Quantum Game Theory – II: A Comprehensive Study

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Quantum Game Theory – II

* A Comprehensive Study

Indranil Ghosh

Indranil Ghosh is a graduate student doing post graduation

in physics specialising in condensed matter physics,

from the Department of Physics, Jadavpur University,

Kolkata. His research interests include computational physics,

numerical computing, quantum mechanics and

quantum computing.

The following article is a continuation the article titled Quan- tum Game Theory – I: A Comprehensive Study [1], that ap- peared in the May issue ofResonance.

1. Quantum Game Theory

Classical game theory, when analysed from the perspective of quantum algorithms and quantum probability amplitudes, is called

‘quantum game theory’ [2]. This provides us with the extra abil- ities to utilize the entanglement between players’ moves and also the linear superpositions of strategy actions. The players perform local unitary operations on their qubits analogous to the classical moves they play in classical game theory. For dynamic games like quantum penny flip, the property of superposition between qubits may be sufficient for analysis, but static games like quantum prisoner’s dilemma generally require quantum entanglement.

Next, we analyse the game theory models in details.

1.1 Penny Flip Game 1.1.1 Classical version

This game is played between two players. Let us assume them to be Alice and Bob. Here Alice always plays the first move followed by Bob. The rules of the game are as follows:

• Out of the two equally probable states of the unbiased penny, i.e, {H,T}, Alice sets the initial state of the penny in eitherHorT.

*Vol.26, No.6, DOI: https://doi.org/10.1007/s12045-021-1180-6

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• Now both of them have access to two possible strategies, i.e, S₁ ≡flip the penny,S₂ ≡keep the penny as it is. After Alice’s initial setting of the state, Bob, without knowing this initial state, plays eitherS1orS2.

• After this, again Alice, without knowing Bob’s move, plays ei- therS1orS2.

• At last Bob, without knowing Alice’s move again, plays either S1orS2. Now, the final state is checked.

• If the final state is inHstate, Alice looses 1 point and Bob gains 1 point. Otherwise, if inT state, the reverse happens. This is a two-playerzero-sum game, where their payoffs in every situation are exactly opposite to each other and add up to give 0.

1.1.2 Quantum version

TheHandT states are replaced by the up,uand down,dstates, i.e,|0iand|1ieigenstates of a quantum system (like an electron) respectively. The up state is defined as, u = |0i =





 1 0





and the down state as, d = |1i =





 0 1





. The strategies are replaced by unitary operatorsS₁ =σ_x, i.e, the spin flip operator andS₂ = I, i.e, the identity operator. The initial quantum state is set to be either up, u or down, d. The rules for the sequence of player moves remain the same as that of the classical version. Finally the final quantum state is measured to analyse the game. We later introduce some complicated situations where either Alice or Bob cheats or where both of them do.

1.1.3 Representation and the game frame

For our analysis, let’s suppose Alice sets the initial state to be always in up,ustate and further build upon the situation. Instead, if Alice sets the initial state to be in down,dstate, the outcome of the game remains same, with the change that the payoffs to Alice and Bob just get reversed. The game can be represented by the binary tree:

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Alice,u

d

u

d Bob,σ

x

u Bob

,I Alice,σ^x

d

u Bob,σ

x

d Bob

,I Alic

e,I Bob,σ^x

u

d

u Bob,σ

x

d Bob

,I Alice,σ^x

u

d Bob,σ

x

u Bob

,I Alic

e,I Bob,I

The simple operations that have been utilized here are, u = Iu, d = Id,u= σ_xdandd =σ_xu. The sequence of the penny states can be jotted down in the form of a matrix, given inTable1.

Table 1. Sequence for penny states.

Alice/Bob I,I I, σ_x σ_x,I σ_x, σ_x I u,u,u d,d,d d,u,u u,d,d σ_x d,d,u u,u,d u,d,u d,u,d

This as a result gives us the payoffmatrix of the game as inTable 6.

Table 2. _Payoff_{matrix for} Alice and Bob.

Alice/Bob I,I I, σ_x σ_x,I σ_x, σ_x I (−1,1) (1,−1) (1,−1) (−1,1) σ_x (1,−1) (−1,1) (−1,1) (1,−1)

The first element in each cell of the matrix is Alice’s payoff, and the second element is Bob’s. Alice and Bob both have access to mixed strategies. Both of them chooses moves σ_x and I with equal probability of ¹₂. It is evident from the payoffmatrix that, despite what Bob’s move is, Alice’s expected payoffis π_Alice = ¹₂((1)+(−1))= 0. Similarly, Bob’s expected payoff, de-

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spite Alice’s move, is,π_Bob= ¹₂((1)+(−1)+(1)+(−1))=0. For

‘N’ such games and ‘x’ wins for Alice, as given in [3], the actual payoffto Alice would be a member of the payoffset,

Y={f(x;N)}{2x−N,x=0,1, ...,N}. (1) The probability of these payoffs,

P(Y

)={f(x;N,p)} ={ N x

!

p^xq^N−x,∀x}. (2) For Bob too, we will have the same computations. In our case, we have takenN=1 giving usQ

={−1,+1}andP(Q

)={¹₂,¹

2}.

1.1.4 Simulation results

A simple quantum simulator named, QSimulator.py. is written in Python version 3.6.8. All the Python files written hereafter are present in the Github repository [4]. The above file is further used in analysing all the quantum games in this article. The program is designed for a maximum of an 8-qubit system. The qubit|0i is represented byQ0and qubit|1iis represented byQ1. Quantum state,|01110iis represented byQ01110and so on. Also,

|01110i=|0i ⊗ |1i ⊗ |1i ⊗ |1i.⊗ |0i (3) Python’s numpy package lets us create these quantum systems.

The ⊗ in the above equation is called the tensor product or the outer product and is implemented bynumpy’s outer()function.

TheQSimulator.py file provides with all the necessary quantum logic gates required in our further studies. For example,

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>>> Q0

a r r a y ( [ 1 . , 0 . ] )

>>> Q1

a r r a y ( [ 0 . , 1 . ] )

>>> Q111

a r r a y ( [ 0 . , 0 . , 0 . , 0 . , 0 . , 0 . , 0 . , 1 . ] )

>>> I 2

a r r a y ( [ [ 1 . , 0 . ] , [ 0 . , 1 . ] ] )

>>> P a u l i X ( I 2 ) a r r a y ( [ [ 0 . , 1 . ] ,

[ 1 . , 0 . ] ] )

>>> Hadamard ( I 2 )

a r r a y ( [ [ 0 . 7 0 7 1 0 6 7 8 , 0 . 7 0 7 1 0 6 7 8 ] , [ 0 . 7 0 7 1 0 6 7 8 , −0 . 7 0 7 1 0 6 7 8 ] ] )

Here, we have simulated|0i,|1i,|111i, identity operator (I), σ_x operator and the Hadamard operator (H). The whole game tree has been simulated by writing a file penny_flip_1.py. Running this file, lets us verify that there is an equal probability of ¹₂ for both the ‘up’ and ‘down’ states to be the final state. Thus, the probability of winning for Alice is equal to the probability of winning for Bob. Look atFigure1 andFigure2.

These two cases occur with equal probability for a large number of samplings. This can be plotted as given inFigure3.

Next, we perform the analysis of few more complicated case studies, where the players may cheat, and the property of quantum superposition is exploited.

1.1.5 Case 1: Alice cheats

Let us suppose Alice takes the help of the quantum property of superposition to cheat. She sets the initial state to be a superposition of both the ‘up’ and ‘down’ states instead of the ‘up’ state as in the previous case. Let the initial state set by Alice be,|ψi= ^u+d^√

2.

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Figure 1. Probability distribution for the final state after measurement (Bob wins).

Figure 2. Probability distribution for the final state after measurement (Alice wins).

We note the following equations,

|φi= I|ψi= u+d

√2

, (4)

and

|φi=σ_x|ψi= d+u

√ 2

= u+d

√ 2

. (5)

where,|φiis the final quantum state. A simple simulation of the above equations is carried out withQSimulator.py.

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Figure 3. Plot that shows that both the possible states have an equal probability of 0.5 to be the final state after collapsing due to measurement.

>>> u = Q0

>>> d = Q1

>>> p s i = ( u + d )/np . s q r t ( 2 )

>>> p s i

a r r a y ( [ 0 . 7 0 7 1 0 6 7 8 , 0 . 7 0 7 1 0 6 7 8 ] )

>>> np . d o t ( I2 , p s i )

a r r a y ( [ 0 . 7 0 7 1 0 6 7 8 , 0 . 7 0 7 1 0 6 7 8 ] )

>>> P a u l i X ( p s i )

a r r a y ( [ 0 . 7 0 7 1 0 6 7 8 , 0 . 7 0 7 1 0 6 7 8 ] )

Keeping the above in mind, the game tree is represented as,

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Alice, ^u+d^√

2

u+d√ 2

Bob,σx/IAlice,σx/IBob,σx/I

The above game tree is simulated by writing a Python file,penny_flip_2.py.

This file is then run to generate the plot given inFigure4.

On measurement of the final state it is discovered by Alice that once again, she loses or wins with equal probabilities. ForN =1, again,Q

= {−1,1}and P(Q

) = {¹₂,¹

2}. Measurement yields the basis eigenstate|0iwith the probability,| h0|φi |²= ¹₂and the basis eigenstate|1iwith the probability,| h1|φi |²= ¹₂.

1.1.6 Case 2: Bob cheats

Now, let’s suppose Alice remains innocent and plays fair, but Bob commits trickery in his move. So, |ψi = |0i again. Instead of playing the classical movesI orσ_x, Bob plays a quantum move, i.e., he plays the Hadamard operation. We note the equations,

H|ψi= u+d

√ 2

, (6)

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Figure 4. Probability densities showing equal probabilities.

and

H(H|ψi)= 1 2





 2 0





=





 1 0





=u. (7)

If the game tree is drawn, it will be seen that after Bob’s first move, the state of the penny will be in the superposed state given by (12) and after his second move, the Hadamard operation brings back the penny state to ‘up’, i.e,|φi=u=|0iagain. This has been simulated with the filepenny_flip_3.py, which generates the plot given inFigure5.

The payoff set for both Alice and Bob is {−1,+1}. Measure- ment yields state ‘up’ with probability,| h0|φi |²= 1 and the state

‘down’ with probability, | h1|φi |² = 0. So, for Alice, P(Q ) = {1,0}and for Bob,P(Q)={0,1}.

1.1.7 Case 3: Both of them cheat

Let us consider two conditions here, allowing Alice to cheat by setting the initial penny state to be ‘down’ or the superposed state instead of the traditional ‘up’ situation. Bob always plays the Hadamard operation.

Alice sets the initial state to be ‘down’, so, |ψi = |1i and Bob

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Figure 5. Probability density showing|0i appears with probability 1.

playsH,

Hd= H|1i= u−d

√ 2

. (8)

Now, Alice playsσ_x,

σ_x(Hd)= d−u

√ 2

, (9)

or she plays I,

I(Hd)= u−d

√2

. (10)

After this, Bob again plays theHoperation,

H(σ_x(Hd))=−d, (11)

or,

H(I(H(d))=d. (12)

The tree that models the game is,

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Figure 6. Probability density showing|1iappears with probability 1.

Alice, d

u−d√ 2

d

Bob,H

Alice,I

d−u√ 2

−d

Bob,H

Alic e,σ

x

Bob,H

This has been simulated with penny_flip_4.py which generates the plot as shown inFigure6.

The payoffset for Alice and Bob is{−1,+1}. When the final state

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is|φi = −d, measurement yields base state ‘down’ with probability 1 and the base state ‘up’ with probability 0 and the similar result emerges for the case when|φi= d. So, the probability set for Alice isP(Q

)={0,1}and for Bob isP(Q

)={1,0}. So, Alice always wins in this case.

In the next case, Alice sets the initial state of the penny in the superposed state as incase 2. Now, Bob playsH,

H|ψi=H[u+d

√

2 ]=u. (13)

Next, Alice plays eitherσ_xorI. So,

σ_xu=d, (14)

or

Iu=u. (15)

Finally Bob again playsH, so,

|φi=Hd= u−d

√2

, (16)

or

|φi=Hu= u−d

√ 2

. (17)

This can be represented by a similar game tree and can be simulated with the filepenny_flip_5.py, that generates the plot given inFigure7

The payoffset for both Alice and Bob isQ

= {−1,+1}. When

|φi = ^u−d^√

2, measurement yields base state ‘up’ with probability,

| h0|φi |²= ¹₂and the base state ‘down’ with probability,| h1|φi |²=

1

2. Similar results appear for|φi = ^u+d^√

2. The probability set for both Alice and Bob is{¹₂,¹

2}. So, Alice and Bob has the equal probability for winning. In the next sub-section, we will be dis- cussing both classical and quantum versions.

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Figure 7. Probability density showing|0iand|1i appearing with equal probability of 0.5.

1.2 Prisoner’s Dilemma 1.2.1 Classical version

Prisoner’s dilemma [5] [6] is a paradoxical situation occurring in game theory or decision analysis problems, which shows why two rationalagents might not cooperate with each other, even though it appears that it is in their best interest to do so. This is a very interesting model having a wide range of applications in computer science, biology, economics and philosophy. The game was developed in 1950 by Merrill Flood and Melvin Dresher while working at RAND cooperation.

Prisoner’s dilemma is a 2 X 2 non-zero sumgame (i.e., a game where the payoffs of each player do not add up to give 0), generally presented in a static form. The game reads in the following way: there are two suspects Alice and Bob, suspected of commit- ting a crime together, who have been brought in for interrogation.

Both of them has access to two strategy moves, i.e., either to “cooperate” (C) with each other by remaining silent or to “defect”

(D), i.e., confess to the crime. None of them knows what the other does. Both the rational players try to maximize their expected utility with mixed strategies. We introduce four payoffs,

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w,x,yandz, which follow the chain of inequalities,

z>w> x>y. (18) Now, the payoffmatrix of the game model is structured as:

Table 3. _Payoff_{matrix for}

prisoner’s dilemma. Alice/Bob C D

C (w,w) (y,z) D (z,y) (x,x)

Assuming that the payoffmatrix is symmetric, it is inferred that the payoffs w, x, y, and zare the same for each player, and the payoffs have ordinal significance too. Here, strategies, S_Alice = {C,D},S_Bob = {C,D}and the payoffs, Q

= {π_Alice, π_Bob}. Now, looking at the payoffmatrix, we make some analyses.

1. First, consider the case of Alice cooperating. If now, Bob too cooperates, he gets awand if he defects, he gets az. Asz>w, Bob has an advantage by defecting rather than cooperating with Alice.

2. For Alice defecting, if Bob cooperates, he gets a y, and if he cooperates, he gets ax. Asx>y, Bob again has an advantage if he defects.

So, the dominant strategy for Bob is the moves_Bob, such that the payoffπ_Bobto Bob has the property,

π_Bob(sⁱ_Alice,s_Bob)≥π_Bob(sⁱ_Alice,s_Bob^j ) (19)

∀sⁱ_Alice ∈S_Alice and ∀s_Bob^j ∈S_Bob. So, the dominant strategy for Bob is to defect, i.e.,sBob = D. For Alice too, if the same analyses are done, it can be seen that the dominant strategy for Alice is to defect, i.e.,s_Alice =D. Thus both the rational players defect to receive a payoffofxeach, whereas two irrational players can cooperate and receive a greater payoffofweach. The game will be in equilibrium with{sAlice,sBob} ={D, D}. This paradoxical outcome is the core problem of the prisoner’s dilemma because

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both Alice and Bob would have a more desirable position, if both of them had cooperated, yielding π_Alice = π_Bob = w, as w > x.

For a player having a strictly dominant strategy, it is irrational for that player to choose any other strategy guaranteeing him/her a lower payoff. In the prisoner’s dilemma, individual rationality leads to the dilemma even though both the players would be better offcooperating. On application of theIDSDSalgorithm (section 2.4) on the payoff matrix of the game, we end up with {D, D}

as the only strategy profile that can be played, on availability of common knowledge of rationality (Table8). For our simulation experiments, we consider w = 3,x = 1,y = 0 and z = 5 as con- sidered in most of the literature. A python file IDSDS_PD.pyis written that simulates the IDSDS algorithm on the payoffmatrix with the assigned values and generates the desired result.

Table 4. Iterated strictly dominant-strategy equilibrium for the prisoner’s dilemma game.

Alice/Bob D D (x,x)

Also, it is noted that the only strong Nash equilibrium of the game is {D, D} because neither Alice nor Bob can increase his/her payoff by one-sidedly departing from the given equilibrium point.

Also, in the payoffmatrix, the point{w,w}isPareto optimal. In the next section, the quantum version of the game is discussed, and it is shown how the addition of quantum moves changes the result of the game and ultimately lets us break out of the dilemma.

1.2.2 Quantum version

The quantum version of prisoner’s dilemma has been studied in details by Eisert, Wilkens, and Lewenstein [7]. Du et al. [8] re- alised the game experimentally on a nuclear magnetic resonance quantum computer. The quantum version of the game makes use of entanglement between the shared states of the two agents be- sides the concept of quantum superposition.

The quantum game, GQ = GQ(H,G⁰_Q,{si}_j,{π_i}_j) is a list of 4 elements, whereHis a Hilbert space, G⁰_Q is the initial prepared

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state of the game, {s_i}_j is the set of strategy moves of player j, while {π_i}_j is the set of payoffs to player j. The two goals of the quantum version are—maximizing the payoffs to player jand finding a way to escape the dilemma. Alice and Bob both possess a qubit each and are allowed to manipulate their qubits. Each of their qubits lie inH²consisting of the basis vectors{|0i,|1i}. The game, thus, lies inH² ⊗H², with basis vectors |00i, |01i, |10i and|11i. As the convention is followed, the left qubit belongs to Alice and the right to Bob.

The initial state of the game is set as,

G⁰_Q=U|00i. (20) U is a unitary operator known to both Alice and Bob. U oper- ates on both of their qubits. U serves the purpose of maximally entangling Alice’s and Bob’s qubits. This operator brings the dif- ference of the quantum version of the game from the classical version. WithoutU, the payoffs to both remain the same as that of the classical version. Now, letUAlice andUBobare the unitary operators that can be applied on the respective players’ qubits.

UAlice,UBob ∈S whereS is the strategy set available to both the players. After the moves played by both of them, the state of the game is represented as:

G¹_Q=|ψ_si=(UAlice⊗UBob)G⁰_Q=(UAlice⊗UBob)U|00i. (21) Next, to get the final state,

G_Q^f =|ψ_fi=U^†G¹_Q=U^†(U_Alice⊗U_Bob)U|00i. (22) Here, U^† = (U^∗)^T. As, U is unitary, so, U^† = U⁻¹. On the measurement of the final state of the system, we get the expected payoffto Alice as,

π_Alice =w| hψ_f|00i |²+y| hψ_f|01i |²+z| hψ_f|10i |²+x| hψ_f|11i |². (23)

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and the expected payoffto Bob is,

π_Bob=w| hψ_f|00i |²+z| hψ_f|01i |²+y| hψ_f|10i |²+x| hψ_f|11i |². (24) Considering,

U = I⊗I+iσ_x⊗σ_x

√2

. (25)

The inverse thus calculated is,

U⁻¹= I⊗I−iσ_x⊗σ_x

√2

. (26)

The initial state has the superposed form:

G⁰_Q=U|00i= |00i+i|11i

√2

. (27)

The classical move “cooperating” is mapped to the operator I, and the move “defecting” is mapped to the operatorσ_x. A Python file namedQPD.pyis written which consists of a functionqpd() that takes in 6 parameters: the operatorsUAlice andUBob, and the payoffsw,x,yandz. This function plots the probability densities of the basis eigenstates along with the expected payoffvalues of Alice and Bob as a tuple. Considering for example, a simple case, where Alice defects and Bob cooperates, i.e., UAlice = σ_x andU_Bob=I,

|ψ_si=(σ_x⊗I)G⁰_Q= |10i+i|01i

√ 2

, (28)

and

|ψ_fi=U^†|ψ_si=|10i. (29) We get|10iwith probability 1 andπ_Alice = 5 andπ_Bob = 0. Few lines of codes can simulate these as follows,

>>> s i g m a _ x=P a u l i X ( I 2 )

>>> qpd ( sigm a_x , I2 , 3 , 1 , 0 , 5 ) ( 4 . 9 9 9 9 9 9 9 9 9 9 9 9 9 9 8 , 0 . 0 )

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Figure 8. Probability density showing |10iappearing with probability 1.

The plot generated is given inFigure8

Now, two less traditional quantum moves represented by the Hadamard matrix,Handσ_zare introduced. So, bothU_Alice,U_Bob ∈ {I, σ_x,H, σ_z}.

We consider few case studies here,

Alice playsσ_x and Bob playsH:U_Alice=σ_xandU_Bob =H. So,

|ψ_si=(σ_x⊗H)G⁰_Q= |10i+|11i+i|00i −i|01i

√2

, (30)

and

|ψ_fi=U^†|ψ_si= |11i −i|01i

√2

. (31)

A measurement of the final state yields an equal probability for a payoff ofy = 0 or a payoff of x = 1 to Alice and an equal probability for a payoffofz= 5 or a payoffofx= 1 to Bob. We obtain,π_Alice =0.5 andπ_Bob=3. Codes to simulate this situation,

>>> s i g m a _ x=P a u l i X ( I 2 )

>>> H=Hadamard ( I 2 )

>>> qpd ( sigm a_x , H , 3 , 1 , 0 , 5 )

( 0 . 4 9 9 9 9 9 9 9 9 9 9 9 9 9 9 8 , 2 . 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 ) The plot generated is given inFigure9.

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Figure 9. Probability density showing |01i and |11i appearing with equal probability of 0.5.

Next both Alice and Bob playsH:U_Alice= HandU_Bob =H. So,

|ψ_si=(H⊗H)G⁰_Q

= |00i+|01i+|10i+|11i+i|00i −i|01i −i|10i+i|11i 2√

2

,

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|ψ_fi=U^†|ψ_si= |00i+|11i −i|01i −i|10i 2

. (33)

Final state measurement yields an equal probability of 0.25 for a payoff ofw = 3 or a payoff ofy = 0 or a payoff ofz = 5 or a payoffof x = 1 to Alice and an equal probability of 0.25 for a payoff ofw = 3 or a payoff ofy = 0 or a payoff ofz = 5 or a payoff ofx = 1 to Bob. Hereπ_Alice = 2.25 and π_Bob = 2.25.

Codes to simulate this situation:

>>> qpd (H , H , 3 , 1 , 0 , 5 )

( 2 . 2 4 9 9 9 9 9 9 9 9 9 9 9 9 8 7 , 2 . 2 4 9 9 9 9 9 9 9 9 9 9 9 9 8 7 ) The plot generated is given inFigure10.

Finally, the case, Alice plays σ_z and Bob playsH: U_Alice = σ_z

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Figure 10. Probability density showing |00i, |01i,

|10iand|11iappearing with equal probability of 0.25.

andU_Bob= H, so,

|ψ_si=(σ_z⊗H)G⁰_Q= |00i+|01i −i|10i+i|11i

√ 2

, (34)

and the final state,

|ψ_fi=U^†|ψ_si= |00i −i|10i

2 . (35)

A measurement of the final state yields an equal probability of 0.5 for a payoffof x = 3 or a payoffofz = 5 to Alice and an equal probability of 0.5 for a payoffof x = 3 or a payoffofy = 0 to Bob. Here,π_Alice = 4 andπ_Bob = 1.5. Simple codes to simulate the situation:

>>> s i g m a _ z=P a u l i Z ( I 2 )

>>> H=Hadamard ( I 2 )

>>> qpd ( s i g m a _ z , H , 3 , 1 , 0 , 5 )

( 3 . 9 9 9 9 9 9 9 9 9 9 9 9 9 9 8 2 , 1 . 4 9 9 9 9 9 9 9 9 9 9 9 9 9 9 3 ) The plot generated is given inFigure11.

This way all the 2⁴ =16 combinations can be worked out to generate the payoffmatrix. A Python file namedPayoff_matrix_QPD.py is written to serve this purpose.

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Figure 11. Probability density showing |00i and

|10i appearing with equal probability of 0.5.

Table 5. _Payoff _ma- trix for quantum prisoner’s dilemma.

Alice/Bob I σ_x H σ_z

I (3,3) (0,5) (0.5,3) (1,1) σ_x (5,0) (1,1) (0.5,3) (0,5) H (3,0.5) (3,0.5) (2.25,2.25) (1.5,4) σ_z (1,1) (5,0) (4,1.5) (3,3)

Applying the IDSDS algorithm on the payoff matrix given by Table 9, the iterated strictly dominant-strategy equilibrium has shifted from {σ_x, σ_x} in the classical version to {σ_z, σ_z} in the quantum case.

Table 6. Iterated strictly dominant-strategy equilibrium for quantum prisoner’s dilemma.

Alice/Bob σ_z σ_z (3,3)

{σ_z, σ_z}is the Nash equilibrium of the game and is alsoPareto optimal. We now notice that individual rationality leads both Alice and Bob to chooseσ_zas the ideal move, yielding the payofftuple (3,3) ≡ (w,w), which was not the case in the classical version.

So, we see that in the quantum version rational moves by players, in fact, help us get out of the dilemma that arises in the classi-

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cal version. In the final section, we will delve into the quantum version of thetwo person duel gameand make a comprehensive study before concluding our three-part article.

A Software Dependencies

As mentioned before, Python version 3.6.8 have been used. To use the softwares [4] for simulating the results for the games presented here, certain independent libraries to be used are: numpy, pandas,random,itertools,pylab,math,matplotlib,tkinter and mpl_toolkits. If any library is missing, it can be down- loaded by writing the following command on the terminal window:

$ p i p 3 i n s t a l l numpy

The above command downloads thenumpy library. If the user wants to run the quantum duel GUI, use the following command by opening the terminal window at the location whereQDuel.py is located:

$ p y t h o n 3 QDuel . py