1-way quantum finite automata: strengths, weaknesses and generalizations

Transcription

1 1-way quantum finite automata: strengths, weaknesses and generalizations arxiv:quant-h/ v3 30 Se 1998 Andris Ambainis UC Berkeley Abstract Rūsiņš Freivalds University of Latvia We study 1-way quantum finite automata (QFAs). First, we comare them with their classical counterarts. We show that, if an automaton is required to give the correct answer with a large robability (greater than 7/9), then any 1-way QFA can be simulated by a 1-way reversible automaton. However, quantum automata giving the correct answer with smaller robabilities are more owerful than reversible automata. Second, we show that 1-way QFAs can be very sace-efficient. We construct a 1- way QFA that is exonentially smaller than any equivalent classical (even randomized) finite automaton. We think that this construction may be useful for design of other sace-efficient quantum algorithms. Third, we consider several generalizations of 1-way QFAs. Here, our goal is to find a model which is more owerful than 1-way QFAs keeing the quantum art as simle as ossible. 1 Introduction It is quite ossible that the first imlementations of quantum comuters will not be fully quantum mechanical. Instead, they may have two arts: a quantum art and a classical art with a communication between two arts. In this case, the quantum art will be considerably more exensive than the classical art. Therefore, it will be useful to make the quantum art as small as ossible even if it leads to some (reasonable) increases in the size of the classical art. This motivates the study of systems with a small quantum mechanical art. Quantum finite automata (QFA) is a theoretical model for such systems. [12] introduced both 1-way and 2-way QFAs, with emhasis on 2-way automata because they are more Address: Comuter Science Division, University of California, Berkeley, CA 94720, ambainis@cs.berkeley.edu. Suorted by Berkeley Fellowshi for Graduate Studies. Part of this work done during 1998 I.S.I.-Elsag Bailey research meeting on quantum comuting. Address: Institute of Mathematics and Comuter Science, University of Latvia, Raina bulv. 29, Riga, LV-1459, Latvia, rusins@cclu.lv. Suorted by Latvia Science Council Grant

2 owerful. However, the model of 2-way QFAs is not quite consistent with the idea of a system with a small quantum mechanical art. [12] allows suerositions where different arts of suerosition have the head of QFA at different locations. (Even more, using such suerositions was the main idea in the roof that 2-way QFAs are more owerful than classical finite automata.) This means that the osition of the head must be encoded into quantum state. Hence, the number of quantum states necessary to imlement a 2-way QFA is not a constant but grows when the size of the inut increases. This also makes state transformations more comlicated (and more difficult to imlement). Hence, we think that more attention should be given to the study of simler models like 1-way QFAs. A 1-way quantum automaton is a very reasonable model of comutation and it is easy to see how it can be imlemented. The finite dimensional state-sace of a QFA corresonds to a system with finitely many articles. Each letter has a corresonding unitary transformation on the state-sace. A classical device can read symbols from the inut and aly the corresonding transformations to the quantum mechanical art. Results about 1-way QFAs in [12] were quite essimistic. It was shown that the class of languages recognized by 1-way QFAs is a roer subset of regular languages. We continue the investigation of 1-way QFAs and show that, desite being limited in some situations, they erform well in other situations. Our first results consider relations between 1-way QFAs and 1-way reversible automata. Clearly, a 1-way reversible automaton is a secial case of a QFA and, therefore, cannot recognize all regular languages. It is a natural question whether 1-way QFAs are more owerful than 1-way reversible automata. Interestingly, the answer deends on the acceting robability of a QFA. If a QFA gives a correct answer with a large robability (greater than 7/9), it can be relaced by a 1-way reversible automaton. However, this is not true for and smaller robabilities. Then, we show that QFAs can be much more sace-efficient than deterministic and even robabilistic finite automata. Namely, there is a 1-way QFA that can check whether the number of letters received from the inut is divisible by a rime with only O(log) states (this is equivalent to log log bits of memory). Any deterministic or robabilistic finite automaton needs states (log bits of memory). We think that this sace-efficient quantum algorithm may be interesting for design of other quantum algoritms as well. Finally, we consider modifications of 2-way quantum automata where the head is always at the same osition for all arts of suerosition. Modified 2-way QFAs can be imlemented with a quantum system of constant size. Several modifications are roosed. In one of our models (1-way QFAs with a robabilistic rerocessing), some non-regular languages can be recognized. 2 Definitions

3 2.1 Quantum finite automata We consider 1-way quantum finite automata (QFA) as defined in [12]. Namely, a 1-way QFA is a tule M = (Q, Σ, δ, q 0, Q acc, Q rej ) where Q is a finite set of states, Σ is an inut alhabet, δ is a transition function, q 0 Q is a starting state and Q acc Q and Q rej Q are sets of acceting and rejecting states. The states in Q acc and Q rej are called halting states and the states in Q non = Q (Q acc Q rej ) are called non-halting states. /c and $ are symbols that do not belong to Σ. We use /c and $ as the left and the right endmarker, resectively. The working alhabet of M is Γ = Σ {/c, $}. A suerosition of M is any element of l 2 (Q) (the sace of maings from Q to C with l 2 norm). For q Q, q denotes the unit vector with value 1 at q and 0 elsewhere. All elements of l 2 (Q) can be exressed as linear combinations of vectors q. We will use ψ to denote elements of l 2 (Q). The transition function δ mas Q Γ Q to C. The value δ(q 1, a, q 2 ) is the amlitude of q 2 in the suerosition of states to which M goes from q 1 after reading a. For a Γ, V a is a linear transformation on l 2 (Q) defined by V a ( q 1 ) = q 2 Q δ(q 1, a, q 2 ) q 2. (1) We require all V a to be unitary. The comutation of a QFA starts in the suerosition q 0. Then transformations corresonding to the left endmarker /c, the letters of the inut word x and the right endmarker $ are alied. The transformation corresonding to a Γ consists of two stes. 1. First, V a is alied. The new suerosition ψ is V a (ψ) where ψ is the suerosition before this ste. 2. Then, ψ is observed with resect to the observable E acc E rej E non where E acc = san{ q : q Q acc }, E rej = san{ q : q Q rej }, E non = san{ q : q Q non }. This observation gives x E i with the robability equal to the amlitude of the rojection of ψ. After that, the suerosition collases to this rojection. If we get ψ E acc, the inut is acceted. If ψ E rej, the inut is rejected. If ψ E non, the next transformation is alied. We regard these two transformations as reading a letter a. Another definition of QFAs. Indeendently of [12], quantum automata were introduced in [13]. There is one difference between these two definitions. In [12], a QFA is observed after reading each letter (after doing each V a ). In [13], a QFA is observed only after all letters have been read. Any language recognized by a QFA according to the definition of [13] is recognized by a QFA according to [12]. The converse is not true. Any finite language can be recognized in the sense of [12]. However, no finite non-emty language can be recognized in the sense of [13]. Everywhere in this aer, we will use the more general definition

4 of [12]. However, our results of section 4.1 which show that 1-way QFAs can be more saceefficient than deterministic or robabilistic automata are true in the more restricted model of [13] as well. 2.2 Examle To exlain our notation, we give an examle of a 1-way QFA. To kee it simle, we use a one letter alhabet Σ = {a}. The state sace is Q = {q 0, q 1, q acc, q rej } with the set of acceting states Q acc = {q acc } and the set of rejecting states Q rej = {q rej }. The starting state is q 0. The transition function can be secified in two ways: by secifying δ or by secifying V x for all letters x Γ. These methods are equivalent: all V x are determined by δ and equation (1). We shall define the automaton by describing V x. V a ( q 0 ) = 1 2 q q q rej, V a ( q 1 ) = 1 2 q q q rej, V $ ( q 0 ) = q rej, V $ ( q 1 ) = q acc. It can be also defined by describing δ. For examle, V a ( q 0 ) = 1 2 q q q rej would be δ(q 0, a, q 0 ) = 1 2, δ(q 0, a, q 1 ) = 1 2, δ(q 0, a, q acc ) = 0, δ(q 0, a, q rej ) = 1 2. As we see, this is much longer. For this reason, we will mainly use V x notation. There are some transitions that we have not described. For examle, V a (q acc ) has not been secified. These values are not imortant and can be arbitrary. We need them to be such that V a is unitary but this is not difficult. As long as all secified V a (q i ) are orthogonal, the remaining values can be assigned so that the whole V a is unitary. In the sequel, we will often shorten descritions of QFAs by leaving out transitions that can be defined arbitrarily. Next, we show how this automaton works on the word aa. 1. The automaton starts in q 0. Then, V a is alied, giving 1 2 q q q rej. This is observed. Two outcomes are ossible. With robability (1/ 2) 2 = 1/2, a rejecting state is observed. Then, the suerosition collases to q rej, the word is rejected and the comutation terminates. Otherwise (with robability 1/2), a non-halting state is observed and the suerosition collases to 1 2 q q 1. In this case, the comutation continues.

5 2. A simle comutation shows that 1 2 q q 1 is maed to itself by V a. After that, a non-halting state is observed. (There are no acceting or rejecting states in this suerosition.) 3. Then, the word ends and the transformation V $ corresonding to the right endmarker $ is done. It mas the suerosition to 1 2 q rej q acc. This is observed. With robability (1/2) 2 = 1/4, the rejecting state q rej is observed. With robability 1/4, the acceting state q acc is observed. The total robability of acceting is 1/4, the robability of rejecting is 1/2 + 1/4 = 3/ Reversible automata A 1-way reversible finite automaton (RFA) is a QFA with δ(q 1, a, q 2 ) {0, 1} for all q 1, a, q 2. Alternatively, RFA can be defined as a deterministic automaton where, for any q 2, a, there is at most one state q 1 such that reading a in q 1 leads to q 2. We use the same definitions of accetance and rejection. States are artitioned into acceting, rejecting and non-halting states and a word is acceted (rejected) whenever the RFA enters an acceting (rejecting) state. After that, the comutation is terminated. Similarly to quantum case, endmarkers are added to the inut word. The starting state is one, acceting (rejecting) states can be multile. This makes our model different from both [3] (where only one acceting state was allowed) and [14] (where multile starting states with a non-deterministic choice between them at the beginning were allowed). We define our model so because we want it to be as close to our model of QFAs as ossible. Generally, it s hard to introduce robabilism into finite automata without losing reversibility. However, there are some tyes of robabilistic choices that are consistent with reversibility. For examle, we can choose the starting state robabilistically. The next examle shows that such robabilistic choices increase the ower of an automaton. Examle. Consider the language L = {a 2n+3 n IN}. It cannot be recognized by a 1- way RFA. However, there are 3 1-way RFAs such that each word in the language is acceted by 2 of them and each word not in the language is rejected by 2 out of 3. Hence, if we choose one of these three automata equirobably, L will be recognized with the robability of correct answer 2/3. Probabilistic choices of this tye can be easily done in our model of QFAs. This may lead to a claim that QFAs are more owerful than classical reversible automata because they can do such robabilistic choices. We wish to avoid such situations and to searate robabilistic choices from real quantum effects. Therefore, we define 1-way finite automata with robabilistic choices (PRFAs) and comare caabilities of QFAs with them. A PRFA is a robabilistic finite automaton such that, for any state q 1 and any a Γ, there is at most one state q 2 such that the robability of assing from q 2 to q 1 after reading a is non-zero. Definitions of accetance and rejection

6 are similar to QFAs and RFAs. Now, the robabilistic automaton from the examle above becomes a 1-way PRFA. Theorem 1 1. If a language is acceted by a 1-way RFA, it is acceted by a 1-way PRFA. 2. If a language is acceted by a 1-way PRFA, it is acceted by a 1-way QFA with the same robability of correct answer. Proof: Easy. In section 3.2 we will comare the ower of 1-way QFAs and PRFAs and show that 1-way quantum automata can actually do more than just robabilistic choices. 3 Caabilities of RFAs and QFAs 3.1 QFAs with robability of correct answer above 7/9 We characterize the languages recognized by 1-way QFAs in terms of their minimal automata. The minimal automaton of a language L is a 1-way deterministic finite automaton recognizing it with the smallest number of states. (Note: the minimal automaton can be non-reversible, even for some languages L that can be recognized by a 1-way RFA. The extreme case of this is our Theorem 12 where the smallest 1-way RFA is exonentially bigger than the minimal nonreversible automaton.) It is well known[9] that the minimal automaton is unique and can be effectively constructed. Theorem 2 Let L be a language and M be its minimal automaton. Assume that there is a word x such that M contains states q 1, q 2 satisfying: 1. q 1 q 2, 2. If M starts in the state q 1 and reads x, it asses to q 2, 3. If M starts in q 2 and reads x, it asses to q 2, and 4. q 2 is neither all - acceting state, nor all - rejecting state. Then L cannot be recognized by a 1-way QFA with robability at least 7/9 + ǫ for any fixed ǫ > 0. Proof: We rove the result for a slightly smaller robability of correct answer 5/6 + ǫ (instead of 7/8 + ǫ). The roof for 7/8 + ǫ is technically more comlicated. Let L be a language such that its minimal automaton contains the forbidden construction and M be a QFA. We show that, for some word y the robability of M giving the correct answer to y L? is less than 5/6 + ǫ. This imlies that L cannot be recognized with robability of correct answer being 5/6 + ǫ.

7 x q1 q 2 x Figure 1: The forbidden construction of Theorem 2. For simlicity, we assume that q 1 is the starting state of M. We introduce some notation. Let P non (ψ) be the non-halting art of ψ and P halt (ψ) be the halting art of ψ. V a = P non V a is a transformation that mas ψ to the non-halting art of V a (ψ). If x is a word consisting of letters a 1...a k, then V x denotes V a k...v a1. ψ x denotes the non-halting art of the QFA s configuration after reading x. It is easy to see that, for any word x and letter a, ψ xa = V a(ψ x ). We recall that l 2 (Q) denotes the state-sace of M with l 2 norm ψ. l 2 (Q) = E acc E rej E non. First, we rove that the state-sace of M can be decomosed into two arts with different behavior. Lemma 1 There are subsaces E 1, E 2 such that E non = E 1 E 2 and (i) If ψ E 1, then V x (ψ) E 1, (ii) If ψ E 2, then V x k (ψ) 0 when k. Proof. We define two sequences of subsaces E1 1, E2 1,... and E1 2, E2 2,... such that E non = E1 i E2. i Let E1 1 = {ψ ψ E non and V a (ψ) E non } (i.e., the subsace of all ψ such that both ψ and V a (ψ) have only non-halting comonents). E2 1 consists of all vectors in E non orthogonal to E1 1. For i > 1, Ei 1 = Ei 1 1 {ψ V a (ψ) E1 i 1 } and E2 i consists of all vectors in E non orthogonal to E1. i Clearly, E1 1 E If Ei+1 1 is a roer subsace of E1 i, then the dimensionality of 1 is smaller than the dimensionality of E1. i This can haen only finitely many times E i+1 because the original E1 1 is finite-dimensional. Hence, there is i 0 such that E i 0 1 = E i We define E 1 = E i 0 1, E 2 = E i 0 2. Next, we check that both (i) and (ii) are true. Let ψ E 1. Then, V a (ψ) E i = E i 0 1 = E 1 and V a (ψ) E non by E 1 E1 1 and the definition of E1 1. It remains to rove that E 2 satisfies (ii) condition of Lemma 1. Claim 1 If ψ E i 1, then P halt (V a (V a l (ψ))) = 0 for all l i 1. Proof: By induction. If i = 1, then P halt (V a (ψ)) = 0 by definition of E1 1. If i > 1 and l = 0, then P halt (V a (ψ)) = 0 because E1 i E1 1. The only remaining case is i > 1 and l > 1. Let ψ = V a (ψ). By definition of E1, i V a (ψ) E1 i 1. We also have V a (ψ) E non because E1 i 1 E non. Hence, V a (ψ) = P non(v a (ψ)) = V a (ψ) = ψ and P halt (V a (V a l(ψ))) = P halt(v a (V a l 1(ψ ))) = 0 by ψ E1 i 1 and inductive assumtion.

8 Claim 2 Let ψ = ψ 1 + ψ 2, ψ 1 E i 1, ψ 2 E i 2. Then, for all l i 1, Proof: By linearity of P halt, V a, V a, P halt (V a (V a l(ψ))) = P halt(v a (V a l(ψ 2))). P halt (V a (V a l(ψ))) = P halt(v a (V a l(ψ 1))) + P halt (V a (V a l(ψ 2))). Claim 1 imlies that P halt (V a (V a l (ψ 1 ))) = 0. Claim 3 Let j {1,..., i 0 }. There is a constant δ j > 0 such that for any ψ E2 i l {0,..., j 1} such that P halt (V a (V a (ψ))) δ l j ψ. there is Proof: By induction. Base Case. Consider the linear transformation T 1 : E2 1 E acc E rej that mas ψ E to the halting art of V a (ψ). T 1 (the norm of T 1 ) is the minimum of T 1 (x) over all x such that x = 1. If T 1 = 0, there is x E2 1 such that x = 1 but T 1 (x) = 0. This means that T 1 (x) = 0, imlying that x E1 1. However, E1 1 E1 2 = { 0 }, leading to a contradiction. Hence, T 1 > 0. Also, T 1 1 because V a is unitary and rojection to the halting subsace can only decrease the norm. We take δ 1 = T 1. Then, the halting art of V a (ψ) is at least T 1 ψ = δ 1 ψ. Inductive Case. We assume that the lemma is true for E2 i and rove it for E2 i+1. We consider the transformation T i+1 maing ψ E2 i+1 to the rojection of V a (ψ) on E2 i. If T i+1(ψ) = 0, then ψ E1 i+1 by the definition of E1 i+1. Similarly to the revious case, E1 i+1 E2 i+1 = { 0 }. Hence, if T i+1 (ψ) = 0 and ψ E2 i+1, then ψ = 0. This means that T i+1 > 0. We can also rove that T i+1 1. We define δ i+1 = T i+1 δ 4 i. Let E3 i = {x Ei+1 2 and x E2 i }. Then, Ei+1 2 = E2 i Ei 3. We also note that Ei 3 is a subsace of E1. i (This follows from definitions of E1 i and E3.) i To show that one of P halt (V a (V a (ψ))) is large enough, we reresent ψ E i+1 l 2 as ψ 2 + ψ 3, ψ 2 E2, i ψ 3 E3. i There are two cases: 1. ψ 2 T i+1 4 ψ. Then, P halt (V a (V a l(ψ T i+1 2))) δ i ψ 2 δ i ψ = δ i+1 ψ 4 for some l i 1 because ψ 2 E2 i and we can aly the inductive assumtion. Claim 2 imlies that this is also true for P halt (V a (V a (ψ))). l

9 2. ψ 2 < T i+1 4 ψ. Then, by triangle inequality, ψ 3 ψ ψ 2 (1 T i+1 4 ) ψ 3 T i+1 ψ. 4 Let ψ, ψ 2 and ψ 3 be the rojections of V a (ψ), V a (ψ 2 ), V a (ψ 3 ) on E2. i Clearly, ψ = ψ 2 + ψ 3. Triangle inequality gives us ψ ψ 3 ψ 2 ψ 3 ψ 2 3 T i+1 4 ψ T i+1 4 We have ψ = P non (ψ ) + P halt (ψ ). Again, we have two cases. ψ = T i+1 ψ. 2 If P halt (ψ ) T i+1 ψ, then P 4 halt (ψ ) δ i+1 ψ because δ i+1 = T i+1 δ 4 i and δ i 1 because all T i are at most 1. Otherwise, by triangle inequality, P non (ψ ) ψ P halt (ψ ) T i+1 ψ. By 4 inductive assumtion, there is l i 1 such that P halt V a (V a (ψ )) δ l i ψ. Therefore, P halt (V a (V a l+1(ψ))) T i+1 δ i ψ = δ i+1 ψ. 4 Claim 4 There is c such that 0 < c < 1 and, for any ψ E 2, t IN, V a i 0 t (ψ) c t ψ. Proof: We take c = 1 δi 2 0. By Claim 3, one of P halt (V a (V a (ψ))) is at least δ l i0 ψ. P non (V a (V a (ψ))) is orthogonal to l this vector. Hence, P non (V a (V a (ψ))) is at most l ψ 2 δi 2 0 ψ 2 = 1 δi 2 0 ψ. V a can be only smaller because V i 0 a is unitary and P non does not increase vectors. We have shown that V a (ψ) c ψ. Reeating this t times, we obtain Claim 4. i 0 Clearly, c t 0 if t. This comletes the roof of Lemma 1. Let ψ /c = ψ /c 1 + ψ2 /c, ψ1 /c E 1, ψ /c 2 E 2. We consider two cases. Case 1. ψ 2 /c > 1/ 3. Then, ψ 1 /c < 1 (1/ 3) 2 = 2/ 3. This also means V x i (ψ 1 /c ) < 2/ 3. For sufficiently large i, V x i (ψ 2 /c ) becomes negligible (art (ii) of Lemma 1). Then, ψ /cx i < 2/ 3. The robability of M halting after this moment is less than 2/3. Hence, M has already halted with robability more than 1/3 and acceted (or rejected) with robability more

10 than 1/6. This means that M cannot reject (accet) any continuation of x i with robability 5/6. However, x i has both continuations in L and continuations not in L. Hence, M does not recognize L. Case 2. ψ 2 /c 1/ 3. q 1 and q 2 are different states in the minimal automaton of L. Therefore, there is a word y Σ such that y leads to accetance from one of q 1, q 2 but not from the other one. We consider the distributions of robabilities on M s answers accet and reject on y and x i y. On one of these words, M must accet with robability at least 5/6 + ǫ and reject with robability at most 1/6 ǫ. On the other word, M must accet with robability most 1/6 ǫ and reject with robability at least 5/6 + ǫ. Therefore, both the robabilities of acceting and the robabilities of rejecting must differ by at least 2/3 + 2ǫ. This means that the variational distance between two robability distributions (the sum of these two distances) must be at least 4/3 + 4ǫ. We show that it cannot be so large. First, we select an aroriate i. Let m be so large that V x m(ψ2 ) δ for δ = ǫ/4. /c ψ /c 1, V x(ψ /c 1 ), V x (ψ 1 ),... is a sequence in a finite-dimensional sace. Therefore, it has a limit 2 /c oint and there are i, j such that V x j(ψ1 /c ) V x i+j(ψ1 /c ) < δ. We choose i, j so that i > m. The difference between two robability distributions comes from two sources. The first is difference between ψ /c and ψ /cx i (the states of M before reading y). The second source is the ossibility of M acceting while reading x i (the only art that is different in two words). We bound the difference created by each of these two sources. The difference ψ /c ψ /cx i can be artitioned into three arts. ψ /c ψ /cx i = (ψ /c ψ 1 /c ) + (ψ1 /c ψ1 /cx i) + (ψ1 /cx i ψ /cx i ). The first art is ψ /c ψ 1 /c = ψ2 /c and ψ2 /c 1 3. The second and the third arts are both small. For the second art, notice that V x is unitary on E 1 (because V x is unitary and V x (ψ) does not contain halting comonents for ψ E 1 ). Hence, V x reserves distances on E 1 and ψ /c 1 ψ1 /cx i = ψ1 /cx j ψ1 /cx i+j δ. The third art is ψ /cx i ψ 1 = /cx i ψ2 and /cx i ψ2 /cx i δ because i > m. Next, we state two lemmas relating differences between two suerositions and differences between robability distributions created by observing these suerositions. The first lemma is by Bernstein and Vazirani[4]. Lemma 2 [4] Let ψ and φ be such that ψ 1, φ 1 and ψ φ ǫ. Then the total variational distance resulting from measurements of φ and ψ is at most 4ǫ.

11 The second lemma is our imrovement of lemma 2. Lemma 3 Let ψ 1 and ψ 2 be such that ψ 1 ψ 2. Then the total variational distance resulting from measurements of ψ 1 and ψ 1 + ψ 2 is at most ψ 2 4ψ ψ 2 2. Proof. Omitted. We aly lemma 3 to ψ /c 1 and ψ2. This gives that the variational distance between /c distributions generated by ψ /c 1 and ψ1 /c + ψ2 is at most 1. Then, we aly lemma 2 to two /c other arts of ψ /c ψ /cx i. Each of them influences the variational distance by at most 4δ. Together, the variational distance between distributions obtained by observing ψ /c and ψ /cx i is at most 1 + 8δ. The robability of M halting while reading x i is at most ψ /c 2 2 = 1/3. Adding it increases the variational distance by at most 1/3. Hence, the total variational distance is at most 4/3 + 8δ = 4/3 + 2ǫ. However, if M distinguishes y and x i y correctly, it should be at least 4/3 + 4ǫ. Hence, M does not recognize one of these words correctly. Theorem 3 Let L be a language and M be its minimal automaton. If M does not contain the forbidden construction of Theorem 2, then L can be recognized by a 1-way reversible finite automaton. Proof. We define a non-reversibility as a tule q 1, q 2, q, a where q 1, q 2, q Q, a Σ, q 1 q 2 and reading a in q 1 or q 2 leads to q. Let m be the number of non-reversibilities in M. We show how to modify M so that the number of non-reversibilities decreases. A reversible automaton is obtained by reeating this modification several times. We define a artial ordering < on non-reversibilities. q 1, q 2, q, a < q 1, q 2, q, a if and only if one of q 1 and q 2 is reachable from q. It is easy to see that < is transitive. Lemma 4 < is anti-reflexive. Proof. For a contradiction, assume there is q 1, q 2, q, a such that q 1, q 2, q, a < q 1, q 2, q, a. We also assume that q 2 is reachable from q by reading a word y. (Otherwise, q 1 is reachable from q and we can just exchange q 1 and q 2.) Then, reading x = ay leads from q 1 to q 2 and from q 2 to q 2. This contradicts our assumtion that M does not contain such q 1, q 2. Hence, there is a tule q 1, q 2, q, a that is maximal with resect to <. We create two coies for state q and all states reachable from q. If M reads a in q 1, it asses to one coy of q, if it reads a in q 2, it asses to the second coy. This eliminates this non-reversibility. Other non-reversibilities are not dublicated because they are not reachable from q. Hence, the number of non-reversibilities is decreased.

12 Corollary 1 A language can be recognized by a 1-way QFA with robability 7/9 + ǫ if and only if it can be recognized by a 1-way reversible finite automaton. Proof. Clearly, a RFA is a secial case of a QFA. The other direction follows from Theorems 2 and 3. This immediately imlies the same result about 1-way reversible automata with robabilistic choices. For this tye of automata, a stronger result can be roved. Theorem 4 A language can be recognized by a 1-way PRFA with robability 2/3 + ǫ (for arbitrary ǫ > 0) if and only if it can be recognized by a 1-way reversible finite automaton. Proof: Omitted. The examle in Section 2.3 shows that Theorem 4 is tight. 3.2 QFAs with robability of correct answer below 7/9 For smaller robabilities, QFAs are slightly more owerful than RFAs or even PRFAs. Theorem 5 The language a b can be recognized by a 1-way QFA with the robability of correct answer = where is the root of 3 + = 1. Proof. We describe a 1-way QFA M acceting this language. The automaton has 4 states: q 0, q 1, q acc and q rej. Q acc = {q acc }, Q rej = {q rej }. The initial state is 1 q 0 + q 1. The transition function is V a ( q 0 ) = (1 ) q 0 + (1 ) q 1 + q rej, V a ( q 1 ) = (1 ) q 0 + q 1 1 q rej, V b ( q 0 ) = q rej, V b ( q 1 ) = q 1, V $ ( q 0 ) = q rej, V $ ( q 1 ) = q acc. Case 1. The inut is x = a. It is straightforward that δ mas 1 q 0 + q 1 to itself while it receives a from the inut. Hence, after reading a the state remains 1 q 0 + q 1 and, after reading the right endmarker, it becomes 1 q rej + q acc. This means that the automaton accets with robability. Case 2. The inut is x = a b +. Again, the state remains 1 q 0 + q 1 while inut contains a. Reading the first b changes it to 1 q rej + q 1. The non-halting art of this state is q 1. It is left unchanged by next bs and maed to q acc after reading the right endmarker. Again, the acceting robability is. Case 3. The inut is x / a b.

13 Then, the initial segment of x is a b + a +. After reading the first b, the state is 1 q rej + q1. The automaton rejects at this moment with robability (1 ). The non-halting art q1 is maed to 1 q 0 +(1 ) q 1 (1 ) q rej by the next V a. Then, the automaton rejects with robability (1 ). The non-halting art 1 q 0 +(1 ) q 1 is unchanged by as. However, either b or right endmarker follows as and then q 0 is maed to q rej and the automaton rejects with robability 2 (1 ). We add the robabilities of rejecting at different moments together and get that M rejects x / a b with robability at least (1 ) + (1 ) + 2 (1 ) = ( )(1 ) = (1 ) = 1 3 =. It is easy to see that the minimal automaton of a b contains the forbidden construction of Theorem 2. Therefore, we have Corollary 2 There is a language that can be recognized by a 1-QFA with robability but not with robability 7/9 + ǫ. Proof: Follows from Theorems 2 and 5. For robabilistic comutation, the roerty that the robability of correct answer can be increased arbitrarily is considered evident. Hence, it was not surrising that [12] wrote with error robability bounded away from 1/2 about QFAs, thinking that all such robabilities are equivalent. However, mixing reversible (quantum comutation) and nonreversible (measurements after each ste) comonents in one model makes it imossible for QFAs. It is oen whether a counterart of Corollary 2 is true for 2-way QFAs. Corollary 3 There is a language that can be recognized by a 1-QFA with robability but not by a classical 1-way reversible FA. This corollary can be imroved by showing that even a 1-way robabilistic reversible automaton cannot recognize this language (and even with robability 1/2 + ǫ). Theorem 6 Let L be a language and M be its minimal automaton. Assume that there are words x, y and M s states q 1, q 2 such that 1. none of q 1 and q 2 is all-acceting or all-rejecting state; 2. reading x in q 1 leads to q 1 ; 3. reading y in q 1 leads to q 2 ; 4. reading y in q 2 leads to q 2 ;

14 5. there is no i > 0 such that reading x i leads from q 2 to q 2. Then L cannot be recognized by a 1-way PRFA with robability 1/2 + ǫ, for any ǫ > 0. Proof. Without the loss of generality, we assume that q 1 is the starting state of M. Let M be a 1-way robabilistic reversible automaton. We are going to show that, for some word x, the robability of M giving the right answer on the inut x is less than 1/2 + ǫ. Lemma 5 For any state q and a Σ +, there is k such that 0 < k Q and, for any sequence of robabilistic choices, one of the following haens: 1. After reading a k in state q, M returns to q; 2. After reading a Q +1 in state q, M accets or rejects. Proof. Let q 0, q 1,... be any sequence of non-halting states such that q 0 = q and the robability that reading a causes M to go from q i to q i+1 is non-zero. If the length of the sequence is greater than Q, then some state aears twice in this sequence. We consider the first state which aears twice. If it is not q 0, then it has two receding states: the state receding it when it aears in the sequence for the first time and the state receding it when it aears in the sequence for the second time. This contradicts the definition of a robabilistic reversible automaton. We have shown that q 0 is the first state which aears twice. Next, assume we have two such sequences: q 0, q 1,... and q 0, q 1,.... Let k 1, k 2 be the smallest numbers such that k 1 > 0, q k1 = q 0 and k 2 > 0, q k 2 = q 0, resectively. We show that k 1 = k 2. For a contradiction, assume that k 1 > k 2 (k 2 > k 1 case is similar.). Then, q k1 1 = q k 2 1 (because the state q k 1 = q k 2 cannot have two receding states), q k1 2 = q k 2 2 and so on, q k1 k 2 = q k 2 k 2 = q 0 = q 0. This contradicts the assumtion that k 1 is the smallest number such that k 1 > 0 and q k1 = q 0. Hence, k 1 = k 2. Lemma 6 For any state q and a Σ +, one of the following haens: 1. There is k such that, after reading a k in state q, M returns to q for any sequence of robabilistic choices; 2. The robability of halting after reading a k in state q tends to 1 when k. Proof. Let k be as in Lemma 5. If M always returns to q after reading a k, Lemma 6 is true. It remains to consider the case if there is a sequence of robabilistic choices for which M does not return to q. Then, by Lemma 5, this sequence causes M to halt. Let be the robability of returning to q after reading a k. Then, the robability of returning to q after reading a ik is i. With robability 1 i, M does not returns to q at some moment and (this is the only alternative) terminates after reading a ik+ Q (or some its refix). Clearly, 1 i 1, if i.

15 We note that one can use the same k = Q! for all q and a. (For any q, a, k Q and Q! is a multile of any such k.) We shall call the states of the first tye return states for a. Let i be the robability of non-halting after reading x i and = lim i i. We select i so that i < ǫ. Let j be the robability of non-halting after reading xi y j and = lim j j. We select j so that j is a multile of Q! and j < ǫ. Next, we comare the behaviour of M on x i y j and x i y j x Q!. These words corresond to different states in the minimal automaton. Hence, there is a continuation z such that exactly one of x i y j z and x i y j x Q! z is in L. If M had acceted or rejected after x i y j (without seeing the right endmarker), it accets (rejects) both x i y j z and x i y j x Q! z. It remains to consider the sequences of robabilistic choices where M does not accet until x i y j. Let q x be the state of M after reading x i. We consider three cases: 1. q x is not a return state for x. Then, reading more x s cause M to halt with robability 1. However, the robability of halting after reading more than i x s is less than ǫ (by the definition of i). Hence, the robability of this case is less than ǫ. 2. q x is not a return state for y. Then, reading y s cause M to halt with robability 1. If it does not haen before reading y j, it haens later with robability 1. The definition of j imlies that, if M does not halt before reading y j, then the robability of it halting later is less than ǫ. Hence, the robability that q x is not a return state and M does not halt before reading x i y j is less than ǫ. 3. q x is return state for both x and y. Then, reading y j causes M to return to q x because j is a multile of Q! and reading x Q causes it to return to q x as well. In both cases, it is in the same state after reading x i y j and after reading x i y j x Q and, hence, does the same thing on both x i y j z and x i y j x Q! z. We see that the third case causes M to react similarly on x i y j z and x i y j x Q! z and the robability of the other two cases together is less than 2ǫ. Hence, the robabilities of acceting these two words differ by less than 2ǫ. However, one of them is in L and must be acceted with robability 1/2 + ǫ and the second is not in L and must be acceted with robability at most 1/2 ǫ. This means that M does not recognize L with robability 1/2 + ǫ. The forbidden construction of Theorem 6 is also resent in the minimal automaton of a b. Therefore, we have Corollary 4 There is a language that can be recognized by a 1-QFA with robability but cannot be recognized by a 1-PRFA with robability 1/2 + ǫ, for any ǫ > 0.

16 We do not know whether all languages with minimal automata not containing the construction in Theorem 6 can be recognized by 1-way PRFAs. Another oen question is characterizating the languages recognized by 1-way QFAs in terms of forbidden constructions. 4 Comlexity 4.1 Divisibility by a rime All revious work on 1-way QFAs ([12, 13] and the revious sections of this aer) considers the question what languages can be recognized by quantum automata. However, there is another interesting and imortant question: how efficient are QFAs comared to their classical counterarts? For 1-way finite automata, the most natural comlexity measure is the number of states in the automaton. We can follow the roof in [12] that any language recognized by a 1-way QFA is regular ste by ste and add comlexity bounds to it. Then, we get Theorem 7 Let L be a language recognized by a 1-way QFA with n states. Then it can be recognized by a 1-way deterministic automaton with 2 O(n) states. So, transforming a QFA into a classical automaton can cause an exonential increase in its size. Our next results show that, indeed, 1-way QFAs can be exonentially smaller than their classical counterarts. Let be a rime. We consider the language L = {a i i is divisible by }. It is easy to see that any deterministic 1-way finite automaton recognizing L has at least states. However, there is a much more efficient QFA! Theorem 8 For any ǫ > 0, there is a QFA with O(log ) states recognizing L with robability 1 ǫ. Proof. First, we construct an automaton acceting all words in L with robability 1 and acceting words not in L with robability at most 7/8. Later, we will show how to increase the robability of correct answer to 1 ǫ for an arbitrary constant ǫ > 0. Let U k, for k {1,..., 1} be a quantum automaton with a set of states Q = {q 0, q 1, q acc, q rej }, a starting state q 0, Q acc = {q acc }, Q rej = {q rej }. The transition function is defined as follows. Reading a mas q 0 to cosφ q 0 +i sinφ q 1 and q 1 to i sin φ q 0 +cosφ q 1 where φ = 2πk. (It is easy to check that this transformation is unitary.) Reading the right endmarker $ mas q 0 to q acc and q 1 to q rej. Lemma 7 After reading a j, the state of U k is ( ) ( ) 2πjk 2πjk cos q 0 + i sin q 1.

17 Proof. By induction. If j is divisible by, then 2πjk is a multile of 2π, cos( 2πjk ) = 1, sin(2πjk ) = 0, reading aj mas the starting state q 0 to q 0 and the right endmarker $ mas it to q acc. Therefore, all automata U k accet words in L with robability 1. For a word a j / L, call U k good if U k rejects a j with robability at least 1/2. Lemma 8 For any a j / L, at least ( 1)/2 of all U k are good. Proof. The suerosition of U k after reading a j is cos( 2πjk maed to cos( 2πjk ) q acc +sin( 2πjk of U k acceting a i is cos 2 ( 2πjk haens if and only if 2πjk ) q 0 + sin( 2πjk ) q 1. This is ) q rej by the right endmarker. Therefore, the robability ). cos2 ( 2πjk ) 1/2 if and only if cos(2πjk) 1/ 2. This is in [2πl + π/4, 2πl + 3π/4] or in [2πl + 5π/4, 2πl + 7π/4] for some l IN. 2π(jk mod ) [π/4, 3π/4] if and only if 2πjk [2πl+π/4, 2πl+3π/4] for some l. is a rime and j is relatively rime with. Therefore, j mod, 2j mod,..., ( 1)j mod are just 1, 2,..., 1 in different order. Hence, it is enough to count k such that 2πk [5π/4, 7π/4]. or 2πk We do the counting for = 8m + 1. (Other cases are similar.) Then 2πk [π/4, 3π/4] [π/4, 3π/4] if and only if m + 1 k 3m and 2πk [5π/4, 7π/4] if and only if 5m + 1 k 7m. Together, this gives us 4m = ( 1)/2 good k s. Next, we consider sequences of 8 ln k s. A sequence is good for a j if at least 1/4 of all its elements are good for a j. Lemma 9 There is a sequence of length 8 ln which is good for all a j / L. Proof. First, we show that at most 1/ fraction of all sequences is not good for any fixed a j / L. We select a sequence randomly by selecting each of its elements uniformly at random from {1,..., 1}. The robability of selecting a good k in each ste is at least 1/2. By Chernoff bounds, the robability that less than 1/4 = 1/2 1/4 fraction of all elements is good is at most e 2(1/4)2 8ln = 1. Hence, the fraction of sequences which are bad for at least one j {1, 2,..., 1} is at most ( 1)/ and there is a sequence which is good for all j {1,..., 1}. This sequence is good for a j / L with j > as well because any U k returns to the starting state after reading a and, hence, works in the same way on a j and a j mod. Next, we use a good sequence k 1,...,k 8ln to construct a quantum automaton recognizing L. The automaton consists of U k1, U k2,..., U k 8 ln and a distinguished starting state. Uon reading the left endmarker /c, it asses from the starting state to a suerosition where q 0 states of all U kl have equal amlitudes.

18 Words in L are always acceted because all U k accet them. For any a j / L, at least 1/4 of the sequence is good. This means that at least 1/4 of all U kl reject it with robability at least 1/2 and the total robability of rejecting any a j / L is at least 1/8. Finally, we sketch how to increase the robability of correct answer to 1 ǫ for an arbitrary ǫ > 0. We do it by increasing the robability of correct answer for each U k. Namely, we consider an automaton U k with 2 d non-halting states where d is a constant deending on the required robability 1 ǫ. The states are labelled by strings of 0s and 1s of length d: q , q and so on. The starting state is the state q corresonding to the all-0 string. The transition function is defined by d δ(q x1...x d, a, q y1...y d ) = δ(q x1, a, q y1 ). j=1 It is easy to see that this is just the tensor roduct of d coies of U k. The automaton also has one acceting state and 2 d 1 rejecting states. After reading the right endmarker, the automaton asses to the acceting state from q and to a rejecting state from any other state q x1...x d. (To ensure unitarity, one-to-one corresondence between q x1...x d and rejecting states is established.) A counterart of Lemma 7 is Lemma 10 The state of U k after reading aj is ( ) ( ) ( ) ( ) 2πjk 2πjk 2πjk 2πjk (cos q 0 + i sin q 1 )... (cos q 0 + i sin q 1 ). }{{} d times The amlitude of q 0... q 0 = q in this suerosition is cos d ( 2πjk ). If j is a multile of, then this is 1, meaning that words in L are always acceted. For a j / L, we call U k δ-good if it rejects a j with robability at least 1 δ. We formulate a counterart of Lemma 8. Lemma 11 For a suitable constant d, at least 1 δ of all U k are δ-good. Then, we define a δ-good sequence of automata as a sequence such that, for any a j / L, at least 1 2δ of all automata in the sequence are δ-good. Similarly to Lemma 9, we show that there is a δ-good sequence U k 1, U k 2,... of length O(logn). Then, we consider an automaton consisting of U k 1, U k 2,... and a distinguished starting state. Uon reading the left endmarker /c, it asses from the starting state to a suerosition where q 0 states of all U k l have equal amlitudes. Again, it accets a j L with robability 1 because all U k accet a j L. Words a j / L are rejected by at least 1 2δ of U k l with robability 1 δ. Therefore, the robability of rejecting any a j / L is at least (1 2δ)(1 δ) > 1 3δ. Taking δ = ǫ/3 and choosing d so that it satisfies Lemma 11 comletes the roof.

19 We have shown an exonential ga between deterministic and quantum 1-way finite automata. Next, we comare quantum and robabilistic finite automata. Generally, robabilistic finite automata can recognize some languages with the number of states being close to the logarithm of the number of states needed by a deterministic automaton[1, 7]. However, this is not the case with L. Here, adding robabilism does not hel to decrease the number of states at all. Theorem 9 Any 1-way robabilistic finite automaton recognizing L with robability 1/2+ǫ, for a fixed ǫ > 0, has at least states. Proof. Assume that there is a 1-way robabilistic finite automaton with less than states recognizing L with robability ǫ, for a fixed ǫ > 0. Since the language L is in a singleletter alhabet, the automaton can be described as a Markov chain. We use the classification of Markov chains described in Section 2 of [10]. According to this classification, the states of the Markov chain (the automaton) are divided into ergodic and transient states. An ergodic set of states is a set which cannot be left once it is entered. A transient set of states is a set in which every state can be reached from every other state, and which can be left. An ergodic state is an element of an ergodic set. A transient state is an element of a transient set. If a Markov chain has more than one ergodic set, then there is absolutely no interaction between these sets. Hence we have two or more unrelated Markov chains lumed together. These chains may be studied seerately. If a Markov chain consists of a single ergodic set, the chain is called an ergodic chain. According to results in Section 2 of [10], every ergodic chain is either regular or cyclic. If a Markov chain is regular, then sufficiently high owers of the state transition matrix P of the Markov chain are with all ositive elements. Thus no matter where the rocess starts, after sufficient lase of time it can be in any state. Moreover, by Theorem of [10] there is limiting vector of robabilities of being in the states of the chain, not deendent of the initial state. If a Markov chain is cyclic, then the chain has a eriod d, and its states are subdivided into d cyclic sets (d > 1). For a given starting osition, it moves through the cyclic sets in a definite order, returning to the set of the starting state after d stes. Hence the d-th ower of the state transition matrix P describes a regular Markov chain. We have assumed that is rime, and the automaton has less than states. Hence for every cyclic state of the automaton the value of d is strictly less than, and because of rimality of, d is relatively rime to. By D we denote the least common multile of all such values d. Hence D is relatively rime to, and so is any ositive degree D n of D. Since 1 Dn / M but 1 Dn M, the total of the robabilities to be in an acceting state exceeds ǫ for 1Dn and is less than 1 2 ǫ for 1Dn. Contradiction with Theorem of [10]. Corollary 5 For the language L, the number of states needed by a classical (deterministic or robabilistic) 1-way automaton is exonential in the number of states of a 1-way QFA.

20 Proof: Follows from Theorems 8 and Equality Divisibility by a rime is quite natural roblem and we exect that our algorithm can be used as a subroutine, making other quantum algorithms more sace-efficient. Here, we show how to use our quantum automaton for another roblem as well. This roblem is checking whether the length of the inut word is equal to some constant n. Theorem 10 [7] Let L n be a language consisting of one word a n in a single-letter alhabet. 1. Any deterministic automaton that recognizes L n has at least n states. 2. For any ǫ > 0, there is a robabilistic automaton with O(log 2 n) states recognizing L n with robability 1 ǫ. The first art is evident. To rove the second art, Freivalds[7] used the following construction. O( log n ) different rimes are emloyed and O(log n) states are used for every loglog n emloyed rime. At first, the automaton randomly chooses a rime, and then the remainder modulo of the length of the inut word is found and comared with the standard. Additionally, once in every stes a transition to a rejecting state is made with a small robability const. The number of used rimes suffices to assert that, for every inut of length n less than n, most of rimes give remainders different from the remainder of n modulo. The small robability is chosen to have the rejection robability high enough for every inut length N such that both N n and an ǫ-fraction of all the rimes used have the same remainders mod as n. This 1-way robabilistic automaton is reversible according to the definition of section 2. We can use Theorem 1 to transform it into quantum automaton with the number of states increasing at most twice. Then, we obtain a counterart of Theorem 10 for quantum case. However, we can do better by counting modulo rime as in Theorem 8. For that, we need O(log ) states for each rime (instead of states in the robabilistic case). Each rime is O(log n) and there are O(log n/ log log n) of them. Therefore, the number of states in quantum case will be We have shown log n O( loglog n log ) = O( log n log log n) = O(log n). loglog n Theorem 11 L n can be recognized by a 1-way QFA with O(log n) states. Again, the QFA is exonentially smaller than the corresonding deterministic automaton.

21 4.3 Are QFAs always sace-efficient? Subsections 4.1 and 4.2 showed cases when 1-way QFAs are more sace-efficient than their classical counterarts. There can be examles of different kind where deterministic finite automata are exonentially smaller than 1-way QFAs. The construction of theorem 3 which transforms the minimal automaton into a 1-way RFA can increase the size of the automaton exonentially. The next theorem shows that this is inevitable. Theorem 12 Let L m = (xy zy) m {(xy zy) i xx 0 i m 1}. Then, 1. L m can be recognized by a 1-way deterministic finite automaton with 3m + 2 states; 2. L m can be recognized by a 1-way reversible automaton but it requires at least 3(2 m 1) states. After first version of this aer aeared, Ambainis, Nayak and Vazirani[2] showed that, for a different language, the number of states needed by a 1-way QFA is almost exonentially bigger than the number of states of a 1-way deterministic finite automaton. 5 Modifications of 2-way QFAs The advantage of 1-way quantum automata is the simlicity of this model. However, we saw that 1-way automata are quite limited in several situations (desite being good in others) while [12] shows that 2-way QFAs are strictly more owerful than classical finite automata. It would be interesting to come u with a model having both advantages, i.e. being both owerful and simle. In the remainder, we roose several modifications of quantum automata which are intermediate between 1-way QFAs and 2-way QFAs. Quantum art is ket finite in all of these models. Questions about exact ower of these models are mostly oen but we have shown that, in most of these models, all regular languages can be recognized and, in at least one of them, non-regular languages can be recognized as well. 5.1 Scanning the tae multile times The simlest modification is to allow a 1-way QFA to scan its inut tae several times (after the right endmarker it goes to the left endmarker and so on). This is enough to make the roof from [12] that 1-way QFAs recognize only regular languages fail. If we allow the automaton to reject words by non-halting, a nonregular language can be recognized. Theorem 13 Let L = {a n b n n IN}. There is a 1-way QFA M scanning tae several times such that 1. If x / L, M stos with robability 1 after O( x ) scans of the tae.

22 2. If x L, M never stos. If we require M to sto in a rejecting state for rejection, a similar question is still oen. It is also oen whether multile scans of the tae can be used by a 1-way QFA to recognize an arbitrary regular language. (However, known roofs that 1-way QFAs do not recognize some regular languages also fail in this case.) 5.2 Passing information back to environment Another ossibility is introducing more comlicated observables. We can artition all nonhalting states into 2 or 3 classes: moving-left states, moving-right states and (may be) non-moving states. Then, after each ste we observe whether the automaton is in acceting, rejecting, moving-right or moving-left state. If it is in a halting state, we terminate the comutation. If it is in a moving-right state, we feed it the next letter (do the transformation on the quantum system corresonding to the next letter). If it is in a moving-left state, we feed it the revious letter. The model of section 5.1 is a secial case of this model where all non-halting states are classified as moving-right states. 5.3 Prerocessing the inut word In this model, we have two automata M 1 and M 2 instead of one. M 1 is a 2-way deterministic (or robabilistic) finite automaton with outut and M 2 is a 1-way QFA. The inut word is given to M 1 and M 2 is run on the outut of M 1. (This can be viewed as M 1 rerocessing the inut word.) Again, the model of section 5.1 can be viewed as a secial case of this model where M 1 moves from left to right all the time and oututs all letters that it reads. Any regular language can be recognized in a trivial way because we can recognize it by M 1 and give the result as an inut to M 2. If the rerocessing is done by a robabilistic automaton, we can do more. Theorem 14 For any ǫ > 0, there is a 2-way robabilistic finite automaton M 1 and a 1-way QFA M 2 such that, with robability at least 1 ǫ, 1. M 1 stos in time quadratic in the length of the inut and, 2. M 2 accets the outut of M 1 if and only if x {a n b n n IN}. Any 2-way robabilistic automaton that recognizes a non-regular language has an exonential exected running time[5, 6, 8, 11]. So, neither olynomial time 2-way robabilistic finite automata nor 1-way QFAs can recognize non-regular languages. However, their combination can do that! Acknowledgments. We thank Isaac Chuang, Ashwin Nayak, Alistair Sinclair, Amnon Ta-Shma and Umesh Vazirani for useful discussions.