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Grover's Algorithm

In this section, we introduce Grover's algorithm and how it can be used to solve unstructured search problems. We then implement the quantum algorithm using Qiskit, and run on a simulator and device.

## 1. Introduction

You have likely heard that one of the many advantages a quantum computer has over a classical computer is its superior speed searching databases. Grover's algorithm demonstrates this capability. This algorithm can speed up an unstructured search problem quadratically, but its uses extend beyond that; it can serve as a general trick or subroutine to obtain quadratic run time improvements for a variety of other algorithms. This is called the amplitude amplification trick.

Suppose you are given a large list of $N$ items. Among these items there is one item with a unique property that we wish to locate; we will call this one the winner $w$. Think of each item in the list as a box of a particular color. Say all items in the list are gray except the winner $w$, which is purple.

To find the purple box -- the marked item -- using classical computation, one would have to check on average $N/2$ of these boxes, and in the worst case, all $N$ of them. On a quantum computer, however, we can find the marked item in roughly $\sqrt{N}$ steps with Grover's amplitude amplification trick. A quadratic speedup is indeed a substantial time-saver for finding marked items in long lists. Additionally, the algorithm does not use the list's internal structure, which makes it generic; this is why it immediately provides a quadratic quantum speed-up for many classical problems.

### Creating an Oracle

For the examples in this textbook, our 'database' is comprised of all the possible computational basis states our qubits can be in. For example, if we have 3 qubits, our list is the states $|000\rangle, |001\rangle, \dots |111\rangle$ (i.e the states $|0\rangle \rightarrow |7\rangle$).

Grover’s algorithm solves oracles that add a negative phase to the solution states. I.e. for any state $|x\rangle$ in the computational basis:

U_\omega|x\rangle = \bigg\{ \begin{aligned} \phantom{-}|x\rangle \quad \text{if} \; x \neq \omega \\ -|x\rangle \quad \text{if} \; x = \omega \\ \end{aligned}

This oracle will be a diagonal matrix, where the entry that correspond to the marked item will have a negative phase. For example, if we have three qubits and $\omega = \text{101}$, our oracle will have the matrix:

U_\omega = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{aligned} \\ \\ \\ \\ \\ \\ \leftarrow \omega = \text{101}\\ \\ \\ \\ \end{aligned}

What makes Grover’s algorithm so powerful is how easy it is to convert a problem to an oracle of this form. There are many computational problems in which it’s difficult to find a solution, but relatively easy to verify a solution. For example, we can easily verify a solution to a sudoku by checking all the rules are satisfied. For these problems, we can create a function $f$ that takes a proposed solution $x$, and returns $f(x) = 0$ if $x$ is not a solution ($x \neq \omega$) and $f(x) = 1$ for a valid solution ($x = \omega$). Our oracle can then be described as:

$$U_\omega|x\rangle = (-1)^{f(x)}|x\rangle$$

and the oracle's matrix will be a diagonal matrix of the form:

$$U_\omega = \begin{bmatrix} (-1)^{f(0)} & 0 & \cdots & 0 \\ 0 & (-1)^{f(1)} & \cdots & 0 \\ \vdots & 0 & \ddots & \vdots \\ 0 & 0 & \cdots & (-1)^{f(2^n-1)} \\ \end{bmatrix}$$
Circuit Construction of a Grover Oracle (click to expand)

If we have our classical function $f(x)$, we can convert it to a reversible circuit of the form:

If we initialise the 'output' qubit in the state $|{-}\rangle$, the phase kickback effect turns this into a Grover oracle (similar to the workings of the Deutsch-Jozsa oracle):

We then ignore the auxiliary ($|{-}\rangle$) qubit.

For the next part of this chapter, we aim to teach the core concepts of the algorithm. We will create example oracles where we know $\omega$ beforehand, and not worry ourselves with whether these oracles are useful or not. At the end of the chapter, we will cover a short example where we create an oracle to solve a problem (sudoku).

### Amplitude Amplification

So how does the algorithm work? Before looking at the list of items, we have no idea where the marked item is. Therefore, any guess of its location is as good as any other, which can be expressed in terms of a uniform superposition: $|s \rangle = \frac{1}{\sqrt{N}} \sum_{x = 0}^{N -1} | x \rangle.$

If at this point we were to measure in the standard basis $\{ | x \rangle \}$, this superposition would collapse, according to the fifth quantum law, to any one of the basis states with the same probability of $\frac{1}{N} = \frac{1}{2^n}$. Our chances of guessing the right value $w$ is therefore $1$ in $2^n$, as could be expected. Hence, on average we would need to try about $N/2 = 2^{n-1}$ times to guess the correct item.

Enter the procedure called amplitude amplification, which is how a quantum computer significantly enhances this probability. This procedure stretches out (amplifies) the amplitude of the marked item, which shrinks the other items' amplitude, so that measuring the final state will return the right item with near-certainty.

This algorithm has a nice geometrical interpretation in terms of two reflections, which generate a rotation in a two-dimensional plane. The only two special states we need to consider are the winner $| w \rangle$ and the uniform superposition $| s \rangle$. These two vectors span a two-dimensional plane in the vector space $\mathbb{C}^N.$ They are not quite perpendicular because $| w \rangle$ occurs in the superposition with amplitude $N^{-1/2}$ as well. We can, however, introduce an additional state $|s'\rangle$ that is in the span of these two vectors, which is perpendicular to $| w \rangle$ and is obtained from $|s \rangle$ by removing $| w \rangle$ and rescaling.

Step 1: The amplitude amplification procedure starts out in the uniform superposition $| s \rangle$, which is easily constructed from $| s \rangle = H^{\otimes n} | 0 \rangle^n$.

The left graphic corresponds to the two-dimensional plane spanned by perpendicular vectors $|w\rangle$ and $|s'\rangle$ which allows to express the initial state as $|s\rangle = \sin \theta | w \rangle + \cos \theta | s' \rangle,$ where $\theta = \arcsin \langle s | w \rangle = \arcsin \frac{1}{\sqrt{N}}$. The right graphic is a bar graph of the amplitudes of the state $| s \rangle$.

Step 2: We apply the oracle reflection $U_f$ to the state $|s\rangle$.

Geometrically this corresponds to a reflection of the state $|s\rangle$ about $|s'\rangle$. This transformation means that the amplitude in front of the $|w\rangle$ state becomes negative, which in turn means that the average amplitude (indicated by a dashed line) has been lowered.

Step 3: We now apply an additional reflection ($U_s$) about the state $|s\rangle$: $U_s = 2|s\rangle\langle s| - \mathbb{1}$. This transformation maps the state to $U_s U_f| s \rangle$ and completes the transformation.

Two reflections always correspond to a rotation. The transformation $U_s U_f$ rotates the initial state $|s\rangle$ closer towards the winner $|w\rangle$. The action of the reflection $U_s$ in the amplitude bar diagram can be understood as a reflection about the average amplitude. Since the average amplitude has been lowered by the first reflection, this transformation boosts the negative amplitude of $|w\rangle$ to roughly three times its original value, while it decreases the other amplitudes. We then go to step 2 to repeat the application. This procedure will be repeated several times to zero in on the winner.

After $t$ steps we will be in the state $|\psi_t\rangle$ where: $| \psi_t \rangle = (U_s U_f)^t | s \rangle.$

How many times do we need to apply the rotation? It turns out that roughly $\sqrt{N}$ rotations suffice. This becomes clear when looking at the amplitudes of the state $| \psi \rangle$. We can see that the amplitude of $| w \rangle$ grows linearly with the number of applications $\sim t N^{-1/2}$. However, since we are dealing with amplitudes and not probabilities, the vector space's dimension enters as a square root. Therefore it is the amplitude, and not just the probability, that is being amplified in this procedure.

In the case that there are multiple solutions, $M$, it can be shown that roughly $\sqrt{(N/M)}$ rotations will suffice.

## 2. Example: 2 Qubits

Let's first have a look at the case of Grover's algorithm for $N=4$ which is realized with 2 qubits. In this particular case, only one rotation is required to rotate the initial state $|s\rangle$ to the winner $|w\rangle$[3]:

1. Following the above introduction, in the case $N=4$ we have $$\theta = \arcsin \frac{1}{2} = \frac{\pi}{6}.$$
2. After $t$ steps, we have $$(U_s U_\omega)^t | s \rangle = \sin \theta_t | \omega \rangle + \cos \theta_t | s' \rangle ,$$where $$\theta_t = (2t+1)\theta.$$
3. In order to obtain $| \omega \rangle$ we need $\theta_t = \frac{\pi}{2}$, which with $\theta=\frac{\pi}{6}$ inserted above results to $t=1$. This implies that after $t=1$ rotation the searched element is found.

We will now follow through an example using a specific oracle.

#### Oracle for $\lvert \omega \rangle = \lvert 11 \rangle$

Let's look at the case $\lvert w \rangle = \lvert 11 \rangle$. The oracle $U_\omega$ in this case acts as follows:

$$U_\omega | s \rangle = U_\omega \frac{1}{2}\left( |00\rangle + |01\rangle + |10\rangle + |11\rangle \right) = \frac{1}{2}\left( |00\rangle + |01\rangle + |10\rangle - |11\rangle \right).$$

or:

$$U_\omega = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \\ \end{bmatrix}$$

which you may recognise as the controlled-Z gate. Thus, for this example, our oracle is simply the controlled-Z gate:

#### Reflection $U_s$

In order to complete the circuit we need to implement the additional reflection $U_s = 2|s\rangle\langle s| - \mathbb{1}$. Since this is a reflection about $|s\rangle$, we want to add a negative phase to every state orthogonal to $|s\rangle$.

One way we can do this is to use the operation that transforms the state $|s\rangle \rightarrow |0\rangle$, which we already know is the Hadamard gate applied to each qubit:

$$H^{\otimes n}|s\rangle = |0\rangle$$

Then we apply a circuit that adds a negative phase to the states orthogonal to $|0\rangle$:

$$U_0 \frac{1}{2}\left( \lvert 00 \rangle + \lvert 01 \rangle + \lvert 10 \rangle + \lvert 11 \rangle \right) = \frac{1}{2}\left( \lvert 00 \rangle - \lvert 01 \rangle - \lvert 10 \rangle - \lvert 11 \rangle \right)$$

i.e. the signs of each state are flipped except for $\lvert 00 \rangle$. As can easily be verified, one way of implementing $U_0$ is the following circuit:

Finally, we do the operation that transforms the state $|0\rangle \rightarrow |s\rangle$ (the H-gate again):

$$H^{\otimes n}U_0 H^{\otimes n} = U_s$$

The complete circuit for $U_s$ looks like this:

#### Full Circuit for $\lvert w \rangle = |11\rangle$

Since in the particular case of $N=4$ only one rotation is required we can combine the above components to build the full circuit for Grover's algorithm for the case $\lvert w \rangle = |11\rangle$:

### 2.1 Qiskit Implementation

We now implement Grover's algorithm for the above case of 2 qubits for $\lvert w \rangle = |11\rangle$.

#initialization
import matplotlib.pyplot as plt
import numpy as np

# importing Qiskit
from qiskit import IBMQ, Aer, assemble, transpile
from qiskit import QuantumCircuit, ClassicalRegister, QuantumRegister
from qiskit.providers.ibmq import least_busy

# import basic plot tools
from qiskit.visualization import plot_histogram


We start by preparing a quantum circuit with two qubits:

n = 2
grover_circuit = QuantumCircuit(n)


Then we simply need to write out the commands for the circuit depicted above. First, we need to initialize the state $|s\rangle$. Let's create a general function (for any number of qubits) so we can use it again later:

def initialize_s(qc, qubits):
"""Apply a H-gate to 'qubits' in qc"""
for q in qubits:
qc.h(q)
return qc

grover_circuit = initialize_s(grover_circuit, [0,1])
grover_circuit.draw()


Apply the Oracle for $|w\rangle = |11\rangle$. This oracle is specific to 2 qubits:

grover_circuit.cz(0,1) # Oracle
grover_circuit.draw()


We now want to apply the diffuser ($U_s$). As with the circuit that initializes $|s\rangle$, we'll create a general diffuser (for any number of qubits) so we can use it later in other problems.

# Diffusion operator (U_s)
grover_circuit.h([0,1])
grover_circuit.z([0,1])
grover_circuit.cz(0,1)
grover_circuit.h([0,1])
grover_circuit.draw()


This is our finished circuit.

### 2.1.1 Experiment with Simulators

Let's run the circuit in simulation. First, we can verify that we have the correct statevector:

sim = Aer.get_backend('aer_simulator')
# we need to make a copy of the circuit with the 'save_statevector'
# instruction to run on the Aer simulator
grover_circuit_sim = grover_circuit.copy()
grover_circuit_sim.save_statevector()
qobj = assemble(grover_circuit_sim)
result = sim.run(qobj).result()
statevec = result.get_statevector()
from qiskit_textbook.tools import vector2latex
vector2latex(statevec, pretext="|\\psi\\rangle =")


### 5.3 The Full Algorithm

All that's left to do now is to put this oracle into Grover's algorithm!

var_qubits = QuantumRegister(4, name='v')
clause_qubits = QuantumRegister(4, name='c')
output_qubit = QuantumRegister(1, name='out')
cbits = ClassicalRegister(4, name='cbits')
qc = QuantumCircuit(var_qubits, clause_qubits, output_qubit, cbits)

# Initialize 'out0' in state |->
qc.initialize([1, -1]/np.sqrt(2), output_qubit)

# Initialize qubits in state |s>
qc.h(var_qubits)
qc.barrier()  # for visual separation

## First Iteration
# Apply our oracle
sudoku_oracle(qc, clause_list, clause_qubits)
qc.barrier()  # for visual separation
# Apply our diffuser
qc.append(diffuser(4), [0,1,2,3])

## Second Iteration
sudoku_oracle(qc, clause_list, clause_qubits)
qc.barrier()  # for visual separation
# Apply our diffuser
qc.append(diffuser(4), [0,1,2,3])

# Measure the variable qubits
qc.measure(var_qubits, cbits)

qc.draw(fold=-1)

# Simulate and plot results
aer_simulator = Aer.get_backend('aer_simulator')
transpiled_qc = transpile(qc, aer_simulator)
qobj = assemble(transpiled_qc)
result = aer_sim.run(qobj).result()
plot_histogram(result.get_counts())


There are two bit strings with a much higher probability of measurement than any of the others, 0110 and 1001. These correspond to the assignments:

v0 = 0
v1 = 1
v2 = 1
v3 = 0

and

v0 = 1
v1 = 0
v2 = 0
v3 = 1

which are the two solutions to our sudoku! The aim of this section is to show how we can create Grover oracles from real problems. While this specific problem is trivial, the process can be applied (allowing large enough circuits) to any decision problem. To recap, the steps are:

1. Create a reversible classical circuit that identifies a correct solution
2. Use phase kickback and uncomputation to turn this circuit into an oracle
3. Use Grover's algorithm to solve this oracle

## 6. References

1. L. K. Grover (1996), "A fast quantum mechanical algorithm for database search", Proceedings of the 28th Annual ACM Symposium on the Theory of Computing (STOC 1996), doi:10.1145/237814.237866, arXiv:quant-ph/9605043
2. C. Figgatt, D. Maslov, K. A. Landsman, N. M. Linke, S. Debnath & C. Monroe (2017), "Complete 3-Qubit Grover search on a programmable quantum computer", Nature Communications, Vol 8, Art 1918, doi:10.1038/s41467-017-01904-7, arXiv:1703.10535
3. I. Chuang & M. Nielsen, "Quantum Computation and Quantum Information", Cambridge: Cambridge University Press, 2000.
import qiskit.tools.jupyter
%qiskit_version_table


### Version Information

Qiskit SoftwareVersion
Qiskit0.27.0
Terra0.17.4
Aer0.8.2
Ignis0.6.0
Aqua0.9.2
IBM Q Provider0.14.0
System information
Python3.7.7 (default, May 6 2020, 04:59:01) [Clang 4.0.1 (tags/RELEASE_401/final)]
OSDarwin
CPUs8
Memory (Gb)32.0
Wed Jun 16 13:45:01 2021 BST