyze the subtraction
    game with
  • subtraction set S 1, 3, 4 by finding its
    P-positions.

There is exactly one terminal position, namely 0.
x 0 1 2 3 4 5 6 7 8 9 10 11 12
13 14. . . position P N P N N N N P N
P N N N N P...
The pattern PNPNNNN of length 7 repeats forever.
23
  • Who wins the game with 100 chips, the first
    player or the second?

The P-positions are the numbers equal to 0 or 2
modulus 7.
Since 100 has remainder 2 when divided by
7, 100 is a P-position the second player to
move can win with optimal play.
24
  • Example n chips on the table. Player I and II
    takes turn to remove k2 chips, kgt0. Find the
    P-positions and N-positions.
  • Solution
  • 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
  • P N P N N P N P N N P N P N N P

25
  • 2. The Game of Nim

The most famous take-away game is the game of
Nim, played as follows. There are three piles
of chips containing x1, x2, and x3 chips
respectively. (Piles of sizes 5, 7, and 9 make a
good game.)

Two players take turns moving.
Each move consists of selecting one of the piles
and removing chips from it.
You may not remove chips from more than one pile
in one turn, but from the pile you selected you
may remove as many chips as desired, from one
chip to the whole pile. The winner is the
player who removes the last chip.
26
  • Can we find P and N positions of Nim by labeling
    algorithm?
  • Consider the simple case of two-pile Nim of one
    chip and two chips.
  • (1,2)
  • (0,2) (1,1)
  • (0,1) (1,0)
  • (0,0)

27
  • 2.2 Nim-Sum. The nim-sum of two non-negative
    integers is their addition without carry in base
    2. This operation is also called the
    exclusive-or-operation (XOR).

Definition. The nim-sum of (xm x0)2 and
(ym y0)2 is (zm z0)2 , and we
write (xm x0)2 ? (ym y0)2 (zm
z0)2, where for all k, zk xk yk (mod 2),
that is, zk 1 if xk yk 1 and zk 0
otherwise.
For example, (10110)2 ? (110011)2 (100101)2.
This says that 22 ? 51 37.
28
  • This is easier to see if the numbers are written
    vertically (we also omit the parentheses for
    clarity)

22 101102 51 1100112 nim-sum 1001012 37

29
  • Nim-sum is associative (i.e. x ? (y ? z) (x?
    y)? z) and commutative (i.e. x? y y ? x), since
    addition modulo 2 is.
  • Thus we may write x ? y ? z without specifying
    the order of addition.

Furthermore, 0 is an identity for addition (0?x
x), and every number is its own negative (x ?
x 0), so that the cancellation law holds x ?
y x ? z implies y z.
(If x ? y x ? z, then x ? x ? y x ? x ? z,
and so y z.)
30
Nim Addition Table
31
  • Theorem 1. (Bouton)A position, (x1, x2, x3), in
    Nim is a P-position if and only if the nim-sum of
    its components is zero, x1 ? x2 ? x3 0.

As an example, take the position (x1, x2, x3)
(13, 12, 8). Is this a P-position? If not, what
is a winning move?
We compute the nim-sum of 13, 12 and 8 13
11012 12 11002 8 10002
nim-sum 10012 9 Since the nim-sum is not
zero, this is an N-position according to Theorem
1.
32
  • Can you find a winning move?

You must find a move to a P-position, that is,
to a position with an even number of 1s in each
column.
33
  • One such move is to take away 9 chips from the
    pile of 13, leaving 4 there. The resulting
    position has nim-sum zero

4 1002 12 11002 8 10002 nim-sum
00002 0
Another winning move is to subtract 7 chips from
the pile of 12, leaving 5.
34
  • 2.4 Proof of Boutons Theorem. Let P denote the
    set of Nim positions with nim-sum zero, and let N
    denote the complement set, the set of positions
    of positive nim-sum.

We check the three conditions of the definition
in Section 1.3.
(1) All terminal positions are in P. Thats
easy. The only terminal position is the position
with no chips in any pile, and 0 ? 0?
0.
35
  • (2) From each position in N, there is a move to
    a position in P.

Heres how we construct such a move.
Form the nim-sum as a column addition, and look
at the leftmost (most significant) column with an
odd number of 1s. Change any of the numbers that
have a 1 in that column to a number such that
there are an even number of 1s in each column.
This makes that number smaller because the 1 in
the most significant position changes to a 0.
Thus this is a legal move to a position in P.
36
  • (3) Every move from a position in P is to a
    position in N. If (x1, x2, . . .) is in P and x1
    is changed to x1lt x1 , then we cannot have x1 ?
    x2 ? 0 x1? x2 ? ,
  • because the cancellation law would imply that
    x1 x1. So x1 ? x2? ?0, implying that
    (x1, x2, . . .) is in N.

These three properties show that P is the set of
P-positions.
37
  • Northcotts game Players can shift the counters
    left or right any number of squares but no
    counter may jump over another counter.

38
  • Nimble In the following strip, each square can
    contain at most one coin. On a players turn, he
    must pick a coin and shift it an arbitrary number
    of squares to the left, without jumping over any
    other coin. Who wins?

39
  • Analyze the following Chinese Chess
    configuration.

40
Misère Version of Nim
  • P-positions and N-positions of Misère version of
    Nim
  • All nim heaps have exactly one chip
  • P-position Odd number of heaps.
  • N-position Even number of heaps.
  • (2) At least one heap has more than one chips
  • P-position Nim-sum0
  • N-position Nim-sum?0
  • Note If exactly one heap has more than one chips
    then the nim-sum is not zero. Hence, it is an
    N-position. It has a move to the P-position in
    (1).

41
  • 3. Graph Games.
  • We now give an equivalent description of a
    combinatorial game as a game played on a directed
    graph.

3.1 Games Played on Directed Graphs.
Definition. A directed graph, G, is a pair
(X,F) where X is a nonempty set of vertices
(positions) and F is a function that gives for
each x ? X a subset of X, F(x) ? X. For a given x
? X, F(x) represents the positions to which a
player may move from x (called the followers of
x). If F(x) is empty, x is called a terminal
position.
42
  • A two-person win-lose game may be played on such
    a graph G (X,F) by stipulating a starting
    position x0 ? X and using the following rules

(1) Player I moves first, starting at x0.
(2) Players alternate moves.
(3) At position x, the player whose turn it is
to move chooses a position y ? F(x).
(4) The player who is confronted with a terminal
position at his turn, and thus cannot move, loses.
43
  • We first restrict attention to graphs that have
    the property that no matter what starting point
    x0 is used, there is a number n, possibly
    depending on x0, such that every path from x0 has
    length less than or equal to n.

Such graphs are called progressively bounded.
44
  • As an example, the subtraction game with
    subtraction set
  • S 1, 2, 3, analyzed in Section 1.1, that
    starts with a pile of n chips has a
    representation as a graph game.

Here X 0, 1, . . . , n is the set of
vertices. The empty pile is terminal, so F(0)
Ø, the empty set. We also have F(1) 0, F(2)
0, 1, and for 2 k n, F(k) k-3, k-2,
k-1. This completely defines the game.
45
  • 3.2 The Sprague-Grundy Function.
  • Definition. The Sprague-Grundy function of a
    graph, (X,F), is a function, g, defined on X and
    taking non-negative integer values, such that

g(x) min n 0 n ? g(y) for y ? F(x). (1)
In words, g(x) the smallest non-negative integer
not found among the Sprague-Grundy values of the
followers of x.
46
  • If we define the minimal excludant, or mex, of a
    set of non-negative integers as the smallest
    non-negative integer not in the set, then we may
    write simply
  • g(x) mex g(y) y ? F(x). (2)

47
  • Note that g(x) is defined recursively. That is,
    g(x) is defined in terms of g(y) for all
    followers y of x. Moreover, the recursion is
    self-starting. For terminal vertices, x, the
    definition implies that g(x) 0, since F(x) is
    the empty set for terminal x.

48
  • Positions x for which g(x) 0 are P-positions
    and all other positions are N-positions. The
    winning procedure is to choose at each move to
    move to a vertex with Sprague-Grundy value zero.

(1) If x is a terminal position, g(x) 0.
(2) At positions x for which g(x) 0, every
follower y of x is such that g(y) ? 0, and
(3) At positions x for which g(x) ? 0, there is
at least one follower y such that g(y) 0.
49
(No Transcript)
50
  • 2. Subtraction Game
  • S 1, 2, 3? The terminal vertex, 0, has
    SG-value 0. The vertex 1 can only be moved to 0
    and g(0) 0, so g(1) 1. Similarly, 2 can move
    to 0 and 1 with g(0) 0 and g(1) 1, so g(2)
    2, and 3 can move to 0, 1 and 2,
  • with g(0) 0, g(1) 1 and g(2) 2, so g(3)
    3. But 4 can only move to 1, 2 and 3 with
    SG-values 1, 2 and 3, so g(4) 0.

51
  • Continuing in this way we see
  • x 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14. . .
  • g(x) 0 1 2 3 0 1 2 3 0 1 2 3 0 1 2. . .
  • In general g(x) x (mod 4), i.e. g(x) is the
    remainder when x is divided by 4.

52
  • 3. At-Least-Half. Consider the one-pile game
    with the rule that you must remove at least half
    of the counters. The only terminal position is
    zero.

We may compute the Sprague-Grundy function
inductively as x 0 1 2 3 4 5 6 7 8 9 10 11
12 . . . g(x) 0 1 2 2 3 3 3 3 4 4 4 4 4 . . .

We see that g(x) may be expressed as the
exponent in the smallest power of 2 greater
than x g(x) min k 2k gt x.
53
One Pile Nim
  • For the Nim game with only one pile of n chips,
    the player can remove 1 to n chips from the pile.
    Therefore,
  • X 0 1 2 3 4 5 6
  • g(x) 0 1 2 3 4 5 6
  • Question How about Two Piles Nim?

54
  • Rook move game

55
  • 4. Sums of Combinatorial Games

Given several combinatorial games, one can form
a new game played according to the following
rules.
A given initial position is set up in each of
the games. Players alternate moves. A move for a
player consists in selecting any one of the games
and making a legal move in that game, leaving all
other games untouched.
56
  • Play continues until all of the games have
    reached a terminal position, when no more moves
    are possible.

The player who made the last move is the winner.
The game formed by combining games in this
manner is called the (disjunctive) sum of the
given games.
57
  • Copycat Principle Let G be a graph game with a
    specified starting position. Then, GG is a
    losing game.

58
  • 4.1 The Sum of n Graph Games. Suppose we are
    given n progressively bounded graphs, G1 (X1,
    F1), G2 (X2, F2), . . . ,Gn (Xn, Fn). One can
    combine them into a new graph, G (X,F), called
    the sum of G1,G2,,Gn, denoted by
  • G G1 Gn as follows.

The set X of vertices is the Cartesian product,
X X1 Xn. This is the set of all n-tuples
(x1, . . . , xn) such that xi ? Xi for all i.
59
  • For a vertex x (x1, . . . , xn) ? X, the set
    of followers of x is defined as
  • F(x) F(x1,...,xn) F1(x1) x2 xn
  • ? x1 F2(x2) xn
  • ?
  • ? x1 x2 F n(xn).
  • Thus, a move from x (x1, . . . , xn) consists
    in moving exactly one of the xi to one of its
    followers (i.e. a point in Fi(xi)).

60
  • The graph game played on G is called the sum of
    the graph games G1, . . . , Gn.
  • Theorem 2. If gi the Sprague-Grundy function of
    Gi , i 1, . . . , n, then
  • G G1 Gn has Sprague-Grundy function
    g(x1, . . . , xn)
  • g1(x1) ? ?gn(xn).

61
Nim
  • For a Nim game with n piles such that the pile
    1 has x1 chips, pile 2 has x2 chips etc then the
    Sprague-Grundy function is then
  • g(x1,x2,,xn)g(x1) ? g(x2) ? ? g(xn)
  • x1 ? x2 ? ? xn

62
  • Proof We prove by induction.
  • The Theorem is true for the terminal position
    (t1,,tn), where ti is the terminal position of
    the ith game, because
  • g(t1,,tn)00? ?0g(t1) ? ?g(tn)
  • Assume the Theorem is true for all followers
    of (x1,xn).
  • We will prove that the theorem is true for
    (x1,xn).

63
  • Let bg(x1) ? ?g(xn).
  • Let altb. We will show in the following that there
    is a follower of (x1,xn) such that its
    Sprague-Grundy value is a.
  • Let da?b and let (d)2 have k digits. Then,
  • kth
  • b 1
  • a 0
  • ________________
  • 0001

64
  • As bg(x1)? ?g(xn), one of the (g(xi))2 must
    have a 1 at the kth place. We suppose that this
    is x1.
  • Thus, d?g(x1)ltg(x1). By definition of g(x1),
    there is a follower x1 of x1 such that
  • g(x1) d?g(x1).
  • Now (x1,x2,xn) is a follower of (x1,xn).
  • By the inductive assumption
  • g (x1,x2,xn)g(x1)?g(x2)??g(xn)
  • d?g(x1)?g(x2)??g(xn)d?ba
  • This completes the proof that a is the
    Sprgue-Grundy value of some follower of (x1,,xn)

65
  • (2) We claim that no follower of (x1,xn) has
    Sprague-Grundy value b.
  • Assume otherwise that (x1,x2,xn), a follower
    of (x1,,xn) has Sprague-Grundy value b. By the
    inductive assumption,
  • bg (x1,x2,xn)g(x1)?g(x2)??g(xn)
  • g(x1)?g(x2)??g(xn).
  • Then, by cancellation, g(x1)g(x1). As x1 is a
    follower of x1, this is a contradiction. This
    completes the proof.
  • Combining (1) and (2), we see that g (x1,xn)b.
  • This completes the proof of the Theorem.

66
  • Essentially, all impartial finite combinatorial
    games are Nim games!
  • Question Why Nim-Sum?

67
Nim Addition Table
68
  • Green Hackenbush on Trees
  • A rooted tree is a graph with a distinguished
    vertex called the root, with the property that
    from every vertex there is a unique path to the
    root. Essentially this means there are no cycles.
  • A move consists of hacking away any segment and
    removing that segment and anything not connected
    to the ground.

69
  • Colon Principle When branches come together at
    a vertex, one may replace the branches by a
    non-branching stalk of length equal to their nim
    sum.

70
  • Assignment 1
  • Exercise I.1.5 1, 2, 4
  • Exercise I.2.6 1(a), 1(b), 2, 3, 4
  • Exercise I.3.51, 2, 3, 5
  • Exercise I.4.5 2, 3, 6, 8.
  • Mission ? Exercise I.1.5 6(b), 8.
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