The Deutsch-Jozsa Problem: De-quantization and entanglement

Transcription

1 The Deutsch-Jozsa Problem: De-quantization and entanglement Alastair A. Abbott Department o Computer Science University o Auckland, New Zealand May 31, 009 Abstract The Deustch-Jozsa problem is one o the most basic ways to demonstrate the power o quantum computation. Consider a boolean unction : {0,1} n {0,1} and suppose we have a black box to compute. The Deutsch-Jozsa problem is to determine i is constant [i.e. (x = const, x {0,1} n ] or i is balanced [i.e. (x = 0 or exactly hal the possible input strings x {0,1} n ] using as ew calls to the black box computing as is possible, assuming is guaranteed to be constant or balanced. Classically it appears that this requires at least n 1 +1 black box calls in the worst case, but the well known quantum solution solves the problem with probability 1 in exactly one black box call. However, it has been ound that in some cases equivalent classical, deterministic solutions also exist. We attempt to clariy the power o the quantum solution in order to gain better insight into where the power o quantum computation comes rom, as well as the ability to determine when classical, deterministic counterparts can be ound. 1 Introduction Deutsch s problem and the more general Deutsch-Jozsa problem were some o the irst problems tackled in the ield o quantum computing. They are simple, but are suiciently non-trivial to be o interest. The generally accepted quantum solutions contain aspects o quantum parallelism, intererence and entanglement, which are commonly cited as the main tools which give quantum computing its power. In this paper, we build on a previous paper [4] that shows Deutsch s original problem is solvable classically. We extend this note to the general Deutsch-Jozsa problem in an attempt to explore what is undamentally important to quantum computation.

2 1.1 Deinitions In order to be able to talk about the dierences between classical and quantum algorithms, we need deine them in a way which captures their dierences in a constructive manner. We will call an algorithm a classical algorithm i it can be computed on a Turing machine and a quantum algorithm i it can be computed by a sequence o unitary operators, U = U L U L 1...U 1. Many quantum algorithms have a trivial classical counterpart: all the operations in the matrix mechanics ormulation o quantum mechanics can be easily computed by classical means. As long as one is careul (i.e. quantum eatures such as randomness are not undamentally important to the algorithm equivalent classical algorithms can be obtained by these means [9]. However, the dimension o Hilbert space grows exponentially with the number o qubits used in a quantum algorithm, so a classical counterpart obtained by the trivial means takes space and time that is exponentially larger than the quantum algorithm does, assuming we measure space in undamental basis elements (i.e. 0,1, 0, 1 take equal space to within a constant actor. In this paper we examine the existence o classical counterparts to quantum algorithms that are not exponential in time or space compared to the quantum algorithm. Throughout this paper we will use the standard orthogonal basis kets 0 and 1. We will use the shorthand notion + and to represent the symmetric and antisymmetric equal superpositions o the basis states, so that the Hadamard gate H has the ollowing eect [3]: H 0 = 1 ( = +, and H 1 = 1 ( 0 1 =. 1. Oracle Quantum Computations and Embeddings The main subject o this paper, the Deutsch-Jozsa problem (and also Deutsch s problem is a orm o an oracle computational problem [, 1]. This means that the input is given to us as a black box and the goal is to determine something about this black box. It is important that this inormation can only be obtained by asking the oracle allowable questions. It must not be the case that examination o the structure alone o the black box allows insight into the nature o the black box [8, p. 554]. In the current literature it is usually implicitly assumed that solving a problem with a classical black box is o the same diiculty as solving it with a quantum, or alternative kind o black box. In general, a standard classical black box can operate on classical bits only (0 or 1, while a quantum black box can operate on any state in dimensional Hilbert space (H. This dierence in some sense appears to add extra complexity to a quantum black box. It is not clear that solving a problem given a quantum black box is the same as solving a problem given a classical black box. However, this is an issue to be explored in the uture. In this paper we will assume the

3 normal situation that we are given a certain type o black box and asked to ind inormation about what the black box represents. Deutsch s Problem The original problem proposed by Deutsch [7] is4 ormulated as ollows. Consider a boolean unction : {0,1} {0,1}, and suppose we are given a black box (oracle to solve. Deutsch s problem is to determine i is constant [i.e. (0 = (1] or balanced [i.e. (0 (1] in as ew as possible calls to the black box computing..1 Quantum Solution A standard quantum solution or Deutsch s problem is briely presented, as all urther analysis will stem rom this. This is based on the ormulation given in [5] which solves Deutsch s problem with probability 1 using only one call to the quantum black box computing. A traditional classical algorithm would require two calls to a classical black box in order to determine i is constant or balanced. The quantum black box extends the classical black box to operate on superpositions o basis states. The quantum black box can be described by the unitary operator U representing an -controlled-not ( -cnot gate such that Noting that U x y = x y (x. U (U x y = U x y (x = x y (x (x = x y, we see that U is its own inverse and U U = 1. Hence U is unitary and our quantum black box is valid. In order to see how the quantum solution works, it is beneicial to observe the ollowing: U x 1 ( 0 1 = x 1 ( 0 (x 1 (x = ( 1 (x x 1 ( 0 1. From this observation we can ormulate the quantum solution. Taking the initial state 0 1 and operating on it with a -qubit Hadamard gate H : H 0 1 = H 0 H 1 = 1 ( Next, operating on the state with U : 1 U ( = 1 (( 1 (0 0 + ( 1 (1 1 = ( 1 (0 ( 0 + ( 1 (0 (1 1 3

4 and applying H one more time we get ( 1 (0 H ( 0 + ( 1 (0 (1 1 = ( 1 (0 (0 (1 1. Measuring the irst qubit we obtain 0 with probability 1 i is constant and 1 with probability 1 i is balanced. This quantum solution is correct with probability 1 using only one call to the quantum black box represented by U. An important note is that this computation involves no entanglement. This can be seen by noting that the second qubit acts as an auxiliary bit and remains unchanged by U, and as a result the two qubits remain separable throughout the algorithm. This is evident in the presentation o the algorithm and as a result this quantum computation gains its power only rom quantum parallelism and intererence.. Classical Solutions As noted, the quantum solution contains no entanglement. It uses the quantum properties o superposition and intererence in order to achieve the desired result. However, these qualities (unlike entanglement are not inherently quantum mechanical, but are rather due to the two dimensionality o qubits compared to the one dimensionality o classical bits. Hence, a classical two-dimensional system possesses these qualities, and should thus be able to achieve the same result as the quantum algorithm. The irst method, presented in [4], embeds classical bits in complex numbers. The set {1, i} with i = 1 acts as a computational basis in the same way that { 0, 1 } does or quantum calculations. It is worth noting that while we are not labeling the basis bits 0 and 1, they represent the classical bits 0 and 1 in the same way that 0 and 1 do. A general complex number may be written as z = a + bi, so a general complex number z is a natural superposition o the basis in the same way that a general qubit is. We are now given a classical black box that computes our unction. This black box can be represented by a unction C, a direct analogue o U (although the requirement o unitarity is no longer necessary. The eect o C (by direct correspondence with U, although the normalization actors and auxiliary bit are no longer necessary is ( C (a + bi = ( 1 (0 a + ( 1 (0 (1 bi. I is constant, C is the identity operation to within a actor o 1 (C (x = ±x. I is balanced, C is the conjugation operation (C (x = ±x. The black box represented by C rotates the input in the complex plane depending on the nature o, which is in direct analogy o the quantum black box U operating on input x in H. In order to measure the output, we need a way to project our complex numbers back on to the computational basis. This is easily done 4

5 by multiplying by the input so the output is either purely imaginary or purely real (i.e. on the computational basis. I z = 1 + i (an equal superposition o basis states, 1 zc (z = ±1 z = ±i i is constant or 1 zc (z = ±1 zz = ±1 i is balanced. In this manner, i the output is imaginary then is constant, i it is real then is balanced. Importantly, this is a deterministic result. This is only one method o solving the problem as well as the quantum algorithm does through classical, deterministic methods. The above method places emphasis on mathematical correspondence with the quantum solution. A dierent solution is presented by Arvind in [1] which draws physical similarities which are more visible than in the above solution. Such a method o taking a quantum algorithm and inding a classic counterpart which works just as eiciently is called a de-quantization. This term was irst used by Calude [4], but the methods used by [1, 10] or example would also classiy as de-quantizations. Arvind uses the (classical polarization o a photon as the computational basis (x-pol, y-pol, and any polarization in the x-y plane is physically valid. It is noted that all transormations in the group SU( can be realized by two quarter-wave plates and a single hal-wave plates orientated suitably. Clearly the transormations required to solve Deutsch s problem are included in this, so the problem can be solved classically with one photon using wave plates. Written in matrix orm the solution is mathematically identical to the quantum one. This corresponds to the ollowing physical process: Preparing a photon in the y-polarization, rotating anti-clockwise in the x-y plane by 45, applying the black box, and applying the anti-clockwise rotation once more beore measuring the y-polarization o the photon. The correspondence here relies not on embedding classical bits in a dierent, classical - dimensional basis but on directly implementing the transormations used in the quantum solution through classical means. In other words, the quantum algorithm does not take advantage o non-classical eects, so the same result can be obtained through purely classical optics. 3 The Deutsch-Jozsa Problem This problem was extended by Deutsch and Jozsa [8] to unctions on n-bit strings. In this case, the problem is ormulated as ollows. Let : {0,1} n {0,1}, and suppose we are given a black box computing with the guarantee that is either constant [i.e. x {0,1} n : (x = a,a {0,1}] or balanced [i.e. (x = 0 or exactly hal o the possible inputs x {0,1} n ]. The Deutsch-Jozsa problem is to determine i is constant or is balanced. An important note is that unlike in Deutsch s problem, where there are exactly two balanced and two constant unctions, the distribution o constant and balanced unctions is asymmetrical in the Deutsch- Jozsa problem. In general, there are N = n possible input strings, each with two possible outputs (0 or 1. Hence, or any given n there are N possible unctions. O these, exactly two are constant and ( N N/ are balanced. Evidently the probability that our (promised to be balanced 5

6 or constant is constant tends towards zero very quickly. Furthermore, the probability that any randomly chosen unction o the N possible unctions is either balanced or constant (i.e. is valid is given by ( N N/ + N which also tends to zero as n increases. This is evidently not an ideal problem to work with, however this does not mean that we cannot gain useul inormation rom studying it. 4 Adaptation or n = In this section we will provide a ormulation o the solution or the Deutsch-Jozsa problem with n = which makes the separability o the states clearly evident and draws obvious parallelism with the n = 1 solution presented in the previous section. 4.1 Quantum Solution For n = the quantum black box we are given takes as input three qubits and is represented by the ollowing unitary operator U, just as it was or n = 1: U x y = x y (x, where x {0,1}. For n = there are 16 possible boolean unctions. Two o these are constant and another six are balanced. All these possible unctions are listed in Table 1. (x Constant Balanced Other (00 = (01 = (10 = (11 = Table 1: All possible Boolean unctions : {0,1} {0,1} Evidently, hal o these unctions are simply the negation o another. I we let (x = (x 1, we have: U x = ( 1 (x x ( = ( 1 (x x = U x. 6

7 In this case the result obtains a global phase actor o 1. Since global phase actors have no physical signiicance to measurement (a result is obtained with probability proportional to the amplitude squared, the outputs o U and U are indistinguishable. I we initially prepare our system in the state 00 1, operating on this state with H 3 gives H = 1 x = ++. (1 x {0,1} In the general case, ater applying the -cnot gate U we have U c x x = ( 1 (x c x x x {0,1} x {0,1} [ = ( 1 (00 c ( 1 (01 c ( 1 (10 c ] +( 1 (11 c ( From the well known rule (see [11] about -qubit separable states, we know that this state is separable i and only i ( 1 (00 ( 1 (11 c 00 c 11 = ( 1 (01 ( 1 (10 c 01 c 10. While there are various initial superpositions o -qubit states which satisy this condition, we only need to consider the equal superposition (shown in Equation (1 that is used in this algorithm. In this case, the separability condition is (00 (11 = (01 (10. By looking back at Table 1 it is obvious this condition must hold or all balanced or constant unctions or n =. We can now rewrite Equation ( as ollows: U ++ = ±1 ( ( 0 + ( 1 (00 ( ( 1 (10 (11 1. (3 Indeed, ( 1 ( ( 1 ( ( 1 ( ( 1 (11 11 = ( 1 ( ( 1 (00 (10 ( ( 1 ( ( 1 (11 ( 11 = ( 1 ( ( 1 (10 ( ( 1 (00 ( ( 1 (00 (11 11 ( ( = ± 0 + ( 1 (00 ( ( 1 (10 (11 1, as desired. By applying a inal 3-qubit Hadamard gate to project this state onto the computational basis we obtain ±1 H 3 ( 0 + ( 1 (00 (10 1 ( 0 + ( 1 (10 (11 1 = ± (00 (10 (10 (11 1. By measuring both the irst and second qubits we can determine the nature o with probability 1. I both qubits are measured as 0, then is constant, otherwise is balanced. 7

8 4. Classical Solutions Because the quantum solution contains no entanglement, the problem can be de-quantized in a similar way to the n = 1 case, but this time using two complex numbers as input. However, in order to retain enough inormation to solve the problem, the de-quantized black box must also output complex numbers (C is a transormation rather than a unction. We will deine C by analogy to U just as we did or the n = 1 case. Let z 1, z be complex numbers, C ( z1 z = C ( a1 + ib 1 a + ib = ( 1 (00 ( a1 + ( 1 (00 (10 ib 1 a + ( 1 (10 (11 ib. (4 It is important to note that, just as in the quantum case where the output o the black box was two qubits that can be independently measured, the output o C is two complex numbers that can be independently manipulated, rather than the complex number resulting rom their product. This is airly intuitive because the ability to measure speciic bits (o any kind is undamental to computation. Note however, that in a quantum system it is impossible to measure entangled qubits independently o each other. The analogue to applying a Hadamard gate to each qubit in order to project it onto the computational basis is to multiply each o the complex numbers that the black box outputs by their respective inputs (in similar ashion to that in the n = 1 case. I we let z 1 = z = 1 + i, we get the ollowing: ( (1 + i(1 + i (1 + i(1 + i = ( i, i is constant, i ( ( ( (1 + i z1 ( 1 (00 (1 + i(1 i 1 C = =, z ((1 + i(1 + i ( i (1 + i(1 + i i =, i is balanced. ((1 + i(1 i ( 1 (1 + i(1 i 1 =, (1 + i(1 i 1 By measuring both o the resulting complex numbers, we can determine whether is balanced or constant with certainty. I both complex numbers are imaginary then is constant, otherwise it is balanced. Because the quantum solution is separable, it is possible to write the output o the black box as a list o two complex numbers, and hence we can ind a solution equivalent to the one obtained via a quantum computation. Writing the output in this orm would not have been possible i the state was not separable, and inding a classical solution in this ashion would have required a list o complex numbers exponential in the number o input qubits. As with the n = 1 case, an alternative classical approach can be presented using two photons. I a transormation on two qubits can be written as a transormation on each bit independently 8

9 (e.g. H H then the transormation is trivially implemented classically. It only remains to the -bit transormation U can be implemented classically on -photons. Equation (3 shows that the quantum black box or n =, U can be written as a product o two 1-bit gates 1 : U (1 + = 1 ( 0 + ( 1 (00 (10 1, U ( + = 1 ( 0 + ( 1 (11 (10 1. Each o these are valid unitary operators, and the transormation describing the black box may be written U = U (1 U (. This means that the operation o can be computed by applying a 1-bit operation (implemented as wave-plates to each photon independently, and thus a classical solution or n = is easily ound in this ashion. The photons need not interact with each other at any point during the algorithm, not even inside the black box implementation. This classical, optical method is equivalent to both the quantum solution and the previously described classical solution. The dierence is in how it is represented, bringing emphasis on the act that or n = the quantum solution does not take advantage o uniquely quantum behaviour and is thus classical in nature. Further, it shows that the solution can be obtained without any interaction or sharing o inormation between qubits. 5 Work on n 3 We have already seen that or n = 1 and, the Deutsch-Jozsa problem can be de-quantized to give a classical algorithm that is just as eicient as the quantum algorithm. To examine what happens or larger n is a little more diicult. For n 3, it is no longer the case that the output o U is separable or all balanced or constant. This can be easily seen by taking into account the results o [11]. We will briely state their result and show how it is applicable. In a system o n qubits, the general system state can be written as ψ n = N 1 i=0 c i i, where N = n and c i is the amplitude o the ith possible state o the system. Pair product invariance is deined as: ψ n is pair product invariant k [1,n], i [0,K 1] : c i c K i 1 is constant where K = k. Pair product invariance can also be reormulated recursively as ollows. Let P n be the set o all tuples (c i c K i 1,α k such that c i c K i 1 must equal a non-zero constant α k or ψ n to be pair 1 So ar we have been considering the case where U operates on n input qubits and one auxiliary qubit,. It has been shown (see [6] that the auxiliary qubit is not necessary i we restrict ourselves to the subspace spanned by. However, we have presented the algorithm with the auxiliary qubit present because it is more intuitive to think o the input-dependent phase actor being an eigenvalue o the auxiliary qubit which is kicked back. The de-quantized solutions however bear more resemblance to this reduced version o U operating only on n qubits. 9

10 product invariant, i.e. P n = {(c i c K i 1,α k, k [1,n], i [0,K 1]}. Recursively, this can be written by breaking up the iteration over all k [1,n]. We ind that P n = P n 1 {(c i c n i 1,α n, i [0, n 1 1]}, with the base case P = {(c 00 c 11,α,(c 01 c 10,α }. The main theorem o relevance to us (Theorem 1 in [11] is: Provided all c i 0, a state ψ n is ully separable i and only i it is pair product invariant. In order to determine i our state ψ n is separable we only need to check to see i all elements o P n evaluate to the corresponding α k. Note that any constant unction (or all n is pair product invariant as all c i are equal and the conditions are trivially satisied. The output o the quantum black box or constant can be separated as 1 n/u x {0,1} n x = 1 n/ x {0,1} n ( 1 (x x = ±1 n/ x {0,1} n x = ± + n. For n = 3 we are able to ind balanced unctions that are not pair product invariant and thus entangled. I we choose such that ( (000, (001, (010, (011, (100, (101, (110, (111 = (0,0,0,1,1,1,1,0, this is obviously balanced. For this choice o, (000 (011 (001 (010. Since the amplitudes o a state x are on a 1-1 mapping with (x, this shows that c 000 c 011 c 001 c 010 and hence the output o U in the standard quantum algorithm is not pair product invariant and is thus entangled. The recursive deinition o pair product invariance allows us to determine exactly how many separable states exist or any given n. For a unction n+1 acting on n+1 bits to be separable, we know that all pairs in P n+1 evaluate to some constant. Since P n P n+1, all pairs in P n must also evaluate to a constant. But the set o n bit unctions n or which each element in P n is constant is exactly the set o separable unctions n. Hence we see that all separable unctions n+1 satisy n+1 (0 x n = n (x n, x n {0,1} n, where 0 x n denotes the concatenation o x n onto 0. This determines the action o n+1 on hal o the possible input strings, and we must determine the possible actions on the other hal. Because our c i = ( 1 (i, the requirement or all pairs in P n+1 to be equal means the parity o all pairs must be the same, i.e. c i c K i 1 = c j c K j 1 (i (K i 1 = ( j (K j 1. 10

11 The action o n+1 on the remaining N = n inputs is determined the pair product invariance condition. This requires that c N 1 c N = c N c N+1 = = c 0 c N 1. Since c N 1 is already determined, c N can take on two possible values. However, once it takes on this value, all c i or i > N are uniquely determined. I we let a n be the number o separable unctions n, then n+1 can have a n possible conigurations or acting o n+1 (0 x n, and or each o these conigurations, there are two conigurations or the remaining N input strings. Hence, a n+1 = a n. This is a simple linear recursion with the known initial condition a 1 = 4. This gives us an explicit result or the number o boolean unctions n such that U n x is separable: a n = n+1. We see that the number o separable states increases exponentially with the number o qubits being used. We also know the number o possible unctions n which are either balanced or constant is ( ( N n b n = + = N/ n +. 1 The raction o possible boolean unctions which can be separated is (( a n = n+1 n / +. b n n 1 This tends towards zero extremely quickly even or small n. The result o this observation is that even i we are promised is balanced or constant, we can no longer be sure the output o the black box is separable, and or n 3 the probability that it is separable tends to zero very quickly. This means that the method o de-quantization used or n = 1, will not scale directly to higher n and in general yields very little inormation about a unction. Intuitively it would look like separability is a requirement or de-quantization to be possible. Looking at this rom the view o computation with classical photons, there is no physical, classical equivalent o entangled photons as this is a purely quantum-mechanical eect. However, in general it is very hard to show that no de-quantization exists or a quantum algorithm which does not introduce exponential increase in space or time. In most cases, as earlier mentioned, a trivial method o de-quantization is possible, but to show no better dequantization exists is very hard. 6 General de-quantization While de-quantization appears to be very hard in the Deutsch-Jozsa problem or n 3, one can investigate the ability to de-quantize algorithms not involving entanglement. So ar dequantization in this kind o ashion has only been explored or Deutsch s problem. The main 11

12 task in trying to de-quantize a quantum algorithm is to de-quantize the black box. Using the methods previously looked at in this paper, this step would initially require showing that both the input and output o the black box is separable. We will assume both the input and output o the black box U is separable, and look at what can be said about the ability to de-quantize the algorithm. The simplest case is that U itsel can be expressed in terms o n 1-qubit gates. I it is possible to do this we can write U = U (1 U ( U (n where U (i acts on the ith qubit only. This is easy to compute classically using the classical photon method used earlier. Since each U (i SU(, they can be implemented using wave-plates. Further, the photons do not interact and are independent o each other throughout the computation. Hence, we can prepare the n photons as required, operate on each one with the appropriate wave-plate arrangements beore measuring the polarization o each photon. Because U can be separated in this way, the de-quantized algorithm clearly implements the same transormation and hence yields the same result as the quantum algorithm, but with a classical deterministic result. The next case to consider is the case that U cannot be written in terms o 1-qubit gates, but still does not entangle the input. We know that any unitary gate can be separated into a sequence o 1-qubit gates and cnot gates (see [1], and since we have already shown all 1-qubit gates are de-quantizable we need only examine thoroughly the cnot gate. We note that any de-quantization o a cnot gate is more complicated than that o any 1-qubit gate. The cnot gate clearly involves some kind o interaction between the two input qubits. The classical photon de-quantization would appear to all apart here, because it relied on the photons being treated independently. However, this does not mean that de-quantization is not possible. In general, when an n-qubit gate is decomposed into 1-qubit gates and cnot gates, we cannot assume the input o the cnot gates is not entangled, even i the input and output o U as a whole is separable. This is because the intermediate steps in the circuit could entangle and then unentangle. For de-quantization to be easy, we want the qubits to remain unentangled throughout the whole computation. The cnot gate is deined as C x y = x y x. We can derive an expression or the requirement that the output o the cnot gate is separable. C(a 0 + b 1 (c 0 + d 1 = C(ac 00 + ad 01 + bc 10 + bd 11 For this to be separable, we require that = ac 00 + ad 01 + bd 10 + bc 11. abc = abd c = ±d, 1

13 subject to the normalization constraint c + d = 1, which tells us that y = ± i c = 1, or y = ± i c = 1. I this condition holds true at every cnot gate in the decomposition o U onto the basis o unitary gates, then de-quantization is possible. More precisely, i we have c = ±d, C x y = C(a 0 + b 1 (c 0 + d 1 = ±(a 0 + b 1 (d 0 + c 1 = ± x (d 0 + c 1. Hence we need only consider the possible cases or y : { C x ± = ± x ±, i c = 1, C x y C x ± = x ±, i c = 1. Importantly, we see that i the conditions are met or the cnot gate to not entangle the input, it can be treated as a 1-qubit gate which introduces a actor o ±1 depending on the makeup o the second input qubit. In this situation, the gate can be treated as a 1-qubit gate that we know can be implemented classically, and de-quantization is possible. I the condition that c = ±d holds at every cnot gate in the decomposition o U, then separability is maintained throughout the computation, and every gate, including the cnot gates, can be represented as 1-qubit gates, thus de-quantization o the complete black box is possible. However, the ability to determine i this condition holds could be relatively hard in general, and this is an issue that remains to be explored urther. I this condition does not hold, it does not necessarily tell us that de-quantization is not possible. There may be an alternative decomposition o U into a dierent set o basis gates or which separability is maintained. For de-quantization to be successul, we must only ind one such suitable decomposition. Exploring this in order to get stronger conditions or the ability to de-quantize an algorithm is a subject o uture research. 7 Conclusion We have examined the ability to de-quantize the Deutsch-Jozsa problem or various values o n in order to gain a better understanding o quantum algorithms and the ability or them to give exponential improvements over classical algorithms. We have extended the method o dequantization presented in [4] to the n = case, and by showing separability o the quantum algorithm or n = have obtained a similar de-quantized solution. We have shown that or n > there exist many balanced boolean unctions or which the output o U is entangled. The raction o balanced unctions which are separable has been shown to approach zero very rapidly. This tells us that i we were to pick a random boolean unction which is constant or balanced, the probability o being able to learn inormation about the nature o the unction through classical means in one black box call tends to zero. The systematic method o tackling quantum algorithms and searching or classical counterparts 13

14 makes it easier to see where quantum algorithms get their power rom. Trying to understand this is an extremely important step in the process o trying to devise new quantum algorithms. In order to make good quantum algorithms with ease, we need to have a much better understanding than we currently do about where their power comes rom, and how to use this eectively. In our investigation we obtained some conditions which, i satisied, indicate de-quantization is possible. These kind o conditions, i explored urther are a step towards better measures o the useulness o a quantum algorithm. An area that still needs to be looked into much urther is that o the black box complexity. We have assumed that we are given a black box, as is usual in oracle computations. However, comparing algorithms which use dierent oracles should take into account the dierence in complexity in these oracles. This has been largely ignored to date, but needs to be given systematic treatment i we really wish to compare quantum oracle algorithms with classical ones. Reerences [1] Arvind. Quantum entanglement and quantum computational algorithms. Pramana - Journal o Physics, 56( & 3: , Jan 001. [] A. Berthiaume and G. Brassard. Oracle quantum computing. Journal o Modern Optics, 41(1:51 535, Dec [3] E. Biham, G. Brassard, D. Kenigsberg, and T. Mor. Quantum computing without entanglement. Theoretical Computer Science, 30:15 33, 004. [4] C. S. Calude. De-quantizing the solution o Deutsch s problem. International Journal o Quantum Inormation, 5(3: , Jun 007. [5] R. Cleve, A. Ekert, C. Macchiavello, and M. Mosca. Quantum algorithms revisited. Proceedings o the Royal Society o London A, 1998(454: , Jan [6] D. Collins, K. W. Kim, and W. C. Holton. Deutsch-Jozsa algorithm as a test o quantum computation. Physical Review A, 58(3:R1633 R1636, Sep [7] D. Deutsch. Quantum theory, the church-turing principle and the universal quantum computer. Proceedings o the Royal Society o London A, 400:97 117, Jan [8] D. Deutsch and R. Jozsa. Rapid solution o problems by quantum computation. Proceedings o the Royal Society o London A, 439: , Jan 199. [9] A. Ekert and R. Jozsa. Quantum algorithms: Entanglement enhanced inormation processing. Philosophical Transactions o the Royal Society A, 356(1743: , [10] M. S. Hannachi, F. Dong, Y. Hatakeyama, and K. Hirota. On the use o uzzy logic or inherently parallel computations. In 3rd International Symposium on Computational Intelligence and Intelligent Inormatics,

15 [11] P. Jorrand and M. Mhalla. Separability o pure n-qubit states: two characterizations. International Journal o Foundations o Computer Science, 14(5: , 003. [1] A. Y. Kitaev, A. H. Shen, and M. N. Vyalyi. Classical and Quantum Computation. American Mathematical Society,