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Quadratic Unconstrained Binary Optimization
Quadratic Unconstrained Binary Optimization
Published: February 28, 2024. Last updated: November 05, 2024.
Solving combinatorial optimization problems using quantum computing is one of those promising applications for the near term. But, why are combinatorial optimization problems even important? We care about them because, in fields such as logistics, finance, and engineering, there exist useful applications that can be translated into combinatorial optimization problems. But useful applications are not enough to justify the use of quantum devices. It is here where the second ingredient comes in—many combinatorial optimization problems are difficult to solve! Finding good solutions (classically) for large instances of them requires an enormous amount of computational resources and time 😮💨.
In this demo, we will be using the quantum approximate optimization algorithm (QAOA) and quantum annealing (QA) to solve a combinatorial optimization problem. First, we show how to translate combinatorial optimization problems into the quadratic unconstrained binary optimization (QUBO) formulation. In the first part of this notebook, we will show how to encode the Knapsack problem as a target Hamiltonian and solve it using the optimization-free version of QAOA and QA on D-Wave Advantage quantum annealer.
Combinatorial Optimization Problems
Combinatorial optimization problems involve finding the best way to arrange a set of objects or values to achieve a specific goal. The word ‘combinatorial’ refers to the fact that we are dealing with combinations of objects, while ‘optimization’ means that we are trying to find the best possible arrangement of them.
Let’s start with a basic example. Imagine we have 5 items ⚽️, 💻, 📸, 📚, and 🎸 and we would love to bring all of them with us. Unfortunately, our knapsack does not fit all of them 😔. So we need to find the best way to bring the most important items with us.
This is an example of the infamous Knapsack Problem. From our problem statement, we know that we need to maximize the value of the most important items. So we need to assign a value based on the importance the items have to us:
items_values = {"⚽️": 8, "💻": 47, "📸": 10, "📚": 5, "🎸": 16}
values_list = [8, 47, 10, 5, 16]
Additionally, we know that we the knapsack has limited space. For simplicity, let’s assume there is a limit to the weight it can hold. So we need to assign an estimate of the weight of each item:
items_weight = {"⚽️": 3, "💻": 11, "📸": 14, "📚": 19, "🎸": 5}
weights_list = [3, 11, 14, 19, 5]
Finally, we need to know the maximum weight we can bring in the knapsack:
maximum_weight = 26
Now we have well-defined optimization problem to work with. Let’s start with the easiest way to solve it, i.e., by trying all possible combinations of the items. But the number of combinations is equal to \(2^n\) where \(n\) is the number of items. Why is this the case? For each item, we have two options—“1” if we bring the item and “0” otherwise. With 2 options for each item and 5 items to choose from, we have \(2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 2^5 = 32\) combinations in our case. For each of these cases, we calculate the sum of the values and the sum of the weights, selecting the one that fulfills the maximum weight constraint and has the largest sum of values (this is the optimization step). Now, let’s write some code to solve the Knapsack problem with this brute-force method!
import numpy as np
def sum_weight(bitstring, items_weight):
weight = 0
for n, i in enumerate(items_weight):
if bitstring[n] == "1":
weight += i
return weight
def sum_values(bitstring, items_value):
value = 0
for n, i in enumerate(items_value):
if bitstring[n] == "1":
value += i
return value
items = list(items_values.keys())
n_items = len(items)
combinations = {}
max_value = 0
for case_i in range(2**n_items): # all possible options
combinations[case_i] = {}
bitstring = np.binary_repr(
case_i, n_items
) # bitstring representation of a possible combination, e.g, "01100" in our problem means bringing (-💻📸--)
combinations[case_i]["items"] = [items[n] for n, i in enumerate(bitstring) if i == "1"]
combinations[case_i]["value"] = sum_values(bitstring, values_list)
combinations[case_i]["weight"] = sum_values(bitstring, weights_list)
# save the information of the optimal solution (the one that maximizes the value while respecting the maximum weight)
if (
combinations[case_i]["value"] > max_value
and combinations[case_i]["weight"] <= maximum_weight
):
max_value = combinations[case_i]["value"]
optimal_solution = {
"items": combinations[case_i]["items"],
"value": combinations[case_i]["value"],
"weight": combinations[case_i]["weight"],
}
print(
f"The best combination is {optimal_solution['items']} with a total value: {optimal_solution['value']} and total weight {optimal_solution['weight']} "
)
The best combination is ['⚽️', '💻', '🎸'] with a total value: 71 and total weight 19
That was easy, right? But what if we have larger cases like 10, 50, or 100? Just to see how this scales, suppose it takes 1 ns to try one case.
def time_to_solution(n, time_single_case):
"""
n (int): number of variables
time_single_case (float): time to solve a single case
"""
return time_single_case * 2 ** n
time_per_case = 1e-9 # time to execute a single case in seconds
sec_day = 3600 * 24 # seconds in a day
sec_year = sec_day * 365 # seconds in a year
print(
f"- For 10 items, 2^10 cases, we need {time_to_solution(2, time_per_case)} seconds."
)
print(
f"- For 50 items, 2^50 cases, we need {round(time_to_solution(50, time_per_case) / sec_day)} days."
)
print(
f"- For 100 items, 2^100 cases, we need {round(time_to_solution(100, time_per_case) / sec_year)} years."
)
- For 10 items, 2^10 cases, we need 4e-09 seconds.
- For 50 items, 2^50 cases, we need 13 days.
- For 100 items, 2^100 cases, we need 40196936841331 years.
Guess we don’t have the time to try all the possible solutions for 100 items 😅! Thankfully, we don’t need to — there are algorithms to find good solutions to combinatorial optimization problems, and maybe one day we will show that one of these algorithms is quantum. So let’s continue with our quest 🫡.
Our next step is to represent our problem mathematically. First, we represent our items by binary variables \(x_i\) that take the value \(1\) if we bring the \(i\)-th item and \(0\) otherwise. Next, we know that we want to maximize the value of the items carried, so let’s create a function \(f(\mathrm{x})\) with these characteristics. To do so, we assign the variables \(x_i\) to each of the items \(\mathrm{x} = \{x_0:⚽️ , x_1:💻, x_2:📸, x_3:📚, x_4:🎸\},\) multiply each variable by the corresponding item value, and define a function that calculates the weighted sum value of the item:
This function, called the objective function, represents the value of the items we can
transport. Usually, solvers minimize functions, so a simple trick in our case is to minimize the
negative of our function (which ends up being maximizing our original function 🤪)
We can write our equation above using the general form of the QUBO representation, i.e., using an upper triangular matrix \(Q \in \mathbb{R}^{n \ \mathrm{x} \ n}:\)
where \(\mathrm{x}\) is a vector representing the items of our problem. Note that \(x_i x_i = x_i\) for binary variables. Let’s look at an example of how to calculate the \(\mathrm{x}^TQ \mathrm{x}\) above:
Q = -np.diag(list(items_values.values())) # Matrix Q for the problem above.
x_opt = np.array(
[[1 if i in optimal_solution["items"] else 0] for i in items_values.keys()]
) # Optimal solution.
opt_str = "".join(str(i[0]) for i in x_opt)
min_cost = (x_opt.T @ Q @ x_opt)[0, 0] # using Equation 3 above
print(f"Q={Q}")
print(f"The minimum cost is {min_cost}")
Q=[[ -8 0 0 0 0]
[ 0 -47 0 0 0]
[ 0 0 -10 0 0]
[ 0 0 0 -5 0]
[ 0 0 0 0 -16]]
The minimum cost is -71
But just with this function, we cannot solve the problem. We also need the weight restriction. Based on our variables, the weight list (items_weight = {“⚽️”: \(3\), “💻”: \(11\), “📸”: \(14\), “📚”: \(19,\) “🎸”: \(5\)}), and the knapsack maximum weight (maximum_weight \(W = 26\)), we can construct our constraint
Here comes a crucial step in the solution of the problem: we need to find a way to combine our objective function with this inequality constraint. One common method is to include the constraint as a penalty term in the objective function. This penalty term should be zero when the total weight of the items is less or equal to 26 and large otherwise. So to make them zero in the range of validity of the constraint, the usual approach is to use slack variables. There is an alternative method that has shown to perform better, called unbalanced penalization , but we present this method later 😉.
The slack variable is an auxiliary variable that allows us to convert inequality constraints into equality constraints. The slack variable \(S\) represents the amount by which the left-hand side of the inequality falls short of the right-hand side. If the left-hand side is less than the right-hand side, then \(S\) will be positive and equal to the difference between the two sides. In our case
where \(0 \le S \le 26.\) But let’s take this slowly because we can get lost here, so let’s see this with some examples:
Imagine this case. No item is selected {\(x_0\): \(0\), \(x_1\): \(0\), \(x_2:\) \(0,\) \(x_3\): \(0,\) \(x_4:\) \(0\)}, so the overall weight is zero (a valid solution) and the equality constraint Eq.(4) must be fulfilled. So we select our slack variable to be 26.
Now, what if we bring ⚽️ and 📚 {\(x_0\): \(1\), \(x_1\): \(0\), \(x_2\): \(0\), \(x_3:\) \(1,\) \(x_4:\) (a valid solution) and the equality constraint is fulfilled if \(22 + S = 26 \rightarrow S = 4.\)
Finally, what if we try to bring all the items {\(x_0\): \(1\), \(x_1\): \(1\), \(x_2:\) \(1,\) \(x_3\): \(1,\) \(x_4:\) \(1\)}, the total weight, in this case, is \(3+11+14+19+5=52\) (not a valid solution), to fulfill the constraint, we need \(52 + S = 26 \rightarrow S=-26\) but the slack variable is in the range \((0,26)\) in our definition, so, in this case, there is valid solution for \(S.\)
Excellent, now we have a way to represent the inequality constraint. Two further steps are needed. First, the slack variable has to be represented in binary form so we can cast it as a sum
where \(N = \lfloor\log_2(\max S)\rfloor + 1.\) In our case \(N = \lfloor\log_2(26)\rfloor + 1 = 5.\) We need 5 binary variables to represent the range of our \(S\) variable.
To compact our equation later, let’s rename our slack variables by \(s_0=x_5\), \(s_1=x_6,\) \(s_3=x_7\), \(s_4=x_8,\) and \(s_5=x_9.\) Then we have
For example, if we need to represent the second case above (⚽️, 📚), \(S = 4 \rightarrow\{x_5:0, x_6:0,x_7:1,x_8:0, x_9:0\}.\)
We are almost done in our quest to represent our problem in such a way that our quantum computer can manage it. The last step is to add the penalty term, a usual choice for it is to use a quadratic penalty
Note that this is simply the difference between the left- and right-hand sides of equation \((4).\) With this expression, the condition is satisfied only when the term inside the parentheses is zero. \(\lambda\) is a penalty coefficient that we must tune to make that the constraint will always be fulfilled.
Now, the objective function can be given by:
or, compacted,
where \(v_i\) and \(w_i\) are the value and weight of the \(i\)-th item. Because of the square in the second term, some \(x_i x_i\) terms show up. We can apply the property \(x_i x_i = x_i\) (if \(x_i = 0 \rightarrow x_ix_i = 0\cdot0 = 0\) or \(x_i = 1 \rightarrow x_ix_i = 1\cdot1 = 1\)).
The quadratic term on the right-hand side of equation \((7)\) can be rewritten as
where \(w_i\) represent the weights for the items and \(2^k\) for the slack variables. We can combine equations \((7)\) and \((8)\) to get the terms of the matrix \(Q.\) So we end up with
The term \(\lambda W^2\) is only an offset value that does not affect the optimization result and can be added after the optimization to represent the right cost. Let’s see how it looks like in our particular example.
N = round(np.ceil(np.log2(maximum_weight))) # number of slack variables
weights = list(items_weight.values()) + [2**k for k in range(N)]
QT = np.pad(Q, ((0, N), (0, N))) # adding the extra slack variables at the end of the Q matrix
n_qubits = len(QT)
lambd = 2 # We choose a lambda parameter enough large for the constraint to always be fulfilled
# Adding the terms for the penalty term
for i in range(len(QT)):
QT[i, i] += lambd * weights[i] * (weights[i] - 2 * maximum_weight) # Eq. 10
for j in range(i + 1, len(QT)):
QT[i, j] += 2 * lambd * weights[i] * weights[j] # Eq. 9
offset = lambd * maximum_weight**2
print(f"Q={QT}")
# optimal string slack string
slack_string = np.binary_repr(maximum_weight - optimal_solution["weight"], N)[::-1]
x_opt_slack = np.concatenate(
(x_opt, np.array([[int(i)] for i in slack_string]))
) # combining the optimal string and slack string
opt_str_slack = "".join(str(i[0]) for i in x_opt_slack)
cost = (x_opt_slack.T @ QT @ x_opt_slack)[0, 0] + offset # Optimal cost using equation 3
print(f"Cost:{cost}")
# At this point, we have encoded the problem in a format that we can use to solve it on quantum
# computers. Now it only remains to solve it using quantum algorithms!
Q=[[ -302 132 168 228 60 12 24 48 96 192]
[ 0 -949 616 836 220 44 88 176 352 704]
[ 0 0 -1074 1064 280 56 112 224 448 896]
[ 0 0 0 -1259 380 76 152 304 608 1216]
[ 0 0 0 0 -486 20 40 80 160 320]
[ 0 0 0 0 0 -102 8 16 32 64]
[ 0 0 0 0 0 0 -200 32 64 128]
[ 0 0 0 0 0 0 0 -384 128 256]
[ 0 0 0 0 0 0 0 0 -704 512]
[ 0 0 0 0 0 0 0 0 0 -1152]]
Cost:-71
QAOA
We use QAOA [1] to find the solution to our Knapsack Problem (read this demo for a more detailed explanation of the QAOA algorithm). In this case, the cost Hamiltonian, \(H_c(Z),\) obtained from the QUBO formulation is translated into a parametric unitary gate given by
where \(\gamma_i \in {1,..., p}\) is a set of \(p\) parameters to be optimized, the term \(e^{-i\gamma_i J_{ij}Z_iZ_j}\) is implemented in a quantum circuit using a \(RZZ(2\gamma_iJ_{ij})\) gate, and \(e^{-i\gamma_i h_iZ_i}\) using a \(RZ(2\gamma_ih_i)\) gate.
The mixer operator applied is
where \(\beta_i\) is the second parameter that must be optimized and \(X = \sum_{i=1}^n \sigma_i^x\) with \(\sigma_i^x\) the Pauli-\(X\) matrix. We implement Eq. \((12)\) with \(R_X(-2\beta_i) = e^{i \beta_i \sigma_x}\) gates applied to each qubit. We repeat this sequence of gates \(p\) times.
# ----------------------------- QAOA circuit ------------------------------------
from collections import defaultdict
import pennylane as qml
shots = 5000 # Number of samples used
dev = qml.device("default.qubit", shots=shots)
@qml.qnode(dev)
def qaoa_circuit(gammas, betas, h, J, num_qubits):
wmax = max(
np.max(np.abs(list(h.values()))), np.max(np.abs(list(h.values())))
) # Normalizing the Hamiltonian is a good idea
p = len(gammas)
# Apply the initial layer of Hadamard gates to all qubits
for i in range(num_qubits):
qml.Hadamard(wires=i)
# repeat p layers the circuit shown in Fig. 1
for layer in range(p):
# ---------- COST HAMILTONIAN ----------
for ki, v in h.items(): # single-qubit terms
qml.RZ(2 * gammas[layer] * v / wmax, wires=ki[0])
for kij, vij in J.items(): # two-qubit terms
qml.CNOT(wires=[kij[0], kij[1]])
qml.RZ(2 * gammas[layer] * vij / wmax, wires=kij[1])
qml.CNOT(wires=[kij[0], kij[1]])
# ---------- MIXER HAMILTONIAN ----------
for i in range(num_qubits):
qml.RX(-2 * betas[layer], wires=i)
return qml.sample()
def samples_dict(samples, n_items):
"""Just sorting the outputs in a dictionary"""
results = defaultdict(int)
for sample in samples:
results["".join(str(i) for i in sample)[:n_items]] += 1
return results
The second thing we must consider is the initialization of the \(\beta_i\) and \(\gamma_i\) parameters and the subsequent classical optimization of these parameters. Alternatively, we can think of QAOA as a Trotterization of the quantum adiabatic algorithm. We start in the ground state \(|+\rangle ^{\otimes n}\) of the mixer Hamiltonian \(X\) and move to the ground state of the cost Hamiltonian \(H_c\) slowly enough to always be close to the ground state of the Hamiltonian. How slow? In our case the rate is determined by the number of layers \(p.\) We can adopt this principle and initialize the \(\beta_i\) and \(\gamma_i\) in this way, moving \(\beta_i\) from \(1\) to \(0\) and \(\gamma_i\) from \(0\) to \(1.\) With this approach, we can skip the optimization part in QAOA.
import matplotlib.pyplot as plt
# Annealing schedule for QAOA
betas = np.linspace(0, 1, 10)[::-1] # Parameters for the mixer Hamiltonian
gammas = np.linspace(0, 1, 10) # Parameters for the cost Hamiltonian (Our Knapsack problem)
fig, ax = plt.subplots()
ax.plot(betas, label=r"$\beta_i$", marker="o", markersize=8, markeredgecolor="black")
ax.plot(gammas, label=r"$\gamma_i$", marker="o", markersize=8, markeredgecolor="black")
ax.set_xlabel("i", fontsize=18)
ax.legend()
fig.show()
This Figure shows the annealing schedule we will use in our QAOA protocol. The y-axis represents the angle in radians and the x-axis represents the i-th layer of QAOA, from \(0\) to \(9\) for a total of \(p=10\) layers.
I know this is a lot of information so far, but we are almost done! The last step to represent the QUBO problem on QPUs is to change the \(x_i\in \{0, 1\}\) variables to spin variables \(z_i \in \{1, -1\}\) via the transformation \(x_i = (1 - z_i) / 2.\) We also want to set the penalty term, so a value of \(\lambda = 2\) will be enough for our problem. In practice, we choose a value for \(\lambda\) and, if after the optimization the solution does not fulfill the constraints, we try again using a larger value. On the other hand, if the solution is suspected to be a valid but suboptimal, then we will reduce \(\lambda\) a little. Eq.(3) can be represented by an Ising Hamiltonian with quadratic and linear terms plus a constant \(O,\) namely
Here, \(J_{ij}\) are interaction terms and \(h_i\) are linear terms, all of them depending on the combinatorial optimization problem.
def from_Q_to_Ising(Q, offset):
"""Convert the matrix Q of Eq.3 into Eq.13 elements J and h"""
n_qubits = len(Q) # Get the number of qubits (variables) in the QUBO matrix
# Create default dictionaries to store h and pairwise interactions J
h = defaultdict(int)
J = defaultdict(int)
# Loop over each qubit (variable) in the QUBO matrix
for i in range(n_qubits):
# Update the magnetic field for qubit i based on its diagonal element in Q
h[(i,)] -= Q[i, i] / 2
# Update the offset based on the diagonal element in Q
offset += Q[i, i] / 2
# Loop over other qubits (variables) to calculate pairwise interactions
for j in range(i + 1, n_qubits):
# Update the pairwise interaction strength (J) between qubits i and j
J[(i, j)] += Q[i, j] / 4
# Update the magnetic fields for qubits i and j based on their interactions in Q
h[(i,)] -= Q[i, j] / 4
h[(j,)] -= Q[i, j] / 4
# Update the offset based on the interaction strength between qubits i and j
offset += Q[i, j] / 4
# Return the magnetic fields, pairwise interactions, and the updated offset
return h, J, offset
def energy_Ising(z, h, J, offset):
"""
Calculate the energy of an Ising model given spin configurations.
Parameters:
- z: A dictionary representing the spin configurations for each qubit.
- h: A dictionary representing the magnetic fields for each qubit.
- J: A dictionary representing the pairwise interactions between qubits.
- offset: An offset value.
Returns:
- energy: The total energy of the Ising model.
"""
if isinstance(z, str):
z = [(1 if int(i) == 0 else -1) for i in z]
# Initialize the energy with the offset term
energy = offset
# Loop over the magnetic fields (h) for each qubit and update the energy
for k, v in h.items():
energy += v * z[k[0]]
# Loop over the pairwise interactions (J) between qubits and update the energy
for k, v in J.items():
energy += v * z[k[0]] * z[k[1]]
# Return the total energy of the Ising model
return energy
# Our previous example should give us the same result
z_exp = [
(1 if i == 0 else -1) for i in x_opt_slack
] # Converting the optimal solution from (0,1) to (1, -1)
h, J, zoffset = from_Q_to_Ising(QT, offset) # Eq.13 for our problem
energy = energy_Ising(
z_exp, h, J, zoffset
) # Caluclating the energy (Should be the same that for the QUBO)
print(f"Minimum energy:{energy}")
samples_slack = samples_dict(qaoa_circuit(gammas, betas, h, J, num_qubits=len(QT)), n_qubits)
values_slack = {
sum_values(sample_i, values_list): count
for sample_i, count in samples_slack.items()
if sum_weight(sample_i, weights_list) <= maximum_weight
} # saving only the solutions that fulfill the constraint
print(
f"The number of optimal solutions using slack variables is {samples_slack[opt_str_slack]} out of {shots}"
)
Minimum energy:-71.0
The number of optimal solutions using slack variables is 22 out of 5000
As you can see, only a few samples from the 5000 shots give us the right answer, there are only \(2^5 = 32\) options. Randomly guessing the solution will give us on average \(5000/32 \approx 156\) optimal solutions. Why don’t we get good results using QAOA? Maybe we can blame the algorithm or we look deeper— it turns out our encoding method is really bad. Randomly guessing using the whole set of variables (\(5\) items + \(5\) slack) \(2^{10} = 1024\) options, \(5000/1024 \approx 5.\) So in fact we have a tiny improvement.
Unbalanced penalization (An alternative to slack variables)
Unbalanced penalization is a function characterized by a larger penalty when the inequality constraint is not achieved than when it is. So we have to modify Eq. 7 to include a linear term in the following way:
where \(\lambda_{1,2}\) are again penalty coefficients. Here [2] and [3] some details about unbalanced penalization. The method is already implemented in OpenQAOA and D-Wave Ocean so we don’t have to code it ourselves. The cliffnotes are that you don’t need slack variables for the inequality constraints anymore using this approach.
from openqaoa.problems import FromDocplex2IsingModel
from docplex.mp.model import Model
def Knapsack(values, weights, maximum_weight):
"""Create a docplex model of the problem. (Docplex is a classical solver from IBM)"""
n_items = len(values)
mdl = Model()
x = mdl.binary_var_list(range(n_items), name="x")
cost = -mdl.sum(x[i] * values[i] for i in range(n_items))
mdl.minimize(cost)
mdl.add_constraint(mdl.sum(x[i] * weights[i] for i in range(n_items)) <= maximum_weight)
return mdl
# Docplex model, we need to convert our problem in this format to use the unbalanced penalization approach
mdl = Knapsack(values_list, weights_list, maximum_weight)
lambda_1, lambda_2 = (
0.96,
0.0371,
) # Parameters of the unbalanced penalization function (They are in the main paper)
ising_hamiltonian = FromDocplex2IsingModel(
mdl,
unbalanced_const=True,
strength_ineq=[lambda_1, lambda_2], # https://arxiv.org/abs/2211.13914
).ising_model
h_new = {
tuple(i): w for i, w in zip(ising_hamiltonian.terms, ising_hamiltonian.weights) if len(i) == 1
}
J_new = {
tuple(i): w for i, w in zip(ising_hamiltonian.terms, ising_hamiltonian.weights) if len(i) == 2
}
samples_unbalanced = samples_dict(
qaoa_circuit(gammas, betas, h_new, J_new, num_qubits=n_items), n_items
)
values_unbalanced = {
sum_values(sample_i, values_list): count
for sample_i, count in samples_unbalanced.items()
if sum_weight(sample_i, weights_list) <= maximum_weight
} # saving only the solutions that fulfill the constraint
print(
f"The number of solutions using unbalanced penalization is {samples_unbalanced[opt_str]} out of {shots}"
)
-- cannot find parameters matching version: , using: 22.1.1.0
-- cannot find parameters matching version: , using: 22.1.1.0
The number of solutions using unbalanced penalization is 1947 out of 5000
We have improved the QAOA solution by encoding our QUBO wisely, with almost 2000 out of the 5000 samples being the optimal solution. Below, we compare the two different methods to encode the problem. The x-axis is the value of the items we bring based on the optimization (the larger the better) and the y-axis is the number of samples with that value (in log scale to observe the slack variables approach). In this sense, QAOA is pointing to the optimal and suboptimal solutions.
fig, ax = plt.subplots()
ax.hist(
values_unbalanced.keys(),
weights=values_unbalanced.values(),
bins=50,
edgecolor="black",
label="unbalanced",
align="right",
)
ax.hist(
values_slack.keys(),
weights=values_slack.values(),
bins=50,
edgecolor="black",
label="slack",
align="left",
)
ax.vlines(-min_cost, 0, 3000, linestyle="--", color="black", label="Optimal", linewidth=2)
ax.set_yscale("log")
ax.legend()
ax.set_ylabel("counts")
ax.set_xlabel("values")
fig.show()
Quantum Annealing Solution
Quantum annealing is a process that exploits quantum mechanical effects to find low energy states of Ising Hamiltonians. We will use the quantum annealer D-Wave Advantage, a quantum computing system developed by D-Wave Systems Inc that has more than 5000 qubits.
from dwave.system import DWaveSampler, EmbeddingComposite
from dwave.cloud import Client
import dimod
import pandas as pd
bqm = {}
# BQM - Binary Quadratic Model
# This creates the QUBO model of our Knapsack problem using the slack variables approach
# offset is the constant term in our QUBO formulation
# ----------- SLACK METHOD -----------
bqm["slack"] = dimod.BQM.from_qubo(QT, offset=lambd * maximum_weight**2)
bqm["slack"].relabel_variables({i: f"x_{i}" for i in range(bqm["slack"].num_variables)})
# ----------- UNBALANCED METHOD -----------
lagrange_multiplier = [0.96, 0.0371] # Again values from the paper
bqm["unbalanced"] = dimod.BQM.from_qubo(Q) # This adds the objective function to the model
bqm["unbalanced"].add_linear_inequality_constraint(
[(n, i) for n, i in enumerate(weights_list)], # This adds the constraint
lagrange_multiplier,
"unbalanced",
ub=maximum_weight,
penalization_method="unbalanced",
)
bqm["unbalanced"].relabel_variables({i: f"x_{i}" for i in range(bqm["unbalanced"].num_variables)})
# If you have an account you can execute the following code, otherwise read the file.
account = False
df = {}
if account:
# Replace with your client information
sampler = DWaveSampler(region="eu-central-1")
sampler_qpu = EmbeddingComposite(sampler)
for method in ["slack", "unbalanced"]:
samples = sampler_qpu.sample(bqm[method], num_reads=5000) # Executing on real hardware
df[method] = (
samples.to_pandas_dataframe().sort_values("energy").reset_index(drop=True)
) # Converting the sampling information and sort it by cost
df[method].to_json(f"QUBO/dwave_results_{method}.json") # save the results
else:
df = {}
for method in ["slack", "unbalanced"]:
df[method] = pd.read_json(f"QUBO/dwave_results_{method}.json")
# Loading the data from an execution on D-Wave Advantage
samples_dwave = {}
values = {}
for method in ["slack", "unbalanced"]:
samples_dwave[method] = defaultdict(int)
for i, row in df[method].iterrows():
# Postprocessing the information
sample_i = "".join(str(round(row[q])) for q in bqm[method].variables)
samples_dwave[method][sample_i] += row["num_occurrences"]
values[method] = {
sum_values(sample_i, values_list): count
for sample_i, count in samples_dwave[method].items()
if sum_weight(sample_i, weights_list) <= maximum_weight
}
The histogram below shows the results of both encodings on D-Wave Advantage. Once again, we prove that depending on the encoding method for our problem, we get good or bad results.
fig, ax = plt.subplots()
bins = {"unbalanced": 5, "slack": 40}
for method in ["unbalanced", "slack"]:
ax.hist(
values[method].keys(),
weights=values[method].values(),
bins=bins[method],
edgecolor="black",
label=method,
align="right",
)
ax.vlines(-min_cost, 0, 5000, linestyle="--", color="black", label="Optimal", linewidth=2)
ax.set_yscale("log")
ax.legend()
ax.set_ylabel("counts")
ax.set_xlabel("value")
fig.show()
Conclusion
We have come to the end of this demo. We have covered the definition of combinatorial optimization problems and how to formulate one of them, the Knapsack Problem, using QUBO, and two different encodings: slack variables and unbalanced penalization. Then, we solved them using optimization-free QAOA and QA. Now, it’s your turn to experiment with QAOA! If you need some inspiration:
Look at the OpenQAOA set of problems. There are plenty of them like bin packing, traveling salesman, and maximal independent set, among others.
Play around with larger problems.
References
[1] Farhi, E., Goldstone, J., & Gutmann, S. (2014). A Quantum Approximate Optimization Algorithm. http://arxiv.org/abs/1411.4028
[2] Montanez-Barrera, A., Willsch, D., A., Maldonado-Romo, & Michielsen, K. (2022). Unbalanced penalization: A new approach to encode inequality constraints of combinatorial problems for quantum optimization algorithms. http://arxiv.org/abs/2211.13914
[3] Montanez-Barrera, J. A., Heuvel, P. van den, Willsch, D., & Michielsen, K. (2023). Improving Performance in Combinatorial Optimization Problems with Inequality Constraints: An Evaluation of the Unbalanced Penalization Method on D-Wave Advantage. https://doi.org/10.1109/QCE57702.2023.00067
About the author
Alejandro Montanez
Alejandro is a postdoctoral researcher at the Jülich Supercomputer Center (JSC) at Forschungszentrum Jülich. At JSC, he contributes to benchmarking, large-scale simulation, and the development of applications that use quantum computing.
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