Exceptional quantum walk search on the cycle

Abstract

Quantum walks are standard tools for searching graphs for marked vertices, and they often yield quadratic speedups over a classical random walk’s hitting time. In some exceptional cases, however, the system only evolves by sign flips, staying in a uniform probability distribution for all time. We prove that the one-dimensional periodic lattice or cycle with any arrangement of marked vertices is such an exceptional configuration. Using this discovery, we construct a search problem where the quantum walk’s random sampling yields an arbitrary speedup in query complexity over the classical random walk’s hitting time. In this context, however, the mixing time to prepare the initial uniform state is a more suitable comparison than the hitting time, and then, the speedup is roughly quadratic.

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Notes

  1. 1.

    \(f(x) = {\Theta }(g(x))\) means f is asymptotically bounded both above and below by g, so there exist positive constants \(c_1, c_2\), and \(x'\) such that \(c_1 g(x) \le f(x) \le c_2 g(x)\) for all \(x \ge x'\). It is related to the more familiar big-O notation, which is only an upper bound.

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Acknowledgements

T.W. thanks the quantum computing group at the University of Texas at Austin for useful discussions. T.W. was supported by the U.S. Department of Defense Vannevar Bush Faculty Fellowship of Scott Aaronson. R.S. was supported by the RAQUEL (Grant Agreement No. 323970) project and the ERC Advanced Grant MQC.

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Correspondence to Thomas G. Wong.

Appendix: Hitting time on the cycle

Appendix: Hitting time on the cycle

In this appendix, we derive the hitting time HT of a classical random walk on the cycle of L vertices with a single marked vertex and show that it is \(\mathrm{HT} = (L^2 - 1)/6\).

Say the random walker is currently at distance i from the marked vertex. Then, let \(h_i\) be the expected number of steps needed to move one step closer to the marked vertex from this position. Trivially, \(h_0 = 0\), so let us find an expression for \(h_i\) with \(i > 0\), depending on if L is even or odd.

If L is even, then i is at most L / 2, and

$$\begin{aligned} h_{L/2} = 1 \end{aligned}$$

since the walker can only move closer to the marked vertex. We also have in general for \(i < L/2\) that

$$\begin{aligned} h_i = \frac{1}{2} (1) + \frac{1}{2} \left( 1 + h_{i+1} + h_i \right) , \end{aligned}$$

where first term comes from immediately moving closer to the marked vertex, and the second term comes from moving a step away first. Solving this for \(h_i\),

$$\begin{aligned} h_i = 2 + h_{i+1}. \end{aligned}$$

By recursion,

$$\begin{aligned} h_i = 2 \left( \frac{L}{2} - i \right) + h_{L/2} = L - 2i + 1. \end{aligned}$$

If L is odd, then i is at most \((L-1)/2\), and there are two vertices at this distance from the marked vertex. Then,

$$\begin{aligned} h_{(L-1)/2} = \frac{1}{2} (1) + \frac{1}{2} \left( 1 + h_{(L-1)/2} \right) . \end{aligned}$$

The first term comes from immediately moving closer to the marked vertex, and the second term comes from moving to the other vertex that is a distance of \((L-1)/2\) away from the marked vertex. Solving for \(h_{(L-1)/2}\),

$$\begin{aligned} h_{(L-1)/2} = 2. \end{aligned}$$

Another way to arrive at this result is noting that

$$\begin{aligned} h_{(L-1)/2} = \frac{1}{2} (1) + \frac{1}{4} (2) + \frac{1}{8} (3) + \frac{1}{16} (4) + \cdots = \sum _{i=1}^{\infty } \frac{i}{2^i} = 2, \end{aligned}$$

where the infinite series was shown to equal 2 by Oresme in the mid-fourteenth century [27]. When \(i < (L-1)/2\), we again have that \(h_i = \frac{1}{2} (1) + \frac{1}{2} \left( 1 + h_{i+1} + h_i \right) \), and applying this recursively yields

$$\begin{aligned} h_i = 2 \left( \frac{L-1}{2} - i \right) + h_{(L-1)/2} = L - 2i + 1. \end{aligned}$$

So whether L is even or odd, \(h_i\) takes the same value.

Now let \(H_i\) denote the expected time to reach the marked vertex if the walker starts a distance i from it. Trivially, \(H_0 = 0\). Using the above formula for \(h_i\), we find \(H_i\) when \(i > 0\):

$$\begin{aligned} H_i= & {} \sum _{j=1}^i h_j = \sum _{j=1}^i \left[ 2\left( \frac{L}{2}-i\right) + 1 \right] \\= & {} i(L+1) - 2 \sum _{j=1}^i i = i(L+1) - i(i+1) = i(L-i). \end{aligned}$$

An alternative approach to these results so far is given in [28].

Finally, we find the hitting time HT of the random walk, assuming that the walker begins in a uniform distribution over the vertices. If L is even,

$$\begin{aligned} \mathrm{HT}= & {} \frac{1}{L} \left[ H_0 + 2 \sum _{i=1}^{L/2-1} H_i + H_{L/2} \right] = \frac{1}{L} \left[ 0 + 2 \sum _{i=1}^{L/2-1} i(L-i) + \frac{L}{2} \left( L - \frac{L}{2} \right) \right] \\= & {} \frac{1}{L} \left[ \frac{L^2}{4} + 2 \left( L \sum _{i=1}^{L/2-1} i - \sum _{i=1}^{L/2-1} i^2 \right) \right] . \end{aligned}$$

Using

$$\begin{aligned} \sum _{i=1}^n i = \frac{n(n+1)}{2}, \quad \sum _{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}, \end{aligned}$$

we get

$$\begin{aligned} \mathrm{HT} = \frac{L^2 - 1}{6}. \end{aligned}$$

If L is odd, the hitting time is

$$\begin{aligned} \mathrm{HT}= & {} \frac{1}{L} \left[ H_0 + 2 \sum _{i=1}^{(L-1)/2} H_i \right] = \frac{1}{L} \left[ 0 + 2 \sum _{i=1}^{(L-1)/2} i(L-i) \right] \\= & {} \frac{2}{L} \left( L \sum _{i=1}^{(L-1)/2} i - \sum _{i=1}^{(L-1)/2} i^2 \right) = \frac{L^2 - 1}{6}. \end{aligned}$$

So whether the cycle is even or odd length, its hitting time is \((L^2-1)/6\).

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Wong, T.G., Santos, R.A.M. Exceptional quantum walk search on the cycle. Quantum Inf Process 16, 154 (2017). https://doi.org/10.1007/s11128-017-1606-y

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Keywords

  • Quantum walk
  • Quantum search
  • Spatial search
  • Exceptional configuration
  • Random walk
  • Markov chain
  • Hitting time
  • Mixing time