Combinatorial Games with Restricted Options under Normal and Misere Play

by

Paul Ottaway

Submitted in partial fulfillment of the requirements for the degree of Doctor of Philosophy

at

Dalhousie University Halifax, Nova Scotia June 2009

© Copyright by Paul Ottaway, 2009 Library and Archives Bibliotheque et 1*1 Canada Archives Canada Published Heritage Direction du Branch Patrimoine de Tedition

395 Wellington Street 395, rue Wellington OttawaONK1A0N4 OttawaONK1A0N4 Canada Canada

Your file Votre reference ISBN: 978-0-494-56441-7 Our file Notre rSterence ISBN: 978-0-494-56441-7

NOTICE: AVIS:

The author has granted a non­ L'auteur a accorde une licence non exclusive exclusive license allowing Library and permettant a la Bibliotheque et Archives Archives Canada to reproduce, Canada de reproduire, publier, archiver, publish, archive, preserve, conserve, sauvegarder, conserver, transmettre au public communicate to the public by par telecommunication ou par I'lntemet, prefer, telecommunication or on the Internet, distribuer et vendre des theses partout dans le loan, distribute and sell theses monde, a des fins commerciales ou autres, sur worldwide, for commercial or non­ support microforme, papier, electronique et/ou commercial purposes, in microform, autres formats. paper, electronic and/or any other formats.

The author retains copyright L'auteur conserve la propriete du droit d'auteur ownership and moral rights in this et des droits moraux qui protege cette these. Ni thesis. Neither the thesis nor la these ni des extraits substantiels de celle-ci substantial extracts from it may be ne doivent etre imprimes ou autrement printed or otherwise reproduced reproduits sans son autorisation. without the author's permission.

In compliance with the Canadian Conformement a la lor canadienne sur la Privacy Act some supporting forms protection de la vie privee, quelques may have been removed from this formulaires secondaires ont ete enleves de thesis. cette these.

While these forms may be included Bien que ces formulaires aient inclus dans in the document page count, their la pagination, il n'y aura aucun contenu removal does not represent any loss manquant. of content from the thesis.

1*1 Canada DALHOUSIE UNIVERSITY

To comply with the Canadian Privacy Act the National Library of Canada has requested that the following pages be removed from this copy of the thesis:

Preliminary Pages Examiners Signature Page (pii) Dalhousie Library Copyright Agreement (piii)

Appendices Copyright Releases (if applicable) Table of Contents

List of Tables vi

List of Figures viii

Abstract ix

Acknowledgements x

Chapter 1 Theory of Combinatorial Games 1 1.1 Overview 1 1.2 Properties 4 1.3 Game Representations 4 1.4 Outcome Classes 5 1.5 Structure of Games 8

1.6 NIM 15 1.7 Subtraction Games 19

Chapter 2 Option-Closed Games 22 2.1 Definitions and Examples 22 2.2 Reduced Canonical Form 23

2.3 ROLL THE LAWN 29

2.4 CRICKET PITCH 30

Chapter 3 Recent Results in Misere Play 35 3.1 Introduction 35 3.2 The Disjunctive Sum of Two Misere Games 36 3.3 Equivalence Classes of Misere Games 38

3.4 END-NIM 44

iv Chapter 4 Consecutive Move Ban 47 4.1 Motivation and Definitions 47 4.2 Normal Play 48 4.3 Misere Play 50 4.4 Two-handed Games 59 4.5 A Coin-Flipping Game 66

Chapter 5 The Short Disjunctive Sum 74 5.1 Definitions 74 5.2 Results 75 5.3 Regular Games 80 5.4 CLOBBER 87

5.5 MAZE and MAIZE 89

Chapter 6 Discussion 92 6.1 Option Closed Games 92 6.2 Misere games 92 6.3 Consecutive Move Ban 93 6.4 Short Disjunctive Sum 93

Appendix A Tables 95

Bibliography 100

v List of Tables

Table 2.1 A short cricket pitch dictionary 31

Table 3.1 Comparison of normal and misere outcome classes 37

Table 4.1 Examples from each of the eight classes of one-handed games . 52 Table 4.2 Examples of all possible two-handed games 59 Table 4.3 The outcome class of a CMB game denoted by [a,b,c] 61 Table 4.4 The outcome of a CMB game denoted by [a, b, c, d, e] with (c — b) even and d = 0 63 Table 4.5 The outcome of a CMB game denoted by [a, 6, c, d, e] with (c — b) even and d = 1 63 Table 4.6 The outcome of a CMB game denoted by [a, 6, c, d, e] with (c — b) odd and d = 0 . 63 Table 4.7 The outcome of a CMB game denoted by [a, b, c, d, e] with (c — b) odd and d = 1 64 Table 4.8 The effect of a player's move on score and control values in a CMB game (part 1) 67 Table 4.9 The effect of a player's move on score and control values in a CMB game (part 2) 68

Table 4.10 The period lengths for CMB games with SL = SR = {a, 6, a + 6} 70

Table 4.11 The CMB class sequence for SL = SR = {2, 4, 7} 70

Table 4.12 The CMB class sequence for SL = SR = {3, 5, 8} 70

Table 4.13 The CMB class sequence for S^ = SR = {any subset of odd numbers} 71 Table 4.14 Periods and pre-periods for various subtraction sets 71

Table 4.15 The CMB class sequence for SL = {1, 2, 3}, SR = {2, 3, 4} . . . 72

Table 4.16 The CMB class sequence for SL = {1, 2, 4}, SR = {1, 3, 4} ... 73

Table 4.17 The CMB class sequence for SL = {1, 8}, SR = {3, 5} 73

Table 4.18 The CMB class sequence for SL = {1, 2, 5}, SR = {2, 3, 4} . . . 73

vi Table 5.1 The possible outcome classes of the short misere sum G + H . 77 Table 5.2 Games played with a short misere sum which are born by day 1 85 Table 5.3 A brief dictionary of misere clobber positions 88

Table A.l Examples of all possible outcomes of a misere sum (part 1) . . 96 Table A.2 Examples of all possible outcomes of a misere sum (part 2) . . 97 Table A.3 Examples of all possible outcomes of a short misere sum .... 98 Table A.4 A partial dictionary of Normal Play game values 99

vn List of Figures

Figure 1.1 A typical game tree 5 Figure 1.2 Normal Play outcome analysis of a game tree 8 Figure 1.3 Misere Play outcome analysis of a game tree 9 Figure 1.4 The partial order of outcome classes 9 Figure 1.5 Normal Play value of a game tree 15 Figure 1.6 Example of the nim-sum operation 17 Figure 1.7 Finding a winning move in a game of NIM with four piles ... 18 Figure 1.8 The nim-sequence for S = {1, 3,4} 20

Figure 1.9 The sequence of values for SL = {1},SR = {2, 3} ...... 21

Figure 2.1 A game of ROLL THE LAWN 23

Figure 2.2 A game of CRICKET PITCH 23

Figure 3.1 Outcome analysis of NIM with one heap of size 2 35

Figure 4.1 Imposing a consecutive move ban 48 Figure 4.2 A consecutive move ban applied to a sum of components ... 50

Figure 5.1 A sample misere game with value 002 79 Figure 5.2 The games G and T(G) 81 Figure 5.3 Finding regular K where T(G) + T(H) = T(K) 83 Figure 5.4 The partial order of games played with a short misere sum born by day 1 86 Figure 5.5 A game of MAZE 89 Figure 5.6 Outcome analysis of misere MAZE 90 Figure 5.7 Values for MAZE played with a short misere sum 90

Figure 5.8 A game of MAIZE with T(G) = oo4 91

vm Abstract

This thesis presents new results with respect to the analysis of combinatorial games played under both Normal and Misere Play ending conditions. We begin by demon­ strating how the concept of the reduced canonical form of a game can be applied to the analysis of option closed games such as CRICKET PITCH and ROLL THE LAWN. We continue with Misere Play games and show why their analysis is much more difficult than that of their Normal Play counterparts. In particular, we show that they do not form a group and admit no partial order which preserves outcome classes. In an effort to understand more regarding Misere Play games, we consider a specific restriction where a given player cannot play twice in the same component of a sum unless their opponent plays between those moves. This restricts the universe of different games to a finite number and we present the analysis of arbitrary sums under both Normal and Misere play rules. Finally, we consider the short disjunctive sum of games which is an alternate interpretation of disjunctive sum which is consistent with the traditional Normal Play interpretation but different when considering Misere Play games. Using this approach, we show that overriding moves are possible and develop a system for dealing with them in an arbitrary sum. We also show how the short disjunctive sum affects familiar games such as MAZE, MAIZE and CLOBBER when played using Misere Play rules.

IX Acknowledgements

I would like to thank Richard Nowakowski for his friendship, advice and support. I would also like to thank my family, friends and especially my wife, Heather, whose love has seen me through and made this possible.

x Chapter 1

Theory of Combinatorial Games

1.1 Overview

Combinatorial games have been formally studied for over a hundred years but have only recently become more popular since the publication of Winning Ways [4] and On Numbers and Games [8]. Combinatorial games involve two players who alternately make moves with the winner being determined by who makes the last legal move in the game. We assume that both players are perfect in the sense that they never make a losing move when they have a move which allows them to win. Under Normal Play rules, the last player to move wins while under Misere Play rules, the last player to move loses. This small difference in play convention can have a dramatic effect on the analysis of a game. In particular, they are not, as one might expect, simply opposites of each other and we find that the winner using one convention cannot be used to de­ termine the winner using the other. The majority of research in the field has focussed on Normal Play games. The mathematical structure we find in the study of games comes from the disjunctive sum of games which can be thought of as playing two games simultaneously. On each player's turn, they pick in which game they would to play. In this thesis, we will examine the similarities and differences between the two play conventions and attempt to discover why so little is known about Misere Play games.

After a brief overview of the notable results regarding Normal Play Games in this chapter, we proceed to examine in detail how the Misere Play convention affects the mathematical structure of the set of games. In particular, we will show that this seemingly minor change in the rules removes much of the structure which we have grown accustomed to in Normal Play games. In both cases, we have a method which allows us to simplify games but in the case of Misere Play games, it cannot be applied

1 2

in as many cases. We also demonstrate that there is a unique game which can act as an identity with respect to the sum whereas using the Normal Play convention, we have infinitely many which are all equivalent. In short, Misere Play games are much more complicated and it is almost hopeless to expect to find a way to analyze them in general.

In an effort to gain a greater understanding regarding Misere Play games, we make our task easier by placing restrictions on the types of options which are allowed to each player. In this way, we reduce the size of the set of games we are considering. The proof that there is a unique identity with respect to the sum of Misere Play games uses a game which has the property that if a player can only play once, they lose but if they are able to play twice consecutively, they win. The existence of this type of position is viewed as a main reason why the structure of Misere Play games is so poorly understood. This naturally suggests a way to reduce the set of games which we may want to consider.

In Chapter 3, we look at games where any position a player can obtained after two consecutive moves could have been obtained in a single move. These option-closed games unfortunately have not provided us with much insight regarding Misere Play games but we include them anyway since many interesting Normal Play results were forthcoming. Using the concept of reduced canonical form, we show that all option closed games must have a particular kind of structure which drastically limits the types of values which can occur. We demonstrate these results by looking at the games CRICKET PITCH and ROLL THE LAWN.

After failed attempts at obtaining useful results regarding option-closed games under the Misere Play convention, we turn our attention to a further restricted set of games. In this case, we simply ban a player from being able to make two consecutive moves in the same component. In Chapter 4, we analyze these games under both play con­ ventions. The Normal Play analysis is quite straight forward and while the Misere 3

Play analysis is more involved, we succeed in determining the winner based on the number and type of the components which make up an arbitrary sum. In particular, we find that many games end up being equivalent based on a parity rule and we are able to compute an invariant which in turn determines the outcome of the game. We demonstrate how these concepts can be applied to specific games by considering one involving stacks of coins.

We conclude the thesis with a more detailed look at what is meant by the disjunctive sum of games in Chapter 5. In particular, we show that there are in fact two different interpretations which are both consistent with our view of how they should work with Normal Play games. We call this new interpretation the short disjunctive sum since the sum of games ends which the first component finishes rather than when the last component finishes as would be the case under our usual interpretation. We consider the implications of playing the short disjunctive sum of Misere Play games and find that in many cases we can draw parallels to Normal Play games. We borrow the concept of overriding moves to describe options a player may have which allow him to win the sum regardless of which games are being played in the sum. We demonstrate how the short disjunctive sum affects some familiar games like CLOBBER and MAZE. Finally, we show that the algebra which is developed can be extended to Normal Play games which can allow for overriding moves as well. 4

1.2 Properties

I shall be examining a variety of combinatorial games which are characterized by the following properties:

• There are two players, denoted Left and Right, who move alternately during the course of the game. By convention, Left is female and Right is male.

• There are clearly defined rules which specify the moves allowed to each player.

• There is complete information. That is, all information regarding the game is equally available to both players.

• There is no element of chance. Therefore, there is never any use of dice, spinners or similar devices.

• The last move determines the winner. Under Normal Play rules, a player who cannot make a legal move loses. Under Mis ere Play rules, a player who cannot make a legal move wins.

• There cannot be an infinite sequence of moves in the game. Therefore, the game must eventually end with one winner and one loser.

While the bulk of the material regarding combinatorial games deals with Normal Play rules, this thesis will deal primarily with Misere Play rules and how that change affects the analysis of such games. A more complete introduction to combinatorial game theory can be found in any of the three standard texts on the subject: On Numbers and Games, Winning Ways and Lessons in Play [8] [4] [2]. All non-referenced results are original work.

1.3 Game Representations

Throughout this document, it will be necessary to refer to specific games and for clarity, they will often be described using game trees. The root of the tree will be given at the top of the diagram. From that node, Left's options will be to move along any edge which goes down and left to a new node. Likewise, Right's options will be 5

Figure 1.1: A typical game tree to move along any edge which goes down and right to a new node. A sample game tree is shown in Figure 1.1.

Another way to specify a game is using the notation G = {GL\GR} where QL and QR represent the sets of games which are the left and right options of the game G. When we want to specify a particular left or right option we use GL and GR respectively. To make things a little easier to read, we use the symbol • to represent the empty set and use an increasing number of longer vertical bars to represent outer levels. In essence, this represents the depth-first representation of the game tree. The tree shown in Figure 1.1 would be denoted as follows:

G = {{{{•!•} II {•!•}},{• I-} |||-} I {{•!•} II •},{•!•}}

1.4 Outcome Classes

The outcome of a combinatorial game is completely dependent on the moves allowed to each player as well as which player is designated as having to move first. Since combinatorial games have perfect information and no element of chance it is possible 6

(though possibly time-consuming) to determine the winner before any moves have been made by examining all possible sequences of play. To this end, we can partition all games into four outcome classes:

• £ - The set of all games where Left can win playing first or second.

• 71 - The set of all games where Right can win playing first or second.

• A/" - The set of all games where both players can win playing first (Next Player).

• V - The set of all games where both players can win playing second (Previous Player).

The following result is well known but is included for completeness and as an introduction to the material in general. We assume throughout that both players play perfectly and always choose a winning line of play if one exists.

Lemma 1. [2][Thru 2.1] Given a game G, a player can force a win playing first if and only if the other player cannot force a win playing second.

Proof. We proceed by induction and assume without loss of generality that Left plays first. Once Left makes a move, she will be the second player from the new position and Right will be the first player. If Left can win playing first from G, she must move to a position where she can win playing second. By induction, this means that Right cannot win playing first from that position. Therefore, Right cannot win G playing second. For the opposite direction, if Right cannot win G playing second then it must be the case that Left has some move from G to a position where Right cannot win playing first. Again, by induction, this implies that Left can win playing second from that position. Therefore, Left can win G playing first by making that move. •

Theorem 2. [2][Thm 2.1] Every game G is a member of exactly one outcome class.

Proof. Due to the previous lemma, if we know that a player can win playing first we know the other player cannot win playing second. So, we can classify any given game by knowing if Right can win playing first and if Left can win playing first. This gives us the four possible outcome classes and ensures that each game is in exactly one of them. • 7

Definition 3. An impartial game G is one in which Left and Right have the same set of options and these options are also impartial. When G can have any set of games as left and right options, we refer to it as partizan.

Due to the symmetry of impartial games, if Left can win by playing first then Right can win by playing first as well by adopting Left's strategy. Likewise, if Left cannot win by playing first, then Right cannot win by playing first either. Therefore, impartial games are completely contained within the M and V outcome classes. Also note that this definition implies that impartial games are a strict subset of partizan games. We now introduce the following two functions:

Definition 4. The outcome function o(G) gives the outcome class of the game G under Normal Play rules: o(G) = C where C £ {£, H,J\f, V}.

Definition 5. The misere outcome function o~(G) gives the outcome class of the game G under Misere Play rules: o~{G) = C where C £ {£,1Z,J\f,V}.

Theorem 6. [2][Obs 2.4] The outcome class of a game G can be determined by the outcome classes of its left and right options.

Proof. If Left's options contain a game which is in £ U V then she can win playing first by moving to such an option since Right loses when playing first from such a position. If Left has no such option, then her options must be in 1Z U J\f from which Right will win playing first. Therefore, Left loses G playing first.

Likewise, if Right's options contain a game which is in TZUV then he can win playing first by moving to such an option since Left loses when playing first from such a position. If Right has no such option, then his options must be in C U M from which Left will win playing first. Therefore, Right loses G playing first.

These two observations determine whether or not each player can win playing first which in turn determines the outcome class of G. •

We can now analyze games by assigning outcome classes to the nodes of their game trees, starting with the leaves which are V positions for Normal Play games 8

AM V V V V

Figure 1.2: Normal Play outcome analysis of a game tree and M positions for Misere Play games. Examples of this analysis are shown in Figure 1.2 and Figure 1.3.

We can also note that there is an order of preference among the outcome classes with respect to each player. Left will always prefer a game in C to any other outcome class while Right will always prefer a game in 1Z. Neither player prefers V over H or vice versa since they depend on who plays first. If we arbitrarily assign Left as preferring greater items, we can construct a small partial order among the outcome classes as shown in Figure 1.4.

We may now compare outcome classes by making statements such as "G E V and o(G) > o{H) which implies H E TZ U V".

1.5 Structure of Games

Definition 7. The disjunctive sum of two games G and H is recursively defined as follows: 9

n

M n M

VMM

M M

Figure 1.3: Misere Play outcome analysis of a game tree

£

\ / K

Figure 1.4: The partial order of outcome classes 10

G + H= {gL + H,G + 7iL | GR + H,G + HR}

Note that QL + H is the abbreviated way of writing {G{ + H, G^ + H,...} where QL — {Gf,G%,...}. The disjunctive sum is both commutative and associative.

Often a game can be analyzed by breaking it down into a disjunctive sum of simpler games. Each summand is referred to as a component.

Another important concept when dealing with games is that of equality. Clearly, if two games have the identical game trees then they are going to be equal.

Definition 8. Given games G and H, G = H if G and H have identical game trees.

If we take one of the game trees and add a right option which is in some sense worse than another of Right's options, we have different game trees even though the two games might behave similarly when played in isolation or even in a sum with other games. To make this more formal, we would like a definition for equality which gives us an equivalence class of games which behave in the same way regardless of other games with which it might be played in a sum.

Definition 9. Given games G and H, we say that G is equal to H under Normal Play rules if o(G + K) = o(H + K) for all K. In this case, we write G = H.

We can likewise define equality for games played under Misere Play rules by mak­ ing use of the misere outcome function.

Definition 10. Given games G and H, we say that G is equal to H under Misere Play rules if o~(G + K) = o~{H + K) for all K. We again write that G — H. 11

Although we use the notation G = H in both cases, it should be clear with context which we are dealing with. The set of games under Normal Play rules is different from the set of games under Misere Play rules and equality as defined above gives us the equivalence classes in each group. We will see in Chapter 3 that these differences can be significant with respect to our ability to analyze games.

One last concept which allows us to compare games is that of inequality. We would like to say that if a game G is at least as good for Left as H then G > H. Since we want this to be true regardless of other summands, we get the following similar pair of definitions:

Definition 11. Given games G and H, G > H if o(G + K) > o(H + K) for all K under Normal Play rules.

Definition 12. Given games G and H, G > H if o~[G + K) > o~~(H + K) for all K under Misere Play rules.

To further simplify our analysis of Normal Play games, we can assign them values as first described in On Numbers and Games [8]. The simplest game we can play is the one where neither player has an option. Under Normal Play rules, this game is assigned the value 0. The motivation for this is that when playing a disjunctive sum of games, if there is a component which has no options for either player, it has no bearing on the outcome of the game and is therefore an additive identity.

We may now begin to build up more games by using 0 as an option for either player. This gives us three new games to consider:

{0|-},{-|0},{0|0}

These games have values of 1, -1 and * respectively. We use the values 1 and -1 to describe the fact that these games give one free move to Left and Right respectively. In general, positive values are an advantage to Left and negative values are an ad­ vantage to Right. 12

Definition 13. The game G = \QL | GR} is born on day n if it can be represented with the elements ofQL and QR being games born by day n—1 but cannot be represented with the elements of QL and QR being games born by day n — 2 and the game {• | •} is born on day 0. We say the birthday of G is n.

Definition 14. A game is born by day n if it is born on day k for k < n.

Definition 15. A follower of a game G is any game H which can be reached after some sequence of (possibly non-alternating) moves.

Due to the recursive nature of games, analysis is often done in an inductive man­ ner using the birthday of a game. The negative of a game is motivated by the idea of playing the game with the Left and Right roles reversed. Formally, we get the following definition.

Definition 16. The negative of a game G is given by: —G = { — QR | —<3L}

In Normal Play games, this allows us to develop some algebraic identities which will be useful in the analysis. These results can be found in any of the standard texts on combinatorial game theory [2, 4, 8].

Lemma 17. [2][Lemma 4..11] For all games G using Normal Play rules, G G P if and only if G = 0.

Proof. Given G E P, we must show that o(G + K) = o(0 + K) for all games K. First we assume that Left can win K. Then Left can win G + K by playing a winning strategy in K and responding locally in G if Right plays there. We know that Left can always make a response in G because there is a winning strategy by playing second in that component.

Likewise, if Right can win K then Right can win G + K in the same manner. There­ fore, o(G + K) = o(0 + K) and we can conclude that G — 0. •

Lemma 18. [2][Lemma 4-M] For all games G using Normal Play rules, G — G = 0. 13

Proof. By making use of the previous lemma, we must simply show that G-\- (-G) G V. Using the definition of —G we are considering the sum \QL | GR} + {—QR \ —GL}- If Left plays first to some GL in the first component, then Right can respond by playing to — GL in the other component. If Left instead plays to some — GR in the second component then Right can respond to GR. In either case, after the pair of moves we are again at a position in the form H — H which is inductively equal to 0 which means that this is a winning strategy for Right. We can likewise show that Right has no good move playing first so we can conclude that G — G G P and therefore G-G = 0. • •

Lemma 19. [2][Lemma J^.18] For any games G and H using Normal Play rules, G = H if and only if G — H = 0.

Proof Since G — H we know that o(G+K) = o(H + K) for all K. If we add the com­ ponent —H to each side then we see that o{G — H + K) = o(H — H + K) = o(0 + K) for all K and we can conclude that G — H = 0.

If we instead assume that G — H = 0 then we know that o(G — H + K) = o(0 + K) = o(H - H + K). Then by adding H to each side we get o(G + K) = o{H + K) for all K and we conclude that G — H. •

Under Normal play rules, we also get rather important results that games form an abelian group and that the relation > gives a partial order. The proofs for these results can be found in Lessons in Play [2]. As we will see in Chapter 3, > is still a par­ tial order on games under Misere Play rules but they no longer form an abelian group.

It is worthwhile to note that when dealing with the group of games, equality is an equivalence relation as defined above. That is to say, for each group element there may be many different game trees which represent that element. Fortunately, there is a way to determine a "simplest" member of the equivalence class which we refer to as its canonical form. 14

Definition 20. Given a game G, a left option GLl is dominated if there exists some other left option GLi where GLl < GL2. Likewise, a right option GRl is dominated if there is some other right option GR2 where GRl > GR2.

Definition 21. Given a game G, a left option GL is reversible if there is some GLR where GLR < G. Likewise, a right option GR is reversible if there is some GRL where GRL > G.

Given an arbitrary game, its game tree may be simplified in one of two ways:

1. Remove dominated options.

2. Bypass reversible moves.

In the first case, if a player has a dominated option, it cannot improve the outcome since we know there is some other option which is at least as good, regardless of any other summands. Therefore, it does no harm to remove that option.

In the second case, if Left plays to an option which is reversible, Right may re­ spond in the same component to a game that is at least as good for him as G was. Therefore, we can simplify the game tree by bypassing this pair of moves and replac­ ing GL by the set of all the left options of GLR. We can likewise bypass a reversible Right option GR by replacing it with all right options of GRL.

Theorem 22. [2j[Thm 4-33] Repeated application of these two methods of simplifying a game tree result in the game's unique canonical form.

For most analysis of games, we may restrict ourselves to the canonical forms of the games in question. To further streamline the process, we assign values to many of the important equivalence classes of games. We have already seen that in Normal Play, 0 is an important game since it represents all games where the second player has a winning strategy and it acts as an additive identity in the group of games. 15 /{110w} i i o

* 0 0

0 0

Figure 1.5: Normal Play value of a game tree

On Numbers and Games [8], Winning Ways [4] and Lessons in Play [2] offer a more complete introduction to the naming of games and show the development of numbers, ups, tinies and many others. A brief dictionary of game values used in this thesis can be found in the appendix. Another group of games called nimbers is discussed in the next section.

Now that we have values to represent specific games under Normal Play rules, we can calculate the value of a game based on its game tree by assigning values to each node, starting with the leaves which must have value 0. An example is shown in Figure 1.5. Unfortunately, we are unable to assign values when dealing with games with Misere Play rules if we want those values to have a useful algebraic meaning like they do under Normal Play rules. We will examine why in the next chapter.

1.6 NIM

The game of NIM is a game played with heaps of counters. On each player's turn, they select one of the heaps and removes a positive number of counters from that heap. Since each player's options are the same, this is an impartial game. Under Normal Play rules, whoever takes the last counter wins while under Misere Play rules 16 whoever takes the last counter loses. The game is well understood under both ver­ sions and is the first combinatorial game that was studied in a mathematical sense [6] .

Definition 23. The nimber */c, k > 0 is recursively defined as the game

{0, *1, *2,..., *(k - 1) | 0, *1, *2,..., *(k - 1)}

We more succinctly refer to *1 as * and *0 as 0.

Note that a game of NIM played with a single heap of size n will have value *n. This is how these game values got their name.

Lemma 24. [2][Thru 7.7] For any n, *n = — *n.

Proof. We proceed by induction and note that trivially 0 = — 0 since 0 has no options for either player. For a given k, we get:

*k = {0,*1,*2, ...,*(k- 1) | 0, *1,*2, ...,*(k- 1)} = {-0, -*1, -*2,..., -*(k - 1) | -0, -*1, -*2,..., -*(k - 1)} = — *k

D

Definition 25. Following the Sprague-Grundy theory of impartial games, a game with value *n is said to have Q-value n.

Definition 26. Let A be a set of nimbers and let B be the set {k : *k E A}. Then MEX(A) = *i where i is the minimal excluded non-negative integer in B.

Theorem 27. [2][Thm 7.7] Let G be a game where the options to both players are a set S of nimbers. Then G = MEX(S).

Proof. Let MEX(S) = *n. To show G = MEX(S) we can equivalently show that G — *n = G + *n = 0. Assume the first player plays in *n to some *k. Then the second player can play in G to *k to win since we know k < n and n was the minimal 17

3 -> (011)2 ® 5 -> (101)2 6 <- (110)2

Figure 1.6: Example of the nim-sum operation

excluded option of G. Assume instead that the first player plays in G. If his move is to some *TO with m > n then the second player can respond in the same component to *n to win. If his move was to some *m with m < n then the second player can play in *n to *m to win. Therefore, the second player has a winning strategy and we conclude that G = MEX(S). •

Theorem 28. [2][Thm 7.8] Every impartial game G is equivalent to some nimber.

Proof. By induction the options of every impartial game are impartial and therefore equivalent to some nimber. The previous theorem then shows us how to compute the nimber which is equal to G. •

Definition 29. The nim-sum of two non-negative integers is equal to the exclusive or (XOR) sum of their binary representations.

Theorem 30. [2][7.12] A game of NIM with Normal Play rules is a V-position if and only if the nim-sum of the heaps is 0.

Proof. It will suffice to show that from any position where the nim-sum is 0, every move must be to a position where the nim-sum is non-zero and from every position where the nim-sum is not 0, there must be some move to a position which has nim- sum 0.

For the first case, if the nim-sum is 0 then when a move is made exactly one of the heaps must reduce in size. This in turn implies that some digit in its binary representation must now be different. Therefore, when computing the new nim-sum, that column will have an odd number of Is and therefore the nim-sum is not 0.

For the second case, we begin with a game where the nim-sum is not 0. To find 18

9 -> (1001)2 (1001)2 -> 9 13 -> (1101)2 -> (1010)2 -»• 10 5 -> (0101)2 (0101)2 -+ 5 e 6 -^ (0110)2 (0110)2 -» 6 e 7 <- (0111)2 (0000)2 -»• 0

Figure 1.7: Finding a winning move in a game of NIM with four piles a move which returns to a position which has nim-sum 0 we must first examine the binary representations of the heaps and choose a heap which has a 1 in the highest order column where the nim-sum also has a 1. We then reduce that heap to create a 0 in that column and ensuring that there are an even number of Is in all lower order columns. This can always be done because the resulting binary string represents a heap size smaller than before the move was made. Therefore, the nim-sum after this move will be 0. •

When playing the misere version of NIM, we first note that our simplest case of having no counters in any heap is an J\f position rather than a V position as it was in Normal Play. Fortunately, the analysis for the Misere Play version is strikingly similar to that of the Normal Play version.

Theorem 31. Let G be a game of NIM. Then o~(G) = V if and only if

1. There are an odd number of heaps and each heap has exactly one counter, or

2. There is some heap of size at least 2 and o{G) = V.

Proof. We again must show that for every game that satisfies the above property, every legal move creates a position which does not have that property and that from every position which does not have the property, there exists some move to a position which does.

We will begin by considering the positions which are composed of only heaps of size 1. On each player's turn, they must remove one of the heaps so the game will end after a number of moves equal to the number of heaps. If there are an even number of heaps, the first player will win since his opponent will take the last counter 19 leaving him with no move. Likewise, if there are an odd number of heaps, the second player will win.

If there is some heap of size 2 or more, then we will proceed by induction on the sum of counters in all heaps. If the nim-sum is 0 then there is no move to a game of nim-sum 0 as we've already shown. The only possibility for a winning move would be to an odd number of heaps of size 1. In order to have such a move while having nim-sum 0 implies that all heaps are already size 1 which contradicts that there is a heap of size 2. Therefore, the first player cannot win from such a position.

Finally, consider a position where the nim-sum is not 0 and there is some heap of size 2 or more. If there is some option which has nim-sum 0 and includes some heap of size 2 or more, then this is a winning move by induction. If the only move to a position of nim-sum 0 has only heaps of size 0, we should instead move to create an odd number of heaps. We can always do this since we must be removing counters from the single pile of size 2 or more. We should leave a single counter if there are an even number of other heaps and remove the entire heap otherwise. Therefore, the first player can win from such a position.

We can therefore conclude that the condition given describes the set of all misere NIM positions from which the second player can win. •

This demonstrates that most games of NIM have the same outcome class when played under Normal or Misere Play rules but care must be taken when we approach the end of the game. We will see that a similar phenomenon occurs with other more complex games as well. The strategy used in the game is one of maintaining control until a critical decision must be made to ensure victory. In that sense, maintaining control in a game does not necessarily depend on the winning condition but rather the rules of the game. 20

n 0 1 2 3 4 5 6 7 8 9 10 11 12 13 ... Rvalue I 0 1 0 1 2 3 2 0 1 0 1 2 3 2 ~

Figure 1.8: The nim-sequence for S = {1,3,4}

1.7 Subtraction Games

Subtraction Games are a generalization of NIM. The game is still played with heaps of counters except that we specify a subtraction set (which may be finite or infinite)

S = {si, s2, •••} where each Sj is a positive integer. On each player's turn they select one of the heaps and removes k counters from the heap where k G S. Clearly, if we allow S = Z+ we would be playing NIM since we could remove any number of counters.

When analyzing a subtraction game, you would need to start by computing the Q- value for each heap size. When doing this by hand, it is easiest to construct a Grundy scale which consists of two rows of values. The top row shows the heap size and the bottom row shows the associated ^-value and is also referred to as the nim-sequence. To find the Rvalue of a heap of size n using such a scale, you need only look at the

(/-values of the heaps of size n — s1; n — S2, ••• for each s* G S then take their MEX. Figure 1.8 shows an example of such a nim-sequence derived from a Grundy scale where S = {1,3,4}.

Theorem 32. [2][7.33] All nim-sequences for subtraction games where S is finite are periodic.

Proof. Since we know that S is finite, then we know each player has at most \S\ options from any given position. Since we use the MEX rule to determine the next Q-value we know that there is no Q-value greater than \3\. We also know that there is a maximum s* G S which means that for any given n, the (5-value of a heap of size n only depends on the

n | 0 1 2 3 4 5

value | 0 1 {1 | 0} -x 0 {0 | -J ...

Figure 1.9: The sequence of values for SL = {1},SR = {2, 3} have such repeating blocks the term following such blocks must also be the same. Therefore, the sequence has become periodic. •

Now that we know that all subtraction games are periodic, we would like to find a way to determine what the period is for a given finite S = {si, s2, •••}• If, when computing the nim-sequence we have a conjecture for the length of the pre-period I and period p then we can be sure that it will continue indefinitely if we have checked up to heaps of size I + p + s — 1 where s — max{sj}.

To see why this is the case, consider the Q-value of the heap of size / + p + s. We know that a player's options are all of the form I + p + s — Sj which we know have the same value as 1 + s — S; since we've already checked for periodicity from I on. We then observe that these are exactly the options of the game starting with a heap of size I + s which implies that Q{1 + s) = Q{1 + p + s). Applying induction from this point shows that the observed pattern will continue.

Subtraction games may be further generalized if we designate subtraction sets for each player SL and SR. A player can only remove k counters from a given heap if k is an element of their subtraction set. Since this might allow options to one player that are not allowed to the other it becomes a partizan game. Although we are no longer restricted to nimbers, we can still construct a scale which will contain the game values for each heap size. Figure 1.9 shows sequence of values obtained in the game where SL = {1},5R = {2,3}. The values need not be periodic when each players' subtraction sets are different. It has been shown, however, that the sequence of out­ come classes is ultimately periodic for any partizan subtraction game [13]. We will see a similar result in chapter 4. Chapter 2

Option-Closed Games

2.1 Definitions and Examples

In this chapter we will examine a specific subset of games where if a player can get to a given position in two moves, he can get to that position in one move. In other words, the set of options available after a player's move is a subset of the options be­ fore his move. This is of interest because we would ultimately like to analyze misere games. By first restricting ourselves to smaller sets of games, we may gain insight as to how misere games interact. As it turns out, no significant results were found regarding option-closed games under Misere Play rules but quite a bit was discovered when considering Normal Play rules. In Chapter 4, we will see a further restriction on the options allowed to each player which does yield some interesting Misere Play results.

Definition 33. A game G is called option-closed if GLL C GL, GRR C GR and all of the options of G are option-closed.

It is important to note that option-closed is a property of a game which depends on its representation. In particular, any specific equivalence class of games may have some elements which are option-closed while others are not. Moreover, the canon­ ical form of a game need not be option-closed even if its equivalence class contains option-closed games. When referring to a game G which is option closed, we assume that G is represented in a way which satisfies the definition given above.

The game of ROLL THE LAWN uses a row of bumps (nonnegative integers) and a roller that is either between two bumps or at one end. Left moves the roller to the left flattening each bump by 1 unless the bump has been flattened to 0 in which case

22 23

[5, 0, 2, 4] -^ [5,1, 3, 0] ^ [5, 0, 0, 2] -^ [5, 0,1, 0] -^ [5, 0, 0, 0]

Figure 2.1: A game of ROLL THE LAWN

[5, 0,1, 2, 4] -^ [5, 0,1, 0, 4] -^ [5, 0, 0, 0, 4]

Figure 2.2: A game of CRICKET PITCH

nothing happens, however, at least one bump must be flattened by one on a move. Right moves the roller to the right, with the same effect and constraint. We represent a position by a string of nonnegative integers with a roller, 0, in the string. We also use —>• to indicate a Left move and a Right move by —>. Figure 2.1 shows an example of play where Right plays first and Left wins. Note that the impartial version of this game where each player can move the roller in either direction is not option closed. To see why, we consider the game with a single bump. On a player's turn their only option is to reduce it by one but can further reduce it on a subsequent turn by rolling over it again. The option of reducing it by two did not exist from the initial position.

In ROLL THE CRICKET PITCH, or CRICKET PITCH for short, there is an extra constraint: the roller cannot go over a bump which has already been flattened to 0. Figure 2.2 shows an example of play where Right plays first and Left wins.

2.2 Reduced Canonical Form

Although the canonical form of an option-closed game can be complicated, we will show in this section that they can be expressed in simpler terms. For example, in CRICKET PITCH;

G=[1,1,3,0,5,3,1] = {1, {1|0}|0, {1|0}, {1,{1|0}|0,{1|0}}}, but this is {1|0} + G' where G' is an infinitesimal (see Definition 34). Left's winning move is to [1, 0, 0, 2, 5, 3,1] and Right's is to [1,1, 3, 4, 2, 0, ©]. It can be shown that all other moves lose. 24

The difference G — {1 | 0} is an infinitesimal, indeed, — i < G — {1 | 0} < +i . As we shall show in the next section, the reduced canonical form (see Definition 37) of any option-closed game is simple and the difference between a game and its reduced canonical form is not very large.

For many situations, a player would like to ignore the infinitesimal values since they only determine the parity of the number of moves once the associated non- infinitesimal value has reached zero. Grossman & Siegel [14] showed that the idea of a simplest game infinitely close to a given game, called the reduced canonical form, is well-defined. (Calistrate [7] first introduced the idea but the proof contained a flaw and a different approach was required.) We need a brief development of the tools so that we can apply them to option-closed games.

Definition 34. A game G is an infinitesimal if, for every positive number x, we have —x < G < x. Let Inf denote the set of infinitesimals. When G — H is infinitesimal, we say that G and H are infmitesimally close, and write G =inf H. We will sometimes say that H is G-ish (G Infinitesimally SHifted).

Definition 35. [14] G >/n/ H if G > H + e for some infinitesimal e; G

Definition 36. [14] Let G be any game.

L L L • A Left option G is Inf -dominated if G

L LR LR • A Left option G is Inf -reversible if G

The definitions for Right options are similar.

For example, let G = {1, {1|0}|0}. Then {1|0} is an Inf-dominated Left option of G. since {1|0} < 1 + T- If this 'dominated' option could be eliminated, the resulting simpler game would be {1|0}. In the reduced canonical form, the Inf-dominated options are removed. 25

Definition 37. [14][Def 4.2] A game G is said to be in reduced canonical form provided that, for every follower H of G, either:

• H is a number in simplest form; or

• H is not a number or a number plus an infinitesimal, and contains no Inf- dominated or Inf-reversible options.

Theorem 38. [14][Thm 4-3] For any game G, there is a game G in reduced canonical form with G =inf G.

Theorem 39. [14j[Thm 4-4] Suppose that G and H are in reduced canonical form.

IfG=ln{H, thenG = H.

This then shows that the reduced canonical form of a game G is well-defined and unique. It is denoted G. The next lemma and theorem are needed to prove results about the reduced canonical form of option-closed games.

Lemma 40. [14][Lemma 4-6] If G is not a number and G' is obtained from G by eliminating an Inf-dominated option, then G' =inf G.

Theorem 41. [UjfThm 2.10] If G = {GL\GR} is not a number and G' = {GL'\GR'} L L R R is a game with G =inf G and G =inf G , then G' =inf G.

We are now in a position to consider option-closed games.

Lemma 42. For any numbers a and b, if b < a then a >inf {a\b}.

Proof Consider a — {a\b} + n- | = a + {—b\ — a} + n • f- By the Number Avoidance Theorem [2], if Right can win then she must have a good move in either {—b\ — a} or in n • |. But the former move leaves a — a + n • | > 0 while the latter leaves a + {—b\ — a} + (n — 1) • f *, in which case Left wins for n > 3 by responding to a-6+(n-l)-T*>0. D

Theorem 43. Let G be a option-closed game, then the reduced canonical form of G is either a number or a switch, {a\b}, where a and b are numbers. 26

Proof. We proceed by induction and use Lemma 40 and Theorem 41. Suppose G is the simplest option-closed game whose reduced canonical form is not a switch or a number. Then the reduced canonical forms of all the Left options of G are all switches or numbers, so by Lemma 40 and Theorem 41 any Left option can be replaced by its reduced canonical form. If the switch {a\b} is a Left option

then so is a. Now by Lemma 42, a >inf {a\b} and Lemma 40 and Theorem 41 the switch is Inf-dominated and can be removed. •

Definition 44. [2, 4] Denote the left stop and right stop of a game G by LS(G) and RS(G), respectively. They are defined in a mutually recursive fashion:

G if G is a number, LS(G) (2.1) L ' max(RS(G )) if G is not a number,

G if G is a number, RS(G) = (2.2) " min(LS(GR)) if G is not a number.

The proofs of the next two Corollaries follow directly from Theorem 43 and from the definitions of stops and option-closed games.

Corollary 45. Suppose G is option-closed.

1. If G = x then LS{G) = RS(G) = x. Moreover, if G ^ x then there exists left and right options with GL = GR = x.

2. If G = {a\b} then LS(G) = a, RS(G) = b. Moreover, there exists left and right options with GL = a and GR — b.

Corollary 46. Suppose G is option-closed and suppose LS(G) = x. If the left option GLl has a right option then there is some right option GLlR2 = y < x for some number y.

Note that Corollary 45 gives information about the infinitesimals when G is a number. 27

Corollary 47. Let G be a option-closed game. IfG is a number and G ^ G then G — G is confused with 0.

However, G — G is infinitesimal and can also be bounded.

Theorem 48. Let G be a option-closed game then JJ-* < G — G < f*

Proof. If G = G then the result is true. First, let's assume that G — G is positive. Then we must show that G — G + JJ-* < 0.

We have two cases. First, we assume G = x for some number x. Now we have to show that Right can win G — x + JJ-* going first or second. The situation, is highlighted by the following game tree. G - x + JJ.*

G'-x + ty* G-x + i ^*

\ \ JJ-* i When Right plays first, we know that she can play in G to x since x must be a Right stop of G. Using the fact that G is option-closed tells us she can get to this stop in one move. That leaves us with 4* < 0 so Right wins.

If Left plays first, his only good moves can be in G or Jj.*. Suppose Left plays in G to G'. If Right has a legal response in G then by Corollary 46, Right has a move to a number which is at most x. After this pair of moves, it will be Left's turn and in a position no more than JJ,* < 0 so he again loses. On the other hand, if Right has no legal move in G\ and since G ^ x then, by Corollary 45, G' must be a number with value at most x. This will leave a position with value at most JJ-* which Left loses. Therefore, his only possible winning move would have to be in JJ-* to [. As before, Right wins by replying in G to x. So, Left has no good move and loses regardless of who plays first. 28

Now we assume that G = {a\b} where a > b. So, we need to now show that G + {—b\ — a] + 4* < 0. If Right plays first, she can move to b + {—b\ — a} + JJ-* by Corollary 45. From this position, Left's best move must be in the switch to b — a + J. (by the Number Avoidance theorem and since a > b) but this is negative since b < a so Right wins playing first. The moves are highlighted in the following game tree. G + {-b\ - a} + ij-*

\ b + {-b\ -a} + ty*

/ b — a + -Ij.*

If Left plays first in G + {—b \ —a} + 4* he has three different components he could play in. These are illustrated in the last game tree. To begin, if he plays to G — b + -Ij-*, Right responds to JJ-* and wins.

If instead he plays to G + {—b\ — a} + J,, by Corollary 45, Right can respond to b + {—b\ — a} + 1. From this position, Left's best move must be in the switch which leaves | < 0 and he loses.

The final case to consider is when Left plays in G from G + {—b\ — a} + JJ-*. As before, we will call this position G' and we are left with G' + {—b\ — a} + JJ-*. Right has a winning strategy from here by replying to G' — a + -JJ-* since now Left must either play in JJ-* or G" if a move exists for him in that component. In the first case, we are left with G' — a + j. As before, by Corollary 46, if Right has a legal move in G' we know it is to a number no larger than a and this is a winning move. If there is no legal move for Right, then we are already in such a position which must have a value of at most J. and it is again a win for Right. In the second case, we assume Left has a legal move in G' and plays to G" leaving us with the position G" — a + -JJ.*. As before, by Corollary 46, either Right can move in G" in which case it leaves a game with value at most -Ij-* or Right has no move in G" in which case we are already in a position with value at most JJ-*. Both cases lead to a Right win. 29 G + {-b\ - a} + #*

G + {-b\-a} + i G-b + ty* G' + {-b\ -a}+ 4*

b + {-b\-a} + i -It* G' - a + ^*

G" - a + ^* G' - a + j

-U-* a

2.3 ROLL THE LAWN

In our analysis of ROLL THE LAWN, the main observation is that if the roller starts on one side of a bump then it will finish on the same side if the bump is even and on the opposite side if the bump is odd. The winner will be the player that has more odd bumps in their direction.

Lemma 49. Let G = [ai, a2, •.., %, 0, Oj+1, a*;] be a ROLL THE LAWN position and let bi = a,j (mod 2) then

[al5 a2,..., a,j, 0, ai+i,..., ak] = [bu b2, • • •, fy, 0, &j+i, • • •, &fc]-

Proof. We consider the difference game

[au a2,..., aj, 0, aj+1,..., ak] - [bu b2,..., bj, 0, bj+1, ...,bk] and proceed by induction on the sum over all <2j. If all a* are 0 or 1, then <2j = 6j for all i and the difference is 0. Otherwise, without loss of generality, consider any move by Left. If Left moved the roller over an odd bump then Right can make the same move in the other component and we are left with a position which by induction is a second player win. If, on the other hand, Left did not move the roller over any 30 odd bump, then Right can play in the same component by moving the roller back to where it was before Left played. All of the bumps in that component which have been reduced are reduced by exactly 2 and since there was no bump of size 1, we are again in a position which is a second player win by induction. Therefore, Left cannot win playing first. A similar argument shows that Right cannot win playing first and therefore the difference game is in V and we conclude that the two games are equal. •

It is an easy to prove that the actual value is a number and is given by the next result.

Corollary 50. Let G — [ai, a2,..., %, 0, a,-+1,..., ak] be a ROLL THE LAWN position and let fcj = at (mod 2) with hi e {0,1} then

Proof. We proceed by induction over the sum of all 6j. By induction, Left's options are to integers from — Y2i>j ^ to (^2ij h) — 1- Likewise, Right options are to integers from f Yliij &i) + 1 to X^7 ^ which is therefore the value of the game. •

2.4 CRICKET PITCH

Before considering a game of CRICKET PITCH we first introduce some notation. If

X = (XJ)"=1 then we put X — 1 = (xi — l)^=i- In the analysis of the CRICKET PITCH position [X, 0, Y] where X and Y are strings of non-negative integers, a typical left move will be to [Xu 0, X2 - 1, Y].

Since CRICKET PITCH is an option-closed game, then by Theorem 43 the reduced canonical form for [X, 0, Y] is either a number or a switch {a|6}.

Observation 51. If a position of CRICKET PITCH has a bump of size 0 then anything on the other side of the bump from the roller is irrelevant since the roller can never 31

[i,e,i] = * [1,3,0,3,1] = *2 [1,3,5,0,5,3,1] = *3 [1,3,5,7,0,7,5,3,1] = *4 [1,1,0,3,3,2] = 5/4 [1,2,0,3,3,1] = | + i2* [1,1,0,3,3,1] = {0,{0,{0|-1}|-1}|-1} [1,3,0,3,1,2] = *2 [1,2,0,3,3,3] = 1/8

Table 2.1: A short cricket pitch dictionary

reach it, and so can be dropped from the position without changing the game. For example, [2, 3, 0,3,4, 0,1, 2, 0,56] = [3, 4, 0,1, 2]

Lemma 52. For any CRICKET PITCH position,

[ai, a2,..., dj, 0, aj+1,..., ak] = [ai + 2, a2 + 2,..., a,- + 2, 0, aj+i + 2,..., ak + 2]

Proof. Let A, 5 be sequences of non-negative numbers. We let (A + 2) denote the sequence where every number in A has been increased by 2. Consider the position [A, 0, B] and consider the game [(A + 2), 0, (S + 2)] - [A, 0, B] with Left moving first. We claim that Left cannot win. If Left moves to [(A + 2),Q,(B + 2)]-[A,B',Q,B"] then Right responds with the 'mirror' move in the other component to

[(A + 2), {B1 + 1), 0, (B" + 2)] - [A, B\ 0, B"\ which is 0 by induction. If Left moves to

[{A + 2), 0, (A" + 1), (B + 2)] - [A, 0, B] then if possible, Right mirrors by playing to

[(A' + 2), 0, (A" + 1), (B + 2)] - [A1, ©, (A" -1),B] 32 which is zero by induction. If this move is not possible then we know that Left moved over a 2 in (A + 2). That is, A = C, 0, D where D does not contain any 0s. Then Left must have initially made the following move:

[(C + 2), 2, (D + 2), 0, (B + 2)] - [C, 0, D, 0, B] -+

\{C + 2), e, (C" + 1), 1, (D + 1), (B + 2)] - [C, 0, D, 0, B).

We claim that Right wins by moving to

[(C + 2), 0, (C" + 1), 1, (D + 1), (B + 2)] - [C, 0, 0, (£> - 1), B).

From here, if Left moves in the second component of

[(C + 2), 0, (C" + 1), 1, (D + 1), (B + 2)] - [C, 0, 0, (15 - 1), B]

Right mirrors Left's move in the other component and the resulting position is 0 by induction. Otherwise, Left moves in [((7 + 2), 0, (C" + 1), 1, (D + 1), (B + 2)] then Right responds in the same component moving over but not further than the bump of size 1, i.e. ignoring the bumps that are no longer reachable. She moves to

[0,Q,(D + l),(B + 2)]-[0,Q,(D-l),B] and this is 0 by induction. •

Definition 53. Let G be the Cricket Pitch position [AQB]. The Left odd, low point, ldip(G), is

ldip{G) = min{al\ai is odd and a* < a,j, Vjf > i},

If there is no such heap then ldip(G) = oo. Similarly,

rdip(G) = min{bi\bi is odd and bj > bi,\/j > i},

For example, in [1,2,3,4,0,1,2,2,3], the heaps in position 1 and 3 are both smaller than the heaps between themselves and the roller but the heap of size 1 is the least so that ldip(G) = 1. Note that rdip(G) = 1 since the heap immediately to the right of the roller is odd and there is no smaller heap between it and the roller. In [3, 2, 3, 4, 0, 2, 2, 2, 3], ldip(G) = 3 and rdip(G) = oo. 33

Lemma 52 and Observation 51 give an algorithm to simplify CRICKET PITCH po­ sitions. If we apply Lemma 52 repeatedly, we eventually reach an equivalent position with a heap of size either 0 or 1, we say that this position has been reduced. Prom Observation 51, if a position has a 0 heap, it can be replaced by a simpler, we say pruned, position. A position has been reduced and pruned if these two operations are repeated until either there are no heaps left, or there is a heap of size 1 but none of size 0. For example:

[3,2,5,8,0,6,2,5,7] -• [1, 0, 3, 6, 0, 4, 0, 3, 5] reduce —> [3, 6, 0, 4] prune -» [1,4,0,2] reduce

Theorem 54. Let G = A,Q,B be a CRICKET PITCH position. The outcome classes are determined by the odd low points:

1. if ldip{G) < rdip(G) then G e C;

2. if ldip{G) > rdip(G) then G e TZ;

3. if ldip(G) = rdip{G) < oo then G £ J\T;

4. if ldip(G) = rdip(G) = oo then G e V.

Proof. For part 1, since ldip(G) exists it is an odd number and is also smaller than all the heaps between it and the roller, therefore we can reduce and prune to get an equivalent position G' with ldip(G') = 1, rdip(G") > 1 and no heap of size 1 to the right of the roller. Now Left can win G', and so G, playing first or second since he can always move past the 1 leaving Right with no move. Similarly for part 2. For part 3, the reduced-and-pruned G' will have ldip(G) = rdip(G) = 1 and now both players have a winning first move. Suppose ldip(G) = rdip(G) = oo. Now any odd-sized heap has at least one smaller even-sized heap between it and the roller then all odd heaps will get pruned. Therefore the reduced-and-pruned position is 0 with no heaps. D 34

Intuitively, low odd numbers and high even numbers are preferred. We can make this statement more exact. For playing CRICKET PITCH in isolation, Theorem 54 is good enough. But how does one play a combination of several games of CRICKET PITCH, or CRICKET PITCH with other games? For example, in the following disjunctive sum where every position is in J\f

[1, 2, 3,1, 0, 4,1, 3] + [1,1, 3, 2, 0,1, 3,1] + [3, 3, 2,1, 2, 0, 2, 2,1] + [1, 2, ©, 3,1, 2]

but who wins? The next result is a step in the right direction.

Theorem 55. Let G be a CRICKET PITCH position.

• If oo= ldip{G) = rdip(G) then G = 0;

• // ldip(G) = rdip(G) then G = {a \ b} for integers a and b, a>0>borG = 0 if a = b = 0;

Proof. Case 1 follows since G G V.

If ldip(G) = rdip(67) then the reduced-pruned game has heaps of size 1 to the left and to the right of the roller. Let a be the non-negative integer obtained by Left moving just past the closest heap of size 1 and b the non-positive obtained by Right moving just past the closest heap of size 1.

On Left's turn, he must move at least far enough to go past the closest heap of size 1, otherwise Right will win on her next turn. Also, we know that LS(G) > a since Left has an option to a. If LS(G) = x > a then it must correspond to a move beyond his move to a. This is a contradiction since it would imply that there is a left option of a that is to a number greater than a. • Chapter 3

Recent Results in Misere Play

3.1 Introduction

Since the difference between Normal Play and Misere Play only involves a change in the winning condition it would be easy to think that games that were Left wins are now Right wins, games that were first player wins are now second player wins and vice versa. This is not the case. A simple counterexample is NIM played with a single heap of size. 2 which is a first player win regardless of the play convention. The outcome analysis in both Normal Play and Misere Play is shown in Figure 3.1.

In [8, 4] it is mentioned that misere games are harder to analyze. A survey of the literature reveals only a few papers which are either devoted to the analysis of a single game [16, 18] or to a restricted class of games [3, 11, 12, 21]. Recently, however, Plambeck [17] and Siegel [20] have made major steps forward in analyzing impartial misere games. It is clear that relatively speaking, there is very little known regarding misere games compared to their Normal Play counterparts. Following this work, we

Normal Play Misere Play

M

V V V

Figure 3.1: Outcome analysis of NIM with one heap of size 2.

35 36

first try to answer the question: "What makes misere games so hard?" from a math­ ematical point of view. Our first goal will be to narrow down what aspects of Misere Play and Normal Play games are different and what aspects are the same.

A hint that the analysis of Misere Play games will be more difficult than Normal Play is that there is no equivalence class that G — G lies in for all G. Recall in impartial games that — G — G, and again, in misere NIM 2 + 2 is in V but 1 + 1 is in J\f. We shall show that

1. Misere Play games are partitioned into equivalence classes, but they may be small in comparison to those induced by Normal Play—as we shall prove, {-|-} is in a class by itself, see Corollary 59;

2. there are concepts similar to dominated and reversible options that can be used to simplify games, thus making a start towards finding a canonical form; and

3. misere games form a partial order, but many more games are incomparable which makes the partial order less useful than in Normal Play games.

3.2 The Disjunctive Sum of Two Misere Games

Despite the earlier NIM example of a game being in two different classes under the two conventions, there still might be some relationship between Normal Play and Misere Play outcomes. The first result shows that this is false.

Theorem 56. Let C\ and C2 be outcome classes. Then there exists a game G for which o(G) = C\ and o~(G) = C2.

Proof. Table 3.2 demonstrates an example of a game G for each of the 16 possible combinations for C\ and C2- •

Despite there not being an implication between the outcome class in Normal Play and the outcome class in Misere Play, does Misere Play preserve some of the structure of the outcome classes present in Normal Play? For example, if G,H E C does it follow that G + H G CI This is false and in the worst way. With G and H as given below, 37

Table 3.1: Comparison of normal and misere outcome classes

Normal —•>• Misere J. V c n M

• • V J // V /' \ \ • V • •

// \ // ' • \ • • * *\ c \ \

• • * V ^ / / n /' / •

<\ x ' / X O* \' / JV 38

G H

both are in C but G -\- H 6 1Z. In fact, there is no implication between the individual outcomes and the outcome of the sum, as can be seen from the next result.

Theorem 57. Let Cy, C% and C3 be outcome classes. There exists games G and H

with o~{G) = Ci, o~(H) = C2 and o~(G + H) = C3.

Proof. The Appendix contains a dictionary of positions demonstrating that the out­ come class of the disjunctive sum of G and H does not depend on the outcome classes of G and H played in isolation. •

3.3 Equivalence Classes of Misere Games

Without a partial order it is difficult to build a reduction scheme for Misere Play game's similar to that of Normal Play games but it is not impossible.

We begin by looking at the equivalence class of the empty game {• | •}.

Theorem 58. If H ^ {• \ •} then H ^ {• | •}.

Proof. We begin by letting H be some non-trivial game and therefore having a game tree different from the empty game. Without loss of generality we can assume it has some left option. We then consider G:

G 39

We can now show that H ^ {• | •} by showing that o~(H + G) ^ o~({- \ -} + G). The string of consecutive Left moves in G is longer than the depth of the game tree of H. We can easily verify that G G C. Now, when we play G + H, Right can win playing first by playing in G. Since Left has no moves in GR, he must play in H. Right can now play again to GRR + HL. Since GRR is a string of left moves longer than any sequence of play in HL we can deduce that Right will win this game. Since Right can win G + H playing first we know G + H 0 C and hence o~(G + H) ^ o~(G). •

Corollary 59. The equivalence class under = of {• | •} contains only one element.

Proof. This follows directly from the previous Theorem. •

As we can see, this is a major difference from normal play games. Rather than having all games within an outcome class being equivalent to {• | •} = 0, we discover that in misere play there are no other games which are equivalent to {• | •}. This naturally brings up the question of whether or not all equivalence classes are trivial. The following theorem demonstrates that this is not the case.

Theorem 60. There exists games G and H with G ^ H such that G = H.

Proof. Consider the following games G and H:

G H

Let K be a game and assume Right can win G + K playing first. Then Right can also win H + K playing first by following the same strategy while ignoring the string of moves starting at x. Likewise, if Right can win G + K playing second, he can win H + K playing second.

Now assume Left can win G + K playing first. We would like to show that Left can win H + K playing first. This is equivalent to showing that Right cannot win 40

H + K playing second. We proceed by contradiction and assume that Right can win playing second. If Right never plays to x, then we can use the same strategy when playing G + K, giving Right a winning strategy playing second which gives a con­ tradiction since we originally assumed Left would win playing first. So, Right must play to x at some point in her winning strategy. But then Right can win G + K playing second by ignoring the option to y. This again gives a contradiction so we can conclude that Right cannot win H + K playing second.

So, if Left can win G + K playing first, Left can win H + K playing first. Using a similar argument we can show that if Left can win G + K playing second, then Left can win H + K playing second.

Therefore, putting the results together, we can conclude that H + K has the same outcome class as G + K. •

This result tells us that misere games have non-trivial equivalence classes under =. We can also see from the construction that there is something similar to domination for misere games. Is there also an operation analogous to reversible options? And if so, does this lead to a canonical form? We make precise a notion of a dominated and of a reversible option.

Using the definition for G > H , we can find a couple general constructions for misere games which are are members of the same equivalence class. Much like normal play games, we find rules similar to domination and reversibility.

We note that there are misere games G and H with G ^ H for which G > H. As an example, we can choose the following two games: 41

G > H

/

The following three theorems are motivated by the process of simplifying games in normal play. In particular, applying Corollary 62 is similar to removing a dominated option while applying Theorems 63 and 64 are similar to bypassing reversible moves.

Theorem 61. If(HR D GR ^ 0 or HR = GR = 0) and (GL D HL ^ 0 orGLHL = 0) then G>H.

Proof. Let K be a game so that Left wins H + K. If GL D HL ^ 0 or GL = HL = 0 then playing first, he can adopt the same winning strategy when playing G + K. If right plays first, he cannot win H + K. Since his moves in G + K are a subset of the moves available in H + K, he cannot win G + K either. Therefore, Left wins G + K and we get that G > H. •

Corollary 62. Given the game G = {GL\GL\ GL\GR}, if we have that GLl > GL2 L L R thenG={G \jG \G }.

Proof. This follows from the definition of >. In any line of play, Left can always do at least as well by playing to GLl as he would by playing to GL2. Therefore, the option to GL2 will never be played and we get that G = {GL\ GL\GR}. •

Theorem 63. Let H = {A|5} and G = {A, C\B} where A is non-empty and A D RR RR CRL ^ 0_ Furthermore, B C C if B non-empty and C = 0 if B = 0. Then G = H.

G = H

B A /^ ^^\B

CR

Proof. To begin, we can note that G > H > CR by applying Theorem 62. Therefore, for any game K, if Left has a winning strategy in H + K, she also has a winning strategy in G + K.

Now we assume that Right has a winning strategy in H + K for some K. When playing G + K, if Left never plays to C, we know she will lose since Right will follow their winning strategy from H + K. So, if Left can win G + K she must at some point play to C. Right can immediately play to CR. Since CR < H and we know Right can win H + K, Right can also win CR + K. So, Left has no good move in G + K and Right wins. Therefore, we can conclude that G = H. D

Theorem 64. Let C be any set of games and let G and H be given as in the diagram below. Then G = H.

G = H

C

Proof. Assume Left wins H + K for some K. Since she has a move in H, it must at some point be a move she makes during her winning strategy. From this point, either Left will win before any further moves have been made in H, or Right will eventually move in K to some C + K' which we know is a winning position for Left. Moreover, any possible K' during Left's winning strategy makes C + K' a winning position for Left.

Now we consider the game G + K. Left follows the same strategy she would as if he were playing H + K moving in H when he normally would have moved in G. 43

Now, if Right eventually plays in G, Left immediately responds in G as well. After Left's move, we are in a winning position for Left because it is identical to a position along his winning strategy in H + K. This shows that G > H.

Now we assume Right can win H + K for some K and we note that GLR = H. In the game G + K, Right follows the same strategy except that if Left plays in G, Right immediately responds in G to give the position H + K'. Since K' was arrived at by following a winning strategy as if playing H + K, we know it is a winning strategy for Right. Thus we get that H > G and we can conclude that G = H. •

Naturally, we also get reversibility with Right's moves as well. This can be seen by taking the negatives of the games in the previous two theorems.

Theorem 65. Repeated application of Theorems 63, 64 and Corollary 62 to a game A leaves a game B with A = B.

Proof. Removing a dominated left option from a subgame G involves finding GLl and GL2 with GLl > GL2. We can then simplify G by removing his option to GLz. Like­ wise, this can be done with Right's options by finding GRl and GR2 with HRl > GR2 in which case GRl is removed.

For some game K, consider A + K. By our definition of >, we know in each case that a winning strategy for either player will never use a dominated option in A since they can always do at least as well by playing to the other option. Moreover, a player can­ not do any better by having removed the dominated option since it was never part of any winning strategy in the first place. Therefore, after removing a dominated option from A to get B, we know that A + K and B + K have the same outcome, i.e. A — B.

Bypassing a reversible move involves finding a subgame G of A with the form demon­ strated in Theorems 63 or 64. We then replace G by the game H from the appro­ priate theorem to get the new game B. Now, if playing B we end up at H, we know that o~{G + K) = o~(H + K) for all K. Therefore, we can conclude that o~(A + K) = o-(B + K) for all K, i.e. A = B. 44

Since = is an equivalence relation, repeated use of these steps always leaves us within the same equivalence class. We also can see that each step leaves us with a simpler game tree with fewer vertices. •

This leaves us with the problem of determining if repeated use of this new interpre­ tation of domination and reversibility leave us with a unique representative from each of the equivalence classes. The main difficulty is that we cannot apply the equivalent proof used for normal play games. The method used there is to consider two games G and H which are both arrived at by repeated use of domination and reversibility from a common game K. Now, we know that G = H and therefore G — H = H — H. In normal play, this implies G — H e V but this is not the case in misere play, as was mentioned in the introduction.

To resolve this issue, we must first determine if there are any other ways to sim­ plify misere games while preserving equivalence. Then, we need a technique which says that if games G and H have no dominated or reversible options and G = H, then G = H. Siegel has since expanded on this work to show that canonical forms of misere games do exist [19].

3.4 END-NIM

The game of END-NIM is played on a number of heaps much like NIM except that the heaps are in order and players may only take counters from the first or last heap. They must take at least one counter on their turn and can take up to the entire heap if they so choose. The impartial version of the game allows either player to take from either end heap on their turn. The partizan version restricts Left to taking counters only from the left end heap while Right may only remove counters from the right end heap. For simplicity, a game will be denoted by a string of positive integers repre­ senting the number of counters in each heap. 45

Both the impartial and partizan versions of END-NIM have been analyzed under Nor­ mal Play rules by examining the outcome classes of positions rather than their values which are much more difficult to analyze [1, 9, 13]. To that end, there is a complete solution to the impartial version and a simple strategy which describes how to find a winning move if one exists.

The partizan version has been analyzed using the threshold functions L(w) and R(w) for a given string w which in turn is used to determine the triple point associated with w and finally construct its phase diagram. Finally, the authors give an exponential algorithm for determining the outcome class for w [1].

Since the Normal Play analysis relies entirely on outcome classes rather than game values, a good deal of the work may be borrowed to analyze the case where we play with Misere Play rules.

The first and most important issue to deal with is that the simplest case of any game, where neither player has a move, is a V position in Normal Play, but is an M position in Misere Play. This introduces a minor complication which ends up not posing too great a problem. The normal play analysis of partizan END-NIM introduces the remarkable result that there are no M positions of even length. We already know this will be false for the misere version, but we get the similar result that the only M positions of even length are of the form l2fc. This can been seen due to a more general result:

Theorem 66. Given a string w representing a game of END-NIM the outcome of the Misere Play version is related to the outcome of the Normal Play version in the following way:

• o~{w) = Vifw = l2k+1, k>0

• o~(w) =M ifw = l2k, k>0

• o~(w) = o(w) otherwise. 46

Proof. The first two cases are obvious given that each player has no choice on their turn but to remove a single heap of size 1. Thus, the game is "She-loves-me-she-loves- me-not" with the parity of the number of heaps determining the winner.

For the last case, we start with w which is not a string of all ones and we must show that if a player has a winning move in Normal Play, they must still have a winning move in Misere Play. We will proceed by induction on the sum of all heaps and note that the base case of 0 is already covered. If a player has a winning move which takes them to some other string v which is not of the form ln then this is still a winning move in Misere Play by induction.

If instead their winning move is to a string v = ln in Normal Play then, we know that n is even. Therefore, if the move was to reduce a heap of size 2 or more to a heap of size 1, then in the misere game we have the winning move of removing the entire heap which would leave the position ln~l which is in V since n — 1 is odd. Likewise, if the winning move to v in Normal Play was achieved by removing an entire heap, we should instead leave 1 counter in that heap to reach the position ln+1 which is in V in Misere Play since n + 1 is odd. We can always leave one counter because if our move was from a heap of size 1, then we would have that w was already of the form lm which contradicts our choice of w. •

The most striking thing about this result is that it applies to both impartial and partizan versions of END-NIM and we can conclude that computationally, it is equally difficult to determine the outcome class of an END-NIM position in normal and misere play since their outcome classes are the same except in the simplest cases. It is also interesting to note that this generalizes the known strategy for playing the misere version of NIM. Chapter 4

Consecutive Move Ban

4.1 Motivation and Definitions

As we've now seen in the previous chapter, the Misere Play ending condition causes many problems that did not exist when we restricted ourselves to the analysis of games with Normal Play ending conditions. The reason the equivalence class of the empty game was a singleton in misere was at least in part due to the fact that a player could make two moves in a component by moving through a position which had no options for their opponent. Such a move would lose if there were no other options for your opponent in any other component but could be a very good move otherwise.

Rather than try to find a complete theory for Misere Play games, we will instead try to find ways in which we can change or restrict the game to make analysis easier as we did in Chapter 2. In this chapter we will examine the set of games where a player is banned from making two consecutive moves in the same component thus eliminating one aspect of misere games which causes them to be difficult to analyze.

Definition 67. A game G has a consecutive move ban if GLL = GRR = 0 and all the followers of G have a consecutive move ban.

This definition guarantees that after a player has made a legal move in G, he cannot play in G again until his opponent has made a move first. Note that it is possible to impose a consecutive move ban on an arbitrary game G by removing all Left (Right) options from any position obtained by a Left (Right) move in the game tree of G.

Definition 68. A game G with a consecutive move ban is left-handed if GR = 0. Likewise, it is right-handed if GL = 0.

47 48

Figure 4.1: Imposing a consecutive move ban

4.2 Normal Play

Theorem 69. For normal play games with a consecutive move ban, the only possible values are 0, *, 1, —1, |, — \.

Proof. To begin, we examine all possible game trees with depth at most 1. Since it is impossible for a player to play twice consecutively in such a game, all the associated values are possible. These include 0,1,-1 and *.

For game trees of depth 2, we know that our options must be from this set. More­ over, we know that any Left option must be to a position which has no left option and hence a non-positive integer. Therefore, it can only have value 0 or -1. Simi­ larly, any Right option will have value 0 or 1. This follows from the fact that any other choices would allow for consecutive moves for one of the two players. It is also possible that a player has no option. This gives us the following possibilities: o,i,-i,*,f = {o | i},-| = {-i |-o}.

We now assume that all games with a tree depth at most k have only these val­ ues. For a game with tree depth k+l we know that its options are games from those listed above. Since we cannot allow consecutive moves, we know that left's options can only include 0 or -1 while right's options can only include 0 or 1. It is also possi­ ble that a player's set of options is empty. As shown above, this gives exactly the 6 possible values we have already found. 49

Therefore, for normal play games with a consecutive move ban, the only values which are possible are 0, *, 1, —1, \,—\- •

Since these values are well understood, it is easy to now play sums of normal play games with a consecutive move ban. Unfortunately, imposing a consecutive move ban on a normal play game does not, in general, preserve the value of the game and it may not even preserve its outcome class. In fact, it is not very difficult to show that for a game in a given outcome class, it is possible to be in any other outcome class if you impose a consecutive move ban. There is, however, a quick way of determining the value of a component that has had a consecutive move ban imposed based on its previous value and whether or not each player has a legal move from the initial position.

If a single component game G £ V and we impose a consecutive move ban, its value remains 0.

If a single component game G £ J\f and we impose a consecutive move ban, its value becomes *.

If a single component game G £ £ and we impose a consecutive move ban, its value becomes 1 if Right has no legal move and \ otherwise.

If a single component game G £ 7Z and we impose a consecutive move ban, its value becomes —1 if Left has no legal move and — | otherwise.

Figure 4.2 shows an example of a game in three components which is a first player win. After applying a consecutive move ban to each component by pruning the circled vertices, we see that the game is now a Left win. 50

2 + -2* + {T I -1} ={T*|-H

>

1 + ~\ + = ^*

Figure 4.2: A consecutive move ban applied to a sum of components

4.3 Misere Play

The situation is somewhat similar for Misere Play games but we must be more careful since we don't have values. Instead, we will rely on showing that games are equivalent to each other in the following way:

Definition 70. Let G and H be games with a consecutive move ban. G and H are equivalent if o~ [G + K) = o~(H + K) where K is an arbitrary sum of games with a consecutive move ban.

This is essentially the same as the definition for equality between misere games except that we are restricting ourselves to the universe of games with a consecutive move ban. This is similar to the technique used by Siegel in his paper analyzing impartial misere games [20].

To begin, we will examine only those games which are left-handed.

We will partition these games into 4 sets much like we did when we found outcome classes. We will ultimately show that these classes are more useful to deal with than 51

outcome classes and lead us to a complete solution for arbitrary sums of consecutive move ban games. We will again proceed by considering each game and asking two questions:

1. Is it possible for Left to guarantee that the game will end after an even number of moves when she plays first?

2. Is it possible for Left to guarantee that the game will end after an odd number of moves when she plays first?

This gives us the following 4 classes of games:

• Even - Left can force an even number of moves, but not an odd number of moves when playing first.

• Odd - Left can force an odd number of moves, but not an even number of moves when playing first.

• Both - Left can force both an even number of moves, or an odd number of moves when playing first.

• Neither - Left can force neither an even number of moves, nor an odd number of moves when playing first.

We can see that these same 4 classes could have been arrived at by asking whether or not Right could force an even or odd number of moves when playing second. For example, if Left cannot force an even number of moves when playing first, we know that Right can force an odd number of moves when playing second. We can deduce this because we know that all of Left's options are to right-handed games. Since Left cannot force an even number of moves playing first, Right has the option to avoid an odd number of moves when playing from GL.

We can also follow this line of reasoning to partition all right-handed games in the same way. This gives us 8 possibilities and examples of each are shown in Table 4.1. Even Odd Both Neither

Left-handed / < < >•

Right-handed \ > > \

Table 4.1: Examples from each of the eight classes of one-handed games

Theorem 71. Using Misere Play rules, each of the examples in table 4.1 are equiv­ alent to every game in their respective classes.

Proof. For all cases, let G be a sum of games with a consecutive move ban and let if be a left-handed game with a consecutive move ban. The result for when H is a right-handed game will follow by symmetry. We begin with the Odd class. Let H £ Odd and let Go = {0 | •}. We must show that o~(G + H) = o~{G + Go).

Assume that Left can win G + Go- Then, when playing G + H he follows his same winning strategy, moving in H when he would move in Go- He plays in H in such a way that he can force an odd number of moves. Now, if Right ever plays in H, Left can reply in H because he can force an odd number of moves. Following this line of play, if Right eventually wins we could use this exact sequence of moves, omitting the pairs of moves in H after the initial move to show that Right has a winning strategy in G + Go- Since we know this is false, we conclude that Right cannot win G + H. 53

Now assume that Right can win G + Go- We begin by noting that since Left can­ not force an even number of moves in H, we can conclude that Right can force an odd number of moves in H. Now, Right will play her winning strategy from G + Go- If Left ever plays in H and Right has a legal move in that component, she immediately plays in H so as to force an odd number of moves. If this does not lead to a winning strategy for Right, then Left can employ this same strategy in G + Go by ignoring the pairs of moves in H to win. Since this is impossible, we conclude that Right must win G + H.

Therefore, we conclude that o~(G + H) = o~(G + Go) as desired.

The proof regarding the other classes is similar. Using the Even class, we let

H G Even and let GE = {• | 0 || •}. We show that o~(G + H) = o~(G + GE).

For the Both class, we let H G Both and let GB = {0, {• | 0} || •}. We show that o-(G + H) = o~(G + GB).

Finally, for the Neither class and let H e Neither while GN = {• \\ 0, {0 | •} ||| •}.

We show that o~(G + H) = o~(G + GN).

For each case, each player can use their strategy of how to win G + Gx and use it to construct a winning strategy for G + H by playing pairs of moves in H if needed. For the other direction, a player can ignore these pairs of moves in H to construct a winning strategy for G + Gx if they have a winning strategy in G + H. •

We can see that the previous Theorem would also hold if we were to replace all of the games by ones in which only Right had a move from the initial position. So, by including these "right-handed" versions, we have only 8 different components which we must consider and they are exactly the ones shown in Table 4.1.

For the following four lemmas, I will refer to these games using the form GxY where X will denote the class and Y will denote if it is left-handed or right-handed. We 54 would like to show that for each of the basic components, the left-handed and right- handed versions of each component are negatives of each other. It is important to note that for each lemma, G can be any arbitrary sum of components, each of which has a consecutive move ban but is not required to be one-handed.

Lemma 72. Let GoR = {• | 0} and GoL = {0 | •}. Then for all sums of consecutive

move ban games G, o~(G + GoR + GoL) = o~{G).

Proof. We begin by assuming Left can win G. Then when playing G + GoR + GoL, he follows his winning strategy in G. If Right plays in GoR to 0, then Left should respond in GoL to 0 as well. In this case, Left will win since we can omit these two moves to see it is the same as his winning strategy for G. If Left runs out of moves before either player has played in these components, Left simply plays in GoL- Now Right has at least one move remaining. Prom this point, whenever Right makes a move, if Left can respond in the same component, he does so and if not, he wins since there are no moves remaining. Either way, Left has won the sum. By symmetry, we can show that Right can win the sum if we begin by assuming that Right can win G. U

Lemma 73. Let GER = {• || 0 | •} and GEL — {• | 0 || •}. Then for all sums of consecutive move ban games G, o~~(G + GER + GEL) = o~(G).

Proof. We begin by assuming Left can win G. Then when playing G + GER + GEL, he follows his winning strategy in G. If Right plays in GER, then Left should respond in GEL • This leaves us with G + GoR + GoL which has the same outcome class as G by the previous lemma. If Left runs out of moves before either player has played in these components, he must play to H + GER + GOR where H is a sum of right-handed games. From this point, Left must respond in the component that Right plays in.

If such a move does not exist, he wins. Since Right must eventually play in GoR-, Left will win. By symmetry, we can show that Right can win the sum if we begin by assuming that Right can win G. •

Lemma 74. Let GBR = {• || {0 | •}, 0} and GBL = {0, {• | 0} || •}. Then for all sums of consecutive move ban games G, o~(G + GBR + GBL) = o~{G). 55

Proof. We begin by assuming Left can win G. Then when playing G + GBR + GBL , he follows his winning strategy in G. If Right plays in GBR, then Left should respond in GBL • This leaves us with either G + GoR + GoL or simply G depending on which Right option was chosen. In either case, we have the same outcome class as G by the previous lemmas. If Left runs out of moves before either player has played in these components, Left must move to H + GBR + GoR where H is a sum of right-handed games. From this point, Left can only have a move in the same component that Right plays in. Eventually, Right must play in a component which leaves no move for Left and he will win. By symmetry, we can show that Right can win the sum if we begin by assuming that Right can win G. •

Lemma 75. Let GNR = {• ||| {0 | •}, 0 || •} and GNL = {• || 0, {• | 0} || •}. Then for all sums of consecutive move ban games G, o~{G + GNR + GNL) = o~(G).

Proof. We begin by assuming Left can win G. Then when playing G + GNR + GNL , he follows his winning strategy in G. If Right plays in GNR, then Left should respond in GNL . This leaves us with G + GBR + GBL which has the same outcome class as G by the previous lemma. If Left runs out of moves before either player has played in these components, he must play to H + G^R + GBR where H is a sum of right handed games. From this point, Left will only be able to play in the same component that

Right plays in. Eventually, Right must play in G^R and Left will respond to GoR which guarantees he will win, as before. By symmetry, we can show that Right can win the sum if we begin by assuming that Right can win G. •

Thanks to the previous four lemmas, we get the following relations for all possible classes X:

GxL + GxR = 0

GxL = -GXR 56

This allows us to represent all possible sums of the 8 available components as:

a • GoL + b • GEL + c • GBL + d • GNL

where a, b, c, d are integer coefficients.

Now we would like to find out which outcome class a particular sum is in if we know the four coefficients. If we specify a game by the four-tuple (a, 6, e, d) we get the following legal moves for Left:

(a — 1, b, c, d) if a > 0 (a-l,b-l,c,d) if 6 > 0 (a,b,c,d)->{ (a,b,c-l,d) if c > 0 (a—l,b,c — l,d) if c> 0 (a,b,c-l,d-l) if d > 0 Likewise, the legal moves for Right are given by:

(a + 1,6, c, d) if a < 0 (a+l,6+l,c,d) If fc» < 0 (a, b, c, d) —> { (a,b,c+l,d) if c < 0 (a+l,b,c+l,d) ifc<0 (a, fe,c + l,d+ 1) ifd<0 We now note that since we are playing a misere game, we'd like to be the first person who has no legal move. So, given a particular game given by its characteristic four- tuple, that means Left would like to make all the entries non-positive while Right will win if they are all non-negative.

Theorem 76. Let the game G be represented by (a, b, c, d). Then G££ifa + d<0, GeKifa + d>0 andGeMifa + d^O.

Proof. We begin by noting that a player will always alter the sum a + d by 0 or 1 on each of their turns. If we begin with a + d < 0 then we must show that Left can win 57 by playing first or second. On each player's turn, then can alter the sum a + d by at most 1. So, Left can guarantee that on Right's turn the sum a + d will be negative. Thus Right must have some legal move. Since the game must eventually end we know Left will win. Likewise, if we begin with a + d > 0 then playing first or second, Right can guarantee that on Left's turn a + d will always be positive. This means he will have a legal move and therefore Right will win. Finally, if we begin with a + d = 0, if Right plays first she has a move which will make a + d = 1 from which she can win playing second. Likewise, if Left were to play first, he can always make a move so that a + d = — 1. We've shown this is a Left win so we can conclude that when a + d = 0, the first player to move will win. •

This theorem now gives us a complete characterization of partizan misere games played with a consecutive move ban with the caveat that only one player has a legal move from the starting position of every component in a sum.

In order to play these games, however, we need to determine what a player will do from a given position depending on the options he has available to him. We have already shown that there are at most five distinct options available at any time.

Theorem 77. Let G be a single component game with a consecutive move ban. Among all options available to the next player, there is a order of preference of op­ tions, illustrated as follows, with higher elements being preferred over lower ones:

Neither / \ Even Odd Both

Proof. To begin, it is straightforward to see why an option to Neither is preferred above any other option. If you can win only when the game ends after an odd number of moves, then we see that moving to Neither is at least as good as moving to Even since we will eventually have the option of forcing an odd number of moves in either case. Likewise, the option to Neither is at least as good as moving to Odd if we would 58 like the game to end in an even number of moves. Therefore, Neither is always at least as good as any other option.

We can see that Both is always your worst option since it gives your opponent the choice of ending the game in an even or odd number of moves. Even if we still lose, an option to Even or Odd cannot be worse than moving to Both.

Finally, Even and Odd are incomparable since the parity of the number of moves required to win the game in a sum might depend on the rest of the summands. So, there can be no way to prefer Even over Odd or vice versa in all cases. •

Using this theorem, we can determine the class of a game based on the classes of its options. In particular:

• If G has no options, G G Even.

• If G has an option to Even but not Odd nor Neither, G G Odd.

• If G has an option to Odd but not Even nor Neither, G G Even.

• If G has options to Even and Odd, G 6 Both.

• If G has an option to Neither, G G Both.

• Otherwise, if G has only options to Both, G G Neither. 59

Right - Even No Move Both Even Odd Neither Left | or Odd No Move

Both

Even

Odd

Even or Odd

Neither

Table 4.2: Examples of all possible two-handed games

4.4 Two-handed Games

It is possible to have a game which has a consecutive move ban, but in which both players have a legal move from the initial position. In such a game, once every compo­ nent has been played in at least once, we are left with a game which we can analyze easily since each component is one-handed. The initial positions, however, can be much more complicated. From an arbitrary initial position, we have 6 possibilities for each player's options. A player can have a option to Even, Odd, Both and Neither as before, but they could also have no option or two options, one of which is to Even and the other to Odd. It is important to note that having an option to Even and Odd is not the same as having an option to Neither since from the initial position, the other player may have a move which could prevent you from taking the preferred path if you do not take your move first. All of these possibilities are summarized in Table 4.4.

To continue the analysis of consecutive move ban games, we need to consider how this larger set of games interacts. The first thing to note is that we cannot use the simplification GL + GR = {-|-} in all cases as we did with the one-handed games. 60

In the Neither/Neither case, for example, we find that Gi + GR G P instead of AT as we might expect. We begin by modifying our notation for consecutive move ban games to include these additional games.

Definition 78. Let Ga^ be a consecutive move ban game where a represents the Left side of the game tree and (3 represents the Right side of the game tree where a,P G {-,B,E,0,X,N} (Respectively: No Move, Both, Even, Odd, Even and Odd, Neither).

We have seen in Lemmas 72 through 75 that for games where one player has no legal move we get the identity Ga?. + G.

Lemma 79. Ga^ + G0>a = {•]•} unless Gafi 6 {G0,o, GO,N, GNJ0, GN,N}.

Proof. Let G be an arbitrary sum of games with a consecutive move ban and assume without loss of generality that a 6 {B,E,X}. We must now show that o~(G +

Ga,/3 + Gp^a) = o~(G). Assume that Left can win G. Then, when playing G + Ga^ +

GpiCt she follows her winning strategy in G unless Right makes a move in one of the other two components. In that case, she can make the corresponding move in the other component. Both of these components are now one-handed and by Lemmas 72 through 75 we know they cancel each other. Therefore, we know Left can win this sum. If Right does not play in either of these components then we know he must eventually play to a follower of G which has no Left option. In that case, Left plays in Ga>p so as to force an even number of moves. From this point, Left responds to any of Right's moves locally if possible. If no such move exists, she has either won the game or her only legal move will be to play in Gp

G + Ga,j3 + Gg

b> c+l b=x+l b = c b = c- 1 b 0 £ C V K n

Table 4.3: The outcome class of a CMB game denoted by [a, b, c]

In the case of the four "bad" types of components, we will now deal with them individually.

Lemma 80. The game a • G^N £ V for all a > 1.

Proof. We proceed by induction and consider the base case with a = 1. Whoever plays first gives their opponent the opportunity to force an odd number of total moves which allows the second player to win. Now for a = k we can see that whoever plays first must leave a position where their opponent can force an even number of moves which means they must again play first with a = k — 1. By induction, this is a V position and the second player wins. •

Lemma 81. When playing the game a • G^tN + b • GO,N + c • GAT,O, neither player's best move will be in a G^,N component unless there are no other possible moves.

Proof. Left will win playing second when 6 = 0 and likewise, Right will win playing second when c = 0 so it is to both player's advantage to reduce b and c respectively before ever playing in a GN,N component. •

We can use the above lemmas to determine all outcome classes for games of. the form a • G^,N + b • GO,N + c • GN,O- They are completely described by Table 4.3.

Lemma 82. Consider the game a • GN,N + b • GO,N + c • GJV,O + d • Go,o- If d is even then the outcome class is the same as given in Table 4-3- If d is odd, then the outcome class is as in Table 4-3 with the V and M positions interchanged.

Proof. When d is even, whoever has a winning strategy can simply play that strategy while ignoring the GQ,O components. If their opponent plays in one such component, 62

they can respond in another. When d is odd and the remainder of the position is V, the first player can win by playing in a Go,o component leaving d even. If d is odd and the remainder of the position is in J\f, then we have a = 0, b = c. The second player has a winning strategy by always maintaining b = c on their turn which will force the first player to eventually play in a Go,o component leaving an J\f position and thus losing. •

This completes the analysis of arbitrary sums of these four "bad" components. In arbitrary sums of CMB games, however, we will see that a player can win in one of two ways. The first is that they have so few moves that the game will end with poten­ tially many components unplayed. If, on the other hand, there are too many of these "bad" components, then each player must try to accumulate one-sided components which will allow them to control parity. These can be used to force their opponent to play first when only "bad" components remain.

So, it is worthwhile to continue the analysis with positions where we add a num­ ber of GB,- positions. For the following two lemmas, we consider positions of the form a e • GN,N + b • GO,N + c • Gpito + d • Go,o + • GB,- which we will denote by the five-tuple [a,b,c,d,e].

Lemma 83. If [a, b, c, d, 0] is in J\f, V or C, then for [a, b, c, d, e] when e > 0 it is in £ with only one exception: [0, 0, 0, 0, e] is in M for all e.

Proof. When e > 0, Left has the option of reducing e by 1 and forcing Right to play first in the remainder of the game. If Left can win [a, 6, c, d, 0] playing first then she can use the following strategy. When she runs out of moves, she can play in each of the GB,- components so as to force an even number of moves. Since Right must make the last move in each component, Left will win the game. If on the other hand Left can win [a, 6, c, d, 0] playing second then she has a winning strategy in [a, 6, c, d, e] by first reducing e by 1 and playing her winning strategy from that point. The only time Right can win such a position is when he plays first and has no moves in any component which can only occur in a position of the form [0, 0, 0, 0, e}. • 63 e e e<^ >¥ 2 a> 0 £ P ?e a = 0 A/" A/" 71

Table 4.4: The outcome of a CMB game denoted by [a, 6, c, d, e] with (c — b) even and d = 0

p — c-b e > — c 2 e<^ c ^ •?. a > 0 C N n a = 0 C V n

Table 4.5: The outcome of a CMB game denoted by [a, 6, c, d, e] with (c — b) even and d= 1

When [a. b, c, d, 0] is in 7Z, then for [a, b, c, d, e] when e > 0 we can get any of the outcome classes as described by Tables 4.4 through 4.7. In short, this demonstrates that if the game will end with only these "bad" com­ ponents, each player should try to accumulate as many one-sided Both components as possible. Of course, the game may end much earlier if one player has the advan­ tage of having many fewer components where they must play both first and last. We formalize this notion with the concept of a player's score.

Definition 84. Let an arbitrary sum of CMB games be grouped together to be given in the form i • G.to + j • G.tN + k • Go,0 +1 • Gu,p + rn • Ga,o + n • Ga^ +p • GC,D where a, (3 G {£, E, O, X, N} and C,D E {•, B, E,X}. The Left Score of the game is given byi + j — k — I and the Right Score is given by —i — j — m — n.

c h l p - C-b-l e > ~ ~ C 2 e<^ a > 0 C n n a = 0 £ V n

Table 4.6: The outcome of a CMB game denoted by [a, 6, c, d. e] with (c — b) odd and d=0 64

_ c-b-l e > -5-1 C 2 e<^ a > 0 £ ft n a = 0 £ M K

Table 4.7: The outcome of a CMB game denoted by [a, b, c, d, e] with (c — 6) odd and d= 1

There are a number of important things to note regarding this definition. First, we assume that we've canceled one-handed components with their negatives as much as possible. Although this doesn't technically need to be done, it gives fewer types of components to consider.

Secondly, some components are counted twice in the expression. In particular, Go,o contributes to both k and m while GN,N contributes to both I and n. Essentially, a player's score is the difference between the number of one-handed components where their opponent will move first and last and the total number of two-handed compo­ nents where you will play first and last should you opt to play in that component.

Finally, all of the coefficients in the expression are non-negative except possibly i and j since they represent one-handed components. So, if Left has a positive score, for example, then we conclude that i + j > k + I. This then implies that — i — j is negative and we can conclude that Right's score must be negative. Likewise, if Right's score is positive we can deduce that Left's score must be negative.

Lemma 85. If a player's score is non-negative they can win playing first.

Proof. On your turn, simply play in any component which has an Odd or Neither side for you which will have the effect of increasing your score by 1. On your opponent's turn, they can decrease your score by at most 1. Eventually, on your turn, none will remain and your score will still be at least 0. If you have no moves remaining, you win.

If you have moves remaining, then in any such component you can force that it will end after an even number of moves. Therefore, you can ensure that there will 65 be at least one component where your opponent must make the last move after you have run out. •

Once we make the observation that a player can only change their score (and their opponent's score) by at most one on their turn, we can use this lemma to determine the outcome class of some positions. Namely, if Left's score is positive the game must be in C while if Right's score is positive, the game must be in 1Z. We can also note that for positions where both player's scores are negative, the winner will be determined by parity in the sense that the first player who must play in one of the "bad" components will lose. As we've seen, one can control parity by having a large number of one-handed components where you can force both an even or odd number of moves.

Definition 86. The control value for a sum of games with a consecutive move ban is equal to the number of left-handed Both components plus the number of right-handed Neither components.

Note that the control value for a particular sum can be negative since a right-handed position is the negative of a left-handed position. When the control value is positive, Left has the ability to control parity while Right has the ability to control parity if it is negative. At the end of the game we can refer to Tables 4.4 through 4.7 to determine the outcome since the control value will give us the value of e.

So, the only thing remaining is to examine all possible types of components and see how a Left or Right move will affect each player's score as well as the control value. Tables 4.4 and 4.4 summarize this information. The positions noted with a * indicate that the player has chosen an option which will force the game to end after an odd number of moves. We can see from these tables that it is always better to play in some positions than others. For example, either player would choose to play in a component of type GB,B over anv other available component since it has the greatest possible benefit while at the same time preventing their opponent from getting that same benefit. 66

On each player's turn, they can refer to the tables to determine the best possible component to play in. When choosing where to play, you must keep in mind that the most important goal is to ensure that your opponent's score is negative after your turn. Failure to do so gives the opponent a winning strategy in the remainder of the game. Among options which keep the opponent's score negative, if you have no direct winning strategy (your score is negative) then among available options you should choose one which alters the control value in your favour (positive for Left, negative for Right). If you still have multiple equivalent options, it is best to choose one in which your opponent would most prefer to play.

4.5 A Coin-Flipping Game

We can now use these results to analyze particular instances of games with a consec­ utive move ban. To illustrate this, consider a game involving stacks of coins. Each stack consists of a number of coins, all of which are either heads up or tails up. We assign each player a subtraction set, SL and SR respectively. On Left's turn, she may select any non-empty stack which is heads up and an element from her subtraction set which can be no larger than the number of coins in that stack. She then removes that number of coins from the stack and if there are any remaining, she then flips them over so that the stack is now tails up. Since we would like to play the misere version of this game, when a player cannot make a legal move, they win. Clearly this game will involve a consecutive move ban since no player can play twice consecutively using the same stack. We also note that only one player has a legal move from a stack in its initial position.

To analyze the game, we first consider the case where SL = SR = N. This game resembles nim in that if a player has a legal move in a given component, he may take as many coins as he would like. Trivially, a stack of size 0 will end after an even num­ ber of moves since neither player can make a legal move. We then note that a stack with only 1 coin must end after an odd number of moves. Any stack with 2 or more p l-i

oo H a> Position o {a, 13) 03 O

P

i—i i—> >—i i—i i—i i—i i—> i—i i—i i—i i—i i—ii—i i—ii—i i_i i—i i_i i—i i—i i—i i—i i—i i—ii—i to

CD II I I I I I I l OH-'OOOI-^I—' O O »—' 1—>0>—' I—> O >—'OH-' O »—' O •—' O I—' I—' Co 7£ o < Contri CD o CO 1 1 1 1 1 O 1—'(—'(—' i—'OOOOOOI—'OOOOOOOOOl—' h-' I—i i—' 1—' o O CD P II 1 1 1 1 1 1 1 O t—' O O O I—' I—' O O I—' i—> O I—' I—' O i—' O I—' O I—' O I—' O I—' I—' Co a o o cS». ES l-i 1—' 1—1 1—1 1—1 I—1 1—1 1—1 1—1 1—1 1—1 1—' 1—1 1—' 1—' 1—1 1—' 1—' 1—1 I—' 1—' 1—» >—' 1—' 1—» 1—' Co S Contrt

CD CO 1 1 1 1 1 B' OOOH-'OOH^OH-' \—i I—* i—» O I—' O O I—'OOOH-' O O O >—i ^ p o

td p en CD 68

Position Left Right (a,P) LS RS Control LS RS Control BX* 1 -1 1 0 0 0 EX* 1 -1 0 0 0 0 OX* 1 0 0 1 0 0 XX* 0 0 0 0 0 0 NX* 1 0 -1 1 0 0 XB* 0 0 0 -1 1 -1 XE* 0 0 0 -1 1 0 xo* 0 1 0 0 1 0 XN* 0 1 0 0 1 1 •B — — — -1 1 1 •B* — — — 0 0 1 •E — — — -1 1 0 •0 — — — -1 1 0 •N — — — -1 1 0 B- 1 -1 -1 — — — B-* 0 0 -1 — — — E- 1 -1 0 — — — 0- 1 -1 0 — — — N- 1 -1 0 — — —

Table 4.9: The effect of a player's move on score and control values in a CMB game (part 2) 69 coins will end after an even or odd number of moves depending on whether the player able to move removes all of the coins or leaves 1. Therefore, each stack is either right- handed or left-handed and is either a member of the Both class if it has at least 2 coins and a member of the Odd class if it has exactly 1 coin. Applying the previous results we see that the outcome of a game involving multiple stacks simply depends on the number of stacks of size 1. Furthermore, a winning strategy will be to remove all but one coin from any stack with two or more coins or remove a singleton stack otherwise.

If we continue to let Sz, = SR but remove some elements from the subtraction sets, we get a game which is similar to an impartial subtraction game. Along these lines, we can construct tables similar to Grundy scales to describe the types of positions which occur. In a standard Grundy scale, the entries are natural numbers which represent the G-value of a given game with a starting size of n. In our case, the entries will be of the set {B, E, O, N} which represent the four different types of games. The next value in a Grundy scale is determined using the MEX of the options. Naturally, we have a different propagation rule based on the following rules:

1. If your options include A" or both of E and O then the game is of type B

2. If your options include an O but no E then the game is of type E

3. If your options include an E but no O then the game is of type O

4. Otherwise, the game is of type N (your only options are to games of type B)

To illustrate, consider the game where S^ = SR = {1,2}.

Number of coins: 0 1 2 3 4 5 6 7 8 Game Type: E 0 B E 0 B E 0 B

In this case, we note that the pattern of game types is periodic with a period length of 3 and no pre-period. Indeed, when SL and SR are finite it is always the case that the pattern will eventually become periodic since we only have finitely many types of games. Table 4.10 shows the period lengths of the games where the subtraction set is of the form {a, b, a + b} for small values of a and b. Interestingly, none of the sequences 70

a —> 1 2 3 4 5 6 bi 1 3 4 7 8 11 12 2 4 6 7 8 22 14 3 7 7 9 10 11 12 4 8 8 10 12 13 14 5 11 22 11 13 15 16 6 12 14 12 14 16 18

Table 4.10: The period lengths for CMB games with SL = SR = {a, b,a + b}

Number of coins: 0 1 2 3 4 5 6 7 8 9 10 11 Game Type: E E 0 0 B B E B 0 E B 0 period length = 3, pre-period = 6

Table 4.11: The CMB class sequence for SL = SR = {2, 4, 7} have a pre-period. Tables 4.11 through 4.13 show further examples of sequences for various subtraction sets. There are various ways of getting arbitrarily long periods by including large ele­ ments in the subtraction sets. Without the corresponding class sequences, Table 4.14 shows a few computed period and pre-period lengths based on subtraction sets. Note that fn denotes the nth fibonacci number.

During the construction of these tables, I noticed two curious things. First and most striking, the game type N never occurs. Secondly, whenever you have a game of type E if you increase n by the smallest element of the subtraction set, you always get a game of type O. It is interesting to note that the second observation is very similar to the Ferguson Pairing Principle [4] for subtraction games. It turns out that

Number of coins: 0 1 2 3 4 5 6 7 8 9 10 Game Type: E E E O O O B B B B B period length =11, pre-period = 0

Table 4.12: The CMB class sequence for SL = SR = {3, 5, 8} 71 Number of coins: I 0 I 1 2 I 3 Game Type: E 0 E 0 period length = 2, pre-period = 0

Table 4.13: The CMB class sequence for SL = SR = {any subset of odd numbers}

SL — SR — pre - period period 1,4,8 5 7 3,5,8,13 0 38 k,k+l 0 2k+ 1 k,k+l,k + 2 0 2k+ 2 Jni Jn+li Jn+2 0 2fn + /ra+1

Table 4.14: Periods and pre-periods for various subtraction sets the these two observations are related to each other as demonstrated by the following results.

Lemma 87. Let sx denote the smallest element of the subtraction set and assume

that G(k) ± N for k < n. Then G{n) = 0 iffG{n-s1) = E.

Proof. We will proceed by contradiction in which case there are two cases to consider.

For the first case, we assume G(n) = O but G(n — sx) ^ E. We can deduce that G(n — Si) cannot be N or O so it must be the case that it is of type B. Since there are no options of type TV we in turn find that there must exist some s^ where G(n — Si — Sfc) = E. By induction we find that Gin — Sk) = O which gives the contradiction since it is an option from n which is also of type O. For the second case we assume G(n) ^ O when G(n — Si) = E. We can deduce that G(n) = B since it is not 0 and has an option to E. Since we cannot have an option to TV, there must be some k for which G[n — Sk) = O. By induction, this gives us that G(n — s^ — Si) — E which gives the contradiction since it is an option from n — Si which is also of type E. •

Theorem 88. When SL = SR, G{n) ^ N for all n.

Proof. We will proceed by contradiction and assume that G(n) = N where n is minimal. By the definition of games of type N, we know that for all k, G(n — Sk) = B. 72

Coins: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 LH: E 0 0 B B E 0 0 B B E 0 0 B B E RH: E E 0 B B E E 0 B B E E 0 B B E period length = 5, pre-period = 0

Table 4.15: The CMB class sequence for SL = {1, 2, 3}, SR = {2, 3, 4}

In particular, G(n — s\) — B. Since n is minimal, there must exist some i,j such that G{n — s± — Sj) = E and G(n — S\ — Sj) = O. Again, since n is minimal, we can apply the previous lemma to conclude that G{n — Sj) = O which contradicts that

G[n -sk)=B for all k. •

Corollary 89. G(n) = O iff G{n - sx) = E

Proof. Thanks to the previous theorem, we can omit the assumption in the above lemma regarding the absence of games of type N since we now know that none exist. •

Of course, there is no need to restrict ourselves to the case where Left and Right's subtraction sets are the same. By having different sets, we must construct two se­ quences simultaneously to completely describe the game. The propagation rule still applies except that all of your options are taken from the other player's sequence. Naturally, a stack of size 0 is always of type E. In the following example, we let SL = {1, 2} and SR = {1}:

Number of coins: 0 1 2 3 > 4 Left-Handed: E 0 B B B Right-Handed: E 0 E N N

Since we now have a game type for the left and right handed stack for each number of coins, 16 different ordered pairs are possible. This in turn gives us more possibilities when it comes to periods that eventually arise. Tables 4.15 through 4.18 show further examples which demonstrate longer and more interesting periods and pre-periods. 73

Coins: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 LH: E 0 B B B B 0 B B B B 0 B B B B RH: E 0 E 0 E E N E N E E N E N E E period length = 5, pre-period = 4

Table 4.16: The CMB class sequence for SL = {1, 2, 4}, SR = {1, 3, 4}

Coins: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 LH: E 0 0 0 E 0 N 0 0 0 0 B 0 0 0 0 RH: E E E 0 E B E B E B E B E E E E period length = 1, pre-period =12

Table 4.17: The CMB class sequence for SL = {1, 8}, SR = {3, 5}

Coins: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 LH: E 0 0 B E 0 0 B N N 0 0 N N 0 N RH: E E 0 B B E B B B E B B B B B B period length = 1, pre-period = 15

Table 4.18: The CMB class sequence for SL = {1, 2, 5}, SR = {2, 3, 4} Chapter 5

The Short Disjunctive Sum

5.1 Definitions

In previous chapters, we have been considering misere games using the standard interpretation for disjunctive sums. There is, however, another interpretation which is consistent with our view of disjunctive sums of Normal Play games. Therefore, it might be that this alternative view will allow for easier analysis of games played with Misere Play ending conditions. In each case, a move consists of choosing a component to play in, then making a legal move in that component. The difference comes from whether or not the choice of component is part of the requirement to find a legal move. The standard interpretation is a player wins in a sum of misere games if and only if there are no moves for that player in all components. If, however, we allow the player to first pick a component then make a legal move in that component, he or she could win by choosing a component where no legal move exists. In this case, a player wins if there is no legal move in at least one component. You can think of a sum of games as being a set of boards on which the games are played. In this way, we see that there is a difference between an empty board where neither player can make a move as compared to the lack of a board where the component itself doesn't exist. To differentiate between these two interpretations, we will for this chapter refer to them as short and long disjunctive sums.

Definition 90. The short disjunctive sum of games is played by first choosing a component then making a legal move in that component if such a move exists. The game ends when a player chooses a component in which he has no legal move. We will often abbreviate this as the short sum.

Definition 91. The long disjunctive sum of games is played by first choosing a component then making a legal move in that component if such a move exists. The

74 75

player must choose a component where he may make a legal move if possible. The game ends when no such component exists.

For each interpretation, we can play using normal or misere rules. It is important to note that under Normal Play rules, the short and long disjunctive sums are equiv­ alent. It would always be a bad move to choose a component in which you have no legal move (and lose immediately) if you could make a move elsewhere.

The main motivation for considering this alternative view for disjunctive sums is to see if it aids in the analysis of misere games. As demonstrated in chapter 3, the long disjunctive sum is very difficult to analyze in general. The previous chapter analyzed the case where we place some restrictions in which components each player could play.

5.2 Results

We will begin in much the same way that we did when analyzing the long disjunctive sums of misere games. Although we are still playing a disjunctive sum of games, the outcome class of a sum of games could differ when playing the short or long version. Therefore, we will introduce the following notation for clarity:

Definition 92. The function os~(G) gives the outcome class of the game G under Misere Play rules with a short disjunctive sum: os~{G) = C where C £ {£, H,Af, V}-

Definition 93. Given games G and H, we say that G is equal to H under Misere Play rules with a short disjunctive sum if os~(G + K) = os~(H + K) for all K. In this case we write G = H

Definition 94. Given games G and H, G > H ifos~(G + K) > os~(H + K) for all K under Misere Play rules with a short disjunctive sum.

Note that we could equivalently define the function os(G) and the related com­ parisons for a short sum using Normal Play rules, but as we've seen this is the same 76

as the long sum and is therefore not needed. We can now consider the possible out­ come classes of a compound game which can arise based on the outcome classes of the summands.

Lemma 95. Given os~ (G) = V and any other game H, then os~(G + H) = os~(H).

Proof. If a player has a winning strategy in H, then that player can play that same winning strategy in G + H by responding in G whenever their opponent plays in G. D

Lemma 96. Given os~(G) = C and os~(H) = N, then os-(G + H) = C or N.

Proof. Left makes a winning move in H then responds locally to Right's moves, playing a winning strategy in that component. Since Left wins playing first, we know that the outcome class of the sum must be either C or M. Furthermore, it is possible to find examples of G and H such that their sum is in C or M. In the latter case we can simply allow H to be the game with no options for either player. •

Lemma 97. Given os~(G) = C, os~(H) = £, then os~{G + H) = C

Proof. Left has a winning strategy in both components individually playing first or second. He can therefore follow that same winning strategy by locally responding to Right's moves. Once he wins in one component, he wins the short sum by definition. •

Putting together these results with their counterparts when G e 1Z, we get Table 5.1 which shows the possible outcome classes for G + H given the outcome classes of G and H played in isolation. For the cases where multiple outcome classes are possible, examples of each are given in the Appendix.

Note that this table is exactly the same as what we have when playing a disjunc­ tive sum of games under Normal Play rules. It is quite different than what was found under the long disjunctive sum using Misere Play rules which allowed every possible outcome class regardless of the outcome class of the summands. This alone gives 77 G-> V M £ n Hi V V M C K M M Any AfUC Afun C C MuC L Any K K Nun Any n Table 5.1: The possible outcome classes of the short misere sum G + H

us reason to believe that the short misere sum might be easier to analyze than its counterpart.

We can now see that all V positions act like 0 as they do in Normal Play games. Also, if we follow the convention of games in C having positive value, and games in 1Z having negative value, we find no contradictions to how we would expect summation to work.

Despite many similarities to the disjunctive sum of Normal Play games, the one biggest difference is with respect to the empty game {-|-}. In Normal Play, this is the canonical form of 0. In Misere Play, however, it is an M position and hence cannot be in the equivalence class of 0 if we wish to reserve that symbol as an additive identity. In Normal Play, if this game exists in a summand, we simply discard it because no player would ever want to move in it. This is true for V positions in misere as well, but we cannot do this with the empty game. In fact, if one of the summands is the empty game, both players can win playing first regardless of the rest of the summands. We will refer to a move in a component which wins for a given player regardless of the rest of the summands as an overriding move as first introduced in volume 2 of Winning Ways [5]. Likewise, we will sometimes refer to suiciding moves which refer to a player's option which causes them to lose immediately regardless of any other summands. We will continue to borrow some portion of the notation regarding tallies and timers for the following definitions [5].

Definition 98. The game denoted COQ represents a win for Left regardless of any 78

other summands. Likewise ooo represents a win for Right regardless of any other summands.

It is important to note that these games do not naturally occur as individual components but instead are used to represent overriding moves for each player.

Definition 99. Given any game G, G is an overriding move for Left if G = ooo • R R Likewise, G is an overriding move for Right if G = oo0.

Definition 100. Given any game G, GL is a suiciding move for Left if GL = ooo or LR LR R if there exists a G with G = oo0. Likewise, G is a suiciding move for Right if R RL RL G = oo0 or if there exists a G with G = oo0.

Definition 101. Given k e Z+, oo& is a game where Left has an option to ook-i> Left has no options to OOJ for i < k — 1 and for every GR there exists j < k such that RL G = ooj.

Definition 102. 00^ is the negative of oo^.

The definition for cofc, while similar to the definition for timers, only counts the unanswered moves a player must make before winning the game rather than the sum of moves made by both players. The reason for this is that timers were originally used to analyze games played in a conjunctive sum where each player would play in every component where it was advantageous to do so. When playing in a disjunctive sum, we are only allowed to play in one component at a time and situations arise where you must decide if it is more important to get yourself closer to winning or to hinder your opponent from doing the same. In a Misere Play game with a short sum, any component which has no option for a player in the game tree instead has an overriding move for that player. Figure 5.1 shows a Misere Play game whose value is 002.

We will now continue by examining some properties of these new games.

Lemma 103. OOJ > OOJ for i < j 79

oo2

OOi/ »±OO0

±oo0

Figure 5.1: A sample misere game with value 002

Proof. Assume that Left has a winning strategy in 00 j + K for some game K. When playing OOJ + K she will initially follow the same strategy. If she wins by eventually playing to oo0 in the first component, then she will still do so since i < j and it will take fewer unanswered moves in that component. If instead she wins by having no options in K on her turn, then following the same strategy will still result in a win.

Assume that Right has a winning strategy in OOJ + K for some game K. When playing OOJ + K he can follow the same strategy. Since Right cannot have a winning strategy by playing in the first component, we know that he can win by playing only in K. Since Left could not win by responding locally and it would take him i moves to win by playing only in OOJ then we know he still cannot win when playing in OOJ since i < j.

Putting these two facts together gives us that O5~(OOJ + K) > OS~(OOJ + K) for all games K. Since ooj 7^ OOJ, we get the desired result. •

Lemma 104. OOJ + OOJ = OOJ if i < j.

Proof. This is true when i — 0 since 000 is idempotent. We then proceed by induction on i. Left has an option to OOJ_I + oo,- which is equal to OOJ_X by induction. By definition, we also know that she has no move to any oo^ for k < i — 1. Any right option of OOJ must have some left option to oo^ for k < j. Likewise, any right option of 00j must have some left option to oo^ for k < i. Therefore, this sum satisfies all the requirements of the definition of the game ooj as desired. •

Lemma 105. oo; + oa, = OOJ if i < j. 80

Proof. We again begin by noting that when i = 0 the statement holds. Proceeding by induction on i, Left can play in OOJ to OOJ_I which makes the sum oOi_i. If Right plays first and chooses a move in the first component, Left can respond in that component to a position no worse than OOJ. Assume then that Right plays in the second component from 06j to OOJ_I. Left then responds by playing in the other component to OOJ_I + OOJ_I = c©;_i by induction. This gives us the result. •

Corollary 106. OOJ + ooj = {OOJ-JJOOI-I} = ±OOJ_I.

Proof. Applying the previous Lemma, we can see that Left has an option to OOJ_I and Right has an option to oo;_i which gives the result. •

5.3 Regular Games

Our next goal is to assign values to games in much the same way that is done for Normal Play games. We begin by introducing a transformation to the game tree which preserves how the game behaves when played in any short misere sum.

Definition 107. For any game G, T{G) is the game which is constructed by adding an option to coo from every vertex of its game tree where Left has no option and adding an option to oo0 from every vertex of its game tree where Right has no option.

Figure 5.2 shows how applying T affects the game tree of a single component game G. Note that T does not preserve the outcome class of a sum in the sense that in general os~(T(G + H)) ^ os~(T(G) +T(H)). It does, however, have some other nice properties dealing with our various outcome class functions.

Lemma 108. For any game G, os~{G) = os~(T(G)) = o~(T(G)) = o(T(G)).

Proof We first note that the only nodes in the game tree of T{G) which have no options for either player are the leaves which have value oo0 or oo0 depending on if it was a Left or Right option. Therefore, for either player, the only way to win T(G) is to make an overriding move. This gives us the last two equalities. The first equality follows from the fact that an overriding move in T{G) corresponds to a position in 81

G T{G)

O°0 OOQ °°0 OOQ °°0 COo

Figure 5.2: The games G and T{G)

G where the player in question has no options. If it is that player's turn, they win regardless of any other components due to the short misere sum. So we can conclude that it doesn't matter if we play T(G) using Misere Play or Normal Play rules. •

Definition 109. A game G is regular if it has no overriding followers.

The following lemmas will now demonstrate how arbitrary games of the form T(G) behave with respect to regular games using Normal Play rules.

Lemma 110. Let G be a game such that T(G) = oo^ for some non-negative integer k. Then for any regular game H, T(G) > H using Normal Play rules.

Proof. Since we are using Normal Play rules, we can examine the difference game cofc — H. If k = 0 we are done since Left has already won the sum. Otherwise, Left will ignore —H and instead play in co^ at every opportunity. The best Right can do is postpone the inevitable by playing in that component as well but Left will always be able to play to a position no worse than it was before Right played. Eventually, since the game is finite, Left will move to oo0 in that component and win regardless of the state of the other component. Therefore, the difference game is in C and we conclude that oo^ > H as desired. •

Lemma 111. Let G be a game such that T(G) = oo^ for some non-negative integer k. Then for any regular game H, T(G) + H = T{G) using Normal Play rules.

Proof. We first note that if k = 0 the statement holds. We proceed by induction on k. If Left plays first from oo^ + H then she has an option to oo^-i + H = 0Ofc_i. If 82

Right plays in H, then Left can respond as before to oo^-i. If Right instead plays in oofc then by definition, Left can respond in that component to some OOJ for i < k. Since the game is finite, Left will eventually be able to play to some oo^ for i < k. Therefore, oo/j + H = 00^ as desired. •

We can now compare games of the form oofc to any of the games we are familiar with. This in turn allows us to find the canonical form of T{G) by removing domi­ nated options and bypassing reversible moves if we restrict ourselves to Normal Play rules.

Lemma 112. Let G and H be regular games. There exists a regular game K such thatT(G)+T(H)=T(K).

Proof. When playing the sum T(G) + T(H), we can only win by making some over­ riding move. Therefore we can construct a single game tree to represent T(G)+T(H) where all of the winning moves are overriding. If a player has an overriding move, it dominates all other options so we may delete them from the tree. We now have a game tree where all winning moves are overriding and a player has no other moves when an overriding move exists. Thus, we construct K to be the game whose game tree is the same as above with all of the overriding moves deleted. •

Figure 5.3 shows a small example of the application of this lemma. We pick any regular G and H then construct T{G) and T(H). Then we construct a single game tree to represent playing the sum T(G) + T(H) making sure to delete any options which are dominated by coo or oo0 for Left and Right respectively. We can then delete all overriding options to recover the game K.

Lemma 113. Let G and H be regular games. Thenos~(G+H) = os~(T(G)+T(H)).

Proof. Assume Left can win the short misere sum G+H. Then we know she has some strategy to end up in a position where she cannot make a legal move in at least one of 83

G T(G)

oo0 / ^o0 oo0

oo0 oo0

H T{H)

oo0

o°o oo0

T(G)+T(H) = T{K) K

oo0

OO0 OOQ OOo

OO0 OO0

Figure 5.3: Finding regular K where T(G) + T(H) = T(K) 84 the components on her turn. Following this same strategy in the sum T{G) + T(H) she will end up in a position where she has an option to ooo which she takes to win. A similar argument shows that if Right can win one sum, then he can win the other as well. This gives us the result. •

Theorem 114. Let G, H be regular games. Then os~(G + H) = o(T(G) + T(H)).

Proof. We apply Lemma 113 to get that os~(G + H) = os~(T{G) + T(H)). We can then apply Lemma 112 and say there is some K where T(K) = T(G) + T(H). Finally, we can apply Lemma 108 that tells us os~(T(K)) = o(T(K)) which gives us the result. •

Note that when T(G) and T(H) are regular games with T(G) = A and T(H) = B respectively, then the winning strategy for playing G + H as a short misere sum is the same as the winning strategy for playing A + B using Normal Play rules. From this point on, if a game G is such that T(G) is regular with T(G) = A then we will refer to that game using its common Normal Play symbol.

We can now construct a partial order with respect to the short misere sum of games born by day n. To do this, for each regular game G, we first compute T{G) then find the canonical form of T(G) under Normal Play rules. As usual, we will begin with the empty game, but for convenience we will start at day -1. The reason for this is so that the familiar game values we have grown accustomed to are born on the same day as they would be for Normal Play games. Table 5.2 shows the construction of games born by day 0 as well as listing all unique games born by day 1.

Figure 5.4 shows the partial order formed by the games born by day 1. As we can see, there are sixteen different games rather than the usual four we would get if we restricted ourselves to regular games. Another interesting feature is that the game ±c©o does not compare with any other game since any sum which contains it is a first player win. The number of misere games played with a short sum born by day n grows exponentially and it is currently not known how many unique games there are born by day k > 1. 85

Day-1: G = {•!•} ^ T(G) = {ooolooo} = ±oo0

Day 0: G = {{•[}[} - T(G) = {{ooo|«)o}|^o} = ooi G = {-|{-|-}} -+ T(G) = {oo0|{oo0|ooo}} = ™i G = {{-|-}|{-|-}} -+ T(G) = {{cx)o|ooo}|{ooo|oo0}} = 0.

Day 1: {0|00!} = 1 {0|0} = * {O0!|0} = -1

{oOxloOi} = 002 {ooi|ooi} = oo2

{oo0|0}

{oo0|ooi}

{oox|0} {00i|00].} = ±OOi {OlooJ {cx)i|oo0} {0|oo0} Table 5.2: Games played with a short misere sum which are born by day 1 86

ooi {oo0|0} {oo0|ooi}

±OOi ±OOo

ooi {0|oo0} {cx)i|oo0}

Figure 5.4: The partial order of games played with a short misere sum born by day 1 87

5.4 CLOBBER

The game of CLOBBER is played on a rectangular board with the initial position being an alternating pattern of black and white stones. On Left's turn, she may move any black stone onto an orthogonally adjacent square which contains a white stone. The white stone is clobbered and removed from the game. Right's options are similar but he moves the white stones. Clearly the game is partizan but it is also all-small.

Definition 115. An all-small game is one where if one player has a legal move in a given component, the other player does as well.

Lemma 116. Let G be a non-trivial component of a short misere sum of all-small games. Then the canonical form ofT(G) is regular.

Proof. We will show, without loss of generality, that the position ooo does not occur as a left option of any follower of T(G) after it has been simplified. Given any position, if Left can win by making an overriding move in T(G), it means he has no move in G. This in turn means that Right also has no move in G due to the fact that it is all-small. Therefore G = {-|-} which contradicts that G was non-trivial. So, such an overriding move does not exist and the canonical form of T(G) is regular. •

This shows us that the only way to lose a short misere sum of all-small games is to be the first to create a trivial component. It also tells us that the short misere sum is equivalent to some Normal Play game. In this case, the transformation is fairly straightforward. Since creating a trivial component immediately loses, we may instead ban such moves and play under Normal Play rules. If a player would be left with no moves in this game, he would be left with only suiciding moves in the short misere sum.

When playing CLOBBER under a short misere sum, we must first take care to de­ fine a component of the game. It makes the most sense to define a component as being any (orthogonally) connected set of pieces. The only time we get a trivial component in which neither player has an option is when a component is completely composed of one player's pieces. Therefore, we may transform the short misere sum 88

Position Value Position Value t\ A A A t\ t\ t\ A Afo 0 ±1 A t\ At\ /&. I* 0 * At\ A& AtsAt\ 0 AAt\At\t\ *2 At\

0 AAt\ iififi A fik C\ T*

A Ht A ft Atsfo 1 t\AAAt\ T ts tsA ftfi ISA t\ At±t\t± 2 +i t\AAAAt\ ti tsA Atit\At\ i Atst\t\ts 3 ts A 2

Table 5.3: A brief dictionary of misere clobber positions

to a normal play game by banning players from creating a trivial component. Note that while the game started as an all-small game (and still is under the misere in­ terpretation), after transforming it to a Normal Play game by applying T we have introduced positions in which one player can move but the other cannot.

Table 5.3 shows a brief dictionary of values which can occur when playing the short misere sum of CLOBBER. Note that it is possible to generalize some of the construc­ tions. Using these as a basis, we can find positions with any integer value, any switch of the form {x\y} with x > 0 and y < 0 as well as +k and — /. for any positive integer k. 89

/@\

Figure 5.5: A game of MAZE

5.5 MAZE and MAIZE

The game of MAZE is played on a rectangular grid oriented in such a way as to have the edges at a 45 degree angle from horizontal. Some of the interior edges are des­ ignated as walls which cannot be crossed and there is at least one token, denoted by "©", which players may move on their turn. From any given position, Left may move the token any number of spaces in a south-westerly direction as long as he does not cross a wall. Likewise, Right may move the token any number of spaces in a south-easterly direction. If there are multiple tokens, they do not interact and thus create a disjunctive sum. MAIZE is a variant of MAZE where each player can only move his piece one square on their turn. To differentiate from an instance of MAZE, the token is denoted by "O". Figure 5.5 shows a sample game of MAZE.

It is easy to see that MAZE is a game which is option closed. When we play the .short misere sum of MAZE positions, however, we would like to first transform them using T(G) which unfortunately breaks our option closed condition since a player can only make an overriding move when no options existed in G. We do, however, LL L get the similar relations G \ {oo0} C G for any position in which an overriding RR R move cannot be made by Left and G \ {oo0} C G for any position in which an overriding move cannot be made by Right.

We can analyze the short misere sum of MAZE positions in the same way that was done for CLOBBER except that in this case we find that some non-trivial positions do not give us a regular game. Figure 5.6 shows the outcome class analysis for a given game of MAZE. Figure 5.7 shows the same game of MAZE with values computed by 90

Figure 5.6: Outcome analysis of misere MAZE

Figure 5.7: Values for MAZE played with a short misere sum first taking T(G) for each position and finding its canonical form.

Lemma 117. There is no MAZE position G such that T(G) has value oo^ for k > 3.

Proof. For any position of value oo^ there must be an option to oOfc_i but no option to oofc_2- Let G be a position of MAZE where T{G) = oo3. T{G) must have a left option to 002 which in turn must have a left option to 001. Due to the fact that the transform of MAZE is almost an option closed game as described by the relations above, we can deduce that T{G) must have an option to 00! which gives a contradiction. • 91

Figure 5.8: A game of MAIZE with T(G) = oo4

It is possible, however, to have positions of value oofc for any k in the game of MAIZE since from each position of the token, each player has exactly one option. An example for k = 4 is given in Figure 5.8. Chapter 6

Discussion

6.1 Option Closed Games

In chapter 2 we saw how the reduced canonical form can aid in the analysis of cer­ tain games. In particular, option closed games always have a reduced canonical form which is a number or a switch. Given a particular game, like CRICKET PITCH, is it possible to determine if it is a number or a switch based on some other property of the game?

Very little work has been done using reduced canonical forms and there are other games which are almost option-closed games in the sense that except for a few bound­ ary cases, most of a player's options have the property that GLL C GL or GRR C GR. For example, END NIM [1]; PARTIAL NIM [15] (played on one heap) is option-closed for one-player but not the other; also in both KONANE [10] and TOPPLING DOMINOES [2] the moves have a linear directional aspect. For each of these cases, is there any way the reduced canonical form can help simplify these games?

6.2 Misere games

Chapter 3 showed how the structure of Misere Play games differ from Normal Play games. Are there other ways to find more information regarding the structure of misere games in an effort to bring the analysis up to par with what is known regarding Normal Play games. Although we've shown that misere games don't have the nice group structure that is so useful in our analysis of Normal Play games there may be other avenues which could be explored.

92 93

Besides proving that canonical forms exist for misere games [19], Siegel has begun looking at other ways to analyze misere games by reducing the universe of possible games and considering the monoid they generate [20].

6.3 Consecutive Move Ban

One way of simplifying misere games is to introduce a restriction which limits the number and type of games to be analyzed. In Chapter 4, this was done by considering the set of all games where a given player could not play twice in the same component without their opponent playing in that component first. The complete analysis of this game is incomplete in the sense that although we know how to play the game, we do not have a closed form expression for determining the outcome class of a misere sum. This is clearly an aspect which could be addressed in the future.

Following along the same line, are there other non-trivial restrictions we could place on games so that the analysis under Misere Play rules is more approachable. One such idea would be to consider the set of all games born by day n. Our proof that the empty game was the only game with an additive inverse breaks down when you have game trees with a maximum depth. Unfortunately, we still do not have a group structure for large n because we won't be able to construct inverses for small game trees. Even still, the analysis may be much simpler and further our understanding of misere games in general.

6.4 Short Disjunctive Sum

The short disjunctive sum as described in Chapter 5 gives us a new way to consider misere games. Since this is the first place it has appeared, there is very little known. In particular, future work might include a closer look at the structure of games born by day n. More importantly, though, we demonstrate that the concept of overriding moves can be incorporated into the analysis of Normal Play and Misere Play games. This will allow the analysis of games under a disjunctive sum where one player may have overriding moves while the other does not. Clearly, this concept could be applied 94

to connection games where a game ends as soon as one player has achieved some desired configuration on the board. TIC-TAC-TOE, GO MOKU and HEX are examples of such games. Another natural game to consider is SEEPAGE where one player is trying to prevent sewage from reaching a clean water source while the other player makes moves as the sewage. The sewage player can win a sum if they can win in any component (in this case, the water becomes contaminated). Therefore, any such winning move is an overriding move. The defending player, on the other hand, can only succeed if he prevents the water from being contaminated in every component and therefore has no overriding moves. This type of game is similar to maker-breaker style games where one player is attempting to achieve a particular configuration and his opponent can only win by perpetually preventing him from doing so. I'm sure many other variants could be considered and analyzed using the work started here. Appendix A

Tables

Tables A.l and A.2 shows that the misere outcome class of the sum of two games does not depend on outcome classes of the games themselves. Each cell in the table shows two games G and H. The outcome class of G is C\ and the outcome class of H is Ci- Its position shows the outcome class of G + H according to the diagram at the top of table A.l. Some of the 10 possible combinations are not shown as they can be derived by taking the negatives of some other entry.

95 96

V c Ci + C : 2 K M

/\ /' \ X\ "\ V + V: /\ y\ ,-'N-

./\ \ A \ \ V + C: \ ./\ \ /\

./\ ./\ r> • V+M: ./\ o /\ O

Table A.l: Examples of all possible outcomes of a misere sum (part 1} 97

X ,<> \ \ C + C: \ \ x> \ x> X

\ \ / X • C + K: \ /' \ /

\ \ A / \ / C + Af: \ /\ / \ • • • \ /

A A A^A \• •/ \• o N + N: \ X \ / \ / '• /

Table A.2: Examples of all possible outcomes of a misere sum (part 2) 98 V C Ci + C : 2 K M

X^ X\ x 'X- JC + K:

\ x -"

x« Not Possible \ / C+N:

x Not Possible "\ * \ /

/\ /\ /\ ./\ XX

Table A.3: Examples of all possible outcomes of a short misere sum

Table A.3 shows that the short misere sum of two games can be in any of possible outcome classes described in chapter 5. Each cell in the table shows two games G and H. The outcome class of G is Cx and the outcome class of H is C2. Its position shows the outcome class of G + H according to the diagram at the top of the table. Symbol Canonical Form 0 {•I-} 1 {0|-} n {n-l|-} * {0|0} T {0|*} i {*|0} it* {0|T} -!)-* {110} +k {0||0|-A:} ~k {A;|0||0}

Table A.4: A partial dictionary of Normal Play game values Bibliography

[1] M. H. Albert and R. J. Nowakowski. The game of end-nim. Electr. J. Combm., 8(2):#R1, 12pp., 2001. Volume in honor of Aviezri S. Praenkel.

[2] Michael Albert, Richard J. Nowakowski, and David Wolfe. Lessons in Play. A. K. Peters, 2007.

[3] D. T. Allemang. Generalized genus sequences for misere octal games. Internat. J. Game Theory, 30(4):539-556 (2002), 2001.

[4] Elwyn R. Berlekamp, John H. Conway, and Richard K. Guy. Winning ways for your mathematical plays. Vol. 1. A K Peters Ltd., Natick, MA, second edition, 2001.

[5] Elwyn R. Berlekamp, John H. Conway, and Richard K. Guy. Winning ways for your mathematical plays. Vol. 2. A K Peters Ltd., Natick, MA, second edition, 2003.

[6] C. L. Bouton. Nim, a game with a complete mathematical theory. Annals of Mathematics, 3(2):35-39, 1902.

[7] Dan Calistrate. The reduced canonical form of a game. In Richard Nowakowski, editor, Games of No Chance: Combinatorial Games at MSRI, 1994, pages 409- 416. Cambridge University Press, Mathematical Sciences Research Institute Pub­ lications 29, 1996.

[8] John H. Conway. On Numbers and Games. A K Peters, Ltd., Natick, Mas­ sachusetts, 2nd edition, 2001. First edition published in 1976 by Academic Press.

[9] Adam Duffy, Garrett Kolpin, and David Wolfe. Ordinal partizan end nim. In preparation, 2003.

[10] Michael D. Ernst. Playing konane mathematically: A combinatorial game- theoretic analysis. UMAP Journal, 16(2):95-121, Spring 1995.

[11] T. S. Ferguson. On sums of graph games with last player losing. Internat. J. Game Theory, 3:159-167, 1974.

[12] T. S. Ferguson. Misere annihilation games. J. Combin. Theory (Ser. A), 37:205- 230, 1984.

[13] A. S. Fraenkel and A. Kotzig. Partizan octal games: partizan subtraction games. Internat. J. Game Theory, 16:145-154, 1987.

100 101

[14] J. P. Grossman and A. A. Siegel. Reductions of partizan forms. In Games of No Chance 3. Cambridge University Press, 2008.

[15] Chu-WeeLim. Partial nim. INTEGERS, Electr. J. of Combinat. Number Theory, 5:#G02, 9pp., Comb. Games Sect., 2005.

[16] T. E. Plambeck. Daisies, kayles, and the sibert-conway decomposition in misere octal games. Theoret. Corn-put. Set. (Math Games), 96:361-388, 1992.

[17] T. E. Plambeck. Taming the wild in impartial combinatorial games. INTEGERS, Electr. J. of Combinat. Number Theory, 5:#G5, 36pp., Comb. Games Sect., 2005.

[18] W. L. Sibert and J. H. Conway. Mathematical Kayles. Internat. J. Game Theory, 20(3):237-246, 1992.

[19] Aaron N. Siegel. Misere canonical forms of partizan games.

[20] Aaron N. Siegel. Misere games and misere quotients. Notes from a short course.

[21] Walter Stromquist .and Daniel Ullman. Sequential compounds of combinatorial games. Theoret. Comput. Sci., 119(2):311-321, 1993.