Tit-For-Tat and Win Stay-Lose Shift strategies via memory-two

Abstract

Decisions in many dilemmas are based on a combination of factors, including as incentive, punishment, reputation, and memory. The impact of memory information on cooperative evolution in multi-round games is a decision-making process in group evolution. The iterated prisoner’s dilemma is an excellent model for the development of cooperation amongst the payoff-maximizing individuals. Since tit-for-tat proved successful in Axelrod’s repeated prisoner’s dilemma tournaments, there has been a great deal of interest in creating new strategies. Every iterative prisoner’s dilemma method bases its decision-making on a specific duration of past contacts with the opponent, which is referred to as the memory’s size. This study examines the impact of strategy memory size on the evolutionary stability of n-person iterated prisoner’s dilemma strategies. In this paper, we address the role that memory plays in decision-making. We interested in the model of the Iterated Prisoner’s Dilemma game for three players with memory two, and we will look at strategies with similar behavior, such as Tit-For-Tat (TFT) strategies as well as Win Stay-Lose Shift (WSLS) strategies. As a result of this paper, we have shown that the effect of memory length is almost non-existent in the competitions of strategies that we studied.

Keywords

Memory-Two Tit-For-Tat strategies (TFT)three-players iterated prisoner’s dilemma game (3P-IPD)transition matrix Win Stay-Lose Shift strategies (WSLS)

1 Introduction

Game theory does not tell us how to play the game; rather, it discusses some possible strategies for playing that you could find appealing. Additionally, the area of applied arithmetic is utilized to develop the best possible approach for contending with uncertainty and inadequate knowledge like most real-life scenarios. It is the mathematical analysis of conflict modeling and decision-making that occurs in all kinds of fields and companies on a daily basis.

Since the game theory is the theory of “trategic thinking”, it has been developed for military purposes and defense. In the past, it has also been used as an alternative and complementary approach to deal with robustness in the presence of worst-case uncertainties or disturbances in many areas such as economics [1, 4], engineering [5], computer science, politics [8 , 15], and sociology. As well as to the biological sciences such as evolutionary biology [3 , 12]. Moreover, scientists tend to use game-theoretic tools to avoid traffic flows or predict blackouts in power networks. In the 21^st century, game theory applies to a wide range of behavioral relations.

The mathematician John Von Neumann and the economist Oskar Morgenstern wrote a book [6] in 1944 that contained the first mention of game theory. This book’s second edition offered an axiomatic theory of expected utility that enabled economists and statisticians to study decision-making in the face of uncertainty. In the 1950s, a large number of academics greatly expanded the field of game theory. In the 1970s, it was specifically used in relation to evolution.

There are two types of games: simultaneous games [2, 9] and alternating games [10, 13]. In a simultaneous game, both players make their choices in the same time without knowing the other player’s choice. While in the alternating game, the first player makes his choice and then the other player makes his choice after knowing the first player’s choice. In this paper, we will discuss one of these two types, which is the simultaneous game through a three-player competition model in the iterative game.

The Prisoner’s Dilemma is one of the most popular and simple strategy models developed by Merrill Flood and Melvin Dresher in 1950. In this model, no matter how many participants, each participant has only two decisions. Either he chooses to cooperate and plays C or defect and then plays D. Accordingly, in two-player Iterated Prisoner’s Dilemma games, we have four outcomes (C, C) , (C, D) , (D, C) and (D, D), so there are 2⁴ possibilities for the transition arrows made a total of 16 different strategies denoted by S₀, S₁, . . . , S₁₅. Whereas, in three-player Iterated Prisoner’s Dilemma games, we have eight outcomes (C, (C, C)), (C, (C, D)), (C, (D, C)), (C, (D, D)), (D, (C, C)), (D, (C, D)), (D, (D, C)) and (D, (D, D)). In our study, we look at symmetric games, so the outcomes become six only (C, (C, C)), (C, (C, D)), (C, (D, D)), (D, (C, C)), (D, (C, D)) and (D, (D, D)). Wherefore, there are 2⁶ possibilities for the transition arrows made a total of 64 different strategies denoted by S₀, S₁, . . . , S₆₃. We can categorize strategies based on their behavior, such as Tit-For-Tat, Win Stay-Lose Shift Strategies and other Strategies.

Tit for tat is a highly effective strategy in game theory. An agent using this strategy will first cooperate, then repeat opponent’s previous decision. If the opponent previously was cooperative, the agent is cooperative. If the opponent previously was defective, the agent is defective. In Win Stay-Lose Shift strategies, if the game in the previous round has a succeeds, then the player plays the same strategy in the next round. Alternatively, if the game resulted in a failure, the player switches to another action.

Previous studies focused on studying the repeated prisoner’s dilemma game for two players only, and recently we studied the repeated prisoner’s dilemma game for three players but with memory-one [2,17, 2,17], meaning that the player relies in his decision on the previous round only. In this paper, we will discuss the repeated prisoner’s dilemma game for three players but with memory-two. We will focus on Tit-For-Tat and Win Stay-Lose Shift strategies and study the competition between them. At the end of this paper, we will answer the question of what does memory two do in the competition of Tit-For-Tat strategies with each other, and also the competition of Win Stay-Lose Shift strategies with each other.

2 Payoff and transition matrices

We will study the simultaneous game on Iterated Prisoner’s Dilemma model. We assume that both players play simultaneously without knowing the other player’s decision. Both players get a reward, $R$ , when they cooperate and get a punishment, $P$ , when they defect. While if one of them cooperates and the other defect, then the defector gets $T$ and the cooperator gets $S$ . So, the corresponding payoff matrix for (2P-IPD) [9] is given by $\begin{matrix} \begin{matrix} C & D \end{matrix} \\ \begin{matrix} C \\ D \end{matrix} [\begin{matrix} R & S \\ T & P \end{matrix}] \end{matrix},$ (1) where: $S < P < R < T and R > \frac{T + S}{2} .$ (2)

Based on the aforementioned, we have four outcomes O₁ = (C, C) , O₂ = (C, D) , O₃ = (D, C) and O₄ = (D, D) and if we assume that player I plays with probability P = (p₁, p₂, p₃, p₄) and player II with probability Q = (q₁, q₂, q₃, q₄), where p_i is the probability of player I for playing C after outcome O_i. Therefore, the Markov transition matrix for two players [9] is given by

$M_{2} = [\begin{matrix} p_{1} q_{1} & p_{1} (1 - q_{1}) & (1 - p_{1}) q_{1} & (1 - p_{1}) (1 - q_{1}) \\ p_{2} q_{3} & p_{2} (1 - q_{3}) & (1 - p_{2}) q_{3} & (1 - p_{2}) (1 - q_{3}) \\ p_{3} q_{2} & p_{3} (1 - q_{2}) & (1 - p_{3}) q_{2} & (1 - p_{3}) (1 - q_{2}) \\ p_{4} q_{4} & p_{4} (1 - q_{4}) & (1 - p_{4}) q_{4} & (1 - p_{4}) (1 - q_{4}) \end{matrix}],$ (3)

where p₁ (1 - q₁) is the transition probability from state O₁ to O₂, where player I plays C by probability p₁ after state O₁ and (1 - q₁) probability of player II playing D after state O₁.

As well, in Three-player Iterated Pensioner’s Dilemma games, the player’s payoffs are $R, K, S, T, L$ , or $P$ and the correspondence payoff matrix for (3P-IPD) can be represented by $\begin{matrix} \begin{matrix} CC & CD & DD \end{matrix} \\ \begin{matrix} C \\ D \end{matrix} [\begin{matrix} R & K & S \\ T & L & P \end{matrix}] \end{matrix},$ (4) where $S < P < K < L < R < T .$ (5)

The three-player Iterated Prisoner’s Dilemma, we have six outcomes O₁ = (C, (C, C)), O₂ = (C, (C, D)), O₃ = (C, (D, D)), O₄ = (D, (C, C)), O₅ = (D, (C, D)) and O₆ = (D, (D, D)) and if we assume that player I plays with probability P = (p₁, p₂, p₃, p₄, p₅, p₆), player II with probability Q = (q₁, q₂, q₃, q₄, q₅, q₆) and player III with W = (w₁, w₂, w₃, w₄, w₅, w₆). Therefore, the Markov transition matrix for three-player [17] is given by $M_{3} = (\begin{matrix} p_{1} q_{1} w_{1} & p_{1} [q_{1} (1 - w_{1}) + (1 - q_{1}) w_{1}] & p_{1} (1 - q_{1}) (1 - w_{1}) & (1 - p_{1}) q_{1} w_{1} & (1 - p_{1}) [q_{1} (1 - w_{1}) + (1 - q_{1}) w_{1}] & (1 - p_{1}) (1 - q_{1}) (1 - w_{1}) \\ p_{2} q_{2} w_{4} & p_{2} [q_{2} (1 - w_{4}) + (1 - q_{2}) w_{4}] & p_{2} (1 - q_{2}) (1 - w_{4}) & (1 - p_{2}) q_{2} w_{4} & (1 - p_{2}) [q_{2} (1 - w_{4}) + (1 - q_{2}) w_{4}] & (1 - p_{2}) (1 - q_{2}) (1 - w_{4}) \\ p_{3} q_{5} w_{5} & p_{3} [q_{5} (1 - w_{5}) + (1 - q_{5}) w_{5}] & p_{3} (1 - q_{5}) (1 - w_{5}) & (1 - p_{3}) q_{5} w_{5} & (1 - p_{3}) [q_{5} (1 - w_{5}) + (1 - q_{5}) w_{5}] & (1 - p_{3}) (1 - q_{5}) (1 - w_{5}) \\ p_{4} q_{2} w_{2} & p_{4} [q 2 (1 - w_{2}) + (1 - q_{2}) w_{2}] & p_{4} (1 - q_{2}) (1 - w_{2}) & (1 - p_{4}) q_{2} w_{2} & (1 - p_{4}) [q_{2} (1 - w_{2}) + (1 - q_{2}) w_{2}] & (1 - p_{4}) (1 - q_{2}) (1 - w_{2}) \\ p_{5} q_{3} w_{5} & p_{5} [q_{3} (1 - w_{5}) + (1 - q_{3}) w_{5}] & p_{5} (1 - q_{3}) (1 - w_{5}) & (1 - p_{5}) q_{3} w_{5} & (1 - p_{5}) [q_{3} (1 - w_{5}) + (1 - q_{3}) w_{5}] & (1 - p_{5}) (1 - q_{3}) (1 - w_{5}) \\ p_{6} q_{6} w_{6} & p_{6} [q_{6} (1 - w_{6}) + (1 - q_{6}) w_{6}] & p_{6} (1 - q_{6}) (1 - w_{6}) & (1 - p_{6}) q_{6} w_{6} & (1 - p_{6}) [q_{6} (1 - w_{6}) + (1 - q_{6}) w_{6}] & (1 - p_{6}) (1 - q_{6}) (1 - w_{6}) \end{matrix}) .$ (6)

3 Strategies competitions with memory two

The purpose of this section is to explain the significance of the Tit-For-Tat and Win Stay-Lose Shift strategies with memory-two in the context of the three-player Iterated Prisoner’s Dilemma game. A two-state automata is adopted to represent the strategies where each player is automaton which can be in one of two states, namely C and D. The sixty-four strategies will be represented by six coordinates of zeros and ones using a binary scheme. Each digit denotes the player’s response when any of the six known round outcomes occur $(R, K, S, T, L, P)$ . Ones and zeros show that the player will move D for a zero and C for a one in the future. Whereas the binary representation of S₃₆ is (1, 0, 0, 1, 0, 0) and known as TFT, the representation of S₆₃ is (1, 1, 1, 1, 1, 1) which is call All C. For example, after S₃₆ has just defected D, he will remain in this D stat if at least one of the opponent just played D, but will shift to its C state if the both opponents just play C as shown in Fig. 1. For another example, after S₃₅ has just defected D, he will remain in this D state if the both opponents just played C, but will shift to its C state if at least one of the opponents just played D as shown in Fig. 2.

Fig. 1

Automates of TFT strategies.

Fig. 2

Automates of WSLS strategies.

3.1 TFT strategies with memory-two

We will introduce four different strategies S₃₆, S₃₈, S₅₂ and S₅₄, which are called TFT 1, TFT 2, TFT 3 and TFT 4, respectively. These four strategies are presented by the automatons in Fig. 1, such that:

We study an example for the three players S₃₆, S₃₈ and S₅₂, such as player I (S₃₆) against player III (S₅₂) with player II (S₃₈). We will get the following eight cases, and every case has eight sub-cases as follows:

Case 1: in round 1, players I, II and III play C, and in round 2, we discuss all eight possible cases.

Case 2: in round 1, players I and II play C, but player III plays D and in round 2, we discuss all eight possible cases.

Case 3: in round 1, players I and III play C, but player II plays D and in round 2, we discuss all eight possible cases.

Case 4: in round 1, player I plays C, but players II and III play D and in round 2, we discuss all eight possible cases.

Case 5: in round 1, player I plays D, but players II and III play C and in round 2, we discuss all eight possible cases.

Case 6: in round 1, players I and III play D, but player II plays C and in round 2, we discuss all eight possible cases.

Case 7: in round 1, players I and II plays D, but player III plays C and in round 2, we discuss all eight possible cases.

Case 8: in round 1, players I, II and III play D, and in round 2, we discuss all eight possible cases.

•Case 1.1: If in state 1, the three players (S₃₆, S₃₈, S₅₂) start with C and also in state 2, the three players (S₃₆, S₃₈, S₅₂) start with C.

•Case 1.2: If in state 1, the three players (S₃₆, S₃₈, S₅₂) start with C and in state 2, players I and II (S₃₆, S₃₈) start with C but player III (S₅₂) starts with D.

•Case 1.3: If in state 1, the three players (S₃₆, S₃₈, S₅₂) start with C and in state 2, players I and III (S₃₆, S₅₂) start with C but player II (S₃₈) starts with D.

•Case 1.4: If in state 1, the three players (S₃₆, S₃₈, S₅₂) start with C and in state 2, players I (S₃₆) starts with C but player II and III (S₃₈, S₅₂) start with D.

•Case 1.5: If in state 1, the three players (S₃₆, S₃₈, S₅₂) start with C and in state 2, players II and III (S₃₈, S₅₂) start with C but player I (S₃₈) starts with D.

•Case 1.6: If in state 1, the three players (S₃₆, S₃₈, S₅₂) start with C and in state 2, players II (S₃₈) starts with C but players I and III (S₃₆, S₅₂) start with D.

•Case 1.7: If in state 1, the three players (S₃₆, S₃₈, S₅₂) start with C and in state 2, player III (S₅₂) starts with C but players I and II (S₃₆, S₃₈) start with D.

•Case 1.8: If in state 1, the three players (S₃₆, S₃₈, S₅₂) start with C and in state 2, the three players (S₃₆, S₃₈, S₅₂) start with D.

From the previous case, we have three regimes $R$ , $\frac{P + R}{2}$ and $\frac{2 R + K + T}{4}$ . Therefore, by use the same Technique using the algorithm in the appendix section, we get that in the cases 2, 4, 6, 7 and 8, there are three regimes $P$ , $\frac{P + R}{2}$ and $\frac{2 P + K + T}{4}$ . In cases 3 and 5, also there are three regimes $\frac{2 R + K + T}{4}$ , $\frac{T + K}{2}$ and $\frac{2 P + K + T}{4}$ . Finally, we have six regimes $R_{1} = R,$ (7) $R_{2} = \frac{R + P}{2},$ (8) $R_{3} = \frac{2 R + T + L}{4},$ (9) $R_{4} = P,$ (10) $R_{5} = \frac{2 P + T + K}{4},$ (11) $R_{6} = \frac{T + K}{2} .$ (12)

When we do the perturbation, we get the following: In regime R₁, if S₃₆ and S₃₈ play D instead of C, then regime R₁ will transfer to R₃, and if S₅₂ plays D instead of C, then regime R₁ will transfer to R₂. In regime R₂, if S₃₆ and S₃₈ play D instead of C in column 1, then regime R₂ will transfer to R₅, and if S₅₂ plays D instead of C in column 1, then regime R₂ will transfer to R₄. But in column 2, if S₃₆, S₃₈ and S₅₂ play C instead of D, then regime R₂ will not change in all these cases. In regime R₃, if S₃₆ and S₃₈ play D instead of C in columns 1 and 3, then regime R₃ will transfer to R₆, and if S₅₂ plays D instead of C in columns 1 and 3, then regime R₃ will transfer to R₅. But in column 2, if S₃₆ and S₅₂ play D instead of C, then regime R₃ will transfer to R₂ and if S₃₈ plays C instead of D, then regime R₃ will transfer to R₁. While in column 4, if S₃₆ plays C instead of D, then regime R₃ will transfer to R₁ and if S₃₈ and S₅₂ play D instead of C, then regime R₃ will transfer to R₂. In regime R₄, S₃₆, S₃₈ and S₅₂ play C instead of D, then regime R₄ will not change in all these cases. In regime R₅, if S₃₆ plays C instead of D in column 1, then regime R₅ will transfer to R₂, and if S₃₈ and S₅₂ play D instead of C in column 1, then regime R₅ will transfer to R₄. In columns 2 and 4, S₃₆, S₃₈ and S₅₂ play C instead of D, then regime R₅ will not change in all these cases. While in column 3, if S₃₆ and S₅₂ play D instead of C, then regime R₅ will transfer to R₄, and if S₃₈ plays C instead of D, then regime R₅ will transfer to R₂. In regime R₆, if S₃₆ and S₅₂ play D instead of C in column 1, then regime R₆ will transfer to R₅, and if S₃₈ plays C instead of D in column 1, then regime R₆ will transfer to R₃. But in column 2, if S₃₆ plays C instead of D, then regime R₆ will transfer to R₃, and if S₃₈ and S₅₂ play D instead of C, then regime R₆ will transfer to R₅. Thus, the corresponding transition matrix becomes $\begin{matrix} \begin{matrix} R_{1} & R_{2} & R_{3} & R_{4} & R_{5} & R_{6} \end{matrix} \\ \begin{matrix} R_{1} \\ R_{2} \\ R_{3} \\ R_{4} \\ R_{5} \\ R_{6} \end{matrix} [\begin{matrix} 0 & 1 / 3 & 2 / 3 & 0 & 0 & 0 \\ 0 & 1 / 2 & 0 & 1 / 6 & 1 / 3 & 0 \\ 1 / 6 & 1 / 3 & 0 & 0 & 1 / 6 & 1 / 3 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 / 6 & 0 & 1 / 3 & 1 / 2 & 0 \\ 0 & 0 & 1 / 3 & 0 & 2 / 3 & 0 \end{matrix}] \end{matrix}$ (13)

By calculate the left eigenvectors for the eigenvalue 1, we get the following equations: $- v_{1} + \frac{1}{6} v_{3} = 0,$ (14) $\frac{1}{3} v_{1} - \frac{1}{2} v_{2} + \frac{1}{3} v_{3} + \frac{1}{6} v_{5} = 0,$ (15) $\frac{2}{3} v_{1} - v_{3} + \frac{1}{3} v_{6} = 0,$ (16) $\frac{1}{6} v_{2} + \frac{1}{3} v_{5} = 0,$ (17) $\frac{1}{3} v_{2} + \frac{1}{6} v_{3} - \frac{1}{2} v_{5} + \frac{2}{3} v_{6} = 0,$ (18) $\frac{1}{3} v_{3} - v_{6} = 0,$ (19)

By solving the linear system of the previous equations from Equations (14) to (19) with the equation v₁ + v₂ + v₃ + v₄ + v₅ + v₆ = 1, then we obtain the eigenvector V as $V = (v_{1}, v_{2}, v_{3}, v_{4}, v_{5}, v_{6}) = (0, 0, 0, 1, 0, 0)$ (20)

Now, we can get the payoff values by

$\begin{matrix} E (S_{36}, S_{38}, S_{52}) \\ = & v_{1} R_{1} + v_{2} R_{2} + v_{3} R_{3} \\ + v_{4} R_{4} + v_{5} R_{5} + v_{6} R_{6} \\ = & P \end{matrix}$ (21)

According to Equation (21), the payoff vector Π = (π₁, π₂, π₃, π₄, π₅, π₆) is equal to (0, 0, 0, 0, 0, 1). Using the same approach depending on the algorithm in the Appendix section; we examined all possibilities for studying three strategies from the four TFT strategies, which are 64 possibilities times eight sequences. Therefore, we obtained the results shown in the following tables from Tables 1 to 4.

Table 1

The payoff for player I against player III when player II fixed with the strategy TFT 1 (S₃₆)

Player II (S₃₆)	TFT 1	TFT 2	TFT 3	TFT 4
is fixed	S ₃₆	S ₃₈	S ₅₂	S ₅₄
TFT 1	(0, 0, 0, 0, 0, 1)	(0, 0, 0, 0, 0, 1)	(0, 0, 0, 0, 0, 1)	(0, 0, 0, 0, 0, 1)
S ₃₆
TFT 2	(0, 0, 0, 0, 0, 1)	(1, 3, 6, 0, 9, 5)	(0, 0, 0, 0, 0, 1)	(2, 3, 3, 3, 3, 2)
S ₃₈
TFT 3	(0, 0, 0, 0, 0, 1)	(0, 0, 0, 0, 0, 1)	(1, 0, 0, 0, 0, 1)	(1, 0, 0, 0, 0, 1)
S ₅₂
TFT 4	(0, 0, 0, 0, 0, 1)	(2, 6, 3, 0, 3, 2)	(1, 0, 0, 0, 0, 1)	(1, 0, 0, 0, 0, 0)
S ₅₄

Table 2

The payoff for player I against player III when player II fixed with the strategy TFT 2 (S₃₈)

Player II (S₃₈)	TFT 1	TFT 2	TFT 3	TFT 4
is fixed	S ₃₆	S ₃₈	S ₅₂	S ₅₄
TFT 1	(0, 0, 0, 0, 0, 1)	(1, 0, 3, 3, 12, 10)	(0, 0, 0, 0, 0, 1)	(2, 3, 0, 3, 6, 2)
S ₃₆
TFT 2	(0, 0, 0, 0, 0, 1)	(1, 2, 1, 1, 2, 1)	(1, 2, 1, 1, 2, 1)	(1, 0, 0, 0, 0, 0)
S ₃₈
TFT 3	(0, 0, 0, 0, 0, 1)	(1, 2, 1, 1, 2, 1)	(1, 0, 0, 0, 0, 1)	(1, 0, 0, 0, 0, 0)
S ₅₂
TFT 4	(2, 6, 3, 0, 3, 2)	(0, 0, 0, 0, 0, 1)	(1, 0, 0, 0, 0, 0)	(1, 0, 0, 0, 0, 1)
S ₅₄

Table 3

The payoff for player I against player III when player II fixed with the strategy TFT 3 (S₅₂)

Player II (S₅₂)	TFT 1	TFT 2	TFT 3	TFT 4
is fixed	S ₃₆	S ₃₈	S ₅₂	S ₅₄
TFT 1	(0, 0, 0, 0, 0, 1)	(0, 0, 0, 0, 0, 1)	(1, 0, 0, 0, 0, 1)	(1, 0, 0, 0, 0, 1)
S ₃₆
TFT 2	(0, 0, 0, 0, 0, 1)	(1, 2, 1, 1, 2, 1)	(1, 0, 0, 0, 0, 1)	(1, 0, 0, 0, 0, 0)
S ₃₈
TFT 3	(1, 0, 0, 0, 0, 1)	(1, 0, 0, 0, 0, 1)	(1, 0, 0, 0, 0, 1)	(1, 0, 0, 0, 0, 1)
S ₅₂
TFT 4	(1, 0, 0, 0, 0, 1)	(1, 0, 0, 0, 0, 0)	(1, 0, 0, 0, 0, 1)	(1, 0, 0, 0, 0, 0)
S ₅₄

Table 4

The payoff for player I against player III when player II fixed with the strategy TFT 4 (S₅₄)

Player II (S₅₄)	TFT 1	TFT 2	TFT 3	TFT 4
is fixed	S ₃₆	S ₃₈	S ₅₂	S ₅₄
TFT 1	(0, 0, 0, 0, 0, 1)	(1, 0, 0, 0, 0, 1)	(1, 0, 0, 0, 0, 1)	(1, 0, 0, 0, 0, 0)
S ₃₆
TFT 2	(2, 3, 3, 3, 3, 2)	(5, 9, 0, 6, 3, 1)	(1, 0, 0, 0, 0, 0)	(1, 0, 0, 0, 0, 0)
S ₃₈
TFT 3	(1, 0, 0, 0, 0, 1)	(1, 0, 0, 0, 0, 0)	(1, 0, 0, 0, 0, 1)	(1, 0, 0, 0, 0, 0)
S ₅₂
TFT 4	(1, 0, 0, 0, 0, 0)	(1, 0, 0, 0, 0, 0)	(1, 0, 0, 0, 0, 0)	(1, 0, 0, 0, 0, 0)
S ₅₄

3.2 WSLS strategies with memory two

We will introduce five different strategies S₃, S₃₅, S₃₃, S₄₉ and S₄₈, which are called WSLS 1, WSLS 2, WSLS 3, WSLS 4, and WSLS 5, respectively. These five strategies are presented by the automatons in Fig. 2, such that:

Now, we study an example for the three players S₃, S₃₅ and S₃₃, such as player I (S₃) against player III (S₃₃) with player II (S₃₅). We will get the following eight cases and every case has 8 sub-cases as follows: