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Exploring the computational frontier of combinatorial games

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Exploring the computational frontier of combinatorial games

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Content EXPLORING THE COMPUTATIONAL FRONTIER OF COMBINATORIAL GAMES by Matthew Ferland A Dissertation Presented to the FACULTY OF THE USC GRADUATE SCHOOL UNIVERSITY OF SOUTHERN CALIFORNIA In Partial Fulfillment of the Requirements for the Degree DOCTOR OF PHILOSOPHY (COMPUTER SCIENCE) August 2024 Copyright 2024 Matthew Ferland Dedication Dedicated to my family, friends, and colleagues that supported me along the way. A list in no particular order: • Thanks to my mom and dad who put up with me regularly being too busy to answer her phone calls. • Thanks to my friends for putting up with me missing things due to my PhD work, such as missing several Dungeons and Dragons sessions in a row due to a variety of conference travel. • Thanks to Shanghua for being a wonderful advisor, who was always willing to have conversations about anything, whether teaching me about Chinese philosophy, debating modern politics, or listening to my complaints about air travel. • Thanks to Kyle for being both a great friend and a great coauthor, especially while continuing to be excellent while dealing with his own chaotic life. • Thanks to Svenja for joining our research, and for putting up with my own bratty behavior in Italy. • Thanks to Michael for being both a great mentor in CS Education and friend in general. • Thanks to all of my other coauthors who I don’t have room here to mention, as I have enjoyed working with all of you, and hope to continue doing so in the future! ii Acknowledgements Much of this work (in particular, the content of chapters 3, 4, and 6) in a large part is derived from joint work done with Kyle Burke, Shanghua Teng, and Svenja Huntemann [12, 9, 14, 10]. Much of this work was supported by Shanghua through his receipt of NSF Grant CCF-2308744 and the Simons Investigator Award from the Simons Foundation. I would also like to extend thanks to David Kempe, who left many (over 800) very detailed comments for an earlier version of this dissertation, pointing out both typos and grammatical areas, along with giving several suggestions for improvement. iii Table of Contents Dedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii List of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vii Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ix Chapter 1: Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Chapter 2: CGT Homeland . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 2.1 Fundamentals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 2.2 Impartial Games . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.2.1 Nim . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.2.2 Extension to General Impartial Games . . . . . . . . . . . . . . . . . . . . . . . . . 10 2.3 Partizan games . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 2.3.1 Canonical Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 2.3.2 Game Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 2.3.2.1 Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 2.3.2.2 Infinitesimals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 2.3.2.3 Switches and More . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 2.3.3 Other Forms of Measurement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 2.3.3.1 Mean Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 2.3.3.2 Confusion Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 2.3.3.3 Temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 2.4 Game value complexity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 2.5 A result for very short games . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 Chapter 3: The Impartial Glacial Geography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 3.1 Generalized Geography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 3.2 PSPACE-Complete Grundy Values of Undirected Geography . . . . . . . . . . . . . . . . 27 3.3 A Polynomial-Time Branch-and-Bound . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 3.4 Finding Grundy Values By Branch-And-Bound . . . . . . . . . . . . . . . . . . . . . . . . . 33 3.4.1 Nimber-Winnability Reduction: Degree of Phase Transition to Infeasibility . . . . 34 3.4.2 The Impact of Large-Degree Nodes in Branch-and-Bound . . . . . . . . . . . . . . 35 3.5 Misère-Play Winnability v.s. Grundy Values . . . . . . . . . . . . . . . . . . . . . . . . . . 35 3.6 Undirected Geography with Polynomial Grundy Values . . . . . . . . . . . . . . . . . . 37 iv 3.6.1 Logarithmic Intuition and Polynomial Challenge . . . . . . . . . . . . . . . . . . . 37 3.6.2 Polynomially High Nimber Constructability . . . . . . . . . . . . . . . . . . . . . . 39 3.6.3 Complexity Implication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 3.7 Polynomial-Time Nimber-Preserving Reduction . . . . . . . . . . . . . . . . . . . . . . . . 45 3.7.1 Star Atlas: A Complete Generalized Geography . . . . . . . . . . . . . . . . . . . . 47 Chapter 4: The Slushy Frontier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 4.1 Superstars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 4.2 From Bits to Superstars: Hardness Reduction . . . . . . . . . . . . . . . . . . . . . . . . . . 54 4.2.1 Reduction to Superstars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 4.2.2 EPMX Hardness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 Chapter 5: The Ocean of Game Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 5.1 A Three Oracle Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 5.2 Algorithm 1: Impartial Games . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 5.3 Algorithm 2: Nearest Number or Confusion Interval for Partizan Games . . . . . . . . . . 64 5.3.1 Phase 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 5.3.2 Phase 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 5.3.3 Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 5.4 An Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 Chapter 6: The “Quantum” World . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 6.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 6.1.1 Intuition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 6.1.2 Quantum Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 6.1.3 Highlights of Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 6.2 Exploring Quantum Undirected Geography . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 6.2.1 Strategic Graph Contraction: Polynomial-Time Solution for Classical Starts . . . . 75 6.2.2 Intractability of Quantum Geography Positions . . . . . . . . . . . . . . . . . . . . 79 6.3 Complexity Leap in Quantum Nim: Logic Connection . . . . . . . . . . . . . . . . . . . . . 83 6.3.1 Proof Outline: The Logic Connection of Quantum Nim . . . . . . . . . . . . . . . 83 6.3.2 Isomorphism Between Quantum Avoid True and Quantum Boolean Nim . . . . 85 6.3.3 Nim Encoding and Structural Witness of Σ p 2 -Hardness in Quantum Games . . . . . 87 6.3.4 Tower Nim and algorithms which do not need modification . . . . . . . . . . . . . 89 6.4 Quantum Transformation of PSPACE-Complete Games . . . . . . . . . . . . . . . . . . . . 90 6.4.1 Quantum Preservation and Quantum Collapse . . . . . . . . . . . . . . . . . . . . . 90 6.4.2 The Depth of Complexity Collapse in the Quantum Setting . . . . . . . . . . . . . 102 6.4.3 Quantum Lift of Schaefer’s Partisan-Impartial Reduction . . . . . . . . . . . . . . . 105 6.4.4 Natural PSPACE-Complete Games . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 6.5 A Complexity Barrier to a Quantum Leap . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 6.5.1 Quantum Game Trees Are No Taller . . . . . . . . . . . . . . . . . . . . . . . . . . 116 6.5.2 Quantum Node Kayles in PSPACE . . . . . . . . . . . . . . . . . . . . . . . . . . 117 Chapter 7: Homecoming . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 7.1 Putting things together . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 7.1.1 Impartial Games and the Oracles . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 7.1.2 Hardness and the Oracles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 7.1.3 Quantum CGT and other sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 v 7.2 Implications and Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 7.2.1 Historical Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 7.2.2 Future Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 7.3 Future Directions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 7.3.1 Impartial Games . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 7.3.2 Tepid Games . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 7.3.3 Game value computation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 7.3.4 Quantum Combinatorial Games . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130 vi List of Figures 3.1 The gadget for each directed edge (x, y). . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 3.2 Prelude Gadget. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 3.3 The overall schema for Nimber Constructability. The value of the position with the token at Nn is ∗n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 3.4 Each vertex Ni is connected to all other N-vertices, as well as its own ∗3 gadget, its own ∗2 gadget, its own ∗ gadget, and its own Ri vertex (with value 0). Ri is also connected to the Pk and Mk gadgets with k > i as shown in the following figures. . . . . . . . . . . . . 38 3.5 A ∗ gadget, which has value ∗ unless a lower-rank-N vertex is removed. If all of the Nk vertices remain in the graph, then moving to any one of them from Rk results in a 0-board (by Lemma 3.6.4). This causes the move Mi → M (a) i,k to be equal to ∗2. Otherwise, one Rk has value 0, so Mi → M (a) i,k is a move to ∗, so Mi instead has value ∗3. . . . . . . . . . . . 39 3.6 The ∗2 gadget . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 3.7 Result of the classic QSAT and Generalized Geography reductions of the question, Qi , “Does G = ∗i?”, with the added vertices ai and bi . . . . . . . . . . . . . . . . . . . . . . . . 49 3.8 vertices t0 through tg−2. Each vertex ti has edges to t0, t1, . . . , ti−1. Thus, the nimber value of the Geography position at vertex ti is ∗i. . . . . . . . . . . . . . . . . . . . . . . . 50 3.9 The gadget combining all of the f(Qi) gadgets into a single Generalized Geography instance. The vertices bi , ci , and di are in rows and columns indexed by the letters on the left and the numbers along the top. Each bi has an edge to ai as in Figure 3.7. . . . . . . . . 50 4.1 Multi-state variable xa with four possible states: s1, s2, s3, s4. The overall color indicates that the chosen state is s2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 6.1 Winning strategy for current player inQuantum Nim (2, 2), showing that quantum moves impact the game’s outcome. (There are four additional move options from ⟨(1, 2) | (2, 1)⟩ that are not shown because they are symmetric to moves given.) . . . . . . . . . . . . . . . 72 vii 6.2 Gadgets for reducing from a Geography edge (a, b). The left is the gadget as it will appear in the main board and all other boards except from a (a, b)-board. The right is the edge as it appears in a (a, b)-board. All other parts of the two boards will be the same. . . . . . . . 79 6.3 Gadget for a Directed Geography edge (a, b) in Undirected Geography. Prior to the current position, there were moves ab−1 → ⟨ab1 | stopab⟩ → ab−2. . . . . . . . . . . . . . 81 6.4 PSPACE Reduction to Quantum Geography is a simple transformation on the edges. . . 83 6.5 Example of the PSPACE reduction from QBF to Quantum Node Kayles. QBF: ∃x1 : ∀x2 : ∃x3 : (x1 ∨ x2 ∨ x3) ∧ (x2 ∨ x3) ∧ (x3 ∨ x3) . . . . . . . . . . . . . . . . . . . 111 6.6 Example of the PSPACE reduction from QBF to Q-BiGraphNodeKayles, using the same formula as in Figure 6.5. Here Blue is going first as the True player. . . . . . . . . . . . . . 114 6.7 Graph for Quantum BiGraphNodeKayles, followed by the result of the reduction to Quantum Snort on that graph. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 viii Abstract People have been playing games since before written history, and many of the earliest games were combinatorial games, that is to say, games of perfect information and no chance. This type of game is still widely played today, and many popular games of this type, such as Chess and Go, are some of the most studied games of all time. This work proposed revolves around a game-independent systemic study of these games. More specifically, it involves computational properties involving evaluating mathematical analysis tools for combinatorial games, such as Grundy values and confusion intervals, as well as identifying what can be determined about these games using simple oracle models. ix Chapter 1 Introduction Combinatorial games form some of the oldest games created by humans. Go was created somewhere between 2500–4000 years ago in ancient China [47]. Mancala may be even 8000 years old, as some scholars believe it was played by the ancient Nabataeans in 6000 BCE, based on an archaelogical site in modern day Jordan [44]. The tradition extends to modern-day, where not only are old games like Chess and Go played even at the tournament level, and also modern games such as Domineering and Slime Trail∗ . There are also popular games like Fjords that, while not strictly combinatorial, are influenced by combinatorial games. For Fjords, its initial phase is a combinatorial game [13]. A more complete definition will be given in the next chapter, but in short, combinatorial games are games of perfect information and no chance. For example, there are no chance-based elements like dice or a shuffled deck of cards, and no non-public information like an opponent’s hand of cards or secret moves by one’s opponent. Thus, some ancient games, like the Royal Game of Ur, are not combinatorial games, because the game used dice [25]. Stratego, a more modern game which has no randomness, is not a combinatorial game since neither player knows the ranks of their opponent’s pieces. [51]. Several combinatorial games are combined in modern variants with the aim of making the games either more social or interesting. For example, Bughouse is a 2v2 team-variant of chess that is sometimes thought ∗ https://www.di.fc.ul.pt/ jpn/cnjm17/ 1 of as a “party” variant due to its zany and social nature [57]. Ultimate Tic-Tac-Toe (and also Tic-Tac-Ku† ) takes the overly simple (and solved) Tic-Tac-Toe and replaces it with the combination of 9 tic tac toe games [6]. Combinatorial Game Theory (CGT) is an area of mathematics dedicated to studying these kinds of games, and is especially concerned with studying in how they interact when they are combined. Mathematicians have developed a robust mathematical framework to tackle this challenge [48], which has led to deeper understanding of how these games operate, and have contributed to real-world computation of games [54, 31, 3]. Simultaneously, computational results involving these games have produced many results [45, 40, 35]. But notably, computational results directly interacting with the theory, rather than the games themselves, have been historically rare [26]. Our body of research attempts to remedy that, and this dissertation collects the results we have published during the past few years, as well as a few additional results. We also note possible future related work. The following are brief descriptions of what each chapter entails: • Chapter 2: Covers the very fundamentals of combinatorial game theory (CGT), which is essential for understanding the research that follows. It also introduces some notation which will be used throughout the dissertation. • Chapter 3: Covers results involving impartial games, which are games in which both players have the same options at every position. • Chapter 4: Covers a result involving tepid games, which are a type of game in which neither player can get very far “ahead” † https://www.colorku.com/product-page/tic-tac-ku 2 • Chapter 5: Focuses on results involving hot games. Specifically, contains some preliminary results in how to derive the CGT analytical tools (game values) by some computational oracles, and some implications thereof. • Chapter 6: Covers results involving “quantum combinatorial games”, a kind of non-deterministic combinatoral game. • Chapter 7: Covers miscellaneous results, examples of how many of our results synergize to produce other interesting conclusions, some applications of these synergies, and directions for future research 3 Chapter 2 CGT Homeland This chapter acts as a primer to Combinatorial Game Theory (CGT), but only so far as it applies to this dissertation. Most things described are presented as definitions rather than theorems with proofs. Many concepts are left out, and some concepts are only discussed at a very high level. For a more comprehensive treatment of the field, we suggest one of many excellent introductory books [48, 5, 1]. 2.1 Fundamentals Combinatorial Game Theory is an area of mathematics focused on studying combinatorial games. Combinatorial games are typically defined by the presence of two properties: • Perfect information: Perfect information games are typically defined as games in which all parts of “events” (i.e.: moves) made so far in the game are known to all players, including the “initialization event” which creates the game. • No chance: No chance means that the game is deterministic. For a given state of the game, the possible moves (also called options) are fixed, and always have the same outcome if selected. 4 While these two properties are all that is needed for a game to be a combinatorial game, and there is a large amount of work for a wide variety of “‘non-standard” contexts [38, 39, 20, 23, 30], for the purposes of this dissertation, we will be focused on games with the following properties (which we consider “standard”): • Two Player: The game is played between exactly two players. • Finite: The games have a finite description of the game state. So, there are a finite number of possible positions of the game. • Loopfree: Once a position has been reached, it is not possible to reach that position again. • Normal Play: The game only ends when a player is unable to make any moves, and in particular, when a player is unable to make a move, they lose the game. Any time this dissertation breaks any of these assumptions, it will be explicitly stated, and carefully written to avoid any confusion. Also, note that the Finite and Loopfree properties together characterize what is known as short combinatorial games. While most of our results are about combinatorial games following the normal-play convention (as listed above), we also want to note that there is also the concept of misère-play which will appear briefly in Chapter 3. For misère games, instead of the player who cannot make a move losing, the player who cannot make a move wins. The analysis of these games is significantly more complicated, and the theory is incomplete even for very simple classes of games, and is an area of active research. The primary driver behind CGT is the analysis of outcomes of games under simultaneous play. While there are many definitions of “simultaneous play” that have been studied, the primary one for the field, and the one that this dissertation will exclusivly focus on, is that of disjunctive sums of games. We will formally define this term later once other notation has been introduced, but intuitively, in a disjunctive sum, on a player’s turn, they choose exactly one game to move on, and make exactly one move on that 5 game, and the outcome of the sum follows normal play convention (i.e.: when one cannot make a move on any game in the sum, one loses the game). First, we need to distinguish between rulesets and positions. A ruleset is a description of the rules of a game. For example, one could describe the ruleset of chess by describing how the pieces can be moved, and how one wins the game. A position is a specific instantiation of a game in question. For example, a position in chess can be given by describing the locations of the pieces. The two definitions are introduced together here, since the term game itself is an overloaded term. “Game” can be used to refer to either the ruleset, or a position. And this overloaded term is regularly used in the literature, including in this dissertation. Generally, if we use “game,” then the intended usage should be clear in context; when it is not, we will switch to using the terms “ruleset” or “position.” The options or moves of a position are the positions to which a player can move. For example, in the starting position of chess, a player has 20 options. They can move any of their 8 pawns forward 1 space or 2 spaces (which totals 16 options), or either of their two knights to two possible locations (representing the other 4 options). Each of these options is represented by the position reached if the option is selected (i.e.: a representation of the game board that will be reached if one selects that option). At the start of the game, both players are assigned an identity, Left or Right. Left options represent the possible moves that can be taken by the Left player and Right options are the moves that can be taken by the Right player. Sometimes other player names are used, such as blue and red or black and white, respectively. But these are then mapped onto the formal left and right identities in formal analysis. For example, again with the chess starting position, white (formally: right) has 20 options, and black (formally: left) has 20 options. Formally, a ruleset R is a mapping X → 2 X × 2 X ∪ {∅}, where X = {G1, G2, . . . } is the set of all possible positions that a game can be in at any point of play, and the mapping indicates what positions can 6 be moved to and from, with the domain being the position moved from, and the range being the position moved to. For loop-free games, as we focus our attention on, this mapping will not form a cycle. A position G is composed of two sets of options: left options GL ⊆ X and right options GR ⊆ X . Then, we can represent a position G as its options {GL|GR}, which describes the mapping of the position to its options as given by the ruleset. One may be concerned that, in the formal definition, since X is very large, possibly infinite in size, R may be an infinite mapping, and not the “intuitive” description of the rules as presented earlier. To resolve this, one can think of the intuitive description as a succinct oracle, such that for any pair of positions G, H, given the succinct description, one can determine if H ∈ GL, and determine if H ∈ GR. Now, we can formally define what a disjunctive sum is. We can represent the disjunctive sum of two games G and H (notated G + H). If G = ∅ and H = ∅, then G + H = ∅. Otherwise, G + H = {GL + H, G + HL|GR + H, G + HR}. In other words, Left can play on G and leave H untouched, or move on H and leave G untouched, and similarly Right can either move on G or H while leaving the other one untouched. Now, we can define the game tree. Simply put, it is a tree which describes all possible positions the game can reach from the current position. The position {|} is a leaf of the game tree, since it has no positions it can be moved to, and thus has no child nodes. Otherwise, for other positions {GL|GR}, there is a corresponding node of the game tree, with edges to each G ∈ GL and G ∈ GR. We use size(G) to denote the number of nodes in the game tree. A definition we will use throughout the paper is as follows: Definition 2.1.1. A ruleset R is said to be polynomially short if, for any game G ∈ R, where G is given in n bits, the game tree height can described as poly(n). If for any position G ∈ X , the position has GL = GR, then we say the game is impartial. Otherwise, we say the game is partizan. In other words, impartial games are those in which player identities don’t 7 affect gameplay, while partizan games are ones where it does. For example, chess is partizan because the white player cannot move the black pieces. The result of play on a position under optimal play by both players is called the outcome class. We denote the function for determining the outcome class of game G by o(G). There are four possible outcome classes: • L: Left can win, regardless of whether they go first or second. • R: Right can win, regardless of whether they go first or second. • N : The first player to play on the position can win, regardless of whether they are Left or Right. • P: The second player to play on the position can win, regardless of whether they are Left or Right. P, as an outcome class, can be confused with the computational complexity class P. So, in this dissertation, we will always use “polynomial-time solvable” or another similar phrase rather than refer to the complexity class P. Positions can be inverted by recursively swapping Left’s and Right’s options for every position in the game tree. We represent the inversion of G as −G. So, for G = {GL 1 , GL 2 , . . . , GL n |GR 1 , GR 2 , . . . , GR m}, we have −G = {−GR 1 , −GR 2 , . . . , −GR m| − GL 1 , −GL 2 , . . . , −GL n}. Note that since we are also negating each position, that means they recursively have their own game trees swapped. Inversion naturally also affects the outcome. Inverting a position with outcome class N or P doesn’t change the outcome class, but inverting positions with outcome class L or R have their outcome classes change to the other (i.e.: If o(G) = L, then o(−G) = R, and vice versa). Two games G and H can be compared by examining the outcome class of the disjunctive sums of the inversions. G + −H (commonly written as G − H) will have 4 possibilities of outcome class. If o(G − H) = P, then we say G = H. If o(G − H) = L, then we write G > H. If o(G − H) = R, then G < H. Finally, if o(G − H) = N , then we say G is confused with H, denoted by ̸≷. 8 Names for ruleset of games in CGT are normally written in smallcaps (e.g.: Chess or Mancala). From this point onwards, we will follow this convention for denoting games. With all of this initial information out of the way, let’s examine an initial simple class of games: impartial games. 2.2 Impartial Games Impartial games, from an analytical point of view, are simpler than partizan games in structure to analyze, with their symmetric nature leading to several convenient properties. First of all, rather obviously, the outcome classes L and R cannot arise in impartial games, since the symmetric nature means that any strategy Left has is also shared by Right. Thus, instead of referring to the outcome class, we may instead refer to the problem of indentifying winnability of a game. Second, it turns out that an efficient method for determining o(G + H) (with some caveats) for disjuntive sums of imparital games is well understood. We will first look at a simple game known as nim and later expand our viewpoint to other impartial combinatorial games. 2.2.1 Nim Nim is a game whose roots can be traced back to the ancient Chinese game 捡石子 ("picking stones"). The rules are simple: There is a set of piles of sticks. On a player’s turn, they may choose a single pile and take as many sticks as they wish from that pile. Then, the turn alternates to the other player who does the same. When a player is unable to select any pile (because there are no sticks remaining), they lose the game. The single pile case is trivial, since the first player can just take all of the sticks to win the game. The two pile case is also simple, and is a fairly common pedagogical tool when demonstrating inductive proofs to undergraduate students [33]. One player wins by maintaining the invariant that whatever option they 9 select will have both piles containing an equal number of sticks, since regardless of their opponent’s move on one pile, they can mimic that move on the other pile, and eventually the equality will hold with the number of sticks in each pile being 0, causing the other player to lose. The proof for the general case, as created by Bouton in 1901 [7], is just a generalization of this two pile proof. Instead of maintaining an invariant that the two piles are equal at the end of a player’s turn, the player wishes to maintain the invariant that the bitwise xor of the size of every pile equals 0. If the bitwise xor isn’t 0, a move can be made to a position with bitwise xor of 0, and eventually the invariant will hold with the number of sticks in all piles at 0, implying the player wins the game. 2.2.2 Extension to General Impartial Games Taking a step back, it is interesting to note that Nim itself is already a disjunctive sum of single piles. Players choose exactly one pile to move on in an alternating fashion. This isn’t a one-time coincidence: many natural games can easily be decomposed into disjunctive sums, which is one of the many motivating factors for the popularity of studying disjunctive sums. However, Nim is notable for its simplicity, and for the fact that it inspired a general theory for disjunctive sums of impartial games. R.P Sprague and P.M. Grundy independently in the late 1930’s realized that the intuition driving Bouton’s Nim algorithm was not exclusive to Nim, but rather, could be extended to all impartial games [50, 32]. A (modified) way to present their theorem is as follows: Theorem 2.2.1 (Nim Equivalence). Any impartial game G has an equivalent nim pile which defines how it interacts in a sum. Let nim(G) be a function that outputs this nim-equivalent. Then, a sum of games G1 + G2 + · · · + Gn has a winning move if and only if nim(G1) ⊕ nim(G2) ⊕ · · · ⊕ nim(Gn) ̸= 0. Definition 2.2.2 (Nim function). nim(G) is defined as follows: • nim(G) = 0 if and only if o(G) = P 10 • nim(G) = k if and only if there exists at least one option G′ of G with nim(G′ ) = i for all i < k, but no option with nim(G′ ) = k (i.e.: k is the smallest number that is not the nim equivalent of any option) For example, if game G1 has a “nim equivalence” of 1, G2 of 2, and G3 of 3, then o(G1+G2+G3) = P, since 1 ⊕ 2 ⊕ 3 = 0. Note that since all leaf nodes of the game tree will have value 0, every position will have exactly one nim-equivalent. This theorem, known today as the Sprague-Grundy Theorem, forms the backbone, and part of the impetus, for modern CGT. These “‘nim equivalents” are called Grundy values, nim-values, or nimbers, and all of these terms are interchangable. Typically, in modern CGT writing, instead of writing these values as 0, 1, 2, . . . k, these values are written as 0, ∗, ∗2, ∗3, . . . , ∗k. In this dissertation, the modern notation will be used, and they will be referred to as any of the three terms interchangeably. In general, throughout this dissertation, we may equivalently use one of nim(G) or nimber(G) to refer to the nimber of a game, if the game is an impartial game. 2.3 Partizan games Impartial games have a succinct game value (through Grundy values) to summarize how the games affect the outcome of disjunctive sums. So, one may hope that a similar simple characterization can lead us to strategies for sums of parizan games. Unfortunately, the characterization for partizan games is significantly more complicated. In the worst case, the characterization may be comparable in size to the entire game tree. As we will later discuss, determining who will win on a sum, even with succinct game values, can be PSPACE-complete. 2.3.1 Canonical Form Before we dive into game values in earnest, we will (informally) define the canonical form. Put simply, the canonical form is the simplest representation of a game. When we introduced the {GL|GR} notation 11 previously, we assumed that representations of the games (notably, a representation of the game board) were written for each option, or substituting those positions with their own options (perhaps recursively). Eventually, this will terminate at {|}, where neither player has options, which is a leaf node in the game tree. However, with the canonical form, there are a few additional simplifications. First, a game has succinct game value representations, as we will discuss later, then we may use those instead of explicit positions themselves. Second, we use the simplest game that has the same outcome no matter what sum it is a part of. One way this manifests itself is that any options that are strictly worse than some other option (meaning an optimal player would never choose them, regardless of the disjunctive sum the game is a part of), then the option can be discarded and not included in the canonical form. These options are known as dominated options. One final point about canonical forms to mention is that we need to define the notation for when games are explored several moves down in the game tree. All that happens in this case is that additional vertical bars are used to highlight the order of moves. Commas are used to list options. Take for example, the following position: {3|0 2| − 1, ∗} Here, the two vertical bars represent the first decision point. Left can move to the game {3|0}. Meanwhile, Right can move to the game {2| − 1} or move to the game ∗. In general, this continues to extend, so if the canonical form needs to represent, say, 5 different “levels,” then the first move (or top of the game tree) is represented with 5 vertical bars, the next level with 4 vertical bars, and so on, until the final level is represented with a single vertical bar. 12 2.3.2 Game Values Game Values for partizan games do not have a single form like with impartial games. Rather, they have several forms. We in general can refer to the game value of G with the function value(G). The habitat of a ruleset is the range of game values that are possible for a position in that ruleset. For example, the habitat for Nim is the set of all nimbers. If the habitat has all possible combinatorial game values, we refer to that game as a Universial Ruleset. In this subsection, we will explore common game values that appear in the habitat of games. 2.3.2.1 Numbers Numbers form the simplest game values in partizan games. 0 is a number, which remains defined the same as it was for impartial games. That is to say, whoever moves first on a game with value 0 will lose that game (if the game is played alone, instead of in a sum). The canonical form representation is {|}. That is to say, neither player has a move. While some 0 positions may have options, they act the same as if one cannot move on the game in a sum, since they act the same in disjunctive sums. Intuitively, the positive and negative numbers represent how biased the game is toward a particular player. High positive numbers represent strong favoritism towards Left, while high magnitude negative numbers represent favoritism towards Right. Let’s start with the definition of the integer 1. 1 is defined by the canonical form {0|}. That is, Left can move to 0, while Right cannot move. Similarly, −1 is defined by {|0} (as expected by the definition of negation). So, the integer 1 acts as a sort of “stored move” for Left (and −1 a stored move for Right). In a sum, Left may avoid playing on the game with value 1 until they have no other moves, letting themselves win unless Right has stored moves for themselves. 13 This naturally extends to higher integers. 2 is defined by {1|}, 3 by {2|}, and so on. Similarly, −2 is defined by {| − 1}, −3 by {| − 2}, and so on. Thus, the magnitude of the integer reflects the number of stored moves. Beyond integers, games can also have values that are dyadic rationals. That is to say, values which have the form c 2 k , where c can be any integer and k can be any positive integer. These have canonical forms where both players have moves to integers or dyadic rationals, and when GR > GL. The rational value can be found by simply taking GR−GL 2 Let’s look at a few examples. • 1 2 = {0|1} • 3 4 = { 1 2 |1} • 1 1 2 = {1|2} Similarly, the negative versions of all of these would just involve flipping the left and right options recursively (e.g.: −1 1 2 = {−2| − 1}). In sums, numbers act as you would expect from normal arithmetic. For example, 1 2 + 1 2 + −1 = 0, meaning that whoever plays first in that sum will lose the game. Let’s trace through this to illustrate the fact. Let’s suppose Left goes first. Left has two options, but both are symmetrical since they are just playing on a 1 2 . After moving one of them to 0, Right can move the other half to 1, which leaves Left’s only option to bring 1 to 0. Right can then move the −1 to 0, winning the game since Left has no remaining moves. Similarly, if Right goes first, they can go either on the 1 2 or −1. If they bring 1 2 to 1, then Left can move the other 1 2 to 0, leaving 1 + −1, where Right brings −1 to 0 and Left brings 1 to 0, winning the game and causing Right to lose. If Right instead made their first move to bring −1 to 0, then Left brings one of the 1 2 to 0, Right moves the other to 1, and Left takes the final 1, winning the game. In general Berlekamp, Conway, and Guy, in their foundational book [5], proved 14 Theorem 2.3.1. On a sum of numbers, let the arithmetic sum (i.e.: the arithmetic sum one learns in elementary school) of those numbers be x. If x > 0, then Left wins the sum. If x < 0, then Right wins the sum. Finally, if x = 0, then whoever plays second wins. In fact, you may notice that the outcome of that example game is deterministic, in that even if the winning player wanted to lose, or the losing player wanted their opponent to have extra moves remaining, they cannot. This is not a coincidence, since numbers have no decisions that can affect the outcome of the game. They are called, as we will later discuss, cold games, since there is no reason to play on them earlier than one must, because no advantage is gained. The players are simply exhausting their stored moves. So, one can imagine they are “cold” from never being played. 2.3.2.2 Infinitesimals Infinitesimals, at an intuitive level, are values that are smaller in magnitude than all positive or negative numbers, but aren’t 0. One example of an infinitesimal value already introduced are nimbers, which favor neither Left nor Right. However, there are infinitesimals which do favor one side. Take, for example, the value ↑= {0|∗}. This game is in L, since if Left goes first, they bring to game to 0 and win, and if Right goes first, they bring the game to ∗ then Left brings it to 0 and wins. However, if we add any negative number to ↑ (via a disjunctive sum), then Right will win, since if Left plays on ↑, it will result in just the negative number remaining. Left cannot win by playing on the negative number, since either Left cannot play on it at all, or if they can, any move will result in it being another negative number. If Left brings the negative number to another negative number, Right can bring the ↑ to ∗, and then to 0 if Left plays on a negative number again, which results in just a negative number, which is a Right win. Definition 2.3.2 (infinitesimally close). Games G and H are called infinitesimally close if G − H is an infinitesimal value. 15 One type of infinitesimal games are dichotic games. Dichotic games are combinatorial games in which, if one player has a move at any position, the other player also has a move (removing any dominated options from consideration). So, the game will always end at a 0 value position (rather than a non-zero number). Related to considering “how the game ends,” we will also define left stops and right stops. Intuitively, these concepts encapsulate what number will be reached if only the game in question is played. For example, in the game {4|2 |3| − 2}, the left stop is 2, since Left will go to a game where Right can go to 2, arriving at an integer, and the right stop is 3, since Right can go to a game where Left will go to 3. More formally, the left stop function L(G) and right stop function R(G) are defined as follows: L(G) =    G if G is equal to a number max GL R(GL) otherwise R(G) =    G if G is equal to a number min GR L(GR) otherwise If L(G) = R(G), then G is infinitesimally close to L(G). So a formal way to define infinitesimal games is that L(G) = R(G) = 0 (this is an “if and only if”). There are many types of infinitesimal values. We will briefly introduce a type of infinitesimals known as superstars in Chapter 4, but discussing these value further isn’t relevant for the dissertation. 2.3.2.3 Switches and More Switches are games that have the form {x|y}, where x and y are both numbers, and x > y. Dyadic rationals, by contrast, have the same form, but with x < y instead. Notice that because x > y here, moving on these games is more valuable than the types of games have explored so far, since moving on these games results in a swing in the number of stored moves. Take, for example: 16 {4|2} + −3 + ∗ If Left goes first, and they play on the switch, the resulting game will be 1 + ∗, which will be a win for Left, since the number dwarfs the infinitesimal. But if Right is first to move on the switch, then the game will have value −1 + ∗, which will be a win for Right. We will elaborate on this increased incentive in the Section 2.3.3.3. But beyond Switches, there are (infinitely) more types of combinatorial game values, as some canonical forms cannot be simplified to a nice, succinct representation. For example, we will use a position from Subtraction. Essentially, it it is a variant of nim, except where the players are restricted in how much they can take every turn. For example, in Subtraction{(1, 3)|(2, 3)}, the Left player can subtract 1 or 3 from a pile, and Right can subtract 2 or 3. The following is the canonical form of Subtraction{(1, 3)|(2, 3)} with only a single nim pile of 6, in its simplest form: ( 1 1, {1|0} 0, {1|0} 0 0, {1|0} 0 ) Notice that this is a game that can’t last longer than four turns, and it is already incredibly complicated compared to what we have seen. In general, there is no limit to how complicated the canonical form can be other than the game tree of the game. For a game with an exponentially large game tree, the canonical form could also be exponentially large. So, while game values are a complete summary of how games operate in disjunctive sums, they are also often unsatisfactory in achieving their descriptive property over a full game tree. Nevertheless, they often still do provide summaries which are sometimes succinct with practical uses in applications [3, 2]. 17 2.3.3 Other Forms of Measurement Following the limitations of game values (most notably, their lack of succintness), other measures have been used. Broadly speaking, they sacrifice the precision and accuracy of game values for succinctness and intelligibility, each in their own ways. As with all other sections, this list is not meant to be in any way complete (e.g.: we don’t discuss reduced canonical form), but only sufficient to understand the results of this dissertation. 2.3.3.1 Mean Value One intuitive approach of alternative measurement is to take what the game “feels like on average”. In other words, we repeatedly add the game to itself, and take the limit of the sum of n games divided by n as n goes to infinity. This is called the mean value of the game. More formally, mean value is defined as limn→∞ L(n · G) n = limn→∞ R(n · G) n Note that here, n · G is defined as the disjunctive sum of n games of G (i.e.: G + G + · + G, with a total of n − 1 summations). The equality holds for all games, so the Left stop and Right stop eventually converge. Recall that when the left stop and right stop are equal, the game value is infinitesimally close to L(G). So, intuitively, the expression is indicating that after a game is added to itself enough times, the value of the sum becomes infinitesimally close to the mean value. The full proof is more involved, but can be found in most introductory CGT textbooks. The limitations of mean value as an approach are apparent. First, it ignores infinitesimal factors, which limits its utility when games are “close”. Second, the “average” is only based on interactions with itself. 18 A game might have a mean value of 5, but with most games, “feel” like -10, and with others, act like 7. Nevertheless, it achieves its goal of being a succinct description, and proves itself a useful analytical tool, and is used in definitions and proofs for a variety of concepts in CGT. 2.3.3.2 Confusion Intervals One of the issues with mean value, as just mentioned, is that it doesn’t tell one the range of how the game may act in different sums. Confusion intervals are a direct answer to this, representing the range of numbers the game can “feel like” in a sum. The definition is simply the range of numbers that the game is confused with. More formally, let D be the set of Dyadic rationals. Then, for a game G, and for all k ∈ D, the confusion interval is the set of k for which G ̸≷ k. This will produce a single range of numbers, since if G ̸≷ x, and G ̸≷ y, then for all z with x ≤ z ≤ y, G ̸≷ z. Thus, all that is required to find a confusion interval are the two end points. And in fact, the two endpoints for the confusion interval for a game G are just L(G) and R(G). However, the endpoints can be open or closed, depending on infinitesimal factors. Thus, given a confusion interval, if a game G is summed with a value v above or below the interval, one can still identify the outcome class, since G’s impact is insufficient to “counter” v. Furthermore, it can be paired with the mean value to provide a more comprehensible picture on how the game acts in different contexts. However, the confusion interval is still limited. First, as with mean value, infinitesimal factors are ignored (outside of determining whether the endpoints are open or closed). Second, the range, even when paired with the mean value, still can be unhelpful in a sum where the exact point in the range matters. It nevertheless is a useful analysis tool, and alongside the mean value, is an instrumental definitions within CGT. 19 2.3.3.3 Temperature Temperature takes a different approach than the two we’ve seen so far. Rather than attempt to be a descriptor on the result of the game, it focuses on how much a player is incentivized to play on a game. Usually, optimal play on a sum of games will involve playing games from highest temperature to lowest (though this is not always the case). To illustrate the case, let’s reexamine the sum {4|2} + −3 + ∗. In this sum, {4|2} has temperature 2, ∗ has temperature 0, and -3 has temperature -1. So, the first player to play will want to play on the switch game, then the nimber, and finally will play on the numbers last. The three games in the sum also illustrate the three broad temperature classifications of games. Games with positive temperature, such as switches, are called hot games. All infinitesimals have temperature 0 and are called tepid games. And finally, all numbers have temperature -1 and are called cold games. Temperature is defined by a process known as cooling. Essentially, cooling is a process by which one examines how much one could decrease the incentive to move to a location before it becomes infinitesimally close to a number. This is done by taking a number t, and subtracting it from the left option and adding it to the right option. Then, t(G), the temperature of G, is defined as the smallest t to make the game infinitesimally close to a number. The formal definition is as follows (taken from [48]): Definition 2.3.3. Let t ≥ −1. We define G cooled by t, denoted by Gt , as follows. If G is equal to an integer n, then simply Gt = n. Otherwise, put G˜ t = {G L t − t | G R t + t} Then Gt = G˜ t , unless there is some t ′ < t such that G˜ t ′ is infinitesimally close to a number x. In that case, fix the smallest such t ′ and put Gt = x. 20 Definition 2.3.4. The temperature of G, dented by t(G), is the smallest t ≥ −1 such that Gt is infinitesimally close to a number. Let’s take a look at the example {8| − 6}. Here, the smallest t value to achieve this is t = 7, which gives us {1|1}, which is equivalent to 1 + ∗. Thus, {8| − 6} has temperature 7. Note that tepid games are always infinitesimally close to 0, so they, by definition, always have temperature 0. Numbers are cold games with temperature −1. While playing on the greatest temperature first is not always a winning strategy, it is (anecdotally), often a good heuristic. Furthermore, a weaker statement that does always hold is that one will always play hot games first, and one will always play on tepid games before one plays on cold games. And while the other two descriptions in this section are useful, temperature in particular is an active area of research in CGT, and often results is new insights about classically studied games. It is worth briefly noting before moving on that all three concepts mentioned in this section (mean values, confusion intervals, and temperature) are used in the analysis tool of thermographs, which combines all three ideas mentioned here into a single representation. Thermographs provide one of the strongest bases for summarizing games that lack succinct game values [4]. 2.4 Game value complexity Before diving into our results, we must first discuss what computational results already existed. First, there is a series of results that apply combinatorial game theory in order to develop algorithms, such as, but not limited to, [54] and [3]. Second, there are results which are in service of CGT. One example, among others, is [2]. Third, there are results that just happen to involve combinatorial games, but don’t involve the actual mathematical theory. These results are numerous, including game heuristics such as AlphaGo[17], computational complexity of individual combinatorial games[35], or other possible areas, 21 such as raw computational search to “solve” the game, such as the recently completed Othello tree search [53]. The final group of results, and the most important for this dissertation, is the group of computer science theory results about properties of combinatorial game theory. Unfortunately, this also appears to be the least explored area. The first computational result in this area is from 1981 by Morris, where the disjunctive sum of a polynomial number of games listed in canonical form was proven to be PSPACE-complete [41]. It is worth noting that a wide variety of games were used in this summation, including both hot partisan games and infinitesimal games, and that each game could use up to a polynomial number of bits to describe in canonical form, with the game trees being possibly exponentially tall. Specifically, it uses polynomial bits to represent numbers in the game values, which correspond to exponentially long games. A related result [58] applied Morris’ idea to Go end games, but required an exponentially large Go board. The next major improvement was introduced in 2002 by Wolfe, where, in order to extend the Go end-game results to a polynomially-sized board, and thus show the game was PSPACE-complete, improved Morris’ result to have each game use only a logarithmic number of bits [55]. Beyond these mentioned results, we are unaware of any other results that directly tackle computational CGT questions. As a final note, we will use poly(n) throughout the paper to denote some polynomial function of n. This is assumed to be some fixed polynomial which is problem-dependent. 2.5 A result for very short games Before continuing with more involved results, in this section, we will provide a result about games with constant-sized game trees which isn’t game-specific. We first need to introduce the concept of a game’s birthday. 22 Definition 2.5.1. The birthday of a game G, denoted by b(G), is the height of the game tree in canonical form. We say a game G is born on day n if b(G) = n, and we say G is born by day n if b(G) ≤ n. Lets trace through some examples to illistrate the concept. There is only one game born by day 0, since there is exactly one canonical form with tree height of 0 ({|}, i.e.: 0). There are three games with birthday 1: 1 = {0|}, −1 = {|0}, and ∗ = {0|0}. Thus, there are four games born by by day 1 (the four games previously mentioned). In general, the best known upper bound for the number of games born by day n is g(n) ≤ g(n − 1) + 2g(n−1) + 2 [56]. We will call this function for the upper-bound b(n). Note that because birthdays only refer to games in the canonical form, we will use “height” when we are referring to the game tree of a specific game that may or may not be in canonical form. For example, for a game with value 0, the birthday of that game is 0, even though the height of the game could be very high, perhaps even exponentially so in the input size. Theorem 2.5.2. Identifying the outcome class of a disjunctive sum of n constant-sized games is solvable in polynomial time. Proof. First, note that since the sizes of the games are bounded by a constant, there is only a constant number of unique games. Let H be the game with the highest birthday. Since it has a constant birthday c, we can enumerate every game born by day c (in order of increasing birthday). Thus, we can represent the sum as n1G1 + n2G2 + · · · + nkGk where k is the number of games born by day c (as a reminder, at most k = g(c) ≤ g(c − 1) + 2g(c−1) + 2), and Pk i=1 ni = n. Then, that sum can form the basis for dynamic programming. Let the sum be represented by a kdimensional vector. There is a partial order for the vector, where each dimension is reliant only on lower sorted dimensions (namely, the ones with smaller birthdays). We can then have a recurrence for each possible move, with all recurrences having lower rank in the partial order. So, one can table the outcome class of all O(n k ) combinations of assignments of n, starting with n instances of a 0 birthday game, and expand from there. Since there are the number of states of the vector is no more than O(n k ), and each 23 entry can be filled in constant time (by examining the outcome class of each each possible move, which there are only a constant number of). 24 Chapter 3 The Impartial Glacial Geography In this chapter, we will explore a variety of results we have obtained about impartial games. Historically, when impartial games have been studied, results have been exclusively about algorithms or the computational complexity of individual combinatorial games. At first, this may makes sense, since the algorithm for winning on sums of impartial games is simple and well known, so seemingly there is little computational work to do on the mathematical underpinnings of these games. However, several other major questions remained unexplored: Question 1. Does there exist an impartial game whose outcome class is polynomially solvable for every position, but where identifying the nimber is intractable? Question 2. If the answer to Question 1 is “yes”, then what nimber classifications can be intractable and which must be polynomial-time solvable? Question 3. If the answer to Question 1 is “yes”, are there minimum tree dimensions (i.e.: height or fanout) to be intractable for games that are polynomially solvable? Question 4. Is there any computational link between nimber computation and misère outcome computation? Question 5. Is there a way to perform polynomial-time reductions that preserve the nim-value, even for intractable games, to a certain family of games? 25 To answer all of these questions, we study one of the most influential impartial combinatorial games, Geography. 3.1 Generalized Geography Generalized Geography is a graph game created in the 1970s to generalize a common road trip game. In the road trip game, players alternate turns naming cities. After a city is named, players must name a new city whose first letter is the same as the last letter of the previous city named. A city can never be named twice. When a player is unable to name a new city, the game ends with that player’s loss. For the graph generalization, the game is played on a directed graph with one of the vertices containing a token. Players alternate turns moving the token to adjacent vertices. The token cannot be moved to any vertex the token was previously on. A player loses when they cannot move the token anymore.∗ Generalized Geography is known to be PSPACE-complete, and was one of the first PSPACE-hardness reductions ever performed [45]. Since then, it has been a common tool used in reductions for PSPACE-hard problems due to its simple ruleset and graphical nature. One of the best known examples (of many) is how it was used to prove Go PSPACE-hard [40] Another variant of Generalized Geography is restricted to undirected graphs, known as Undirected Geography. This variant is, surprisingly, polynomial-time solvable, with a simple algorithm. First, remove any vertices that have already been traversed. Second, check if the current vertex is in all possible maximum matchings. If so, then following an incident edge in any of the matchings is a winning move. Otherwise, there are no winning moves. So, the running time for the algorithm is just that for finding a maximum cardinality matching, since the algorithm involves running it twice, once on the current graph, and again after removing the vertex the token is on, and checking if the cardinality is the same or not. ∗Of course, for the real world game, the difficulty comes with not knowing all the nodes, so the graph version is quite different. 26 Formally, we will say that Geography and Undirected Geography are composed of two inputs: a graph G and a starting node s, which we can denote by the pair (G, s). As a last point before moving to the main results, we will briefly discuss nomenclature. There are four major variants of Geography. We have discussed two already, but there are two others: Edge Geography and Undirected Edge Geography. These variants are as they sound: instead of not being able to travel through a vertex a second time, one is only restricted to being unable to travel through edges a second time; this distinction is analogous to the difference between hamiltonian paths and eulerian paths. We mention this because the literature is inconsistent in its naming. For this dissertation, for the vertex-based variants, we will use Geography, Directed Geography, or Generalized Geography to refer to the game on directed graphs, and Undirected Geography to refer to the game on undirected graphs. Similarly, we will use Edge Geography and Undirected Edge Geography to refer to the edge variants, respectively. 3.2 PSPACE-Complete Grundy Values of Undirected Geography We are able to demonstrate the PSPACE-hardness of calculating the nim-value of Undirected Geography, even on a planar bipartite graph of degree no more than 4. In section 3.6 we will extend this result by dropping the graph characteristics to extend the nimber classification hardness to almost any classification. Theorem 3.2.1 (Complexity Separation of Winnability and Grundy Values). Distinguishing between ∗ and not ∗ for the polynomial-time solvable Undirected Geography is PSPACE-hard to compute. Our reduction, r, takes a Generalized Geography position (G, s) and yields an Undirected Geography position r(G, s) = (G′ , s) where: (G′ , s)    = ∗, if (G, s) = 0 (∈ P) ∈ N \ {∗}, if (G, s) ∈ N The reduction itself contains two parts: 27 1. Modify G so that each vertex v ∈ V is given an adjacent leaf vertex, v0, adjacent to no other vertices. In other words, forall v ∈ V we will add a vertex v0 and (undirected) edge (v, v0). 2. For each of the directed edges in G, (x, y), we replace it with the gadget shown in Figure 3.1. x a a0 b c c0 f d d0 y Figure 3.1: The gadget for each directed edge (x, y). The correctness of the reduction hinges on this gadget acting like a directed edge from x to y. We assert that fact with two lemmas. We will use some notation to represent a graph, G, after moves have been made. For a subset S ⊂ V , we will use GS to be the graph obtained from G by removing S and all edges incident to S. Thus, from a position (G, s), if a player chooses to traverse edge (s, t), the resulting position is (G{s} , t). Since we are often considering sets S with just one vertex, we will shortcut examples like that by dropping the braces, as (Gs, t). Lemma 3.2.2 (Wrong Way). Moving from y to any vertex d results in a value of ∗2 or ∗3 (depending on the value of x). In other words, (G′ y , d) = ∗2 or ∗3. Proof. We prove this by examining the three options from d. Moving to d0 is clearly a move to 0. Moving to c will be non-zero, because c0 is zero. It remains to show that moving d → f results in a ∗-position. We can see this by considering the following necessary move f → b. Since both of b’s remaining neighbors, a and c, have terminal neighbors (a0 and c0), they are non-zero. Thus, the move to b must be a zero position, and the move to f must be ∗. 28 Lemma 3.2.3 (Correct Way). Moving from d to y results in a value of ∗ exactly when moving from x to a in the same gadget results in ∗. Proof. To prove both directions for the if and only if, we will use the fact that moving from b to f results in a 0-position, because d is always non-zero and it is f’s only neighbor. For the first direction, we will assume that moving from d to y results in ∗. This means that moving c to d has value ∗2, as the option to f is 0 (since b has already been visited) and so is d0. Thus, b to c has value ∗. Since b has options to both 0 and ∗, moving to b has value ∗2, and moving x to a has value ∗. In the other direction, assume that moving d to y does not have value ∗. (Either it is ∗2 or higher or y has already been removed.) Thus, moving c to d results in a value of ∗, because d’s other options are 0. b to c then has a value of ∗2, meaning that a to b has a value of ∗. This means that moving x to a has a value of ∗2 ̸= ∗, completing the proof. With our reduction and our two lemmas, we can now prove Theorem 3.2.1. Proof. Determining the winnability of Generalized Geography position (G, s) is PSPACE-hard [45]. Thus, it remains to be shown that for the Undirected Geography position resulting from the reduction, (G ′ , s)    = ∗, if (G, s) = 0 (∈ P). ∈ N \ {∗}, if (G, s) ∈ N . Consider any Generalized Geography position (H, t), k moves after (G, s) and the analogous Undirected Geography position (H′ , t) = r(H, t), reached 5k moves after (G′ , s) by traversing the gadgets corresponding to the directed edges traversed to reach (H, t) (if the player chooses otherwise, they will lose the game, so we can assume they played “the correct way”). If there are no options from (H, t), then (H′ , t) has options to t0, which has value 0; possibly to gadget vertices d, which have value either ∗2 or ∗3 by Lemma 3.2.2; and possibly to gadget vertices a where the corresponding y vertex has already been 29 removed (since we are assuming there are no moves in Generalized Geography position (H, t)), which have a non-∗ value (specifically ∗2) by Lemma 3.2.3. Thus, there is a move to zero and there might be moves to ∗2 or ∗3. (H′ , t) = ∗. Therefore, if there are options from (H, t), then the position is either in P or N . We can complete our proof inductively by assuming that the piece-wise formula above is true for all options of (H, t) and showing that this mean it holds for (H, t). It holds for the base case of all terminal positions of Generalized Geography as demonstrated in the previous paragraph. Let (Ht , p) be the notation for some option of (H, t). Then, we deal with two cases for the inductive case: • If (H, t) ∈ P, then each option, (Ht , p) is in N . Thus, by our induction hypothesis, (H′ t , p) = ∗z, where z ≥ 2. This means that (H′ , t) doesn’t have any options equal to ∗. Since it does have a move to zero (t0), (H′ , t) = ∗. • If (H, t) ∈ N , then some option, (Ht , p) ∈ P. Thus, (H′ t , p) = ∗. Since (H′ , t) has a move to zero (t0) and ∗, the value is ∗z where z ≥ 2. Currently our reduction creates positions in which the initial value (at (G′ , s)) is either ∗ or another nimber higher than 1. We can refine our reduction so that the question is whether we can distinguish between ∗ and ∗2, specifically, by appending a Prelude gadget (see Figure 3.2) before s and then asking what the value of the overall game position is when starting at the “start” vertex. start s Figure 3.2: Prelude Gadget. 30 Corollary 3.2.4. Determining whether an Undirected Geography position equals ∗ or ∗2 is PSPACEcomplete, even on bipartite, planar graphs with a maximum degree of four. Proof. Since the height of the game tree is at most n, the Grundy-value can be computed in polynomial space using the standard DFS technique [42]. Adding our prelude gadget to the reduction from Theorem 3.2.1, the value at vertex “start” will either be ∗ or ∗2. By calculating the nimber backtracking from the value at s, we see that it will be ∗ exactly when the value at s is ∗, and ∗2 for any of the other values of s. From Lichtenstein and Sipser [40], we know that the winnability of Generalized Geography is PSPACE-hard on bipartite, planar graphs with a maximum degree of three. Our reduction preserves the planarity, and, since there is no odd cycle in the gadget, the bipartite property as well. We increase the degree by one because we add an extra vertex adjacent to the original vertices in V . Thus, identifying the nimber of Undirected Geography is still hard on graphs with a maximum degree of four. 3.3 A Polynomial-Time Branch-and-Bound We first show that for any undirected graph G with maximum degree at most three, the Grundy value of Undirected Geography is polynomial-time computable. In this case, we present a polynomial-time reduction from the Grundy-value computation to the outcome class. More broadly, in Section 3.4.1, we generalize this finding of nimber-outcome reduction abstractly to impartial games characterized by their degrees and heights. In Section 3.4.2, we strengthen the result by analyzing the impact of high-degree nodes in the process. For Undirected Geography at a position G = (V, E) and s ∈ V , the degree of s in G is equal to the number of its feasible moves, and hence serves as a upper bound on the Grundy value of the position. Similarly, the maximum degree in G characterizes the maximum branching factor of the game tree at position (G, s): If the maximum degree of G is ∆, then the branching factor of every node except the root 31 is at most ∆ − 1, because the current Undirected Geography path entering the node will take away at least one edge incident to the node. The root may have branching factor ∆ but no more. Theorem 3.3.1 (Following the Winning Way in Undirected Geography). For any undirected graph G = (V, E) with maximum degree 3, and node s ∈ V , the Grundy value at the Undirected Geography position (G, s) can be computed in polynomial time in n = |V |. Proof. We first consider the case when the degree of s is 1 or 2. We then extend to the case of 3. Single Option: When the degree of s is 1, the Grundy value of Undirected Geography at (G, s) is ∗ if and only if (G, s) is a winning position. So, the Grundy-value can be directly reduced to the decision of outcome, which has a polynomial-time matching-based solution [28]. Double Options: When the degree of s is 2 (say with neighbors v1 and v2), the maximum branching factor of the game tree for position (G, s) is 2. Let Gs denote the graph obtained from G by removing s and edges incidents to s. Then, the position with move to v1 is (Gs, v1) and with move to v2 is (Gs, v2). Note that the degree of v1 and v2 in Gs is at most 2. We run the polynomial-time matching-based winnability algorithm to determine whether or not (Gs, v1) and (Gs, v2) are winning positions in Undirected Geography, and consider the four cases: 1. (N , N ) - both (Gs, v1) and (Gs, v2) are winning positions: nimber(G, s) = 0. 2. (P,P) - both (Gs, v1) and (Gs, v2) are losing positions: nimber(G, s) = ∗. 3. (N ,P) - (Gs, v1) is a winning position and (Gs, v2) is a losing position: nimber(G, s) = ∗(3 − nimber(Gs, v1)) 4. (P, N ) - (Gs, v1) is a losing position and (Gs, v2) is a winning position: nimber(G, s) = ∗(3 − nimber(Gs, v2)) 32 In the last two cases, one of v1 and v2 is 0, and the other has value x = ∗ or x = ∗2. By the mex rule, (G, s) will be the other of those values (∗2 or ∗, respectively) which is exactly 3−x (or 3⊕x), so the above derivation works. In the first two cases, we have found the Grundy value of (G, s) in polynomial time. Crucial to the tractability, in the last two cases, we reduce the Grundy-value computation of (G, s) to a single Grundyvalue computation of either (Gs, v1) or (Gs, v2). Because Gs has one less node than G, the depth of the branch-and-bound process is O(n). In total, we made O(n) calls to the decision-of-winnability algorithm in order to compute the Grundy value of position (G, s). Three Options: Finally, we consider the case when the degree of s is three, say with u1, u2, and u3 as neighbors. The maximum degree in Gs remains at most three. Then: nimber((G, s)) = mex ({nimber((Gs, u1), nimber((Gs, u2), nimber((Gs, u3)}) . Note that ui has degree at most two in Gs, for all i ∈ {1, 2, 3}. Thus, the nimbers of (Gs, ui) can be computed, by our method above for degree one or two, in polynomial time. 3.4 Finding Grundy Values By Branch-And-Bound In this section, we extend the branch-and-bound Grundy-value evaluation process to the abstract setting as well as analyze the impact of high-degree nodes in the process. But first, as an illustration of what has been been established in Section 3.3, consider the following family of fun planar Undirected Geography games: Corollary 3.4.1 (Alternation Paths Through Fully-Triangulated Planar Maps). For any fully triangulated planar graph G, and a face f in G, the Grundy value of the Undirected Geography game on the dual of G, starting at f, can be computed in polynomial time. 33 3.4.1 Nimber-Winnability Reduction: Degree of Phase Transition to Infeasibility Our dichotomy characterization of Grundy-value computation can be extended from the concrete Undirected Geography to an abstract nimber-winnability reduction in general impartial games, characterized by their degrees and heights. • Degree: For a positive integer ∆, we say that an impartial game G is a degree ∆ game if G and all positions reachable from G have at most ∆ feasible moves. • Height: For a positive integer h, we say an impartial game G is a height-h game if the height of G’s game tree is at most h. For example, consider an undirected graph G = (V, E) with n nodes and maximum degree ∆, and a node s ∈ V with degree less than ∆. Then, Undirected Geography at position (G, s) is a game with degree (∆ − 1) and height n. Theorem 3.4.2 (Dichotomy of Nimber-Outcome Reduction). For any degree-two, polynomially short impartial game, Grundy-value computation can be reduced in polynomial-time to the decision of the outcome class. In contrast, there exists a family of degree-three, polynomially short impartial games for which the outcome class can be solved in polynomial time, but Grundy-value computation is PSPACE-hard. Proof. In the branch-and-bound process at each node of the game tree encountered, we first run a winnability test for the position. If it is “Zero”, then return 0. Otherwise, run a winnability test for each of their children in the game tree; we know that one of them must be “Zero”. If both are “Zero”, then return ∗. Otherwise, we follow the winning way to determine whether the other child is ∗ or ∗2, and return ∗2 or ∗ accordingly. 34 3.4.2 The Impact of Large-Degree Nodes in Branch-and-Bound We can extend the analysis of Section 3.3 to show that a few nodes with large degrees will not stop the polynomial-time branch-and-bound: Theorem 3.4.3 (A Tractable Nimber Terrain in Undirected Geography). For any constants D and ∆, c, if G = (V, E) is an undirected graph with n = |V | nodes, in which at most c log2 n nodes with degree in the range [4, ∆ + 1], and at most D nodes with degree more than ∆, then the Grundy value of the Undirected Geography game over G can be computed in time O(n D+c log ∆+3). Proof. We will apply the “following the winning way” technique at all degree-two game-tree nodes in the standard DFS-based recursive evaluation methods. (1) Each time when we evaluate a node with degree more than ∆, the branching factor is at most n. (2) Each time when we evaluate a node with degree in the range [4, ∆ + 1], the branching factor is at most ∆ (or ∆ + 1 for such a starting node). (3) Otherwise, at a degree three node, the branching factor is one. If we use an O(n 3 ) time algorithm for maximum matching, then we can bound the total time by: O n D · ∆c log n · n 3 = O n D+c log ∆+3 . 3.5 Misère-Play Winnability v.s. Grundy Values We now demonstrate that the connection between misère rule and normal-play Grundy values is very subtle. Indeed, the strategy to play for termination positions and the strategy to play to avoid termination positions can be different. The following theorem provides an answer to the question about the computational relationship between misère play and nimber value identifiability posed at the beginning of the chapter. Theorem 3.5.1 (Complexity Independence). There are non-empty sub-classes PP, HP, PH, and HH of Undirected Geography positions with the following properties: (1) For class PP, both the misère outcome class and Grundy value are polynomial-time computable. (2) For class HP, Grundy values are polynomial-time computable, but the misère outcome class is PSPACE-complete to determine in the worst case. (3) For class PH, the misère outcome class is polynomial-time computable, but Grundy values are PSPACE-complete to compute in the worst case (4) For class HH, both the misère outcome class and Grundy values are PSPACE-complete in the worst case. Proof. Undirected Geography on trees is one of many examples of PP. By Theorem 3.2.1 and RenaultSchmidt’s result [43], Undirected Geography itself is an example of HH. We now show that Renault-Schmidt’s construction in [43] for the misère-play setting provides an example of HP. Thus, Renault-Schmidt’s construction cannot directly tell us about Grundy value-hardness. The main reason is that the Grundy value (in the normal setting) of the construction in [43] can be computed in polynomial time. More specifically, for a starting vertex u, the starting position has value ∗, ∗2, or ∗3, depending on the following properties (below we use the vertex naming conversion of the arc gadget of [43]): (1) If u is adjacent to some uv1 and some ru7, then it has value ∗3. (2) If u is only adjacent to some uv1 but no ru7, then it has value ∗2. (3) Otherwise, it has value ∗. Our proof for Theorem 3.2.1 provides an example for PH, because for all these games, the winnability in the misère setting is polynomial-time solvable. The main reason is that our gadget of Figure 3.1 has simple winnability for the current player at x or y in the misère setting without needing to get out of the gadget (i.e. it localizes the decision). Particularly, (1) Any move to a is a winning move for the first player, since a0 is a losing move for the second player (in the misère setting), and thus they move to b, the first player to f, the second player to d, and then the first player to c. Then, the second player can only move to c0 and lose the game. (2) Any move to d for the first player is a losing move, as the second player can move to f, then the first player must move to d, and the second player can move to c and thus win, since the first player must move to c0 and lose. (3) Any position with moves to neither a nor d must have no moves, and thus is a winning position for the first player. 36 3.6 Undirected Geography with Polynomial Grundy Values A fundamental problem in combinatorial game theory is that of nimber constructability. That is to say— when specialized to the game of our focus—the question of whether a game of Undirected Geography can actually have a certain Grundy value (equivalent to determining the habitat for impartial games), and if it can, whether it can be succinctly encoded. The existence is important primarily from a pure mathematical standpoint. The succinct encoding is needed for sums of games with high Grundy values to actually be shown intractable. In Section 3.6.3, we extend Theorem 3.2.1 to distinguishing between a larger range of nimbers using the (polynomially) succinct encoding of high nimbers to support our complexity analyses. 3.6.1 Logarithmic Intuition and Polynomial Challenge The habitat going up to the maximum degree in the graph is simple. We will present it in the next construction to motivate our more advanced proof. Observation 3.6.1 (Logorithmic Nimber in Trees). Through a simple tree structure, one may create an Undirected Geography position with nimber ∗n, highest degree n, and 2 n vertices. Proof. We will use a binomial tree structure. This is recursively defined a tree t(n) which has moves to t(n − 1), t(n − 2), . . . , t(0). For the base case, t(0) is a single isolated vertex, which clearly has Grundy value 0. The size of this tree is 2 0 for the base case, and we can inductively assume each of the smaller t(i) have 2 i vertices, so for tree with k + 1 levels we have its size as 2 0 + 21 + · · · + 2k + 1, where the final 1 is the new root. Thus, we can get a logarithmic nimber in the number of vertices (and certainly any constant nimber, which we will use for gadgets up to ∗3). The exponential size of the basic construction comes from the fact that we repeat each tree in each subtree. This is necessary, since if we attempt to combine the subtrees, being able to move "back up" those 37 trees could change the Grundy values. Note here that in Generalized Geography, one can use directed edges to prevent undesired “up” moves to share the lower nimber nodes. Thus, it is in fact straightforward to achieve nimber n with n + 1 vertices. However, we cannot just replace these with our directed-edge gadget from Figure 3.1, because the inner degree on the gadget is constant and will prevent us from getting arbitrarily large nimbers. We need a more sophisticated mechanism to get nimbers of any size. Nn ∗ ∗2 Nn−1 ∗ ∗2 · · · ∗ N5 ∗2 ∗ N4 ∗2 Figure 3.3: The overall schema for Nimber Constructability. The value of the position with the token at Nn is ∗n. Ni N4 · · · Nn ∗3 Ri Other ∗, ∗2 Pi ∗2 gadget Mi ∗ gadget Figure 3.4: Each vertex Ni is connected to all other N-vertices, as well as its own ∗3 gadget, its own ∗2 gadget, its own ∗ gadget, and its own Ri vertex (with value 0). Ri is also connected to the Pk and Mk gadgets with k > i as shown in the following figures. 38 Mi 0 ∗2 Ni M (a) M i,5 (a) i,4 M (b) i,4 M (c) i,4 M (d) i,4 R4 N4 M (b) i,5 M (c) i,5 M (d) i,5 R5 N5 · · · M (a) i,i−1 M (b) i,i−1 M (c) i,i−1 M (d) i,i−1 Ri−1 Ni−1 Figure 3.5: A ∗ gadget, which has value ∗ unless a lower-rank-N vertex is removed. If all of the Nk vertices remain in the graph, then moving to any one of them from Rk results in a 0-board (by Lemma 3.6.4). This causes the move Mi → M (a) i,k to be equal to ∗2. Otherwise, one Rk has value 0, so Mi → M (a) i,k is a move to ∗, so Mi instead has value ∗3. 3.6.2 Polynomially High Nimber Constructability To attain nimber n, we create vertices N4, . . . , Nn, which form a clique, as in Figure 3.3. Each Ni has Grundy value ∗i so long as all Nk with k < i remain. (We will say these vertices have a lower rank.) We argue that starting with the token on vertex Nn is a ∗n-position. (We do not have N0, N1, N2, or N3, since we use 0 through ∗3 as mechanisms in our structure to ensure the player is unable to move “up” in rank.) After any move from Nn to Nk, we no longer want vertices with higher-rank than k to retain their nimber value. To attain this, we create ∗ and ∗2 gadgets for each Ni , which have their named values if and only if no vertex of a lower rank has been removed from the graph. So, a later move to a higher-rank Ni will have value either ∗ or ∗2 instead of ∗i. 39 Pi 0 ∗ Ni P (a) P i,5 (a) i,4 P (b) i,4 P (c) i,4 P (d) i,4 P (e) i,4 P (f) i,4 R4 N4 P (b) i,5 P (c) i,5 P (d) i,5 P (e) i,5 P (f) i,5 R5 N5 · · · P (a) i,i−1 P (b) i,i−1 P (c) i,i−1 P (d) i,i−1 P (e) i,i−1 P (f) i,i−1 Ri−1 Ni−1 Figure 3.6: The ∗2 gadget We present these designs in Figures 3.4, 3.5, and 3.6 for illustration. In addition to the figures, we also include a formal algorithmic formulation in Algorithm 1: Observation 3.6.2. The Nimber Generation Algorithm runs in polynomial time. Proof. By simple observation, each loop has no more than n iterations, and there are never more than two nested loops. Each line in each loop runs in constant time. The running time, in fact, is O(n 2 ). Lemma 3.6.3 (Grounded). If there is a game where the only vertices removed are some Ni vertices along with some of their associated Mi and Mip vertices or Pi and Pip vertices, then traversing edge (Nk, Rk) always results in a move to 0. 40 Algorithm 1: Nimber Generation Algorithm if n ≤ 3 then Return a tree representation of ∗n with the corresponding start position end else Let the graph we are working on be G = (V, E) Add vertex N4 and R4 to V for i = 5 to n do Add vertices Ni , Mi , Pi , Ri , M(0) i , and P (0) i to V Add edges (Ni , Ri),(Mi , M(0) i ), and (Pi , P(0) i ) to E Create tree versions of ∗3 to V and E, and connect Ni to it for j = 4 to i − 1 do Add vertices P (a) ij , P(b) ij , P(c) ij , P(d) ij , P(e) ij , P(f) ij , M(a) ij , M(b) ij , M(c) ij , and M (d) ij to V Add edges (Pi , P(a) ij ),(P (a) ij , P(b) ij ),(P (a) ij , P(c) ij ),(P (c) ij , P(d) ij ), and (P (c) ij , P(e) ij ) to E Add edges (P (e) ij , P(f) ij )(P (f) ij , Rj ), and (Ni , Nj ) to E (Mi , M(a) ij ),(M (a) ij , M(b) ij ),(M (a) ij , M(c) ij ),(M (c) ij , M(d) ij ) and (M (d) ij , Rj ) to E end Add a tree representation of ∗2 to V and E, and add an edge to Mi to E Add a tree representations of ∗ to V and E, and an edge connect this to Pi end Add the tree versions of ∗ and ∗2 the edge to N4 to V and E Return G and starting node Nn end Proof. Any move to an M (a) i,k vertex has nimber at least ∗ since it has a move to M (b) i,k which is 0. Thus, moving to M (c) i,k from M (d) i,k results in a 0, so all M (d) i,k moves from Rk result in ∗. The same is true of moving from Rk to P (f) i,k because P (c) i,k is also non-zero. Since Rk only has ∗-options, its value is zero when moving from Nk. Lemma 3.6.4. If there is a game where the only vertices removed are some Ni vertices along with some of their associated Mi and Mip vertices or Pi and Pip vertices, then traversing the edge (Rp, Np) is a move to 0. Proof. There is a move to ∗3 (and, if p = 4, ∗2 and ∗), moves to Mp and Pp, which both have a move to 0 by construction, and to various other Ni , which have moves to Ri , which are moves to 0 by Lemma 3.6.3. Lemma 3.6.5 (Skip ∗2). So long as only Ni vertices are removed from the graph, the position from moving from Nk to Mk has nim-value ∗ if and only if no Ni have been removed with i < k. Otherwise, it is ∗3. 41 Proof. Consider the result of moving M (d) i,k → Rk. All moves to other M (d) j,k vertices are losing moves as established in Lemma 3.6.3. There is only a winning move if Nk still exists, so the position at Rk ̸= 0 iff Nk still exists. Let’s consider these two cases: (1) If Nk exists, then moving to Rk is not a 0 move. Thus, moving to M (d) i,k yields 0, so moving to M (c) i,k yields ∗, and moving to M (a) i,k yields ∗2. If all Nk exist, then Mi does not have a move to ∗, so its value is ∗ from Ni . (2) On the other hand, if Nk does not exist, then moving to Rk from M (d) i,k yields 0. Thus, moving to M (d) i,k yields ∗, so moving to M (c) i,k yields 0, and moving to M (a) i,k yields ∗. Since Mi has a ∗-option, moving there from Ni now yields a position with value ∗3. Lemma 3.6.6 (Skip ∗). So long as only Ni vertices are removed from the graph, a token on Pk has nim-value ∗2 if and only if no Ni have been removed with i < k. If the value is not ∗2, it is ∗3. Proof. By the same logic as in Lemma 3.6.5, we see that Rk has value 0 exactly when Nk no longer exists. We examine the two cases to complete the proof: (1) If Nk exists, Rk is non-zero, so moving to P (f) i,k from above yields 0. Working back up, moving to P (e) i,k yields ∗, moving to P (c) i,k yields ∗2, and moving to P (a) i,k yields ∗. If these all exist, then Pi has moves to only 0 and ∗, so it is ∗2. (2) If some Nk doesn’t exist, Rk is zero, so moving to P (f) i,k from above yields ∗. Again, working back up, moving to P (e) i,k yields 0, moving to P (c) i,k yields ∗, and moving to P (a) i,k yields ∗2. Now that Pi has an ∗2-option, it is instead ∗3 from Ni . Lemma 3.6.7. If the token is on Nk, and only Ni vertices with i > k have been removed from the graph, then the nim-value must be at least ∗4. Proof. It is trivial for k = 4, since by construction it has moves to 0 through ∗3. 42 For larger k, they have a move to 0 through Rj (by Lemma 3.6.3). They also have a move to ∗3 by construction. The move to Mk is ∗, and the move to Pk is ∗2 by Lemmas 3.6.5 and 3.6.6, respectively. Lemma 3.6.8 (Parity). If the only vertices removed are Ni vertices, the token is currently on Nk, Nj is of lower rank than Nk and is the lowest rank that has been removed, and if there are an odd number of Np vertices remaining, where Np are higher rank than Nj , then the nim-value of the game is ∗. If there are an even number of those vertices remaining, then the game has value ∗2. Proof. We will do this by induction on the number of remaining Np. Base Cases: If the token is on an Np vertex, when it is the only one remaining, the value is ∗, and when there are two Np vertices remaining, the value is ∗2. To establish this: Nk has a move to ∗3 by construction, along with moves to Mk and Pk (∗3 by Lemmas 3.6.5 and 3.6.6), a move to Rk, which is a move to 0 by Lemma 3.6.3, and to any Ni with rank no more than j, which is at least ∗4 by Lemma 3.6.7. Thus, there are no moves to ∗ and a move to 0, so the value is ∗. In the case where there is one other Np vertex remaining, we have the exact same analysis, but there is a move to ∗ as well, by the previous base case. Thus, the position has value ∗2. Inductive Hypothesis: If there are x Np remaining, where x is odd, then the value is ∗, and if x is even, the value is ∗2. Inductive Step: If there are x + 1 Np remaining, where x + 1 is odd, then the value is ∗, and if x + 1 is even, the value is ∗2. To establish this: by the same argument as in the base cases, there is a move to 0, three moves to ∗3, and a collection of moves with value at least ∗4. The rest of the moves are to various Np. By the IH, each of these options have value ∗ if x is odd, and value ∗2 if x is even. Thus, if x is odd, then the value with x + 1 is ∗2, and if x is even, then the value with x + 1 is ∗. Theorem 3.6.9 (Right Amount of Stars). When the token is on Nn, the resulting game has nim-value ∗n. 43 Proof. For 0 through ∗3, it clearly works as we just build a tree as described by Observation 3.6.1. For larger Grundy values, we have the token on vertex Nn. We will prove this has value ∗n through induction on the values of a starting token on Ni . Base Case: As long as the only vertices removed from the graph are Ni vertices, the token on N4 will have value ∗4. To establish this: there are moves to ∗, ∗2, and ∗3, each by construction. There is a move to 0 through R4 by Lemma 3.6.3. The only other available moves are some subset of of the Nj , which by Lemma 3.6.8, have value ∗ or ∗2. Inductive Hypothesis: As long as the only vertices removed from the graph are various Ni vertices where i > k, Nk has value ∗k. Inductive Step: We need to show that as long as the only vertices removed from the graph are various Ni vertices where i > k + 1, Nk+1 has value ∗k + 1. To establish this: Nk+1 has moves to ∗3 and ∗4, by construction, and to Mk+1, Pk+1, Rk+1, all of N4 through Nk, and some of Nk+2 to Nn. Moves to N4 to Nk are ∗4 to ∗k by induction. The move to Rk+1 is a move to 0 by Lemma 3.6.3. The move to Mk+1 is ∗, by Lemma 3.6.5 since all Nj remain. The move to to Pk+1 is ∗2, by Lemma 3.6.6, again since no Nj is removed. 3.6.3 Complexity Implication We now use our polynomially high nimber constructability result to extend Theorem 3.2.1, establishing other than the polynomial-time “Zero” v.s. “Fuzzy” classifier, every classifier of the Grundy values in Undirected Geography is PSPACE-hard. Theorem 3.6.10 (Too “Fuzzy” to Classify). In Undirected Geography, determining whether or not the Grundy value of (G, s), where s has degree ∆, is in any given set S ⊂ [∆], is PSPACE-hard. 44 Proof. Recall that the proof of Corollary 3.2.4 shows that distinguishing ∗ from ∗2 is PSPACE-hard. We will first use this fact to prove that distinguishing between ∗(k−1) and ∗k is PSPACE-hard, for any k ≥ 2. We prove this by taking a position (G2, v2) that is hard to distinguish between ∗ and ∗2. We introduce a new vertex v3 with moves to its own 0 and ∗ and add edge (v3, v2) to create G3. Then we will create a new vertex v4 with moves to its own 0, ∗, ∗2, and connect (v4, v3) to create G4, and so on, until we create a vertex vk with moves to its own 0 to ∗(k − 1), and add edge (vk, vk−1) to create Gk. These vertices vi and their associated gadgets have size polynomial in i due to Theorem 3.6.9. Now, if (G2, v2) = ∗, then (G3, v3) doesn’t have a move to ∗2, so (G3, v3) = ∗2. Similarly, (G4, v4) = ∗3, (G5, v5) = ∗4, . . . ,(Gk, vk) = ∗(k − 1). If instead, (G2, v2) = ∗2, then (G3, v3) = ∗3, because there is a move to ∗2. Likewise, (G4, v4) = ∗4, . . . ,(Gk, vk) = ∗k. Thus, it is PSPACE-hard to distinguish between ∗k and ∗(k − 1). Next, we prove that distinguishing between any ∗k and ∗p is PSPACE-hard. (We will assume p > k, without loss of generality.) We first create a (Gk, vk) where distinguishing ∗(k − 1) from ∗k is hard, then add a new vertex vp which has moves to its own 0 to ∗(k − 1), vk, and ∗(k + 1) to ∗(p − 1). We name this graph G′ p ; the position (G′ p , vp) has value ∗p exactly when (Gk, vk) has value ∗k. (G′ p , vp) has value ∗k otherwise. Thus, it is PSPACE-hard to distinguish between ∗p and ∗k. Finally, we use this to show that distinguishing between any possible fixed set of Grundy values is hard. For any possible set S, there must be at least one Grundy value x ∈ S and one Grundy value y ∈ S¯ := [∆] \ S. Then, we can, as described above, create a position where it is PSPACE-hard to distinguish between ∗x and ∗y. Thus, if one could classify the game to be within that set of Grundy values, one could solve a PSPACE-hard problem. 3.7 Polynomial-Time Nimber-Preserving Reduction In this section, we consider the following natural concept of reduction among impartial games. 45 Definition 3.7.1 (Nimber-Preserving Reduction). † A map ϕ is a nimber-preserving reduction from impartial ruleset R1 to impartial ruleset R2 if for every position G of R1, ϕ(G) is a position of R2 satisfying nimber(G) = nimber(ϕ(G)). Because an impartial position is a losing position if and only if its nimber is zero, nimber-preserving reductions enhance winnability-preserving reductions in traditional complexity-theoretical characterizations of combinatorial games [24, 45]: Polynomial-time nimber-preserving reductions introduce the following natural notion of “universal” impartial rulesets. Definition 3.7.2 (Sprague-Grundy Completeness). For a family J of impartial rulesets, we say R ∈ J is a Sprague-Grundy-complete ruleset for J if for every ruleset R′ in J , one can construct a mapping f of positions of R′ to positions of R such that for every position G of R′ , where f(G) is evaluated in time polynomial in |G| and nimber(G) = nimber(f(G)). As the main technical contribution of this section, we prove the following theorem regarding the expressiveness of Generalized Geography. The natural family of rulesets containing Generalized Geography is I P , the family of all impartial rulesets whose positions have game trees with height polynomial in the sizes of the positions. We call games of I P polynomially short games. In addition to Generalized Geography, I P contains many combinatorial rulesets studied in the literature, including Node Kayles, Chomp, Proper-K-Coloring, Atropos, and Avoid True, as well as Nim and Graph Nim with polynomial numbers of stones. Theorem 3.7.3 (A Complete Geography). Generalized Geography is a Sprague-Grundy complete ruleset for I P . †This natural concept of reduction in combinatorial game theory can be viewed as the analog of functional-preserving reductions in various fields. To name a few: approximation-preserving, gap-preserving, structure-preserving reductions in complexity and algorithmic theory, hardness-preserving and security-preservation in cryptography, dimension-preserving, metric-preserving, and topology-preserving reductions in data analytics, parameter-preserving reductions in dynamic systems, counterexample-preserving reductions in model checking, query-number-preserving, sample-preserving and high-ordermoment-preserving in statistical analysis, and modularity-preserving reductions in network modeling. We are inspired by several of these works. 46 In other words, for example, given any Node Kayles or Avoid True game, we can, in polynomial time, generate a Generalized Geography game with the same Grundy value, despite the fact that the Grundy value of the input game could be intractable to compute. Because nimber-preserving reductions generalize winnability-preserving reductions, every SpragueGrundy complete ruleset for I P must be PSPACE-complete to solve. However, for a simple mathematical reason, we have the following observation: Proposition 3.7.4. Unless P = PSPACE, not every PSPACE-complete ruleset in I P is Sprague-Grundy complete for I P . In particular, Atropos [15] is PSPACE-complete but not Sprague-Grundy-complete for I P . Thus, together Theorem 3.7.3 and Proposition 3.7.4 highlight the fundamental difference between winnabilitypreserving reductions and nimber-preserving reductions. Our result further illuminates the central role of Generalized Geography—a classical PSPACE-complete ruleset instrumental to Lichtenstein-Sipser’s PSPACE-hard characterization of Go [27]—in the complexity-theoretical understanding of combinatorial games. We will note this brief corollary (without proof) Corollary 3.7.5. Edge Geography, formulated in [45] and studied in [28], is also Sprague-Grundy-complete for I P . 3.7.1 Star Atlas: A Complete Generalized Geography Mathematically, one can view ∗ as a map from Z + ∪ {0} to (infinite) subfamilies of impartial games in I, such that for each k ∈ Z + ∪ {0}, nimber(G) = k, for all G = ∗k. In other words, ∗ is nature’s game encoding of non-negative integers. 47 Sprague-Grundy Theory establishes that each impartial game’s strategic relevance in disjunctive sums is determined by its nimber (i.e., its star value). In this section, we prove Theorem 3.7.3, showing how to use Generalized Geography to efficiently “map out” games’ nimbers across I P . For readability, we restate Theorem 3.7.3 to make the needed technical component explicit: Theorem 3.7.6 (Sprague-Grundy-Completeness of Generalized Geography). There exists a polynomialtime algorithm ϕ such that for any game G ∈ I P , ϕ(G) is a Generalized Geography position satisfying nimber(ϕ(G)) = nimber(G). The high level idea is as follows: we will reduce the question (which is in PSPACE) of “Is nim(G) = k?” to a QSAT instance, for all k <height(G). Then, we will each of those QSAT instances into a Geography instance. Finally, we will use teach of those Geography instances to create a single Geography game where the nimber of the game is equal to the nimber of the game we reduced from. Now, we will dive into the details. Our proof starts with the following basic property of nimbers, which follows from the recursive definition Proposition 3.7.7. For any position G of an impartial game, nimber(G) is bounded above by both the height of its game tree, h, and the number of options at G, l. In other words, G = ∗k, where k ≤ min(h, l). To simplify notation, we let g = min(h, l). To begin the reduction, we will need a reduction for each decision problem Qi = “Does G = ∗i?” (for all i ∈ [g]). Since, for any polynomially short impartial ruleset R and a position G in R, nimber(G) can be computed in space polynomial in size(G), the decision problem Qi is in PSPACE. Thus, we can reduce each Qi to an instance in the PSPACE-complete QSAT (Quantified SAT), then to a Generalized Geography instance using the classic reduction, f, from [45, 40]. Referring to the starting node of f(Qi) as si , we add two additional vertices, ai and bi , with directed edges (bi , ai) and (ai , si). Now, 48 • si has exactly two options, so the value of f(Qi) is either 0, ∗, or ∗2. By the reduction, it is 0 exactly when G ̸= ∗i, and in {∗, ∗2} when G = ∗i. • ai has exactly one option (si ), so the value of the Generalized Geography position starting there (instead of at si ) is 0 when G = ∗i and ∗ otherwise. • bi has exactly one option (ai ), so the value of the Generalized Geography position starting there is ∗ when G = ∗i and 0 otherwise. Each of these constructions from Qi is shown in Figure 3.7. bi ai si x1T x1F · · · f(Qi) Figure 3.7: Result of the classic QSAT and Generalized Geography reductions of the question, Qi , “Does G = ∗i?”, with the added vertices ai and bi . We will combine these g + 1 Generalized Geography instances into a single instance, but first we need some utility vertices each equal to one of the nimber values 0, . . . , ∗(g − 2). We can build these using a single gadget as shown in Figure 3.8. We previously used a more complicated gadget to construct nimbers in Undirected Geography, but we include the figure to demonstrate just how much easier it is to construct values in Generalized Geography. This gadget consists of vertices t0, t1, . . . , tg−2 with edges (ti , tj ) for each i > j. Thus, each vertex ti has options to tj where j < i and no other options, exactly fulfilling the requirements for ti to have value ∗i. Now we build a new gadget to put it all together and combine the f(Qi) gadgets, as shown in Figure 3.9: 49 t0 t1 t2 t3 · · · tg−2 Figure 3.8: vertices t0 through tg−2. Each vertex ti has edges to t0, t1, . . . , ti−1. Thus, the nimber value of the Geography position at vertex ti is ∗i. 0 b c d 1 2 t1 3 · · · t1 . . . ti−2 b1 c2 . . . ci−1 i · · · t1 . . . tg−2 b1 c2 . . . cg−1 g start Figure 3.9: The gadget combining all of the f(Qi) gadgets into a single Generalized Geography instance. The vertices bi , ci , and di are in rows and columns indexed by the letters on the left and the numbers along the top. Each bi has an edge to ai as in Figure 3.7. • For all i ≥ 1 : add a vertex ci , as well as edges (ci , bi) and for all j ∈ [1, i − 2] : (ci , tj ). • For all i ≥ 2 : add a vertex di as well as edges (di , b1) and for all j ∈ [2, i − 1] : (di , cj ). • Finally, add a vertex start with edges (start, b0), (start, c1), and for all j ∈ [2, g] : (start, dj ). Lemma 3.7.8. The Generalized Geography position starting at each vertex ci has value ∗(i−1) if G ̸= ∗i, and value 0 otherwise. 50 Proof. The Generalized Geography position starting at ci has options to tj , for all t ∈ [1, i − 2]. That means that ci has options with values ∗, . . . , ∗(i − 2). If the move to bi has value 0, then there are moves to 0, ∗, . . . , ∗(i − 2), so the value at ci is ∗(i − 1). Otherwise, there is no option from ci to a zero-valued position, so the value at ci is 0. Lemma 3.7.9. If G = 0, then the Generalized Geography position starting at each vertex di has value ∗i. Proof. di has moves to b1, c2, . . . , ci . Since G = 0, by Lemma 3.7.8 none of the vertices cj have values 0 (and b1 does have value 0), so the options have values 0, ∗, ∗2, . . . , ∗(i−1), respectively. The mex of these is ∗i, so di has value ∗i. Lemma 3.7.10. Let G = ∗k, where k > 0. Then the Generalized Geography position starting at each vertex di has value ∗i if i < k, and value ∗(k − 1) if i ≥ k. Proof. We need to prove this by cases. We’ll start with k = 1, then show it for k ≥ 2. When k = 1, di has moves to b1, c2, . . . , ci . (There is no d0 or d1, so i > k.) The value at b1 is ∗, and by Lemma 3.7.8, the remainder have values ∗, ∗2, . . . , ∗(i − 1), respectively. 0 is missing from this list, so di = 0 = ∗(1 − 1) = ∗(k − 1). For k ≥ 2, we will split up our analysis into the two cases: i < k and i ≥ k. We will next consider the case where k ≥ 2 and i < k. From di there are moves to b1, c2, . . . , ci . Since k > i, these have values 0, ∗, ∗2, . . . , ∗(i − 1), respectively, by Lemma 3.7.8. The mex of these is ∗i, so di has value ∗i. Finally, when k ≥ 2 and i ≥ k, di has options to b1, c2, . . . , ck−1, ck, ck+1, . . . , ci . By Lemma 3.7.8, these have values 0, ∗, ∗2, . . . , ∗(k − 2), 0, ∗k, . . . , ∗(i − 1), respectively. ∗(k − 1) doesn’t exist in that list, so that’s the mex, meaning the value of di is ∗(k − 1). Theorem 3.7.11. Let G = ∗k. Then the Generalized Geography position beginning at start equals ∗k. 51 Proof. The options from start are b0, c1, and for all i ∈ [2, g] : di . If G = 0, then b0 is ∗, c1 is ∗, and, by Lemma 3.7.9, each di is ∗i. Since 0 is missing from these options, the value at start is 0 = ∗k. If G = ∗ (and thus k = 1), then the move to b0 is 0, the move to c1 is also 0, and, by Lemma 3.7.10, the moves to di are each also 0, because each i > k = 1 and ∗(k − 1) = ∗(1 − 1) = 0. All the options are to 0, so the value at start is ∗ = ∗k. Finally, if G = ∗k, where k ≥ 2, then the moves are to b0, c1, d2, . . . , dk−1, dk, dk+1, . . . , dg. These have values, respectively, 0, ∗, ∗2, . . . , ∗(k − 1), ∗(k − 1), ∗(k − 1), . . . , ∗(k − 1), by Lemma 3.7.10. The mex of these is k, so the value of start is ∗k. Interestingly, performing a nimber-preserving reduction to Undirected Geography is actually PSPACEhard. We will discuss this more in Chapter 7. 52 Chapter 4 The Slushy Frontier In this chapter, we move beyond the questions about impartial games from the previous chapter, and instead focus our attention on tepid games. As a reminder, tepid games are games containing only infinitesimal positions. They may be partizan, but they don’t have stored moves, since, in canonical form, they will end with a nimber value or 0 rather than integers or fractions. In this chapter specifically, we will focus on a game known as Superstars to explore the following question: Can a sum of Superstars, written in canonical form, be intractable to compute? 4.1 Superstars Superstars are minor generalization of nimbers, and are also a generalization of a well-studied family of games known as Flowers. Definition 4.1.1. In a Superstar, all of Left’s options are nimbers, and so are all of Right’s options. So, they are of the form {∗L1, ∗L2, . . . , ∗Lc | ∗R1, ∗R2, . . . , ∗Rk}. Note that the number of options for Left and the number of options for Right do not need to be equal (but both sides do need to have options, since otherwise the game would not be infinitesimal). Also, a superstar in which the options for both players are the same is just a nimber. Even more generally we have the following: 53 Proposition 4.1.2 ([19]). The superstar {0, ∗, . . . , ∗(n − 1), ∗x1, . . . , ∗xk | 0, ∗, . . . , ∗(n − 1), ∗y1, . . . , ∗yl}, where xi , yj > n for all i and j, is equal to the nimber ∗n. When a sum consists of only superstars of this form, the sum is reduced to a sum of nimbers, and is thus solvable in polynomial time. As we show in the main result of this chapter in section 4.2, solving a sum of superstars more generally is NP-hard. As with the previous chapter, a note must be taken about the overloading of terms within the CGT community. Superstars can also refer to a related concept by Conway[21] (a modified form to account for a concept known as atomic weight). After discussion with several members within the CGT community, we hope that from now on the Conway concept can be referred to as Comets instead. Throughout this dissertation, Superstars refers exclusively to the concept explained above, as introduced by [5]. One other minor note is that while a sum of Superstars is sometimes called Supernim, we will not use that term in this chapter (or elsewhere). 4.2 From Bits to Superstars: Hardness Reduction In order to show that sums of superstars are NP-hard, we need to introduce some additional computational problems. XOR-SAT [46] is a classical logical satisfiability problem consisting of a conjunction of clauses, each of the XOR of boolean literals. That is to say, it takes this form: (xi ⊕ xj ⊕ · · · ⊕ xk) ∧ (xl ⊕ · · · ⊕ xp) ∧ · · · ∧(xq ⊕ · · · ⊕ xr). Negations are allowed, but unimpactful, since a negation is just equivalent to adding an XOR of 1. It is known that XOR-SAT is polynomial-time solvable, since it exactly describes a set of simultaneous linear equations over F2 [46]. 54 Our next problem, which is NP-hard, is motivated by XOR-SAT∗ . It uses multi-state variables, which can be assigned to one of many states instead of just True and False. Each literal of a variable is labelled with one of those states (e.g. xa,si ) and is only true if the variable is assigned to that state. More formally, let xa be a multi-state variable with possible states s1, s2, . . . , si , . . . , sk, then for all states si of xa we have xa,si =    True if xa is set to si , False if xa is set to sj and j ̸= i. Figure 4.1 displays a multi-state variable. xa s1 s2 s3 s4 Figure 4.1: Multi-state variable xa with four possible states: s1, s2, s3, s4. The overall color indicates that the chosen state is s2. Definition 4.2.1. Multistate XOR-SAT is a ruleset where a position is a conjunction of clauses consisting of the XOR of multi-state literals instead of boolean literals. In other words, the clauses are of the form (xi,si ⊕ · · · ⊕ xj,sj ). To represent current variable assignment, variables can represented as literals. Variables are divided between the two players, X, and Y , and clauses may contain variables from both players, e.g. (xi,si ⊕ yj,sj ⊕ · · ·). On their turn, the current player selects one of their unassigned variables and picks a state to assign it to. Once both players have assigned all variables, X wins if the formula is true, and Y wins if the formula is false. If both players have the same number of variables, then we call the game Equal-Partitioned Multistate-XORSAT, or EPMX. ∗At the end of this chapter, we will briefly discuss why we only achieve NP-hardness 55 We will show that EPMX is NP-hard after we show the reduction from EPMX to a sum of super stars. To reduce from EPMX, we must first discuss elementary strategies in a sum of superstars. To aid in this, we partition superstars into five classes: • nimbers, • no-0: neither player has 0 as one of their options, • left-0: only Left has a move to 0, • right-0: only Right has a move to 0, and • both-0: both players have moves to 0. First, an observation that follows directly from Theorem 4.1.2: Corollary 4.2.2 (No-0 games). A No-0 game has value 0. Then, we will prove the following lemma: Lemma 4.2.3 (0 game win). Consider a sum of superstars with no both-0 games. If at the start of the Left player’s turn there are more left-0 games than right-0 games, then Left wins. Similarly, if at the start of the Right player’s turn there are more right-0 games than left-0 games, then Right wins. It is possible to prove Lemma 4.2.3 using atomic weights and the two-ahead rule, as shown in [49]. We provide the following proof that avoids use of atomic weights. Proof. We will call the winning player A and the losing player B, so there are more A-0 games than B0. We will prescribe the following algorithm for A to win: they can “eliminate” 0s in the B-0 games by making a move on the game (by choosing one of their nimber options arbitrarily). They repeatedly do this until no B-0 games remain. Now, after B’s following turn, the remaining games include at least one A-0 game, some nimbers (maybe none), and some no-0 games (maybe none). At this point, A should avoid 56 playing A-0 games until there are only A-0 games remaining, or there is exactly one A-0 game left (along with the other types of games), whatever comes first. If there are only A-0 games remaining, then for the first of those games, A can just bring a game to 0, and then if there are any games left, B has to make one into a nimber, which A can just bring to 0. This will repeat until the last A-0 game is taken this way, in which case A wins. If there is exactly one A-0 game (along with potential no-0 games and nimbers), then A can identify the value of the sum of everything but the single game by XORing the nimbers (by observation Corollary 4.2.2, the others have value 0). If the nim-sum is 0, then A may take the move to 0 and thus wins the game. Otherwise, they can bring the nim-sum to 0 and inductively keep it so until B plays on A-0, bringing the game to a non-zero nim sum, which A can then win from. 4.2.1 Reduction to Superstars With the previous lemma, we can prove the following theorem: Theorem 4.2.4. There exists a polynomial time reduction from the game Equal-Partitioned MultistateXORSAT (EPMX) to a sum of superstars, such that if True wins going first on EPMX, then the outcome class of a sum of superstars is L or P (i.e. Left wins going second). Proof. Let X be the player whose goal is to make the formula true in EPMX, Y be the player whose goal is to make it false, and m be the number of clauses. We assume that the EPMX formula contains literals for each state of each variable. (If it doesn’t, we can create a dummy clause that will always be true for each missing variable missing a state. That clause contains one copy of each state that variable can have.) We will use the following construction: First, we will assign the t th clause an identity zt = 2t . In other words, we use powers of twos {1, 2, 4, . . . , 2 m−1} to identify clauses. For the i th variable assigned to X, we will create a right-0 game which we will call xi , and for the i th variable assigned to Y , we will create a left-0 game we will call yi . Right’s options for each xi contain only a single option of 0, and similarly, 57 Left’s options for yi contain only 0. Then, for each possible state of the i th variable for X, there will be a Left option in xi to a nimber whose value is, for each clause that contains the variable at that state, the sum of their corresponding identity values (i.e., if the t th clause is involved, then zt is included in the sum). Similarly, for each state of variable yi for Y , there will be a Right option in yi to a nimber with value equal to the sum of the corresponding identity values of the involved clauses. In addition, for each game xj , there will be an additional option of ∗2 m+j . (The yi positions do not have this extra option.) The game we consider is then G = x0 + · · · + xk + y0 + . . . + yℓ + ∗(2m − 1). For example, the position (x0,a ⊕x1,a ⊕y0,a ⊕y1,c)∧(x0,b ⊕x1,a ⊕y0,b ⊕y1,a)∧(x1,a ⊕x1,b)∧(y1,a ⊕ y1,b ⊕ y1,c) with states x0 : {a, b}, x1 : {a, b}, y0 : {a, b}, y1 : {a, b, c} reduces to x0 + x1 + y0 + y1 + ∗15, where • x0 =    ∗ |{z} a , ∗2 |{z} b , ∗16 |{z} 2m+0 0    • x1 =    ∗7 |{z} a , ∗4 |{z} b , ∗32 |{z} 2m+1 0    • y0 = ( 0 ∗ |{z} a , ∗2 |{z} b ) • y1 = ( 0 ∗10 |{z} a , ∗8 |{z} b , ∗9 |{z} c ) In the example above, the identities of these four clauses are respectively, 1, 2, 4, and 8. Now we demonstrate the correctness of the reduction. As mentioned in the theorem statement, the union of L and P is equivalent to proving that Left wins going second. We will show that the game should progress by alternating moves of Right playing on left-0s and Left playing on right-0s. Since the options of those components are all nimbers, each of these plays changes the whole game by removing 58 that component and modifying the nimber term. If this pattern is followed, then after Left plays on the final right-0, they win if and only if the nimber term has been reduced to zero. Otherwise, Right can bring the nin-sum to 0, and then win through following the nim strategy. If both continue to hold to that pattern, at the beginning of each of Left’s turns, there is one more right-0 component than left-0. Thus, Left must play on a right-0 component or they will lose by Lemma 4.2.3. Right starts each turn with balanced left-0s and right-0s, but they still need to follow the pattern. If Right deviates by playing on an x game, then they will lose by Lemma 4.2.3. If Right plays on the nimber term instead, this switches the roles of the players with respect to their ending conditions; now Right will win if and only if the nimber term is reduced to zero when they make their final move. Left, however, can avoid this by playing on one of the large nimber (with value at least 2 m) options included in any x. Right doesn’t have any options that can cancel out that large nimber, so the final nimber term will always be non-zero and Right can no longer win. Following the prescribed sequence of play, Left wins if the nimber term equals zero. Since it started at ∗2 m − 1, the sum of all nimber options chosen must also equal ∗2 m − 1. If Left has a winning strategy in EPMX, they can play on the nim values corresponding to each variable state in their winning strategy, which must then result in a final nimsum of 0. If they do not have a winning strategy, then note that playing on the 2 m+j values can’t give them a chance to win, since Right can stick with their EPMX strategy and the nimsum can’t equal 0. Thus, Right can follow Y’s winning strategy of assignments to result in a non-zero value. 4.2.2 EPMX Hardness Now we will show NP-hardness for EPMX. Theorem 4.2.5. There exists a polynomial-time reduction from 3SAT to EPMX. 59 Proof. Let X be the player whose goal is to make the formula true in EPMX, Y be the player whose goal is to make it false. In our reduction, Y will not make meaningful decisions in the course of the game (due to problems where Y easily dominates the game). Doing so could result in a PSPACE-hardness reduction, but we were unable to identify one, despite a serious amount of effort put towards this goal. Let n be the number of variables and m be the number of clauses from the 3SAT instance. Note that 3SAT is hard even if every variable appears only at most 3 times, at least once negated and at least once unnegated [46]. It is also hard adding on a further restriction that there are an odd number of variables in the formula. We will assume both of these are true in this reduction. For each clause in 3SAT, we will create a clause in EPMX. We will also have two separate clauses cx and cy. For each variable (xi ) in 3SAT, we will create a variable (xi ) in EPMX with five different states: {xi,a, xi,b, xi,c, xi,d, xi,e}. WLOG, we assume that xi appears once unnegated and twice negated, in clauses r, s, and t in 3SAT respectively. Every state appears in cx, with each state being XORed together. In the EPMX formula, the first state (xi,a) appears in no other clauses. The second state, xi,b, corresponds to the unnegated appearance, and appears in clause r. The third state, xi,c, corresponds to both negated 3SAT literals, and appears in clauses s and t. The fourth state, xi,d, corresponds to only the first negated 3SAT literal, and appears only in clause s. Finally, the fifth state, xi,e, corresponds to only the second negated literal, and only appears in clause t. In order to be an EPMX position, we need to include the same number of y variables as x. We can do this by creating n dummy variables with two states each: a and b. We can include all of the states of each of the variables into cy. Note that since there is an odd number of y variables, cy will always be true (the same is true of cx). If there is a solution to the 3SAT formula, then a solution to EPMX can be constructed by iterating over the variables. For each xi , if the 3SAT assignment is true, then X chooses xi,b, unless the EPMX clause r 60 has already been satisfied, in which case X chooses xi,a. If xi is assigned to false, then X selects the correct choice of a, c, d, and e, depending on which of clauses r and t have already been satisfied. The inverse direction is simpler: an assignment of xi,a means it doesn’t matter what X picks. xi,b means xi must be true to satisfy the 3SAT formula, and any of the others means it must be assigned to false. Note the weak point here is that we only demonstrate EPMX to be NP-hard rather than PSPACE-hard. This is not due to a lack of effort, but because the problem is difficult to work with, since Y is “too strong” of a player. If Y goes second, they can play arbitrarily, and use their final variable to make a clause false. If Y has a variable in which no X variable is in all of the same clauses (or more), then Y can win by saving that variable as their last one, and then flipping a clause to False that X’s final variable is not in. So, one must work within this constrained space for the reduction, which proves to be difficult. There are more complicated situations as well that give these sort of degenerate wins to Y as well. So, it is difficult to even get “interesting” games where Y and X actually interact with each other. 61 Chapter 5 The Ocean of Game Values This chapter is about a novel oracle model for combinatorial games, and the computational implications of this model. Broadly speaking, the idea is that if we have an algorithm to determine the game’s outcome class, an algorithm to sum two games in the same ruleset into a third game of that ruleset, and an algorithm to create instances of certain game values, then we can generate different game values, add them to the game, and then check the outcome class to “probe” what the game value is. Then, through employing search techniques, we can identify the game value. First, in this chapter, we only consider games that are closed under summation. That is to say, if for some ruleset R, there are positions G1 ∈ R and G2 ∈ R, then we also have G1+G2 ∈ R. Second, we also only consider games where all numbers and all multiples of ↑ and ↓ are part of the habitat of the game, if the game is partizan. We can relax this second condition, but one would need a full characterization of the habitat of the game, and also modify the algorithm to account for these factors, but there isn’t an intrinsic barrier. Third, we only consider games where all values in the game have a polynomially succinct encoding. Definition 5.0.1 (Polynomially succinct encoding). A ruleset R has a polynomially succinct encoding if and only if, for every G ∈ R, there exists a position G′ such that value(G′ ) = value(G) and size(G′ ) = O(poly(value((G)). 62 Note that like the second assumption, we can relax this assumption if we know that no polynomially succinct encoding exists, and have a tight characterization of what the most succinct encoding is. 5.1 A Three Oracle Model The three oracles used for the two algorithms we devise are as follows: • Outcome Oracle: Given a position G, this oracle outputs o(G). • Value Oracle: Given a ruleset R and a game value v (possibly polynomial), outputs a position G ∈ R where size(G′ ) = O(poly(v) • Summation Oracle: For any two positions G1 ∈ R and G2 ∈ R, outputs a new position G3 = G1 + G2 ∈ R, where size(G3) = poly(size(G1)+ size(G2)). Using these three oracles, we derive two algorithms. Theorem 5.1.1 (Nimber identification). If all 3 oracles can be implemented in polynomial time (in the number of bits to represent a position) for some impartial ruleset R, then for any position G ∈ R, if G = ∗n, the nimber can be identified in time polynomial in n. Note that the above algorithm is only pseudo-polynomial time, since if the nimber had ℓ = log n bits, then the value would be Θ(∗2 ℓ ) and the algorithm would take exponential time. Fortunately, many impartial games do not have an exponential nimber for a polynomial size game board, since it requires both exponential fanout and exponential height. Theorem 5.1.2 (Confusion Interval Identification through Binary Search). If all three oracles can be implemented in polynomial time (in the number of bits to represent a position) for some ruleset R, for any position G ∈ R, either a nearest number can be found (i.e.: whatever number the game is infinitesimally close to), or 63 the confusion interval can be identified in time polynomial in the number of bits of the position and output size. We will describe each algorithm in its own section. For both algorithms, the idea is similar: use the value oracle to generate a “test” value, use the summation oracle to add it to the game, and use the outcome oracle for classification, then iterate and continue to search until the algorithm finds the proper value. 5.2 Algorithm 1: Impartial Games Conceptually, the idea is simple: do a linear search of the nimbers using the oracles as described before. For any nimber k1, the only nimber k2 that will cause k1 ⊕ k2 = 0 is when k2 = k1. So then, we first use the outcome oracle to check if the game is in P. If so, then the game has value 0. Otherwise, we use the value oracle to generate a position with value ∗, then use the summation oracle to add it to the game, and finally check the outcome class. If the outcome class is P, then ∗ is the Grundy value of G. Otherwise, we continue and repeat with ∗2, ∗3, and so on, until we identify the nim-value. The high level idea behind this is that, through doing linear search, every value is exhaustively checked for equality, and will continue until one is found. However, as described in Theorem 5.1.1, this is only pseudo-polynomial, since n is unary. 5.3 Algorithm 2: Nearest Number or Confusion Interval for Partizan Games The algorithm for finding the nearest number or confusion interval employs the principle of binary search that uses the oracles alongside some CGT properties to search for the nearest number, or for the endpoints of the confusion interval. As opposed to the previous algorithm, which only runs in polynomial time with 64 a game value written in a logarithmic number of bits, this algorithm, because of binary search, will work when the game value can be written in an exponential number of bits. We are able to now employ binary search since, intuitively, we now have 4 outcome classes to work with, where L and R, which can tell us “higher” or “lower” in addition to P for “hit,” while with impartial games we only have P and N which are just “hit” and “miss,” respectively. Extracting that intuition, we have Observation 5.3.1. If we take a position G1 with value v1, and sum it with a position G2 with value v2 to get position G3, then: • if G3 ∈ L, then v1 > −v2 • if G3 ∈ R, then v1 < −v2 • if G3 ∈ P, then v1 = −v2 The algorithm runs in two phases (and each phase has 3 steps). The first phase is the “number finding” phase. This is where the algorithm is looking for the nearest number. The algorithm remains in the first phase until either it runs into a value that is confused with G, or finds the number nearest to G and returns it. If it runs into a confused value, the algorithm will proceed to the second phase. We split the discussion into the two phases. 5.3.1 Phase 1 Step 1: Use the Outcome Oracle on G to identify the outcome class. If it is P, then the value is 0, and can be returned. If it is N , then the position is confused with 0, and thus we will proceed to the second phase (as described in Subsection 5.3.2). If it is L, then the game value is greater than 0, and we enter the second step with given range (0, ∞). If it is R, then the value is less than zero, and we enter the second step with range (−∞, 0). 65 Step 2: As input for this step, we are given a range (B, U), which is a range of numbers the game value must lie between. If |U − B| = 1, then we move to Step 3. Otherwise, we first need to identify a test value t. If B = −∞, then t = 2U − 1. If U = ∞, then t = 2B + 1. Otherwise, we let t = B + ⌊(U + B)/2⌋. Then, we make t ′ = −t. Next, we create G′ =ValueOracle(R, t′ ). Then, we let G∗ =AdditionOracle(G, G′ ). Finally, we use the Outcome Oracle to identify the outcome class of G∗ . If the outcome class is P, then G has value t, and the algorithm finishes. If the outcome class is N , then G is confused with t, and we move on to phase 2. If the outcome class is L, then we repeat Step 2, but with interval (t, U). If the outcome class is R, then we repeat Step 2 with interval (B, t). Step 3: Now, we know the nearest number is between integers B and U (possibly B or U themelves due to infinitesimal factors). If this is the first time Step 3 has been entered, create a variable s = 1. Otherwise, double s. First, we let G′ =ValueOracle(−B) and G′′ = ValueOracle(s· ↑). Then, as always, apply the addition oracle for G′′′ = AdditionOracle(G′ , G′′). Finally, we let G∗ = AdditionOracle(G, G′′′) and, like before, use the outcome oracle to determine the outcome class. If the outcome class is P or R, then we stop the algorithm and return B as the nearest number. Otherwise, we continue. We then repeat this process using the upper bound instead. That is to say, we let G′ =ValueOracle(−U) and G′′ =ValueOracle(s· ↓). Then we have G′′′ = AdditionOracle(G′ , G′′). Finally, we again use the addition oracle to let G∗ = AdditionOracle(G, G′′′) and like before, use the outcome oracle to determine the outcome class. If the outcome class is P or L, then we stop the algorithm and return U. Otherwise, we continue. Finally, we let t = B + (B − U)/2 and t ′ = −t. Just as with Step 2, we create G′ =ValueOracle(R, t’), then G∗ =AdditionOracle(G, G′ ) and then use Outcome Oracle on G∗ . If the outcome class is P, then G has value t, and the algorithm returns t. If the outcome class is N , then G is confused with t, and we move 66 on to phase 2. If the outcome class is L, then we repeat Step 3, but with interval (t, U). If the outcome class is R, then we repeat Step 3 with interval (B, t). 5.3.2 Phase 2 Step 1: From phase 1, we are given a t that G is confused with, and an associated range B, U. Step 2 of this phase will be repeated twice. Once for range (B, t), which gives the lower endpoint (i.e.: R(G)), and again with range (t, U), which will give us the upper endpoint (i.e.: L(G)). When both endpoints have been found, they can be returned, along with whether they are inclusive endpoints, or exclusive endpoints. Step 2 Exactly the same as Steps 2 through 3 from phase 1, except: • Outcome class P will never appear, since the game can’t equal a number while being confused for another. • Outcome class N will be used as a “hit”, while outcome classes L and R will be treated as misses. • If a number isn’t confused with G, but it added with some s· ↑ or s· ↓ is confused with G, then an open endpoint has been found, and should be returned as such. 5.3.3 Proof Now we prove Theorem 5.1.2. First, to show the correctness of the algorithm, we use the following lemma: Lemma 5.3.2. At the end of each step of the algorithm for either phase, either the correct number or confusion interval endpoint has been returned, or the correct number/interval endpoint lies in the range (B, U). Proof. We trace through the logic of each step. For the first step, note that if the outcome class is P, then 0 is the value by definition, so it is correctly returned. If it is N , then it is confused with 0, so proceeding to 67 the confusion phase is appropriate. Otherwise, it is greater than 0 or less than 0 based on outcome classes L and R, respectively, which is represented in the intervals at the end of the step. Step 2 again only returns on an exact “hit,” corresponding to outcome class P. It goes to Phase 2 with an N result. Otherwise, the algorithm determines it is below or above some target value, and the bounds are updated accordingly. Step 3 has 4 cases: • The outcome class is P when subtracting a fraction (in which case it is correctly returned). • The algorithm identifies that the value is an infinitesimal amount above or below one of the bounds (in which case, it is correctly returned). • The game value is confused with some number (in which case the algorithm proceeds to phase 2). • Otherwise, the range updates as before. Step 1 for phase 2 establishes 2 ranges for the two end points. Since the two ranges are (−∞, t) and (t, ∞), which, since t is confused with G, must be the two possible ranges of endpoints. Step 2 has two cases. Either the outcome class is N , or the outcome class is any of the other 3. If the outcome class is N , then the the confusion interval can’t be between B and t, so having t as the lower bound for the endpoint must be correct. For the other 3 outcome classes, it must be that t isn’t part of the confusion interval. Thus, replacing U with t as the upper bound must be correct. For Step 3, there are 4 cases: • One of the endpoints summed with s· ↑ or s· ↓ results in the outcome class moving from N to a different outcome class, in which case that is a closed endpoint. • One of the endpoints summed with s· ↑ or s· ↓ results in the outcome class moving from some outcome class to N , in which case that endpoint is an open endpoint. 68 • G when summed with t is in the range (B, U), where, WLOG, G − B ∈ N and G − U ∈ R is in N . Then, the range (t, U) is the correct range for the endpoint. • G when summed with t is in the range (B, U), where, WLOG, G − B ∈ N and G − U ∈ R is in R. Then, the range (B, t) is the correct range for the endpoint. The algorithm never returns a false solution by Lemma 5.3.2, so the correctness immediately follows. Then, what remains is to show that the algorithm terminates in polynomial time. We will do this by bounding the length of time of each step. Each step for all phases relies on only on a few cases, basic arithmetic, and a constant number of oracle calls, which must be in polynomial time. So, all that needs to be shown is that the number of times each step is called is at most a polynomial number of times. Step 1 for both phases happens at most once. Step 2 is only repeated until an exact integer match is found, or until the two closest integers are found. Because the output must be a polynomial number of bits (we will call the number x bits), the number has magnitude less than 2 x . So, after x steps, both a proper upper and lower bound will be reached, and through binary search, cut into half each time, results in Θ(x) steps to find the nearest two integers in the worst case (that is, an exact integer match isn’t found). For Step 3, by the same notion, we know that for a closest number k/2 p , p must be no more than x. So, after at most x steps, the closest pair of fractions will be arrived at. But, one may wonder if multiples of ↑ and ↓ will be able to distinguish the correct nearest fraction in at most a polynomial number of calls. Fortunately, there is a useful theorem (originally from Winning Ways[5]): Theorem 5.3.3. For any infinitesimal G, there is some n such that n· ↑> G > n· ↓. More specifically, n is at most 2 more than the height of the game tree for the game value. Given that we assume the game value to be written in x bits, this means that s = 2 + h, where h is the height of the game tree, should suffice (since h is at most 2 x , this is still polynomial time in output bits). 69 5.4 An Application One main application of Algorithms 1 and 2 is the contrapositive: if the game value is PSPACE-complete, then at least one oracle is PSPACE-complete. So, if one can show that the summation oracle and value oracles can be implemented in polynomial time, and show that some classification of game value is intractable, then one also obtains that even identifying the outcome class is intractable. As an example, for the game Domineering, the summation oracle is already known. Furthermore, a value oracle for numbers and sums of ↑ and ↓ in their most succinct form is known. However, an algorithm to determine outcome class is unknown. Algorithm 2 would allow even demonstrating identifying the confusion interval to be intractable as sufficient to show identifying the outcome class to be intractable. We will discuss more usages in Chapter 7. 70 Chapter 6 The “Quantum” World This chapter takes a different focus than the previous chapters. Rather than exploring computation involving combinatorial game values, we instead explore a relatively recent “quantum” modification introduced in 2017 by Paul Dorbec [22]. 6.1 Preliminaries Despite what the name might suggest, quantum combinatorial games don’t actually involve quantum computation, at least in the traditional sense. Rather, it is the name given to a nonderministic variation of the games inspired by Quantum Tic-Tac-Toe, which was created for the purpose of physics education [29]. This was later generalized to be used for any combinatorial game by Paul Dorbec, as we desribe below. 6.1.1 Intuition Essentially, the idea is that instead of making a single move, the player makes a set of moves (in other words, a “quantum move”). This results in several possible positions of the game board (called “realizations”). When the player makes a move when there are several realizations, only the combinations of moves and realizations that are possible continue, while any other realizations “collapse.” 71 To more easily illustrate this concept, let’s examine the Nim position (2, 2). Classically, this position is in P. If the first player takes both from one pile, the other player will take both from the other pile. If the first player takes 1 from a pile, the second player will take 1 from the other pile, leaving the first player to take the last one in a pile, and letting the second player taking the last one from the other pile. So the first player loses. (2, 2) N ⟨ (1, 2) | (2, 1) ⟩ P ⟨ (0, 2) | (1, 0) | (1, 1) ⟩ N ⟨ (0, 2) | (1, 1) ⟩ N (0, 1) N ⟨ (0, 2) | (1, 1) | (0, 1) ⟩ N ⟨ (0, 2) | (1, 1) | (2, 0) ⟩ N (0, 0) P ⟨ (−1, 0) | (0, −1) ⟩ ⟨ (−1, 0) | (0, −2) ⟩ (−1, 0) (−2, 0) ⟨ (−1, 0) | (−2, 0) ⟩ ⟨ (−1, 0) | (0, −1) ⟩ (0, −2) (0, −2) (0, −1) (0, −2) (−2, 0) Figure 6.1: Winning strategy for current player in Quantum Nim (2, 2), showing that quantum moves impact the game’s outcome. (There are four additional move options from ⟨(1, 2) | (2, 1)⟩ that are not shown because they are symmetric to moves given.) However, this is not true in the quantum case. As seen in Figure 6.1, if the first player makes the quantum move ⟨(−1, 0) | (0, −1)⟩, then this results in the position ⟨(1, 2) | (2, 1)⟩, which is a P position. One can think of this position as “either the current position is (1, 2) or the position is (2, 1).” In the figure, one can see five of the different possible responses (ignoring the symmetrical moves, as well as a few others for space). All of those five moves will result in an immediate winning move by the first player. 72 • If the second player takes 2 from a pile, then the realization where the first player took 1 from the same pile collapse, leaving us with the position (0, 1) which the first player can then win on. • If the second player takes 1 from a pile, there are still two possible realizations (0, 2) and (1, 1). But then, the first player can take 2 from the right pile, which will collapse the (1, 1) realization, winning the game. • If the second player tries taking the superposition of 1 from the left pile, 1 from the right pile, than there are 3 possible realizations: (0, 2), (1, 1), and (2, 0). But then, the first player can choose to take 2 from either pile to win the game. • If the second player tries the superposition of 1 from a pile, or 2 from the same pile, then we still get the same 3 realizations from before, so the first player can take 2 from a pile and win. • Finally, if a player tries taking 1 from a pile, or 2 from another, then there are 3 realizations: (0, 2), (1, 0), and (1, 1). Either way, the first player can, again, take 2 from a pile to collapse the other realizations and win. Not included in the above analysis or image is taking a quantum move involving 3 or 4 possible moves, but the same strategy of taking 2 from one of the piles is sufficient to win regardless. The rest of this section will be spent formalizing this intuitive structure to set up the later proofs. 6.1.2 Quantum Notation Formally, if we have some ruleset R, then ruleset RQ is the quantum variant of the game. We denote a w-wide superposition of classical moves by M := ⟨m1|m2| . . . |mw⟩. An s-wide superposition of positions is similarly denoted G := ⟨g1|g2| . . . |gs⟩. So, the classical ruleset can be defined as corresponding to w = 1 and s = 1. We can denote a ruleset that only admits width up to w by RQ(w) . 73 In RQ, M is said to be feasible for G if for all i ∈ [1, w], mi is feasible for a non-empty subset of realizations in G, according to R. M takes G to a new quantum position with “nondeterministic” game states resulting from feasible transitions in the “Cartesian product” of {m1, ..., mw} and {g1, ..., gs}, according to R: Realizations in G, according to [29, 22], are “entangled” in that each move mi applies to the entire nondeterministic composition. If mi is not feasible for a realization, then the realization collapses, and thus, can no longer be factored in for making future moves. As a quantum game progresses, the number of realizations in the quantum state of the board can go up and down. When all possible moves of a player would cause a realization to collapse, we call it a terminal realization for that player. We will analyze games in this chapter by analyzing three different types of starting positions. The first kind is a classical start, where all starting positions are assumed to be that of the original, non-quantum game. The second kind is a reachable quantum position, where a starting with a superposition is allowed so long as it can be reached with a classical start. In other words, the input will be a classical board and a polynomial number of moves (in the size of the game board representation). Notably, this could result in an exponential number of realizations. Finally, there is the poly-wide start, where any starting superposition is allowed. In other words, the input is a description of each super position (and thus, the number of realizations is a function of the input size), 6.1.3 Highlights of Results In Section 6.2, we examine the quantum variant of undirected geography in three contexts. First, in Subsection 6.2.1 we examine what happens when play is started from a classical start (ie: a position with width s = 1). Then, in Section 6.2.2, we demonstrate two PSPACE-completeness results. The first is deciding the outcome of a polynomially describable position (ie: an arbitrary position where s is polynomial). The second is restricted to a reachable position (ie: a position that can be reached after an arbitrary amount of play starting on a position with s = 1). This requires exponential s to achieve. 74 In Section 6.3, we outline a reduction chain proving that arbitrarily entangled Nim positions (with polynomial s) is Σ p 2 -hard. In Section 6.4, we trace through many different examples of classically PSPACEcomplete games. Finally, in Section 6.5, we give a game-independent proof that most games in PSPACE remain in PSPACE. 6.2 Exploring Quantum Undirected Geography 6.2.1 Strategic Graph Contraction: Polynomial-Time Solution for Classical Starts Theorem 6.2.1 (Tractability of Classical Starts). For any Undirected Geography position R, one can solve RQ(2) with the same starting position in polynomial time. We will prove the theorem by establishing that for any undirected graph G = (V, E) and starting vertex s ∈ V , position (G, s) is a winning position (of the current player) in the quantum setting with 2- wide quantum moves if and only if (G, s) is a winning position in the classical Undirected Geography. The classical case can be solved by the Fraenkel-Scheinerman-Ullman algorithm guided by an elegant matching theory: (G, s) is a winning position if and only if s is in every maximum matching of G [28]. Our graph-contraction-based algorithm extends this matching theory to the quantum setting. In order to prove this, we will first present the algorithm that the winner in the classical game (which we will call the hero for easier redability) will use to win under quantum play. The hero will always make classical moves, but we need to show how they respond to quantum moves by their opponent, the villain. If the villain makes a quantum move, the hero will try to make a winning collapsing move (described further below). If they cannot, they can still win by keeping track of the quantum superposition, treating the two quantumly-chosen vertices as one combined (contracted) vertex. The hero keeps track of an overlaid graph G′ = (V ′ , E′ ): (A) Initially, V ′ = {{v} | v ∈ V }. We refer to c(v) (the contraction with v) as the element of V ′ that contains v. (B) E′ will be updated so that 75 (X, Y ) ∈ E′ ⇔ X, Y ∈ V ′ and ∀x ∈ X, y ∈ Y : (x, y) ∈ E. (C) Whenever a player makes a classical move from a → b (meaning the current player makes a classical move after the previous player makes a classical move) the hero will remove c(a) from V ′ and all incident edges from E′ . (D) Whenever the villain makes a quantum move a → ⟨v1 | v2⟩, the hero will again remove c(a) from V ′ and all incident edges from E′ . (E) Whenever the hero follows a quantum move with a classical move, ⟨v1 | v2⟩ → b, the hero will update G′ based on whether they make a collapsing move: (E.1) If the hero collapses that quantum move, then they can remove the remaining vi as though it had been a classical move by the villain. (E.2) If the hero does not collapse, but c(v1) = c(v2), then all of those vertices have all been visited in all realizations. The hero can remove c(v1) from V ′ and all incident edges from E′ . (E.3) If the hero does not collapse and c(v1) ̸= c(v2), then the hero removes both c(v1) and c(v2) from V ′ and replaces them with c(v1) ∪ c(v2). Then the hero will reset E′ to match the definition given above. Next we describe how the hero chooses their move, using maximum matchings on G′ . (In our notation, we consider a matching, M, as both a set of pairs and a function. So, (a, b) ∈ M ⇔ M(a) = b ⇔ M(b) = a.). In the classical winning position, the current vertex is in all maximum matchings on the graph [28]. This means that after the classical winner’s turn, the loser must start from a vertex not contained in some maximum matching. Our hero will maintain a similar invariant on G′ : there is a maximum matching on G′ such that the villain will be starting their turn on a vertex not in that matching. Our algorithm is as follows: • If the villain makes a classical move to v, the hero considers any maximum matching, M on G′ , and then moves to x ∈ M(c(v)) such that (v, x) ∈ E. This leaves us with a maximum matching, M \ {(c(v), c(x))} on the remaining graph that does not include c(x), thus upholding the invariant. The hero removes c(v) and c(x) from V ′ . • If the villain makes a quantum move to ⟨ a | b ⟩, and c(a) = c(b), then the hero acts as though the villain moved classically to only a, finding a maximum matching, M on G′ , then moving to 76 x ∈ M(c(a)) such that (a, x) ∈ E. Subtracting (c(a), c(x)) from M results in a maximum matching without c(x), upholding the invariant. The hero removes c(a) = c(b) and c(x) from the partition V ′ they are keeping track of. • If the villain moves to ⟨ a | b ⟩, and c(a) ̸= c(b), then the hero has additional work to do. Notably, they find a matching, M, such that there is an x ∈ M(c(a)) where (a, x) ∈ E, but (b, x) ∈/ E; or there is an x ∈ M(c(b)) where (b, x) ∈ E, but (a, x) ∈/ E, if one exists. There are three cases: 1. If there is an M with x ∈ M(c(a)) where (a, x) ∈ E, but (b, x) ∈/ E, then the hero moves to x. Since (b, x) ∈/ E, the realization where the villain moved to b collapses out. Subtracting (c(a), c(x)) from M yields a maximum matching without c(x), upholding the invariant. The hero removes c(a) and c(x) from V ′ . 2. If there is an M with x ∈ M(c(b)) where (b, x) ∈ E, but (a, x) ∈/ E, then the hero moves to x as in the previous case, with a and b swapping roles, removing c(b) and c(x) from V ′ . 3. If no maximum matching exists with those requirements, then the hero can just use any maximum matching, M, to make their move. The following must be true of M: ∀x ∈ M(c(a)) : (b, x) ∈ E and ∀y ∈ M(c(b)) : (a, y) ∈ E. The hero can now move to any x ∈ M(c(a)) and update V ′ by removing c(x) and contracting c(a) and c(b) into one element c(a)∪c(b). In order to continue safely, the hero needs this new contracted vertex to be adjacent to c(y), for all y ∈ M(c(b)). Thus, the hero needs both that (c(a), c(y)) ∈ E′ and (c(b), c(y)) ∈ E′ . The latter is already true because y ∈ M(c(b)). We show the former using some extra lemmas. By Lemma 6.2.2, (a, x) and (b, y) are in a maximum matching on G. Thus, there is another maximum matching with the swapped edges (a, y) and (b, x). Then, by Lemma 6.2.3, (c(a), c(y)) ∈ E′ . 77 Lemma 6.2.2. If (c(a), c(b)) is in a maximum matching of G′ , then in any realization where a and b haven’t been used, (a, b) is part of a maximum matching on the unvisited graph. Proof. Let M be the maximum matching on G′ containing (c(a), c(b)). Also let H be the remaining subgraph of G that hasn’t been visited in the given realization. Then, for each element X ∈ V ′ , there is exactly one vertex in H remaining. Due to the definition of E′ , that vertex must be adjacent to the vertex of H inside the contraction M(X). Thus, each matched pair in M corresponds to exactly one unique pair of neighbors in H, which creates a maximum matching on that graph. Thus, a and b must be neighbors and (a, b) is in the maximum matching on H. Lemma 6.2.3. If (a, b) is in a maximum matching on G and c(a) ̸= c(b), then at any point in the game where a and b are still included in contractions, (c(a), c(b)) ∈ E′ . Proof. We prove by proving the contrapositive. Assume (c(a), c(b)) ∈/ E′ . Thus, there is (a ′ , b′ ) ∈/ E, where a ′ ∈ c(a) and b ′ ∈ c(b). Without losing generality, we assume that the most recent contraction breaks the statement of the lemma; all prior contraction graphs G′ contained all edges from all maximum matchings in G. Assume that the villain’s last quantum move was to ⟨x | y⟩ and the hero had to respond with a non-collapsing move to vertex z. Thus: 1. In the prior contraction, c ′ , c ′ (x) ̸= c ′ (y). 2. c ′ (x) ∪ c ′ (y) = c(a). 3. There exists M, a matching on G′ that the hero used to choose z. 4. WLOG, a ′ ∈ c ′ (x) and b ′ ∈ M(c ′ (y)) 5. Thus, (a ′ , b′ ) ∈ E, contradicting our assumption. 78 Proof. (of Theorem 6.2.1) The invariant the hero maintains is: at the end of the hero’s turn, after having moved to x, there is a maximum matching on G′ that does not contain the contracted vertex c(x). Thus, either (A) There are no more edges leaving x and the villain loses immediately, or (B) If there is a move and the villain moves from x to y, then y must be contained in one of the other matched pairs, meaning that the hero will be able to move to y’s match. (Lemma 6.2.2 requires that y is part of one of those matches, because if it wasn’t, then (x, y) would be part of a maximum matching on G and that edge will be represented as an edge in a maximum matching in G′ , which won’t work with the invariant.) When there are no moves left in G′ , there are no moves left in G, since if the edge (x, y) has c(x) = c(y) then only one vertex is remaining in each realization, and if c(x) ̸= c(y), then (c(x), c(y)) must be in G′ by Lemma 6.2.2. The invariant will be maintained because each turn the hero will start on a vertex in all maximum matchings of G′ and will traverse the edge from one of them. Since the hero will always have a move to make on their turn, they will never lose the game. For higher widths, the invariant of exactly one node per contraction being unvisited doesn’t hold, so this proof structure no longer holds, but we conjecture that the same strategy is still a winning one, as we have found no counterexamples. 6.2.2 Intractability of Quantum Geography Positions Theorem 6.2.4 (Intractability of Quantum Starts). Quantum Undirected Geography with a polynomialwide quantum start is PSPACE-complete (with any quantum width). a ab1 ab2 b a ab1 ab2 stop b Figure 6.2: Gadgets for reducing from a Geography edge (a, b). The left is the gadget as it will appear in the main board and all other boards except from a (a, b)-board. The right is the edge as it appears in a (a, b)-board. All other parts of the two boards will be the same. 79 Proof. Because the classical game tree has height at most n, by Theorem 6.5.2, the quantumization with poly-wide quantum start is in PSPACE. For hardness, we reduce from Quantum Generalized Geography, the original version which includes directed edges (and which we prove to be PSPACE-hard in Theorem 6.2.6. Consider a Generalized Geography instance where the underlying graph has n nodes and m edges. We will create an (undirected) superposition that consists of m + 1 entangled realizations. • One realization, main board, has a very similar structure toGeography, but each arc (a, b)is replaced by a three-edge path with two new vertices, ab1 and ab2: (a, ab1), (ab1, ab2), and (ab2, b). • For each edge (a, b), we include (a, b)-board, which is exactly the same as main board except there is an extra vertex, stop, and the edge (ab1, ab2) is replaced with (stop, ab2). Intuitively, the main board will be the one where play happens on, while the other m boards will be used to enforce that the players follow the correct edge direction. We show the two relevant edge gadgets in Figure 6.2. The main board can never collapse out unless a player moves to stop, since all other edges are also in the other realizations. If a player does move to stop, the game immediately ends, as all other realizations with stop collapse. Thus, the game either ends when there are no moves left in the main game board, or stop is entered. Additionally, a player can move to stop if and only if their opponent reached the ab2 vertex by moving from b, corresponding to going backwards on arc (a, b) in the Quantum Directed Geography game. This cannot happen when moving from ab1 since traversing the (ab1, ab2) edge collapses out the realization with an edge to stop. Thus, traversing an edge backwards is always a losing move. Thus, winning strategies forQuantum Undirected Geography correspond exactly to those strategies for Quantum Directed Geography. 80 Theorem 6.2.5 (Intractability of Reachable Quantum Positions). Quantum Undirected Geography is PSPACE-complete for positions reachable with polynomial number of moves after a classical start. ab−1 a ab1 stopab ab−2 ab2 b Figure 6.3: Gadget for a Directed Geography edge (a, b) in Undirected Geography. Prior to the current position, there were moves ab−1 → ⟨ab1 | stopab⟩ → ab−2. Proof. To prove this, we use the reduction from Theorem 6.2.4, which contains what we refer to as the core realizations. Our new usage differs in that we have separate vertices STOPab instead of a common STOP. We call the remaining (exponential number of) realizations redundant realizations, because the players can ignore them. This is because, in each redundant realization, R: 1. There is a “core” realization such that if it collapses, so does R, and 2. At any position in the game, there is a core realization that contains all available moves in R. By the first property, no redundant realization can ever be the only one remaining. By the second property, no core realization can be collapsed using moves only available due to the redundant realizations. Thus, the redundant realizations never provide any additional moves to either player. Since this construction is just the one from Theorem 6.2.4 with additional realizations that don’t affect how the game is played, hardness is obvious, so long as it is reachable. Thus, to complete the proof, we show that we can reach a superposition consisting only of (1) all the necessary core boards and (2) redundant boards. For our new starting position, we take all vertices and edges in our main board from the Poly-Wide reduction, and add vertices and all new edges. As shown in Figure 6.3, for each arc (a, b) in QDG we 81 can insert vertices ab1, ab2, and stopab, along with edges (a, ab1), (ab1, ab2), (stopab, ab2), and (ab2, b). In addition, we include two vertices, ab−1 and ab−2, which will have already been previously visited in all realizations by the time we reach the start. These vertices are connected by edges (ab−1, ab1), (ab−1, stopab), (ab1, ab−2), and (stopab, ab−2). For some other edge, (c, d), we can connect the gadgets by setting ab−2 = cd−1. Now we prescribe a series of prior moves across all edges {(ai , bi) | i ∈ {1, . . . , m}}: (a1b1)−1 → ⟨((a1b1)1 | stopa1b1 ⟩ → (a1b1)−2 = (a2b2)−1 → · · · → (am−1bm−1)−2 = (ambm)−1 → ⟨(ambm)1 | stopambm⟩⟩(ambm)−2 → x were made, where x is the starting vertex of QDG. By construction, no realizations were ever collapsed in these prior moves. Additionally, there is a realization for each of the core realizations, by doing the following: • For the main board, one simply needs to examine he branch where, across every edge, stop was chosen. • For each core board, one simply needs to examine the branch where, across every edge, stop was chosen except exactly one edge e = (a, b) where ab1 was chosen instead to give us the (a, b)-board. To complete the proof, we only need to show that the rest of the realizations are redundant. All realizations have only the edges in either the main board or an (a, b)-board fulfilling the first property for redundancy. For the second property, we note that each of the non-core realizations must contain at least two stop vertices; its current vertex is either adjacent to one stop (all its moves are in the core realization with the stop) or not (all its moves are in the main board). Theorem 6.2.6. Quantum Generalized Geography with a classical start is PSPACE-complete. Proof. By Theorem 6.5.2, the game is in PSPACE, since the game is polynomially short. We have a very simple reduction from classic Generalized Geography. We replace each edge in the Generalized Geography graph as shown in Figure 6.4 with a path through two new vertices. 82 A B A AB1 AB2 B Geography Edge Resulting Quantum Geography Gadget Figure 6.4: PSPACE Reduction to Quantum Geography is a simple transformation on the edges. Now, if a player ever makes a quantum move from a classical move, e.g. from A to the superposition of AB1 and AC1, then the opponent can immediately collapse to either AB2 or AC2, effectively choosing which of B and C will be moved to. Thus, a player X making a quantum move only gives the next player the power to choose X’s move and will never give a classically losing player a winning quantum strategy. 6.3 Complexity Leap in Quantum Nim: Logic Connection In this section and the next section, we prove that superpositions of moves and positions profoundly impact the complexity of Nim. Theorem 6.3.1 (Quantum Nim Leap Over NP). Quantum Nim with quantum starts is Σ p 2 -hard. 6.3.1 Proof Outline: The Logic Connection of Quantum Nim Our proof of Theorem 6.3.1 is somewhat involved. We first discuss its high-level steps, connecting Quantum Nim to the quantumized Avoid True. The impartial ruleset Avoid True is a logic game introduced in Schaefer’s landmark paper [45]. A starting position is defined by a positive CNF formula (meaning all variables are unnegated in all clauses), with all variables set to false. Two players take turns selecting a variable, and changing the assignment from false to true. Once a variable is assigned to true, it can never be assigned false again. A move is feasible if setting the selected variable true will not make the whole CNF evaluate to true. 83 Boolean Nim is a variant of Nim where each pile has either zero or one stone. LetQuantum Boolean Nim denote the poly-wide variant of the game. We will first prove: Theorem 6.3.2. Quantum Boolean Nim and Quantum Avoid True are isomorphic rulesets. In Section 6.4, we will demonstrate the complexity of Quantum Avoid True through another logic game, Partition-Free QBF, which is a partisan game also played on a formula. The goal of the True player is to make the formula true and the goal of False is to make it false. Boolean variables of the formula are partitioned into two sets, one for each player. At their turn, players can “freely” set any of their unassigned variables. Using a delicate argument to counter an “unwelcome” quantum impact, in Section 6.4.3 we extend Schaefer’s reduction from Partition-Free QBF to Avoid True in the quantum setting as well. Theorem 6.3.3 (Partizan-Impartial Reduction in Quantum Setting). There exists a polynomial-time reduction from Quantum Partition-Free QBF to Quantum Avoid True. In Section 6.4.1, we study the quantumized complexity of the family of PSPACE-complete QBF games as a whole, addressing the subtlety of transforming QBF-games into normal-play games in the quantum setting. In particular, in Theorem 6.4.8, we establish that Quantum Partition-Free QBF with a classical start is Σ p 2 -complete when there are an even number of variables, and Π p 2 -complete when there are an odd number of variables. Motivated by complexity characterizations of Lander’s NP-intermediate problems [37] and Ko’s intricate, meticulous separation of levels of the polynomial-time hierarchy [36], in Section 6.4.2, we refine the complexity collapses in the last theorem to prove the following: Theorem 6.3.4 (Complexity Collapses into Polynomial-Time Hierarchy). For any integer k > 0, for complexity class Σk, Πk, or both Σk and Πk, there exists a classically PSPACE-complete combinatorial game whose quantum generalization is complete in that class. 84 6.3.2 Isomorphism Between Quantum Avoid True and Quantum Boolean Nim Proof of Theorem 6.3.2. Crucial to our complexity analysis, this isomorphism betweenQuantum Boolean Nim and Quantum Avoid True is polynomial-time computable. First note that Quantum Avoid True has the following interesting self-isomorphic property. Quantum Avoid True with quantum starts is isomorphic with Quantum Avoid True with classical starts. This is because the superposition of any two classical Avoid True positions is “equivalent” to the classical position defined by the and for their CNFs. Thus, focusing on the logic-to-nim direction, we consider a classical Avoid True position (F, V, T), where V = {x1, ..., xn}, F is the formula with m clauses, C1, ..., Cm, and T ⊆ V denotes the subset of variables set to true. (T is empty at the start.) We now reduce this position to a Boolean Nim superposition B(F,V,T) with m realizations—one for each clause—and n piles (encoding the Boolean variables). In the realization for Ci , we set piles corresponding to variables in Ci to zero to set up the mapping between collapsing the realization with making the clause true. We also set all piles associated with variables in T to zero, to set up the mapping between collapsing the realization with selecting a selected variable. We now use these two mappings to inductively establish that the game tree for Quantum Boolean Nim at B(F,V,T) is isomorphic to the game tree for Quantum Avoid True at (F, V, T). We first demonstrate this reduction on the following Avoid True position, with formula: (x1 ∨ x2 ∨ x3 ∨ x4) | {z } A ∧ (x1 ∨ x5 ∨ x6 ∨ x7) | {z } B ∧ (x1 ∨ x3 ∨ x6) | {z } C ∧ (x2 ∨ x5 ∨ x8) | {z } D and already-chosen variables, T = {x8}. Our reduction produces the Quantum Boolean Nim position with following three realizations: A = (0, 0, 0, 0, 1, 1, 1, 0), B = (0, 1, 1, 1, 0, 0, 0, 0), and C = 85 (0, 1, 0, 1, 1, 0, 1, 0). Note that the clause corresponding to D was already true, so the corresponding realization has already collapsed. To prove the correctness of the reduction encoding, we prove that the current player has a winning strategy at (F, V, T) in Quantum Avoid True if and only if the next player has a winning strategy at B(F,V,T) in Quantum Nim. We will prove inductively that the following invariant always holds if we play our Quantum Nim encoding and its original Quantum Avoid True position in tandem: For every active clause in each realization of Quantum Avoid True there is exactly one realization in the Quantum Nim encoding with the corresponding realization having piles of 0 for each variable in the clause and variables that are no longer false for that position. As the basis case of the induction, the stated invariant is true at the start. There is only one realization of Quantum Avoid True. In its Quantum Nim encoding constructed above, each realization has piles of 0 for the variables in the corresponding clause, and all variables are false, so the rest of the piles are at 1. Now, we will show that the invariant still holds after any move in Quantum Nim by examining each effect of its “coupled move” on any possible clause. • Any classical or quantum move targeting a pile of size 1 in the Quantum Nim position does not collapse the Nim realization, consistent with the fact that selecting the corresponding variable results in the corresponding clause remaining active in theQuantum Avoid True instance. After the move, the pile will now be at 0 in theQuantum Nim realization, consistent with the fact that after selecting the corresponding variable in the realization of Quantum Avoid True, the variable is no longer false. • Any classical or quantum move targeting a pile of 0 in the Quantum Nim instance, that is not in the corresponding Quantum Avoid True clause can’t be taken. Since this variable must have already been flipped, that means that it is 0 in all realizations, preventing it from being selected. 86 • If a classical or quantum move is made in the Quantum Nim position on a pile of 0 that corresponds to a variable in the associated Quantum Avoid True clause, then the clause is satisfied, and the corresponding realization collapses. This matches the Quantum Avoid True case where the clause is no longer active. This invariant establishes the desired structural morphism between playing Quantum Avoid True and playing its encoding Quantum Nim. The inverse encoding from Quantum Boolean Nim to Avoid True is the following: Given a position B in Quantum Boolean Nim, we can create a or-clause from each realization in B. Suppose B has m realizations and n piles. We introduce n Boolean variables, V = {x1, ..., xn}. For each realization in B, the or-clause consists of all variables corresponding to piles with zero pebbles. The reduced CNF FB is the and of all these or-clauses. Taking a stone from a pile collapses some set of realizations. This move is mapped to selecting the corresponding Boolean variable making the or-clauses associated with those realizations true. Thus, playing Quantum Boolean Nim at position B is isomorphic to playing Avoid True starting at position (FB, V, ∅). Note that the reduction can be set up in polynomial time. 6.3.3 Nim Encoding and Structural Witness of Σ p 2 -Hardness in Quantum Games In our reduction from Σ p 2 -hard Quantum Avoid True, our Quantum Nim game uses a poly-wide superposition of perhaps the simplest Nim positions: all Nim positions that we used in our encoding are from the family of Boolean Nim. So, our Σ p 2 -hard intractability proof of Quantum Nim holds for Quantum Boolean Nim and can be extended broadly to the quantum extension of Subtraction games (as defined in Chapter 2). In this section, we present a more systematic theory to extend these hardness results. Particularly, we can use the Σ p 2 -hardness result of Quantum Nim to get results for several other games in the quantum setting at poly-wide quantum positions. We show that if a game is able to classically properly embed 87 a binary game with properties that we will describe, then its quantumized complexity at a poly-wide superposition is at least Σ p 2 -hard. Definition 6.3.5 (Robust Binary-Nim Encoding). A ruleset R has a robust binary-Nim encoding if R has a position b with the following properties for any n. • b has a set M of n = |M| distinct feasible moves. • For each σ ∈ M, selecting σ moves the game from b to a position bσ ∈ P, such that position bσ has (n − 1) feasible moves given by M \ {σ}. In addition, bσ recursively induces a robust binary-Nim encoding for (n − 1). • When n = 0, the game has no available moves. Lemma 6.3.6. If some fixed classical ruleset R has a robust binary-Nim encoding, then it is Σ p 2 -hard to determine the winability of poly-wide quantum positions in its quantum setting. Proof. We reduce from Quantum Boolean Nim at a poly-wide superposition. Suppose there are n piles and a superposition with m realizations ⟨b1 | · · · |bm⟩. Now we focus on the position b in R that induces the robust binary-Nim encoding for n, with set of moves M = {σ1, . . . , σn}. For each pile i, we associate it with move σi . Each realization, b ∈ {b1, . . . , bm}, in Quantum Boolean Nim defines a set S ⊆ [n], corresponding to piles with value 1. In the reduced quantum position for R, we make a realization with the position whose feasible moves are {σi |s ∈ S}. This is a direct encoding, where we are just relabeling move i to σi , setting up the desired morphism between playing Quantum Boolean Nim and playing its encoding in the quantum setting of R. We can easily embed Boolean Nim in several games, even games with fixed outcomes like Brussels Sprouts. For “interesting” games, such as Go and Domineering, this process is often simple, as one needs only to be able to make a board state that is a sum of combinatorial game value ∗ with only a single move in each of them. 88 6.3.4 Tower Nim and algorithms which do not need modification There is a variant of Nim where the winning strategy does not change at all under the quantum transformation. Tower Nim is a game defined on a stack with elements S = (s1, s2, . . . , sn), where s1 is the top of the stack and sn the bottom. Each pile si ∈ N \ {0}. Players alternate turns by decreasing the pile size of the top element to a smaller non-negative integer. Whenever the size of the top element is reduced to 0, the top cell of the stack is removed, leaving the remaining n − 1. From a classical start, the algorithm for the untransformed game[8] is still an optimal one. Finding the next move is broken up into four cases for the algorithm: 1. If s1 = 1, then bring the game to 0 (this is the only move available) 2. If s1 > 1 and n = 1 (there is only one cell) then set s1 = 0. 3. If s1 > 1 and s2 > 1, then set s1 = 1. 4. Otherwise, s1 > 1 and s2 = 1. Identify the first si ̸= 1 where i ̸= 1. Let the last bit of i be p. Then, identify whether si+1 ∈ S. If so, let q = 1, otherwise let q = 0. Then, reduce the first element to p Lq. One may read the proof of the algorithm in the original paper[8], but the gist is that the winner forces the opponent to always play on stacks where s1 = 1. If the next pile is at least 2, one may set the current pile to 1, allowing themself to be the first person to play on the next pile. The only time when they don’t do this is when there is a chain of 1s and whether there are any moves available after the chain, which determines whether they take all or leave 1. Notice that this algorithm has the property that so long as the current player has at least more than one option (ie: the pile size is strictly more than 1), then the current player wins the game. This means that a losing position will never have more than 1 option, thus the losing player is unable to benefit at all from 89 quantum moves so long as the classical winner follows the original winning algorithm. Thus, the original algorithm is still correct with exactly the same proof. Of course, any game with the property of the losing player having only one option can have the same argument applied to it. This is really just a specific case of the more general rule below: Theorem 6.3.7. If an algorithm for a game does not take the opponent’s moves into account (ie: you will play the same regardless), then the quantum transformation does not in any way alter optimal algorithms. Clearly if a game has only a single option for your opponent, you will always make the same move since their choices can’t affect anything. 6.4 Quantum Transformation of PSPACE-Complete Games So far, we’ve seen that when introducing quantumness into classically-tractable combinatorial games, they may remain tractable (e.g., Undirected Geography with classical starts with width-2 moves) or become intractable (e.g., Quantum Nim and Quantum Undirected Geography with quantum starts). The next logical direction to explore is the complexity landscapes of the quantum transformation of classically intractable games. Is their intractability always preserved? Is their complexity ever lost? In this section, we shed some light on these questions with various concrete games. We also complete the proof of theorems stated in Section 6.3.1. 6.4.1 Quantum Preservation and Quantum Collapse It is well known that the Quantified Boolean formula problem (QBF)—determining whether a quantified Boolean formula is true or false—can be viewed as a classical combinatorial game. Technically, combinatorial games must also fulfill the “normal play” requirement, meaning that a player losesif and only if they are unable to make a move. There are several equivalent ways to do this (transforming logical QBF decisions 90 into combinatorial games) which are all classically polynomial-time reducible to each other. Complexitytheoretically, QBF is the canonical complete problem for PSPACE, and thus all these QBF variants are PSPACE combinatorial games; they form the bedrock of PSPACE reductions. However, because quantum games are more subtly dependent on the explicit move definitions in the ruleset, these known reductions from the “classical world” don’t necessarily continue to hold. In fact, these QBF-based games have richer stories in the quantum world, revealing the sensitive interactions between quantum moves and logic positions. Our analysis exhibits the fundamental difference among various QBF variants in the quantum setting, illustrating the reasons they are dependent on the explicit move definitions in the ruleset. In this Subsection, we discuss various “end-of-QBF” game variants and give intractability results for all “natural” Quantum QBF variants, even though some “end-of-QBF” transformations contribute to the complexity collapse in the quantum setting. This is critical to our intractability analysis of Quantum Nim, and provides the needed technical reduction to complete the proof of the leap in Nim’s quantumized complexity from polynomial-time to the second level of the polynomial-time hierarchy. In the literature, the Quantified Boolean formula problem is also known as the Quantified Satisfiability Problem (QSAT). So, QBF and QSAT are used interchangeably. However, in this dissertation, we will—for clarity of presentation—use the following ruleset naming convention for QBF and QSAT, in order to denote two different ruleset families of combinatorial games rooted in the Quantified Boolean formula problem. • The QBF Family: We will use QBF to denote the family of the combinatorial games that textually implements the Quantified Boolean formula problem: An instance of QBF is given by a CNF f, whose clauses may contain both positive and negative literals, over two ordered lists of Boolean variables. We have two cases depending on whether or not one of the lists has one more variable than the other one: – Case Even: (T1, ..., Tn) and (F1, ..., Fn) 91 – Case Odd (T1, ..., Tn, Tn+1) and (F1, ..., Fn) So, the quantified boolean formulae are, respectively: Case Even: ∃T1∀F1 · · · ∃Tn∀Fn f(T1, ..., Tn, F1, ..., Fn) Case Odd: ∃T1∀F1 · · · ∃Tn∀Fn∃Tn−1 f(T1, ..., Tn, Tn+1, F1, ..., Fn) In this game, one player—Player True—aims to satisfy the CNF formula, while the other player— Player False—wants the formula to be unsatisfied. Starting with True, the two players alternate turns setting their next variables to True or False. In other words, in their respective i th turn, Player True sets variable Ti , then Player False sets variable Fi . The variants in this family, as we shall define later, differ in the details of the termination condition in transforming logic QBF into a combinatorial game. • The QSAT Family: We will use QSAT to denote the family of partition free assignment QBF games, with relaxations on the variables the players can choose to set at each turn. As in QBF, each instance of QSAT includes a CNF formula f, and two players—Player True and Player False—who take turns to set their own variables; Player True can only set variables labeled Ti and Player False can only set variables labeled Fi . Again, we have two cases—Case Even and Case Odd—depending on whether Player True has one more variables than Player False. For the game, Player True aims to satisfy the CNF formula, while Player False wants the formula to be unsatisfied. Unlike in QBF, however, the players can “freely” set any of their unassigned variables to True or False, as opposed to setting variables according to the prescribed order. The variants in this family also differ in the details of termination conditions in transforming logical QBF into normal-play combinatorial games. We now summarize the varying termination details of our QBF/QSAT games. 92 • Termination Rules: In the first three variants, players continues to play the game until all variables are assigned. Then the following rules govern how games terminate. – In Phantom-Move QBF/Phantom-Move QSAT, after all variables have been assigned, the next player has a feasible (phantom) move available if and only if they have a winning assignment. This variant of the game is implicitly invoked in several classic reductions, including Schaefer’s Avoid True reduction [45]. In the next two variants, after setting all of the variables, there are additional moves based on clauses and literals in the formula. In both, in the even case, there is an additional move for Player True directly after setting variables. After that, in both odd and even cases, the false player is the current player. At this point, they may be given the chance to select a clause they believe True failed to satisfy. – In ClauseSelector QBF/ClauseSelector QSAT, Player False may make a final move by selecting any clause with all false literals. (If none exist, then Player True wins.) – In LiteralSelector QBF/LiteralSelector QSAT, Player False may choose any clause, then Player True may make a final move by selecting a True literal in that clause. (If no True literals exist in that clause, then Player False wins.) Both are common interpretations used in reductions. For example, ClauseSelector QBF is used in the original Node-Kayles reduction [45], and LiteralSelector QBF is used in the Geography reduction [40]. • The Mercy Rules: The last three variants define when the game terminates early after a player has achieved a victory condition (and to stop the losing player to continue playing in a completely hopeless position). The 93 victory condition for Player True is having a true literal in every clause, and for Player False is having a clause with all false literals in it. The only difference between the following three is how the mechanic for ending the game early works. – In TKO QBF/TKO QSAT, the game ends automatically as soon as a victory condition is reached. This means that a player may neither make an assignment that results in their opponent’s victory condition, nor make an assignment once their opponent’s victory condition has been satisfied. – In Haymaker QBF/Haymaker QSAT, each turn, a player may, instead of a normal variable assignment (or after all variables have been assigned) make a “game ending blow” which is only available after that player’s victory condition has been met. After the move is made, the opponent can no longer move. – In KO QBF/KO QSAT, while a player makes a move, if they have a achieved a winning state, they can also choose to end the game, forcing the other player to be unable to make a move. We will also briefly a few problems related to the polynomial hierarchy. • Tautology is the canonical co − NP-complete problem. It is the problem of, given a CNF, determining whether the the CNF is a tautology; i.e.: determining if ∀x1, ∀x2, · · · , ∀xnf(x1, . . . xn) evaluates to true, where f is a CNF formula. • Σ p 2 SAT is the problem of determining whether ∃x1, ∃x2, · · · , ∃xn/2 , ∀xn/2+1, · · · ∀xnf(x1, . . . xn) evaluates to true • Π p 2 SAT is the problem of determining whether ∀x1, · · · , ∀xn/2 , ∃xn/2+1, · · · ∃xnf(x1, . . . xn) evaluates to true 94 For complexity-theoretical analysis, we first focus on the more restrictive QBF family, in which players must set their variable according to the prescribed order. Theorem 6.4.1 (Quantum Collapses of QBF: Classical Starts). Quantum Phantom-Move QBF with a classical start is Σ p 2 -complete in Case Even, and is Π p 2 -complete in Case Odd. Proof. Notice that the player who makes the final move (the “phantom move”) will be allowed to choose to collapse any quantum moves so that the variables are assigned in a way that they will win (if it is possible). Since quantum moves cannot collapse before the end of the game, they can choose the assignments for any variables made during assignments by either player. So, the optimal move for this collapsing player will be to make exclusively quantum moves, and the optimal move for the other player will be to never make a quantum move. So, the non-phantom player makes a variable assignment of half of the variables, and then only wins if all other assignments of the other variables are winning positions for them. Otherwise, the phantom move player wins. If the phantom move player is False, then this is exactly the Σ p 2 SAT problem. Otherwise, it is exactly the Π p 2 SAT problem. For formal completeness, we go through the trivial two way reduction. In Case Even, we can create a Σ p 2 SAT problem, giving True’s variables as the “exists” variables, and False’s variables as the “for all” variables. In Case Odd, we can create a Π p 2 SAT by similarly giving False’s variables as the “for all” variables, and True’s variables as the “exists” variables. As previously described, both of these will output which player wins correctly. For the other direction, we can do the same with both, only in the other direction. In other words, make all “exist” variables into True’s variables and all “for all” variables into False’s variables. Then, we have a corresponding Quantum Phantom Move QBF instance. We can also extend the result to quantum starts: 95 Corollary 6.4.2 (Quantum Collapses of QBF: Quantum Starts). Quantum Phantom-Move QBF with a quantum start is Σ p 2 -complete in Case Even and is Π p 2 -complete in Case Odd. Proof. For Case Even, given a poly-wide quantum position B in which the i th realization is defined by CNF fi , the players make decisions as if they play Quantum Phantom-Move QBF with the classical start for logic function W fi . Case Odd is similar. Theorem 6.4.3. Quantum Literal Selector QBF with a classical start is co-NP complete. Proof. Since Player True wins so long as a variable in the selected clause has a quantum variable assignment, and can’t be collapsed before then, they will always make a quantum move, and Player False will never make a quantum variable assignment. So we can, WLOG, remove every clause that has a True variable in it, which will always be satisfied. Now, False wins if and only if there exists an assignment to make a clause unsatisfied, and thus make the CNF evaluate to false. This is is exactly Tautology, the canonical co − NP-complete problem. Reducing between the two is trivial, since it involves doing nothing one way, and creating dummy clauses and variables for Player True when reducing from Tautology to Quantum Literal Selector QBF. Similarly, we can prove: Theorem 6.4.4. Quantum Clause Selector QBF with a classical start is NP complete. Proof. This closely follows the proof of Quantum Literal Selector QBF. We assume WLOG that no clauses have a False variable appear both negated and unnegated (since those clauses will always be satisfied and thus not selected by False in the end). False just needs a single realization to have a clause with all false literals, so they will always make quantum moves, since any classical play they could make could also be reached by playing quantum and choosing the same clause in the end. Since any quantum literals can be assumed to be false, we can WLOG remove all of False’s variables, and True can win if and only if 96 there exists a satisfying assignment for the rest of the variables, which is to say, only True’s literals. This is exactly a SAT problem. Theorem 6.4.5. Quamtum TKO QBF with a classical start is PSPACE-complete. Proof. By Section 6.5, Quantum TKO QBF is a PSPACE-solvable game. We will show hardness of the quantum game by reducing from its classical TKO variant. Consider any given TKO QBF instance R′ , in which, for clarity of presentation, we assume that True’s variables are labeled T1, T9, T17, . . . , T8n+1 with False’s variables labeled F4, F12, F20, . . . , F8n−4, where the order of play is: T1, F4, T9, F12, . . . , F8n−4, T8n+1. Then, we reduce R′ to another TKO QBF instance R, and analyze the complexity for determining the outcome class for its quantum lift RQ. In R, we will create variables such that the order of play is: T1, F2, T3, F4, . . . , F8n, T8n+1, F8n+2. For each classic variable for False, Fi , we create clauses to add to the CNF: (Fi ∨ Ti+1) ∧ (Fi+2 ∨ ¬Fi+2). For each classic variable for True, Ti , we create clauses (Ti ∨ ¬Fi+1 ∨ Ti+2) ∧ (¬Ti ∨ Fi+1 ∨ ¬Ti+2) ∧ (Fi+1 ∨ Ti+4 ∨ F8n+2) ∧ (¬Fi+1 ∨ ¬Ti+4 ∨ F8n+2) ∧ (Fi+3 ∨ ¬Fi+3). If Player False makes a quantum move on any of the classic False variables, Fi , then Player True can set Ti+1 to false to collapse False’s move to true. So, if the QBF evaluates classically to true, then False can’t make a quantum move, and thus True wins. Now we will show that if Player False is winning the classic QBF, then False will win this quantum QBF. Notice that if True makes any classical move on classical variable Ti , they will not be hurt by the additional clauses, as no matter what False does, they can satisfy those clauses by an appropriate response. 97 If Player True makes a quantum move on any of the classic true variables Ti , then Player False has a winning strategy, by setting Fi+1 to false. If Ti+2 is set to quantum, then Fi+1 can never collapse until the end, even if Ti collapses. So, no matter what Ti+4 is set to, there exists a possibility that not all clauses are satisfied, and by setting Fn+1 to false, all the possibilities where Player True had satisfied all clauses collapse, and thus one of the extra clauses must be all false, giving Player False the win. If Ti+2 is set to true, then there are three possibilities for what could happen to Ti . In the first case, it never collapses, which has the same outcome as above, giving Player False the win. The second is that Ti collapses to false. In this case, Fi+1 never collapses, and thus the previous argument that Player False will win still holds. The last is that Ti collapses to true. This causes Fi+1 to collapse to true as well. If Ti+4 is set to true as well, then Player False will win as there is one clause with only falses and a future False’s variable in it. If it was set to quantum, then the old argument applies where eventually Player False will win when setting F8n+2. If it was set to false, then Player False can just play as if it was set to true, since if it collapses to false, Player False wins. If Ti+2 is set to false, then the same general arguments hold. If Ti never collapses, then we follow the old argument to win. If it collapses to true, then Fi+1 never collapses, and the old argument holds. If it collapses to false, then Fi+1 collapses to false. If Ti+4 is set to false, then there is an unsatisfiable clause for Player True, causing the player to lose. If it was set to quantum, than the old argument still holds. If it was set to false, then Player False can just play as if Ti was set to false, since if it collapses to true, Player False wins. The reduction is polynomial space, since we are just quadrupling the number of variables and adding a constant number of clauses for each variable. We can similarly prove the following two theorems: Theorem 6.4.6. Quantum KO QBF is PSPACE-complete. 98 Proof. Again, by our complexity characterization in Section 6.5, Quantum KO QBF is a PSPACE-solvable game. For hardness of the quantum game, we will reduce from the classical KO QBF. Consider any given KO QBF instance R′ . For clarity of presentation we assume that True’s variables have labelings of the form T1, T9, T17, . . . , T8n−7 and False’s variables are labeled F5, F13, F21, . . . , F8n−3, so the order of play on the variables is: T1, F5, T9, F13, . . . , T8n−7, F8n−3. Then, we reduce R′ to another KO QBF instance R, and analyze the complexity for determining the outcome class for its quantum lift RQ. In R, we will add variables such that the order of play is: T1, F2, T3, F4, . . . , F8n−1, T8n, F8n+1. First, in all of the existing clauses, we add an additional ∨T8n ∨ F8n+1. We then create three new “gadget” clauses for each of True’s variables Ti : (Ti ∨ Ti+2) ∧ (¬Ti ∨ ¬Ti+2) ∧ (Fi+1 ∨ ¬Fi+1) Then, for each False’s variable Fi , we add the following three “gadget” clauses: (Fi ∨ Fi+2 ∨ ¬T8n) ∧ (¬Fi ∨ ¬Fi+2 ∨ ¬T8n) ∧ (Ti+1 ∨ ¬Ti+1) . The gadget clauses are designed to satisfy that the logic value of the formula for R preserves the logic value of the formula for R′ . First, suppose R′ has value false, so R also has value false. Note that if Player True makes a quantum move for any of the classic variables Ti , then no matter what they assign Ti+2 to, there exists a realization where either (Ti ∨ Ti+2) or (¬Ti ∨ ¬Ti+2) evaluates to false, so Player False can win with a 99 KO move. Player True will not be able to deliver a KO move to win because there will exist clauses that have not yet been satisfied. False will simply play their classic strategy for classic variable Fi and echo that assignment for Fi+2. If True plays purely classically until T8n, then the classic variables of one of the classic clauses will evaluate to false when they select T8n. So, they will need to set it to true to avoid having that clause evaluate to false the next turn. However, since Player False is echoing their choices into Fi+1, at least one clause of the form (Fi ∨ Fi+2 ∨ ¬T8n) currently has Fi ∨ Fi+2 = 0, and the T8n variable will be set to 1, which will allow Player False to make the KO move on their next turn. If Player True makes a quantum move for T8n, then False can still deliver the KO move because there exists a realization with all false. Now suppose that R′ has value true, so R also has value true. True’s strategy is to play their classical strategy on the old variables, and satisfy the new clauses with only True’s variables in them by choosing appropriately. If there is any point where Player False makes a quantum move, True plays as if False selected true for the variable. Notice that Player False is unable to KO until Player True sets T8n, since T8n is in every clause that could possibly evaluate to false. And then when Player True sets T8n, they can set it to false and KO, satisfying all of the new clauses, and there exists a realization where all of the classical clauses are true (when Player False chose true for the quantum move). The reduction is in polynomial space. All we do is create three more variables and three new clauses for each existing variable, so the reduction is linear in the QBF input size. Theorem 6.4.7. Quantum Haymaker QBF is PSPACE-complete. Proof. Like the other Quantum QBF variants, Quantum Haymaker QBF is still a polynomial-space solvable game. For hardness analysis, the reduction is exactly the same, and the proof follows the same guidelines as for Quantum KO QBF. Notice that if Player True makes a quantum move on Ti , no matter what they select for Ti+2, Player False can make the game ending blow on their next turn. If False makes a quantum move, False can’t make the game stopping move until after F8n+1 has been set. But if Player 100 True was the classical winner, setting T8n to false, and was playing optimally to one of False’s moves, True can make the game-ending blow after that to win the quantum game. We now turn our attention to Partition-Free QBF and its quantumized complexity. In contrast to standard QBF games, the partition-free version relaxes the order requirement on setting the variables. As we discussed before, we will refer to this family of QBF-based games as the QSAT family. Classically, all variants of QSAT are PSPACE-complete games. Below, we consider the complexity of QSAT games in the quantum setting. Most importantly, for our proof of quantum leap in Nim’s complexity, we prove the following theorem. Theorem 6.4.8 (Quantum Collapses of Partition-Free QBF). Phantom Move QSAT with a classical start is Σ2-complete in Case Even and Π2-complete in Case Odd. Proof. We will call the player that makes the phantom move the PM Player and the other player the NM Player, for short. Then, the theorem follows from the following observations: Observation 6.4.9. If the PM Player has a winning strategy, then assigning each variable to ⟨true | false⟩ is also a winning strategy. Proof. Suppose for the sake of contradiction that the PM player has a winning strategy S1 but the above described strategy S2 isn’t a winning strategy. Then, that means that the NM Player has a strategy against S2, which must be a sequence of moves such that in all realizations, the PM hasn’t fulfilled their winning condition, since otherwise they could move into the phantom move. But then, that exact sequence of moves is a winning sequence of moves against S1, and any possible deviations of it, so S1 isn’t a winning strategy. Observation 6.4.10. If the NM payer has a winning strategy, then they have a winning strategy of selecting classic moves independent of the PM player’s choices. 101 Proof. If the NM Player has a winning strategy against the PM player, then they must also have one against the PM Player’s strategy of only selecting quantum assignments as mentioned in Observation 6.4.9. Regardless of what the NM player’s strategy is, the fact that it wins means that it must win against all realizations of the PM player’s strategy, since otherwise the PM Player could collapse to that realization on the phantom move, then win. The NM Player can do this with only classic moves, as any realization of a quantum strategy must also not have any winning realizations for the NM Player either. From here, we can form a trivial reduction to and from Σ p 2 -SAT and Π p 2 -SAT for instances where False gets the phantom move and True gets the phantom move, respectively. The proof can natually be extended to handle any superposition width. It can also be extended to poly-wide quantum starts. Corollary 6.4.11 (Quantum Collapses of Partition-Free QBF: Quantum Starts). Quantum Phantom-Move QSAT with a quantum start is Σ p 2 -complete in Case Even, and is Π p 2 -complete in Case Odd. Proof. For Case Even, given a poly-wide quantum position B in which the i th realization is defined by CNF fi , the players make decisions as if they are playing on Quantum Phantom-Move QSAT with the classical start for logic function W fi . Case Odd is similar. 6.4.2 The Depth of Complexity Collapse in the Quantum Setting We now have seen instances of classically PSPACE-complete games becoming complete in the first or second level of the polynomial hierarchy. This leads us to a natural question of whether games exist that collapse to any level of the polynomial hierarchy. For all levels other than 0, we have an answer: Theorem 6.4.12 (PSPACE-Intermediate Quantum Games). For any complexity class Σ p k , Π p k , or Σ p k ∪ Π p k , there exists a classically PSPACE-complete game, whose quantum generalization is complete in that class. 102 Proof. We create games called QuantumLevel(ℓ, C), where ℓ and C are replaced with different values. The value of ℓ is the level of hierarchy that we want the quantum complexity to be, and C is one of “Σ”, “Π”, and “Σ∪Π”. This game will have classic complexity PSPACE-complete and quantumized complexity under RQ of Cℓ . We will first be focusing on only the cases where C is Σ or Π. The game of QuantumLevel(ℓ, C) is played over a CNF formula with variables that are labeled T1, T2, T3, . . . , Tn and F1, F2, . . . , Fn, where ℓ 2 divides n. If ℓ is odd, there are an additional x = n ⌊ ℓ 2 ⌋ variables. If C = Σ then they are labeled Tn+1, Tn+2, . . . Tn+x. If C = Π, then they are labeled Fn+1, Fn+2, . . . Fn+x. The T variables and the F variables are further grouped into collections of size x. We will thus denote T1, T2, . . . Tx by GT,1, Tx+1, Tx+2, . . . , T2x as GT,2, and so on, and the same for GF,i. We will use the term “echo” in our description below. What this term means is that the player must repeat a previous move, either that player’s previous move, or their opponent’s (which move to be echoed will be specified when the term is used). For example, if True makes a move of “Tx1 = true”, and if the false player is to echo it, they then must make their move “Tx1 = true.” In this game, the same move can be made multiple times, and it is even required by the rules. So, if a player makes a move σ, then the echoing player makes the same move σ. Intuitively, the idea is as follows: we will have a classically PSPACE-complete game, as it is just a QBF problem, with a bunch of arbitrary echoing steps. However, the added echoing forces the player to collapse their own moves, and if they echo for a second time, they will give the other player a chance to collapse the move into whatever they want. Thus, through controlling echo round, we can choose the order of variable assignment to be non-alternating in the echoed version, and thus, through careful control of this, collapse the complexity to any level of the polynomial hierarchy. Classically, the game is played as follows (with i starting at 1): 1. Both players alternate turns assigning the variables in Gi , in order of the variables’ index. 103 2. Upon all variables in the group being assigned, the turn proceeds to the starting player of the group. This is True if C = Σ, and False otherwise. 3. The starting player must make a move echoing the assignment of the first variable in the group that they assigned. The other player then echoes the starting player’s variable as well. Then the starting player echoes their second variable, and the other player again echoes the second variable. Play like this continues until the same player echoes the last assignment in the group. 4. After the last of the echoing above, the starting player echoes one final time, then the other begins to echo their variables, and the process repeats for the other player’s variables in the group now. 5. Repeat step 1, now for the variables in Gi+1 6. If there is a group in the end for one player with no variables for the other player, then the player with the variables plays and the other player just echoes their assignment. 7. Once all variables are assigned, the player whose turn it is may move if and only if they have achieved their win condition in all clauses. It is not difficult to see classically that these games are PSPACE-complete. Under classical player, it is simply a QBF problem with additional echoing moves, making no changes at all to the assignment. Thus a trivial two way reduction between QBF and these constructed games exists. For the quantum game, we can reduce to and from the SAT that is canonical for the desired complexity class. Note for this CNF game, the optimal strategy for the first phase is to just take a quantum move of both variable assignments, as one can assign them later, and there is no benefit to assigning them earlier. Then, if a player chooses a quantum move during phase 1 or 2, the opponent gets to make an assignment for the player, so it is always in the player’s favor to make a classical assignment here. But then, when done for all variables, this means that this is exactly a problem on the ℓ level of the hierarchy, and thus there is a trivial reduction between the Σ p ℓ and Π p ℓ SAT problems. 104 Finally, we briefly note that for QuantumLevel(ℓ, Σ and Π), we simply make the ruleset contain all positions in QuantumLevel(ℓ, Σ) and QuantumLevel(ℓ, Π). Since the games in the first are Σ p ℓ -complete canonically and for the second are Π p ℓ -complete canonically, this game is now Σ p ℓ∪Π p ℓ complete, completing the proof. 6.4.3 Quantum Lift of Schaefer’s Partisan-Impartial Reduction We now prove Theorem 6.3.1, showing that Quantum Boolean Nim is intractable, with complexity between NP and PSPACE. We will reduce to it—via Quantum Avoid True—from the quantum generalization of Phantom-Move QSAT. In Phantom-Move QSAT, one can equivalently define the “phantom move” to be included with the final variable selection, such that the player that would assign the final variable may only do so if they will have reached their win condition upon playing that variable. One can reduce from the other Phantom move variant by introducing a variable vn+1 which only appears in a clause as (vn+1 ∨ ¬vn+1) given to the phantom move player. So the hardness remains the same. We will be reducing from this variant of the game. Classically, this game, as proved by Schaefer [45], is PSPACE-complete. We proved in Section 6.4.1 that its quantum extension is Σ p 2 -complete or Π p 2 -complete, depending on which player is making the final variable assignment. When restricted to the cases where the final variable is assigned by the False Player, it is Σ p 2 -complete. When assigned by the True Player, it is Π p 2 -complete. Below, we will use this fact to complete the proof of Theorem 6.3.1 by establishing the following theorem. Theorem 6.4.13 (Baseline Complexity of Quantum Avoid True). Quantum Avoid True is Σ p 2 -hard. Proof. As a reminder, in Avoid True, variables appear only unnegated in the CNF, and start as all assigned to false. Players alternate flipping these variables to true. Player are not allowed to make a move which will make the CNF evalute to true. 105 Consider a Phantom-Move QSAT instance R. In the proof, we will focus on the case where both players have the same number of variables, which is an even number. This is acceptable since the hardness results remain even when fixed to any arbitrary parity. Recall that the player that goes first will be called the True player and the player that goes second the False player. Below in both R and its quantum generalization RQ, we will denote the True Player’s variables by {T1, ..., Tm} and the False Player’s variables by {F1, ..., Fm}. In our proof, we will show that Schaefer’s reduction from Phantom-Move QSAT to Avoid True in the classical setting can be quantum lifted into a Quantum Avoid True instance to encode RQ: Schaefer’s reduction (in our notation) is the following: • For each True Player’s variable Ti , we introduce two new variables Ti1 and Ti2 , and create a positive clause (Ti1 ∨ Ti2 ). We will collectively call these clauses TV clauses. Ti1 will represent assigning a true literal, and Ti2 will represent assigning a false literal. • For each False Player’s variable Fi , we introduce three new variables Fi1 , Fi2 , and FiG, and create a clause (Fi1 ∨ Fi2 ∨ FiG). We will collectively call these clauses FV clauses. As before, Fi1 will represent assigning a true literal, and Fi2 will represent assigning a false literal. The FiG variable is simply an arbitrary variable for the purposes of ensuring a certain clause parity (which we will give the motivation for later). It will function as an alternate truth literal assignment, as it appears in every clause Fi1 does. • For each of the clauses in R, we replace each instance of positive literals Ti and Fi with Ti1 and Fi1 , respectively; we replace each instance of negated variables ¬Ti and ¬Fi with Ti2 and Fi2 , respectively. We will collectively call these clauses QBF clauses. • For each instance of a variable Ti1 , Fi1 , Ti2 , and Fi2 in the QBF clauses, we add, to the same clause, variables T ′ i1 , FiG T ′ i2 , or F ′ i2 , respectively. We will collectively call these new variables duplicate 106 variables. These serve both as a way to ensure the QBF clauses all have even parity, and for player strategy, as we will cover later. Through our proof, we will let X denote the Avoid True instance obtained from Schaefer’s reduction of Phantom-Move QSAT instance R. We will call the first player in X Player True and the second player Player False. Schaefer proved that True can win (the impartial) X if and only if True can win (the partizan) R. Below, we will extend Schaefer’s proof to show that the quantum generalizations RQ and XQ have the same winner (when optimally played). Before going on with our proof, we first recall one of the key properties, formulated by Schaefer [45] (theorem 3.12 in his paper), and extend it to the quantum setting. Lemma 6.4.14 (Avoid-True Destiny). For any classical position created by Schaefer’s reduction, if all unsatisfied clauses have an even number of variables, then Player True will win, under arbitrary play by both players for the remainder of the game. If instead all unsatisfied clauses have an odd number of variables, then Player False will win, under arbitrary play by both players for the remainder of the game. Proof. ∗ Let’s first note that in this game, there are always an even number of total unassigned variables remaining on True’s turn and an odd number of unassigned variables remaining on False’s turn. This is because the total number of variables in the game is even at the start, and True goes first. Thus, since exactly one variable is assigned each turn, the parity of remaining variables is maintained. The proof then is simple. If there is an even number of unassigned variables remaining, and all the unsatisfied clauses have an odd number of variables in them, then there must exist a clause and an unassigned variable such that the variable is not in that clause. Similarly, if there is an odd number of variables remaining, with all remaining clauses having an even number of variables, than there must exist an unassigned variable not in at least one of the clauses. Since this property holds under induction for arbitrary ∗We include a proof here both in order to provide readers more intuition about Schaefer’s reduction and to add that it holds under arbitrary play (which wasn’t explicitly stated by Schaefer). 107 play, one can’t lose if there is a move remaining, the game length is finite, and as previously stated, the parity of number of variables remaining is equivalent to the player, this completes the proof. Lemma 6.4.14 captures the “parity” design used in Schaefer’s reduction. It also defines a scenario satisfying the condition of “quantumness doesn’t matter.” Lemma 6.4.15 (When Quantum Doesn’t Matter). If all realizations of a game classically have only one possible winner, no matter what sequence of moves either player makes, then quantumness doesn’t matter. Proof. In order for a game to classically have only a single possible winner, all paths to a leaf node in a game tree must be of the same parity. For a game to end, all realizations must have no moves remaining. Otherwise, the ruleset would allow for some kind of move to be made. So, all games end with all remaining realizations at leaf nodes. And since all of the realization’s leaf nodes have the same parity, the quantum game will end after the same parity of moves. Then, that same player wins no matter what moves either player makes. Therefore, quantumness doesn’t matter. Combining Lemma 6.4.14 and Lemma 6.4.15, with the facts that all TV clauses have length 2, FV clauses have length 3, and all QBF clauses have an even length, we get: Corollary 6.4.16 (Quantum-Avoid True Destiny). For any position reached in Schaefer’s reduction, if in all realizations, there are only TV clauses and/or QBF clauses unsatisfied, then False wins; if in all realizations, there are only FV clauses unsatisfied, than True wins. To prove the correctness of this reduction in the quantum setting, we simply need to prove that if True has a winning strategy in the instance of RQ that we are reducing from, then True has a winning strategy in XQ, and that if False has a winning strategy in RQ, then False has a winning strategy in XQ. In the proof below, we will crucially use the fact that we established in Observation 6.4.10: True has a winning strategy in Quantum Phantom-Move QSAT if and only if True has a single assignment (of only classical moves) that can always win, regardless of what moves False makes. 108 So, if True is the winner in RQ, we prescribe the following strategy for True in XQ: 1. Assign variables in the TV clauses according the winning strategy for RQ. 2. Assign all of the F ′ i2 variables. Our proof is based on the following key observation: If True is able to follow this strategy to completion, then all TV and QBF clauses must be satisfied, resulting in a True win by Corollary 6.4.16. To proceed, we just need to show that True’s strategy is always a legal set of moves. First, True must be able to satisfy all m TV clauses, because False can’t satisfy all m FV clauses before True, thus whatever TV clause true assigns last can’t be the final assignment of the game. Then, because that variable assignment in R must have satisfied all clauses regardless of how the FV variables were assigned, that means in no realization is it legal to make a move that would satisfy all FV clauses, as all QBF clauses must be satisfied at the same time. Since all realizations still have at least one FV clause unsatisfied, True can then assign all F ′ i2 variables, since none of them appear in an FV clause (or, if False assigned them early, then equivalently True can simply take a move not made by False in the FV clauses). So, that strategy is always achievable. Now, for the second case. Suppose that False wins the instance RQ. Then, False applies the following strategy: 1. For False’s move i ≤ m − 1, if Fi1 and Fi2 are not yet classically assigned, make a quantum move of ⟨Fi1 | Fi2 ⟩. Otherwise, play arbitrarily on the FV clauses not yet assigned in all realizations. 2. For the final move, assign the variable in final FV clause to either true or false, whichever is a legal move. If both are, choose arbitrarily. Note that if False is able to carry out this strategy, then by Corollary 6.4.16, they will win the game, as they will have satisfied all FV clauses leaving only TV and/or QBF clauses remaining. 109 For the proof of the correctness of this strategy, first note that, as we proved in Observation 6.4.9, if False has a winning strategy in Quantum Phantom-Move QSAT, then repeatedly choosing, for an arbitrary variable, a quantum assignment of ⟨true | false⟩ is also a winning strategy. If True has only played consistently (i.e.: following their prescribed strategy), then False will be able to make a move on the final unsatisfied FV clause, as there exists a realization where not all QBF clauses are satisfied. If True ever makes a move with at least one realization that isn’t on a variable in a TV clause, then when False moves to the final FV clause, there exists a realization where True hasn’t assigned all of the TV clauses, so False can make their final move arbitrarily on the clause. 6.4.4 Natural PSPACE-Complete Games In this Subsection, we consider quantum transformation of several well-studied PSPACE-complete combinatorial games. In addition to Generalized Geography, we will also analyze the following games: • Node Kayles: An impartial game where players alternate turns placing tokens on vertices of a given graph. A player is only able to place a token on vertex if that vertex does not already contain a token and is not adjacent to any vertex with a token. So, a player is unable to move when the tokens form a maximal independent set. • Bigraph Node Kayles: A variant of Node Kayles in which nodes are partitioned into red nodes and blue nodes, where the blue player can only play on blue vertices and the red player can only play on red vertices. • Snort: A game where one player is a blue player, and the other is a red player. Players alternate placing tokens of their color onto vertices of a given graph. Players can’t place tokens on vertices adjacent to vertices with a token of the opponent’s color. Starting positions can contain some tokens already placed into some vertices. 110 Theorem 6.4.17. Quantum Node Kayles is PSPACE-complete. x1 x1 y1,4 y1,3 y1,2 x2T x2F x2T x2F y2,4 y2,3 x3T x3F x3T x3F C1 C2 C3 level 1 level 2 level 3 level 4 Figure 6.5: Example of the PSPACE reduction from QBF to Quantum Node Kayles. QBF: ∃x1 : ∀x2 : ∃x3 : (x1 ∨ x2 ∨ x3) ∧ (x2 ∨ x3) ∧ (x3 ∨ x3) Proof. By Theorem 6.5.2, the game is in PSPACE, since the game is polynomially short. Our proof of this theorem follows a similar plan to the original proof of the PSPACE-hardness of Node Kayles[45]. We reduce from instances of the classic QBF problem where true assigns both the first and last variable. This assumption is without loss of generality, since if an instance ended with Player False, we can just give Player True a variable xn+1 and add an extra clause (xn+1 ∨ x¯n+1). To create the reduction seen in Figure 6.5, we do the following: 1. For x1, create two vertices x1 and x¯1. 111 2. For each other variable xi , create four vertices xiT , xiF , x¯iT , and x¯iF . We refer to the group of these as Xi . 3. For each original variable xi , create vertices yij for all i < j ≤ n + 1 4. Connect all vertices in each Xi to each other vertex in Xi and to each yij . We will call the clique this forms “level i”. 5. Create a vertex for each clause Ci and add edges between all clause vertices. This clique forms level n + 1. 6. Connect each xiT and each xiF to x(i+1)T and x¯(i+1)T . Connect each x¯iT and x¯iF to x(i+1)F and x¯(i+1)F 7. Connect each xi to every clause Cj where xi appears unnegated, and connect each x¯i to each clause Cj where xi appears negated. 8. Connect each yij to every other vertex in the graph except the vertices on level j. This game of quantum node Kayles is winnable by the first player if and only if the original QBF is a true instance. We refer to the first player as “True” and the second player as “False”. We prescribe the following order of play: on turn i, the current player will play classically on a vertex representing variable xi . When all of these variables are selected, False will then attempt to play on a remaining clause vertex. If they can, then False will win, as each level of vertices has had a vertex played in it, and each level is a clique. If False can’t play on a clause vertex, then they are all covered, so no vertices remain in the graph and True wins. Clearly if both players play on the variable vertices optimally, then False can choose a clause vertex at the end if and only if they could in the QBF. 112 It remains to be shown that if this prescribed play is violated by the classically losing player, they will lose. Assume the order is observed until level i, then on turn i+ 1 it is violated. For each of these violating plays, the classically winning player has a winning response: • If they classically played a vertex on level j, then the other player simply plays on y(i+1)j and wins. • If they classically played on y(i+1)j , then the other player simply needs to play on any vertex on level j to win. • If the player made a quantum move that includes a vertex in a later level j, the other player can play on y(i+1)j , which will collapse all realizations where they didn’t play on level j. In those realizations, there are no moves left for the other player. • For any y(i+1)j chosen in a quantum move, the other player can continue playing normally, since those will collapse immediately upon making the next move. • Since play has been classical up to turn i, we have ruled out any quantum move for the previous xi selection beyond xi and x¯i . This quantum move will collapse immediately after the other player makes their next move. WLOG, the player can have the move collapse to a true assignment, and make a winning response to that move. Thus, quantum moves can’t be made, and the players, in order to win, need to follow a winning strategy for the corresponding QBF. Theorem 6.4.18. Quantum BiGraph Node Kayles is PSPACE-complete . Proof. Again, by Theorem 6.5.2, the game is in PSPACE, since the game is polynomially short. We use the proof from regular Quantum Node Kayles and color the vertices based on who should play on the vertex if the player is following the rules. Since the legal moves are still possible, players have 113 fewer deviating options, and if they do cheat, the legal response is still available, the reduction still holds. See Figure 6.6 for illustration. x1 x1 y1,4 y1,3 y1,2 x2T x2F x2T x2F y2,4 y2,3 x3T x3F x3T x3F C1 C2 C3 Level 1 Level 2 Level 3 Level 4 Figure 6.6: Example of the PSPACE reduction from QBF to Q-BiGraphNodeKayles, using the same formula as in Figure 6.5. Here Blue is going first as the True player. Theorem 6.4.19. Quantum Snort is PSPACE-complete. Proof. Again, by Theorem 6.5.2, the game is in PSPACE, since the game is polynomially short. We reduce from Bigraph Node Kayles. We simply create an extra vertex connected to all red vertices and put a red token on it, and then create a vertex with a blue token on it connected to all blue vertices. See Figure 6.7. 114 Thus, the game now acts as an instance of Bigraph Node Kayles. The Red player can only play on the Red nodes of Bigraph Node Kayles, and the Blue player can only play on the Blue nodes of Bigraph Node Kayles. a b c e d h f g a b c d e f g h Figure 6.7: Graph forQuantum BiGraphNodeKayles, followed by the result of the reduction toQuantum Snort on that graph. 6.5 A Complexity Barrier to a Quantum Leap While the introduction of quantum moves potentially increases the complexity of combinatorial games, we show in this section that essentially all PSPACE-solvable combinatorial games remain in PSPACE with strengthened quantum power. We prove that for a large family of combinatorial games—including many of PSPACE-complete ones such as QSAT, Generalized Geography, Avoid true, Node Kayles, along with many others—the quantum variants do not elevate the complexity of the game to beyond PSPACE. Particularly, this general upper bound applies to any PSPACE-solvable combinatorial games satisfying the following properties: Definition 6.5.1 (Polynomially Short Games). A combinatorial game is called a polynomially short game if for any of its instances, the height of its game tree is polynomial in the instance’s descriptive size. 115 For example, Generalzed Geography is a polynomially short game. Each Generalized Geography instance is defined by a graph G = (V, E), whose number of nodes, |V |, characterizes the instance’s descriptive size (because the number of (directed) edges |E| can be at most |V | 2 ). Note that Generalized Geography is also a polynomially short game because it has only |V | possible moves. Because no node of G can be selected twice in the game, the height of the game tree of Generalized Geograph on G is at most |V |. In contrast, Nim—although polynomial-time solvable—is not a polynomially short game, because for almost all Nim instances, the height of their game trees is exponential in their description sizes. Or, as an alternative phrasing, the games only have pseudo-polynomial height in their input description complexity. We can use standard techniques, with an algorithm generating and evaluating the game trees, to show that all polynomially short games are PSPACE-solvable. In this section, we will give the intuition for the following theorem by proving weaker variants of it. The proof of the full theorem can be found in [11]. Theorem 6.5.2 (Quantumized Polynomially Short Games is inPSPACE). For any polynomially short game R, and superposition width w polynomial in the descriptive complexity of R, RQ(w) is PSPACE-solvable. 6.5.1 Quantum Game Trees Are No Taller Before proving this theorem, let’s start with some high-level intuition of the proof. To establish the PSPACE-solvability—like in the classical case—we need to evaluate the game tree of the quantumized games in polynomial space. A key observation, in this context, is that quantum extension does not increase the height of the game tree: Proposition 6.5.3 (Height of Game Trees). For any game R, and superposition width w, the height of the game tree for RQ(w) is the same as that of R. 116 Proof. Every position associated with the game tree for RQ(w) is a superposition of realizations, each of which is defined by the interaction between the initial game position (which is classical) and a sequence of classical moves. Furthermore, the length of the sequence is equal to the length of the sequence of quantum moves in RQ(w) that defines the superposition. Therefore, each realization must appear as a position in the game tree for R at the same level. Thus, the game tree for RQ(w) can not be taller than the game tree for R. Proposition 6.5.3 is a crucial observation to our algorithmic proof. It implies that we can still apply the DFS-based framework—just like in the classical case—to evaluate the game trees, provided that we can effectively evaluate the outcome at each node in the game trees. Thus, one of the main technical barriers that we need to overcome is to evaluate, in polynomial space, these superpositions, which potentially could have an exponential number of realizations. 6.5.2 Quantum Node Kayles in PSPACE For clarity of presentation, we first focus our proof of Theorem 6.5.2 on the quantum extension of a concrete PSPACE-complete game, namely Node Kayles and with superposition width 2. This proof is in fact quite general, directly extendable to PSPACE-complete games including QSAT, Hex, Geography, Atropos, Avoid true, etc. We then extend our algorithmic construction to complete the proof for Theorem 6.5.2. Theorem 6.5.4 (Solvability of Quantum Node Kayles). Quantum Node Kayles with moves of superposition width 2 is polynomial-space solvable. Proof. Suppose R is defined by a graph G = (V, E) with V = [n] and |E| = m. Consider the game tree of RQ(2). Like its classical counterpart—Node Kayles—this game tree has height at most n (Proposition 6.5.3). Furthermore, each internal node in the game tree has at most O(n 2 ) children because there are at most n possible classical moves and n(n − 1)/2 possible quantum moves. However, unlike in Node 117 Kayles, some nodes in the game tree—either internal or at the leaf level—could be a superposition of a possibly exponential number of realizations, because the number of realizations is usually exponential in the number of quantum moves leading to that superposition. Thus, we cannot simply apply the traditional PSPACE postorder-traversal based (DFS) procedure to evaluate this game tree. In our polynomial-space algorithm for Quantum Node Kayles, we will use a tree-traversal for analyzing the superposition inside the traditional postorder-traversal based DFS evaluation of the game tree. We first prove the following: • Identifying Leaves: We can, in polynomial space, determine if a node in the game tree for Quantum Node Kayles is a leaf node or an internal node. • Outcomes at Leaves: We can, in polynomial space, evaluate the outcome at each leaf in the game tree forQuantum Node Kayles, even if its superposition has an exponential number of realizations. • Outcomes at Internal Nodes: We can, in polynomial space, evaluate the outcome class of each internal node in the game tree, given the outcome classes of its children (in the context of recursive evaluation of the game tree for Quantum Node Kayles, as in the standard PSPACE argument), even if its superposition has exponential number of realizations. Let’s first focus on the task of identifying leaves in the game tree for RQ(2). Our approach to outcome evaluations of super-polynomial superposition at both leaves and internal nodes is similar. We will address them afterwards. The key observation is the following: Although some superposition associated with the game tree for RQ(2) can have an exponential number of realizations, each superposition’s realizations can be organized by another tree of O(n) depth, which we will refer to as a QP-tree (short for “quantum position tree”), in order to differentiate them from the game tree itself. 118 Like its corresponding superposition in the game tree for RQ(2), each QP-tree is formulated by the sequence of “classical” and “quantum” moves defining the superposition. Each move in the sequence grows the QP-tree by one level. At level 0 is the root, corresponding to the starting position of RQ(2). For t = 1 onward, suppose mt is the t th move in the sequence. Then, mt defines level t positions from positions at level (t − 1) in the QP-tree. Consider a QP-tree node with position b at level (t − 1). There are two cases depending on whether mt is a classical or quantum move. 1. If mt = ⟨σt⟩ is classical and feasible for b, then mt creates a single child in the QP-tree, with position labeled b, σt . 2. If mt = ⟨σt,1|σt,2⟩ is a quantum move, then (a) if both σt,1 and σt,2 are feasible for b, then mt creates two children in the QP-tree, with positions labeled b, σt,1 and b, σt,2, respectively; (b) if one of them, say σt,2 is not feasible for b, then mt creates one child in the QP-tree, with position labeled b, σt,1; (c) if both σt,1 and σt,2 are infeasible, then no children are created. Based on this construction, the height of the QP-tree is bounded by the length—let’s denote it by T for now—of the sequence of moves defining it. The superposition is NULL if the QP-tree is empty at level T; otherwise, the superposition consists of all distinct positions at level T of the QP-tree.† Let’s now return to the game tree for RQ(2), in which a node v is a leaf if its associated superposition has no feasible classical or quantum move. If any realization in v’s associated superposition has a feasible move (either classical or quantum), then v is an internal node in the game tree. Thus, to determine if v’s superposition contains any realization with a feasible move, we apply a postorder-traversal based DFS procedure to evaluate its QP-tree. Suppose there are T moves in the sequence defining v. Then we use the DFS procedure to test if any of its position at level T has a feasible move. ‡ Upon encountering the first feasible move at level T, we can conclude that v is not a leaf in the game tree. If the procedure terminates †The QP-trees for other quantum flavors are slightly more subtle, and we will discuss them later in the section. ‡ In fact, it is sufficient to simply test whether any level-T position has a feasible classical move. The reason is the following: For any position, the existence of a feasible quantum move implies the existence of a feasible classical move. 119 without detecting any feasible move out of level T, then we conclude that v is a leaf in the game tree. The DFS procedure for analyzing QT-trees can be implemented in space polynomial in T. Because Node Kayles is impartial, Quantum Node Kayles is also impartial. Thus, each superposition in the game tree can only be in one of the two outcome classes: N or P. Because the superposition associated with every leaf of the game tree is in P, in polynomial space, we can both identify whether or not a node in the game tree for Quantum Node Kayles is a leaf, and determine the outcome class if the node is a leaf. Because this theorem only considers quantum moves with superposition width 2, each node in the game tree has only a polynomial number of possible feasible moves. Thus, we can, in polynomial space, determine the outcome class of each superposition in the game tree, and particularly the outcome class of the starting position (associated with the root of the game tree) by running the post-order-traversal DFS procedure with our polynomial-space QP-tree evaluation algorithm. Therefore, Quantum Node Kayles with superposition width 2 is polynomial-space solvable. 120 Chapter 7 Homecoming This chapter contains three major sections. In Section 7.1, we demonstrate how many of our results don’t exist in isolation, and have interesting corollaries when combined. In Section 7.2, we examine the utility of our work by looking how it simplifies past research questions, and how it can provide answers for future ones. Finally, in Section 7.3, we present some open problems for the future involving computation of combinatorial games. 7.1 Putting things together Something worth highlighting in particular is that our results are not unrelated to one another, and in fact, when combined, produce several interesting views, and even additional results and corollaries. 7.1.1 Impartial Games and the Oracles But beyond these individual narratives, we also have results that synergize to create interesting insights. First, we will examine combined insights between Chapters 3 and 5. Chapter 3 reveals three facts: 1. Undirected Geography is PSPACE-complete to classify the nimber (even distinguishing between any pair of nimbers). 121 2. For any Grundy value ∗k, one, in polynomial time, can output an instance of Undirected Geography with nimber ∗k, where the size of the created position is no more than poly(k). 3. Every polynomially short impartial game, including a disjunctive sum of polynomially short games, can be reduced to an instance of Directed Geography in polynomial time while preserving its nimber. But note a corollary of the third is: Corollary 7.1.1. Given two instances of Directed Geography, one may produce a third instance of Directed Geography which has a nim-value equal to nim-sum of the input instances. In other words, Directed Geography has a polynomial time implementation of an addition oracle. But through Theorem 5.1.1, we obtain the following corollary of the first two points: Corollary 7.1.2. The problem of taking two instances of Undirected Geography and producing a third instance of Undirected Geography which has a nim-value equal to nim-sum of the input instances is PSPACEcomplete. Proof. We know identifying the game-value of Undirected Geography is PSPACE-complete. However, identifying the outcome class is in P, and the second chapter 3 fact listed above gives us an implementation of the value-oracle. Thus, by the contrapositive of Theorem 5.1.1, since identifying the game-value is PSPACE-complete, at least one of the three oracles must be as well. Since every oracle other than the addition oracle is in P, the addition oracle must be PSPACE-complete. This leads to an interesting consequence: Directed Geography has a polynomial time addition oracle, while Undirected Geography does not. Intuitively, this seems rather puzzling, since Directed Geography is a generalization of Undirected Geography, so naturally, one should assume that anything possible for Directed Geography should also be possible with Undirected Geography. 122 The answer to this conundrum is that the addition oracle acts as a sort of reduction. And indeed, performing reductions to more difficult games is easier, as the extra difficulty actually gives additional power to embed other problems with. So, summation oracles are easier to construct for harder games rather than easier ones. 7.1.2 Hardness and the Oracles An interesting corollary emerges from the combination of the main theorems of Chapters 4 and 5. Corollary 7.1.3. For any game which has an addition oracle, and has a known value generation oracle which includes all superstars, identifying the outcome class must be NP-hard. Proof. With the two oracles given, we can construct the setup given by the Chapter 4 theorem, using the addition and value oracles. This is particularly useful since, as we will discussion in Section 7.2, this gives the opportunity to obtain “automatic” hardness results from mathematical results about the habitat of games. In a similar vein, we can use Morris’ Theorem [41] to get stronger hardness results for certain games: Corollary 7.1.4. Any universal combinatorial game with a value oracle and addition oracle is PSPACE-hard 7.1.3 Quantum CGT and other sections Something notable about the intersection of our results about Undirected Geography when paired with our results aboutQuantum Undirected Geography is that the problems of identifying grundy values and the outcome class for the misère variant are both intractable for Undirected Geography. This seems to demonstrate that the game is “just barely” in P, and small perturbation to the problem make it intractable. Despite this intuitive notion, when starting with a classical position inQuantum Undirected Geography, the game remains in P. This, in some sense, casts some doubt on the complexity impact of quantum moves 123 when starting from a classical position, even if we don’t have explicit algorithms for the other games we looked at. 7.2 Implications and Examples The results in this dissertation can impact future work. In particular, the work in Chapter 5, and the corollaries from this chapter, provide a strong foundation for future results. To demonstrate this, we will: 1. Discuss historical hardness proofs which could be simplified with our proofs. 2. Demonstrate how advances in CGT could automatically result in further computational results. 7.2.1 Historical Results Two games have been demonstrated to be universal combinatorial games: Konane [16] and TurningTiles [52]. Let’s first focus our attention on TurningTiles. The game was shown to be a universal combinatorial game before the PSPACE-hardness result was proven [59]. And the proof also included (while not the direct intention) a summation oracle. While the paper for Konane came later than the hardness result [34], the proof of universality is similar in including a summation oracle. So, with our results here, we render these hardness proofs unneeded. Simply by embedding Morris’[41] or Wolfe’s[55] constructions using the summation and value oracles used in the proofs, one can transform that historically manual process of demonstrating hardness into an automatic one. Universal rulesets are automatically PSPACEcomplete, so long as the proofs for these rulesets are constructive. In other words, the mathematical results automatically imply computational ones. 124 7.2.2 Future Results Any game that in the future is shown to be universal is PSPACE-hard. But, perhaps more importantly, a lower standard of just embedding superstars is sufficient to demonstrate NP-hardness. This is particularly relevant since there are many games which have no known hardness results which are candidates to being shown NP-hard. All that is needed is for mathematicians to be able to demonstrate superstar constructability. For example, Clobbers is a well-known combinatorial game, which has competitive tournaments, with no known hardness results when played on a grid. However, all nimbers, and many infinitesimals, have known positions in Clobber [18]. Any result pushing up “just one level” from nimbers will be sufficient to show the grid variant to be NP-hard, since the summation oracle is trivial for the game. But beyond this, being able to classify all possible rulesets generally gives us an equivalence between game value and outcome class, so long as the summation oracle is also in P (which is usually trivial to easy). Thus, for a large set of combinatorial games, such as Domineering, which don’t have known results (despite lots of effort), there is a possible new paradigm to tackle them: demonstrate that certain game values are hard to compute. Then, the computational question is redirected into a mathematical one. Or, people aiming to prove hardness of identifying game values can construct sums where the exact value of the original game determines the outcome class. 7.3 Future Directions For each section, we examine some open problems. 7.3.1 Impartial Games In Section 3.7, we explored how every polynomially short impartial game can be reduced to Geography while preserving the nimber. And indeed, Geography is a rather nice and flexible game to work with for 125 reductions, which is part of the reason it was one of the first ever PSPACE-reductions performed, and is regularly used in reductions to new games. However, it is currently unclear how unique Geography is in terms of it’s Sprague-Grundy completeness. As previously discussed, we know Undirected Geography is not Sprague-Grundy complete, due to the summation oracle being PSPACE-hard to implement for it. Additionally, for games such as Atropos[15], since they have a finite maximum nimber (∗6 for Atropos), they cannot be Sprague-Grundy Complete. So, at the very least, the set of Sprague-Grundy complete games contains several less games than impartial PSPACE-hard games. As such, we have the following natural question: Open Question 1. Are there games other than Directed Geography which are Sprague-Grundy Complete? Next, recall how we demonstrated how distinguishing between any nimber for Undirected Geography was PSPACE-complete. To the best of our knowledge, it is currently unique in being the only known game that is solvable for the winnability problem, but where identifying its nimber is hard. And indeed, due to that property, the game exhibits several other interesting properties, such has having the summation oracle be intractable (which is generally simple for most combinatorial games). It also gives a unique ability for showing hardness results in other games, since instead of using Directed Geography in reductions, researchers can use Undirected Geography with two tokens, where one of them is in any winning game. As such, to get similar useful games for study, it would be useful to gain more hardness results for “easy” games with a twist. Thus, we get the question: Open Question 2. Are there games other than Undirected Geography where identifying the outcome class is polynomial time solvable, but identifying the nimber is intractable? 126 7.3.2 Tepid Games An immediate follow up to our work in Chapter 4, and indeed, when working within the CGT community, what we were immediately asked, is as follows: Open Question 3. Are there other families of infinitesimal games beyond superstars where determining the winner on a disjunctive sum is intractable? In particular, the question of whether a sum of Flowers (which, as mentioned in chapter 4, Superstar is a generalization of) is intractable has been an open problem for many years. It is of great interest to the CGT community, many of whom have attempted hardness results for the game for several years. But really, any family would be useful, especially when combined with the oracle model to give “automatic” hardness proofs. Any additional results can greatly increase the number of games known hard. Additionally, given our results only give NP-hardness, an obvious follow up is the following: Open Question 4. Can one show a sum of superstars to be PSPACE-hard? This can be done either by elevating EPMX to be PSACE-hard, or identifying another game to reduce from. Of course, given that our reduction is rather close to a one-to-one embedding of left-0 and right-0 games, one would presumably have to employ both-0 games into the reduction. 7.3.3 Game value computation Open Question 5. Can mean value be computed with the oracle model? The answer to this seems to be close to yes, but some annoying issues remain about when to terminate, unless one is allowed to look exactly one level down in the game tree. Essentially, one can just apply the definition and recursively add the game to itself, and use Algorithm 2 from Chapter 5 to identify the nearest number. Except, it may not be clear if one is approaching a certain number, and it seems the only way to check is to check the values of the options of the game, violating the oracle model. 127 Open Question 6. Can temperature be intractable to compute? This is especially interesting as temperature is often a good heuristic to use for playing games, so demonstrating that it can’t be computed adds another layer to difficulty of an already hard game. Or, alternatively, it would be especially interesting if it could be shown to be hard even for a polynomial time solvable game. And of course, on the other side, we have the goal of actually finding temperature. Open Question 7. Can Temperature be obtained via the oracle model? Or, can it be intractable to compute even with the oracles? 7.3.4 Quantum Combinatorial Games While in our quantum game chapter, we explored many combinations of how complexity could either increase or decrease with the introduction of quantum moves, there is still one interesting case that we were never able to find: Open Question 8. Can any game that is classically polynomial time solvable become intractable in the quantum model with a classical start? If we were able to get this, we could show that the introduction of quantum moves to a “normal” starting position could actually make games hard. But so far, all we have been able to show is that it doesn’t change the complexity, or makes it easier, which is counter to one’s intuition on what effect the introduction of the additional complexity should have. 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Abstract (if available)

Abstract People have been playing games since before written history, and many of the earliest games were combinatorial games, that is to say, games of perfect information and no chance. This type of game is still widely played today, and many popular games of this type, such as Chess and Go, are some of the most studied games of all time. This work proposed revolves around a game-independent systemic study of these games. More specifically, it involves computational properties involving evaluating mathematical analysis tools for combinatorial games, such as Grundy values and confusion intervals, as well as identifying what can be determined about these games using simple oracle models.

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Creator Ferland, Matthew Thomas (author)

Core Title Exploring the computational frontier of combinatorial games

School Andrew and Erna Viterbi School of Engineering

Degree Doctor of Philosophy

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Degree Conferral Date 2024-08

Publication Date 07/12/2024

Defense Date 05/02/2024

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