J. Robinson

English

Research Papers in Economics

Annals of Mathematics
An Iterative Method of Solving a Game
Author(s): Julia Robinson
Source: The Annals of Mathematics, Second Series, Vol. 54, No. 2 (Sep., 1951), pp. 296-301
Published by: Annals of Mathematics
Stable URL: http://www.jstor.org/stable/1969530
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http://www.jstor.orgANNALS  OF  MATHEMATICS 
Vol. 54, No.  2, September,  1951 
AN ITERATIVE  METHOD  OF  SOLVING  A GAME 
BY JULIA  ROBINSON 
(July  28,  1950) 
A two-person  game' can be represented  by  its  pay-off  matrix  A =  (aij). The 
first  player  chooses  one of  the m rows  and the second player  simultaneously 
chooses  one  of  the  n columns.  If  the  ith row  and the  jth  column  are chosen,  then 
the second  player  pays the first  player  aij. 
If  the  first  player  plays  the  ith row  with  probability  xi  and the  second  player 
plays  the  jth  column  with  probability  yj , where  xi _ 0, EZx =  1,  yj >  0, and 
Eyj  =  1, then the expectation  of the first  player is Zaijxiyj.  Further- 
more, 
(1)  mmin  Eaixi  <  max  Eaiyj,, 
j  i  i  j 
since 
min  Eaij  xi <  E Eai  xi  Ayj  <  max Eaij  y,. 
j  i  i  i  i  j 
The minimax  theorem  of  game  theory  (see [1]  page 153) asserts  that  for  some 
set  of probabilities  X  =  (x1 , ***,  X,)  and  Y  =  (yi ,  .**,  yn) the  equality 
holds  in (1). Such a pair (X, Y) is called a solution  of  the  game. The value v 
of  the game is defined  by 
v  =  min  Eai  xi  = max  Eaijyj, 
i  i  i  j 
where  (X, Y) is a solution  of  the game. 
In this  paper,  we shall  show  the  validity  of  an iterative  procedure  suggested 
by George  W. Brown  [2]. This method  corresponds  to each player  choosing  in 
turn  the best pure strategy  against the accumulated mixed strategy  of his 
opponent  up to then. 
Let A =  (aij) be an m X n matrix.  Ai. will  denote  the  ith row  of  A and A.j, 
the  jth  column.  Similarly,  if V(t) is a vector,  then  vj(t) is the  jth  component. 
Let max V(t) =  maxj vj(t) and min V(t) = mini  vj(t).  In this  notation,  (1) can 
be rewritten  as follows: 
(2)  min  EA .xi <  max  EA.jyj, 
i  .7 
whenever  xi_  0, Exi  =  1, y,  _  0,  and Zyj  =  1. 
DEFINITION  1. A system  (U, V) consisting  of  a sequence of  n-dimensional 
vectors U(O), U(1),  *  *  *  and a sequence of  m-dimensional vectors V(O), V(1),  * 
is called a vector  system  for  A provided  that 
min U(O)  =  max V(O), 
1 More  technically,  a finite  two-person  zero-sum  game.  See [1]  in  the  bibliography  at the 
end  of  the  paper. 
296 AN  ITERATIVE  METHOD  OF  SOLVING  A  GAME  297 
and 
U(t +  1) =  U(t)  +  Ai.,  V(t +  1) =  V(t) +  A.j, 
where  i and  j satisfy  the  conditions 
vi(t) =  max V(t),  uj(t) = min  U(t). 
Thus a vector  system  for  A can be formed  recursively  from  a given U(O) 
and V(O). At each step,  the  row  added to U is determined  by  a maximum  com- 
ponent  of V and the column  added to V is determined  by a minimum  com- 
ponent  of  U. 
An alternate  notion  of vector  system  is obtained if the condition  on j  in 
Definition  1 is replaced  by 
uj(t +  1) =  min  U(t +  1). 
A vector  system  Qf  this  new  type  can also be built  up recursively.  The only  dif- 
ference  is that  here  successive  U and V are determined  alternately  while  in  the 
other  definition  U and V could  be obtained  simultaneously.  In all the  following 
proofs  and theorems,  either  definition  may be used. 
In the  special case U(O) =  0 and V(O) =  0, we see that U(t)/t  is a weighted 
average  of  the  rows  of  A and V(t)/t  is  a weighted  average  of  the  columns.  Hence 
for  every  t  and t', 
min  U(t) <  v <  max  V(t') 
t  ti 
If  for  some  t  and t',  these  two  bounds  are  equal, we  have a solution  of  the  game. 
Unfortunately,  this  is not  always the case. However  George  Brown  [2]  conjec- 
tured  that  as t  and t' tend  to co,  the  two bounds approach  v.  The main  result 
of  this  paper is to prove  this  for  any vector  system.  In numerical  examples, 
vector  systems  of  the second kind appear to converge  more  rapidly  than the 
first. 
2 
THEOREM.  If (U,  V) is a vector  system  for  A, then 
lim mm  U(t)  max V(t)  lim  ~  lim  =V. 
t boo0  t  t ?*0  t 
The proof  will  be divided  into  four  lemmas. 
LEMMA  1.  If (U, V) is a vector  system  for  a matrix  A, then 
lim  inf  max V(t) -  min U(t) >  0. 
t-boo  t 
PROOF.  For each t, 
V(t) =  V(O) +  tEyjA.j  where  y  _ 0,  Ey j =  1, 
i 
2 The solution  to Problem  5 in  the  RAND Mathematical  Problem  Series  II is contained 
as a special case of  this  theorem. 298  JULIA ROBINSON 
and 
U(t) =  U(O) +  tExjAi.  where  xi _ O.  Exi  =  1. 
Hence 
max V(t) >  min  V(O) +  t  max  jZyA.;  >  min  V(O) +  tv, 
and 
min  U(t) <  max U(O) +  t  min  ExiAj.  <  max U(O) +  tv. 
Therefore, 
lim  inf  max V(t) -  min  U(t) >  0. 
t-.oo  t 
DEFINITION  2. If (U, V) is a vector  system  for  A, then  we say that the  ith 
row  is eligible  in  the  interval  (t,  t') provided  that  there  exists  ti  with 
t <  tl <  t' 
and 
vi(t1)  =  max V(t1). 
Similarly,  the  jth  column  is eligible  in  the  interval  (t,  t') if  there  exists  t2  with 
t  t2 ?  t' 
and 
uj(t2) =  min  U(t2). 
LEMMA 2. Given  a vector  system  (U, V) for  A, then  if  all the  rows  and columns 
of  A are  eligible  in the  interval  (s, s +  t), 
max U(s  +  t) -  min U(s  +  t) <  2at 
and 
max V(s  +  t) -  min V(s  +  t) <  2at, 
where 
a  =  maxi,j | aij 
PROOF.  Let j be such that 
uj(s +  t) =  max U(s +  t). 
Choose t'  with  s ?  t' <  s +  t  so that 
uj(t')  =  min U(t'). 
Then 
uj(s +  t) <  uj(t') +  at =  min  U(t') +  at, AN  ITERATIVE  METHOD  OF  SOLVING  A  GAME  299 
since  the  change  in  the  ith  component  in  t  steps  is not  more  than  at.  But 
min  U(s +  t) >  min  U(t') -  at, 
so that 
max U(s +  t)  -  min  U(s +  t) < 2at. 
Similarly, 
max V(s +  t)  -  min  V(s +  t) < 2at. 
LEMMA  3. If all the  rows  and columns  of  A are  eligible  in (s, s +  t)  for  a given 
vector  system  (U, V), then 
max V(s +  t)  -  min  U(s +  t) _ 4at. 
PROOF.  By Lemma 2, 
max V(s +  t)  -  min U(s +  t) < 4at  + min  V(s +  t)  -  max U(s +  t). 
Hence  it  is  sufficient  to  show  that  min  V(s +  t) < max U(s +  t).  Now applying 
(2) to the transpose  of  A, we have 
min IA.,yj  < max  A  i.xi, 
i  i 
whenever  xi _ 0, E  xi =  1, yj >  0, and E  yj =  1. In particular,  choose  xi 
and yj  satisfying 
U(s  + t)  = U(O)  + (s  + t)  ZAi.xi, 
V(s +  t) =  V(O) +  (s +  t)Z  A.yj,. 
Then 
min  V(s +  t) < max V(O) +  (s +  t)  min  A.jyj 
?  min  U(O) +  (s +  t)  max  A  i.xi 
< max U(s +  t). 
LEMMA 4. To every  matrix  A and e >  0, there  exists  to  such  that  for  any vector 
system  (U, V), 
max V(t)  -  min  U(t) <  et  for  t  _  to. 
PROOF.  The theorem  holds  for  matrices  of  order  1 since U(t) =  V(t) for  all 
t. Assume  the theorem  holds for  all submatrices  of  A, then  we will  show by 
induction  that  it  holds  for  A. Choose t*  so that  for  any  vector  system  (U',  V') 
corresponding  to a submatrix  A' of  A, we have 
max V'(t)  -  min  U'(t) <  'et  whenever  t  >  t*. 
We shall prove  that if  in the given  vector  system (U, V) for  A, some row 
or  column  is not  eligible  in  the  interval  (s, s +  t*),  then 
(3)  max V(s +  t*)  -  min  U(s +  t*)  <  max V(s)  -  min  U(s) +  2Ft*. 300  JULIA ROBINSON 
Suppose,  for  example,  that  the  kth  row  is  not  eligible  in  the  interval  (s, s +  t*). 
Then we can construct  a vector  system  (U',  V') for  the matrix  A' obtained 
by deleting  the kth  row  of  A, in the  following  way: 
U'(t) =  U(s +  t) +  C, 
V'(t) =  ProjkV(s  +  t)  for  t = 0,1,  **  , 
where  C is the n-dimensional  vector  all of whose components  are  equal to 
max V(s)  -  min U(s) and ProjkV  is the vector  obtained from  V by omitting 
the kth  component.  The rows  of  A' will  be numbered  1,  2, *  , k  -  1, k +  1. 
m. Notice first  that  min  U'(0)  =  max V'(0). Furthermore,  if 
U(s+t+  1)=  U(s+t)+AA.,  V(s+t+  1) =  V(s+t)+A.;, 
then 
P (t  +  1) =  U(t) + Ai.  V'(t  +  1) =V'(t)  + A.j. 
Also  vM(s  +  t) =  max V(s +  t)  if  and only  if  v'(t) =  max V'(t) and uj(s +  t) = 
min U(s +  t) if  and only  if  u'(t)  min  U'(t) for  0 <  t <  t*  .  Hence we see 
that U' and V' must  satisfy  the recursive  restrictions  of the definition  of a 
vector  system  for  0 <  t <  t*, since U and V do. Naturally,  we may continue 
U' and V' indefinitely  to form  a vector  system  for  A'. 
Now by  the  choice  of  t*  , we  know  that 
max V'(t*) -  min  U'(t*) <  1et*  . 
Hence 
max V(s +  t*)  -  min  U(s +  t*) 
=  max V'(t*) -  min  U'(t*) +  max V(s)  -  min  U(s) 
<max  V(s)  -  min U(s) +  let*. 
We can now  show  that  given  any vector  system  (U, V) for  A, 
max V(t) -  min  U(t) <  et  for  t _ 8at*/e. 
Consider  t >  t*  . Let 0 with  0 ?  0 <  1 and q a positive  integer  be so chosen 
that t =  (0 +  q)t*  . 
CASE  1. Suppose  there  is  a positive  integer  s ?  q  so that  all rows  and columns 
of A are eligible  in the interval  ((0 +  s -  1)t*  , (O  +  s)t*). Take the largest 
such s, then 
(4)  max V(t) -  min U(t) 
-  max V((0 +  s)t*) -min  U((0 +  s)t*) +  le(q -  s)t*. 
We obtain  this  inequality  by repeated  application  of (3), since  in each of  the 
intervals 
((0 +  r-  1)t*, (0 +  r)t*)  for  r =  s +  1, *  *  q, AN  ITERATIVE  METHOD  OF  SOLVING  A  GAME  301 
some  row  or  column  of  A is not  eligible.  From Lemma  3 and the  choice  of  s,  we 
have 
(5)  max V((O +  s)t*) -  min  U((O +  s)t*) ?  4at*. 
From (4) and (5), we obtain 
max V(t)  -  min U(t)  <  4at* +  1e(q -  s)t* <  (4a  +  2eq)t*. 
CASE 2. If  there  is  no  such  s,  then  in  each  interval  ((0 +  r -  1)t*  , (O  +  r)t*) 
some  row  or  column  of  A is not  eligible.  Hence 
max V(t) -  min  U(t) <  max V(Ot*)  -  min U(Ot*)  +  leqt* <  2aOt*  +  leqt*. 
Therefore,  in either  case, 
max V(t) -  min  U(t) <  (4a +  'eq)t* <  4at*  +  let <  et  for  t _ 8at*/e. 
From  Lemmas 1 and 4, we see that 
lim  max  V(t) -  min  U(t) 
t- oo  t 
But from  (1), 
lim  sp  min  U(t) < 
t- oo  t 
lim  inf  max  V(t) >  v. 
tf-oo  t 
Hence 
lm  nin  V  (t) =lmmax  V(t) 
t  b  oo  t  t  bCoo  t 
which  completes  the proof  of  the theorem. 
THE  RAND  CORPORATION, 
SANTA MONICA,  CALIFORNIA 
BIBLIOGRAPHY 
[1] J. VON NEUMANN  and 0.  MORGENSTERN,  Theory  of  Games and Economic  Behavior, 
Princeton  University  Press. 
[2] G.  W.  BROWN,  Some notes on computation of Games Solutions, RAND  Report P-78, 
April  1949,  The RAND Corporation,  Santa Monica,  California. 

Some notes on computation of Games Solutions,

An Iterative Method of Solving a Game

http://www.dklevine.com/archive/refs4422.pdf

An Iterative Method of Solving a Game

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