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Computational complexity and applications of quantum algorithm
Applied Mathematics and Computation 218 (2012) 8019–8028
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Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Computational complexity and applications of quantum algorithm S. Iriyama ⇑, M. Ohya Tokyo University of Science, 2641, Yamazaki, Noda City, Chiba, Japan
a r t i c l e
i n f o
Keywords: Quantum algorithm Quantum Turing machine Computational complexity
a b s t r a c t We have studied on quantum algorithm several years, and introduced a mathematical model of it in order to discuss the computational complexity. Our model of quantum algorithm, called a generalized quantum Turing machine (GQTM) contains not only unitary computation process but also quantum measurement and dissipative process. Moreover, we discovered that a chaos dynamics has very important role in quantum algorithm, that is useful to solve NP complete problem in polynomial time. In this paper, we introduce the GQTM and some applications. Ó 2011 Elsevier Inc. All rights reserved.
1. Introduction A quantum algorithm is known as an effective algorithm to solve searching problems. Usually, a computational complexity of algorithm is given as the number of gates for computation in it. A quantum algorithm is represented by a product of unitary operators usually. Since a product of unitary operators can be written by one unitary operator, in some cases the computational complexity depends on the way of description. To discuss a computational complexity of quantum algorithm, we defined a generalized quantum Turing machine using quantum channel and density operator, so that we can consider the measurement and dissipative process in quantum algorithm. Ohya and Volovich discovered that there exist a quantum algorithm with chaos dynamics solving NP complete problem in polynomial time [1–3,8]. We rewrote this algorithm in Turing machine form, and discussed the computational complexity rigorously [13,14]. Recently quantum information has been actively used in decision making and cognitive science [20,21]. A quantum algorithm has more important role in several fields. In this paper, we introduce a generalized quantum Turing machine and application of it. Moreover, we discuss the universal quantum Turing machine. 2. Quantum algorithm A quantum algorithm is constructed by the following steps: 1. 2. 3. 4. 5. 6.
Prepare a Hilbert space H ¼ Cn . Construct an initial state jwin i 2 H. Construct unitary operators U to solve the problem. Apply them for the initial state and obtain a result state jwouti = Ujwini. If necessary, amplify the probability of correct result. Measure an observable with the result state.
⇑ Corresponding author. E-mail address: [email protected] (S. Iriyama). 0096-3003/$ - see front matter Ó 2011 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2011.05.024
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In the first step, we define the Hilbert space depending on the problem. Let C2 be a Hilbert space spanned by j0i = (1, 0)t and j1i = (0, 1)t, a normalized vector jwi = aj0i + bj1i on this space is called a qubit. Since we can use a superposition of j0i and j1i as an initial state vector, the quantum algorithm is more effective than classical one. One can apply Hadamard transformation
1 U H ¼ pffiffiffi 2
1
1
1 1
to create a superposition. Hadamard transformation has a very important role in a quantum algorithm. Applying (UH)n to the vector jwi ¼ j0i j0i 2 ðC2 Þn , we have n 1 1 2X ðU H Þn jwi ¼ pffiffiffiffiffin jei i; 2 i¼0
where {jeii} is a complete orthnormal system of ðC2 Þn defined as
je0 i ¼ j0i j0i; . . . ; je2n 1 i ¼ j1i j1i: One can represent any positive integers less than 2n by jeii. Here we introduce unitary gates, which are NOT gate, C-NOT gate and CC-NOT gate. We call these gates fundamental gates. We can construct AND and OR gate by the product of fundamental gates and some implementations [11,10]. The NOT gate UNOT is defined on a Hilbert space C2 as
U NOT ¼ j1ih0j þ j0ih1j: C-NOT UCN gate and CC-NOT UCCN are given on two and three qubit Hilbert space as
U CN ¼ j0ih0j I þ j1ih1j U NOT ; U CCN ¼ j0ih0j I I þ j1ih1j j0ih0j I þ j1ih1j j1ih1j U NOT ; respectively. 2.1. Oracle quantum algorithms and non-Oracle ones There are several quantum algorithms known as ‘‘Oracle’’ type quantum algorithms. For example, Deutsch–Jozsa algorithm, Grove’s search and Shor’s factoring algorithms. Oracle is given as the black box which replies for any input in one step, and the quantum version of it is prepared as a unitary operator by a third person. Users cannot know the detail of it. To discuss the computational complexity of these algorithms, we have to mention how to construct the Oracle, and how much information needed to construct it. If we construct the Oracle using the answers of given problem, we have to solve the problem before the construction. This situation is not natural, then we should discuss the process to construct the Oracle. We have proposed the quantum algorithms which do not use any Oracles [1,2,8,11,14], so that we can discuss the computational complexity by counting the number of fundamental gates in it. Originally, Oracle is used for the discussion of relative computational complexity in Classical Turing Machine. This topic is discussed more precisely in [16]. 2.2. Computational complexity of quantum algorithm In order to discuss the computational complexity of quantum algorithm, we introduced fundamental gates above. Here we define the computational complexity of quantum algorithm as the number of fundamental gates in it. For example, we say the computational complexity is n if the unitary operator U is constructed by a product of n fundamental gates. In this paper, we define the computational complexity of Oracle is 1. Moreover, we defined a generalized quantum Turing machine (GQTM) using quantum channel and density operator. Based on GQTM we defined some language classes. BGQPP is the language class which is recognized a GQTM in polynomial time with a halting probability 1/2. We proved that the class NP in included by BGQPP [13,14]. 3. Generalized quantum Turing machine In this section, we discuss the generalized quantum Turing machine (GQTM) [13,14] which contains the unitary quantum Turing machine as a special case. 3.1. Generalized quantum Turing machine Let Q be a set of states including an initial state q0 and a set of final states {qF} and R a set of alphabets including a blank symbol #. We denote a set of all infinite sequences of alphabets in R as
R ¼ R R ¼ R1 :
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Let A 2 R⁄ be a sequence of alphabets representing the tape state. Each tape alphabet has a unique index. We write the kth tape alphabet ak as A(k). The GQTM Mgq is defined by the quadruplet ðQ ; R; H; Kd Þ, where Kd is a quantum transition function from C ¼ Q R Z to C. Q and R are represented by a density operator on Hilbert spaces HQ and HR , which are spanned by canonical basis {jqi; q 2 Q} and {jai; a 2 R}, respectively. A tape configuration A is a sequence of elements of R represented by a density operator on the Hilbert space HR spanned by a canonical basis {jAi; A 2 R⁄}. A position of tape head is represented by a density operator on the Hilbert space HZ spanned by a canonical basis fjii; i 2 Zg. Then a configuration of GQTM Mgq is described by a density operator q in H HQ HR HZ . Let SðHÞ be the set of all density operators in the Hilbert space H. Here, we define the transition function
d1 : R Q R Q R Q R f0; 1g Q R f0; 1g ! C: A quantum transition function is given by a quantum channel [7]
Kd1 : SðHÞ ! SðHÞ; satisfying the following condition. Definition 1. Kd1 is called a quantum transition channel if there exists a transition function d1 such that for any quantum P P P P 2 configuration q ¼ k kk jwk ihwk j; jwk i ¼ l ak;l jqk;l ; Ak;l ; ik;l i; k kk ¼ 1; 8kk P 0; l jak;l j ¼ 1; 8ak;l 2 C it holds
X
Kd1 ðqÞ
0
0
0
d1 ðkk ; qk;l ; Ak;l ðik;l Þ; qm;n ; Am;n ðim;n Þ; p; b; d; p0 ; b ; d Þ jp; B; ik;l þ dihp0 ; B0 ; im;n þ d j;
k;l;m;n;p;b;d;p0 ;b0 ;d0
BðjÞ ¼
b
j ¼ ik;l ;
Ak;l ðjÞ otherwise; (
B0 ðjÞ ¼
0
b
j ¼ im;n ;
Am;n ðjÞ otherwise
so that the RHS is a state.
Definition 2. M gq ¼ ðQ ; R; H; Kd Þ is called a LQTM (Linear Quantum Turing Machine) if there exists a transition function
d2 : Q R Q R Q R f0; 1g Q R f0; 1g ! C such that for any quantum configuration qk ; Kd2 is written as
X
Kd2 ðqk Þ
0 0 0 d2 ðqk;l ; Ak;l ik;l ; qm;n ; Am;n ðim;n Þ; p; b; d; p0 ; b ; d Þ jp; B; ik;l þ dihp0 ; B0 ; im;n þ d j
k;l;m;n;p;b;d;p0 ;b0 ;d0
so that the RHS is a state. For any quantum configuration q ¼
X
Kd2
k
! kk qk
¼
X
P
k kk
qk ; Kd2 is affine;
kk Kd2 ðqk Þ:
k
Definition 3. A GQTM Mgq is called unitary QTM (UQTM), if the quantum transition channel Kd3 is a unitary channel implemented: Kd3 ¼ U d3 U d3 , where U d3 is given by, for any jwi = jq, A, ii,
U d3 jwi ¼ U d3 jq; A; ii ¼
X
d3 ðq; AðiÞ; p; b; dÞjp; B; i þ di;
p;b;d
where
d3 : Q R Q R f0; 1g ! C is a transition function and it satisfies for any q 2 Q, a 2 R, q0 (–q) 2 Q, a0 (–a) 2 R,
X
jd3 jðq; a; p; b; dÞj2 ¼ 1;
p;b;d
X
0
d3 ðq0 ; a0 ; p; b; d Þ d3 ðq; a; p; b; dÞ ¼ 0:
p;b;d;d0
Remark 4. Classical Turing Machine (CTM) is represented as LQTM where the transition channel has only diagonal part. Moreover, for any q, p 2 Q, a, b 2 R, d 2 {0, ±1}, put d3(q, a, p, b, d) = 0 or 1 then UQTM is a reversal CTM.
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3.2. Computation process of GQTM Let M ¼ ðQ ; R; H; Kd Þ and q0 = jw0ihw0j where jw0i = jq0, A, 0i, we call this state an initial state and A an input of M. The computation of GQTM proceeds, applying Kd to q0, till the processor state becomes qf 2 {qF}, then it halts. This process is described by the products of Kd as Kd Kd(q0) = qf. Let Sf ðHÞ ¼ fqjqSðHQ Þ 2 fqf gg be the set of final configurations, where ( means the restriction of a state on SðHQ Þ. q is called a final state if q is in the form
q¼ X
X
kk qk þ
k
kk þ
X
k
where
X
ll rl
l
ll ¼ 1; 8 kk ; ll P 0;
l
rl 2 Sf ðHÞ. We call p ¼
P
l
ll the halting probability.
3.3. Language classes defined by GQTM Let M be a GQTM and L a set of alphabet sequences, we say M recognizes L if M halts with any input x 2 L and does not halt for x R L. We call L a language if there exists GQTM M which recognizes L. Definition 5. If languages are accepted by non-deterministic Classical Turing Machine in polynomial time of input size with a certain probability p, then this class of languages is called a bounded probability polynomial time (BPP). If there is a polynomial time algorithm to solve a NP-complete problem in the above sense, then P = NP. The existence of such an algorithm is demonstrated in [1,2] in an extended quantum domain, as is reviewed in the next section. We will show that the OV SAT algorithm can be written by the GQTM. Let us define the recognition of GQTM and some classes of languages. Definition 6. Given a GQTM Mgq and a language L, if there exists N steps when Mgq recognizes L with the probability p, we say that the GQTM Mgq recognizes L by the probability p and its computational complexity is N. Definition 7. A language L is bounded quantum probability polynomial time GQTM (BGQPP) if there is a polynomial time GQTM Mgq which accepts L with probability p P 12. Similarly, we can define the class of languages BUQPP (=BQPP) and BLQPP corresponding to UQTM and LQTM, respectively. In the paper [13,14], it is pointed out that LQTM includes classical TM, which implies
BPP # BLQPPL # BGQPP: Moreover, if NLQTM accepts the SAT OV algorithm in polynomial time with probability p P 12, then we have the inclusion [13]
NP # BGQPP: 3.4. Quantum algorithm for partial recursive function We can make a superposition of configuration by applying the Hadamard gate. Using the superposition, quantum algorithm is more effective than classical ones to solve searching problem. However, even if there exist a classical algorithm for calculating a partial recursive function, it is not clear to create the quantum algorithm which uses a superposition state. In this section we discuss on the quantum algorithm for partial recursive function and halting scheme of it. Let f : R ? R be a partial recursive function
f ðxÞ ¼
f ðgðxÞÞ hðxÞ ¼ 1; x
hðxÞ ¼ 0;
where g : R ? R and h : R ? {0, 1} are computable functions, and M be a CTM that computes f(x) for the input x. We write as
MðxÞ ¼ f ðxÞ: Suppose that we do not know the exact number of steps for computation M(x) before the computation done. Generally, we cannot know the exact time N because we should just calculate f(g(x)N) if we know it. N is decided by the number of looping. Here, we first consider the UQTM MU and LQTM ML that compute f(x) for the input x. We can construct MU and ML according to M following the Deutch’s Theorem, and it implies there exists TM M0 that yields the code of MU and ML denoted by hMUi and hMLi respectively.
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For certain input x, let Nx be the computational complexity of M with input x. We write this computational complexity as T(M(x)) = Nx. Here, consider another input y – x, let Ny = T(M(y)), it holds
Ny – Nx generally. Let Ud be the unitary operator of MU, for input x, we can obtain the result after Nx steps as
N MU ðxÞ ¼ ðU d ÞNx jq0 ; x; 0ihq0 ; x; 0j U d x ¼ jqf ; f ðxÞ; 0ihqf ; f ðxÞ; 0j: For the input y, we can obtain the result after Ny steps as
N MU ðyÞ ¼ ðU d ÞNy jq0 ; y; 0ihq0 ; y; 0j U d y ¼ jqf ; f ðyÞ; 0ihqf ; f ðyÞ; 0j: Moreover, we consider the input
q0 ¼ jwihwj; 1 jwi ¼ pffiffiffi ðjq0 ; x; 0i þ jq0 ; y; 0iÞ: 2 After Nx steps, the state of UQTM becomes
N ðU d ÞNx q0 U d x ¼ jw0 ihw0 j; 1 jw0 i ¼ pffiffiffi ðjqf ; f ðxÞ; 0i þ ðU d ÞNx jq0 ; y; 0iÞ; 2 where the vector ðU d ÞNx jq0 ; y; 0i is something unknown state vector. While for the same input, we obtain after Ny steps
N ðU d ÞNx q0 U d x ¼ jw00 ihw00 j; 1 jw00 i ¼ pffiffiffi ððU d ÞNy jq0 ; x; 0i þ jqf ; f ðyÞ; 0iÞ; 2 where ðU d ÞNy jq0 ; x; 0i is unknown state vector not same as jqf, f(x), 0i. Here, we consider the input data more generally as
q ¼ j/ih/j; P P where j/i ¼ ni ai jq0 ; xi ; 0i; ni jai j2 ¼ 1; 8ai 2 C; n 2 N; n < þ1. This state is a pure state of the superposition. Let N xi ¼ TðMðxi ÞÞ for i = 1, . . . , n, the result of computation after N xj steps becomes
ðU d Þ
N xj
q U d
N x
j
¼ j/0j ih/0j j;
where
j/0j i ¼ aj jqf ; f ðxj Þ; 0i þ
n X
ai ðU d ÞNxj jq0 ; xi ; 0i:
i–j
For superposition input, UQTM can compute f(xi) just at the time N xi for each xi because the computational step is different from each other. Now, we consider how to construct UQTM which computes f(xi) with same time N. Let MUS be a stationary UQTM which computes f(xi) for any input xi. That is, for the input q = j/ih/j, it outputs after N steps
N ðU d ÞN q U d ¼ j/0 ih/0 j; where
j/0 i ¼
n X
ai jqf ; f ðxi Þ; 0i:
i
Theorem 8. There exists a Turing machine M0 (hMi) that outputs hMUSi and its computational complexity T(M0 (hMi)) is
TðM0 Þ ¼ O
n X
!
TðMðxi ÞÞ ;
i
where T(M(xi)) is the computational time of M for the input xi, and O(n) is a usual big-O notation which hides constant factors and lower order terms. The following theorems are proven in the paper [16].
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Theorem 9. There exists a Turing machine M00 that outputs the code of ML and its computational complexity T(M00 ) is
TðM 00 Þ ¼ OðTðMðxÞÞÞ: 3.5. Universal quantum Turing machine A universal quantum Turing machine is first discussed by Deutsh [4], and Bernstein and Vasirani gave the precise way to construct it [5]. We constructed a novel model of a universal QTM which is an extension of Bernstein and Vazirani’s universal QTM by using a property of Riemann zeta function. Our universal QTM simulates all the steps of a target (stationary normal) UQTM for any accuracy e with a slowdown F (defined later) which is a polynomial function of t and 1/e, and avoids a halting problem. Suppose that M ¼ ðQ ; R; H; Kd Þ and M0 ¼ Q 0 ; R0 ; H0 ; K0d0 are UQTMs with the unitary operators Ud and U d0 , respectively. Let 0 t be a positive integer and e > 0, we say that a UQTM M with its input c00 simulates M and its input c0 for t steps with accuracy e and slowdown f which is a polynomial function of t and 1/e, if the following conditions are satisfied: for all q 2 Q ; T 2 R ; i 2 Z,
ðt;1eÞ 0 2 jhq; T; ijU t jc0 ij2 jhq; T; ijU tþf jc ij d 0 < e: d0 Bernstein and Vazirani proved that there exists a normal form, i.e., for any final states, the quantum transition channel changes them into an initial state, UQTM MBV simulating any UQTM M with any accuracy e for t steps with slowdown f t; 1e which can be computed in polynomial steps of t and e. This UQTM MBV is known as one of the models of universal QTM. The input data of MBV is a quadruplet (x, e, t, c(M)) where x is an input of M, e is accuracy of the simulation, t is a simulation time and c(M) is a code of M. Note that it is necessary there to give a time t as an input of MBV. Now, we consider an another model of universal QTM whose input data is (x, e, c(M)), that is, we do not need a simulated time t as an input. It suggests that we do not need to know when the given UQTM halts. The following theorem is given in our paper [12]. Theorem 10. For any UQTM M, there exists a UQTM M which simulates each step of M for an input data (x, e, c(M)) where x is an input of M, e is accuracy of the simulation and c(M) is a code of M. the following inequality is obtained
2 2 tþF ðt;1e Þ 6e t jcM i < 2 2 ; hq; T; ijU d jcM i hq; T; ijU d0 p t where Ud and U d0 are the unitary operator corresponding to M and M, respectively. Suppose that M halts and gives an outcome with probability p; M gives the same outcome with probability p0 satisfying jp p0 j 6 e. According to a property of Riemann Zeta function, we have
jp0 pj 6
1 X 6e i¼1
p2 i2
6 e:
By inserting the process to calculate the proper accuracy for MBV , we can construct an universal QTM and avoid the halting problem. 4. OV SAT algorithm In this section, we explain the OV SAT algorithm solving SAT problem in polynomial time. It is discussed more precisely in the papers [1–3,9,11]. We do not use any Oracles for computation, so that the unitary operator of this algorithm also can be concretely constructed in polynomial time. Let X fx1 ; . . . ; xn g; n 2 N be a set. xk and its negation xk ðk ¼ 1; . . . ; nÞ are called literals. Let X f x1 ; . . . ; xn g be a set, then the set of all literals is denoted by X 0 X [ X ¼ fx1 ; . . . ; xn ; x1 ; . . . ; xn g. The set of all subsets of X0 is denoted by F ðX 0 Þ and an element C 2 F ðX 0 Þ is called a clause. We take a truth assignment t to all variables xk. If we can assign the truth value to at least one element of C, then C is called satisfiable. Let L = {0, 1} be a Boolean lattice with usual join _ and meet ^, and t(x) be the truth value of a literal x in X. Then the truth value of a clause C is written as t(C) _x2Ct(x). Moreover the set C of all clauses Cj (j = 1, 2, . . . , m) is called satisfiable iff ðCÞ ^m j¼1 tðC j Þ ¼ 1. Thus the SAT problem is written as follows: Definition 11 (SAT Problem). Given a Boolean set X fx1 ; . . . ; xn g and a set C ¼ fC 1 ; . . . ; C m g of clauses, determine whether C is satisfiable or not.
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That is, this problem is to ask whether there exists a truth assignment to make C satisfiable. It is known in usual algorithm that it is polynomial time to check the satisfiability only when a specific truth assignment is given, but we can not determine the satisfiability in polynomial time when an assignment is not specified. In [3] the quantum algorithm of the SAT problem was first studied, which was rewritten in [9] by showing that the Ohya– Masuda SAT algorithm is combinatorial. In [1,2] it is shown that the chaotic quantum algorithm can solve the SAT problem in polynomial time. 4.1. Unitary gates for computation Ohya and Masuda pointed out [3] that the SAT problem, hence all other NP problems, can be solved in polynomial time by quantum computer if the superposition of two orthogonal vectors j0i and j1i is physically detected. However this detection is considered not to be possible in the present technology. The problem to be overcome is how to distinguish the pure vector j0i from the superposed one ajui j0i + bju0 i j1i, obtained by the OM algorithm, if b is not zero but very small. If such a distinction is possible, then we can solve the NPC problem in the polynomial time. In [1,2] it is shown that it can be possible by combining nonlinear chaos amplifier with the quantum algorithm, which implies the existence of a mathematical algorithm solving P = NP. In this section, we explain the quantum algorithm of SAT problem based on [9,3]. Here, we estimate the total number of qubits more rigorously and show how to construct the unitary gates more precisely for any case. We first calculate the total number of qubits and show that this number depends on the input data only and is calculated in polynomial time. Then we decide the Hilbert space for the quantum algorithm and the initial state vector on it. Let C ¼ fC 1 ; . . . ; C m g be a set of clauses on X 0 fx1 ; . . . ; xn ; x1 ; . . . ; xn g. The computational basis of this algorithm is the Hilbert space H ¼ ðC2 Þnþlþ1 where l is a number of dust qubits, it is shown that l is less than 2mn [11]. Let
jv in i j0n ; 0l ; 0i; be an initial state vector where j0n i ¼ j0; 0; . . . ; 0i ¼ j0i j0i. |fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl} For X = {x1, . . . , xn} and a truth assignment t, we put n
tðx1 Þ ¼ e1 ; tðx2 Þ ¼ e2 ; . . . ; tðxn Þ ¼ en ; where e1, e2, . . . , en 2 {0, 1}, and we write t as a sequence of binary symbols:
t ¼ e1 e2 en : A unitary operator U C : H ! H computes tðCÞ for all truth assignment as follows n 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 2X l U C jv in i ¼ U C j0n ; 0l ; 0i ¼ pffiffiffiffiffin jei ; d ; tðCÞi ¼ qjui j0i þ 1 q2 ju0 i j1i jv out i; 2 i¼0
where jdli is dust qubits denoted by l strings of binary symbols, and jeii = j e1, e2, . . . , eni is a binary representation of t. Theorem 12 [11]. For a set of clauses C ¼ fC 1 ; . . . ; C m g on X 0 fx1 ; . . . ; xn ; x1 ; . . . ; xn g, the number l of dust qubits for algorithm of SAT problem is
l ¼ sf 1 n 6 2nm: Theorem 13 ([3,9,11]). For a set of clauses C ¼ fC 1 ; . . . ; C m g, there exists the unitary operator U C to calculate the truth value of C as
UC
m1 Y
U AND ðiÞ
m Y
i¼1
U OR ðjÞHðnÞ;
j¼1
where H(k) is a unitary operator acting by the application of the Hadamard transformation to first k qubits, that is
HðkÞ ¼ Hk INk : The computational complexity of quantum computation depends on the number of unitary operator in the quantum circuit. Let U be the unitary operator, it is written as
U ¼ U n U n1 U 1 ; where Un, . . . , U1 are fundamental gates. The computational complexity T(U) is considered as n. The above unitary operators UAND and UOR can be written as a combination of fundamental gates NOT gate UNOT, controlled-NOT gate UCN and controlled-controlled-NOT gate UCCN. The computational complexity TðU C Þ of SAT algorithm is defined by the number of UNOT, UAND and UOR used.
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Theorem 14 [11]. For a set of clauses C ¼ fC 1 ; . . . ; C m g and literal X 0 ¼ fx1 ; . . . ; xn ; x1 ; . . . ; xn g; TðU C Þ is
TðU C Þ ¼ m 1 þ
m X 0 jC k j þ 2ik 1 6 4mn 1: k1
4.2. Chaos amplifier Here we will briefly review how chaos amplifier can play a constructive role in computation (see [1,2] for the details). Consider the so called logistic map which is given by the equation
xnþ1 ¼ axn ð1 xn Þ g a ðxÞ;
xn 2 ½0; 1 :
The properties of the map depend on the parameter a. If we take, for example, a = 3.71, then the Lyapunov exponent is positive, the trajectory is very sensitive to the initial value and one has the chaotic behavior [2,6]. It is important to notice that if the initial value x0 = 0, then xn = 0 for all n. The state jvouti of the previous subsection is transformed into the density matrix of the form
q ¼ q2 P1 þ ð1 q2 ÞP0 ; where P1 and P0 are projectors to the state vectors j1i and j0i. One has to notice that P1 and P0 generate an Abelian algebra which can be considered as a classical system. Let KC be a quantum channel on one qubit space such that
Þ ¼ KCA ðq
Þr3 Þ ðI þ g a ðq ; 2
k , we have where I is the identity matrix and r3 is the z-component of Pauli matrices. Applying KCA to q
k I þ g ka ðq2 Þr3 qk ¼ KCA ðqÞ ¼ : 2
To find a proper value k we finally measure the value of r3 in the state qk such that
k r3 ¼ g ka ðq2 Þ: Mk trq The following theorems is proven in [1,2,10]. Theorem 15. For the logistic map xn+1 = axn(1 xn) with a 2 [0, 4] and x0 2 [0, 1], let x0 be 21n and a set J be {0, 1, 2, . . . , n, . . . , 2n}. If a is 3.71, then there exists an integer k in J satisfying xk > 12. . Theorem 16. Let a and n be the same in above theorem. If there exists k in J such that xk > 12, then k > log2n1 3:711 Theorem 17. Let jtðCÞj be the cardinality of these assignments, if x0 2rn with r ¼ jtðCÞj and there exists k in J such that xk > 12, then there exists k satisfying the following inequality if C is SAT.
n 1 log2 r 5 6k6 ðn 1Þ : log2 3:71 1 4 From these theorems, for all k, it holds
Mk
¼ 0 iff C is not SAT; > 0 iff C is SAT:
The purpose of this amplification is to see the difference of the states
q ¼ P0 and
q ¼ q2 P1 þ ð1 q2 ÞP0 ; q2 > 0: k If q = P0, it stays q = P0 after applying KCA , and if q – P0, Mk becomes larger than 12 in polynomial of n. Ohya and Accardi found an alternative algorithm solving the SAT problem by means of the stochastic limit [8]. A physical realization of the chaos amplifier is discussed in [10,19]. A photosynthesis is considered as one of possible amplifiers, which uses the environment to amplify the signal from the antenna system.
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5. GQTM expression of OV SAT algorithm In this section, we construct the 3 multi-track GQTM M SAT ¼ ðQ ; R3 ; H; Kd Þ that achieves the OV algorithm. This GQTM does not belong to LQTM and UQTM because the chaos amplification process is described by a non-linear CP channel. The k through q jvfihvfj explained above. The computation from OV SAT algorithm runs from an initial state q0 jv0ihv0j to q k is due to a non-unitary chanq0 jv0ihv0j to q jvfihvfj is due to unitary channel KC UC UC, and that from q jvfihvfj to q nel KkCA KI , so that all computation can be done by KkCA KI KC , which is a completely positive deterministic map. Let us explain our computation by a multi-track GQTM. The first track stores the input data and literals. The second track is used for the computation of f(Ci), (i = 1, . . . , m), and the third track is used for the computation of the result f ðCÞ. This algorithm is represented by the following 5 steps [13]:
Step Step Step Step Step
1: 2: 3: 4: 5:
Apply the Hadamard transform to Track 1. Calculate f(C1), . . .f(Cm) and store them in Track 2. Calculate f ðCÞ, and store it in Track 3. Empty the working space. Apply the chaos amplifier to the result state and repeat this step.
5.1. Computational complexity of the SAT algorithm Let T(K) be a computational complexity of K defined as the number of steps in K. Theorem 18. For a set of clauses C with n Boolean variables, the computational complexity T(KSAT) of the OV SAT algorithm including the chaos amplifier, is
TðKSAT Þ ¼ ðTðKC Þ þ TðKT ÞÞT KkCA ¼ OðpolyðnÞÞ; where k is a positive number smaller than
5 4
ðn 1Þ and poly(n) denotes a polynomial of n.
Proof. The computational complexity of KSAT depends on the computational complexity of unitary calculation and chaos amplifier. The unitary calculation part is written as KC, and its computational complexity is essentially discussed in [3,11]. For X 0 ¼ fx1 ; . . . xn ; x1 ; . . . ; xn g and C ¼ fC 1 ; . . . ; C m g the set of literals, the computational complexity T(KC) is obtained as
TðKC Þ ¼ 3
m m X X ðcardðC k Þ 1Þ þ 2cardðC k \ fx1 ; . . . ; xn gÞ þ m 1 þ n 6 ð8mn 2m 1Þ ¼ OðpolyðnÞÞ: k¼1
k1
After the unitary calculation, we translate it to the chaos amplifier. MSAT empties the work space for calculation in this process, and it is proved in [11] that the size of workspace is exactly in polynomial order of input size. Therefore, the computational complexity of KT is TðKT Þ ¼ OðpolyðnÞÞ. We have to repeat these calculation because we do not know exact time of chaos amplification process when the halting probability is greater than 1/2. Then, the total computational complexity becomes
TðKSAT Þ ¼ ðTðKC Þ þ TðKT ÞÞT KkCA ¼ OðpolyðnÞÞ:
6. Results and topics on quantum algorithms Here we mention our further recent studies. We consider the applications of quantum algorithms not only to mathematical problems, but also problems of Bio-informatics; (1) sequence alignment of amino acid sequence [15] and (2) protein folding problem [18]. There exists a harder language class than the class NP, for example EXPTIME. We consider a quantum algorithm of the Pebble game problem which belongs to EXPTIME. Players move one stone at a time along a given rule alternatively. A player wins when he moves a stone to the winning position on the board given by a certain rule, or he loses when he cannot move any stone. We want to know whether there exist strategies such that a first mover can win every time. The computational complexity of this problem is obviously very large. In some cases depending on a size of board and the rule, it takes exponential time to solve this problem. We proposed a quantum algorithm for Pebble Game with an Oracle, and the computational complexity is polynomial of the size of game board [17]. Note that even if we assume such an Oracle, the computational complexity of classical algorithm is still exponential. References [1] M. Ohya, I.V. Volovich, Quantum computing and chaotic amplification, J. opt. B 5 (6) (2003) 639–642. [2] M. Ohya, I.V. Volovich, New quantum algorithm for studying NP-complete problems, Rep. Math. Phys. 52 (1) (2003) 25–33.
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[3] M. Ohya, N. Masuda, NP problem in quantum algorithm, Open Syst. Inform. Dyn. 7 (1) (2000) 33–39. [4] D. Deutsch, Proc. Roy. Soc. London A 400 (1985) 97–117. [5] E. Bernstein, U. Vazirani, Quantum complexity theory, in: Proceedings of the 25th Annual ACM Symposium on Theory of Computing, ACM, New York, 1993, pp.11–22. SIAM Journal on Computing 26, 1411, 1997. [6] M. Ohya, Complexities and their applications to characterization of chaos, Int. J. Theor. Phys. 37 (1998) 495. [7] L. Accardi, M. Ohya, Compound channels, transition expectations, and liftings, Appl. Math. Optim. 39 (1999) 33–59. [8] L. Accardi, M. Ohya, A stochastic limit approach to the SAT problem, Open Syst. Inform. Dyn. 11 (2004) 1–16. [9] L. Accardi, R. Sabbadini, On the Ohya–Masuda quantum SAT Algorithm, Preprint Volterra, N. 432, 2000. [10] M. Ohya, I.V. Volovich, Mathematical Foundation of Quantum Information and Quantum Computation, Springer Velrag, 2011. [11] S. Iriyama, M. Ohya, Rigorous estimation of the computational complexity for OV SAT algorithm, Open Syst. Inform. Dyn. 15 (2) (2008) 173–187. [12] S. Iriyama, T. Miyadera, M. Ohya, Note on a universal quantum turing machine, Phys. Lett. A 372 (2008) 5120–5122. [13] S. Iriyama, M. Ohya, Language classes defined by generalized quantum turing machine, Open Syst. Inform. Dyn. 15 (4) (2008) 383–396. [14] S. Iriyama, M. Ohya, I.V. Volovich, Generalized quantum turing machine and its application to the SAT chaos algorithm, in: QP-PQ: Quantum Prob. White Noise Anal, Quantum Information and Computing, vol. 19, World Sci. Publishing, 2006, pp. 204–225. [15] S. Iriyama, M. Ohya, On quantum algorithm for multiple alignment of amino acid sequences, QP-PQ 24, Quantum Bio-Inform. II (2009) 92–104. [16] S. Iriyama, M. Ohya, The problem to construct Unitary Quantum Turing Machine computing partial recursive functions, TUS preprint, 2009. [17] S. Iriyama, M. Ohya, Quantum Algorithm for Pebble Game and Its Computational Complexity, TUS preprint, 2010. [18] S. Iriyama, M. Ohya, I. Yamato, Quantum Algorithm of Protain Folding, TUS preprint, 2010. [19] S. Iriyama, M. Ohya, K. Sato, I.V. Volovich, Photosynthetic Antenna and Entropy Decreasing, in preparation. [20] A. Khrennikov, Ubiquitous Quantum Structure: From Psychology to Finances, Springer, Berlin-Heidelberg-New York, 2010. [21] M. Asano, M. Ohya, A. Khrennikov, Quantum-like model for decision making process in two players game A non-Kolmogorovian model, Foundat. Phys. (2010), doi:10.1007/s10701-010-9454-y.
Contents lists available at ScienceDirect
Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Computational complexity and applications of quantum algorithm S. Iriyama ⇑, M. Ohya Tokyo University of Science, 2641, Yamazaki, Noda City, Chiba, Japan
a r t i c l e
i n f o
Keywords: Quantum algorithm Quantum Turing machine Computational complexity
a b s t r a c t We have studied on quantum algorithm several years, and introduced a mathematical model of it in order to discuss the computational complexity. Our model of quantum algorithm, called a generalized quantum Turing machine (GQTM) contains not only unitary computation process but also quantum measurement and dissipative process. Moreover, we discovered that a chaos dynamics has very important role in quantum algorithm, that is useful to solve NP complete problem in polynomial time. In this paper, we introduce the GQTM and some applications. Ó 2011 Elsevier Inc. All rights reserved.
1. Introduction A quantum algorithm is known as an effective algorithm to solve searching problems. Usually, a computational complexity of algorithm is given as the number of gates for computation in it. A quantum algorithm is represented by a product of unitary operators usually. Since a product of unitary operators can be written by one unitary operator, in some cases the computational complexity depends on the way of description. To discuss a computational complexity of quantum algorithm, we defined a generalized quantum Turing machine using quantum channel and density operator, so that we can consider the measurement and dissipative process in quantum algorithm. Ohya and Volovich discovered that there exist a quantum algorithm with chaos dynamics solving NP complete problem in polynomial time [1–3,8]. We rewrote this algorithm in Turing machine form, and discussed the computational complexity rigorously [13,14]. Recently quantum information has been actively used in decision making and cognitive science [20,21]. A quantum algorithm has more important role in several fields. In this paper, we introduce a generalized quantum Turing machine and application of it. Moreover, we discuss the universal quantum Turing machine. 2. Quantum algorithm A quantum algorithm is constructed by the following steps: 1. 2. 3. 4. 5. 6.
Prepare a Hilbert space H ¼ Cn . Construct an initial state jwin i 2 H. Construct unitary operators U to solve the problem. Apply them for the initial state and obtain a result state jwouti = Ujwini. If necessary, amplify the probability of correct result. Measure an observable with the result state.
⇑ Corresponding author. E-mail address: [email protected] (S. Iriyama). 0096-3003/$ - see front matter Ó 2011 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2011.05.024
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In the first step, we define the Hilbert space depending on the problem. Let C2 be a Hilbert space spanned by j0i = (1, 0)t and j1i = (0, 1)t, a normalized vector jwi = aj0i + bj1i on this space is called a qubit. Since we can use a superposition of j0i and j1i as an initial state vector, the quantum algorithm is more effective than classical one. One can apply Hadamard transformation
1 U H ¼ pffiffiffi 2
1
1
1 1
to create a superposition. Hadamard transformation has a very important role in a quantum algorithm. Applying (UH)n to the vector jwi ¼ j0i j0i 2 ðC2 Þn , we have n 1 1 2X ðU H Þn jwi ¼ pffiffiffiffiffin jei i; 2 i¼0
where {jeii} is a complete orthnormal system of ðC2 Þn defined as
je0 i ¼ j0i j0i; . . . ; je2n 1 i ¼ j1i j1i: One can represent any positive integers less than 2n by jeii. Here we introduce unitary gates, which are NOT gate, C-NOT gate and CC-NOT gate. We call these gates fundamental gates. We can construct AND and OR gate by the product of fundamental gates and some implementations [11,10]. The NOT gate UNOT is defined on a Hilbert space C2 as
U NOT ¼ j1ih0j þ j0ih1j: C-NOT UCN gate and CC-NOT UCCN are given on two and three qubit Hilbert space as
U CN ¼ j0ih0j I þ j1ih1j U NOT ; U CCN ¼ j0ih0j I I þ j1ih1j j0ih0j I þ j1ih1j j1ih1j U NOT ; respectively. 2.1. Oracle quantum algorithms and non-Oracle ones There are several quantum algorithms known as ‘‘Oracle’’ type quantum algorithms. For example, Deutsch–Jozsa algorithm, Grove’s search and Shor’s factoring algorithms. Oracle is given as the black box which replies for any input in one step, and the quantum version of it is prepared as a unitary operator by a third person. Users cannot know the detail of it. To discuss the computational complexity of these algorithms, we have to mention how to construct the Oracle, and how much information needed to construct it. If we construct the Oracle using the answers of given problem, we have to solve the problem before the construction. This situation is not natural, then we should discuss the process to construct the Oracle. We have proposed the quantum algorithms which do not use any Oracles [1,2,8,11,14], so that we can discuss the computational complexity by counting the number of fundamental gates in it. Originally, Oracle is used for the discussion of relative computational complexity in Classical Turing Machine. This topic is discussed more precisely in [16]. 2.2. Computational complexity of quantum algorithm In order to discuss the computational complexity of quantum algorithm, we introduced fundamental gates above. Here we define the computational complexity of quantum algorithm as the number of fundamental gates in it. For example, we say the computational complexity is n if the unitary operator U is constructed by a product of n fundamental gates. In this paper, we define the computational complexity of Oracle is 1. Moreover, we defined a generalized quantum Turing machine (GQTM) using quantum channel and density operator. Based on GQTM we defined some language classes. BGQPP is the language class which is recognized a GQTM in polynomial time with a halting probability 1/2. We proved that the class NP in included by BGQPP [13,14]. 3. Generalized quantum Turing machine In this section, we discuss the generalized quantum Turing machine (GQTM) [13,14] which contains the unitary quantum Turing machine as a special case. 3.1. Generalized quantum Turing machine Let Q be a set of states including an initial state q0 and a set of final states {qF} and R a set of alphabets including a blank symbol #. We denote a set of all infinite sequences of alphabets in R as
R ¼ R R ¼ R1 :
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Let A 2 R⁄ be a sequence of alphabets representing the tape state. Each tape alphabet has a unique index. We write the kth tape alphabet ak as A(k). The GQTM Mgq is defined by the quadruplet ðQ ; R; H; Kd Þ, where Kd is a quantum transition function from C ¼ Q R Z to C. Q and R are represented by a density operator on Hilbert spaces HQ and HR , which are spanned by canonical basis {jqi; q 2 Q} and {jai; a 2 R}, respectively. A tape configuration A is a sequence of elements of R represented by a density operator on the Hilbert space HR spanned by a canonical basis {jAi; A 2 R⁄}. A position of tape head is represented by a density operator on the Hilbert space HZ spanned by a canonical basis fjii; i 2 Zg. Then a configuration of GQTM Mgq is described by a density operator q in H HQ HR HZ . Let SðHÞ be the set of all density operators in the Hilbert space H. Here, we define the transition function
d1 : R Q R Q R Q R f0; 1g Q R f0; 1g ! C: A quantum transition function is given by a quantum channel [7]
Kd1 : SðHÞ ! SðHÞ; satisfying the following condition. Definition 1. Kd1 is called a quantum transition channel if there exists a transition function d1 such that for any quantum P P P P 2 configuration q ¼ k kk jwk ihwk j; jwk i ¼ l ak;l jqk;l ; Ak;l ; ik;l i; k kk ¼ 1; 8kk P 0; l jak;l j ¼ 1; 8ak;l 2 C it holds
X
Kd1 ðqÞ
0
0
0
d1 ðkk ; qk;l ; Ak;l ðik;l Þ; qm;n ; Am;n ðim;n Þ; p; b; d; p0 ; b ; d Þ jp; B; ik;l þ dihp0 ; B0 ; im;n þ d j;
k;l;m;n;p;b;d;p0 ;b0 ;d0
BðjÞ ¼
b
j ¼ ik;l ;
Ak;l ðjÞ otherwise; (
B0 ðjÞ ¼
0
b
j ¼ im;n ;
Am;n ðjÞ otherwise
so that the RHS is a state.
Definition 2. M gq ¼ ðQ ; R; H; Kd Þ is called a LQTM (Linear Quantum Turing Machine) if there exists a transition function
d2 : Q R Q R Q R f0; 1g Q R f0; 1g ! C such that for any quantum configuration qk ; Kd2 is written as
X
Kd2 ðqk Þ
0 0 0 d2 ðqk;l ; Ak;l ik;l ; qm;n ; Am;n ðim;n Þ; p; b; d; p0 ; b ; d Þ jp; B; ik;l þ dihp0 ; B0 ; im;n þ d j
k;l;m;n;p;b;d;p0 ;b0 ;d0
so that the RHS is a state. For any quantum configuration q ¼
X
Kd2
k
! kk qk
¼
X
P
k kk
qk ; Kd2 is affine;
kk Kd2 ðqk Þ:
k
Definition 3. A GQTM Mgq is called unitary QTM (UQTM), if the quantum transition channel Kd3 is a unitary channel implemented: Kd3 ¼ U d3 U d3 , where U d3 is given by, for any jwi = jq, A, ii,
U d3 jwi ¼ U d3 jq; A; ii ¼
X
d3 ðq; AðiÞ; p; b; dÞjp; B; i þ di;
p;b;d
where
d3 : Q R Q R f0; 1g ! C is a transition function and it satisfies for any q 2 Q, a 2 R, q0 (–q) 2 Q, a0 (–a) 2 R,
X
jd3 jðq; a; p; b; dÞj2 ¼ 1;
p;b;d
X
0
d3 ðq0 ; a0 ; p; b; d Þ d3 ðq; a; p; b; dÞ ¼ 0:
p;b;d;d0
Remark 4. Classical Turing Machine (CTM) is represented as LQTM where the transition channel has only diagonal part. Moreover, for any q, p 2 Q, a, b 2 R, d 2 {0, ±1}, put d3(q, a, p, b, d) = 0 or 1 then UQTM is a reversal CTM.
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3.2. Computation process of GQTM Let M ¼ ðQ ; R; H; Kd Þ and q0 = jw0ihw0j where jw0i = jq0, A, 0i, we call this state an initial state and A an input of M. The computation of GQTM proceeds, applying Kd to q0, till the processor state becomes qf 2 {qF}, then it halts. This process is described by the products of Kd as Kd Kd(q0) = qf. Let Sf ðHÞ ¼ fqjqSðHQ Þ 2 fqf gg be the set of final configurations, where ( means the restriction of a state on SðHQ Þ. q is called a final state if q is in the form
q¼ X
X
kk qk þ
k
kk þ
X
k
where
X
ll rl
l
ll ¼ 1; 8 kk ; ll P 0;
l
rl 2 Sf ðHÞ. We call p ¼
P
l
ll the halting probability.
3.3. Language classes defined by GQTM Let M be a GQTM and L a set of alphabet sequences, we say M recognizes L if M halts with any input x 2 L and does not halt for x R L. We call L a language if there exists GQTM M which recognizes L. Definition 5. If languages are accepted by non-deterministic Classical Turing Machine in polynomial time of input size with a certain probability p, then this class of languages is called a bounded probability polynomial time (BPP). If there is a polynomial time algorithm to solve a NP-complete problem in the above sense, then P = NP. The existence of such an algorithm is demonstrated in [1,2] in an extended quantum domain, as is reviewed in the next section. We will show that the OV SAT algorithm can be written by the GQTM. Let us define the recognition of GQTM and some classes of languages. Definition 6. Given a GQTM Mgq and a language L, if there exists N steps when Mgq recognizes L with the probability p, we say that the GQTM Mgq recognizes L by the probability p and its computational complexity is N. Definition 7. A language L is bounded quantum probability polynomial time GQTM (BGQPP) if there is a polynomial time GQTM Mgq which accepts L with probability p P 12. Similarly, we can define the class of languages BUQPP (=BQPP) and BLQPP corresponding to UQTM and LQTM, respectively. In the paper [13,14], it is pointed out that LQTM includes classical TM, which implies
BPP # BLQPPL # BGQPP: Moreover, if NLQTM accepts the SAT OV algorithm in polynomial time with probability p P 12, then we have the inclusion [13]
NP # BGQPP: 3.4. Quantum algorithm for partial recursive function We can make a superposition of configuration by applying the Hadamard gate. Using the superposition, quantum algorithm is more effective than classical ones to solve searching problem. However, even if there exist a classical algorithm for calculating a partial recursive function, it is not clear to create the quantum algorithm which uses a superposition state. In this section we discuss on the quantum algorithm for partial recursive function and halting scheme of it. Let f : R ? R be a partial recursive function
f ðxÞ ¼
f ðgðxÞÞ hðxÞ ¼ 1; x
hðxÞ ¼ 0;
where g : R ? R and h : R ? {0, 1} are computable functions, and M be a CTM that computes f(x) for the input x. We write as
MðxÞ ¼ f ðxÞ: Suppose that we do not know the exact number of steps for computation M(x) before the computation done. Generally, we cannot know the exact time N because we should just calculate f(g(x)N) if we know it. N is decided by the number of looping. Here, we first consider the UQTM MU and LQTM ML that compute f(x) for the input x. We can construct MU and ML according to M following the Deutch’s Theorem, and it implies there exists TM M0 that yields the code of MU and ML denoted by hMUi and hMLi respectively.
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For certain input x, let Nx be the computational complexity of M with input x. We write this computational complexity as T(M(x)) = Nx. Here, consider another input y – x, let Ny = T(M(y)), it holds
Ny – Nx generally. Let Ud be the unitary operator of MU, for input x, we can obtain the result after Nx steps as
N MU ðxÞ ¼ ðU d ÞNx jq0 ; x; 0ihq0 ; x; 0j U d x ¼ jqf ; f ðxÞ; 0ihqf ; f ðxÞ; 0j: For the input y, we can obtain the result after Ny steps as
N MU ðyÞ ¼ ðU d ÞNy jq0 ; y; 0ihq0 ; y; 0j U d y ¼ jqf ; f ðyÞ; 0ihqf ; f ðyÞ; 0j: Moreover, we consider the input
q0 ¼ jwihwj; 1 jwi ¼ pffiffiffi ðjq0 ; x; 0i þ jq0 ; y; 0iÞ: 2 After Nx steps, the state of UQTM becomes
N ðU d ÞNx q0 U d x ¼ jw0 ihw0 j; 1 jw0 i ¼ pffiffiffi ðjqf ; f ðxÞ; 0i þ ðU d ÞNx jq0 ; y; 0iÞ; 2 where the vector ðU d ÞNx jq0 ; y; 0i is something unknown state vector. While for the same input, we obtain after Ny steps
N ðU d ÞNx q0 U d x ¼ jw00 ihw00 j; 1 jw00 i ¼ pffiffiffi ððU d ÞNy jq0 ; x; 0i þ jqf ; f ðyÞ; 0iÞ; 2 where ðU d ÞNy jq0 ; x; 0i is unknown state vector not same as jqf, f(x), 0i. Here, we consider the input data more generally as
q ¼ j/ih/j; P P where j/i ¼ ni ai jq0 ; xi ; 0i; ni jai j2 ¼ 1; 8ai 2 C; n 2 N; n < þ1. This state is a pure state of the superposition. Let N xi ¼ TðMðxi ÞÞ for i = 1, . . . , n, the result of computation after N xj steps becomes
ðU d Þ
N xj
q U d
N x
j
¼ j/0j ih/0j j;
where
j/0j i ¼ aj jqf ; f ðxj Þ; 0i þ
n X
ai ðU d ÞNxj jq0 ; xi ; 0i:
i–j
For superposition input, UQTM can compute f(xi) just at the time N xi for each xi because the computational step is different from each other. Now, we consider how to construct UQTM which computes f(xi) with same time N. Let MUS be a stationary UQTM which computes f(xi) for any input xi. That is, for the input q = j/ih/j, it outputs after N steps
N ðU d ÞN q U d ¼ j/0 ih/0 j; where
j/0 i ¼
n X
ai jqf ; f ðxi Þ; 0i:
i
Theorem 8. There exists a Turing machine M0 (hMi) that outputs hMUSi and its computational complexity T(M0 (hMi)) is
TðM0 Þ ¼ O
n X
!
TðMðxi ÞÞ ;
i
where T(M(xi)) is the computational time of M for the input xi, and O(n) is a usual big-O notation which hides constant factors and lower order terms. The following theorems are proven in the paper [16].
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Theorem 9. There exists a Turing machine M00 that outputs the code of ML and its computational complexity T(M00 ) is
TðM 00 Þ ¼ OðTðMðxÞÞÞ: 3.5. Universal quantum Turing machine A universal quantum Turing machine is first discussed by Deutsh [4], and Bernstein and Vasirani gave the precise way to construct it [5]. We constructed a novel model of a universal QTM which is an extension of Bernstein and Vazirani’s universal QTM by using a property of Riemann zeta function. Our universal QTM simulates all the steps of a target (stationary normal) UQTM for any accuracy e with a slowdown F (defined later) which is a polynomial function of t and 1/e, and avoids a halting problem. Suppose that M ¼ ðQ ; R; H; Kd Þ and M0 ¼ Q 0 ; R0 ; H0 ; K0d0 are UQTMs with the unitary operators Ud and U d0 , respectively. Let 0 t be a positive integer and e > 0, we say that a UQTM M with its input c00 simulates M and its input c0 for t steps with accuracy e and slowdown f which is a polynomial function of t and 1/e, if the following conditions are satisfied: for all q 2 Q ; T 2 R ; i 2 Z,
ðt;1eÞ 0 2 jhq; T; ijU t jc0 ij2 jhq; T; ijU tþf jc ij d 0 < e: d0 Bernstein and Vazirani proved that there exists a normal form, i.e., for any final states, the quantum transition channel changes them into an initial state, UQTM MBV simulating any UQTM M with any accuracy e for t steps with slowdown f t; 1e which can be computed in polynomial steps of t and e. This UQTM MBV is known as one of the models of universal QTM. The input data of MBV is a quadruplet (x, e, t, c(M)) where x is an input of M, e is accuracy of the simulation, t is a simulation time and c(M) is a code of M. Note that it is necessary there to give a time t as an input of MBV. Now, we consider an another model of universal QTM whose input data is (x, e, c(M)), that is, we do not need a simulated time t as an input. It suggests that we do not need to know when the given UQTM halts. The following theorem is given in our paper [12]. Theorem 10. For any UQTM M, there exists a UQTM M which simulates each step of M for an input data (x, e, c(M)) where x is an input of M, e is accuracy of the simulation and c(M) is a code of M. the following inequality is obtained
2 2 tþF ðt;1e Þ 6e t jcM i < 2 2 ; hq; T; ijU d jcM i hq; T; ijU d0 p t where Ud and U d0 are the unitary operator corresponding to M and M, respectively. Suppose that M halts and gives an outcome with probability p; M gives the same outcome with probability p0 satisfying jp p0 j 6 e. According to a property of Riemann Zeta function, we have
jp0 pj 6
1 X 6e i¼1
p2 i2
6 e:
By inserting the process to calculate the proper accuracy for MBV , we can construct an universal QTM and avoid the halting problem. 4. OV SAT algorithm In this section, we explain the OV SAT algorithm solving SAT problem in polynomial time. It is discussed more precisely in the papers [1–3,9,11]. We do not use any Oracles for computation, so that the unitary operator of this algorithm also can be concretely constructed in polynomial time. Let X fx1 ; . . . ; xn g; n 2 N be a set. xk and its negation xk ðk ¼ 1; . . . ; nÞ are called literals. Let X f x1 ; . . . ; xn g be a set, then the set of all literals is denoted by X 0 X [ X ¼ fx1 ; . . . ; xn ; x1 ; . . . ; xn g. The set of all subsets of X0 is denoted by F ðX 0 Þ and an element C 2 F ðX 0 Þ is called a clause. We take a truth assignment t to all variables xk. If we can assign the truth value to at least one element of C, then C is called satisfiable. Let L = {0, 1} be a Boolean lattice with usual join _ and meet ^, and t(x) be the truth value of a literal x in X. Then the truth value of a clause C is written as t(C) _x2Ct(x). Moreover the set C of all clauses Cj (j = 1, 2, . . . , m) is called satisfiable iff ðCÞ ^m j¼1 tðC j Þ ¼ 1. Thus the SAT problem is written as follows: Definition 11 (SAT Problem). Given a Boolean set X fx1 ; . . . ; xn g and a set C ¼ fC 1 ; . . . ; C m g of clauses, determine whether C is satisfiable or not.
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That is, this problem is to ask whether there exists a truth assignment to make C satisfiable. It is known in usual algorithm that it is polynomial time to check the satisfiability only when a specific truth assignment is given, but we can not determine the satisfiability in polynomial time when an assignment is not specified. In [3] the quantum algorithm of the SAT problem was first studied, which was rewritten in [9] by showing that the Ohya– Masuda SAT algorithm is combinatorial. In [1,2] it is shown that the chaotic quantum algorithm can solve the SAT problem in polynomial time. 4.1. Unitary gates for computation Ohya and Masuda pointed out [3] that the SAT problem, hence all other NP problems, can be solved in polynomial time by quantum computer if the superposition of two orthogonal vectors j0i and j1i is physically detected. However this detection is considered not to be possible in the present technology. The problem to be overcome is how to distinguish the pure vector j0i from the superposed one ajui j0i + bju0 i j1i, obtained by the OM algorithm, if b is not zero but very small. If such a distinction is possible, then we can solve the NPC problem in the polynomial time. In [1,2] it is shown that it can be possible by combining nonlinear chaos amplifier with the quantum algorithm, which implies the existence of a mathematical algorithm solving P = NP. In this section, we explain the quantum algorithm of SAT problem based on [9,3]. Here, we estimate the total number of qubits more rigorously and show how to construct the unitary gates more precisely for any case. We first calculate the total number of qubits and show that this number depends on the input data only and is calculated in polynomial time. Then we decide the Hilbert space for the quantum algorithm and the initial state vector on it. Let C ¼ fC 1 ; . . . ; C m g be a set of clauses on X 0 fx1 ; . . . ; xn ; x1 ; . . . ; xn g. The computational basis of this algorithm is the Hilbert space H ¼ ðC2 Þnþlþ1 where l is a number of dust qubits, it is shown that l is less than 2mn [11]. Let
jv in i j0n ; 0l ; 0i; be an initial state vector where j0n i ¼ j0; 0; . . . ; 0i ¼ j0i j0i. |fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl} For X = {x1, . . . , xn} and a truth assignment t, we put n
tðx1 Þ ¼ e1 ; tðx2 Þ ¼ e2 ; . . . ; tðxn Þ ¼ en ; where e1, e2, . . . , en 2 {0, 1}, and we write t as a sequence of binary symbols:
t ¼ e1 e2 en : A unitary operator U C : H ! H computes tðCÞ for all truth assignment as follows n 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 2X l U C jv in i ¼ U C j0n ; 0l ; 0i ¼ pffiffiffiffiffin jei ; d ; tðCÞi ¼ qjui j0i þ 1 q2 ju0 i j1i jv out i; 2 i¼0
where jdli is dust qubits denoted by l strings of binary symbols, and jeii = j e1, e2, . . . , eni is a binary representation of t. Theorem 12 [11]. For a set of clauses C ¼ fC 1 ; . . . ; C m g on X 0 fx1 ; . . . ; xn ; x1 ; . . . ; xn g, the number l of dust qubits for algorithm of SAT problem is
l ¼ sf 1 n 6 2nm: Theorem 13 ([3,9,11]). For a set of clauses C ¼ fC 1 ; . . . ; C m g, there exists the unitary operator U C to calculate the truth value of C as
UC
m1 Y
U AND ðiÞ
m Y
i¼1
U OR ðjÞHðnÞ;
j¼1
where H(k) is a unitary operator acting by the application of the Hadamard transformation to first k qubits, that is
HðkÞ ¼ Hk INk : The computational complexity of quantum computation depends on the number of unitary operator in the quantum circuit. Let U be the unitary operator, it is written as
U ¼ U n U n1 U 1 ; where Un, . . . , U1 are fundamental gates. The computational complexity T(U) is considered as n. The above unitary operators UAND and UOR can be written as a combination of fundamental gates NOT gate UNOT, controlled-NOT gate UCN and controlled-controlled-NOT gate UCCN. The computational complexity TðU C Þ of SAT algorithm is defined by the number of UNOT, UAND and UOR used.
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Theorem 14 [11]. For a set of clauses C ¼ fC 1 ; . . . ; C m g and literal X 0 ¼ fx1 ; . . . ; xn ; x1 ; . . . ; xn g; TðU C Þ is
TðU C Þ ¼ m 1 þ
m X 0 jC k j þ 2ik 1 6 4mn 1: k1
4.2. Chaos amplifier Here we will briefly review how chaos amplifier can play a constructive role in computation (see [1,2] for the details). Consider the so called logistic map which is given by the equation
xnþ1 ¼ axn ð1 xn Þ g a ðxÞ;
xn 2 ½0; 1 :
The properties of the map depend on the parameter a. If we take, for example, a = 3.71, then the Lyapunov exponent is positive, the trajectory is very sensitive to the initial value and one has the chaotic behavior [2,6]. It is important to notice that if the initial value x0 = 0, then xn = 0 for all n. The state jvouti of the previous subsection is transformed into the density matrix of the form
q ¼ q2 P1 þ ð1 q2 ÞP0 ; where P1 and P0 are projectors to the state vectors j1i and j0i. One has to notice that P1 and P0 generate an Abelian algebra which can be considered as a classical system. Let KC be a quantum channel on one qubit space such that
Þ ¼ KCA ðq
Þr3 Þ ðI þ g a ðq ; 2
k , we have where I is the identity matrix and r3 is the z-component of Pauli matrices. Applying KCA to q
k I þ g ka ðq2 Þr3 qk ¼ KCA ðqÞ ¼ : 2
To find a proper value k we finally measure the value of r3 in the state qk such that
k r3 ¼ g ka ðq2 Þ: Mk trq The following theorems is proven in [1,2,10]. Theorem 15. For the logistic map xn+1 = axn(1 xn) with a 2 [0, 4] and x0 2 [0, 1], let x0 be 21n and a set J be {0, 1, 2, . . . , n, . . . , 2n}. If a is 3.71, then there exists an integer k in J satisfying xk > 12. . Theorem 16. Let a and n be the same in above theorem. If there exists k in J such that xk > 12, then k > log2n1 3:711 Theorem 17. Let jtðCÞj be the cardinality of these assignments, if x0 2rn with r ¼ jtðCÞj and there exists k in J such that xk > 12, then there exists k satisfying the following inequality if C is SAT.
n 1 log2 r 5 6k6 ðn 1Þ : log2 3:71 1 4 From these theorems, for all k, it holds
Mk
¼ 0 iff C is not SAT; > 0 iff C is SAT:
The purpose of this amplification is to see the difference of the states
q ¼ P0 and
q ¼ q2 P1 þ ð1 q2 ÞP0 ; q2 > 0: k If q = P0, it stays q = P0 after applying KCA , and if q – P0, Mk becomes larger than 12 in polynomial of n. Ohya and Accardi found an alternative algorithm solving the SAT problem by means of the stochastic limit [8]. A physical realization of the chaos amplifier is discussed in [10,19]. A photosynthesis is considered as one of possible amplifiers, which uses the environment to amplify the signal from the antenna system.
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5. GQTM expression of OV SAT algorithm In this section, we construct the 3 multi-track GQTM M SAT ¼ ðQ ; R3 ; H; Kd Þ that achieves the OV algorithm. This GQTM does not belong to LQTM and UQTM because the chaos amplification process is described by a non-linear CP channel. The k through q jvfihvfj explained above. The computation from OV SAT algorithm runs from an initial state q0 jv0ihv0j to q k is due to a non-unitary chanq0 jv0ihv0j to q jvfihvfj is due to unitary channel KC UC UC, and that from q jvfihvfj to q nel KkCA KI , so that all computation can be done by KkCA KI KC , which is a completely positive deterministic map. Let us explain our computation by a multi-track GQTM. The first track stores the input data and literals. The second track is used for the computation of f(Ci), (i = 1, . . . , m), and the third track is used for the computation of the result f ðCÞ. This algorithm is represented by the following 5 steps [13]:
Step Step Step Step Step
1: 2: 3: 4: 5:
Apply the Hadamard transform to Track 1. Calculate f(C1), . . .f(Cm) and store them in Track 2. Calculate f ðCÞ, and store it in Track 3. Empty the working space. Apply the chaos amplifier to the result state and repeat this step.
5.1. Computational complexity of the SAT algorithm Let T(K) be a computational complexity of K defined as the number of steps in K. Theorem 18. For a set of clauses C with n Boolean variables, the computational complexity T(KSAT) of the OV SAT algorithm including the chaos amplifier, is
TðKSAT Þ ¼ ðTðKC Þ þ TðKT ÞÞT KkCA ¼ OðpolyðnÞÞ; where k is a positive number smaller than
5 4
ðn 1Þ and poly(n) denotes a polynomial of n.
Proof. The computational complexity of KSAT depends on the computational complexity of unitary calculation and chaos amplifier. The unitary calculation part is written as KC, and its computational complexity is essentially discussed in [3,11]. For X 0 ¼ fx1 ; . . . xn ; x1 ; . . . ; xn g and C ¼ fC 1 ; . . . ; C m g the set of literals, the computational complexity T(KC) is obtained as
TðKC Þ ¼ 3
m m X X ðcardðC k Þ 1Þ þ 2cardðC k \ fx1 ; . . . ; xn gÞ þ m 1 þ n 6 ð8mn 2m 1Þ ¼ OðpolyðnÞÞ: k¼1
k1
After the unitary calculation, we translate it to the chaos amplifier. MSAT empties the work space for calculation in this process, and it is proved in [11] that the size of workspace is exactly in polynomial order of input size. Therefore, the computational complexity of KT is TðKT Þ ¼ OðpolyðnÞÞ. We have to repeat these calculation because we do not know exact time of chaos amplification process when the halting probability is greater than 1/2. Then, the total computational complexity becomes
TðKSAT Þ ¼ ðTðKC Þ þ TðKT ÞÞT KkCA ¼ OðpolyðnÞÞ:
6. Results and topics on quantum algorithms Here we mention our further recent studies. We consider the applications of quantum algorithms not only to mathematical problems, but also problems of Bio-informatics; (1) sequence alignment of amino acid sequence [15] and (2) protein folding problem [18]. There exists a harder language class than the class NP, for example EXPTIME. We consider a quantum algorithm of the Pebble game problem which belongs to EXPTIME. Players move one stone at a time along a given rule alternatively. A player wins when he moves a stone to the winning position on the board given by a certain rule, or he loses when he cannot move any stone. We want to know whether there exist strategies such that a first mover can win every time. The computational complexity of this problem is obviously very large. In some cases depending on a size of board and the rule, it takes exponential time to solve this problem. We proposed a quantum algorithm for Pebble Game with an Oracle, and the computational complexity is polynomial of the size of game board [17]. Note that even if we assume such an Oracle, the computational complexity of classical algorithm is still exponential. References [1] M. Ohya, I.V. Volovich, Quantum computing and chaotic amplification, J. opt. B 5 (6) (2003) 639–642. [2] M. Ohya, I.V. Volovich, New quantum algorithm for studying NP-complete problems, Rep. Math. Phys. 52 (1) (2003) 25–33.
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