Power of the interactive proof systems with verifiers modeled by semi-quantum two-way finite automata
aa r X i v : . [ c s . CC ] M a y Power of the interactive proof systems with verifiers modeled bysemi-quantum two-way finite automata
Shenggen Zheng , , Daowen Qiu , , ∗ , Jozef Gruska Department of Computer Science, Sun Yat-sen University, Guangzhou 510006, China Faculty of Informatics, Masaryk University, Brno 60200, Czech Republic SQIG–Instituto de Telecomunica¸c˜oes, Departamento de Matem´atica, Instituto Superior T´ecnico, Universidade de Lisboa,Av. Rovisco Pais 1049-001, Lisbon, Portugal
Abstract
Interactive proof systems (IP) are very powerful – languages they can accept form exactly PSPACE. Theyrepresent also one of the very fundamental concepts of theoretical computing and a model of computationby interactions. One of the key players in IP are verifiers. In the original model of IP whose power is thatof PSPACE, the only restriction on verifiers is that they work in randomized polynomial time. Because ofsuch key importance of IP, it is of large interest to find out how powerful will IP be when verifiers are morerestricted. So far this was explored for the case that verifiers are two-way probabilistic finite automata (Dworkand Stockmayer, 1990) and one-way quantum finite automata as well as two-way quantum finite automata (Nishimura and Yamakami, 2009). IP in which verifiers uses public randomization is called
Arthur-Merlinproof systems (AM). AM with verifiers modeled by Turing Machines augmented with a fixed-size quantumregister (qAM) were studied also by Yakaryilmaz (2012). He proved, for example, that an NP-completelanguage L knapsack , representing the 0 - 1 knapsack problem, can be recognized by a qAM whose verifier isa two-way finite automaton working on quantum mixed states using superoperators.In this paper we explore the power of AM for the case that verifiers are two-way finite automata withquantum and classical states (2QCFA) – introduced by Ambainis and Watrous in 2002 – and the commu-nications are classical. It is of interest to consider AM with such “semi-quantum” verifiers because theyuse only limited quantum resources. Our main result is that such Quantum Arthur-Merlin proof systems(QAM(2QCFA)) with polynomial expected running time are more powerful than the models in which theverifiers are two-way probabilistic finite automata (AM(2PFA)) with polynomial expected running time.Moreover, we prove that there is a language which can be recognized by an exponential expected runningtime QAM(2QCFA), but can not be recognized by any AM(2PFA), and that the NP-complete language L knapsack can also be recognized by a QAM(2QCFA) working only on quantum pure states using unitaryoperators. Keywords:
Quantum computing, quantum finite automata, quantum Arthur-Merlin proof systems,two-way finite automata with quantum and classical states. ∗ Corresponding author.
E-mail address: [email protected] (D. Qiu).
Preprint submitted to Elsevier July 11, 2018 . Introduction
An important way to get deeper insights into the power of various quantum resources and operations isto explore the power of various quantum variations of the basic models of classical automata. Of a specialinterest is to do that for various quantum variations of the classical finite automata, especially for those thatuse limited amounts of always expensive quantumness – quantum resources: states, correlations, operationsand measurements. This paper aims to contribute to such a line of research.There are two basic approaches toward how to introduce quantum features to classical models of finiteautomata. The first one is to consider quantum variants of the classical one-way (deterministic) finiteautomata (1FA or 1DFA) and the second one is to consider quantum variants of the classical two-wayfinite automata (2FA or 2DFA). Already the very first attempts to introduce such models, by Moore andCrutchfields [21] as well as Kondacs and Watrous [17] demonstrated that in spite of the fact that in theclassical case, 1FA and 2FA have the same recognition power, this is not so for their quantum variations (incase only unitary operations and projective measurements are considered as quantum operations). Moreover,already the first model of two-way quantum finite automata (2QFA), namely that introduced by Kondacsand Watrous, demonstrated that quantum variants of 2FA are much too powerful – they can recognizeeven some non-context free languages and are actually not really finite in a strong sense [17]. It started tobe therefore of interest to introduce and explore some “less quantum” variations of 2FA and their power[1–3, 5, 6, 18–20, 32, 33].A “hybrid” quantum variation of 2FA, namely, two-way finite automata with quantum and classical states (2QCFA) was introduced by Ambainis and Watrous [3]. Using this model they were able to show, in anelegant way, that already an addition of a single qubit to a classical model can much increase its power. A2QCFA is essentially a classical 2FA augmented with a quantum memory of constant size (for states of afixed Hilbert space) that does not depend on the size of the (classical) input. In spite of such a restriction,2QCFA have been shown to be more powerful than two-way probabilistic finite automata (2PFA) [3, 36, 37].In mid 1980s, Babai [4] and Goldwaser et al. [12], independently, introduced so-called interactive proofsystems with unlimited power provers and polynomial power randomized verifiers. A famous result of [29],stated as IP=PSPACE, that languages recognized by IP are exactly those from PSPACE, demonstratedenormous power hidden in simple interactions of IP.It is therefore natural to explore power also of some weaker variations of IP. Since unlimited power ofprovers seems to be very essential for the whole concept of IP, the research started to focus on the caseswith limited power verifiers. This has been done at first by Dwork and Stockmeyer [9] – they explored thecase that verifiers are two-way probabilistic finite automata (IP(2PFA)). They showed that every languagein the class EXP can be accepted by some IP(2PFA). However, the set of languages recognized by such IPin which verifiers use public randomization (also called Arthur-Merlin proof systems) is a proper subset ofP. Later, Nishimura and Yamakami [24] explored the case that verifiers are modeled by one-way quantumfinite automata as well as two-way quantum finite automata and demonstrated strengths and weaknesses ofboth IP.Of importance is also a variant of IP, called
Arthur-Merlin proof systems (AM). The difference betweenIP and AM is that the prover of IP has at each step only partial information of the configuration of theverifier while the prover of AM always has complete information of the current configuration of the verifier.Also for such interactive proof systems it is of importance to explore their power for the case that verifiershave a more limited power and to find out relations between IP and AM with verifiers of different power.AM with verifiers modeled by Turing Machines augmented with a fixed-size quantum register (qAM) were2tudied also in [34, 35] and it was shown that the an NP-complete language L knapsack , representing the 0 - 1knapsack problem, can be recognized by a qAM whose verifier is a two-way finite automaton working onquantum mixed states using superoperators. In Yakaryilmaz’s notation, two-way finite automata workingon quantum mixed states using superoperators are called 2QCFA. However, 2QCFA as defined originally in[3], are working only on quantum pure states using unitary operators. They can be simulated efficiently by two-way finite automata working on quantum mixed states, but whether two-way finite automata workingon quantum mixed states can be simulated by 2QCFA, or not, is unknown. The model of 2QCFA we useis that of [3] and it is weaker, actually a special case of the model used in [34, 35]. Our results concerningthe acceptance of the language L knapsack are therefore stronger. It is also worth mentioning that a notionof QMA for quantum-automata verifiers was introduced in [23–25] (under the name “public QIP”).Our model will be denoted as QAM(2QCFA). One can see this model also as a classical AM augmentedwith a quantum memory of constant size – to store quantum states of a fixed Hilbert space – that doesnot depend on the size of the (classical) input. Our main results show that such models are more powerfulthan AM(2PFA) – that is AM with 2PFA as verifiers, and the NP-complete language can be recognized byQAM(2QCFA).The paper is structured as follows. In Section 2 all models involved are described in detail. After thatwe show for the language L middle = { xay | x, y ∈ { a, b } ∗ , | x | = | y |} that for any 0 ≤ ε < / A ( P, V ε ) – with the prover P and the verifier V ε that accepts L middle with one-sided error ε in a polynomial expected running time – notation QAM(ptime-2QCFA). This language cannot be recognizedby any AM(2PFA) in polynomial expected running time, as shown in [9]. As we will show in the paper, forthe language L mpal = { xax R | x ∈ { a, b } ∗ } , that for any 0 ≤ ε < / A ( P, V ε ) thatcan recognize L mpal with one-sided error ε in an exponential expected running time. We will prove that thislanguage cannot be recognized at all by an AM(2PFA). These results show that QAM(2QCFA) are morepowerful than AM(2PFA). Afterwards we show that there is an NP-complete language, namely L knapsack ,representing the 0 - 1 knapsack problem, that can be recognized by QAM(2QCFA) in an exponential expectedrunning time. Finally, we discuss languages, L = { w | ∃ s, t, u, v ∈ { a, b } ∗ , w = sbt = ubv, | s | = | v |} and L = { w | ∃ s, t, u, v ∈ { a, b } ∗ , w = sat = ubv, | s | = | v |} , that can be recognized by QAM(ptime-2QCFA).The language L is proved to be nonstochastic. The language L , the set of nonpalindromes, is stochastic[10].The fact that the non-regular language L middle can be recognized by a QAM(ptime-2QCFA) and it seemsthat it can not be recognized by a 2QCFA, indicates that QAM(ptime-2QCFA) are likely more powerfulthan 2QCFA. Interestingly enough, this situation seems to be different for 2PFA. It is still an open problemto find out whether there is a non-regular language that can be recognized by AM(ptime-2PFA), but weknow that any 2PFA needs exponential time to recognize a non-regular language [8, 13] .
2. Basic models
At first we introduce formally the model IP(2PFA) and afterwards also the model QIP(2QCFA). Con-cerning basics of quantum computation we refer the reader to [14, 22], and concerning basics of classicaland quantum automata we refer the reader to [14–16, 26, 28].
Notation: A coin-tossing distribution on a finite set S is a mapping φ : S → { , / , } such that P s ∈ S φ ( s ) = 1. 3 Classical Control Unit C Communication Cell A 2PFA verifier
A prover
Powerful processor An infinite tape Read only tape head Read only tape head Read/write tape head C A source of randomness … Figure 1: A schematic model of an IP(2PFA)
Definition 1.
An IP(2PFA) A = ( P, V ), where P is a so-called prover and V is a so called verifier (thatare specified bellow and communicate through a communication cell as illustrated in Figure 1).An action of A starts with a step of the verifier. The verifier’s head scans the left end-marker, and theverifier writes a symbol to the communication cell. That is followed by an action of the prover who writesits response into the communication cell. Such rounds of steps continue until the verifier decides to end it.The verifier V is a 2PFA specified as follows: V = ( S, Σ , δ, s , S acc , S rej ), where1. S is a finite set (of classical states) partitioned into subsets R, C, H of reading, communication andhalting states, respectively.2. s ∈ R is the initial reading state.3. H is partitioned into sets S acc and S rej of the accepting and rejecting states.4. Σ is a finite input alphabet that is extended into the alphabet Σ ′ = Σ ∪ { | c, $ } , where | c / ∈ Σ will beused as the left end-marker and $ / ∈ Σ will be used as the right end-marker. Γ is a communicationalphabet (shared by both the verifier and the prover).5. δ is a transition mapping defined as follows:(a) For each reading state s ∈ R and σ ∈ Σ ′ , δ ( s, σ ) is a coin-tossing distribution on S × {− , , } ,where − , δ is well defined in the sense that on end-markers it never determines a move outside of thetape region separated by end-markers | c and $. Therefore δ specifies, for each state of the controlunit and each symbol on the tape, the probability that the control unit will be in a particularnew state and that the tape head moves in a particular direction.(b) With each communication state s ∈ C , a unique communication symbol γ s is associated. In thecase the verifier is in such a communication state s , then it writes γ s in the communication celland then the prover gets into an action. It reads the symbol stored in the communication celland, depending on the whole communication history and the input string (that is the same forboth the prover and the verifier), the prover writes a symbol γ ∈ Γ into the communication cell.Afterwards the verifier gets into the action that depends on its current state and on the symbol4n the communication cell. For each pair s ∈ S and γ ∈ Γ a coin-tossing distribution functionon the set S is defined that determines the probability for each state from S that it should bethe next state of the verifier. At such after-communication step the head of the verifier does notmove.(c) The verifier halts when it gets into a halting state. Therefore there is no need to define δ onstates in H .The prover P is all-powerful and at each communication step what the prover writes into the com-munication cell depends on the whole input and the whole communication history up to that point ofcommunication. Namely, it is determined, for an input string x and communication history y ∈ Γ ∗ to thatpoint, by a coin-tossing distribution ρ ( x, y ) defined on Γ.Note. The communication cell always holds a symbol from the communication alphabet Γ. Wheneverthe verifier needs some information from the prover, the verifier writes a request to the communication cellvia a symbol from the alphabet Γ and the prover responds. The verifier then continues, probabilistically,depending on the prover’s respond. Remark 1.
In the above definition, the prover’s understanding of the verifier’s computation is only throughthe communication cell that can contain information only from the finite set Γ. This mode of communicationis called private coins [11] or partial information [7]. One can consider also IP in which the prover hascomplete information on the current configuration of the verifier. Such a communication mode is called public coins [11] or complete information [7] mode. In Babai’s terminology, an IP with the public coinscommunication mode is called Arthur-Merlin proof system (AM). In an AM(2PFA), the verifier in each stepautomatically sends its configuration information, that is the communication symbol corresponding to thecommunication alphabet contains an element of S × {− , , } , that is, the current state and the last moveof the head, to the prover through the communication cell in every computation step. This information issufficient for the all-powerful prover to keep the track of the configuration of the verifier because the proverknows the strategy of the verifier.The computation of ( P, V ) on an input string w starts with the string | cw $ on the input tapes of both theverifier and the prover. The tape head of the verifier is positioned on the left end-marker | c and the verifierbegins to act with the initial state s in its control unit. The action of the verifier and the prover in thenext steps is then governed (probabilistically) by the transition functions δ and ρ , as defined above, untilthe verifier enters a halting state. For a particular input w and a halting state s let P r ( P,V ) ( w, s ) be theprobability that IP ( P, V ) halts its computation on w in the state s . The probability that the verifier haltsin a halting state s on the input w is taken over all random choices of the verifier and the prover. If s is inan accepting (a rejecting) state, then the input is accepted (rejected).The prover-verifier pair ( P, V ε ) is an AM(2PFA) for (accepting) a language L ⊂ Σ ∗ with an errorprobability ε < / (Completeness condition) : for all w ∈ L , P r [( P, V ε ) accepts w ] ≥ − ε , and2. (Soundness condition) : for all w L and any prover P ∗ , P r [( P ∗ , V ε ) rejects w ] ≥ − ε .We say that a language L is recognized by AM(2PFA) if for some ε < /
2, there is an AM(2PFA) (
P, V ε )that accepts the language L with the error probability ε .5 .2. Model QIP(2QCFA) Arthur-Merlin proof system called now quantum Arthur-Merlin proof system (QAM)have drawn a significant attention [31, 34, 35]. In [34, 35], for Arthur-Merlin IP verifiers are augmented bya fixed-size quantum memory. In [23–25] one-way quantum finite automata (both of the measure-many and measure-once types) and two-way quantum finite automata are considered as verifiers.In this paper, such IP are mainly considered in which verifiers are not 2PFA, as in [23–25], but 2QCFAand the verifier and the prover have just classical communication. More exactly, verifiers are augmentedby a quantum memory size of which does not depend on the size of the input. The formal definition is asfollows: C Classical Control Unit C Communication Cell A 2QCFA verifier Powerful processor C Quantum Memory Classical An infinite tape Read only tape head Read only tape head Read/write tape head
A prover … Figure 2: A schematic model of a QIP(2QCFA)
Definition 2.
A QIP(2QCFA) is given by a pair (
P, V ), where P is a prover and V is a verifier, as illustratedin Figure 2, and the prover communicate classically with the verifier through a special communication cell.In such a QIP(2QCFA) ( P, V ), the verifier V is a 2QCFA A = ( Q, S, Σ , Γ , Θ , δ, | q i , s , S acc , S rej ), where1. Q is a finite set of an n basic orthonormal quantum states.2. S is a finite set of classical states that is partitioned into subsets R, C, H of the reading, communicationand halting states..3. Σ is a finite set of input symbols extended to the tape symbols Σ ′ = Σ ∪ {| c, $ } , where | c is used as theleft end-marker and $ is used as the right end-marker.4. Γ is a finite set of communication symbols.5. | q i ∈ Q is the initial quantum state. 6. s is the initial classical state.7. The set H is partitioned into subsets S acc and S rej of the accepting and rejecting states.8. Θ is a mapping Θ : R × Σ ′ → U ( H ( Q )) ∪ O ( H ( Q )), where U(H(Q)) and O(H(Q)) are sets of unitaryoperations and projective measurements on the Hilbert space generated by quantum states from Q .9. δ is a classical transition function defined as follows.(a) For each reading state s ∈ R and a tape symbol σ i. If Θ( s, σ ) ∈ U ( H ( Q )), then the unitary operation Θ( s, σ ) is applied on the current state ofquantum memory to produce a new quantum state and, in addition, if δ : R × Σ ′ → S × {− , , } , then, for the case δ ( s, σ ) = ( s ′ , d ), the new classical state of the verifier is s ′ and its headmoves in the direction d .ii. If Θ( s, γ ) ∈ O ( H ( Q )), then Θ( s, σ ) is a projective measurement with a set of possible eigen-values E Θ( s,λ ) = { λ , . . . , λ n } and projectors { P , . . . , P n } , where P i is the projector onto theeigenspaces generated by eigenvectors corresponding to λ i . In such a case δ ( s,σ ) : E → S × {− , , } and δ ( s,σ ) ( λ i ) = ( s ′ , d ) means that when the projective measurement outcome is λ i , then thenew classical state is s ′ and its head moves in the direction d .(b) With each communicating state s ∈ C a communication symbol γ s is associated. For each s ∈ C and γ ∈ Γ, δ ( s, γ ) ∈ S . This has the following meaning: If the verifier gets into a state s ∈ C , itwrites γ s into the communication cell and then the prover, depending on the whole communicationhistory and on the input w writes a γ ∈ Γ into the communicating cell and the verifier comes intothe action. If δ ( s, γ ) = s ′ , then s ′ is the new state of the verifier and the tape head of the verifierdoes not move.(c) The verifier halts and accepts (rejects) the input when it enters a classical accepting (rejecting)state from H .The prover is a processor of an unlimited power as in the general case of IP. Remark 2.
A QIP(2QCFA) is a QAM(2QCFA) if the verifier sends, at the end of each its communicationround through a communication symbol information about its current configuration, that is an element of S ×{− , , }× E , where E is the set of all possible measurement outcomes, that is the communication symbolcontains the current classical state, the last move of the head and the projective measurement outcome (leta special symbol is used if there was no measurement in the computation step), to the prover through thecommunication cell in every computation step. This information is sufficient for the all-powerful prover tokeep the track of the configuration of the verifier because the prover knows the strategy of the verifier.The computation of QIP(2QCFA) with a prover P and a verifier V on an input w ∈ Σ ∗ starts with thestring | cx $ on the input tapes of both of them. At the start, the tape head of the verifier is positioned onthe left end-marker and the verifier begins the computation in the classical initial state and in the initialquantum state. In each of the next steps, if the current classical state of the verifier is s and the currentquantum state of the verifier is | ψ i and the scanning symbol is σ , then the quantum and classical states arechanged according to Θ( s, σ ) and δ as follows. 7. If s ∈ R , then:(a) If Θ( s, σ ) is a unitary operation U , then U is applied on the current quantum state | ψ i , changingit into the quantum state U | ψ i , and δ ( s, σ ) = ( s ′ , d ) makes s ′ to be a new classical state and thehead moves in the direction d . If s ∈ S acc ( s ∈ S rej ), then the input is accepted (rejected).(b) If Θ( s, σ ) is a projective measurement, then the current quantum state is changed to P j | ψ i / || P j | ψ i|| with the probability || P j | ψ i|| and in such a case δ ( s,σ ) is a mapping from the set of potentialclassical outcomes (eigenvalues) of the measurement to S × {− , , } . In particular, if the mea-surement coutcome is λ i and δ ( s,σ ) ( λ i ) = ( s ′ , d ), then:i. if s ′ ∈ S − H , then s ′ is the new classical state and the head moves in the direction d ;ii. if s ′ ∈ S acc ( s ′ ∈ S rej ), then the verifier accepts (rejects) the input and computation halts.2. If the current classical state s ∈ C , then the verifier sends γ s to the prover through the communicationcell. Depending on this and all previously obtained symbols from the verifier, as well as on the inputstring, the prover sends to the communication cell a symbol γ . If δ ( s, γ ) = s ′ , then s ′ will become thenew classical state of the verifier and the input head does not move. In case s ′ ∈ S acc ( s ′ ∈ S rej ) theinput is accepted.The probability that an input word is accepted is defined in a similar way as in the case of 2QCFA.The prover-verifier pair ( P, V ε ) is a QAM(2QCFA) for (accepting) a language L ⊂ Σ ∗ with one-sidederror 0 ≤ ε < / Completeness condition : for all w ∈ L , P r [( P, V ε ) accepts w ] = 1, and2. Soundness condition : for all w L and any prover P ∗ , P r [( P ∗ , V ε ) rejects w ] ≥ − ε .We say that a language L is recognized by QAM(2QCFA) if for some ε < / P, V ε ) that accepts the language L with one-sided error ε .
3. Examples of languages recognized by QAM(2QCFA)
In this section we provide a detailed proof for five languages that they are accepted by QAM(2QCFA). L middle The importance of the fact that the language L middle can be recognized by a QAM(ptime-2QCFA) isunderlined by the fact that there is no AM(ptime-2PFA) for this language [9]. Theorem 1.
For any ε < / there exists a QAM(2QCFA) A ε with a verifier-prover pair ( P, V ε ) for thelanguage L middle with one-sided error ε in the expected running time O (cid:0) ε (cid:0) n + n log ε (cid:1)(cid:1) , where n is thelength of the input.Proof. At first we present informally the main idea of the proof. A 2QCFA verifier V ε uses quantum memorywith states generated by two orthogonal quantum states | q i and | q i , where | q i will be the initial quantumstate. At the beginning the input is shared by both the verifier and the prover. In the first part of theinteraction/computation process, until the middle symbol is reached, for each input symbol σ the verifierdoes the following.It applies the following unitary transformation U α , α = √ π , U α | q i = cos α | q i + sin α | q i , U α | q i = − sin α | q i + cos α | q i , (1)8 epeat the following steps ad infinity:1. Move the tape head to the symbol that is next to the left end-marker.2. Ask the prover whether the currently read symbol is the middle one.2.1 If the answer is “no”, move the head to the right, apply transformation U α and repeat Step 2.2.2 If the answer is “yes” and the scanned symbol is not a the verifier rejects the input; otherwisethe verifier moves the head one cell to the right and proceeds according to Step 3.3. Until the scanned symbol is the right end-marker, apply U − α to the current quantum state andmove the head one cell to the right.4.0 When the right end-marker is reached, measure the quantum state in the basis {| q i , | q i} .4.1 If the result is | q i , the input is rejected.4.2 Otherwise repeat the following subroutine two times:4.2.1 Move the head to the first symbol right to the left end-marker.4.2.2 Until the currently read symbol is one of the end-markers simulate a coin-flip and movethe head right (left) if the outcome of the coin-flip is “head” (“tail”).5. If the above process ends both times at the right end-marker, simulate k coin-flips and if all outcomesare “head”, then accepts the input. Figure 1: Description of the behavior of of pair (
P, V ) for mpal . The choice of will depend on
Figure 3: Description of the behavior of the verifier V ε in a QAM recognizing the language L middle . The choice of k dependson ǫ , as discussed in the text. to the current quantum state (that is the state is rotated by the angle α ) and asks the prover whether the nextsymbol is the middle one. The all-powerful prover provides the answer. If the next symbol is not the middleone, the verifier moves its head one cell to the right and this process repeats. When the symbol in the middleis reached, the verifier checks whether it is the symbol a . If not, the verifier rejects the input; otherwise itcontinues to read the input, symbol by symbol, and each time it applies the unitary transformation U − α to the current quantum state (and therefore this quantum state is rotated by the angle − α ) until the rightend-marker is reached. (During this process the verifier has no need to get any information from the prover.)When the right end-marker is reached, the verifier measures the current quantum state in the basis {| q i , | q i} . If | q i is the resulting quantum state, the input is rejected, otherwise the verifier proceeds asshown in Figure 3. Lemma 2.
If the input w ∈ L middle , then the quantum states of the verifier V ε evolve with certainty into | q i when the right end-marker is reached in Step 4.Proof. If w ∈ L middle , then there are strings x, y ∈ Σ ∗ such that | x | = | y | and w = xay . Since the all-powerful prover can tell the verifier for sure when the middle symbol is reached, the quantum state whenthe right end-marker is reached is | q i = ( U − α ) | y | ( U α ) | x | | q i = cos α − sin α sin α cos α ! | y | cos α sin α − sin α cos α ! | x | | q i (2)= cos | y | α − sin | y | α sin | y | α cos | y | α ! cos | x | α sin | x | α − sin | x | α cos | x | α ! | q i = ! | q i = | q i . (3) Lemma 3. [37] A coin flipping can be simulated by the verifier V ε using states | q i and | q i . Lemma 4. [3] If the input w ∈ L middle , then the execution of loops in the steps 4.2 and 5 leads to theacceptance with the probability k ( n +1) .
9t may happen that the process described in Figure 3 does not terminate. This happens only if the resultof the last coin-flip is “tail” and this happens either when the left end-marker is reached in Step 4.2 or thishappens in Step 5. In such a case the quantum state of the verifier is | q i and a new iteration of the processstarts. Completeness condition: If w ∈ L middle , then the quantum state of the verifier after Step 4.0 is | q i and the input is never rejected in Step 4.1. After the rest of the steps, in 4.2 and 5, the probability ofaccepting the input is, if we denote n = | w | , P a = 1 / (2 k ( n + 1) ) according to the previous lemma, and theprobability of rejecting the input is P r = 0. If the whole process is repeated for infinitum, the acceptanceprobability is P r [( P ∗ , V ε ) rejects w ] = X i ≥ (1 − P a ) i (1 − P r ) i P a = P a P a + P r − P a P r = P a P a = 1 . (4) Soundness condition : Let the input string w L middle and n = | w | . Observe that the verifier, inits communication with the prover, waits only for the information that the currently read symbol is in themiddle and let this come after reading m (0 < m < n ) symbols in the input string (this m can be differentat different iterations and for different provers). Lemma 5. If w L middle , then the verifier rejects the input for any prover after Step 4.1 with probabilityat least / (2 n + 1) .Proof. Suppose that a prover P ∗ , after m steps, informs the verifier that the next symbol is the middle one.If w does not have the form xay , | x | = m , then w is rejected in Step 2.2.If w = xay with | x | = m , then, before the measurement in Step 4.0 the quantum state of the verifier willbe | q i = ( U − α ) n − m − ( U α ) m | q i = cos α − sin α sin α cos α ! n − m − cos α sin α − sin α cos α ! m | q i (5)= cos( n − m − α − sin( n − m − α sin( n − m − α cos( n − m − α ! cos mα sin mα − sin mα cos mα ! | q i (6)= cos( n − m − α sin( n − m − α sin( n − m − α cos( n − m − α ! | q i (7)= cos (( n − m − α ) | q i + sin (( n − m − α ) | q i . (8)The probability of observing | q i is sin (cid:0) √ n − m − π (cid:1) in Step 4.0. Without a loss of generality, weassume that n − m − >
0. Let l be the closest integer to √ n − m − √ n − m − > l , then2( n − m − > l . So we get 2( n − m − − ≥ l and l ≤ p n − m − −
1. Therefore √ n − m − − l ≥ √ n − m − − p n − m − − √ n − m − − p n − m − − √ n − m −
1) + p n − m − − √ n − m −
1) + p n − m − − √ n − m −
1) + p n − m − − > √ n − m − . (11)10ecause l is the closest integer to √ n − m − < √ n − m − − l < /
2. Let f ( x ) = sin ( xπ ) − x . We have f ′′ ( x ) = − π sin( xπ ) ≤ x ∈ [0 , / f ( x ) is concave inthe interval [0 , / f (0) = f (1 /
2) = 0. So, for any x ∈ [0 , / f ( x ) ≥
0, that is sin( xπ ) ≥ x .Therefore, we have sin ( √ n − m − π ) = sin (cid:16) ( √ n − m − − l ) π (cid:17) (12) ≥ (cid:16) √ n − m − − l ) (cid:17) = 4 (cid:16) √ n − m − − l (cid:17) (13) > (cid:18) √ n − m − (cid:19) = 12( n − m − > n − m − + 1 . (14)If √ n − m − < l , then 2( n − m − < l . So we get 2( n − m − + 1 ≤ l and l ≥ p n − m − + 1. We have therefore √ n − m − − l ≤ √ n − m − − p n − m − + 1 (15)= (cid:16) √ n − m − − p n − m − + 1 (cid:17) (cid:16) √ n − m −
1) + p n − m − + 1 (cid:17) √ n − m −
1) + p n − m − + 1 (16)= − √ n − m −
1) + p n − m − + 1 < − p n − m − + 1 (17)and this implies that l − √ n − m − > p n − m − + 1 . (18)Because l is the closest integer to √ n − m − < l − √ n − m − < /
2. Therefore,sin (cid:16) √ n − m − π (cid:17) = sin (cid:16) ( √ n − m − − l ) π (cid:17) (19)= sin (cid:16) ( l − √ n − m − π (cid:17) ≥ (cid:16) l − √ n − m − (cid:17) (20)= 4 (cid:16) l − √ n − m − (cid:17) > p n − m − + 1 ! = 12( n − m − + 1 . (21)As 0 < m < n , we have | n − m − | < n , and therefore12( n − m − + 1 > n + 1 . (22)So the lemma has been proved.If w L middle , then for any prover P ∗ the above verifier rejects the input after Step 4.1 with theprobability P r > n + 1 (23)according to Lemma 5. Moreover, after the last two steps the verifier accepts w with the probability P a = 12 k ( n + 1) . (24)If k = 1 + ⌈ log /ε ⌉ , then ε ≥ / k − . 11f the whole process is repeated indefinitely, then the probability that the verifier rejects the input w forany prover P ∗ is P r [( P ∗ , V ε ) rejects w ] = X i ≥ (1 − P a ) i (1 − P r ) i P r = P r P a + P r − P a P r > P r P a + P r (25) > / (2 n + 1) ε/ n + 1) + 1 / (2 n + 1) = ( n + 1) / (2 n + 1) ε/ n + 1) / (2 n + 1) (26)If we now denote f ( x ) = xε/ x = 1 − ε ( ε +2 x ) , then f ( x ) is monotonously increasing in (0 , + ∞ ) and since( n + 1) / (2 n + 1) ≥ /
2, we have
P r [( P ∗ , V ε ) rejects w ] > / / ε/ ε > − ε. (27)If | w | = n , then Steps 1 to 4.1 takes O ( n ) time, the loops 4.2 takes O ( n ) time, and Step 5 takes O ( k )time. The expected number of the repetitions of the algorithm is O (2 k n ) in both cases. Hence, the expectedrunning time of ( P, V ε ) is O (cid:0) ε (cid:0) n + n log ε (cid:1)(cid:1) . Remark 3.
Concerning the expected running time, if the algorithm halts with the probability h ( n ) : N → [0 ,
1] in one iteration, then the expected number of repetitions till the algorithm halts is + ∞ X i =1 (1 − h ( n )) i − h ( n ) × i = 1 h ( n ) . (28)The halting probability in one iteration in Figure 3 is Ω (cid:0) k n (cid:1) , so according to Equality 28, the expectednumbers of iterations of the algorithm is O (2 k n ). Theorem 6.
Interactive proof systems QAM(ptime-2QCFA) are more powerful than AM(ptime-2PFA).Proof.
The fact that a coin-flipping can be simulated perfectly by 2QCFA in polynomial time, see [36],implies that 2PFA can be simulated in polynomial time by 2QCFA and therefore QAM(ptime-2QCFA) areat least as powerful as AM(ptime-2PFA). By [9], the language L middle cannot be recognized by AM(ptime-2PFA) and this implies that QAM(ptime-2QCFA) are more powerful than AM(ptime-2PFA). L mpal We show now that the language L mpal = { xax R | x ∈ { a, b } ∗ } , which cannot be recognized at all byAM(2PFA), as shown later, can be recognized by QAM(2QCFA) in an exponential expected running time. Theorem 7.
For any ε there is a QAM(2QCFA) A ε with a verifier-prover pair ( P, V ε ) accepting the language L mpal with one sided error ε in the exponential running time O (cid:16) n log ε · n log ε (cid:17) , where n = | w | is the lengthof the input w .Proof. In the QAM described in the following, the prover is used only to determine, for the verifier, themiddle symbol of the input.At the beginning of the actions of the QAM, both the verifier and the prover share an input string w .The verifier V ε starts with the head at the left end-marker, in the initial classical state s and in the initial12 epeat ad infinity:1. Move the tape head to the right of the left end-marker.2. If the scanned symbol is σ and the prover responds negatively to the question of the verifier whetherthe reading head is in the middle, then the verifier applies the operation U σ to the state in itsquantum memory and the head is moved one cell to the right and Step 2 is repeated. Otherwise, goto the next step.3. If the scanned symbol is b ( a ) and the Prover informs the verifier that the head scans the symbolin the middle of the input string, the input is rejected (the head is moved to the next cell and theprocess goes to Step 4).4. Until the scanned symbol σ is the right end-marker, the operation U − σ is applied to the verifier’squantum state and the head moves one cell to the right.5. When the right end-marker is reached, the quantum state of the verifier is measured in the basis {| q i , | q i , | q i} . If the outcome is not | q i , the input is rejected; otherwise go to the next step.6. Move the head to the cell on the left from the right end-marker and set a special register flag to 0.7. Until the currently scanned symbol is not the left end-marker, do the following: Simulate k coin-flips.Set flag to 1 if not all outcomes are “heads”. Move the head one cell left.8. If flag =0, accept. Figure 1: Description of the behavior of of pair (
P, V ) for mpal . The choice of will depend on
Figure 4: Description of the behavior of the verifier V ε of the QAM ( P, V ε ) when the language L mpal is recognized. The choiceof k depends on ε , as discussed in the text. quantum state | q i and keeps moving its head, cell by cell, till the right end-marker is reached, following therules described informally in Figure 4, where U a = 15 − , U b = 15 − (29)are unitary matrices that will be applied on verifier’s quantum states in the Hilbert space spanned by threeorthonormal basis states | q i , | q i and | q i .An application of the unitary matrices on these basis states can also be described as follows U a | q i = | q i − | q i U b | q i = | q i − | q i U a | q i = | q i + | q i U b | q i = | q i U a | q i = | q i U b | q i = | q i + | q i Let us now describe more formally the actions of the verifier V ε for an input w with | w | = n . Completeness condition : After the start, in each round, the verifier V ε keeps asking the prover whetherthe symbol currently scanned is in the middle of the input, until he gets a positive reply. After that theverifier has no need to get more information from the prover. Lemma 8.
If the input w = xax R , then the quantum state of the verifier V ε after Step 4 is the quantumstate | q i .Proof. Let x = x x . . . x l for all x i ∈ { a, b } . After Step 4 the quantum state will be | ψ i = U − x U − x · · · U − x l U x l · · · U x U x | q i = | q i . (30) Lemma 9.
An execution of Steps 6 to 8 leads to an acceptance with probability − kn . roof. The probability that at the end of Step 8 flag = 0 is 2 − kn . This is therefore the probability that theinput is accepted in Step 8. If this is not the case, a new iteration starts in the state | q i .If the input w ∈ L mpal , then the verifier never rejects the input in Step 5. After Steps 6 to 8 theinput is accepted with the probability P a = 2 − kn and is rejected with the probability P r = 0. If the wholecomputation process is repeated indefinitely, the accepting probability is P r [( P ∗ , V ε ) rejects w ] = X i ≥ (1 − P a ) i (1 − P r ) i P a = P a P a + P r − P a P r = P a P a = 1 . (31) Soundness condition : For a three dimensional vector u , let u [ i ] denote its i -th component. Lemma 10. [37] Let u = Y − . . . Y − l X m . . . X (1 , , T where X j , Y j ∈ { A, B } . If X j = Y j for all ≤ j ≤ m and l = m , then u [2] + u [3] = 0 ; otherwise u [2] + u [3] > − ( m + l ) . Lemma 11.
If the input string w L mpal , then for any prover P ∗ the verifier V ε rejects w after Step 5with the probability at least − n .Proof. Suppose that Prover P ∗ informs the verifier after scanning the ( m + 1)-st symbol that it is the symbolin the middle of the input string.If the input string w is not of the form xay with | x | = m , then the verifier rejects the input in Step 3.Assume now that w = xay , | x | = m, | y | = l and y = x R .Let x = x x · · · x m and y = y y · · · y l . Starting with the state | q i , the verifier V ε changes its quantumstate after the loop 4 to: | ψ i = U − y l · · · U − y U − y U x m · · · U x U x | q i . (32)Let | ψ i = β | q i + β | q i + β | q i . According to Lemma 10, β + β > − ( m + l ) . Therefore, if in Step 5, thequantum state | ψ i is measured, then the input is rejected with the probability P r = β + β > − ( m + l ) =5 − n .If w L mpal , then, according to Lemma 11, for any Prover P ∗ the verifier V ε rejects the input after Step5 with the probability P r > − n (33)and the input is accepted after Steps 6 to 8 with the probability P a = 2 − kn . (34)If the whole computation process is repeated indefinitely, the probability that the verifier rejects the inputis (taking into the consideration that k ≥ max { log 5 , log ε } ) P r [( P ∗ , V ε ) rejects w ] = X i ≥ (1 − P a ) i (1 − P r ) i P r = P r P a + P r − P a P r (35) > P r P a + P r > − n − kn + 5 − n >
11 + ε > − ε. (36)14 ime analysis: Steps 1 to 5 take O ( n ) time and Steps 6 to 8 O ( kn ) time. The halting probability is Ω (cid:0) kn (cid:1) in both cases, so the expected number of repetitions of the above process is O (2 kn ) in both cases.Hence the expected running time of the QAM ( P, V ε ) is O ( kn · kn ) and therefore O (cid:16) log ε · n · n log ε (cid:17) .In order to prove that the language L mpal cannot be recognized by AM(2PFA) the following Lemma willbe used. Lemma 12. [9] Let a language L ⊆ Σ ∗ . Suppose there is an infinite set I of positive integers such that, foreach m ∈ I , there are an integer N ( m ) and multisets W m = { w , w , · · · , w N ( m ) } , U m = { u , u , · · · , u N ( m ) } and V m = { v , v , · · · , v N ( m ) } of words such that(1) | w | ≤ m for all w ∈ W m ,(2) for every integer k there is an m k such that N ( m ) ≥ m k for all m ∈ I such that m ≥ m k , and(3) for all ≤ i, j ≤ N ( m ) , u j w i v j ∈ L iff i = j ,then L cannot be recognized by AM(2PFA). We are now ready to prove the following theorem.
Theorem 13. L mpal cannot be recognized by AM(2PFA).Proof. Let I be the set of all positive integers. For each m ∈ I , let N ( m ) = 2 m , and let w , ..., w N ( m ) bean ordering of all words in { a, b } m . Let W m = { w , w , · · · , w N ( m ) } . By means of W m , u i = λ (the emptyword) and v i = aw Ri for all i , then for all 1 ≤ i, j ≤ N ( m ), u j w i v j ∈ L mpal if and only if i = j . Accordingto Lemma 12, the language L mpal cannot be recognized by AM(2PFA) and theorem has been proved.From the last theorem and from Theorem 7 it follows: Corollary 14.
QAM(2QCFA) are more powerful than AM(2PFA).3.3. Recognition of the language L knapsack In this subsection, we consider a language, over the alphabet { , , } , L knapsack = { b a a · · · a n | such that a , · · · , a n , b ∈ { , } ∗ , and there exists a set I ⊆ { , · · · , N } such that v ( b ) = P i ∈ I v ( a i ) } ,where v ( x ) is the number such that x is its binary representation. L knapsack is actually the 0 - 1 knapsackproblem, which is NP-complete. Yakaryilmaz studied QAM with the verifier augmented with a fixed-sizequantum register in the Arthur-Merlin proof system [34, 35] and proved that L knapsack can be recognized byQAM whose verifier is a 2QCFA which uses superoperators. Using an idea from [34, 35] (coding of binarystrings into the amplitudes of quantum states), we prove that the language L knapsack can also be recognizedby QAM(2QCFA). Our model of 2QCFA is weaker than that of [34, 35] and different tools, comparing thosefrom [34, 35], are needed to prove our result. Theorem 15.
For any ǫ < / , there exists a verifier-prover pair ( P, V ǫ ) of a QAM(2QCFA) to recognizethe language L knapsack with one-sided error ǫ in the exponential running time O (cid:0) ε ( n n + log ε ) (cid:1) , where n is the length of the input. roof. Let us assume that the input string w be of the form b a a · · · a n , where b, a i ∈ { , } ∗ . Thiscan be easily checked by an FA. Otherwise, the input string w is rejected immediately. The main idea of theproof is as follows: we consider a 2QCFA verifier V ǫ that uses a quantum memory with states generated bythe set of orthogonal quantum states {| q i i : i = 0 , , · · · , } , where | q i , | q i , | q i are used to encode the valueof the binary strings. The verifier V ǫ starts to work with the initial quantum state | q i . The tape head ofthe verifier V ǫ moves from the left to right. Firstly, the verifier V ǫ encodes the value of b into the amplitudesof the quantum state | q i . With the help of the prover P , the verifier V ǫ will know whether a i is selected ornot . If a i is selected, the value of a i is calculated and encoded into the amplitudes of the quantum state | q i and then subtracted from the amplitude of the quantum state | q i . When the right end-marker $ isreached, the verifier V ǫ measures the current quantum state. If the resulting quantum state is | q i , the inputstring w is rejected. Otherwise, the verifier continues as shown in Figure 5, where the unitary operators andprojective measurements are as follows: If the input string w is not of the form b a a · · · a n b, a i ∈ { , } ∗ (this can be easilychecked by a FA), w is rejected. Otherwise, repeat the following ad infinitum:1. Move the tape head to the right symbol of the left end-marker.2. While the currently scanned symbol σ is not ‘ U σ on the current quantum state, where U σ is defined on the text.2.2 Measure the current quantum state with M for = { P f , P r } .2.2.1 If the measurement result is f , move the tape head one square to the right.2.2.2 Otherwise, set the quantum state to | q i and start a new iteration.3. While the currently scanned symbol is not the right end-marker, do the following:3.1 If the Prover informs the verifier that a i is not selected, move the tape head to the symbol ‘ ′ that is left of a i +1 .3.2 Otherwise, move the tape head one square to the right.3.2.1 While the currently scanned symbol is not ‘ U ′ σ on the currentquantum state, where σ is the currently scanned symbol. Measure the quantum state with M for .3.2.1.1 If the measurement result is f , move the tape head one square to the right.3.2.1.2 Otherwise, set the quantum state to | q i and start a new iteration.3.2.2 Apply U on the current quantum state and then measure the quantum state with M for .3.2.2.1 If the measurement result is f , move the tape head one square to the right.3.2.2.2 Otherwise, set the quantum state to | q i and start a new iteration.4. Measure the currently quantum state with M fin . If the result is not | q i , reject.5. Simulate k coin-flips and if all outcomes are “head”, then accepts the input. Figure 1: Description of the behavior of of pair (
P, V ) for mpal . The choice of will depend on
Figure 5: Description of the behavior of the verifier V ǫ of the pair ( P, V ǫ ) when recognizing the language L knapsack . The choiceof k depends on ǫ , as discussed in the text. U = 12 − √ √ √ √ − √ √ − √ √ − , U = 1 √ − − − − − − − − , (37) Consider the equation v ( b ) = P i ∈ I v ( a i ) = P ni =1 c i v ( a i ), where c i ∈ { , } . c i = 1 means a i is selected. ′ = 12 − √ √ √ √ − √ √ − √ √ − , U ′ = 1 √ − − − − − − − − , (38) U = − − − √ √ , (39) M for = { P f , P r } and M fin = { P i : i = 0 · · · , } , where P f = X i =0 | q i ih q i | , P r = X i =3 | q i ih q i | and P i = | q i ih q i | . (40)Unitary operators U and U are used to encode the value of b , U ′ and U ′ are used to encode the valueof a i , U is used to subtract the amplitude of | q i from the amplitude of | q i , respectively. The projectivemeasurement M for is used to keep the quantum state in span {| q i , | q i , | q i} . Lemma 16.
Setting the quantum state to the initial state in Step 2.2.2 can be done by a projective mea-surement and unitary operators.Proof.
Let the current quantum state be | ψ i = P i =0 α i | q i i , and let the projective measurement M fin beperformed on | ψ i . If the classical measurement result is i , then the resulting quantum state is | q i i . Byapplying the unitary operator U i = | q ih q i | on the state | q i i , the state | q i is obtained. Completeness condition : If the input string w ∈ L knapsack , then the all-powerful prover can make theright choice of a i and tells that to the verifier during their communications, step by step. Let a j , a j , · · · , a j s be strings selected by the prover (selection is not unique) where j p < j q if p < q and v ( b ) = P si =1 v ( a j i ). Lemma 17.
If the input w ∈ L knapsack , then the quantum state of the verifier V ε will be | q i after Step 4.Proof. Let b = b b · · · b l . The quantum state of the verifier V ε after the string b is read is (see Appendix Afor a detailed proof) | ψ b i = Q li =1 ( P f U b i ) | q ik Q li =1 ( P f U b i ) | q ik = 1 p v ( b ) v ( b )0...0 . (41)17et a j = a j a j · · · a j l ′ . The quantum state of the verifier V ε after the string a j is read is (see AppendixA for a detail proof) | ψ b a j i = Q l ′ i =1 ( P f U ′ a j i ) | ψ b ik Q l ′ i =1 ( P f U ′ a j i ) | ψ b ik = 1 p v ( b ) + v ( a j ) v ( b ) v ( a j )...0 . (42)Afterwards, the unitary operator U and the projective measurement M for are performed. If the outcomeof the measurement is f (that is the quantum state is in span {| q i , | q i , | q i} ), the resulting quantum stateis | ψ b a j i = P f U Q l ′ i =1 ( P f U ′ a j i ) | ψ b ik P f U Q l ′ i =1 ( P f U ′ a j i ) | ψ b ik = 1 p v ( b ) − v ( a j )) v ( b ) − v ( a j )0...0 . (43)Therefore, when the verifier reaches the end of Step 4, the quantum state of the verifier is | ψ w i = 1 q v ( b ) − P si =1 v ( a j i )) v ( b ) − P si =1 v ( a j i )0...0 = = | q i . (44)The resulting quantum state in Step 4 will be | q i with probability 1. After Steps 5 the input is acceptedwith the probability P a = 2 − k and is rejected with the probability P r = 0. If the whole process is repeatedfor infinitum, the accepting probability is P r [( P ∗ , V ε ) rejects w ] = X i ≥ (1 − P a ) i (1 − P r ) i P a = P a P a + P r − P a P r = P a P a = 1 . (45) Soundness condition:
If the input string w / ∈ L knapsack and n = | w | , no matter what choices theprover makes, there is not I ⊆ { , · · · , N } such that v ( b ) = P i ∈ I v ( a i ). Suppose a j , a j , · · · , a j s are thestrings selected by the prover where j p < j q if p < q . Lemma 18.
If the input string w L knapsack , then for any prover P ∗ the verifier V ε rejects w after Step4 with the probability at least . roof. According to the analysis in Lemma 17, the quantum state at the beginning of Step 4 is | ψ w i = 1 q v ( b ) − P si =1 v ( a j i )) v ( b ) − P si =1 v ( a j i )0...0 . (46)Obviously, | v ( b ) − P si =1 v ( a j i ) | ≥
1. Therefore, the probability of getting the state | q i and rejecting theinput w is at leat .If w L knapsack , then, according to Lemma 18, for any prover P ∗ the verifier V ε rejects the input afterStep 4 with the probability P r >
12 (47)and the input is accepted after Steps 5 to 7 with the probability P a = 2 − k . (48)If the whole computation process is repeated indefinitely, the probability that the verifier rejects the inputis (taking into consideration that k = 1 + ⌈ log /ε ⌉ ) P r [( P ∗ , V ε ) rejects w ] = X i ≥ (1 − P a ) i (1 − P r ) i P r = P r P a + P r − P a P r (49) > P r P a + P r > / − k + 1 / ≥ / ε/ / ε > − ε. (50) Time analysis:
Let n = | w | . The probability of getting as the outcome f in the projective measurement M for is at least in the whole process in Figure 5. Hence, the probability for the verifier to reach Step 4in one iteration is at least (cid:0) (cid:1) n . If every outcome of the projective measures M for in all computation stepsin an iteration is f , then the running time from Step 1 to Step 4 is O ( n ). Therefore, the expecting runningtime from Step 1 to Step 4 is less than + ∞ X i =1 (cid:18) − (cid:18) (cid:19) n (cid:19) i − (cid:18) (cid:19) n · i · O ( n ) = 6 n · O ( n ) . (51)The running time of Step 5 is O ( k ). The halting probability is Ω (cid:0) k (cid:1) in both cases, so the expected numberof repetitions of the algorithm is O (2 k ) in both cases. Hence the expected running time of the QAM ( P, V ε )is O (2 k · (6 n · O ( n ) + O ( k ))) and therefore O (cid:0) ε ( n n + log ε ) (cid:1) .Dwork and Stockmeyer [9] proved that the set of languages recognized by AM(2PFA) is a proper subsetof P. However, in the previous theorem we prove that the language L knapsack , which is NP complete, canbe recognized by QAM(2QCFA). This is another example that indicates QAM(2QCFA) are more powerfulthan AM(2PFA). 19 .4. Recognition of other languages In this subsection, we sketch the proofs for the following languages, where Σ = { a, b } : L = { w | ∃ s, t, u, v ∈ Σ ∗ , w = sbt = ubv, | s | = | v |} , (52) L = { w | ∃ s, t, u, v ∈ Σ ∗ , w = sat = ubv, | s | = | v |} , (53)that they can be recognized by QAM(2QCFA). Freivalds et al. [10] proved that the language L is non-stochastic, whereas L , the set of nonpalindromes, is stochastic. We prove that these languages can berecognized by QAM(2QCFA) in polynomial expected running time. Theorem 19.
For any ε < / , there exists a verifier-prover pair ( P, V ε ) of a QAM(2QCFA) that canrecognize L with one-sided error ε in the expected running time O (cid:0) ε (cid:0) n + n log ε (cid:1)(cid:1) , where n is the lengthof the input.Proof. If w ∈ L , then w can be of one of the following three typesType 1. w = sbt = ubv , where | s | = | u | = | v | .Type 2. w = sbt = ubv , where | s | > | u | . If s = ubx , where x ∈ Σ ∗ , then w = ubxbt and | u | = | w | − | v | − | w | − | s | − | t | . Type 3. w = sbt = ubv , where | s | < | u | . If u = sby , where y ∈ Σ ∗ , then w = sbybv and | s | = | v | . Therefore, L = { w | ∃ u, v, z ∈ Σ ∗ , w = ubv or w = ubzbv, where | u | = | v |} .By virtue of the method of the proof of Theorem 1, the quantum state of the verifier will be rotated byan angle α every time a symbol in u is scanned. The prover just tells the verifier the right time to stoprotation and of which type the input is. After the verifier checks the input is of the form b or bzb , the provertells the verifier the right position to resume rotation of the quantum state by the angle − α . The rest ofthe proof is similar to the one in Theorem 1. Theorem 20.
For any ε < / , there exists a verifier-prover pair ( P, V ε ) of a QAM(2QCFA) that recognizesthe language L with one-sided error ε in the expected running time O (cid:0) ε (cid:0) n + n log ε (cid:1)(cid:1) , where n is thelength of input.Proof. Suppose w ∈ L . The input string w can be of one of the following typesType 1. w = sat = ubv where | s | > | u | . Let s = ubx where x ∈ Σ ∗ , then w = ubxat and | u | = | w | − | v | − | w | − | s | − | t | . Type 2. w = sat = ubv where | s | < | u | . Let u = say where y ∈ Σ ∗ , then w = saybv and | s | = | v | . Therefore, L = { w | ∃ u, v, z ∈ Σ ∗ , w = uazbv or w = ubzav, where | u | = | v |} .The proof method of Theorem 1 will be employed. The prover will tell the verifier two positions – theposition of the symbol after u and the position of the symbol before v . The verifier rotates its quantum stateby an angle α every time it scans a symbol in u and then checks whether symbols in these two positions,pointed out by the prover, are different. If they are different, the quantum state of the verifier is rotated byan angle − α every time a symbol in v is scanned. Otherwise, the input string is rejected. The rest of theproof is similar to that in Theorem 1. 20 . Concluding remarks We have explored quantum interactive proof systems with 2QCFA verifiers and classical communication.We have focused on the public coin version of the interactive proof systems – namely QAM(2QCFA).We have showed that QAM(2QCFA) are more powerful than their classical counterparts AM(2PFA). Inparticular, we have shown a number of specific results demonstrating that: (1) The language L middle = { xay | x, y ∈ Σ ∗ , Σ = { a, b } , | x | = | y |} can be recognized by QAM(2QCFA) in a polynomial expectedrunning time, but cannot be recognized by AM(2PFA) in polynomial expected running time. (2) Thelanguage L mpal = { xax R | x ∈ { a, b } ∗ } can be recognized by QAM(2QCFA) in an exponential expectedrunning time, but cannot be recognized by AM(2PFA) at all. (3) The 0 - 1 knapsack language can berecognized by QAM(2QCFA) in an exponential expected running time.A related open problem, first mentioned in [9], is whether AM(ptime-2PFA) are more powerful than2PFA(ptime). Our attempts (and that of others) to show that AM(ptime-2PFA) are indeed more powerfulfailed so far. However, the results of this paper indicate, that the answer is likely positive when it comes tothe quantum case. Indeed, the language L middle can be recognized by QAM(ptime-2QCFA), but it seemsthat it can not be recognized by 2QCFA. That would imply that QAM(ptime-2QCFA) are more powerfulthan 2QCFA. Acknowledgements
The authors would like to thank the referees for helpful suggestions to improve the presentation ofthe paper. This work is supported in part by the National Natural Science Foundation of China (Nos.61272058, 61073054) and supported in part by Employment of Newly Graduated Doctors of Science forScientific Excellence project/grant (CZ.1.07./2.3.00/30.0009) of Czech Republic, and the project of theFCT PEst-OE/EEI/LA0008/2013.
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1, the equality holds, that is | ψ w i = 1 p v ( w ) v ( w )0...0 . (56)We prove the equation holds for b = wσ . If σ = ‘0 ′ , then U | ψ w i = 12 − √ √ √ √ − √ √ − √ √ − · p v ( w ) v ( w )000000 = 12( p v ( w ) ) v ( w )0 − √ √ (57)and | ψ w i = P f U | ψ w ik U | ψ w ik = 1 p v ( w )) v ( w )0...0 = 1 p v ( w v ( w . (58)23f σ = ‘1 ′ , then U | ψ w i = 1 √ p v ( w ) − − − − − − − − v ( w )000000 = 1 p v ( w ) ) v ( w ) + 102 − v ( w ) v ( w )000 (59)and | ψ w i = P f U | ψ w ik U | ψ w ik = 1 p v ( w ) + 1) v ( w ) + 10...0 = 1 p v ( w v ( w ..