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Maximum Matching Problem
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71
For brevity, in the following discussion we say:
matching
, meaning that the node is (or it is not) incident to one of the matching's edges. Also,
we call a node, which does not belong to a matching, an
Denition 6.4.
An alternating path for some matching
a node belongs (or doesn't belong) to a
exposed
M
node.
is said to be an
augmenting
path,
if, additionally,
4. the number of edges,p, is odd and the last node
vpdoes not belong toM.
As an example, consider a matching from g.6.3-a. Easy to see, there is an augmenting
path in the graph (g.6.3-b). Note that switching the edges in the path gives a matching of
greater cardinality (g.6.3-c).
(a) A matching
(b) An augmenting path
(c) Switching
Figure 6.3: Augmenting a matching
Lemma 6.1.
pM Y P qzpM X P q
Proof.
Clearly,
|P X pEzM q| “ |P X M | ` 1
Lemma 6.2.
Given a matching
is a matching of cardinality
M1is a matching, since
, sincePhas odd number of edges. Therefore,
M
and an augmenting path
|M1| “ |M| ` 1
v0, vpR M
andPis an alternating path. Moreover,
P
relative toM. Then
.
|M1| “ |M| ` 1
If there is no augmenting path relative to a matchingM, then
M
is a maximal
M1“
.
matching.
Proof.
H “ pM Y M1qzpM X M1q
ofHare either cycles or paths. The edges of any cycle must alternate between
the cycles are of even length and have equal number of edges from
be of even or odd length, but since
Assume the opposite: there is a matching
. The degree of each node inHis 0, 1, or 2. Thus the connected
|M1| ą |M|
M1such that
|M1| ą |M|
. Consider a subgraph
M1andM, so
M
and
M1. A path may
there is a path that contains more edges from
M1than fromM. Obviously, this path has odd number of edges and its ending nodes do not
belongM, i.e., this is an augmenting path.
That is, to solve the problem we must construct an augmenting path, or ensure that no
such paths exist. Below, we demonstrate how to do it in case the graph is bipartite.
Maximal Matching in a Bipartite Graph
Denition 6.5.
and for each
called
parts
Graph
pi, jq P E
one of nodes
of the graph.
G “ pV, Eq
is called
ti, ju
bipartite
belongs to
, if
V “ V1Y V2such that
V1and the other to
V2. Sets
V1X V2“ H
V1and
V2are
72
Matchings and Assignments
Suppose, we have a matching
M
in a bipartite graphG(the matching may be empty or
include a single edge) and search for an augmenting path. Note that the ending nodes of such
path must belong to dierent parts, since its length is odd. Therefore, we have to nd a route
that
•
starts from an exposed node of part
•
ends in an exposed node of part
•
passes through an edge of
EzM
V1,
V2, and,
when moving from
V1to
V2, and through an edge of
M
when moving in opposite direction.
Such a route can be constructed following a simple labeling procedure. We start by labeling
all the nodes of
augmenting path. Then, we label each node of
with an edge from
nodes of
V1, which are incident to an edge ofMending at some labeled node of
and so on. The process stops when we label an exposed node of
V1, which do not belong toM. These are candidates to be the rst node of an
EzM
V2, which is connected with a labeled node of
(these nodes may be the second on the path). Next, all the unlabeled
V
V2, are labeled,
V2(that means, an augmenting
path exists and can be restored by backtracking), or we cannot label any node (the path does
not exist and the matching is maximal).
1
Algorithm 6.1
Input:
a bipartite graph
Step 1.
1.0.
1.1.
1.2.
Labeling Algorithm for the Bipartite Matching Problem
G “ pV1Y V2, Eq
.
Labeling.
Label with `*' every exposed node in
If there are no unscanned labels, then go to Step 3. Otherwise, choose an unscanned
labeled nodevand go to 1.2, if
Scan node
v P V1: look through all edges
v P V1, and to 1.3, if
then give it labelv. Nodevis scanned. Go to 1.1.
1.3.
Scan node
edges
v P V2: ifvis exposed then go to Step 2. Otherwise, look through all
pv, wq P M
; if
w P V1is unlabeled, then give it labelv. Nodevis scanned.
Go to 1.1.
Step 2.
Step 3.
Augmenting.
the path. Set
M “ pM Y P qzpM X P q
No augmenting path.
An augmenting pathPexists. Use labels to backtrack fromwto nd
. Remove all labels. Go to Step 1.
Stop.
Prove that the algorithm nds a maximal matching.
`
Theorem 6.1.
V1and
V2, respectively. On termination of the algorithm:
Let
V
`
,
V
1
be the sets of labeled, and
2
V1.
pv, wq P EzM
V
v P V2.
; if
w P V2is unlabeled,
´
´
,
V
1
the sets of unlabeled nodes of
2
1.
R “ V
2.
|M| “ |R|
´
Y V
1
, therefore,
`
is a vertex cover forG;
2
M
is a maximal matching.
Assignment Problem
1
2
3
4
5
6
7
8
9
10
1
2
3
4
5
6
7
8
9
10
*
*
*
8
10
3
5
5
1
4
1
2
3
4
5
6
7
8
9
10
*
8
2
73
Proof.
Show that no edge connects nodes in
V zR “ V
toR, i.e.,Ris a vertex cover. Indeed, since no node of
no edge in
from
`
V
1
EzM
that connects
V
has its other endpoint in
`
and
1
`
V
2
´
V
. As for matchingM, an edge incident to a node
2
. ThusRis a vertex cover, and the rst statement holds.
`
1
´
Y V
. If so, all the edges are incident
2
V2can be labeled at Step 1.2, there is
Estimate the cardinality ofM. The algorithm stops when an augmenting path cannot be
constructed, i.e., there are no exposed nodes in
node in
algorithm. Therefore,
Example 6.1.
´
V
is also incident toM, otherwise it would be labeled with `*' in the rst step of the
1
|M| ě |V
´
1
`
Y V
| “ |R| ě |M |
2
Find a maximal matching in the graph in g. 6.4-a.
V
`
: each
2
w P V
, and
`
is incident toM. Also, each
2
|R| “ |M |
.
(a) Initial matching
(b) Augmenting paths
(c) Final matching
Figure 6.4: Implementation of labeling algorithm 6.1
We start with initial matching
of the algorithm, we label exposed nodes
(node 1) and look through its neighborhood:
Consider node 8: it is incident to
is incident to
nd an augmenting path:
this iteration is
(edges not from
p3, 7q R M
: give label `3' to 7, mark node 3 as scanned. Node 7 is exposed, we
t1, p1, 8q, 8, p8, 3q, 3, p3, 7q, 7u
t4, p4, 10q, 10, p10, 5q, 5, p5, 9q, 9u
M
are dashed). AugmentMwith respect to these paths: that gives matching
M1“ tp1, 8q, p3, 7q, p4, 10q, p5, 9qu
Apply the algorithm to
incident to
p2, 8q R M
M1. The only exposed node in
: give 8 label `2', nodes 2 is scanned. The only edge of the matching
M “ tp3, 8q, p5, 10qu
t1, 2, 4uofV1with `*'. Next, we choose one of them
p1, 8q R M
p8, 3q P M
. Node 3 gets label `8', node 8 is scanned. Node 3
. Both of the paths are shown bold in g.6.4-b
(g.6.4-c).
(bold in the gure). At the rst step
, node8gets label `1', node 1 is scanned.
. Another path that can be obtained at
V1is 2. Label it with `*'. It is
having 8 as its endpoint is 1. Label it with `8'. Now, 1 is the only unscanned labeled node, an
it has no incident edges not from
At this point,
cardinality equal to that of
´
V
“ t3, 4, 5u,V
1
M1. That is,
M1. The process stops,
`
2
“ t8u
, and
R “ V
´
1
M1is indeed a maximal matching, andRis a minimal
M1is optimal.
`
Y V
“ t3, 4, 5, 8u
2
is a vertex cover of
vertex cover.
6.3. Assignment Problem
In this section, we discuss the Assignment Problem and demonstrate how it can be solved in
polynomial time using matching technique.
74
Matchings and Assignments
First, recall the problem. A somewhat more general formulation than the one given in
Chapter 1 is as follows. Given a set of workers
t1, 2, . . . , mu(m ď n
associated with assigning workerito task
). Each workerican perform any task of a subset
j P Jiis
I “ t1, 2, . . . , nu
and a set of tasks
JiĂ J
, and the prot
J “
cij. Assign workers to tasks so that each
task is executed by exactly one worker, no worker performs more than one task, and the total
prot is maximal.
Notice that the problem can be reformulated as a search for a maximum-weight maximal
matching in a weighted bipartite graph. Indeed, consider a graph
pi, jq P E
if and only if
j P Ji. The weight of an edge
pi, jqiscij. Any assignment corresponds to
G “ pI Y J, Eq
such that
a matching that covers all the nodes inJ(that means, it is a maximal matching), and the prot
of the assignment equals the matching's weight. If we associate a boolean decision variable
with each
e P E
, we can translate the problem into a linear program:
x
ÿ
cexeÑ max
ePE
ÿ
txeu
,
xeď 1, k P I Y J,
ePδpkq
xeP t0, 1u, e P E.
e
Due to total unimodularity of the constraint matrix, the last requirement,
relaxed to
xeě 0
, so the problem can be solved by means of linear programming. A more
specic approach was devised by Harold Kuhn in 1955 and was called
xeP t0, 1u
, can be
a Hungarian algorithm
by the author. In this section, we, rst, describe the algorithm for the case of complete bipartite
graph with equal parts (that is, we have equal numbers of tasks and workers, and each worker
can be assigned to perform any task), next, we demonstrate how it can be applied to solve the
general maximum-weight matching problem.
6.3.1. Hungarian Algorithm
Consider a linear program:
n
n
ÿ
ÿ
zpxq “
i“1
n
ÿ
xij“ 1, j “ 1, . . . , n,
i“1
n
ÿ
xij“ 1, i “ 1, . . . , n,
j“1
xijě 0, i, j “ 1, . . . , n,
cijxijÑ max
j“1
txiju
,
(6.1)
(6.2)
(6.3)
(6.4)
and its dual counterpart:
wpu, vq “
ui` vjě cij, i, j “ 1, . . . , n.
Note that dual variables
ui,
vjare unrestricted in sign, since corresponding constraints (6.2)
(6.3) of the primal problem are equalities.
n
ÿ
i“1
ui`
n
ÿ
vjÑ min
j“1
tui,vju
,
(6.5)
(6.6)
Assignment Problem
75
Lemma 6.3.
Let
tuiu
n
i“1
,
tvju
n
be some vectors from
j“1
R
n
, and
x “ txiju
n
be a feasible
i,j“1
solution for (6.1)(6.4). The value of assignmentxfor the problem with weights
from its value for the problem with weights
Proof.
Therefore, the two value dier by
Corollary 5.
For every primal feasible solution,
n
ÿ
i“1
n
ÿ
j“1
¯cijxij“
“
n
n
ÿ
ÿ
pcij´ ui´ vjqxij“
i“1
j“1
n
n
ÿ
ÿ
cijxij´
i“1
j“1
n
ř
n
ÿ
i“1
ui`
i“1
An optimal solution of problem (6.1)(6.4) with weights
solution for the problem with weights
Lemma 6.4.
vectors from
1.
¯cij“ cij´ ui´ vjď 0
R
Let
x “ txiju
n
. If
n
be a feasible solution of (6.1)(6.4), and
i,j“1
for each
i, j “ 1, n
¯cij“ cij´ ui´ vjby a constant value.
n
ui´
ř
j“1
n
n
ÿ
ÿ
i“1
vj.
n
ÿ
j“1
cijxij´
n
ÿ
i“1
n
ÿ
u
i
j“1
vj, which is independent onx.
j“1
¯cij“ cij´ ui´ vj.
, and
n
ÿ
ÿ
xij´
j“1
v
j
i“1
cijis also an optimal
n
tuiu
i“1
cijdiers
n
x
ij
,
tvju
n
j“1
be
2.
xij“ 1
thenxis an optimal solution of (6.1)(6.4), and its value is
Proof.If¯cijď 0
and equals 0, due to the second condition. That is,xis an optimal solution for the problem
with
¯cij, and, therefore, an optimal solution for the initial problem of value
Remark
(6.1)(6.4) and (6.5)(6.6): its rst condition states that
when
¯cij“ 0
, then the value of solutionxfor the problem with weights
,
z “
n
ř
i“1
ui`
n
ř
j“1
vj.
¯cijis non-positive
n
ř
i“1
ui`
n
ř
j“1
vj.
6.2.The statement of Lemma 6.4 represents a duality connection between problems
tuiu
and
tvju
is a dual feasible solution,
and the second is, in fact, a complementary slackness condition.
Lemma 6.4 gives us an idea how to solve the problem: we have to nd vectorsuandvsuch
that all the elements of matrix
||¯cij||
are non-positive, and there is at least one zero in each row
and column (in this case, a primal feasible solutionxthat satises the second condition exists).
To nd these vectors, we start with some dual feasible pair
graph with matrix
only if
¯cij“ 0
||¯cij||
: both parts of the graph includennodes, and edge
. Clearly, if the graph has a perfect matching, then the matrix has at least one
zero in a line, and the process stops. If such matching doesn't exist, we change
that leaves the pair feasible and increases the number of zeros in
pu, vq
||¯cij||
and associate a bipartite
pi, jq
exists if and
pu, vq
in a way
.
The correctness of the algorithm follows from the statements below.
Lemma 6.5.
Proof.
is,
therefore,
Each node of
`
|V
| ě |V
1
|V
Each time when Step 1 terminates, the inequality holds:
`
V
is incident to an edge of
`
|
. Moreover, at least one node of
2
`
| ą |V
1
2
`
|
.
2
M˚that has its other endpoint in
`
V
has label `*' and is not incident to
1
|V
`
1
| ą |V
`
|
.
2
`
V
. That
1
M˚,
76
Matchings and Assignments
Algorithm 6.2
Step 0.
Initialization.
0.1.
Find
0.2.
Set
V2, Eq
0.3.
Set
Step 1.
Primal Step.
1.1.
Given a set of edges
labeled nodes
scanning nodes from
1.2.If|M˚| “ n
go to Step 2.
Step 2.
Dual Step.
2.1.
Calculate
2.2.
Change dual variablesu,v:
The Hungarian (
u, v P Rnsuch that
Primal-Dual
) Algorithm for the Assignment Problem
¯cij“ cij´ ui´ vjď 0
V1“ t1, . . . , nu,V2“ t11, . . . , n1u
such that
`
V
“ H,V
1
pi, jq P E ô ¯cij“ 0
`
“ H,M “ H
2
.
.
E “ tpi, jq : ¯cij“ 0u
`
V
`
,
V
1
. Use algorithm 6.1 to construct a maximal matching
2
`
V
and
1
`
V
rst.
2
then Stop. Otherwise, set
δ “ mint´¯cij: i P V
`
, j P V2zV
1
ui“ ui´ δ, i P V
vj“ vj` δ, i P V
for all
i, j P 1, n
.
and construct a bipartite graph
, an initial matching
M
M “ M˚, save current labels
`
u
.
2
`
,
1
`
,
2
G “ pV1Y
and sets of
M˚Ď E
`
V
`
,
V
2
, and
1
and return to Step 1.
Lemma 6.6.
Proof.
From the proof of Theorem 6.1, there is no edge inEthat connects
That is,
Lemma 6.7.
¯cijă 0
At Step 2,
for
i P V
δ ą 0
`
1
and
At the end of Step 2 values
¯cijÐ ¯cij, i P V
¯cijÐ ¯cij, i P V1zV
¯cijÐ ¯cij´ δ, i P V1zV
¯cijÐ ¯cij` δ, i P V
.
j P V2zV
`
.
2
¯cijare changed as follows:
`
, j P V
1
`
, j P V2zV
1
`
2
`
, j P V2zV
1
`
, j P V
1
,
`
2
`
2
`
,
2
,
,
and the new solutionu,vis dual feasible.
Proof.
when
Infeasibility of the new solution yields positivity of some
i P V
`
and
1
j P V2zV
`
2
, but
δ “ mint´¯cij: i P V
`
, j P V2zV
1
¯cij. Note that
`
u
, therefore no
2
positive (though, at least one of the values becomes zero).
The following statement allows to reduce time complexity of the algorithm.
`
V
and
1
V2zV
`
2
¯cijincreases only
¯cijbecomes
.
Lemma 6.8.
The labels of
`
V
and
1
`
V
obtained at the end of the primal step (when no aug-
2
menting path can be found) remain valid after the dual step is completed, and can be used in
the next iteration.
Assignment Problem
77
Proof.
In the dual step, weights
¯cijdo not change for
i P V
`
and
1
j P V
`
. Therefore, all the
2
edges connecting these nodes remain present in the graph after the recalculation of weights,
and the labels remain valid.
From lemmas 6.5, 6.6, 6.7 follows the statement.
Corollary 6.
after changing
At each iteration, the dual objective
uiand
vjat the dual step.
n
ř
i“1
ui`
n
ř
vjincreases by
j“1
δp|V
`
1
|´|V
`
|q ą 0
2
Lemmas 6.7 and 6.8 give us the following.
Corollary 7.
The cardinality of
Lemma 6.9.
Proof.
Since the cardinalities of
After changing
`
V
grows at each iteration by at least one.
2
uiand
vjat the dual step, at least one new edge is added toE.
Algorithm 6.2 solves the assignment problem (6.1)(6.4) in
M
and
`
V
increase each iteration, the algorithm stops after
2
Opn4q
time.
at mostniterations. At each iteration, during the primal step, a number of augmenting paths
are constructed. To nd a single path we need to perform
paths, obviously, does not exceedn, that gives us
1. The dual step takes
the overall complexity is
Opn2q
Opn4q
operations, so, the complexity of a single iteration is
.
Opn|E|q “ Opn3q
Op|E|q
operations, the number of
as the complexity of Step
Opn3q
, and
Example 6.2.
[2] Solve the following instance of the assignment problem:
n “ 4
, the weights
are given in the matrix below.
¨
˛
27 17 7 8
||cij|| “
˚
14 2 10 2
˚
˝
12 19 4 4
‹
‹
‚
.
8 6 12 6
First, we should nd an initial dual feasible solutionu,v. The easiest way to do it is to set
ui“ 0,vj“ maxicij. That is,
u0“ p0, 0, 0, 0q,v0“ p27, 19, 12, 8q,w0“
¨
˛
4
ř
i“1
u
4
ř
0
`
i
j“1
v
0
j
“ 66
,
0 ´2 ´5 0
||¯c
0
ij
|| “
˚
´13 ´17 ´2 ´6
˚
˝
´15 0 ´8 ´4
‹
‹
‚
.
´19 ´13 0 ´2
Note that all the elements of the second row are negative. That means, the initial solution can
be improved by setting
1
u
“ maxj¯c
i
0
:
u1“ p0, ´2, 0, 0q,v1“ p27, 19, 12, 8q,w1“ 64
ij
¨
˛
,
0 ´2 ´5 0
||¯c
1
ij
|| “
˚
´11 ´15 0 ´4
˚
˝
´15 0 ´8 ´4
‹
‹
‚
.
´19 ´13 0 ´2
The graph corresponding to
tp1, 11q, p3, 21q, p4, 31qu
is drawn bold.
1
||¯c
||
if given below in the g.6.5-a. Maximal matching
ij
M1“
78
1
2
3
4
1’
2’
3’
4’
*
2
3’
1
2
3
4
1’
2’
3’
4’
*
2
3’
4
1
2
3
4
1’
2’
3’
4’
*
2
3’
4
Matchings and Assignments
(a) Iteration 1
(b) Iteration 2
(c) Perfect matching
Figure 6.5: Implementation of the Hungarian algorithm
The rst step nishes with
`
V
, j P V2zV
1
`
u “ mint11, 15, 14; 19, 13, 2u “ 2
2
v2“ p27, 19, 14, 8q,w2“ 62
`
V
“ t21, 41u
1
and
,
¨
`
V
2
“ t2u
. Calculate
. Change dual variables:
˛
δ “ mint´¯c
1
ij
: i P
u2“ p0, ´4, 0, ´2q
0 ´2 ´7 0
||¯c
2
ij
|| “
˚
´9 ´13 0 ´2
˚
˝
´15 0 ´10 ´4
‹
‹
‚
.
´17 ´11 0 0
A new zero element appears in the matrix. We add edge
labeling process using labels from the previous iteration (g.6.5-b). Node
p4, 4q
to the graph and continue
41gets label4,
the node is exposed, hence an augmenting path is found. We switch the edges of the path
and get a new matching
gure 6.5-c.
M2is a perfect matching, so the algorithm stops. The optimal assignment is:
x11“ x23“ x32“ x44“ 1
M2“ tp1, 11q, p2, 31q, p3, 21q, p4, 41qu
, its value is 62.
. The matching is shown in the
,
6.3.2. Maximum-Weight Matching Problem
Now, consider the problem of nding a maximum-weight matching in a complete bipartite graph
G “ pV1Y V2, Eq
k ą 0
. Consider a graph
nodes
i1, i2, . . . , ikand
with unequal parts:
G1“ pV1Y V
E1includes edges
|V1| ą |V2|
1
, E1q
2
, and non-negative weights. Let
, where
1
V
is obtained from
2
|V1| “ |V2| ` k
V2by addingkdummy
pv, ilq,l “ 1, . . . , k,v P V1, in addition to edges ofE.
,
Weights of these dummy edges are equal to zero. Now, we have a complete bipartite graph with
parts of equal sizes. Finding a maximum-weight perfect matching in
G1and removing dummy
nodes and edges from the solution constructed by algorithm 6.2 gives us a maximum-weight
matching for the initial graph.
Similarly, by adding dummy edges the problem of nding a maximum-weight matching in
incomplete
bipartite graph can be transformed into the problem for complete graph with equal
parts.
6.4. Exercises
6.1.
Find a maximal matching in the graph below starting from the matching drawn bold.
Exercises
1 2
3
4
5 6
7
8 9 10
11 12
6.2.
Solve the following instance of the assignment problem:
the matrix below.
¨
n “ 5
˛
, the weights are given in
38 35 48 35 30
||cij|| “
˚
40 33 44 43 37
˚
˚
33 41 42 44 38
˚
˝
44 36 31 39 41
‹
‹
‹
‹
‚
45 35 38 40 34
6.3.
Find a maximum-weight matching in a bipartite graph with weight matrix given below.
¨
˛
19 21 32 17 21
||cij|| “
˚
22 20 21 14 24
˚
˚
17 16 21 20 25
˚
˚
18 15 22 18 19
˚
˝
11 14 24 16 24
‹
‹
‹
‹
‹
‹
‚
21 20 24 21 20
79
6.4.
Find a
minimum
-weight matching in a bipartite graph with weight matrix given below.
¨
˚
˚
˚
˚
˝
10 9 7 3 9
6.5.
Solve the instance of TSP (
||cij|| “
minimization
assignment problem as lower bounds in B&B.
3 6 6 8 9
˛
8 7 4 4 10
2 3 2 2 6
‹
‹
‹
‹
‚
3 7 5 5 10
problem) form example 5.2 using solutions of the
Chapter 7.
Introduction to Game Theory
Models and methods of Game Theory proved themselves very useful in qualitative analysis of
conict situations. In this chapter, a basic game-theoretic model, a matrix game, is consid-
ered. We introduce concepts of pure and mixed strategy Nash equilibrium and discuss several
approaches to nd equilibria in games of dierent sizes.
7.1. Concept of a Game
A
game
is an activity that involves two or more participants, or
which aect all of them and result in some gain (positive, negative, or equal to zero) to each
player. Each participant has a predened set of available actions, called
strategies
one of them independently of other players. A game is said to
their strategies and present them simultaneously to each other. A collection of strategies chosen
by players is called an
they consider the outcome and evaluate it according to their utility functions. Each player
would like the game to have an outcome of maximal utility to him/her (though he cannot
aect the whole outcome but only a single component of it).
That is, a game is characterized by:
(as opposed to
mixed
outcome
strategies to be discussed further), and makes a decision to use
of the game. When all the players have made their decisions,
players
occur
, who perform actions,
strategies
, when all the players choose
, or
pure
1.
a set of playersN “ t1, 2, . . . u
a two-person game
2.
sets of strategiesXifor player
sets are nite, the game is called
3.
utility functionsui: X Ñ R
measures the utility, or gain, of an outcome
uipxq
, the better isxfori. A two-person game is called
is loss for the opponent, and, therefore, the players' purposes collide: one of the players
acts to maximize his gain, and the opponent aims to minimize it.
In this chapter, we discuss nite two-person antagonistic games. Such games are often called
matrix
the rst player), which, due to niteness of strategy sets, can be represented by a matrix.
games, for they are completely described by a single utility function (usually, the one of
In further discussion, we denote
, if
|N| ě 3
, which is assumed to be nite. If
, it is ann-person game
i P N
, which can be nite or innite. If all the strategy
nite
, otherwise, it is an
, where
X “
ś
iPN
;
Xiis
x P X
innite
the set of outcomes
for player
antagonistic
|N| “ 2
game;
i P N
, if gain of each player
, the game is
. Value
: the greater is
uipxq
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