# IBM Quantum Challenge 2020 - Part 6

This post is part of a serie of posts on my experience with the IBM Quantum Challenge 2020.

The serie is organised as follow:

- Introduction to the problem.
- Mathematical reformulation of the problem.
- Unoptimised implementation of the solution.
- How to optimise a quantum circuit?.
- Optimising multi-controlled
`X`

gates. - Optimising the
`QRAM`

(this post). - Optimising the oracle

In this blog post I detail the two optimisations I used to lower
down significantly the cost of the `QRAM`

implementation.

## 40k - Re-using computations by using an ancilla qubit

In the non-optimised version, we implemented the QRAM by chaining the following quantum circuit for all the possible indices (from 0 to 15 included) by changing the \(X\) gates sandwiching the 4-controlled \(X\) gates:

But this quantum circuit is far from optimal. Instead of applying independently the six 4-controlled \(X\) gates we could save in one ancilla qubit the result of one of the 4-controlled \(X\) operation and then broadcast this result using six 1-controlled \(X\) gates.

This is done by changing the implementation of the `qram`

and
`encode_problem`

functions:

```
def encode_problem(problem: BoardEncoding) -> QuantumCircuit:
"""
Encode the given problem on the qboard register if the
qstate register is in the |1> state.
This function returns a quantum circuit that encodes the given problem
on the |qboard> register only if the |qstate> register is |1>.
:param problem: a list of asteroid coordinates, given as tuples
of strings.
"""
qboard = QuantumRegister(16)
qstate = QuantumRegister(1)
circuit = QuantumCircuit(qboard, qstate, name="encode_problem")
for x, y in problem:
index = 4 * int(x) + int(y)
circuit.cx(qstate[0], qboard[index])
return circuit
def qram(problem_set: ty.List[BoardEncoding]):
"""
Encodes the given problem set into the |qboard> register.
The returned routine takes as inputs:
- |qindex> that encodes the indices for which we want to encode the
boards. The |qindex> register will likely be given in a
superposition of all the possible states.
- |qboard> that should be given in the |0> state and that is returned
in the state that encodes the problem_set given. For each qindex,
the |qboard> state in |qindex>|qboard> is full of |0> except on the
locations where there is an asteroid, in which case the state is |1>.
- |qancilla> that is a 3-qubit ancillary register.
"""
qindex = QuantumRegister(4)
qboard = QuantumRegister(16)
qstate = QuantumRegister(1)
qancilla = QuantumRegister(2)
circuit = QuantumCircuit(qindex, qboard, qstate, qancilla, name="qram")
for i, problem in enumerate(problem_set):
bin_str = bin(i)[2:].zfill(4)
x_qubits = [j for j in range(4) if bin_str[j] == "0"]
# setting qstate to |1> if we are in the index "i".
if x_qubits:
circuit.x([qindex[j] for j in x_qubits])
circuit.mcx(qindex, qstate, ancilla_qubits=qancilla, mode="v-chain")
circuit.append(encode_problem(problem), [*qboard, *qstate])
# Undo the qstate thing
circuit.mcx(qindex, qstate, ancilla_qubits=qancilla, mode="v-chain").inverse()
if x_qubits:
circuit.x([qindex[j] for j in x_qubits]).inverse()
return circuit
```

The functions using these methods should also be modified to account for the extra qubit added as an ancilla. The final code implementing this optimisation is available on this Gist.

## 28k - Advanced looping through all the integer values on 16 qubits

The optimisation presented in the last section allowed me to
nearly halve the number of calls to the `mcx_vchain`

procedure due
to the QRAM, which is great! But as seen in the call-graph below,
the QRAM still performs nearly 34% of the `mcx_vchain`

calls,
which is too high.

At this point in the optimisation process I had only one optimisation idea left for the QRAM: using Gray codes or something similar. The main idea is the following: as we know for sure we need to loop over all the \(16\) possible integers in the QRAM, maybe I could rearrange the way I loop through them in order to avoid some multiply-controlled gates?

Re-phrasing our QRAM problem, we basically want to implement an
operation known as `Select(V)`

that implements

\[ \mathrm{Select}(V) = \sum_{i=0}^{15} \vert i \rangle \langle i \vert \otimes V_i \]

where \(V_i\) is a quantum circuit encoding the $i$-th board. We
already have an efficient implementation for \(V_i\) with the
`encode_problem`

routine used in the previous section, the only
challenge left is to implement the `Select(.)`

part.

Luckily for me, I was already aware of a very nice paper that was
discussing the implementation of the `Select(.)`

part: Toward the
first quantum simulation with quantum speedup.

The algorithm is perfectly described in Section G.4. of the paper and implemented in https://github.com/njross/simcount. Sadly, because I was quite in a rush, my implementation is not generic at all and is rather specialised to the asteroid problem I was trying to solve. As such, the code is long and not necessarily easy to read.

```
def qram(problem_set: ty.List[BoardEncoding]):
"""
Encodes the given problem set into the |qboard> register.
The returned routine takes as inputs:
- |qindex> (called |x> in the implementation) that encodes the indices for
which we want to encode the boards. The |index> register will likely be
given in a superposition of all the possible states.
- |qboard> that should be given in the |0> state and that is returned in
the state that encodes the problem_set given.
For each qindex, the |qboard> state in |qindex>|qboard> is full of |0>
except on the locations where there is an asteroid, in which case the
state is |1>.
- |qancilla> (called |q> in the implementation) is a 3-qubit ancilla
register.
This implementation is using the "Select V" algorithm from
https://arxiv.org/abs/1711.10980. See Section G.4. and Figure 6.
Note: everything has been unrolled because I did not have the time to write
and debug the generic version of the algorithm. Sorry for this.
"""
x = QuantumRegister(4) # Index
qboard = QuantumRegister(16)
q = QuantumRegister(3)
circuit = QuantumCircuit(x, qboard, q, name="QRAM")
# Trick to have the same indices as in the paper
x = [None, *x[::-1]]
q = [1, x[1], *q]
# First, x[0] positive
circuit.rccx(q[1], x[2], q[2])
circuit.rccx(q[2], x[3], q[3])
circuit.rccx(q[3], x[4], q[4])
# q[4] is |1> only if the index was |1111>
circuit.append(encode_problem(problem_set[0b1111]), [*qboard, q[4]])
# Short-circuit (red)
circuit.cx(q[3], q[4])
# q[4] is |1> only if the index was |1110>
circuit.append(encode_problem(problem_set[0b1110]), [*qboard, q[4]])
# Short-circuit (green)
circuit.cx(q[3], q[4])
circuit.rccx(q[2], x[4], q[4])
circuit.cx(q[2], q[3])
# q[4] is |1> only if the index was |1101>
circuit.append(encode_problem(problem_set[0b1101]), [*qboard, q[4]])
# Short-circuit (red)
circuit.cx(q[3], q[4])
# q[4] is |1> only if the index was |1100>
circuit.append(encode_problem(problem_set[0b1100]), [*qboard, q[4]])
# Go up
circuit.x(x[4])
circuit.rccx(q[3], x[4], q[4])
circuit.x(x[4]).inverse()
# Short-circuit (green)
circuit.cx(q[2], q[3])
circuit.rccx(q[1], x[3], q[3])
circuit.cx(q[1], q[2])
# Go down
circuit.rccx(q[3], x[4], q[4])
# q[4] is |1> only if the index was |1011>
circuit.append(encode_problem(problem_set[0b1011]), [*qboard, q[4]])
# Short-circuit (red)
circuit.cx(q[3], q[4])
# q[4] is |1> only if the index was |1010>
circuit.append(encode_problem(problem_set[0b1010]), [*qboard, q[4]])
# Short-circuit (green)
circuit.cx(q[3], q[4])
circuit.rccx(q[2], x[4], q[4])
circuit.cx(q[2], q[3])
# q[4] is |1> only if the index was |1001>
circuit.append(encode_problem(problem_set[0b1001]), [*qboard, q[4]])
# Short-circuit (red)
circuit.cx(q[3], q[4])
# q[4] is |1> only if the index was |1000>
circuit.append(encode_problem(problem_set[0b1000]), [*qboard, q[4]])
# Go up
circuit.x(x[4])
circuit.rccx(q[3], x[4], q[4])
circuit.x(x[4]).inverse()
# Go up
circuit.x(x[3])
circuit.rccx(q[2], x[3], q[3])
circuit.x(x[3]).inverse()
# Short-circuit (green)
# Small trick here: we are supposed to use q[0] but q[0] is "virtual"
# and always equal to "1", so we can simplify.
circuit.cx(q[1], q[2])
circuit.cx(x[2], q[2]) # circuit.rccx(q[0], x[2], q[2])
circuit.x(q[1]) # circuit.cx(q[0], q[1])
# Go down
circuit.rccx(q[2], x[3], q[3])
# Go down
circuit.rccx(q[3], x[4], q[4])
# q[4] is |1> only if the index was |0111>
circuit.append(encode_problem(problem_set[0b0111]), [*qboard, q[4]])
# Short-circuit (red)
circuit.cx(q[3], q[4])
# q[4] is |1> only if the index was |0110>
circuit.append(encode_problem(problem_set[0b0110]), [*qboard, q[4]])
# Short-circuit (green)
circuit.cx(q[3], q[4])
circuit.rccx(q[2], x[4], q[4])
circuit.cx(q[2], q[3])
# q[4] is |1> only if the index was |0101>
circuit.append(encode_problem(problem_set[0b0101]), [*qboard, q[4]])
# Short-circuit (red)
circuit.cx(q[3], q[4])
# q[4] is |1> only if the index was |0100>
circuit.append(encode_problem(problem_set[0b0100]), [*qboard, q[4]])
# Go up
circuit.x(x[4])
circuit.rccx(q[3], x[4], q[4])
circuit.x(x[4]).inverse()
# Short-circuit (green)
circuit.cx(q[2], q[3])
circuit.rccx(q[1], x[3], q[3])
circuit.cx(q[1], q[2])
# Go down
circuit.rccx(q[3], x[4], q[4])
# q[4] is |1> only if the index was |0011>
circuit.append(encode_problem(problem_set[0b0011]), [*qboard, q[4]])
# Short-circuit (red)
circuit.cx(q[3], q[4])
# q[4] is |1> only if the index was |0010>
circuit.append(encode_problem(problem_set[0b0010]), [*qboard, q[4]])
# Short-circuit (green)
circuit.cx(q[3], q[4])
circuit.rccx(q[2], x[4], q[4])
circuit.cx(q[2], q[3])
# q[4] is |1> only if the index was |0001>
circuit.append(encode_problem(problem_set[0b0001]), [*qboard, q[4]])
# Short-circuit (red)
circuit.cx(q[3], q[4])
# q[4] is |1> only if the index was |0001>
circuit.append(encode_problem(problem_set[0b0000]), [*qboard, q[4]])
# Go up
circuit.x(x[4])
circuit.rccx(q[3], x[4], q[4])
circuit.x(x[4]).inverse()
# Go up
circuit.x(x[3])
circuit.rccx(q[2], x[3], q[3])
circuit.x(x[3]).inverse()
# Go up
circuit.x(x[2])
circuit.rccx(q[1], x[2], q[2])
circuit.x(x[2]).inverse()
# To avoid changing x[1]:
circuit.x(x[1])
return circuit
```

The code including this optimisation is available on this Gist.

The new call-graph is the following:

Waow! The QRAM is now only taking ~13% of the total cost of the
circuit. Compared to the `check_if_match_any_permutation`

subroutine
that takes more than 85% of the total cost, I can say that the
QRAM optimisation has succeeded! At this point, I decided to put
my efforts into optimising the oracle instead of the QRAM because
of the cost difference.

In the
next blog post
I will explain how I optimised the very last subroutine that
contributes greatly to the overall cost:
`check_if_match_any_permutation`

.