F I NI TE
T E R MI NA T I 0 N
WI T H
GAME S
T I E
by
P E T E R
G.
HINMAN
Abstract
We consider the class of finite two-person games with
perfect information in which the last player who can make a
legal move either wins or ends the game in a tie.
We define
an equivalence relation over this class and exhibit a complete set of representatives for the equivalence classes
defined in terms of one-pile Nim games.
AMS 1970
subject classification 90 D05
We shall consider in this paper a class of two-person
games with perfect information in which the players play alternately, at each turn select a move from a finite set of alternatives, and complete a play of the game after finitely many
moves.
If we add the condition that the first player who cannot
make a legal move loses, then every such game may be represented
as follows.
To each
a E G
I must choose some
corresponds the game
a1 E a, player II some
until some player chooses
¢
in which player
9a
a 2 E a1 , and so on
and thus wins the game.
If we
think of a position in a game as being the set of all permissible
succeeding positions, it is clear that any game of the type
described is equivalent to some
The theory of such games
is well-developed (cf. [Ho]).*
We shall enlarge the class of games considered to allow
for the possibility of a tie.
Now a player may be faced with a
position from which he can make no legal move either because he
has lost or because the game has ended in a tie.
all be represented in the form
Sra
for
a E H
Such games can
defined as
follows.
Definition 2.
H0 = [*} ,
Hk+ 1 =[a:
a~
Hk}
In this paper we provide an analysis of such games.
As
motivation for interest in these games, we note that chess is a
*
We are grateful to Dana Scott for suggesting this
problem and this formulation of it.
- 2 finite termination game with tie.
We first define recursively the value
1*1
Definition 3.
\¢1
= 0 ,
Jal
of any
a E H.
= -1 , and otherwise
Jal =-min (lxl : x E a}
Just as in the case of win-lose games,
player I has a winning strategy in
Ia I
iff
= +1
and Ia\ = -1 iff
9a
If lal = 0 , either player
player II has a winning strategy.
can force the game to end in a tie.
We call attention to the
following obvious facts which will be used repeatedly without
reference.
Lemma 4.
For any
a E H
,
a
(i)
lal = +1 <-> ax(x E a
(ii)
la I =
I a I = -1
0
(iii)
Definition 5.
(i)
a ®
I
.
\X I
= -1)
I xl > 0).
<-> Vx(x E a ->
<-> vx(x E a -> lxl
For any
*= *®
*;
= (x
if
® b
a~ b <-> vc E H(!a Ef> cl
a
I * I
r,
';::) aEBb
'
:3:x(x E a •
IxJ = 0);
+1).
b
x E a1 U (a Ef> y
y E b} ;
= lb Ef> cJ).
Of course it is immediate that
The game
=
.
a,b E H:
b =
a ® b
(ii)
*
H is closed under
is played by play1· ng
Ef> •
simul-
taneously with each player at his turn moving in one game or
the other.
The following properties of
derived from the definitions.
Ef)
and
~
are easily
- 3 Lemma
---
6.
(i)
Ef)
is commutative and associative;
(ii)
,..,;
is on equivalence relation;
(iii) a
Ef>
.,
¢ =a
Ib 1
.
c ,... b
Ef>
IaJ =
(iv)
a ""' b ->
(v)
a,... b -> a
Ef>
'
c.
In the analysis of win-lose games, the Nim games play an
important role.
With the usual set-theoretic representation of
a natural number
m as
one-pile Nim with
(0,1, ••• ,m-11
which start with
n •
defined by
and
m2
m is
Vc E G ..• )
m1 , ••• ,mk
a E G is
to a unique natural
There is even a simple process known as Nim
tion for computing
m1
piles having, respectively,
k
(""'G
~
More interesting are games ("'
'JID1<f> •• $Jlk
The main result of [Ho] is that every
equivalent
number
and
m stones: each player may remove any number
of stones from the pile.
stones.
mEG
9
n
such that
m1
Ef>
~­
m2 ,... n- one writes
in binary notation and adds each column modulo 2.
Our aim here is to provide a similar set of representatives for
the equivalence classes of
Lemma 7.
Proof.
For any natural number
For
m
=0
mEf> m
For each
H.
n < m
,
f
lm
this is immediate.
=
[m ® n
.n
< ml
nEen E mEe n
induction hypothesis, so
Corollary 8.
m ,
jm<f>_nl
(i)
m
n -> lm E9 nl
(ii)
m ""' n -> m
=n
•
= +1
;
mJ
For
= -1.
m> 0
'
•
and
= +1.
For any natural numbers
$
In
Ef>
nl
=
Hence
m and
lm
n
by the
-1
(f)
ml
=
-1.
- 4 Lemma 9.
For any
a,b E H and any natural number
(i)
a ..... m -> !a®m\ = -1
(ii)
*
E a -> Ia ® bl > 0
(iii) m E a -> \a® ml = +1
(iv)
as w , m
( i)
Proof.
* E a
.
min {{x\
*
from Lemma 7.
element
.
' { *}
then
'
.$.
0
Ic I
then
Since
so
(iii) is immediate
(a ® bl ?. 0 •
m~ a
aS w and
n
®
for
then
n < m
..
'
IcI
= +.1 by
then
I cl
H = w U (c U (*) : c
Definition 10.
~
S(i)
Proof.
~
For any
Suppose
a E H
a,b E H , if
a = b •
If
a ..... b •
diction.
The remaining case is
b , there exists
Then by
Hence
9(iii)
a
f.
Let
that
X
E a
finite} .
'
If
a = c U (*1
\a® b\ = -1 , but by 9(ii),
I
Ia U (*1 EB ml = 0.
a ..... b, then
by 9(i),
a
'
a EH
is
We show uniqueness first.
a,b E H and
from S(ii) that
If
w, c
.
by (ii)
> 0
H and we aim to show that every
equivalent to a unique
Lemma 11.
Any
0
.
Hence the conditions of 4(ii) are satisfied so
H
,
I*® bl = 0
by definition
= 0
n E a
for
c = a U (*1
Thus
(ii) If
a U {*} ® m is of one of the following forms:
c = * EB m
c = nEE>m
mI = 0 •
(f)
Ia ® ml = \m ® m( = -1.
(iv) Assume that
of
c
'
b E a® b •
~
E a ® b}
X
.
-> Ia U
a,..., m
If
then
'
*a
m
a= c
a= b •
a,b E w , it follows
and
Ia ® b\ ~ 0
u (*]
and
m such that (say)
Ia EB ml = +1
b E w , then
but by 9(iv)
mE a
,
a contrab = d U (*}.
but
m ~ b.
\b ® ml = 0 •
b, which contradicts the hypothesis.
p(a)
be the smallest
implies
p(x) < p(a)
k
.
such that
a E Hk •
Note
- 5 -
Lemma 12.
(i)
For any
= +1 , lbl = 0 ->
I a l = -1 -> fa$ bl =
Ja $ bl
lbl
~
.,
0
•
We prove (i) and (ii) simultaneously by induction on
Proof.
p(a) + p(b) •
=*
Both are vacuous for
Hence we assume
= +1
lal
.
Ia!
(ii)
b
a,b E H
and
=0
lbl
=*
a
and obvious for
p(a) + p(b) > 0 •
Then for some
•
For (i) suppose
x E a ,
=
lxl
and
-1
p(x) + p(b) < p(a) + p(b) , so by (ii) of the induction hypothesis,
so
lx
bj
$
=0
lbf
=
Ia ® bl ' 0 •
Hence
•
min (lei
Fgr_(ii) suppose
Ia!
c
= -1
E
a$ b} ~
0
There are three
•
cases:
Case 1.
= +1
lbl
There exists
•
the induction hypothesis (ii)
a$ y E a$ b
Ia $ bl
Case 2.
•
'
=0
lbl
induction hypothesis,
with
y E b
have
$
$
.
lbl
y E b
'
Yl ' 0 .
Ia $ Yl •
IYI
with
IYI
=
Since
IYI = -1 •
~
,
0
By
-1 •
so by (ii) of the
There is at least one
Since
= -1
lal
,
all
x E a
so by (i) of the induction hypothesis,
lx ® bl ~ 0.
Ja
=0 =
IY\
= +1
lxl
!a
=
Ia $ Yl
= +1 =
For all
y E b
Hence
min (lei : c E a® b}
=0
, and
bl = 0 = lbl .
Case 3.
= -1
lbl
•
For all y E b
the induction hypothesis,
E a ,
all
X
and
Ia $ bl
lx $ bl
=
Corollary 13.
(i)
$
=
ml
lxl
= +1
= IYI
= +1
Hence
min
so by (ii) of
•
Likewise for
(I cl
:c E a$ b)= +1,
a E H and any natural number
-> a~¢ ,
= -1
•
= +1
lbl •
For any
= -1
lal
(ii) Ia
-1
=
Ia $ Yl
IYI
->a-m •
m
- 6 Proof. (i)
By 12 (ii) if
1¢ ~
Ia ~ bl = fbi =
{i) ,
~
a em
Lemma 7 ,
a$ m
bl •
~
(ii)
Proof.
m ,
Case 1.
axE a
hypothesis
b
ml = ja
Case 2.
~
...w
®
mX •
®
x E a
I
and
m .
a
a •
If
x E a ,
x E a
ayE b(lx ® y( = -1).
.QK
®
Then by the induction
yj = -1) •
fa® ml = -1 •
such that
\cl = +1
For every
~
p(a) + p(b) •
m •
Then
By the induction hypo-
there exists a natural number
n = mx •
mx
mx
is unique.
For every
such
Let
n < m there
By 9(i), lx $ nf = lx ® mX l=-1,·.
On the other hand, for every
lx ® mJ = Jmx ® ml = +1.
so
Jm ® bf = Ia $ bJ = -1
Theorem.
that
~
p[p ~ [mx : x E a1] •
m so by 8(i),
c E a® m ,
m •
b
Corollary 8 implies that
Ia ~ nl = +1 •
Hence
a~
so by 13 (ii) , a~ m also.
:Ry E b( Jx
We show first that
x E a
bl = -1)
bJ = -1
m = least
exists
,
Hence
bl = -1, then for some
m for some natural number
vx E a
x
~
But by
®
:Rz E x(lz $ bl = -1).
thesis, for each
mx
m and
~
Hence for all
az E x(lz
that
¢®m= m•
m® m- ¢ .
we have in particular that for all
= +1
fx ® bJ
®
a
Ia
so by
~
We prove the lemma by induction on
(a® bl = -1
Ia
a e m~ m
By transitivity of
Lemma14. Forany a;bEH, if
b ,
)a$ mf = -1
so by (i) again,
a®¢= a •
natural number
Assume
By Lemma 6
lm $ ml = -1
m
®
¢ .
Jal = -1 , then for any
a E H
ja ® ml = -1 •
so
x E a ,
Thus for every
Now by 13 (ii), a~ m
b ~ m also.
there exists a unique
a EH
such
- 7 Proof.
The uniqueness of
a
follows from Lemma 11.
easy to prove by induction that for all
and
~
*
Hence
x
(*}
exists for all
number
x E a •
=
x Ea ,
(x
If
x E w}
a
Ia ® bl ~ 0
H •
=0
•
p{a)> 0
m for some natural
~
Set
a .
a~
Let
b be an
It follows from Lemma 14 that
\a$
and from Lemma 9 (ii) that
it suffices to show that
bl
U (*}
a E H and it remains to show
arbitrary element of
$
Suppose now that
m we are done, so assume otherwise.
a
Then
* = (*}
and we set
1(*1
b
It is
\a~
bl = +1
b\ ~ 0 •
Ia
iff
® bl
Hence
= +1
•
We have
Ja
®
bl
= +1
<->
~z
E a(lz
<->
~X
E a(x E
<->
~X
E a(lx
<-> Ia ® bl
bl
®
w
'
b\
$
= +1
= -1)
lx
=
~ bl
= -1)
-1)
•
The first and fourth equvalences use again the fact that
for any
y ,
\a®
yJ ~ 0
and
Ia ®
valence is just the definition of
follows from
lx ~ bl
= -1
m E H ,
m
Yl
~ 0 •
a .
The second equi-
Half of the third
For the right-to-left direction, if
, then by Lemma 14,
= x-
x ~ m for some
m •
Since
by uniqueness.
Of course the proofs provide us with an algorithm for
computing
a
from
a.
It would be interesting to know if there
is a simple algorithm for computing
a~
b
for
a,b E H
similar to Nim additon.
We have no conjectures as to how much further (if any)
this method can be carried.
An obvious candidate is the class
of games obtained by adding a second atom with value
+1 •
- 8 -
Bibliography
[Ho]
John c. Holladay, Cartesian Products of Termination
Games, Contributions to the theory of games,
Ed. M. Dresher, A.W. Tucker, and P. Wolfe,
Annals of Math. Studies No. 39, Princeton
University Press, Princeton, N.J. 1957, 189-200.
Universitetet i Oslo
and
University of Michigan