OR 4350 Game Theory
March 1, 2012
Prof. Bland
A Complementary Pivot Algorithm to Compute a Nash Equilibrium in
Bimatrix Games
an abbreviated version of notes by Prof MJ Todd
Consider the bimatrix game (
A,B
) with
A
and
B m
×
n
matrices. Then (¯
p,
¯
q
)
∈
P
×
Q
is a Nash equilibrium iﬀ
pA
¯
q
≤
¯
pA
¯
q
for all
p
∈
P,
and
¯
pBq
≤
¯
pB
¯
q
for all
q
∈
Q,
or equivalently,
(
i
)
A
i
¯
q
≤
¯
pA
¯
q
for all
i,
(
ii
)
¯
pB
j
≤
¯
pB
¯
q
for all
j.
Note that, if we add a constant to every entry of
A
to get
A
0
, and a constant to every
entry of
B
to get
B
0
, then (¯
p,
¯
q
) is a Nash equilibrium of (
) iﬀ it is one for (
A
0
,B
0
). This
is because for every
p
∈
P
,
q
∈
Q
,
pA
0
q
is more than
pAq
by the same constant used to form
A
0
from
A
; and similarly for
B
0
and
B
. So we’ll assume that all entries of
A
and
B
are
nonnegative, with a positive entry in each column of
A
and in each row of
B
. This will en-
sure that ¯
pA
¯
q
and ¯
pB
¯
q
will be positive for any Nash equilibrium: ˜
p
= (1
/m,
1
/m,.
..,
1
/m
)
T
guarantees a positive payoﬀ for I for any
q
, and hence for ¯
q
, so a best response like ¯
p
guar-
antees at least as much, and similarly for
B
.
Example:
We take
A
=
"
5 1 1
1 1 2
#
,
B
=
"
0 1 3
2 1 0
#
.
(We could eliminate player II’s second column, since it is strongly dominated by
q
=
(3
/
5
,
0
,
2
/
5)
T
, but we’ll leave it in to better illustrate the algorithm.)
The pair (¯
p
= (2
/
5
,
3
/
5)
,
¯
q
= (1
/
5
,
0
,
4
/
5)
T
) is a Nash equilibrium and gives payoﬀs 9
/
5
and 6
/
5 to players I and II respectively. We’ll ﬁnd this using our complementary pivoting
algorithm below.
We want to ﬁnd a solution to (i) and (ii) above. Unfortunately, while the left-hand sides
are linear in ¯
p
and ¯
q
, the right-hand sides aren’t. Consider instead:
There is some
α >
0,
β >
0 such that
(
i
0
)
A
i
¯
q
≤
α
for all
i
, with equality if ¯
p
i
>
0
,
(
ii
0
)
¯
pB
j
≤
β
for all
j
, with equality if ¯
q
j
>
0
.
1