Playing Quantum Monty Hall Game in a Quantum Computer
PPlaying Quantum Monty Hall Game in a Quantum Computer
Souvik Paul, ∗ Bikash K. Behera, † and Prasanta K. Panigrahi ‡ Department of Physical Sciences,Indian Institute of Science Education and Research Kolkata,Mohanpur 741246, West Bengal, India
Here, we present the quantum version of a very famous statistical decision problem, whose classicalversion is counter-intuitive to many. The Monty Hall game can be phrased as a two person gamebetween Alice and Bob. In their pioneering work, Flitney and Abbott [Phys. Rev. A , 062318(2002)] showed that by using a maximally entangled system for Alice and Bob’s choices, and usingquantum strategies, Bob and Alice can win or lose depending on the strategy chosen by either of theplayers. Here we develop a new quantum algorithm with quantum circuits for playing the quantumMonty Hall game by a user. Our quantum algorithm uses the quantum principles of superpositionand entanglement so that it can be efficiently played on a quantum computer. We present twoschemes, one calculating the probability of winning or loss and the other determining whether aplayer (say Alice) wins or not. Keywords: IBM Quantum Experience, Quantum Monty Hall Problem, Entanglement, Decomposition ofn-bit Networks, Quantum Circuits
I. INTRODUCTION
In the recent past, much interest has been developedin the discipline of quantum information [1, 2] that hasled to the creation of quantum game theory [2, 3]. Quan-tum game theory [2, 4–10] has attracted a lot of atten-tion during the last few years. The quantum Monty Hallproblem [2] is an interesting example in this realm. Thequantum game theory has been shown to be experimen-tally feasible through the application of a measurement-based protocol by Prevedel et al. [11]. They realized aquantum version of the Prisoner’s Dilemma game basedon the entangled photonic cluster states [2]. There aremany paradoxes and unsolved problems associated withquantum information [1] and the study of quantum gametheory is a useful tool to explore this area. Anothermotivation is that in the area of quantum communica-tion, optimal quantum eavesdropping can be treated asa strategic game with the goal of extracting maximal in-formation [10, 12]. It has also been suggested that aquantum version of the Monty Hall problem may be ofinterest in the study of quantum strategies of quantummeasurement [10, 13].The classical Monty Hall problem has raised much in-terest because it is sharply counter-intuitive. From aninformational point of view, it clearly illustrates the casewhere a null operation provides the information aboutthe system. In the classical Monty Hall game [14, 15] thehost Bob secretly selects one door of three behind whichto place a prize. The player Alice picks a door. Bob thenopens another door where there is no prize. Alice nowhas the option of sticking with his current selection orchanging to the untouched door. Classically, the opti- ∗ [email protected] † [email protected] ‡ [email protected] mum strategy for Alice is to alter her choice of door andthis, surprisingly, doubles her chance [14] of winning theprize from to . II. QUANTUM MONTY HALL GAME
A number of researchers have contributed towards thequantization of Monty Hall problem [2, 16, 17]. Forthe Monty Hall game where both participants can applyquantum strategies, it has been observed that maximalentanglement of the initial states results the same pay-offs as compared to the the classical game. The gameis called to be fair, when the host, Bob, has access to aquantum strategy while Alice does not. Even if Bob canadopt a strategy with an expected payoff of for eachperson, Alice can win all the time if she has access toa quantum strategy and Bob does not. Non-entangledinitial states produce the payoffs as the classical case asexpected. Under certain operations, it is also possiblefor Bob to win the game with payoff 1 [10]. Similar re-sults have been recreated by D’Ariano et al. [17] by usingdensity matrices and the concept of quantum notepads.Kurzyk and Glos [18] have used an entangled state be-tween Alice’s first choice and the position of the prize.Using Bayesian networks, they have shown that the en-tanglement of quantum states has influence on the resultsof reasoning, and that there exists a quantum state forwhich the Monty Hall game is fair under the assumptionthat Bob and Alice do not have access to quantum strate-gies. Our present purpose here is to advance a differentquantum version of the Monty Hall problem and to de-velop a quantum circuit [16] so that it can be designedon a quantum computer and any user can play it. a r X i v : . [ qu a n t - ph ] J a n III. SCHEME-1: DETERMINING THEPROBABILITY OF WINNING OR LOSING
Fig. 1 presents the quantum circuit explaining thescheme for playing the quantum Monty Hall game.As illustrated in the circuit, we take three qubits( | (cid:105) D | (cid:105) D | (cid:105) D ) representing the three doors D1, D2,and D3. If there is a prize is in any of the door, thenthe corresponding qubit’s state becomes | (cid:105) . Three morequbits are taken for representing the doors D1, D2, andD3 as at the end of the quantum circuit, three controlledmeasurements needs to performed. Then we take twoqubits ( | (cid:105) B and | (cid:105) B ) for representing Bob’s state forstoring information about the doors and which doorsopen or not. Two qubits ( | (cid:105) A and | (cid:105) A ) are used toAlice’s two-qubit input. Then the two qubits ( | (cid:105) A and | (cid:105) A ) are used to represent Alice’s second choice whichdecides to which two doors to open.The following initial assumptions are taken to proceed; • Door D1 is denoted by | (cid:105) • Door D2 is denoted by | (cid:105) • Door D3 is denoted by | (cid:105) Step-1:
Before the game starts, Bob has classical in-formation with regards to the door, behind which theprize is present. The first three qubits in the quantumcircuit represents the three doors D1, D2, and D3 thatare initialized in state | (cid:105) . If the prize is present in anyof the doors then the state of the corresponding qubitchanges to | (cid:105) state. We denote the Bob’s state by thequbits B B Step-2:
Applying two Hadamard gates ( H ) on Bob’stwo qubits ( | (cid:105) B B ) each prepared in | (cid:105) state, we cre-ate a superposition of | (cid:105) , | (cid:105) , | (cid:105) and | (cid:105) states.Since we need superposition of only three states for rep-resenting the three doors’ states, we remove | (cid:105) state byusing a controlled-Hadamard gate ( CH B B ), where B B Step-3:
Now Alice chooses one of the three doors. Herchoice can be any one from the following cases: • If Alice chooses D1 i.e. | (cid:105) , then | (cid:105) state needs tobe removed from Bob’s state, which can be achievedby using anti-controlled-Z ( ACZ B B ) and anti-controlled-Hadamard ( ACH B B ) gates. Now, thestates with Bob are superposition of | (cid:105) and | (cid:105) . • If Alice chooses D2, i.e. | (cid:105) , an anti-controlled-Hadamard ( ACH B B ) will be applied on the Bob’squbit resulting a superposition of ( | (cid:105) and | (cid:105) )states. • If Alice chooses D3, i.e. | (cid:105) , Bob’s state needs tobe in superposition of | (cid:105) and | (cid:105) states, whichis achieved by applying anti-controlled-Hadamard( H B B ) gate. Step-4:
Here in this step, we make the following as-sumption, i.e., whenever Bob has the option of removingmore than one states (i.e, opening more than one doors),he chooses to remove the state, i.e., opens the lowest dooravailable to him (D1 < D2 < D3). According to the rule ofthe game, at first attempt of Alice, Bob would not openthe door as asked by her, and also would open the doorwhere there is no prize. However, in the second attemptof Alice, Bob has to follow her and exactly open the dooras asked by her. According to the above rule, the fol-lowing three cases can be considered depending upon thepresence of prize in any of the doors;
Case-4A: Alice chooses D1: • The prize is in D3 (the state of the first three qubits( | D D D (cid:105) = | (cid:105) ), then Bob opens the doorD2 i.e., removes | (cid:105) state using anti-controlled-Hadamard ( ACH B B ) operation and Alice is leftwith a superposition state ( | A A (cid:105) ) of | (cid:105) and | (cid:105) . • The prize is in D2 ( | (cid:105) ), then Bob opens thedoor D3, i.e., removes the | (cid:105) state using anti-controlled-Hadamard ACH B B . Hence Alice isleft with superposition of | (cid:105) and | (cid:105) states, i.e, | A A (cid:105) = | (cid:105) + | (cid:105)√ . • The prize is in D1 ( | (cid:105) ), then Bob opens D2i.e., removes the | (cid:105) state using an anti-controlled-Hadamard H B B operation. Thus Alice’s state be-comes, | A A (cid:105) = | (cid:105) + | (cid:105)√ . Case-4B: Alice chooses D2: • The prize is in D3 ( | (cid:105) ), then Bob opens D1 i.e., | (cid:105) is removed using anti-controlled-Z ( ACZ B B )and anti-controlled-Hadamard ( ACH B B ) opera-tions and Alice is left with a superposition state of | (cid:105) and | (cid:105) . • The prize is in D2 ( | (cid:105) ), then Bob opensD1 i.e., removes | (cid:105) state using anti-controlled-Z gate ( ACZ B B ) and anti-controlled-Hadamard( ACH B B ) operations and Alice has the superpo-sition state of | (cid:105) and | (cid:105) . • The prize is in D1 ( | (cid:105) ), then Bob opens the doorD3 i.e., removes | (cid:105) state using anti-controlled-Hadamard ( ACH B B ) and Alice’s state remainsin superposition of | (cid:105) and | (cid:105) . Case-4C: Alice chooses D3: • The prize is in D3 ( | (cid:105) ), then Bob opens D1 i.e., | (cid:105) is removed using anti-controlled-Z ( ACZ B B )and anti-controlled-Hadamard ( ACH B B ) opera-tions and Alice has the superposition of | (cid:105) and | (cid:105) . • The prize is in D2 ( | (cid:105) ), then Bob opens D1, i.e., | (cid:105) is removed using anti-controlled-Z ( ACZ B B ) | i D %| i D %| i D %| i D %| i D %| i D %| i B H Z H H H H | i B H H H H Z H H | i A | i A | i A H H | i A FIG. 1.
Quantum circuit illustrating Scheme-I . The quantum circuit for the implementation of the Monty Hall game. Asuperposition of the initial states is created. Alice then chooses one of the three doors, accordingly Bob opens one of the threedoors, depending on which door Alice chooses to open and the prize in behind which door. Alice then makes her second choice,and Bob opens the doors. and anti-controlled-Hadamard (
ACH B B ) opera-tions and Alice has the superposition of | (cid:105) and | (cid:105) . • The prize is in D1 ( | (cid:105) ), then Bob opens D2, i.e., | (cid:105) is removed using anti-controlled-Hadamard( ACH B B ) and Alice has the superposition of | (cid:105) and | (cid:105) . Step-5:
An entanglement is created between the twostates of Bob and Alice, which allows restricted informa-tion to be communicated to Alice (Bob communicates toAlice the state that has to be removed).
Step-6:
The three super positions left with Alice are: • | (cid:105) and | (cid:105) - 2nd measurement is restricted asBob has already removed | (cid:105) state (i.e, has openedthe door D2). • | (cid:105) and | (cid:105) - 1st measurement is restricted as Bobhas already removed | (cid:105) state (i.e, has opened thedoor D1). • | (cid:105) and | (cid:105) - 3rd measurement is restricted as Bobhas already removed | (cid:105) state (i.e, has opened thedoor D3).Hence controlled measurements are taken. The circuitis simulated and the probabilities of Alice’s win are ob-tained. [See Step 4 ] IV. SCHEME-2: DETERMINING ALICE’S WINOR LOSS
The quantum circuit provided in Fig. 2 depicts thewinning or loss of the player, Alice. The details of thescheme is described in the following steps. As shownin the circuit, we take three qubits ( | (cid:105) D | (cid:105) D | (cid:105) D )representing the three doors D1, D2, and D3 if a prize isthere or not. For example, if the prize is in the seconddoor D2, the three-qubit state becomes, | (cid:105) , i.e., thepresence of prize makes the qubit state to | (cid:105) . Then fourqubits, | (cid:105) I , | (cid:105) I , | (cid:105) I and | (cid:105) I are used for Alice’sinputs, as she has to choose two doors one by one. Herpossible inputs are | (cid:105) , | (cid:105) and | (cid:105) . Then we take twoqubits ( | (cid:105) S and | (cid:105) S ) for creating a superposition of | (cid:105) , | (cid:105) , and | (cid:105) states for representing the initial statewhen no doors are opened. Then three qubits ( | (cid:105) A , | (cid:105) A and | (cid:105) A ) are assigned to Alice’s qubits, whichafter the measurement at the end of the quantum circuit,determines whether Alice wins or loose.The following initial assumptions are taken as thescheme-1 III: • Door D1 is denoted by | (cid:105) • Door D2 is denoted by | (cid:105) • Door D3 is denoted by | (cid:105) Step-1:
We first create a superposition state is created to rep-resent the initial state of the game when no doors areopened. This is done by using Hadamard gates ( H ) onthe qubits ( | (cid:105) S and | (cid:105) S ). Then applying a controlled-Hadamard ( CH S S ) operation, a superposition of threestates | (cid:105) , | (cid:105) and | (cid:105) is prepared. Step-2:
When the prize is in one of the three doorsand Alice chooses to open one of the three doors, thereare following 9 possible cases to be considered before oneof the states is being removed from Bob’s state. Herewe make an assumption that whenever Bob has the op-tion for opening one or more doors, then he choosesto open the door which is the lowest available to him(D1 < D2 < D3). According to the rule of the game, atfirst Bob would not open the door as asked by Alice, andalso would not open the door where the prize is present.However, for the second time when asked by Alice, Bobhas to open the door asked by her. Keeping in mind theabove rule and conditions, the following 9 possible cases(Fig. 3) arise before opening the first door.
Case-2A; Alice chooses D1 | (cid:105) : • Prize is in D3 | (cid:105) . Bob opens D2 | (cid:105) using anti-controlled-Hadamard ( ACH S S ) and Alice has thesuperposition of | (cid:105) and | (cid:105) . • Prize is in D2 | (cid:105) . Bob opens D3 | (cid:105) using anti-controlled-Hadamard, ( ACH S S ) and Alice hassuperposition of | (cid:105) and | (cid:105) . • Prize is in D1 | (cid:105) . Bob opens D2 | (cid:105) using anti-controlled-Hadamard, ( ACH S S ) and Alice hassuperposition of | (cid:105) and | (cid:105) . Case-2B; Alice chooses D2 | (cid:105) : • Prize is in D3 | (cid:105) . Bob opens D1 | (cid:105) using anti-controlled-Z gate ( ACZ S S ) and anti-controlled-Hadamard ( ACH S S ) and Alice has superpositionof | (cid:105) and | (cid:105) . • Prize is in D2 | (cid:105) . Bob opens D1 | (cid:105) using anti-controlled-Z gate ( ACZ S S ) and anti-controlled-Hadamard, ( ACH S S ) and Alice has the superpo-sition of | (cid:105) and | (cid:105) . • Prize is in D1 | (cid:105) . Bob opens D3 | (cid:105) using anti-controlled-Hadamard, ( ACH S S ) and Alice hassuperposition of | (cid:105) and | (cid:105) . Case-2C; Alice chooses D3 | (cid:105) : • Prize is in D3 | (cid:105) . Bob opens D1 | (cid:105) using anti-controlled-Z gate, ( ACZ S S ) and anti-controlled-Hadamard, ( ACH S S ) and Alice has superposi-tion of | (cid:105) and | (cid:105) . • Prize is in D2 | (cid:105) . Bob opens D1 | (cid:105) using anti-Controlled Z-gate, ( ACZ S S ) and anti-controlled-Hadamard, ( ACH S S ) and Alice has superposi-tion of | (cid:105) and | (cid:105) . • Prize is in D1 | (cid:105) . Bob opens D2 | (cid:105) using anti-controlled-Hadamard, ( ACH S S ) and Alice hassuperposition of | (cid:105) and | (cid:105) . Step-3:
Now, for each superposition, there may be four cases.For example, in the superposition of | (cid:105) and | (cid:105) , i.e.doors D1 and D2 remained to be opened, the possiblecases are; • If prize is in D1, and Alice chooses D1, then shewins. • If prize is in D1, and Alice chooses D2, then sheloses. • If prize is in D2, and Alice chooses D1, then sheloses. • If prize is in D2, and Alice chooses D2, then shewins.For our scheme, we only consider the winning cases. Inthe cases Alice loses, the superposition states do not cor-respond to the state of the doors, and hence an Identity-Ioperation is performed on the ancilla state, resulting inno change in their states.
Case-A; When the prize is either in D1 | (cid:105) orD2 | (cid:105) : • When the prize is in D1 | (cid:105) and Alice chooses D1,an anti-controlled-Hadamard ( ACH S S ) is appliedon the superposition state of | (cid:105) and | (cid:105) , afterwhich X gate is applied on the second qubit to get | (cid:105) state. • When the prize is in D2 | (cid:105) and Alice chooses D2,an anti-controlled-Hadamard ( ACH S S ) is appliedon the superposition state of | (cid:105) and | (cid:105) to get | (cid:105) state.Then two four-qubit-controlled-Not operations (the cir-cuit in green in Fig. 2) are applied to the first ancillaqubit ( | (cid:105) A ). If the ancilla qubit after the measurementis changed to | (cid:105) state, then Alice wins. Case-B; When the prize is either in D1 | (cid:105) orD3 | (cid:105) : | i D | i D | i D | i I | i I | i I | i I | i S H H H X H H X H | i S H H H Z H H X H X | i A X X %| i A X X %| i A X X % FIG. 2.
Protocol - Scheme II . The quantum circuit for the implementation of the Monty Hall game. A superposition of theinitial states is done. Alice then chooses one of the three doors. Bob then opens one of the three doors, depending on whichdoor Alice chooses to open and the prize in behind which door. Alice then makes her second choice. Bob opens the doors Alicechooses to open. | i D | i D | i D | i I | i I | i S H H H H H | i S H H H H Z H Z H H H H H Z H
FIG. 3.
Simplified Circuit in Scheme II . The equivalent quantum circuit for creating a superposition state to represent theinitial state of the game when no doors are opened and for obtaining the superposition of two states after Bob opens one of thedoors, depending on where the prize is, Alice’s choice of doors and our assumptions. All the 9 possible cases with 9 controlledoperations are combined into 4 controlled operations as depicted in the right hand side circuit. • When the prize is in D1 | (cid:105) and Alice chooses D1,an anti-controlled-Hadamard ( ACH S S ) is appliedon the superposition state of | (cid:105) and | (cid:105) , afterwhich X gate is applied on the first qubit to get | (cid:105) state. • When the prize is in D3 | (cid:105) and Alice chooses D3,an anti-controlled-Hadamard ( ACH S S ) is appliedon the superposition state of | (cid:105) and | (cid:105) to get | (cid:105) state.Then two four-qubit-controlled-Not operations (the cir-cuit in blue in Fig. 2) are applied to the second ancillaqubit ( | (cid:105) A ). If the ancilla qubit after the measurementis changed to | (cid:105) state, then Alice wins. Case-C; When the prize is either in D2 | (cid:105) orD2 | (cid:105) : • When the prize is in D2 | (cid:105) and Alice chooses D2,a controlled-NOT gate (CNOT) is applied on thesuperposition state of | (cid:105) and | (cid:105) , after whichHadamard gate ( H ) is applied on the first qubitto get | (cid:105) state. • When the prize is in D3 | (cid:105) and Alice chooses D3,a controlled-NOT gate (CNOT) is applied on thesuperposition state of | (cid:105) and | (cid:105) , after whichHadamard gate ( H ) is applied on the first qubitto get | (cid:105) , which is further acted on by X gates onboth the qubits, to get | (cid:105) state.Then two four-qubit-controlled-Not operations (the cir-cuit in purple in Fig. 2) are applied to the third ancillaqubit ( | (cid:105) A ). If the ancilla qubit after the measurementis changed to | (cid:105) state, then Alice wins.After the measurement of all the three ancilla qubits,if the outcomes is any of the states out of | (cid:105) , | (cid:105) and | (cid:105) states, then Alice wins otherwise looses. V. CONCLUSION
To conclude, we have presented a new scheme for thequantum version of the Monty Hall problem. We wishto bring to light the main features in which our presentquantum version of the Monty Hall problem differs fromthe other quantum versions already discussed in the lit-erature. It has similarities with the work of Flitney and Abbott (2002) [10], Khan et al. (2010) [2] and Kurzykand Glos (2016) [18]. However our scheme is differentin the way that we have created a maximally entangle-ment pair between Alice and Bob in the latter part ofthe circuit. While earlier versions of this quantum gameis modelled on the basis of a composite quantum systemconsisting of three qutrits: one associated with the loca-tion of the prize, a second one corresponding to Alice’schoice and a third one associated with Bob’s choice. Inour discussed protocol, we have assigned states to thedoors, while Alice and Bob perform operations on thedoors, depending on certain restrictions.It is shown that a fair two-party zero-sum game canbe carried out if a player is permitted to adopt quan-tum measurement strategy, while in classical situation,the other player can always win with high probability[13]. Since we have been able to design a circuit for aqutrit state using qubits, we have paved the path for fu-ture work in designing algorithms for quantum games andimplementing circuits for the same. Further we wouldlike to work on implementing quantum circuits for othergames like the Prisoner’s dilemna and other Bayesiangames. Also, any two player game can be extended tothree-player games and further to n-player games in gen-eral. Quantum games can be used to understand variousother interesting systems like Optical networks, besidesothers [20]. We are also interested in the experimental re-alization of the quantum Monty Hall problem and othergames [21]. We also wish to increase the efficiency of ourcircuit by working on error correction and better algo-rithms [22–34]. Studying quantum games and quantumproblems like the Monty Hall problem motivates to de-velop advanced techniques more suited to contemporarypractical problems.
VI. ACKNOWLEDGMENTS
S. P. would like to acknowledge the contributions ofN. N. Hegade, K. Haldar and R. K. Singh in the under-taking of this paper. B.K.B. acknowledges the supportof Institute fellowship provided by IISER Kolkata. Theauthors acknowledge the support of IBM Quantum Ex-perience for producing experimental results. The viewsexpressed are those of the authors and do not reflect theofficial policy or position of IBM or the IBM QuantumExperience team. [1] M. A. Nielson and I. L. Chuang, Quantum Computationand Quantum Information, Cambridge University Press,Cambridge (2000).[2] S. Khan, M. Ramzan, and M. K. Khan, Commun. Theor.Phys. , 47 (2010).[3] D. A. Meyer, Phys. Rev. Lett. , 1052 (1999).[4] J. Eisert, M. Wilkens, and M. Lewenstein, Phys. Rev.Lett. , 3077 (1999). [5] L. Marinatto and T. Weber, Phys. Lett. A , 291(2000).[6] A. P. Flitney and D. Abbott, J. Phys. A , 449 (2005).[7] T. Cheon and I. Iqbal, Phys. Soc. Japan , 024801(2008).[8] M. Ramzan, A. Nawaz, A. H. Toor, and M. K. Khan, J.Phys. A.:Math. Theor. , 055307 (2008). [9] A. Iqbal, T. Cheon, and D. Abbott, Phys. Lett. A ,6564 (2008).[10] A. P. Flitney and D. Abbott, Phys. Rev. A , 062318(2002).[11] R. Prevedel, A. Stefanov, P. Walther, and Z. Zeilinger,New J. Phys. , 205 (2007).[12] H. E. Brandt, Prog. Quantum Electron. , 257 (1998).[13] C.-F. Li, Y.-S. Zhang, Y.-F. Huang, and G.-C. Guo,Phys. Lett. A , 257 (2001).[14] M. vos Savant, Am. Stat , 347 (1991).[15] L. Gillman, Am. Math. Monthly , 3 (1992).[16] C. Zander, M. Casas, A. Plastino, and A. R. Plastino,An. Braz. Acad. Sci. , 417 (2006).[17] G. M. D’Ariano, R. D. Gill, M. Keyl, R. F. Werner, B.Kummerer, and H. Maassen, Quantum Inf. Comput. ,355 (2002).[18] D. Kurzyk and A. Glos, Quantum Inf. Process. , 4927(2016).[19] A. Barenco, C. H. Bennett, R. Cleve, N. Margolus, P.Shor, T. Sleator, J. Smolin, and H. Weinfurter, Phys.Rev. A , 3457 (1995).[20] P. B. M. de Sousa and R. V. Ramos, arXiv:1105.2289.[21] C. Schuck, O. Schulz, C. Kurtseifer, H. Wienfurter, Euro.Quantum Elect. Conf. (2003).[22] D. Ghosh, P. Agarwal, P. Pandey, B. K. Behera, andP. K. Panigrahi, Quantum Inf. Process.17