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Methods to Solve Games
91
2 ˆ n
Consider a game with
and
m ˆ 2
games
2 ˆ n
payo matrix
ˆ
A “
a11a12. . . a
a21a22. . . a
1n
2n
˙
.
As before, we assume the matrix has no saddle points, and, therefore, both of the players
have at least two active strategies. Moreover, the observation above leads us to the conclusion
that there exists a solution, in which each player has exactly two active strategies, so, in fact,
to solve a
the resulting
Let
For xedx, the sum
over the unit simplex
2 ˆ n
game, it is sucient to nd two active strategies of the second player and solve
2 ˆ 2
subgame.
p “ pp1, p2q “ p1 ´ x, xq,0 ď x ď 1
αppq “ min
n
ř
Epp, qq “ min
qPQ
n
ra1jp1 ´ xq ` a2jxsqjis a linear function of
j“1
Qn. That means,
n
ÿ
αppq “ min
qPQ
ra1jp1 ´ xq ` a2jxsqj“ min
n
j“1
, be a mixed strategy of the rst player. Then,
n
ÿ
qPQ
ra1jp1 ´ xq ` a2jxsqj.
n
j“1
ra1jp1 ´ xq ` a2jxs,
1ďjďn
tqju
, which is minimized
and the problem of the rst player is to maximize a minorant of a family of linear functions,
min
where
νjpxq “ a1jp1 ´ xq ` a2jx
νjpxq Ñ max
1ďjďn
. The minorant is a convex piecewise-linear function. Since
0ďxď1
,
both of the rst player's strategies are active, the maximum is achieved at an interior point
x˚P r0, 1s
lines,
Example 7.7.
First, write down functions
and can be found in
νjpxq
and
νkpxq
Solve the game with payo matrix
Opn2q
time. Note that
x˚is the point of intersection of two
, which correspond to two active strategies of the second player.
νjpxq
A “
:
ˆ
4 2 3 ´1
´4 0 ´2 2
˙
.
ν1pxq “ 4 ´ 8x, ν2pxq “ 2 ´ 2x, ν3pxq “ 3 ´ 5x, ν4pxq “ ´1 ` 3x.
The graphs of these functions are shown in g.7.2. The minorant of the functions consists of
segments of
ν1pxq
and
ν4pxq
and achieves its maximum at point
x˚such that
ν1px˚q “ ν4px˚q
5
4 ´ 8x˚“ ´1 ` 3x˚, x˚“
.
11
:
That is,
p˚“ p1 ´ x˚, x˚q “ p
11
6
5
,
q
, and
11
ν˚“ ν1px˚q “
11
4
.
92
0
a
12
a
11
a
1j
a
1k
a
22
a
2j
a
21
1
a
2k
ν1(x)
νj(x)
νk(x)
ν2(x)
ν
∗
p
∗
2
1
2
3
4
−1
−2
−3
−4
1
2
3
4
−1
−2
−3
−4
0
1
p
∗
2
=
5
11
ν∗=
4
11
Game Theory
Figure 7.1: Graphical solution of a
To nd the optimal mixed strategy for the second player, we should solve the following
subgame:
From the relations above,
˚
q
“
1
ˆ
6
p˚“
11
Apparently, an
m ˆ 2
game can be solved in a way similar to that above: by reducing it
2 ˆ n
game
Figure 7.2: Diagram for Example 7.7
2ˆ2
ˆ
4 ´1
´4 2
11
3
,
q
8
˚
“
4
. Therefore, the solution of the initial game is:
11
˙
5
,
, q˚“
11
ˆ
3
11
˙
.
, 0, 0,
8
11
˙
, ν˚“
4
.
11
to a problem of minimization of majorant of a family of linear functions corresponding to pure
strategies of the rst player.
3 ˆ 3
games
To illustrate the application of the active strategies theorem to games of larger size, we consider
the
3 ˆ 3
game from example 7.1:
¨
˛
´2 3 ´4
A “
˝
3 ´4 5
‚
.
´4 5 ´6
Write down the rst player's problem. As before, we can assume that for any given
the mean payo
and
q3“ p0, 0, 1q
Epp, qq
. That implies,
achieves its minimum at one of the points
αppq “ min
Epp, qq “ min
qPQ
3
$
´2p1` 3p2´ 4p
&
3p1´ 4p2` 5p
%
´4p1` 5p2´ 6p
p P P
q1“ p1, 0, 0q,q2“ p0, 1, 0q
,
.
3
.
3
-
3
m
,
Methods to Solve Games
1
1
1
2
D
1
D
2
0
1
2
1
1
9
16
R
1
R
2
0
7
12
1
1
11
20
S
1
S
2
0
9
16
93
Substituting
1 ´ p1´ p2for
min
Consider feasible set
D “ tpp1, p2q : 0 ď p1` p2ď 1, p1, p2ě 0u
the subset, where function
(a)
f1and
f
3
Figure 7.3: Sets of pairwise minorization of functions
p3, we get
$
´4 ` 2p1` 7p2“ f
&
%
5 ´ 2p1´ 9p2“ f
´6 ` 2p1` 11p2“ f
,
.
1
2
-
3
Ñ max
fiminorizes two other functions.
(b)
f1and
f
2
p1,p2ě0
0ďp1`p2ď1
and, for each
f1,
.
(c)
f2and
f2, and
i P t1, 2, 3u
f
3
f
3
, nd
Compare
f1and
f3(see g.7.3-a).
„
1
f1ď f3in
D1“"pp1, p2q : p2P
, 1, p1P r0, 1 ´ p2s*,
2
1
f1ě f3in
D2“ tpp1, p2q : p2P r0,
s, p1P r0, 1 ´ p2su.
2
Compare
Compare
f1and
f2ď f1in
f2ě f1in
f2and
f2ď f3in
f2ě f3in
f2(see g.7.3-b).
R1“"pp1, p2q : p1P„0,
R2“"pp1, p2q : p1P„0,
f2(see g.7.3-c).
R1“"pp1, p2q : p1P„0,
R2“"pp1, p2q : p1P„0,
7
12
7
12
9
16
9
16
„
, p1P
, p1P„0,
„
, p1P
, p1P„0,
9 ´ 4p
16
9 ´ 4p
16
11 ´ 4p
20
11 ´ 4p
20
1
, 1 ´ p
1
1
, 1 ´ p
1
1
; p1P
; p1P
*
*
1
,
„
7
, 1, p2P r0, 1 ´ p1s*.
12
,
„
9
, 1, p2P r0, 1 ´ p1s*.
16
Therefore, three lines
six domains,
Ki,
i “ 1, . . . , 6
1
p2“
,
4p1` 16p2“ 9
2
(see g.7.4), such that
, and
4p1` 20p2“ 11
, split feasible setDinto
94
1
1
K
1
K
6
0
9
16
b
1
2
9
16
11
20
K
2
K
3
K
4
K
5
1
4
127
12
Game Theory
Figure 7.4: Sets of minorization of each of the functions
• mintf1, f2, f3u “ f1, if
• mintf1, f2, f3u “ f2, if
• mintf1, f2, f3u “ f3, if
pp1, p2q P K2Y K3,
pp1, p2q P K1Y K4,
pp1, p2q P K5Y K6.
f1,
f2, and
f
3
The lower value of the game is
"
α0“ max
max
pp1,p2qPK2YK
f1pp1, p2q, max
3
pp1,p2qPK1YK
f2pp1, p2q, max
4
pp1,p2qPK5YK
f4pp1, p2q*.
6
Since a linear function achieves its maximal value at a corner point of a feasible set,
max
pp1,p2qPK2YK
max
pp1,p2qPK1YK
max
pp1,p2qPK5YK
ˆ
f1pp1, p2q “ max"f
3
1
1
“ max"´
16
ˆ
f2pp1, p2q “ max"f
4
2
“ max"0, ´
f3pp1, p2q “ max"f3p0, 1q, f
6
“ max"´6, ´
0,
, ´
1
,
4
1
16
1
2
˙
9
, f
16
1
, 0*“ 0,
2
˙
1
, f
2
1
, ´
16
ˆ
3
, 0, ´
ˆ
˙
1
0,
1
2
ˆ
˙
9
0,
2
16
, ´4*“ 0,
˙
1
0,
, f
3
2
1
, ´4*“ 0,
16
ˆ
, f
, f
ˆ
1
2
1
,
4
ˆ
1
2
1
4
˙
9
16
˙*
1
,
2
˙
*
7
,
, f2p0, 1q
16
, f
ˆ
3
16
˙
9
7
,
, f3p0, 1q
16
*
and the value of the game is
ν “ α0“ 0 “ f
Therefore, the optimal mixed strategy of the rst player is
Similar argument for the the second player gives us
ˆ
1
˙
1
1
,
4
“ f
2
ˆ
2
1
4
1
,
2
q˚“
˙
“ f
p˚“
`
1
4
ˆ
3
,
˙
1
1
,
4
`
1
4
1
,
2
.
2
˘
1
1
,
,
.
2
4
˘
1
.
4
Methods to Solve Games
95
7.4.3. BrownRobinson iterative method
Another method to solve two-person zero-sum games was proposed by George W. Brown and
proved by Julia Robinson in 1951 [5]. Within this approach, a game is solved in an iterative
procedure called
¾ctious play¿
. The play consists of a series of steps, or rounds, at each of
which both of the players use their pure strategies. At the rst step, arbitrary pure strategies
are chosen. Then, at further iterations, each player uses his best response strategy with
respect to the empirical mixed strategy of the opponent.
Algorithm 7.1
Given:
Set
payo matrix
N “ 1
BrownRobinson Iterative Method
||aij||
.
Choose arbitrary pure strategies
p1“ p0, . . . , 0, 1i, 0, . . . , 0q, q1“ p0, . . . , 0, 1j, 0, . . . , 0q.
repeat
Calculate values
n
ÿ
αN“ max
1ďiďm
aijq
j“1
Update frequencies
$
&
N`1
p
“
Set
until
i
N “ N ` 1
.
some stopping conditions hold.
%
Np
N`1
.
N
j
Np
N`1
N
`1
i
i1and
ÿ
“
j“1
N
i
, i ‰ i
, i “ i
j1, and set
n
a
i
N`1
N`1
N`1
N
q
, βN“ min
j
j
,
q
N`1
j
,
1ďjďn
“
$
&
%
m
ÿ
i“1
Nq
aijp
Nq
N`1
N
j
N`1
N
i
N
j
, j ‰ j
`1
, j “ j
“
m
ÿ
i“1
a
ij
N`1
N`1
N`1
,
.
N
p
.
i
It was shown in [5] that
N
ÝÝÝÑ
2
NÑ8
ν,
pNÝÝÝÑ
NÑ8
p˚, qNÝÝÝÑ
NÑ8
q˚, νN“
αN` β
but the convergence is rather slow and not monotone, so, in practice, a great number of itera-
tions should be performed.
Example 7.8.
Again, consider a game with payo matrix
¨
˛
´2 3 ´4
A “
˝
3 ´4 5
‚
.
´4 5 ´6
Perform several iterations of BrownRobinson algorithm.
Step 1.
Let
i1“ 1,j1“ 1
. So,
p1“ p1, 0, 0q,q1“ p1, 0, 0q
.
96
Game Theory
Step 2.
Step 3.
Step 4.
Step 5.
Calculate
Calculate
Calculate
Calculate
α1,
β1,
p2, and
q2.
α1“ maxt´2, 3, ´4u “ 3, i2“ 2, p2“
β1“ mint´2, 3, ´4u “ ´4, j2“ 3, q2“
α2,
β2,
p3, and
q3.
α2“ maxt´3, 4, ´5u “ 4, i3“ 2, p3“
β2“ min
α3,
β3,
p4, and
α3“ max ´1,
β3“ min
α4,
β4,
p5, and
1
2
q4.
4
3
q5.
, ´
, ´
(
1
1
,
2
2
4
5
, ´
3
3
5
, 2(“ ´
3
1
“ ´
, j3“ 2, q3“
2
(
4
“
, i4“ 2, p4“
3
5
, j4“ 2, q4“
3
α4“ max t0,0, 0u “ 0, i5“ 1, p5“
β4“ min
7
, ´
4
(
9
11
,
4
4
9
“ ´
, j5“ 2, q5“
4
`
1
1
,
, 0˘;
2
2
`
`
`
1
3
`
`
`
`
1
3
1
2
1
4
1
4
2
5
1
5
, 0,
,
,
,
,
,
,
2
, 0˘;
3
1
,
3
3
, 0˘;
4
1
,
2
3
, 0˘;
5
3
,
5
˘
1
.
2
˘
1
.
3
˘
1
.
4
˘
1
.
5
Step 6.
Step 7.
Step 8.
And so on. It was shown before that the solution is
from gure 7.5 depicting the dynamics of value
Calculate
Calculate
Calculate
α5,
β5,
p6, and
α5“ max
β5“ min 1, ´
α6,
β6,
p7, and
α6“ max 1, ´
β6“ min
α7,
β7,
p8, and
α7“ max
β7“ min ´
q6.
3
4
, ´
, 1(“ 1, i6“ 3, p6“
5
5
(
6
7
,
5
5
6
“ ´
, j6“ 2, q6“
5
q7.
(
2
5
5
“
,
3
, ´
3
(
1
1
,
6
6
1
6
, i7“ 3, p7“
3
1
“ ´
, j7“ 2, q7“
6
q8.
9
, ´
7
3
,
7
4
7
12
7
, ´
(
15
,
7
5
7
(
15
“
, i8“ 3, p8“
7
5
“ ´
, j8“ 3, q8“
7
p˚“ q˚“
N
νN“
αN`β
2
, the process is not monotone, and
convergence doesn't seem to occur in the rst eight iterations.
`
`
`
`
`
`
`
1
4
˘
1
1
1
,
,
3
1
,
6
2
,
7
1
,
7
1
4
1
8
,
;
2
6
˘
2
1
,
.
3
6
˘
3
2
;
,
7
7
˘
5
1
,
.
7
7
˘
3
3
,
,
;
8
8
˘
5
1
,
,
.
8
4
˘
1
1
,
,
4
ν “ 0
2
. As we see
Exercises
ν1= −
1
2
b
ν2=
7
4
b
ν3= −
1
6
b
ν4= −
9
8
b
ν5= −
1
10
b
ν6=
3
4
b
ν7=
5
7
b
ν8= −
1
8
b
0
1
−1
ν
N
97
Figure 7.5: Example 7.8. Dynamics of the empirical value of the game
7.5. Exercises
7.1.
The rst player thinks of a number 1 through 3, which the second player tries to guess. If
the second player succeeds, he gets the value of the number from the rst one. Otherwise, he
pays that value to the opponent.
a) Write down the payo matrix of the game.
b) Calculate the lower and the upper bounds of the game. Does the game have a pure-strategy
solution?
7.2.
Find pure-strategy solutions for the games below.
¨
˛
1 2 ´1 3 0
˝
2 2 0 2 1
aq
‚
; bq
2 1 1 7 1
7.3.
Let
tfi: i “ 1, . . . , mu
game with payo matrix
7.4.
Prove that any
1 ˆ 1
size by sequential removal of dominated strategies. Is that true for games of size
7.5.
Find mixed-strategy solutions for the games below.
2 ˆ n(m ˆ 2
ˆ
2 ´1
aq
1 3
7.6.
Perform 10 iterations of BrownRobinson iterative method to solve the game with payo
matrix
˙
and
||aij||
tgj: j “ 1, . . . , nu
such that
aij“ fi` gj,
) game that has a pure-strategy solution can be reduced to
¨
1 2 3 3 ´1
˝
0 1 1 ´2 ´2
bq
2 3 6 2 1
¨
A “
1 ´1 3
˝
2 1 ´3
¨
2 1 ´1
˚
0 1 ´5
˚
˝
3 1 10
5 2 2
be two nite number sequences. Solve the
i “ 1, . . . , m,j “ 1, . . . , n
˛
‚
; cq
˛
‚
.
˛
‹
‹
.
‚
¨
1 2 1 ´2
˝
3 2 1 ´3
1 2 3 3
˛
‚
.
3 ˆ 3
?
.
2 3 1
Chapter 8.
Network Flows
A wide variety of problems arising in management, industry and other elds of human activity
involve optimization of propagation of some ow in a network. In this chapter, we investigate a
linear network ow model and discuss two optimization problems of great practical importance:
the Maximum Flow Problem and the Minimum Cost Flow Problem.
8.1. Network Flow Model
Let
G “ pV, Aq
s P V
and a single sink
graph, that is, if
which bounds the amount of the ow passing through this arc.
be a connected directed graph without loops and multiarcs, with a single source
t P V
. For simplicity, we also assume no cycles of length 2 exist in the
pi, jq P A
, then
pj, iq R A
. Each arc
pi, jq P A
has a non-negative
capacity
bij,
Denition 8.1.Aow
that
ow conservation conditions hold
(or,
s ´ t
ow
) inGis a set of non-negative values
, that is,
$
´v, j “ s,
ÿ
i: pi,jqPA
xij´
ÿ
xjk“
k: pj,kqPA
&
%
0, j ‰ s, t,
v, j “ t.
and arc ows do not exceed capacities of the arcs:
0 ď xijď bij, pi, jq P A.
Valuevis called
the value of the ow
.
8.2. Maximum Flow Problem
The rst problem to be considered in this chapter is
The Maximum Flow Problem.
capacities
bijě 0
assigned to each arc
The problem can be formulated in terms of linear programming:
ÿ
i: pi,jqPA
Given a network
pi, jq P A
ÿ
v “
j: ps,j qPA
ÿ
xij´
k: pj,kqPA
. Find an
xijÑ max
txu
$
&
xjk“
%
G “ pV, Aq
s ´ t
,
´v, j “ s,
0, j ‰ s, t,
v, j “ t.
xij,
pi, jq P A
, such
with sources, sinkt, and
ow of maximal value inG.
(8.1)
(8.2)
0 ď xijď bij, pi, jq P A.
(8.3)
Maximum Flow Problem
99
Note that problem (8.1)(8.3) can be solved by employing standard linear programming tech-
nique. Moreover, easy to see, the problem's constraint matrix satises total unimodularity
conditions, and, therefore, we can eectively nd an integral solution if all the capacities are
integers.
A more specic approach was developed by Lester R. Ford and Delbert R. Fulkerson in
1956. Below, we describe their method following their classic book ¾Flows in Networks¿ [3].
8.2.1. Cuts and Flows
Denition 8.2.
Let
x “ txiju
be a feasible
s ´ t
ow inG. An
augmenting path
with respect
toxis an undirected pathPin the graph obtained fromGby replacing arcs with edges, such
that for each
1. if
2. if
Denition 8.3.
s P X
and
pi, jq P P
pi, jq P A
pj, iq P A
t P¯X
the following conditions hold:
then
xijă bij(in this case, arc
, then
xjią 0(pi, jq
We call a partition
. The
capacity
of cut
is a
backward
pX,¯Xq
pX,¯Xq
pi, jq
is a
forward
arc ofP) ;
arc of the path).
of nodes of a networkGa
is
cut
(or, an
s ´ t
cut
), if
ÿ
An
CpX,¯Xq “
pi,jqPA,
iPX,jP¯X
s ´ t
cut, the capacity of which is minimal among all the cuts, is called
bij.
minimal
.
From (8.2) and (8.3) we see that the value of a feasible ow does not exceed the capacity of
any cut
pX,¯Xq,s P X,t P¯X:v ď CpX,¯Xq
. Indeed, each ow can be split into a collection of
paths, each delivering a non-negative amount of the ow. Each of these paths includes at least
one arc
and the total amount of the ow passing through the cut is at most
pi, jq
such that
i P X
and
j P¯X
. The total ow passing along this arc is at most
CpX,¯Xq
bij,
. Thus, any cut
is a bottleneck of the whole network, and the capacity of a minimal cut sets an upper bound
on the value of maximal ow. The theorem below states that the bound is exact.
Theorem 8.1
(MaxFlow MinCut Theorem (Ford, Fulkerson)).The value of a maximal ow
is equal to the capacity of a minimal cut.
Proof.
Let
x “ txiju
be a feasible ow. Note that if its value equals the capacity of a cut, the
statement is proved: the ow is maximal, and the cut is minimal, due to observation above.
Assumexis a maximal ow and its value is less then the capacity of any cut. Consider a
subset
X Ď V
constructed using the following procedure.
When the procedure nishes, two cases are possible.
Case 1. The sink
Indeed, for each arc
added toX). Also, for each backward arc
t P¯X.That is,
pi, jq
such that
pX,¯Xq
i P X
is a cut, and its capacity equals the value of the ow.
and
j P¯X,xij“ bij(otherwise,jmust have been
pj, iq,xji“ 0
. That contradicts the assumption that
the value of the ow is strictly less than capacity of any cut.
100
Cut-Constructing Procedure
Network Flows
Place the sourcesintoX:
X “ tsu
.
repeat
if
i P X
if
i P X
until
no new nodes can be placed inX.
Set¯X “ V zX
Case 2. The sink
and
xijă bijfor some
and
xjią 0
for some
j R X
j R X
(the arc is
then
X “ X Y tju
unsaturated)then
.
X “ X Y tju
.
.
t P X.In this case, there is an augmenting path connectingsandt, in which
all the forward arcs are unsaturated and all the backward arcs deliver nonnegative ow. Denote
ε1“ mintbij´ xij: pi, jq
ε2“ mintxji: pj, iq
is a forward arc of
is a backward arc of
P u,
P u,
ε “ mintε1, ε2u ą 0.
Now, increase the ow along the forward arcs of the path byεand reduce the ows along the
backward arcs by the same value. Obviously, the resulting ow remains feasible, and its value
is greater byεthen that of the initial ow, which contradicts the assumption of maximality of
the initial ow.
Corollary 8.
A ow is maximal if and only if there is no augmenting path associated with
this ow.
Note that a network can contain more than one minimal cut. Below, we give without proofs
two statements concerning properties of minimal cuts in a graph.
Theorem 8.2.
pX X Y,
Ğ
X X Y q
Theorem 8.3.
pX,¯Xq
is the cut constructed by the procedure described in Theorem 8.1. Then
Let
pX,¯Xq
are minimal
Let
tpXi,¯Xiq : i “ 1, . . . , mu
and
s ´ t
pY,¯Y q
cuts as well.
be two minimal
be the set of all minimal
s ´ t
cuts. Then
pX Y Y,
s ´ t
Ğ
X Y U q
and
cuts inG, and
m
Ş
X “
Xi.
i“1
8.2.2. FordFulkerson Labeling Algorithm
Theorem 8.1 yields an algorithm to solve the Maximum Flow Problem. The algorithm starts
with a feasible ow (e.g., with the zero ow) and then increases it iteratively by constructing
a series of augmenting
s ´ t
paths.
At each step, the algorithm systematically looks through the nodes and labels the ones that
may occur in an augmenting path. Each node can be in one of three states:
•
not labeled,
•
labeled and not scanned,
•
labeled and scanned.
To label a nodej, a pair
ri˚, εpjqs
is used such that
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