Superiority of exact quantum automata for promise problems

The exact quantum computation has been widely examined for partial (promise) and total functions (e.g. [BH97, BV97, BBC + 98, BCdWZ99, Kla00, BdW03, MNYW05, FI09, YFSA10]). On the other hand, in automata theory, only two results have been obtained: (i) Klauck [Kla00] has shown that realtime quantum finite automata (QFAs) cannot be more concise than realtime deterministic finite automata (DFAs) 1 in case of language recognition, (ii) Murakami et. al. [MNYW05] have shown that there is a promise problem solvable by quantum pushdown automata but not by any deterministic pushdown automata.

In this note, we consider succinctness of realtime QFAs for promise problems. We present an infinite family of promise problems which can be solved exactly by just tuning transition amplitudes of a two-state rtQFAs, whereas the size of the corresponding classical automata grow without bound.

Throughout the paper, (i) Σ denotes the input alphabet not containing left-and right-end markers (¢ and $, respectively) and Σ = Σ ∪ {¢, $}, (ii) ε is the empty string, (iii) w i is the i th symbol of a given string w, and (iv) w represents the string ¢w$, for w ∈ Σ * . Moreover, all machines presented in the paper operate in realtime mode. That is, the input head moves one square to the right in each step and the computation stops after reading $.

A promise problem is a pair A = (A yes , A no ), where A yes , A no ⊆ Σ * and A yes ∩ A no = ∅ [Wat09]. A promise problem A = (A yes , A no ) is solved exactly by a machine M if each string in A yes (resp., A no ) is accepted (resp., rejected) exactly by M. Note that, if A yes = A no , this is the same as the recognition of a language (A yes ).

We give our quantum result with the most restricted of the known QFA model, i.e. Moore-Crutchfield quantum finite automaton (MCQFA) [MC00], (see [YS11] for the definition of the most general QFA model).

A MCQFA is a 5-tuple

where Q is the set of states, q 1 is the initial state, Q a ⊆ Q and is the set of accepting states, and U σ ’s are unitary operators. The computation of a MCQFA on a given input string w ∈ Σ * can be traced by a |Q|dimensional vector. This vector is initially set to |v 0 = (1 0 • • • 0) T and evolves according to

At the end of the computation, w is accepted (resp., rejected) with probability

, where P a = q∈Qa |q q| and P r = I -P a . If we replace the unitary operation with a zero-one left stochastic operator, we obtains a realtime DFA (which we call simply a DFA).

Let A k yes = {a i2 k | i is a nonnegative even integer} and A k no = {a i2 k | i is a positive odd integer} be two unary languages, where k is a positive integer. We will show that a two-state MCQFA can solve promise problem A k = (A k yes , A k no ), but any DFA (and so any PFA) must have at least 2N states to solve the same problem exactly.

) can be solved by a twostate MCQFA M k exactly.

Proof. We will use a well-known technique given in [AF98].

The computation begins with |q 1 and after reading each block of N a’s, the following pattern is followed by M k :

Therefore, it is obvious that M k solves promise problem A k exactly.

Lemma 1. Any DFA solving A k = (A k yes , A k no ) exactly must have at least 2 k+1 states.

Proof. Let N = 2 k and D be a m-state DFA solving A k exactly. We show that m cannot be less than 2N .

Since both A k yes and A k no contain infinitely many unary strings, there must be a chain of t states, say s 0 , . . . , s t-1 such that, for sufficiently long strings, D enters this chain in which D transmits from s i to s (i+1 mod t) when reading an a, where 0 ≤ i ≤ t -1 and 0 < t ≤ m.

Without lose of generality, we assume that D accepts the input if it is in s 0 before reading $. Thus, D rejects the input if it is in s (N mod t) before reading $. Let S a be the set of {s (i2N mod t) | i ≥ 0}. Then, D accepts the input if it is in one of the states in S a before reading $. Note that s (N mod t) / ∈ S a . Let d = gcd(t, 2N ), t ′ = t d , and S ′ be the set

Firstly, we show that each i satisfying (i2N ≡ 0 mod t) must be a multiple of t ′ : For such an i, there exists a j such that i2N = jt. By dividing both sides with t = dt ′ , we get i t ′ 2N d = j. This implies that i must be a multiple of t ′ since left side must be an integer and gcd(t ′ , 2N ) = 1.

Secondly, we show that there is no i 1 and i 2 , i.e. t ′ > i 1 > i 2 ≥ 0, such that (i 1 2N ≡ i 2 2N mod t). If so, we have (i 1 2Ni 2 2N ≡ 0 mod t) and then ((i 1i 2 )2N ≡ 0 mod t). This implies that (i 1i 2 ) must be a multiple of t ′ . This is a contradiction.

Thus, for each i ∈ {0, . . . , t ′ -1}, we obtain a different value of (i2N mod t) and so |S a | contains at least t ′ elements.

If gcd(t, N ) = d, then s (N mod t) also becomes a member of S a . Therefore, gcd(t, N ) must be different than gcd(t, 2N ). This can only be possible whenever t is a multiple of 2N . Therefore, m cannot be less than 2N .

Since a 2 k+1 -state DFA solving promise problem A k exactly can be construc

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