The connection between the $PQ$ penny flip game and the dihedral groups

This paper is inspired by the PQ penny flip game. It employs group-theoretic concepts to study the original game and also its possible extensions. We show that the PQ penny flip game can be associated with the dihedral group $D_{8}$. We prove that within $D_{8}$ there exist precisely two classes of winning strategies for Q. We establish that there are precisely two different sequences of states that can guaranteed Q's win with probability $1.0$. We also show that the game can be played in the all dihedral groups $D_{8 n}$, $n \geq 1$, with any significant change. We examine what happens when Q can draw his moves from the entire $U(2)$ and we conclude that again, there are exactly two classes of winning strategies for Q, each class containing now an infinite number of equivalent strategies, but all of them send the coin through the same sequence of states as before. Finally, we consider general extensions of the game with the quantum player having $U(2)$ at his disposal. We prove that for Q to surely win against Picard, he must make both the first and the last move.


Introduction
It is rather unnecessary to stress the importance of game theory. It has been extensively used for decades now to help researchers and practitioners make sense of situations involving conflict, competition, and cooperation. The abstraction of players who antagonize each other in a specified framework by devising elaborate strategies has been employed in the fields of economics, political and social sciences, biology, and, naturally, to computer science. Game theorists have developed an enormous technical machinery for the quantitative assessment of the players' strategies and their payoffs. Of particular significance is the assumption that the players are rational, which means that they seek to maximize their payoffs. In this paper we use only very basic and easy to grasp notions from game theory. These can be found in all standard textbooks, such as [1], [2], and [3]. The emergence of the quantum era in information and computation also brought about the creation of the field of quantum game theory. This recent field is devoted to the study of classical games in the quantum setting, giving an exciting new perspective and results that are beyond the grasp of the classical realm.

Related work
The year 1999 was an important milestone for the creation of the field of quantum games. In that year two influential works were published. In a seminal paper Meyer [4] introduced the PQ penny flip game, which can be considered the quantum analogue of the classical penny flip game. The other influential work from 1999 was by Eisert et al. in [5]. There the authors presented a novel technique, known now as the Eisert-Wilkens-Lewenstein protocol, that has gained wide acceptance in the field.
In Meyer's PQ penny flip game, the two players are the famous tv characters Picard and Q from the tv series Star Trek. They consecutively "toss" a quantum coin and if at the end of the game the coin is found heads up Q wins, otherwise Picard wins. There is a metaphor behind the two players:

Contribution
This paper is inspired by previous research on the PQ penny flip game. Its novelty is mainly attributed to its use of group-theoretic concepts to study the original game and its possible extensions. We show for the first time, to the best of our knowledge, that the original PQ penny flip game can be associated with the dihedral group D 8 . Interpreting the game in terms of stabilizers and fixed sets, which are basic but helpful group notions, enables us to easily explain and replicate Q's strategy. First, we prove that within D 8 there exist precisely two classes of winning strategies for Q. Each class contains many different strategies, but all these strategies are equivalent in the sense that they drive the coin through the same sequence of states. We establish for the first time in the literature that there are precisely two different sequences of states that can guaranteed Q's win with probability 1.0. We then proceed to show that the same game can be played in the all dihedral groups D 8n , n ≥ 1, with any significant change in the winning strategies of Q. This allows us to conclude that in a way the smallest group that captures the essence of the game is D 8 . Subsequently, we examine what happens when Q can draw his moves from the entire U (2). We provide the definitive answer that, perhaps surprisingly, the situation remains the same. Again, there are exactly two classes of winning strategies for Q, each class containing now an infinite number of equivalent strategies, but all of them send the coin through the same sequence of states as before. In a final analysis, the original PQ penny flip game can be succinctly summarized by saying that there are precisely two paths of states that lead to Q's win and, of course, no path that leads to Picard's win. Finally, we consider general extensions of the game without any restrictions in the number of rounds and with the quantum player having U (2) at his disposal. Our examination, will uncover a very important fact, namely that for the quantum player to surely win against the classical player the tremendous advantage of in terms of available quantum actions is not enough. Q must also make both the first and the last move, or else he is not certain to win.

Organization
The paper is structured as follows. Section 1 sets the stage and gives the most relevant references. Section 2 introduces the notation and terminology used in this article. Section 3 proves the connection of the game with the dihedral group D 8 and Section 4 analyzes Q's strategy in terms of group concepts. Sections 5 and 6 contain the most important results of this work, and, finally, Section 7 provides a summary of our conclusions and sketches some ideas for possible future work.

The P Q penny flip game
In what is now regarded as a landmark paper [4], Meyer defined the penny flip game between the famous television personas Picard and Q from the TV series Star Trek. From now on for brevity we shall refer to it as the P QG (the Picard -Q game). This game is much more than a coin flipping game; its importance lies in the fact that it demonstrates the advantage of quantum strategies over classical strategies. The human player Picard can only employ classical strategies, while the quantum player, Q, is capable of using quantum strategies. This asymmetry is the reason why, no matter what Picard does, Q always wins with probability 1.0. Picard is confined to just the two classical moves available in a 2-dimensional system: he can either do nothing, or he can flip the coin. Doing nothing means that the coin remains in its current state, while flipping the coin changes its state from heads to tails or vice versa. Q's advantage stems from the fact that he can potentially choose from an infinite pool of allowable moves; the only obvious restriction being that his move must be represented by a unitary operator. The game begins with the coin heads up and the two players act on the coin following a predetermined order. Q acts first, then Picard and last Q again. If, after Q's last action, the coin is found heads up, then Q wins. If the coin is found tails up, then Picard wins.
In the context of the P QG and its extensions, it is convenient to employ the terminology outlined in Definition 2.1, adapted from [1] and [3]. Informally, the word strategy implies a rational plan on behalf of each player. This plan ultimately consists of actions, or moves that the player makes as the game evolves. • A strategy is a function that associates an admissible action to every round that the player makes a move. It is convenient to represent strategies as finite sequences of moves from the player's repertoire.
• A strategy σ P for Picard is a winning strategy if for every strategy σ Q of Q, Picard wins the game with probability 1.0.
• Symmetrically, a strategy σ Q for Q is a winning strategy if for every strategy σ P of Picard, Q wins the game with probability 1.0.
• A strategy for Picard or Q is dominated if there exists another strategy that has greater probability to win for every strategy of the other player. A strategy that dominates all other strategies is called dominant.
Of course, in the original P QG, Picard's strategy is just one move, e.g., (F ). Q's strategy on the other hand is a sequence of two moves: (H, H). Moreover, a winning strategy for Q is also a dominant strategy. Meyer proved that the use of the Hadamard operator constitutes a winning strategy for Q. If Q uses the Hadamard operator, he will win with probability 1.0, irrespective of Picard's moves.
In more technical terms, the game takes place in the 2-dimensional complex Hilbert space H 2 . The computational basis of H 2 is denoted by B and consists of the kets |0 = 1 0 and |1 = 0 1 : Typically, |0 and |1 capture the state of the coin being heads up or tails up, respectively. Picard's moves do nothing and flip the coin correspond to the identity operator I and the flip operator F , respectively. As already mentioned, Q's winning strategy is the Hadamard operator H. In H 2 the players' moves are represented by the following 2 × 2 matrices: F is of course one of the famous Pauli matrices, frequently denoted by σ x or σ 1 . In this work we approach the dynamics of the P QG and its extensions by examining the actions available to the players. Consider for instance the composition F H, that can arise in the P QG when Picard replies with F to Q's H. A simple matrix multiplication shows that 3) The operational space M ⋆ contains not only the above operator (2.3), but also every operator that results from a finite composition of the moves in M . Most of them are not realized in the actual P QG because its duration is just 3 rounds; however, this set will provide insight when we consider various extensions of the P QG.
Definition 2.2 can be generalized as follows: Definition 2.3. Given any game V (e.g., an extension of the original P QG game), for which the set of moves is M V , its operational space is M ⋆ V .

Dihedral groups
For the completeness of our presentation we shall recall a few definitions and concepts from group theory. The notation and definitions are based on standard textbooks such as [22] and [23].
Definition 2.4 (Group). A set G equipped with a binary operation • is a group under • if it satisfies the following properties.
1. There exists an element 1 ∈ G, the identity of G, such that 1 • g = g • 1 = g for all g ∈ G.
2. For every g ∈ G there exists an element in G, called the inverse of g and denoted by g −1 , such that 3. For all f, g, h ∈ G : , i.e., the associative property holds.
The number of elements of the group G is called the order of G and is denoted by |G|. It is customary to employ the following notation regarding powers of an arbitrary element g of a group G.
We shall omit the symbol • of the binary operation, particularly in view of the fact that in many occasions the group elements will be represented by 2 × 2 matrices and the operation • will be matrix multiplication. Hence, instead of writing f • g, we will simply use the juxtaposition of the two elements f g.
The groups that capture the symmetries of regular polygons are called dihedral groups. We clarify that by regular polygon it is understood that all the sides of the polygon have the same length and all the interior angles are equal. Furthermore, we assume that the center of the regular polygon is located at the origin of the plane. Definition 2.5 (Dihedral groups). The group of symmetries of the regular n-gon, where n ≥ 3, is called the dihedral group of order 2n and is denoted by D n . 1 Please note that from now on when we refer to an arbitrary dihedral group D n we shall assume that n ≥ 3. The group operation is composition of symmetries, i.e., composition of rotations and reflections. The 2n symmetries of a regular n-gon, where n ≥ 3, can be categorized as follows.
• There are n rotational symmetries. These are the rotations about the center of the n-gon by 2πk n , with k taking the values 0, 1, . . . , n − 1. Figures 1 and 3 show the 7 and 8 rotational symmetries of the regular heptagon and octagon, respectively.
• There are also n reflection symmetries.
-If n is odd these are the reflections in the lines defined by a vertex and the center of the regular n-gon. As an example, see Figure 2 depicting the reflection symmetries of the regular heptagon.
-If n is even, these are n 2 reflections in the lines through opposite vertices and n 2 reflections in the lines passing through midpoints of opposite faces. An example that will play an important role in the rest of our study is given in Figure 4, showing the reflection symmetries of the regular octagon.
By fixing a vertex of the regular n-gon to lie on the x-axis (such a vertex 1 in Figures 2 and 4) and the center of the n-gon at the origin of the plane, we may surmise that the n reflection symmetries correspond to lines through the origin making an angle πk n with the positive x-axis, with k taking the values 0, 1, . . . , n − 1.    The general dihedral group contains the following 2n elements (for details the interested reader may consult [22], [23] or [24]) D n = {1, r, r 2 , . . . , r n−1 , s, rs, r 2 s, . . . , r n−1 s} , (2.4) where r is the rotation by 2π n and s is any reflection. It is evident that each element of D n can be uniquely written as r k s l for some k, 0 ≤ k ≤ n − 1, and l, where l = 0 or 1. Elements 1, r, r 2 , . . . , r n−1 are rotations, i.e., r k is the rotation by 2πk n , and elements s, rs, r 2 s, . . . , r n−1 s are reflections. In particular, the dihedral group D 8 contains the 16 elements D 8 = {1, r, r 2 , . . . , r 7 , s, rs, r 2 s, . . . , r 7 s} , (2.5) where r is the rotation by 2π 8 and s is any reflection. We remark that, referring to Figure 4, s can be taken to be the reflection in the line passing through the vertices 1 and 5, or the reflection in the line passing through the midpoints A and A ′ , or the reflection in the line passing through the vertices 2 and 6, or any of the remaining reflections. Definition 2.6 (Generators). Given a subset X of a group G, the smallest subgroup of G that contains X is denoted by X . The elements of X are called generators for X .
When X is finite, i.e., X = {x 1 , . . . , x n }, as will be the case in this work, it is customary to simply write x 1 , . . . , x n .
A typical way to specify a group is by giving a presentation for the group. This amounts to using generators and relations, with the understanding that all group elements can be constructed as products of powers of the generators, and that the relations are equations involving the generators and the group identity. The following presentation of D n is especially convenient for our analysis: This presentation demonstrates that D n can be generated by two reflections s and t. It appears in [22] and [25], among others, where it is clarified that D n can be generated by two reflections s, t in adjacent axes of symmetry passing though the origin and intersecting in an angle π n . In this case, the product st is a rotation through an angle of ± 2π n . We note though that presentations are not unique. For instance, one other widely used presentation for the dihedral groups is D n = r, s | r n = s 2 = 1, rs = sr −1 .
3 The connection between P QG and D 8

Matrix representations of rotations and reflections
A useful and quite common way to represent rotations and reflections in the plane is to use 2 × 2 matrices. Such matrices, which are often called rotators and reflectors, can be conveniently written in a form that is easy to recognize and manipulate (see [24], [26] and [27] for more details). A rotator representing a counterclockwise rotation through an angle ϕ about the origin is denoted by R ϕ and, similarly, a reflector about a line through the origin that makes an angle ϕ with the positive x-axis is denoted by S ϕ . R ϕ and S ϕ are given by the formulas shown below. Please note that we use capital R and capital S to designate these 2 × 2 matrices, in order to avoid any confusion with the elements of the dihedral group that are denoted by small r and s.
It is now quite straightforward to see that the F and H operators can be written as follows. This form reveals that both are reflectors: F reflects about a line that makes an angle π 4 with the positive x-axis. To be exact this is the line passing through the vertices 2 and 6 in Figure 4. Likewise, H reflects about a line that makes an angle π 8 with the positive x-axis, which is the line passing through the midpoints A and A ′ in Figure 4. Hence, their axes of symmetry intersect in an angle π 8 , as shown in Figure 4. Moreover, their product F H, which is given in (2.3), is just the rotator R 2π 8 , as can be verified by employing formula (3.1). Therefore, by invoking the presentation (P 1 ), associating s to F , and t to H, or vice versa, it becomes evident that F and H generate the dihedral group D 8 . This conclusion is stated as Theorem 3.1.
Please note that in an effort to enhance the readability of this paper, without worrying about the technical details, we have relocated all the proofs in the Appendix.  The above result tells us that Picard and Q's moves generate the group D 8 . This has important ramifications. As long as the two players are allowed to use only the aforementioned actions, no matter what specific game they play, the game will take place in the D 8 group. Every conceivable composition of moves by the players is just an element of D 8 . Therefore, although the rules of the game can change dramatically, e.g., the players' turn, the number of rounds, etc., the available moves will always be elements of D 8 .
Actually, it is a well-known fact that every element of the dihedral group D n can be represented by 2×2 matrices of the form shown in (3.1) and (3.2). One such representation is given below. In the literature it is usually referred to as the standard representation of D n . We remark that, in more technical terms, this is a faithful irreducible representation of dimension 2. To clear any potential misunderstanding, let us emphasize that in the standard representation s corresponds to the reflection in the line passing through the vertices 1 and 5, i.e., the x-axis of Figure 4.  where 0 ≤ k ≤ n − 1.

Orbits and stabilizers
Definition 3.2 (Group action). Let G be a group and let X be a nonempty set. A group action ⋆ of G on X is a function ⋆ : G × X → X that satisfies the following properties.
Under the standard representation of D n , its action on a state of the quantum coin is computed by simply multiplying every matrix corresponding to an element of D n with the ket describing the state of the coin. In what follows, in addition to speaking about an action, we shall occasionally say that G acts on X. Moreover, we shall just write gx instead of g ⋆ x, since the action we study in this paper is that of operators on kets, or, if you prefer, of matrix-vector multiplication.

Definition 3.3 (Orbits and stabilizers).
Suppose that a group G of linear operators, or their corresponding matrix representations, acts on a nonempty set of kets X. We make the next definitions, always taking into account that all kets of the form e iθ |ψ , with θ ∈ R, represent ket |ψ .
1. Given x ∈ X, the G-orbit of x, denoted by G ⋆ x, is the set {g ⋆ x ∈ X : g ∈ G}.
2. Given S ⊂ X, the G-orbit of S, denoted by G ⋆ S, is the union of the orbits G ⋆ x, for each x ∈ S 3. Given x ∈ X, the stabilizer of x, denoted by G(x), is the set {g ∈ G : g ⋆ x = x}.
4. Given g ∈ G, the fixed set of g, denoted by F ix(g), is the set {x ∈ X : g ⋆ x = x}. 5. Given X ⊂ G, the fixed set of X, denoted by F ix(X), is the intersection of the fixed sets F ix(g), for each g ∈ X.
In the next section we shall employ these tools in the analysis of Q's strategy to gain insight from a group theoretic perspective.

Analyzing Q's strategy in terms of groups
We proceed now to interpret the P QG using the aforementioned groups concepts. It will be helpful to utilize the following abbreviations, which are very common in the literature.
Let us first see what is the effect of the action of D 8 on the computational basis B. One easy way to do this is geometrically by consulting Figures 3 and 4 to see where vertices 1 and 3 are sent when being acted upon by the elements of D 8 . Alternatively, one can arrive at the same result algebraically simply by multiplying the matrix representation of every member of D 8 with |0 and |1 . The representations of the elements of D 8 can be readily found by setting n = 8 in the more general formulas (3.7) and (3.8). In any event, for future reference we summarize the action of D 8 on the computational basis B in Proposition 4.1 (recall that e iθ |ψ , with θ ∈ R, and |ψ represent the same state). 1. |0 and |1 have the same orbit: 2. The orbit of B is: Q's first move aims to drive the coin into the state Definition 3.3 is helpful in understanding the advantage of Q's move in terms of group notions. In particular, there are certain elements of D 8 whose action on |+ has no effect whatsoever and which constitute the stabilizer of |+ . These can be easily found either geometrically or algebraically, and are listed in Proposition 4.2. • The stabilizers of |0 and |1 in D 8 are • The stabilizers of |+ and |− are In a complementary manner, we may surmise that Picard's set of moves fixes specific states in H 2 , as demonstrated in Proposition 4.3.    Proposition 4.2 tells us that Picard's set of moves is a subset of D 8 (|+ ) and Proposition 4.3 completes the picture by revealing that ket |+ is among those that are fixed by Picard's moves. Thus, he is completely powerless to change the state |+ of the coin. Under this perspective the progression of the P QG can be abstractly described as by the following "algorithm." Algorithm 1: Q's Winning strategy in the original P QG 1 Q's first move sends the coin to an intermediate target state (in the actual game it happens to be |+ ) that satisfies the following property: this state is fixed by Picard's moves or, equivalently, all of Picard's moves belong to the stabilizer of this state (in the actual game I, F ∈ D 8 (|+ )). 2 Picard acts on the coin, but no matter which move he makes, the quantum coin remains in the same state.  Picard symbolizes the classical player and as such it is quite appropriate to assume that his repertoire is the set M P = {I, F }. This set is also a group, in particular the Z 2 group of two elements 2 . In the rest of this paper we shall always assume that the classical player can only make use of these two actions. In the coming sections we shall employ Algorithm 1 to discover winning strategies for Q in more general situations.

Enlarging the operational space of the game
As we begin this section let us recall that the operational space of the original P QG is indeed a group, and, in particular, the dihedral group D 8 , as established by Theorem 3.1. In this section we shall progressively enlarge the ambient group of the P QG and analyze Q's winning strategies. Our analysis is guided by the belief that the essence of the original P QG is the sharp distinction between the classical and the quantum player. From this perspective, our subsequent investigation relies on the following two assumptions.
1. Picard, who embodies the classical player, can flip the coin. If he is deprived of this ability, then the resulting game becomes trivial and meaningless. He should not be able to do more than that, as this would endow him with quantum capabilities. Formally, we express this by specifying: 2. Q, who stands for the quantum player, must exhibit quantumness. Thus, at least one of his actions must lie outside the classical realm. In more technical terms, his repertoire M Q must contain at least one operator from U (2) other than I and F .
Under the above assumptions, we may state the following properties that are quite general, as they are satisfied by every winning strategy of Q, no matter what the ambient group is. Therefore, we shall invoke these properties when we are examining much larger dihedral groups and the unitary group U (2).
Theorem 5.1 (Characteristic properties of winning strategies). If (A 1 , A 2 ) is a winning strategy for Q, then: We introduce the notion of equivalent strategies in order to simplify the classification of winning strategies. We consider two strategies to be equivalent if, when acting on the same initial state of the coin, they produce the same sequence of states. In view of the extension of the original game that will be undertaken in Section, the next definition is general enough to deal with strategies for games with more than three number of rounds.
Definition 5.1 (Equivalent strategies). Let σ = (A 1 , . . . , A r ) and σ ′ = (A ′ 1 , . . . , A ′ r ) be two strategies of the same player, and let |q 0 be the initial state of the coin. We say that σ and σ ′ are equivalent with respect to |q 0 , denoted by σ ∼ σ ′ , if For example, Q's strategies (H, H) and (R 2π 8 , R 14π 8 ) are equivalent because they send the coin from state |0 first to |+ and then back to |0 . It is obvious that ∼ is an equivalence relation that partitions the set of strategies into equivalence classes of strategies.
Clearly, the state path τ [σ] is well-defined and unique for each class [σ]. The equivalence class [(H, H)] contains 16 strategies, as will be explained in Example 5.1, and the corresponding state path is (|0 , |+ , |0 ).

Inside D 8
Before delving into other groups, we examine the case where Q can chose his moves from the entire D 8 group, i.e., The following Example 5.1 will be instructive.
Example 5.1. In this example, we shall apply Algorithm 1 to study all winning strategies of Q in the original P QG. Let (A 1 , A 2 ) be Q's first and second move in a winning strategy. After Q's first move the quantum coin will in one of the states in the orbit • Let us first establish that if Q leaves the coin at state |0 , or sends it to state |1 , then he will not be able to win with probability 1.0. To see this more clearly, let us recall that, by Theorem The same reasoning shows that if Q's first move sends the coin to state |1 , then he will not be able to win with probability 1.0.
• and R 6π 8 . This means that Q has 16 different winning strategies, which, in view of Definition 5.1, are equivalent. Thus, they constitute one equivalence class of winning strategies. Strategy (H, H) is a representative of this class, but any other strategy would also do. For this class the corresponding state path is (|0 , |+ , |0 ).
• Algorithm 1 enables us to discover one more winning strategy for Q. Q has another option, which is to drive the coin to state |− . This can also be achieved with 4 different ways: S 7π 8 , R 14π 8 , S 3π 8 or R 6π 8 . Picard cannot change this state either because |− is fixed by I and F , according to (4.9). During the final round Q has the opportunity to send the coin back to the |0 state with 4 different ways: S 7π 8 , R 2π 8 , S 3π 8 or R 10π 8 . Hence, Q has 16 more winning strategies, which are equivalent. They make the second equivalence class of winning strategies and any one of them, e.g., (S 7π 8 , S 7π 8 ) can be its representative. For this class, the corresponding state path is (|0 , |− , |0 ).
• Picard, unfortunately for him, has no winning strategy.
According to Definition 2.1, for Q a winning strategy is also a dominant strategy. Hence, Q has precisely two classes of winning and dominant strategies, each containing 16 individual strategies. These two classes correspond to exactly the 2 path states (|0 , |+ , |0 ) and (|0 , |− , |0 ). ⊳ Table 1 and Figure 6 summarize these results.  Figure 6: This figure depicts two different winning strategies for Q that represent the two winning strategy classes, as well as the corresponding path states.
i.e., the ambient group of the P QG is D 8 , then the following hold.
1. Q has exactly two classes of winning and dominant strategies, each containing 16 equivalent strategies: The winning state paths corresponding to C + and C − are 3. Picard has no winning strategy.

5.2
The smaller dihedral groups D 3 , D 4 , D 5 , D 6 and D 7 We may ask whether any of the smaller dihedral groups D 3 , D 4 , D 5 , D 6 and D 7 can be an appropriate operational space for the P QG. The answer is no for the reasons outlined below.
• D 3 , D 5 , D 6 and D 7 do not contain the reflection F . This can be verified by comparing formula (3.3) with formula (3.8) for n = 3, 5, 6 and 7 and k = 1, . . . , n − 1. We have assumed that Picard, the classical player, must be able to flip the coin, as emphasized in (A 1 ).
• D 4 does contain the reflection F . However, the orbit D 4 ⋆ B is {|0 , |1 }. This means that Q can only flip the coin from heads to tails or vice versa. If M Q = D 4 , then the P QG degenerates to the classical coin tossing game. Q is no longer a quantum entity and, as explained in Example 5.1, no longer possesses a winning strategy. From this perspective, it becomes meaningless to play the P QG in D 4 .
These conclusions are contained in Table 2 for easy reference.
Yes (classical coin tossing) No No Therefore, if we accept that the classical player should, at the very least, be able to flip the coin in order to have a nontrivial game, and that the quantum player must exhibit quantumness, then the smallest dihedral group for the P QG is D 8 . This fact is stated as Theorem 5.3.
Theorem 5.3 (The smallest dihedral group for the P QG is D 8 ). D 8 is the smallest of the dihedral groups such that P QG can be meaningful played and in which Q has a quantum winning strategy.

The dihedral groups
The previous subsection demonstrated that the smallest meaningful group for the P QG is D 8 . This subsection examines what happens if we allow Q to choose from a larger repertoire, and, more specifically, if we assume that Let us as first make the helpful observation that when n is odd, then D n does not contain F , which is stated as Proposition 5.4. This result enables us to exclude these groups from now on when considering larger groups where the P QG can be successfully played.
Another useful result about the orbits of B in general dihedral groups is contained in Theorem 5.5.
Theorem 5.5 (The action of D n on B). The action of the general dihedral group D n , n ≥ 3, on the computational basis B depends on whether n is a multiple of 4 or n is even but not a multiple of 4. Specifically, 1. if n is a multiple of 4, then the action of the dihedral group D n on the computational basis B is 2. if n is even but not a multiple of 4, then the action of the dihedral group D n on the computational basis B is In this more complex setting we may resort to Algorithm 1 to establish under what conditions Q still possesses winning strategies and, if so, which are these. This is facilitated by the next Theorem 5.6, which explains what happens to the fixed set of {I, F } in D n .
Theorem 5.6 (The fixed set of {I, F } in D n ). When the general dihedral group D n , n ≥ 3, acts on the computational basis B, the fixed set of M P = {I, F } depends on whether n is a multiple of 8 or not.
1. If n is a multiple of 8, then: 2. In every other case: The significance of Theorem 5.6 is twofold. First, from a somewhat negative perspective, disqualifies most of the dihedral groups as potential ambient groups for the P QG. Simultaneously, in a positive note, ascertains that the P QG can be meaningfully played in every dihedral group D n such that n is a multiple of 8. The next Theorem 5.7 explains what exactly happens in terms of winning strategies when the P QG is played in the aforementioned groups.
Theorem 5.7 (The ambient group of the P QG is D 8n ). If M P = {I, F } and M Q = D 8n , i.e., the ambient group of the P QG is D 8n , where n ≥ 1, then the following hold.
1. Q has exactly two classes of winning and dominant strategies, each containing 16 equivalent strategies: 2. The winning state paths corresponding to C + and C − are 3. Picard has no winning strategy.
We are led to a very important conclusion: nothing substantial will change if the game takes place in much larger groups than D 8 ; the winning strategies remain precisely the same. This realization of course begs the question whether things we will turn to be different when Q has at his disposal the largest group possible, U (2), which is examined in the next subsection.

The entire U(2)
In this section we shall examine the situation when Q is free to choose from all of U (2), which is the largest possible group that Q can draw his moves from. Therefore, without further ado we state our final assumption regarding Q's set of actions.
The major difference now compared to the previous cases is that U (2) contains infinitely many elements, whereas the previous groups were finite. Although, superficially, this might be expected to drastically enhance Q's capabilities, in turns out that in a certain sense everything remains the same. This can be attributed to the following very simple fact. Kets |ψ and e iθ |ψ , where θ ∈ R, physically represent the same state. In turn, this implies that the action of the operator A ∈ U (2) on a ket |ψ is the same as the action of e iθ A ∈ U (2) on |ψ (see [28] for details). If we view e iθ A as denoting a parametric family of operators, it is clear that all these operators can be considered equivalent and any of them, e.g., A, can be taken as the representative of the corresponding equivalence class. In order to simplify the notation, we make the following Definition 5.3.

Definition 5.3 (Families of unitary operators).
If A ∈ U (2), then we define the one-parameter family of unitary operators Similarly, if R ϕ and S ϕ are rotators and reflectors, as given by (3.1) and (3.2), respectively, we define the one-parameter families of operators: where θ ∈ R. Analogously, if H is the Hadamard transform and R 2πk n and S πk n are the matrix representations given by (3.7) and (3.8), we define the following collections of operators: Again it is fruitful to turn to Algorithm 1 to establish under what conditions Q possesses winning strategies and which are these. This approach is guided by the next Theorem 5.8, which establishes the fixed set of {I, F } in U (2).
This result is crucial in discovering and enumerating the winning strategies of Q in U (2). Although, we might have hoped for more variety in discovering winning strategies, the result is not unexpected because the flip operator F cannot fix more that two states. The next Theorem 5.9 explains what exactly happens in terms of winning strategies when the P QG takes place in U (2). Theorem 5.9 (The ambient group of the P QG is U (2)). If M P = {I, F } and M Q = U (2), i.e., the ambient group of the P QG is U (2), then the following hold.
2. The winning state paths corresponding to C + and C − are (5.23)

Picard has no winning strategy.
This final conclusion is illuminating. Although there are infinite winning strategies, they are equivalent to the strategies we determined when we investigated what happens in D 8 . In this perspective nothing is really gained by enabling Q to pick moves from U (2). In a certain sense the spirit of the game is completely captured when it is realized in D 8 . The next Table 3 contains the complete results about Q's classes of winning strategies whether the ambient group belongs to the family of dihedral groups D 8n or is the entire U (2). Table 3: The two classes of winning strategies for Q in the P QG when th game is played in a dihedral group D 8n , n ≥ 1 and when the game is played in the largest possible group U (2).
The ambient group is D 8n , n ≥ 1 Initial state Round 1 Round 2 Round 3 The ambient group is U (2) (θ ∈ R) Initial state Round 1 Round 2 Round 3 In U (2) the strategy classes contain infinite many strategies, but all these strategies are equivalent to the strategies encountered before.

Extending the game
The original P QG can be extended in numerous ways. In each conceivable extension, the precise formulation of the rules of the game is of paramount importance. By drastically changing the rules it is even possible for Picard to win the game. This was accomplished in [9] where the authors exploited entanglement in a clever way, so that whether the system ends up in a maximally entangled or separable state determines the outcome. In [11], it was shown that all possible finite extensions of the P QG can be expressed in terms of simple finite automata, provided that the allowable moves of Picard are either I or F and Q always uses the Hadamard transform H. In this paper, our investigation focused on the enlargement of the operational space of the game. Therefore, as we consider possible extensions of the original P QG, we adhere to the assumptions (A 1 ) and (A 4 ), i.e., M P = {I, F } and M Q = U (2). Additionally, we suppose that: • at the start of the game the coin is in a predefined basis state, which we call the initial state and designate by |q 0 , • Picard and Q alternate turns acting on the coin following a specified order, and • when the game ends, the coin is measured in the computational basis; if it is found in state |q P Picard wins, whereas if it is found in |q Q then Q wins.
It is convenient to refer to |q P as Picard's target state and to |q Q as Q's target state. Obviously, in a zero-sum game the target states |q P and |q Q are different. Furthermore, since both Picard and Q draw their moves from groups, nothing is lost in terms of generality if we agree that neither of them is allowed to make consecutive moves. Two or more successive moves by Q can be composed to give just one equivalent move, and the same holds for Picard. Definition 6.1 (Extended games between Picard & Q). An n-round game, n ≥ 2, is a function that associates one of the players, i.e., Picard or Q, to every round of the game. An n-round game is conveniently represented as a sequence of length n from the alphabet {P, Q}, where the letters P and Q stand for Picard and Q, respectively. According to our previous remark, if (K 1 , K 2 , . . . , K n ) is an n-round game, Definition 2.1 is general enough to also hold for extended games. We state the next important Theorem 6.1, which confirms our suspicion that Picard cannot win any such game with probability 1.0. This negative result implies that for the type of extended games we consider, Picard is at a permanent disadvantage. Theorem 6.1 (Picard lacks a winning strategy). Picard does not have a winning strategy in any n-round game, n ≥ 2, as long as Q makes at least one move.
The next couple of theorems explain in which games Q is unable to formulate a winning strategy. First, rather predictably, if Picard is given the opportunity to act last on the coin, then Q cannot surely win.
Theorem 6.2 (Q lacks a winning strategy when Picard plays last). Q does not have a winning strategy in any n-round game, n ≥ 2, in which Picard makes the last move.
Symmetrically, it is also true that Q is unable to devise a winning strategy when Picard playes first, as the next Theorem 6.3 asserts.    The picture is completed by the next Theorem 6.4 which asserts that Q has a winning strategy if and only if Q makes the first and the last move. This result clarifies that the overwhelmingly larger repertoire of moves of the quantum player by itself is not enough. It has to be combined with the advantage of making both the first and the last move in order to guarantee that the quantum player will surely win. Theorem 6.4 (When Q possesses a winning strategy). In any n-round game, n ≥ 2, Q has a winning strategy iff Q makes the first and the last move.
|0 Figure 12: This figure depicts two winning strategies for Q two an n-round games where Q makes the first and the last move. In the first game the initial state of the coin and the target state for Q are both the same, i.e., |1 . In the second game the initial state of the coin is also |1 , but the target state for Q now is |0 .
One last conclusion we may draw from the above theorems is about the initial state of the coin and the target states of the players. In the original P QG the both the initial state and Q's target state were the same, namely |0 . Clearly, the particular choice of the initial state and the target states is of no importance. If there exists a winning strategy for Q with respect to specific initial and target states, then there exists a winning strategy for every combination of initial and target states.
Corollary 6.5 (The impact of initial and target states). In any n-round game, n ≥ 2, if Q has a winning strategy, then he has a winning strategy for every combination of initial and target states.

Conclusions
Quantum games not only pose many interesting questions, but also motivate research that can have important applications in other related fields such quantum algorithms and quantum key distribution. This work was inspired by the iconic PQ penny flip game that, undoubtedly, helped create the field. We have approached this game using concepts from group theory. This allowed us to uncover the interesting connection of the original game with the dihedral group D 8 . Interpreting the game in terms of stabilizers and fixed sets enabled us to easily explain and replicate Q's strategy. This in turn allowed to prove that there exist precisely two classes of winning strategies for Q. Each class contains many different strategies, but all these strategies are equivalent in the sense that they drive the coin through the same sequence of states. We established that there are exactly two different sequences of states that can guaranteed Q's win with probability 1.0. What is noteworthy is the realization that even when the game takes place in larger dihedral groups or even in the entire U (2), this fact remains true. The essence of the game can be succinctly summarized by saying that there are precisely two paths that lead to Q's win and, of course, no path that leads to Picard's win. When we examined extensions of the game without any restriction, we discovered a very important fact, namely that for the quantum player to surely win against the classical player the tremendous advantage of quantum actions is not enough. Q must also make the first and the last move, or else he is not certain to win.
There still many question to be answered, a lot of important issues have to be addressed. These results were based on the assumption that the initial state of the coin and the target states of the players were one the the basis states. If this is not the case, and, moreover, entanglement comes into play, how does that affect the progression of the game? As this is, in our view, a particularly interesting topic, we expect it to be the subject of a future work.

A.1 Proofs for Section 3
We begin this Appendix by first giving the rather easy proof of Theorem 3.1.
we can readily verify the following facts: 1. For some of the remaining proofs, it will be convenient to explicitly give the standard matrix representation of D 8 by listing for every one of its elements the corresponding 2 × 2 matrix in the following Tables 4 and 5.

A.2 Proofs for Section 4
It is quite straightforward to follow and verify the proofs given below, by keeping in mind that: • under the matrix representation of the dihedral groups, the action of a dihedral group on any state of the quantum coin can be determined by simply multiplying the matrices representing the elements of the group with the ket corresponding to the state, and • a ket of the form e iθ |ψ , with θ ∈ R, represents the same state as the ket |ψ . 1. |0 and |1 have the same orbit: 2. The orbit of B is: Proof of Proposition 4.1. We will make use of the standard matrix representation of the rotations and reflections of D8 as given in Tables 4 and 5. 1. By systematically multiplying all the matrices in Tables 4 and 5   Proof of Proposition 4.2. We will only show how to find the stabilizer of |+ , since the proofs regarding the states |0 , |1 and |− are completely analogous. It suffices to exhaustively multiply every matrix appearing in Tables 4 and 5 with |+ and note for which matrices the outcome is again |+ . The stabilizer of |+ will contain precisely these matrices. These are the two rotations I and Rπ, through angles zero and π (see Figure 3), and the two reflections F , about the line passing through vertices 2 and 6, and S 6π 8 , about the line passing through vertices 4 and 8 (see Figure 4). Therefore, we conclude that D8(|+ ) = {I, Rπ, F, S 6π 8 }.

A.3 Proofs for Section 5
Theorem 5.1 (Characteristic properties of winning strategies). If (A 1 , A 2 ) is a winning strategy for Q, then: Proof of Theorem 5.1. By Definition 2.1 (A1, A2) is a winning strategy for Q if for every strategy of Picard, Q wins the game with probability 1.0. If the coin, just prior to measurement, is in a state a |0 +b |1 , with b = 0, then the probability that Q will win the game is strictly less than 1.0. Therefore, every winning strategy must eventually drive the coin to the state |0 , no matter what Picard plays. This implies that A2IA1 |0 = A2F A1 |0 = |0 . Let us assume in order to reach a contradiction that |ψ = A1 |0 is not fixed by F . Then F |ψ = |ψ ′ , where state |ψ ′ is different from state |ψ . However, according to (A.8), A2IA1 |0 = A2F A1 |0 ⇒ A2I |ψ = A2F |ψ ⇒ A2 |ψ = A2 |ψ ′ ⇒ |ψ = |ψ ′ , which contradicts our assumption that |ψ and |ψ ′ are different states. The last implication is valid because A2, as a group element, has a unique inverse. 1. We may distinguish 4 cases, depending on the state of the coin after Q's first move A1.
(i) If Q leaves the coin at state |0 , i.e., A1 |0 = |0 , then (A.8) implies that A2I |0 = A2F |0 = |0 ⇒ A2 |0 = A2 |1 = |0 ⇒ |0 = |1 , which is absurd. To arrive at this contradiction, we have used the fact that A2, as a group element, has a unique inverse. This result shows that the first move of every winning strategy for Q must drive the coin to a state other than |0 .
2. Based on the above analysis of cases (iii) and (iv) it is straightforward to verify (A.11).
3. By Definition 2.1, Picard has no winning strategy because if Q employs one of his winning strategies, Picard has 0.0 probability to win the game.
Theorem 5.3 (The smallest dihedral group for the P QG is D 8 ). D 8 is the smallest of the dihedral groups such that P QG can be meaningful played and in which Q has a quantum winning strategy.
Proof of Theorem 5.3. Let us first clearly state the two assumptions on which this result is based: 1. Picard's set of moves MP is {I, F }, according to assumption (A1). As a classical player, Picard must certainly be able to flip the coin, or else the game will be meaningless. On the other hand, he should not be able to employ a true quantum move.
2. Q's actions MQ should contain at least one unitary operator other than the classical I and F operators, in order to exhibit quantumness.
With the above clarifications in mind, let us examine whether any of the smaller dihedral groups D3, D4, D5, D6 and D7 can serve as the operational space for a meaningful, or at least nontrivial, realization of the P QG.
• The dihedral group D3 does not contain the reflection F . One can verify this by comparing formula (3.3) with formula (3.8) for n = 3 and k = 0, 1, 2. This shows that D3 does not satisfy assumption (A1) and, hence, is an inappropriate stage for the P QG.
• D4 contains the reflection F . However, the orbit D4 ⋆ B is {|0 , |1 }. This means that Q can only flip the coin from heads to tails or vice versa. If MQ = D4, then the P QG degenerates to the classical coin tossing game. Q is unable to employ a truly quantum strategy, something that contradicts the second assumption at the beginning of Section 5 and goes against the spirit of the P QG. Moreover, in D4 Q no longer possesses a winning strategy. For these reasons, it is meaningless to play the P QG in D4.
• The dihedral groups D5, D6 and D7 do not contain the reflection F either. Once again the formulas (3.3) and (3.8) for n = 5, 6 and 7 and k = 0, 1, . . . , n − 1 can be used to verify this fact. These groups do not satisfy assumption (A1) and are also inadmissible for the P QG. Let us assume to the contrary that there is an odd n such that Dn does contain F . Then there must be a k, 0 ≤ k < n, such that The fact that 0 ≤ k < n, implies that 0 ≤ 2πk n < 2π. Hence, either 2πk n = π 2 or 2πk n = 3π 2 . The former equation leads to k = n 4 and the latter to k = 3n 4 , which are both impossible because n is odd. Thus, we have arrived at a contradiction, which proves that F does not exist in Dn when n is odd.
For completeness of the exposition, we remind the reader of some very familiar notions, that will be invoked in our forthcoming proofs.
Definition A.1 (The unit circle). The circle S 1 of unit radius centered at the origin, which will be henceforth called the unit circle, is defined as The upper semicircle S 1 y≥0 of the unit circle is Symmetrically, the lower semicircle S 1 y≤0 of the unit circle is Given a point x = x y ∈ S 1 , its antipodal point is −x = −x −y ∈ S 1 .  Figure 14: The upper semicircle of the unit circle, its end points and some intermediate points. Figure 15: The lower semicircle of the unit circle, its end points and some intermediate points.
Proof of Lemma A.1.
which shows that all the kets of the |χ k sequence also appear in the |ϕ k sequence.
Another way to arrive at this conclusion is to observe that the Dn-orbits of |0 and |1 consist of kets appearing in the sequences |ϕ k and |χ k , respectively. In view of the fact that |1 = |χ0 appears in the |ϕ k sequence as |ϕm , Lemma A.1 asserts that Dn ⋆ |0 = Dn ⋆ |1 = Dn ⋆ B.
Furthermore, it also happens that only 2m of the kets in the |ϕ k sequence are distinct (|ψ and − |ψ represent the same state). In particular, it holds that that is kets |ϕ k and |ϕ k+2m correspond to antipodal points in the unit circle (Figure 7 gives a geometric depiction of the situation). The latter is easily proved as follows: When k ranges from 0 to 2m − 1, formula ( A.3.ii ) gives the first 2m kets in the |ϕ k sequence , . . . , cos π(2m−1) These are all distinct because each one of them corresponds to a unique different point that lies on the upper semicircle of the unit circle and makes an angle πk 2m , where 0 ≤ k ≤ 2m − 1, with the positive x-axis, as shown in Figure 14. Finally, by noting that 2m − 1 < 2m = n 2 , we verify that (A.22) holds.
Lemma A.4 (The action of D n on B when n = 2m). If n ≥ 3 is even, but not a multiple of 4, then the action of the dihedral group D n on the computational basis B is Proof of Lemma A.4. In this case we know that n = 2m , where m is odd and m ≥ 3 . Once again we encounter the phenomenon that the kets in the above sequences are not all different. Only m of the kets in the |ϕ k sequence and only m of the kets in the |χ k sequence are distinct (as always, we keep in mind that |ψ and − |ψ represent the same state). In particular, it holds that that is kets |ϕ k and |ϕ k+m correspond to antipodal points in the unit circle ( Figure 8 gives a geometric depiction of the situation). This can be shown as follows: The above kets correspond to the m points p0, p1, . . . , pm−1 that lie on the upper semicircle of the unit circle and make angles 0 < π m < 2π m < · · · < π(m−1) m , respectively, with the positive x-axis, as shown in Figure 14. The associated angles lie in the interval [0, π) because π(m−1) m < π and, therefore, the points p0, p1, . . . , pm−1 are all distinct. Finally, by noting that m − 1 < m = n 2 , we verify that (A.23) holds.
The important observation in this case is that To proceed further it is convenient to distinguish the following cases.
• The first case gives the system , which is clearly impossible because there is no ϕ such that sin ϕ = cos ϕ = 0.
• The next system can be rewritten via (A.15) as This is absurd because k1 − k2 is an integer and m is odd, as we recall from ( A.4.i ).
• In the next case we encounter the system , which is clearly impossible because there is no ϕ such that sin ϕ = cos ϕ = 0.
• Finally, we come to the last case concerning the system . This is also absurd because k1 − k2 is an integer and m is odd, as we recall from ( A.4.i ).
Thus, we have shown that the first m kets in the |ϕ k sequence are all different from the first m kets in the |χ k sequence, which establishes the validity of (A.25).
By combining the results of Lemmata A.3 and A.4 we can immediately prove Theorem 5.5. 2. if n is even but not a multiple of 4, then the action of the dihedral group D n on the computational basis B is We may now give the proof of Theorem 5.6.
Theorem 5.6 (The fixed set of {I, F } in D n ). The fixed set of M P = {I, F } in the general dihedral group D n , n ≥ 3, depends on whether n is a multiple of 8 or not.
1. If n is a multiple of 8, then: 2. In every other case: Proof of Theorem 5.6.
1. Let us first consider the case where n is a multiple of 8: According to (A.22) the range of k is 0 ≤ k < 4m. By setting k = m in (A.22) we derive that cos 2πm 8m |0 + sin 2πm 8m |1 = cos π 4 |0 + sin π 4 |1 = |+ belongs to Dn ⋆ B. Likewise, by setting k = 3m in (A.22) we get that cos 2π3m 8m |0 + sin 2π3m 8m |1 = cos 3π 4 |0 + sin 3π 4 |1 = − |− belongs to Dn ⋆ B. The above calculations show that the states |+ and |− belong to the orbit of B. We already know that F fixes these kets (recall Proposition 4.3). What remains is to prove that F fixes no other state in the orbit of B. So, let us suppose to the contrary that F also fixes some ket other than |+ and |− . This means that there exists a k, 0 ≤ k < 4m but k = m, 3m, such that The fact that 0 ≤ k < 4m, implies that 0 ≤ πk 4m < π. Therefore, either πk 4m = π 4 or πk 4m = 3π 4 . The former equation leads to k = m and the latter to k = 3m, which correspond to kets |+ and |− , respectively. No other values for k arise and, thus, F fixes no other state.
2. If n is not a multiple of 8, then we may distinguish the following cases.
• n is a multiple of 4, but not a multiple of 8 The fact that 0 ≤ k < 2m, implies that 0 ≤ πk 2m < π. Therefore, either πk 2m = π 4 or πk 2m = 3π 4 . The former equation leads to k = m 2 and the latter to k = 3m 2 , which are both impossible. Hence, F fixes no state from the orbit of B. According to (A.25) the range of k is 0 ≤ k < m. Let us assume that there exists a k such that πk m = π 4 . But this is absurd because then k must be equal to m 4 . Similarly, the existence of a k such that πk m = 3π 4 is impossible because then k must be equal to 3m 4 , since m is odd. The previous calculations establish that |+ and |− do not belong to the orbit of B. We now show that F fixes no ket in the orbit of B. If F did fix some ket, then there would be a k, 0 ≤ k < m, such that The fact that 0 ≤ k < m, implies that 0 ≤ πk m < π. Hence, either πk m = π 4 or πk m = 3π 4 . The former equation leads to k = m 4 and the latter to k = 3m 4 , which are both impossible because m is odd. Hence, F fixes no state from the orbit of B.
The preceding results allow to easily prove Theorem 5.7. 3. Picard has no winning strategy.
Proof of Theorem 5.7.
1. The two classes of winning strategies of Q, C+ and C−, which were establish by Theorem 5.2, are also present in every dihedral group D8n.
Let us first suppose that in some dihedral group larger than D8 there exists a third class C ′ and consider a strategy σ = (A1, A2) in this class. Then the action of A1 must drive the coin into some state other than |+ or |− . However, (A.9) asserts that A1 |0 ∈ F ix({F }), which, in view of (A.30), implies that A1 |0 ∈ {|+ , |− }, a contradiction. Hence, there are just two classes of winning strategies C+ and C−. It remains to prove that C+ and C− do not contain any new winning strategy. So, let us temporarily suppose that σ = (A1, A2) is a "new winning" strategy, that is other than those established in D8. .
We arrive at similar contradictions if we assume that A1 is a reflection different from H or S 5π 8 or that A1 drives the coin to state |− . Thus, we conclude that, other than those already existing in D8, there are no more winning strategies for Q in the larger dihedral groups D8n.
2. Based on the above analysis it is straightforward to see that (A.33) holds.
3. By Definition 2.1, Picard has no winning strategy because if Q employs one of his winning strategies, Picard has 0.0 probability to win the game.
The next two theorem settle the most general case, where the ambient group is U (2). The general solution is x1 = z and x2 = z, where z ∈ C. In matrix form x1 x2 can be written as z z = z 1 1 .
The normalization condition is satisfied if we take z = 1 √ 2 . Thus, the eigenket corresponding to the eigenvalue 1 is 1 √ 2 1 1 = |+ , which can be viewed as a basis for the eigenspace corresponding to the eigenvalue 1.
Symmetrically, when the eigenvalue is λ = −1 we have The general solution is x1 = z and x2 = −z, where z ∈ C. In matrix form x1 x2 can be written as z −z = z 1 −1 .
Again, the normalization condition is satisfied if we take z = 1 √ 2 . Hence, the eigenket corresponding to the eigenvalue 1 is 1 √ 2 1 −1 = |− , which can be considered as a basis for the eigenspace corresponding to the eigenvalue −1.
Proof of Theorem 5.9. 2. (P, Q, . . . , P, Q, P ), where n is odd, Q makes n 2 moves and Picard make n 2 + 1 moves. In order to arrive at a contradiction, we assume to the contrary that there exists a winning strategy σQ = (A1, . . . , A n 2 ) for Q. In view of Definition 2.1, if Q employs σQ, he will win the game with probability 1.0 no matter which strategy Picard chooses. If Picard uses σP = (I, . . . , I, I), then the fact that σQ is a winning strategy means that after Q's last action the coin must be at the basis state |qQ . On the other hand, if Picard uses σP = (I, . . . , I, F ), then the fact that σQ is a winning strategy means that after Q's last action the coin must be at the opposite basis state. This contradiction proves that Q does not possess a winning strategy. Theorem 6.3 (Q lacks a winning strategy when Picard plays first). Q does not have a winning strategy in any n-round game, n ≥ 2, in which Picard makes the first move.
Proof of Theorem 6.3. Any n-round game in which Picard makes the first move has one of the following two forms.
1. (P, Q, . . . , P, Q), where n is even and both Picard and Q make n 2 moves. Let us assume to the contrary that there exists a winning strategy σQ = (A1, . . . , A n 2 ) for Q. According to Definition 2.1, the fact that σQ is a winning strategy means that for every strategy of Picard, Q wins the game with probability 1.0. Since this holds for every strategy of Picard, it must also hold for the strategies σP = (I, . . . , I, I) and σ ′ P = (F, . . . , I, I). The former implies that A n 2 . . . A1 |q0 = |qQ ⇒ C |q0 = |qQ , ( 6.3.i) where C = A n 2 . . . A1. Since the composition of unitary operators produces a unitary operator, we know that C is unitary. On the other hand, the latter implies that If |q0 = |0 , then F |q0 = |1 , whereas if |q0 = |1 , then F |q0 = |0 . If we combine this last result with ( 6.3.i ) and ( 6.3.ii ), we conclude that which is of course impossible because C is unitary. Thus, Q does not possess a winning strategy.
2. (P, Q, . . . , P, Q, P ), where n is odd, Q makes n 2 moves and Picard make n 2 + 1 moves. In order to arrive at a contradiction, we assume to the contrary that there exists a winning strategy σQ = (A1, . . . , A n 2 ) for Q. In view of Definition 2.1, if Q employs σQ, he will win the game with probability 1.0 no matter which strategy Picard chooses. If Picard uses σP = (I, . . . , I, I), then the fact that σQ is a winning strategy means that after Q's last action the coin must be at the basis state |qQ . On the other hand, if Picard uses σP = (I, . . . , I, F ), then the fact that σQ is a winning strategy means that after Q's last action the coin must be at the opposite basis state. This contradiction proves that Q does not possess a winning strategy.
The next Theorem 6.4 gives a positive answer to the question of whether Q, and in effect the quantum player, can surely win and under what circumstances. Theorem 6.4 (When Q possesses a winning strategy). In any n-round game, n ≥ 2, Q has a winning strategy iff Q makes the first and the last move.
Proof of Theorem 6.4. The one direction, i.e., if Q has a winning strategy then he must make the first and the last move, is an immediate consequence of Theorems 6.2 and 6.3.
It remains to prove the other direction, that is if Q makes the first and the last move, then Q possesses a winning strategy. If the initial state of the coin |q0 is the same as the target state of Q, then σQ = (H, I, . . . , I, H) a winning strategy for Q. Q's first action will drive the coin to either |+ (if the initial state is |0 ) or |− (if the initial state is |1 ). In any case both {|+ , |− } are fixed by {I, F }, as Theorem 5.8 asserts. Q's last move will send the coin back to |q0 . If the initial state of the coin |q0 is different from the target state of Q, then it is trivial to check that σQ = (H, I, . . . , I, F H) a winning strategy for Q. Corollary 6.5 (The impact of initial and target states). In any n-round game, n ≥ 2, if Q has a winning strategy, then he has a winning strategy for every combination of initial and target states.
Proof of Corollary 6.5. An immediate consequence of Theorem 6.4.