On the Semigroup of Positional Games J. Mark Ettinger Los Alamos National Laboratory Mail Stop B-229 Los Alamos, New Mexico 87545 U.S.A.
[email protected] Abstract
We study the partially-ordered semigroup of Milnor's positional games. These games are two-player games of perfect information without randomness and are similar to Conway's combinatorial games. We prove characterization theorems for the partial-order on games. We then prove results which allow for the simpli cation of games and show that all invertible games without atomic reversibilities have a unique canonical form. Certain elements are shown to be noninvertible and the invertible elements are then characterized. Finally we provide natural game-theoretic inverses for the so-called \double atoms" and formalize the question of completing this process of game-theoretically extending the semigroup to a group by de ning a game embedding.
1 Introduction The study of positional games concerns two-player games of perfect information without random elements. The absence of randomness means that there are no dice or other chance elements involved. Perfect information means that both players have full knowledge of the state of the game at all times and that there is no hidden information. The origin of our concept of a game is found in the paper \Sums of Positional Games" by John Milnor [Milnor53]. Milnor de ned a game as a nite sequence of alternating move choices among a nite set of positions. For convenience the two players are referred to as 1
Left and Right. There is a special set of positions, the set of ending positions, from which no further move is possible and which is the domain of two real valued payo functions, one for each of the two players. From any non-ending position it is assumed that both players have non-empty sets of moves. When play reaches an ending position, play halts and both players receive their respective payo s. The incentive of a game, I (G), is de ned as the amount a player gains by moving rst, instead of his opponent moving rst. The most interesting and important concept in the study of positional games is the notion of a sum of games. G + H is a positional game and we de ne it by describing the moves available to each player. G and H are called the components of the sum game, G + H . A move in the sum of games G and H is a move in one and only one of G or H . It is then the other player's turn and the other player may move in his choice of one of the games. Play continues with the players alternating moves in this fashion. When one of the components has reached an ending position then all further moves take place in the remaining non-ended component. When both games have reached an ending position the payo s from each component are added together to determine the total payo . A natural example of this notion of addition of games may be readily seen in endgame positions of the classic game Go1. In the endgame of Go the board is divided into separate, independent regions from which a player must choose one in which to move in a given turn. For a fascinating application of the theory of combinatorial games to analysis of Go endgames beyond the ability of Go experts see [BW94]. Other games which \break up" into components and are therefore natural examples of the utility of this notion of addition are Nim and Dots-And-Boxes. For an extensive treatment of these examples and many others the reader is referred to [BCG82]. Milnor de nes a notion of equivalence of games. This equivalence relation is itself founded on a comparison relation between games, G H , where 1 It is often repeated in military circles that one of the reasons for the poor performance of the United States military forces in Southeast Asia was that the Americans were conducting their campaign according to the analytically based logic of Chess whereas the Vietnamese were operating under the more complex logic derived from a military training utilizing Go. It is also interesting to note that whereas computers now play Chess on the level of the most pro cient humans, the best computers cannot compete in Go with even a moderately practiced novice.
2
G H i G H and H G. Milnor de ned the equivalence relation directly and the comparison relation is not to be found in his original paper. It was John Conway who rst noticed the value of this ner distinction. The idea behind this comparison relation is that G H if and only if the game G is \better" for the Left and H is \better" for Right. Here \better" for player Left means that the outcome when G + X is played results in at least as large a payo for Left as when H + X is played, for any choice of test games, X , in the universe under consideration. Note that this de nition is relative to the universe under consideration, being a universal statement. Therefore it is important to take note that in his paper Milnor restricted his attention to those games with nonnegative incentive. In a game of positive incentive, G, both players desire to move rst as this confers an advantage to the starting player. If G has incentive 0 then neither player cares who moves rst. Milnor explicitly does not consider games whereby a player is forced to move when he would prefer to pass, i.e. games of negative incentive. Milnor's notion of comparison between games is correspondingly a ected by this explicit restriction. In the present paper we consider the set of all positional games, including the games of negative incentive. We rst de ne several classes of games and the binary comparison relation between games which captures the notion of one game being preferable to another. A large portion of this section is devoted to a characterization of this relation. We next consider the simpli cation of games to (almost) unique normal forms. We use simpli cation theorems to exhibit certain games that have no inverse and this shows that the class of games which is our primary interest is not a group. We characterize the invertible games and also discuss the relationships between various natural classes of games. We then proceed to introduce new games that provide inverses for many of the previously non-invertible games and formalize the question of completing this process of game-theoretically extending the semigroup to a group by de ning a game embedding.
2 Games and Sums of Games We consider two-player, non-random games of perfect information where the outcomes are linearly ordered. The typical outcome set will be the set of real numbers, R. 3
De nition 1 Let (P; ; +; 0) be a linearly-ordered, abelian group. Then U , P
the universe of games on P , is de ned inductively as follows: 1. Every pP is a game. 2. If A and B are non-empty sets of games then fAjB g is a game. F , the universe of hereditarily nite games on P , is de ned inductively as follows: 1. Every pP is a hereditarily nite game. 2. If A and B are non-empty, nite sets of hereditarily nite games then fAjB g is a hereditarily nite game. If G = fAjB g then we write G = A and G = B . G is called the set of left options or left moves of G and similarly for the right options of G. Also if HA we write H G and if HB we write H G. So g G i g G and g G i g G . P
L
R
l
L
r
l
l
l
L
R
r
r
r
The construction of the universe of games is very similar to the construction of the cumulative hierarchy in set theory. To each game we assign an ordinal rank (called a \birthday" in [Conway76] ) analogous to rank in set theory. Elements of the outcome group have rank = 0 and we call these games atomic games. Those games G such that GU ? P are called nonatomic games. For non-atomic games we have rank(G) = sup frank(g ) + 1, rank(g ) + 1)g where the supremum is taken over all Left and Right options of G. For example f7j ? 2g UR has rank 1, ff7j ? 2gj9g has rank 2, etc. Proof by induction and de nition by recursion are valid for similar theoretical reasons as in set theory. We will not bother to prove the version of the theorems appropriate for justifying these techniques as they mirror the set-theoretic theorems and this issue is also addressed in [Conway76]. Sometimes the proofs use induction on rank. In [Milnor53] what we call atomic games are called end positions. And similarly to that setting, if G is non-atomic then both players have non-empty sets of options. This is di erent from the situation in [Conway76] where there are games G such that one player may have no moves from G but the other does have at least one move. It is very natural to also see a game as a type of tree. Given a game G we can associate a bi-colored tree T (G) such that the root is G, the left options of G are \bLue" colored descendents, and the right options of G are \Red" colored descendents. The game tree of G, T (G), is de ned by induction as the bi-colored tree with root G. The descendents of the root P
l
r
4
are the game trees of the left and right options of G. For every g there is a blue edge connecting the root G to T (g ) and for every g there is a red edge connecting the root G to T (g ). See [BCG82] for examples. We now de ne an addition operation on games. This operation rst appeared in [Milnor53] and again in [Conway76]. The idea is that to make a play in G + H , one plays in exactly one of the games and leaves the other alone. It is then the other player's turn and he moves in exactly one component and so on until both components have ended. l
l
r
r
De nition 2 We de ne game addition. If p and q are atomic games then we use the underlying group structure to add p and q. Let p + fg 1 ; :::jg 1 ; :::g = fg 1 + p; :::jg 1 + p; :::g: l
r
l
r
Finally, if G and H are both non-atomic then let
fG jG g + fH jH g = fJ jJ g L
where
R
L
L
R
J = fg + H; G + h : g G and h H g L
L
l
and
R
l
l
L
l
J = fg + H; G + h : g G and h H g: R
R
r
r
r
R
r
For the rest of this paper unless otherwise stated, p,q,r will designate elements of the underlying outcome linear order (usually the reals), X ,Y ,Z ,W ,G,H ,J will designate non-atomic games, G will designate the set of left moves of G, g will designate a left move of G (i.e. g G), etc. L
l
l
l
Proposition 1 Game addition is commutative and associative. Proof. Both by induction. The proofs can be found in [Milnor53] and [Conway76]. a Proposition 2 If G and H are hereditarily nite then rank(G + H ) = rank(G) + rank(H ): 5
Proof. By a simple induction. a
This fact actually appeared in another form in [Hanner59]. Hanner de ned l(G) to be the maximal number of steps from G to an atom (not necessarily in alternating play) and then observed that l(G + H ) = l(G) + l(H ). In fact, l(G) = rank(G) from which the previous result follows immediately. From this point on we restrict our attention to FR. Therefore all games are hereditarily nite with real valued outcomes unless stated otherwise. We now de ne the operation of the Left and Right move operators, L, R, on a game. We wish to capture the notion of the determined outcome of a game if both players play completely optimally, i.e. in full knowledge of the entire game tree.
De nition 3 Let G FR. Then we de ne L(G) and R(G), as follows. 1. L(p) = R(p) = p. 2. L(G) = maxfR(g ) : g Gg. 3. R(G) = minfL(g ) : g Gg. l
l
l
r
r
r
Notice that since G FR the maximum and minimum are over nite sets. Thus L(G) and R(G) are the optimal outcomes which must occur if both players play optimally with Left starting and Right starting respectively.
3 Comparing One Game With Another
3.1 Test Games and Inequalities
We now are able to de ne our primary binary relations. The basic idea is that given two games, G and H , G may be preferable over H from the perspective of the left player and H may be preferable over G from the perspective of the right player in sums with test games. First we operate on G + X and H + X , where X is an arbitrary test game, with L and R to produce the optimal outcomes. If the outcome of playing G + X is always greater than the outcome of playing H + X we say G is \greater than" H .
De nition 4 Let G; HF where P is a linearly ordered abelian group. De ne
P
G H iff 8XF L(G + X ) L(H + X ): L
P
6
De ne
G H iff 8XF R(G + X ) R(H + X ): P
R
Finally de ne
G H iff G H and G H: L
R
Notice that if G; HP then this de nition is consistent with the ordering on P . If G H and H G then write G H . Write G > H if G H and not H G. Similarly de ne G > H and G > H . L
L
L
R
Proposition 3 The binary relations and are transitive. Proof Suppose G H J . Then for any X F we have L(G + X ) L(H + X ) L(J + X ): Therefore by transitivity of P we have L(G + X ) L(J + X ) and G J . A similar proof holds for .a Proposition 4 The binary relation is transitive on F . Proof Immediate from transitivity of and . a The next proposition shows that to check if G H it suces to check one of G H or G H . De ne the transitive closure of a game G to be L
L
R
P
L
L
R
P
L
R
all the games which are hereditarily elements of G. Formally, trcls(p) = p S and trcls(G) = G [ G [ trcls(g ) where the union is taken over all options g of G. De ne the support or alphabet of a game G, (G), to be all the set of atoms found in G, i.e. (G) = trcls(G) \ P . For ; P we write >>> to indicate that is much larger than . The exact meaning of \much larger" is dependent on the context in which it is being used. The proofs that utilize this notation could be formulated with precise bounds but this seems to obscure the main ideas so we incorporate this notation. In point of fact if > 0 and we say let >>> then usually choosing = 10 will perform the desired function. This is a useful heuristic. We write >>> (G) if for all (G) we have >>> and (G) >>> (H ) if for all (G) we have >>> (H ). L
R
L
R
s
gs
s
Proposition 5 G H i G H . L
R
7
Proof. By symmetry it suces to prove one direction, say left to right. So we assume G 6 H and show G 6 H . If G 6 H then there exists a game X such that R(G + X ) < R(H + X ). Now let rP such that R
L
R
r >>> (G) [ (H ) [ (X ) >>> ?r: Such an r will exist if P is nontrivial. Now de ne Y = fX j ? rg. Then
L(G + Y ) = R(G + X ) and
L(H + Y ) = R(H + X ) because Left has such a large incentive to prevent Right from choosing ?r in both games. Therefore L(G + Y ) < L(H + Y )
and we have shown G6 H .a L
Proposition 6 If G H then for all J F G + J H + J . Proof. Let W F . Then G H implies G + (J + W ) H + (J + W ). By associativity we obtain (G + J ) + W (H + J ) + W . a Theorem 1 (F = ,,+) is a commutative, partially-ordered, semigroup. Proof Follows from previous propositions. a P
P
L;R
L;R
P
We will see that F is not, in general, a group because many games do not have inverses. A game, G, is called invertible if there exists H such that G + H 0. By the previous theorem we know that all invertible games are cancellable, i.e. if H is invertible and G + H J + H then G J . The questions of whether the cancelation property holds for and generally are open. Clearly an armative answer to the former would imply an armative answer to the latter. P
8
3.2 Characterizing Inequality and Equivalence
We now wish to explore the notion of equivalence. We wish to obtain a direct characterization of the relation. This means that we want a criterion by which we can determine if G H simply by examining G and H (and all hereditary elements if necessary) themselves. Recall that the de nition of game equivalence is a universal statement over the entire universe of hereditarily nite games. De nition 5 Let pP . A game G is left-p-safe (lps) if L(G) p and for all g there exists g such that g is left-p-safe. Example 1 If q is atomic then q is left-p-safe i q p because in this case the second condition is vacuously satis ed. If G = f0jf0j ? 5gg then G is left-0-safe. If H = ff?2j0gjfGj ? 10g; f0j ? 3gg then H is left-0-safe. Lemma 1 If G is not left-p-safe and X = f?sjsg where ?s p then q > Q > p. p;p
p;q
Theorem 3 If p < q then Q has no inverse. Proof. Suppose Q + A 0. We may assume that A is in canonical form. p;q
So for all a either r
p;q
fX jpg + a 0 r
15
or there exists a such that rl
Q + a 0: p;q
rl
In these second case we see that a A by adding A to both sides. We derive this same inequality from the rst case by seeing that if Right moves to p + a then Left must have a good response: p + a 0. But Q p and so Q +a p+a yielding Q +a 0 and again a A. So all Right options of A are reversible and since A is in canonical form these reversible moves must be atomic reversibilities. A similar argument holds for the Left options of A and we deduce that rl
r
rl
p;q
rl
p;q
p;q
rl
rl
rl
A = ffX1jc1g; fX2jc2 g; :::jfd1jY1g; fd2jY2g; :::g: If we let c1 c for all i 6= 1 and d1 d for all j 6= 1 then i
j
A ffX1jc1gjfd1jY1gg by applying the comparison theorem. Now
L(Q + Q 1 1 ) = p + c1 p;q
c ;d
and
R(Q + Q 1 1 ) = q + d1: Since Q + Q 1 1 0 we have c1 = ?p and d1 = ?q. So p;q
p;q
c ;d
c ;d
A ffX1j ? pgjf?qjY1gg but the Left and Right moves are not atomically reversible, contrary to the previous conclusion. a
Corollary 1 (FR= ; +; 0) is not a group. 16
Let us denote the set of invertible games by FIR. Note that (FIR= ; +; 0) is a group and that FMR FIR FR. Now returning brie y to the games, Q , we explain why we introduced them as \double atoms." The proof of the previous theorem showed that we may as well assume that when playing sums of games, no moves are made in Q until all other games have ended. This is because both the (unique) Left and Right moves in Q are atomically reversible. Therefore it never bene ts a player to make these reversible moves as the preceding proof shows. This fact is also implicitly present in the proof of Theorem 14. Therefore in any sum like G + Q , play proceeds in G until G reaches an atom and the Q is nished in two moves yielding either p or q depending on who moved last in G. So Q is like an atom in the sense that it \sits around" until all other play is nished but its value is variable, depending on who begins play in it. Therefore we have shown the following proposition. p;q
p;q
p;q
p;q
p;q
p;q
Proposition 17 L(G + Q ) = max(L(G) + p, (L,L)(G) + q). p;q
Thus L(G + Q ) [L(G) + p, L(G) + q]. p;q
Of course a similar result holds for a Right start with max changed to min and all operators switched. Now consider the game G + Q + Q . The above reasoning applies to both of the double atoms, i.e. play ends in G and only then proceeds to the double atoms. So the outcome in the latter two games will either be p + r or q + s, depending on who moves last in G. Therefore we have shown that Q + Q = Q + + and have proven: p;q
p;q
r;s
p
r;q
r;s
s
Proposition 18 The set of all double atoms Q with p q is a subsemip;q
group of FR.
We now return to a further examination of inverses of games. Our next theorem shows that if a game G is invertible, the inverse must be cm(G). This will also provide another proof of the previous theorem. The proof is given in the appendix.
Theorem 4 If G + H 0 then H cm(G). The proof of the preceding theorem yields more information. Suppose a game G is in canonical form and contains a move (Left say) g such that g l
17
l
is not invertible. We may assume that g is not reversible or else it may as well have the form f0jpg (since the element is irrelevant as we have seen) and is then invertible. G cannot be invertible because there cannot be an adequate Right response to a Left move to g . For if H is a potential inverse to G, we cannot have g + h 0 because g is not invertible and the proof demonstrated that we also cannot have g + h < 0 for this also leads to a contradiction. Therefore we have proved: l
l
l
r
l
l
r
Corollary 2 If G is in canonical form and contains a noninvertible move on either side, then G is not invertible.
Let us now return to the question we left open earlier about the uniqueness of canonical forms.
Theorem 5 Suppose G hereditarily has no atomic reversible moves and is invertible. Then G has a unique canonical form.
Proof. Suppose G and G0 are in canonical form, hereditarily have no atomic reversible moves, G G0, and G (and therefore also G0) is invertible. We wish to show that G = G0 . Let us consider the game G + cm(G0) 0.
Consider a Left option of G, g . Since G has no reversible moves there must exist a Right move in cm(G0), cm(g0), such that g + cm(g0) 0. But as we have seen this implies that g + cm(g0) 0 and thus g g0. By induction we have that not only are g and g0 equivalent but, further, that they are equal, i.e. identical. So for every g there is g0 such that g = g0 and vice-versa. A similar result holds for Right moves in both games. So the moves in the games G and G0 are pairwise equal. Therefore we have shown the equality of G and G0 and the uniqueness of canonical forms.a We now possess a simple algorithm for determining if a game has an inverse. To check if G is invertible we simply compute to see if G+cm(G) 0: If so then ?G = cm(G). If not then G does not have an inverse and we need not try any other game as a potential inverse. l
l
l
l
l
l
l
l
l
l
l
l
l
8 Providing Inverses for Double Atoms Let us now return to investigations of the double atoms, Q . We would like to expand our universe so as provide an inverse to these games. Recall that p;q
18
these games have a unique feature when played in sums. We saw that double atoms are never played until all other component games have ended and then play in a double atom proceeds trivially for two moves until one of the atoms, either p or q is reached, depending on who was forced to begin play in Q . We could use new notation to indicate this situation by creating games with no move options (as is the case with atoms) but with double outcome values depending on who is to move: G = (p)fjg(q). The game G has no move options and its value is p if Left is to move and q if Right is to move. This notation is cumbersome so we write G = p2q to abbreviate the game. Now we have seen that Q has no inverse in F because the only possible inverse, cm(Q ) = Q? ? does not work. What we really need is a game that behaves like (using our new notation) G = (?p)fjg(?q) = ?p2 ? q. Unfortunately there is no such game in F . For example, if H = ffX j1gjf?1jY gg then X and Y are not irrelevant, as in Q?1 1 , and play in H does not always take place after all other play is nished, as in Q?1 1. Thus H does not behave like Q1 ?1 ought to behave. So let us formally introduce all double atoms into the universe. p;q
p;q
p;q
q;
R
p
R
;
;
;
De nition 16 For all pairs of real numbers, p and q (not necessarily dis-
tinct), p2q is a double real number and the set of double reals is denoted by R R. Double real numbers will be called atoms in the present context. The class of hereditarily nite games built on R R, denoted by FRR , includes R R and all games of the form G = fG jG g where G and G are nite sets of games in FRR. De ne (p2q) = fp; qg and (G) is de ned recursively as usual for non-atomic G. Let L
R
L
R
p2q + p02q0 = p + p02q + q0; G + p2q = fg 1 + p2q; :::jg 1 + p2q; :::g; and G + H is recursively de ned as in F . De ne L(p2q) = p and R(p2q) = q. For non-atomic games, Left and Right outcome operators are de ned recursively as in F by taking maximums and minimums over outcomes of l
r
R
R
moves. The move restricted operators have similar modi cations for use in FRR. G H is de ned as in F except we quantify over all test games X FRR. Finally, we set cm(p2q) = ?p2 ? q and extend to non-atomic games as usual. R
19
We clearly hope that in FRR, G = ffX j ? 1gjf1jY g ?121 and will therefore have an inverse, namely 12 ? 1. But none of this is obvious for several reasons. Firstly notice that in FRR, p2p plays the role that p played in F . Namely, neither atom has any moves and the outcome operators on the atoms simply produce the real number, p. In fact we can de ne a semigroup homomorphism f : FR ! FRR with f (p) = p2p for all atoms p and R
f (fg 1 ; :::jg 1 ; :::g) = ff (g 1 ); :::jf (g 1 ); :::g: l
r
l
r
It is easy to check that this function is injective and respects addition. Clearly it is not surjective since p2q is not in the image if p 6= q. Therefore we can consider FR a proper subsemigroup of FRR and we will abbreviate p2p as p. Notice that the identity for the semigroup FRR is 0 = 020 and that all atoms are invertible: p2q + ?p2 ? q = 020. However since the formula determining the comparison relation G H is universal and we have added new games it is possible that G H in FR but G 6 H in FRR. This occurred when we expanded from FMR to FR and it turns out that it occurs again in this context. For example, f0jf0j0gg 0 in FR. But
L(0 + 02 ? 1) = 0 > ?1 = L(f0jf0j0gg + 02 ? 1): So f0jf0j0gg 6 0 in FRR. This also shows why proving that Q?1 1 has an inverse in the extended semigroup FRR does not prove that it is cancellable in FR. For suppose G + Q?1 1 H + Q?1 1 in FR. The necessary and sucient condition for introducing an inverse for a noninvertible element in a semigroup is the cancelation condition. But here we are concerned with inverses for equivalence classes of games. So even if G 6 H in FR (this question is open), introduction of an inverse is not prohibited because introducing a new element may make G 6 H in the expanded semigroup. In this case we split the equivalence class of F into several equivalence classes of FRR. So a natural (open) question is can FR be embedded in a group? We must be careful here about nding a proper notion of an embedding. The idea that we wish to capture is the idea of adding \new" games to a class of games. This action may split equivalence classes, as we have previously seen. We see that a even a negative resolution of the cancelation problem, i.e. that there is a game in FR that is not cancellable, does not resolve the embedding question negatively. ;
;
;
R
20
De nition 17 Suppose (X; + ; ) and (Y; + ; ) are partially-ordered semigroups. Then an injective map f : X ! Y is called a game embedding if f (G + H ) = f (G) + f (H ) and if f (G) f (H ) then G H . X
X
Y
Y
Y
X
Notice that inequalities are not preserved from X to Y but they are preserved from Y to X . So a game embedding preserves addition and it preserves lack of inequality (i.e. if G 6 H then f (G) 6 f (H )). Notice that this latter condition means that equivalence classes will be re ned. X
Y
Open Question 1 Does there exist a partially-ordered group (Y; + ; ) and a game embedding f : FR ! Y such that the elements of Y have a Y
Y
natural game-theoretic interpretation?
The phrase \natural game-theoretic interpretation" is included in order to prevent trivial solutions like the following one. Consider (FR; +; ) where is the trivial ordering in which no two elements are comparable. Then it is easy to show that (FR; +; ) is a commutative semigroup where the cancelation property holds. Therefore it can be embedded in a group, Y , and thus clearly there exists a game embedding of (FR; +; ) into Y . We prefer a di erent sort of solution. It is desirable that the elements of Y \resemble" positional games (consider for example the extension of FR to include double atoms) and that they satisfy certain properties essential to positional games, for example the simpli cation theorems discussed at the beginning of this chapter. Now let us show that in FRR, ?121 Q?1 1 and therefore that the inverse of Q?1 1 is 12 ? 1. First we must update the comparison criteria for this new class of games. The example above shows that the previous one is no longer sucient, though still clearly necessary. In particular we need necessary and sucient conditions to determine if G p2q and if G p2q. Recall that all atoms are invertible and so it suces to determine the criteria for comparing an arbitrary game against 020. T
T
T
;
;
De nition 18 A game G is doubly-left-0-safe i : 1. If G = p2q is atomic then p 0 and q 0. 2. If G is non-atomic then (L,R)(G) 0 and for all g there exists g r
such that g is doubly-left-0-safe. rl
Proposition 19 G 020 i G is doubly-left-0-safe. 21
rl
Proof. Similar to Proposition 7. a
Clearly a symmetric criterion holds for determining G 020. So in summary we may state that G 020 i : 1. (L,R)(G) = 0. 2. (R,L)(G) = 0. 3. For all g , there exists g 0. 4. For all g , there exists g 0. It is now easily checked that ffX j ? 1gjf1jY gg + 12 ? 1 020 by verifying these four conditions. Since inverses are unique up to equivalence we have also shown that ffX j ? 1gjf1jY gg ?121. More generally, the above criterion shows that if p < q then ffX jpgjfqjY gg p2q and the terminology \double atom" is therefore appropriate. An interesting question is whether we can supply game theoretic inverses for other noninvertibles, for example f?1; f?2j0gjf0j2g; 1g, in an analogous way. l
lr
r
rl
Open Question 2 Is there a game embedding of FRR into a partiallyordered group F 0 such that the elements of F 0 have a natural game-theoretic interpretation?
Acknowledgments
The author would like to thank Jerry Keisler, Gian-Carlo Rota, and John Conway for helpful discussions and Los Alamos National Laboratory for nancial support for this research.
Appendix
Proof of Proposition 9. Suppose both of the above conditions hold. We show that
and
L(G + X ) L(H + X )
R(G + X ) R(H + X ) for all games X . First let X = p be atomic. Now suppose L(H + p) = R(h + p); l
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i.e. h is a good opening move for Left. Then by condition 2 we have that either there exists h G in which case L(H + p) = R(h + p) L(h + p) L(G + p) or there exists g h in which case L(G + p) R(g + p) R(h + p) = L(H + p): In both cases we have the result that L(G + p) L(H + p). Now suppose R(G + p) = L(g + p), i.e. that g is a good opening move for Right. By condition 1 we have that either there exists g H in which case R(G + p) = L(g + p) R(g + p) R(H + p) or there exists h g in which case R(G + p) = L(g + p) L(h + p) R(H + p): In both cases we have the result that R(G + p) R(H + p). Now assume X is non-atomic and that the result holds for all X 0 of smaller rank. Suppose L(H + X ) = R(H + x ). By induction R(G + x ) R(H + x ) and so L(G + X ) R(G + x ) R(H + x ) = L(H + X ): Now suppose L(H + X ) = R(h + X ). Then by condition 2 we have that either there exists h G in which case L(H + X ) = R(h + X ) L(h1 + X ) L(G + X ) or there exists g h in which case L(G + X ) R(g + X ) R(h + X ) = L(H + X ): In both cases we have the result that L(G + X ) L(H + X ). Now let R(G + X ) = L(G + x ). By induction L(G + x ) L(H + x ) and so R(G + X ) = L(G + x ) L(H + x ) R(H + X ): Now suppose R(G + X ) = L(g + X ). By condition 1 we have that either there exists g H in which case R(G + X ) = L(g + X ) R(g + X ) R(H + X ) l
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or there exists h g in which case R(G + X ) = L(g + X ) L(h + X ) R(H + X ): In both cases we have the result that R(G + X ) R(H + X ). So we have shown the necessary condition and the proof is complete. a Proof of Theorem 14. We will prove part 1 because part 2 is completely analogous. First assume that L(G0 ) < R(fH jp; J; :::g). Then L(G0 ) < L(G) because fH jp; J; :::g is the best opening move for Left in G and therefore G 6 G0 . Now assume L(G0) R(fH jp; J; :::g). Note that this immediately implies that L(G0) = L(G) p and therefore also G0 p since G and G0 have the same Right options. Now clearly G G0 in any case. We show that G0 G. First let the test game be atomic. By assumption L(G0) = L(G) and therefore for all atoms, p, we have L(G0 + p) = L(G + p) and (trivially) R(G0 + p) = R(G + p): Now assume that for all games J of smaller rank than X we have L(G0 + J ) = L(G + J ) and similarly for a Right start and let us consider the possible opening moves for Left in the games G + X and G0 + X . By our induction assumption we have that for all x R(G0 + x ) = R(G + x ): The only left option available in G + X that is absent from G0 + X is the atomically reversible move to fH jp; J; :::g + X . But R(fH jp; J; :::g + X ) p + L(X ) L(G0 + X ): This last inequality follows from the fact that G0 p. So we have shown that L(G0 + X ) = L(G + X ): Now note that if Right begins in the games G + X and G0 + X that Right's moves in G and G0 are identical. Also by induction for all x we have L(G + x ) = L(G0 + x ): r
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Therefore we have shown
R(G + X ) = R(G0 + X ) and so G0 G. For the next part we need to understand what is happening when G 6 G0 . Basically Left never bene ts from moving from G to fH jp; J; :::g because Right can immediately reverse to p (at least). Therefore it is always the case that Left may as well play elsewhere except when there is no elsewhere! Therefore Left only moves from G to fH jp; J; :::g when the test game has reached an atom and this move is Left's unique best move, i.e. when L(G) = R(fH jp; J; :::g) and for all other g we have R(g ) < L(G). Note L(G) = R(fH jp; J; :::g) p. But recall that p G so L(G) p. Therefore L(G) = R(fH jp; J; :::g) = p and since this sequence of moves only occurs when all other test games have reached an atom the other Right moves in fH jp; J; :::g like J , etc. are never necessary and may be deleted. And clearly H can never be played and so can be replaced by an arbitrary game, K . Finally, if there are two Left options that are atomically reversible, say fH jpg and fK jqg, then if q p we see by the rst part of this section that fK jqg is eliminable. The proof is complete. a Proof of Theorem 4. If G = p is atomic then this is clearly true because then H ?p = cm(G). Suppose we have l
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J = fg 1 ; g 2 ; :::jg 1 :::g + fh 1 ; h 2 ; :::jh 1 :::g 0: l
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We may assume that all dominated options and non-atomic reversible moves have been eliminated from G and H . Now for all Left moves in J , j , there exists j such that j 0 because J 0. So for all g we have g + H 0 or g + h 0. If g + H 0 then by adding G to both sides we get g G. Then g is reversible and therefore g is atomic, say p. First assume that the second condition holds, i.e. that g 1 + h 1 0 and that g 1 is not reversible. We now show that g 1 + h 1 < 0 is impossible and therefore that g 1 + h 1 0 and by induction h 1 = cm(g 1 ). For suppose g 1 + h 1 < 0. By the above argument there also must either exist g 2 such that h 1 + g 2 0 or h 1 1 + G 0. If the second situation holds, by adding H to both sides we again see that h 1 is reversible and so must be atomically reversible. In this case h 1 1 = q, say, and we have q + G 0. Now since g 1 + h 1 < 0 l
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if Left now moves from h 1 to q Right must have an acceptable reply in g 1 . In other words there must exist g 1 1 such that g 1 1 + q 0. But we now have the sequence of inequalities g 1 1 ?q G which yields g 1 1 G and a contradiction because we assumed that g 1 was not reversible. So we must have the rst situation, h 1 + g 2 0. Notice that we have obtained the inequality g 1 + h 1 h 1 + g 2 . Now if h 1 + g 2 0 then h 1 is invertible and the last inequality yields g 1 g 2 . But this is a contradiction because we have eliminated all dominated moves. Therefore h 1 + g 2 > 0. We may continue this process and since G has only nite many moves we obtain a nite subset of moves such that g 1 + h 1 < 0, h 1 + g 2 > 0, g 2 + h 2 < 0,....,g + h < 0, h + g 1 > 0. All these inequalities must be strict and none of these elements are invertible or else we contradict the fact that we have eliminated dominated options, as shown above. Now proposition 10 allows us to add these strict inequalities and obtain another strict inequality. So by adding alternating members of the preceding list we obtain r
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h 1 + g 2 + h 2 + g 3 + ::: + h + g 1 > 0: Since these inequalities are strict this is a contradiction. Therefore the original assumption that g 1 + h 1 < 0 is impossible and we must have had, in fact, that g 1 + h 1 0 and by induction h 1 cm(g 1 ). So we have shown that for every g which is not atomically reversible there is h such that h cm(g ). A similar argument shows that for each g which is not atomically reversible there exists h such that h cm(g ) and likewise for all non-atomically reversible moves of H . Now assume that g 1 is atomically reversible. By Theorem 14 we may assume that g 1 = fsjpg and G has the form G = ffsjpg; g 2 ; :::jg 1 ; :::g where s is a \very large" (compared to anything else in G or H ) atom. We wish to show that there exists an atomically reversible Right move in H and that Left reverses to ?p. This would imply that H has the form H = fh 1 ; :::jf?pjY g; h 2 ; :::g. By applying this result to any atomic reversibles on either side of G and H and then simplifying using Theorem 14 we will complete the proof. Suppose there are no atomically reversible Right moves in H and consider playing G + H with a Right start. We now show that in this case R(G + H ) > 0, contradicting G + H 0. If Right moves to G + r
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h then Left may move to fsjpg + h . Because s is so large Right must move to p + h and the outcome of the play is p + R(H ). Since H has no atomic reversible Right moves, we know that for all h there exists g such that g cm(h ). Now recall that L(X ) = ?R(cm(X )) so for all h , r
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L(h ) = ?R(cm(h )) = ?R(g ): r
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Now Theorem 14 shows that R(g ) < p and so L(h ) > ?p. Therefore R(H ) > ?p and p + R(H ) > 0. So Right does not have a good opening move in H . If Right opens to g + H then there exists g p since G p. Notice then that R(g + H ) p + R(H ) > 0 as before since R(H ) > ?p. So Right does not have a good opening move in G. We have deduced a contradiction and so H must have an atomically reversible Right move. Let H = fh 1 ; :::jfqjtg; h 2 ; :::g where t = ?s. Now suppose q > ?p. Then Right still has no good opening move because if Right opens to G + fqjtg, Left responds to G + q and the outcome is l
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R(G) + q p + q > p + ?p = 0: Therefore q ?p. Now by a symmetrical argument we need p ?q and these two inequalities yield p = q and the proof is complete. a
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References [BCG82] Elwyn Berlekamp and John H. Conway and Richard Guy. Winning Ways Vol. I,II. Academic Press. London, 1982. [BW94] Elwyn Berlekamp and David Wolfe. Mathematical Go: Chilling Gets the Last Point. Ishi Press International. San Jose, London, and Tokyo, 1994. [Conway76] John H. Conway. On Numbers and Games. Academic Press. London, 1976. [Hanner59] Olof Hanner. Mean Play of Sums of Positional Games. Paci c Journal of Mathematics, Vol. 9, 1959. [Milnor53] John Milnor. Sums of Positional Games in Annals of Math. Study. Kuhn and Tucker editors. Princeton, 1953.
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