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.
Unstructured Search
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 timesaver 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 speedup 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:
$$ U_\omegax\rangle = \bigg\{ \begin{aligned} \phantom{}x\rangle \quad \text{if} \; x \neq \omega \\ x\rangle \quad \text{if} \; x = \omega \\ \end{aligned} $$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:
$$ U_\omegax\rangle = (1)^{f(x)}x\rangle $$and the oracle's matrix will be a diagonal matrix of the form:
$$ U_\omega = \begin{bmatrix} (1)^{f(0)} & 0 & \cdots & 0 \\ 0 & (1)^{f(1)} & \cdots & 0 \\ \vdots & 0 & \ddots & \vdots \\ 0 & 0 & \cdots & (1)^{f(2^n)} \\ \end{bmatrix} $$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 DeutschJozsa 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 $N = 2^n$ 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 nearcertainty.
This algorithm has a nice geometrical interpretation in terms of two reflections, which generate a rotation in a twodimensional 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 twodimensional 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 twodimensional 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$ for the case $N = 2^2 = 4$. The average amplitude is indicated by a dashed line.
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 has been lowered.
Step 3: We now apply an additional reflection ($U_s$) about the state $s\rangle$: $U_s = 2s\rangle\langle s  \mathbb{1}$. 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]:
 Following the above introduction, in the case $N=4$ we have $$\theta = \arcsin \frac{1}{2} = \frac{\pi}{6}.$$
 After $t$ steps, we have $$(U_s U_\omega)^t  s \rangle = \sin \theta_t  \omega \rangle + \cos \theta_t  s' \rangle ,$$where $$\theta_t = (2t+1)\theta.$$
 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 controlledZ gate. Thus, for this example, our oracle is simply the controlledZ gate:
Reflection $U_s$
In order to complete the circuit we need to implement the additional reflection $U_s = 2s\rangle\langle s  \mathbb{1}$. 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 only to the state $0\rangle$:
$$U_0 \frac{1}{2}\left( \lvert 00 \rangle + \lvert 01 \rangle + \lvert 10 \rangle + \lvert 11 \rangle \right) = \frac{1}{2}\left( \lvert 00 \rangle  \lvert 01 \rangle  \lvert 10 \rangle  \lvert 11 \rangle \right)$$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 Hgate 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, QuantumCircuit, ClassicalRegister, QuantumRegister, execute
from qiskit.providers.ibmq import least_busy
from qiskit.quantum_info import Statevector
# 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 Hgate 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 initialises $s\rangle$, 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.
sv_sim = Aer.get_backend('statevector_simulator')
job_sim = execute(grover_circuit, sv_sim)
statevec = job_sim.result().get_statevector()
from qiskit_textbook.tools import vector2latex
vector2latex(statevec, pretext="\\psi\\rangle =")
As expected, the amplitude of every state that is not $11\rangle$ is 0, this means we have a 100% chance of measuring $11\rangle$:
grover_circuit.measure_all()
qasm_simulator = Aer.get_backend('qasm_simulator')
shots = 1024
results = execute(grover_circuit, backend=qasm_simulator, shots=shots).result()
answer = results.get_counts()
plot_histogram(answer)
# Load IBM Q account and get the least busy backend device
provider = IBMQ.load_account()
device = least_busy(provider.backends(filters=lambda x: x.configuration().n_qubits >= 3 and
not x.configuration().simulator and x.status().operational==True))
print("Running on current least busy device: ", device)
# Run our circuit on the least busy backend. Monitor the execution of the job in the queue
from qiskit.tools.monitor import job_monitor
job = execute(grover_circuit, backend=device, shots=1024, optimization_level=3)
job_monitor(job, interval = 2)
# Get the results from the computation
results = job.result()
answer = results.get_counts(grover_circuit)
plot_histogram(answer)
We confirm that in the majority of the cases the state $11\rangle$ is measured. The other results are due to errors in the quantum computation.
3. Example: 3 Qubits
We now go through the example of Grover's algorithm for 3 qubits with two marked states $\lvert101\rangle$ and $\lvert110\rangle$, following the implementation found in Reference [2]. The quantum circuit to solve the problem using a phase oracle is:
 Apply Hadamard gates to $3$ qubits initialised to $\lvert000\rangle$ to create a uniform superposition: $$\lvert \psi_1 \rangle = \frac{1}{\sqrt{8}} \left( \lvert000\rangle + \lvert001\rangle + \lvert010\rangle + \lvert011\rangle + \lvert100\rangle + \lvert101\rangle + \lvert110\rangle + \lvert111\rangle \right) $$
 Mark states $\lvert101\rangle$ and $\lvert110\rangle$ using a phase oracle: $$\lvert \psi_2 \rangle = \frac{1}{\sqrt{8}} \left( \lvert000\rangle + \lvert001\rangle + \lvert010\rangle + \lvert011\rangle + \lvert100\rangle  \lvert101\rangle  \lvert110\rangle + \lvert111\rangle \right) $$

Perform the reflection around the average amplitude:
 Apply Hadamard gates to the qubits $$\lvert \psi_{3a} \rangle = \frac{1}{2} \left( \lvert000\rangle +\lvert011\rangle +\lvert100\rangle \lvert111\rangle \right) $$
 Apply X gates to the qubits $$\lvert \psi_{3b} \rangle = \frac{1}{2} \left( \lvert000\rangle +\lvert011\rangle +\lvert100\rangle +\lvert111\rangle \right) $$
 Apply a doubly controlled Z gate between the 1, 2 (controls) and 3 (target) qubits $$\lvert \psi_{3c} \rangle = \frac{1}{2} \left( \lvert000\rangle +\lvert011\rangle +\lvert100\rangle \lvert111\rangle \right) $$
 Apply X gates to the qubits $$\lvert \psi_{3d} \rangle = \frac{1}{2} \left( \lvert000\rangle +\lvert011\rangle +\lvert100\rangle \lvert111\rangle \right) $$
 Apply Hadamard gates to the qubits $$\lvert \psi_{3e} \rangle = \frac{1}{\sqrt{2}} \left( \lvert101\rangle \lvert110\rangle \right) $$
 Measure the $3$ qubits to retrieve states $\lvert101\rangle$ and $\lvert110\rangle$
Note that since there are 2 solutions and 8 possibilities, we will only need to run one iteration (steps 2 & 3).
3.1 Qiskit Implementation
We now implement Grover's algorithm for the above example for $3$qubits and searching for two marked states $\lvert101\rangle$ and $\lvert110\rangle$. Note: Remember that Qiskit orders it's qubits the opposite way round to this resource, so the circuit drawn will appear flipped about the horizontal.
We create a phase oracle that will mark states $\lvert101\rangle$ and $\lvert110\rangle$ as the results (step 1).
qc = QuantumCircuit(3)
qc.cz(0, 2)
qc.cz(1, 2)
oracle_ex3 = qc.to_gate()
oracle_ex3.name = "U$_\omega$"
In the last section, we used a diffuser specific to 2 qubits, in the cell below we will create a general diffuser for any number of qubits.
Details: Creating a General Diffuser (click to expand)
Remember that we can create $U_s$ from $U_0$: $$ U_s = H^{\otimes n} U_0 H^{\otimes n} $$ And a multicontrolledZ gate ($MCZ$) inverts the phase of the state $11\dots 1\rangle$: $$ MCZ = \begin{bmatrix} 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \\ \end{bmatrix} \begin{aligned} \\ \\ \\ \leftarrow \text{Add negative phase to} \; 11\dots 1\rangle\\ \end{aligned} $$ Applying an Xgate to each qubit performs the transformation: $$ \begin{aligned} 00\dots 0\rangle & \rightarrow 11\dots 1\rangle\\ 11\dots 1\rangle & \rightarrow 00\dots 0\rangle \end{aligned} $$ So: $$ U_0 = X^{\otimes n} (MCZ) X^{\otimes n} $$ Using these properties together, we can create $U_s$ using Hgates, Xgates, and a single multicontrolledZ gate: $$ U_s = H^{\otimes n} U_0 H^{\otimes n} = H^{\otimes n} X^{\otimes n} (MCZ) X^{\otimes n} H^{\otimes n} $$ Note that this circuit adds a global phase of 1.def diffuser(nqubits):
qc = QuantumCircuit(nqubits)
# Apply transformation s> > 00..0> (Hgates)
for qubit in range(nqubits):
qc.h(qubit)
# Apply transformation 00..0> > 11..1> (Xgates)
for qubit in range(nqubits):
qc.x(qubit)
# Do multicontrolledZ gate
qc.h(nqubits1)
qc.mct(list(range(nqubits1)), nqubits1) # multicontrolledtoffoli
qc.h(nqubits1)
# Apply transformation 11..1> > 00..0>
for qubit in range(nqubits):
qc.x(qubit)
# Apply transformation 00..0> > s>
for qubit in range(nqubits):
qc.h(qubit)
# We will return the diffuser as a gate
U_s = qc.to_gate()
U_s.name = "$U_s$"
return U_s
We'll now put the pieces together, with the creation of a uniform superposition at the start of the circuit and a measurement at the end. Note that since there are 2 solutions and 8 possibilities, we will only need to run one iteration.
n = 3
grover_circuit = QuantumCircuit(n)
grover_circuit = initialize_s(grover_circuit, [0,1,2])
grover_circuit.append(oracle_ex3, [0,1,2])
grover_circuit.append(diffuser(n), [0,1,2])
grover_circuit.measure_all()
grover_circuit.draw()
backend = Aer.get_backend('qasm_simulator')
results = execute(grover_circuit, backend=backend, shots=1024).result()
answer = results.get_counts()
plot_histogram(answer)
As we can see, the algorithm discovers our marked states $\lvert101\rangle$ and $\lvert110\rangle$.
backend = least_busy(provider.backends(filters=lambda x: x.configuration().n_qubits >= 3 and
not x.configuration().simulator and x.status().operational==True))
print("least busy backend: ", backend)
# Run our circuit on the least busy backend. Monitor the execution of the job in the queue
from qiskit.tools.monitor import job_monitor
job = execute(grover_circuit, backend=backend, shots=1024, optimization_level=3)
job_monitor(job, interval = 2)
# Get the results from the computation
results = job.result()
answer = results.get_counts(grover_circuit)
plot_histogram(answer)
As we can (hopefully) see, there is a higher chance of measuring $\lvert101\rangle$ and $\lvert110\rangle$. The other results are due to errors in the quantum computation.
4. Problems
The function grover_problem_oracle
below takes a number of qubits (n
), and a variant
and returns an nqubit oracle. The function will always return the same oracle for the same n
and variant
. You can see the solutions to each oracle by setting print_solutions = True
when calling grover_problem_oracle
.
from qiskit_textbook.problems import grover_problem_oracle
## Example Usage
n = 4
oracle = grover_problem_oracle(n, variant=1) # 0th variant of oracle, with n qubits
qc = QuantumCircuit(n)
qc.append(oracle, [0,1,2,3])
qc.draw()
grover_problem_oracle(4, variant=2)
uses 4 qubits and has 1 solution.
a. How many iterations do we need to have a > 90% chance of measuring this solution?
b. Use Grover's algorithm to find this solution state. c. What happens if we apply more iterations the the number we calculated in problem 1a above? Why?With 2 solutions and 4 qubits, how many iterations do we need for a >90% chance of measuring a solution? Test your answer using the oracle
grover_problem_oracle(4, variant=1)
(which has two solutions).Create a function,
grover_solver(oracle, iterations)
that takes as input: A Grover oracle as a gate (
oracle
)  An integer number of iterations (
iterations
)
and returns a
QuantumCircuit
that performs Grover's algorithm on the 'oracle
' gate, with 'iterations
' iterations. A Grover oracle as a gate (
5. Solving Sudoku using Grover's Algorithm
The oracles used throughout this chapter so far have been created with prior knowledge of their solutions. We will now solve a simple problem using Grover's algorithm, for which we do not necessarily know the solution beforehand. Our problem is a 2×2 binary sudoku, which in our case has two simple rules:
 No column may contain the same value twice
 No row may contain the same value twice
If we assign each square in our sudoku to a variable like so:
we want our circuit to output a solution to this sudoku.
5.1 Turning the Problem into a Circuit
To create the oracle, we simply need to create a classical circuit that checks whether the state of our variable bits is a valid solution. Since we need to check down both columns and across both rows, there are 4 conditions we need to check:
v0 ≠ v1 # check along top row
v2 ≠ v3 # check along bottom row
v0 ≠ v2 # check down left column
v1 ≠ v3 # check down right column
Remember we are comparing classical (computational basis) states. For convenience, we can compile this set of comparisons into a list of clauses:
clause_list = [[0,1],
[0,2],
[1,3],
[2,3]]
We will assign the value of each variable to a bit in our circuit. To check these clauses computationally, we will use the XOR
gate (we came across this in the atoms of computation).
def XOR(qc, a, b, output):
qc.cx(a, output)
qc.cx(b, output)
Convince yourself that the output0
bit in the circuit below will only be flipped if input0 ≠ input1
:
# We will use separate registers to name the bits
in_qubits = QuantumRegister(2, name='input')
out_qubit = QuantumRegister(1, name='output')
qc = QuantumCircuit(in_qubits, out_qubit)
XOR(qc, in_qubits[0], in_qubits[1], out_qubit)
qc.draw()
This circuit checks whether input0 == input1
and stores the output to output0
. To check each clause, we repeat this circuit for each pairing in clause_list
and store the output to a new bit:
# Create separate registers to name bits
var_qubits = QuantumRegister(4, name='v') # variable bits
clause_qubits = QuantumRegister(4, name='c') # bits to store clausechecks
# Create quantum circuit
qc = QuantumCircuit(var_qubits, clause_qubits)
# Use XOR gate to check each clause
i = 0
for clause in clause_list:
XOR(qc, clause[0], clause[1], clause_qubits[i])
i += 1
qc.draw()
The final state of the bits c0, c1, c2, c3
will only all be 1
in the case that the assignments of v0, v1, v2, v3
are a solution to the sudoku. To complete our checking circuit, we want a single bit to be 1
if (and only if) all the clauses are satisfied, this way we can look at just one bit to see if our assignment is a solution. We can do this using a multicontrolledToffoligate:
# Create separate registers to name bits
var_qubits = QuantumRegister(4, name='v')
clause_qubits = QuantumRegister(4, name='c')
output_qubit = QuantumRegister(1, name='out')
qc = QuantumCircuit(var_qubits, clause_qubits, output_qubit)
# Compute clauses
i = 0
for clause in clause_list:
XOR(qc, clause[0], clause[1], clause_qubits[i])
i += 1
# Flip 'output' bit if all clauses are satisfied
qc.mct(clause_qubits, output_qubit)
qc.draw()
The circuit above takes as input an initial assignment of the bits v0
, v1
, v2
and v3
, and all other bits should be initialised to 0
. After running the circuit, the state of the out0
bit tells us if this assignment is a solution or not; out0 = 0
means the assignment is not a solution, and out0 = 1
means the assignment is a solution.
Important: Before you continue, it is important you fully understand this circuit and are convinced it works as stated in the paragraph above.
5.2 Uncomputing, and Completing the Oracle
We can now turn this checking circuit into a Grover oracle using phase kickback. To recap, we have 3 registers:
 One register which stores our sudoku variables (we'll say $x = v_3, v_2, v_1, v_0$)
 One register that stores our clauses (this starts in the state $0000\rangle$ which we'll abbreviate to $0\rangle$)
 And one qubit ($\text{out}_0\rangle$) that we've been using to store the output of our checking circuit.
To create an oracle, we neeed our circuit ($U_\omega$) to perform the transformation:
$$ U_\omegax\rangle0\rangle\text{out}_0\rangle = x\rangle0\rangle\text{out}_0\oplus f(x)\rangle $$If we set the out0
qubit to the superposition state ${}\rangle$ we have:
If $f(x) = 0$, then we have the state:
$$ \begin{aligned} &= x\rangle0\rangle\otimes \tfrac{1}{\sqrt{2}}(0\rangle  1\rangle)\\ &= x\rangle0\rangle\rangle\\ \end{aligned} $$(i.e. no change). But if $f(x) = 1$ (i.e. $x = \omega$), we introduce a negative phase to the ${}\rangle$ qubit:
$$ \begin{aligned} &= \phantom{}x\rangle0\rangle\otimes\tfrac{1}{\sqrt{2}}(1\rangle  0\rangle)\\ &= \phantom{}x\rangle0\rangle\otimes \tfrac{1}{\sqrt{2}}(0\rangle  1\rangle)\\ &= x\rangle0\rangle\rangle\\ \end{aligned} $$This is a functioning oracle that uses two auxiliary registers in the state $0\rangle{}\rangle$:
$$ U_\omegax\rangle0\rangle{}\rangle = \Bigg\{ \begin{aligned} \phantom{}x\rangle0\rangle\rangle \quad \text{for} \; x \neq \omega \\ x\rangle0\rangle\rangle \quad \text{for} \; x = \omega \\ \end{aligned} $$To adapt our checking circuit into a Grover oracle, we need to guarantee the bits in the second register (c
) are always returned to the state $0000\rangle$ after the computation. To do this, we simply repeat the part of the circuit that computes the clauses which guarantees c0 = c1 = c2 = c3 = 0
after our circuit has run. We call this step 'uncomputation'.
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)
def sudoku_oracle(qc, clause_list, var_qubits, clause_qubits, cbits):
# Compute clauses
i = 0
for clause in clause_list:
XOR(qc, clause[0], clause[1], clause_qubits[i])
i += 1
# Flip 'output' bit if all clauses are satisfied
qc.mct(clause_qubits, output_qubit)
# Uncompute clauses to reset clausechecking bits to 0
i = 0
for clause in clause_list:
XOR(qc, clause[0], clause[1], clause_qubits[i])
i += 1
sudoku_oracle(qc, clause_list, var_qubits, clause_qubits, cbits)
qc.draw()
In summary, the circuit above performs:
$$ U_\omegax\rangle0\rangle\text{out}_0\rangle = \Bigg\{ \begin{aligned} x\rangle0\rangle\text{out}_0\rangle \quad \text{for} \; x \neq \omega \\ x\rangle0\rangle\otimes X\text{out}_0\rangle \quad \text{for} \; x = \omega \\ \end{aligned} $$and if the initial state of $\text{out}_0\rangle = {}\rangle$,:
$$ U_\omegax\rangle0\rangle{}\rangle = \Bigg\{ \begin{aligned} \phantom{}x\rangle0\rangle\rangle \quad \text{for} \; x \neq \omega \\ x\rangle0\rangle\rangle \quad \text{for} \; x = \omega \\ \end{aligned} $$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)
# Initialise 'out0' in state >
qc.initialize([1, 1]/np.sqrt(2), output_qubit)
# Initialise qubits in state s>
qc.h(var_qubits)
qc.barrier() # for visual separation
## First Iteration
# Apply our oracle
sudoku_oracle(qc, clause_list, var_qubits, clause_qubits, cbits)
qc.barrier() # for visual separation
# Apply our diffuser
qc.append(diffuser(4), [0,1,2,3])
## Second Iteration
sudoku_oracle(qc, clause_list, var_qubits, clause_qubits, cbits)
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()
# Simulate and plot results
qasm_simulator = Aer.get_backend('qasm_simulator')
result = execute(qc, backend=qasm_simulator, shots=1024).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! Hopefully it is now clearer as to how we can create Grover oracles from real problems.
6. References
 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:quantph/9605043
 C. Figgatt, D. Maslov, K. A. Landsman, N. M. Linke, S. Debnath & C. Monroe (2017), "Complete 3Qubit Grover search on a programmable quantum computer", Nature Communications, Vol 8, Art 1918, doi:10.1038/s41467017019047, arXiv:1703.10535
 I. Chuang & M. Nielsen, "Quantum Computation and Quantum Information", Cambridge: Cambridge University Press, 2000.
import qiskit
qiskit.__qiskit_version__