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This tutorial solves the max clique problem in graph theory using Classiq. A clique is a subset of vertices in a graph such that each pair is adjacent to one other. Given a graph G=(V,E)G = (V,E), find the maximal clique in the graph. It is known to be in the NP-hard complexity class.

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Defining the Optimization Problem

Encode each node as a binary variable: 
import networkx as nx
import numpy as np
import pyomo.environ as pyo


def define_max_clique_model(graph):
    model = pyo.ConcreteModel()

    # each x_i states if node i belongs to the cliques
    model.x = pyo.Var(graph.nodes, domain=pyo.Binary)
    x_variables = np.array(list(model.x.values()))

    # define the complement adjacency matrix as the matrix where 1 exists for each non-existing edge
    adjacency_matrix = nx.convert_matrix.to_numpy_array(graph, nonedge=0)
    complement_adjacency_matrix = (
        1
        - nx.convert_matrix.to_numpy_array(graph, nonedge=0)
        - np.identity(len(model.x))
    )

    # constraint that 2 nodes without edge in the graph cannot be chosen together
    model.clique_constraint = pyo.Constraint(
        expr=x_variables @ complement_adjacency_matrix @ x_variables == 0
    )

    # maximize the number of nodes in the chosen clique
    model.value = pyo.Objective(expr=sum(x_variables), sense=pyo.maximize)

    return model
Initialize the model with parameters: 
graph = nx.erdos_renyi_graph(7, 0.6, seed=79)
nx.draw_kamada_kawai(graph, with_labels=True)
max_clique_model = define_max_clique_model(graph)
output

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Setting Up the Classiq Problem Instance

To solve the Pyomo model defined above, use the CombinatorialProblem Python class. Under the hood, it translates the Pyomo model to a quantum model of the Quantum Approximate Optimization Algorithm (QAOA) [1], with a cost Hamiltonian translated from the Pyomo model. Choose the number of layers for the QAOA ansatz using the num_layers argument: 
from classiq import *
from classiq.applications.combinatorial_optimization import CombinatorialProblem

combi = CombinatorialProblem(pyo_model=max_clique_model, num_layers=3)

qmod = combi.get_model()

​
Synthesizing the QAOA Circuit and Solving the Problem

Synthesize and view the QAOA circuit (ansatz) used to solve the optimization problem: 
qprog = combi.get_qprog()
show(qprog)
Output:

Quantum program link: https://platform.classiq.io/circuit/38w9cHY6lrCyQi3RXfs4B8gw3Ys
Output:
https://platform.classiq.io/circuit/38w9cHY6lrCyQi3RXfs4B8gw3Ys?login=True&version=15
Set the quantum backend on which to execute: 
execution_preferences = ExecutionPreferences(
    backend_preferences=ClassiqBackendPreferences(backend_name="simulator"),
)
Solve the problem by calling the optimize method of the CombinatorialProblem object. For the classical optimization part of the QAOA algorithm, define the maximum number of classical iterations (maxiter) and the α\alpha-parameter (quantile) for running CVaR-QAOA, an improved variation of the QAOA algorithm [2]: 
optimized_params = combi.optimize(execution_preferences, maxiter=50, quantile=0.7)


import matplotlib.pyplot as plt

plt.plot(combi.cost_trace)
plt.xlabel("Iterations")
plt.ylabel("Cost")
plt.title("Cost convergence")
Output:

Text(0.5, 1.0, 'Cost convergence')
output

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Viewing the Optimization Results

Examine the statistics of the algorithm. The optimization is always defined as a minimization problem, so the Pyomo-to-Qmod translator changes the positive maximization objective to negative minimization. To get samples with the optimized parameters, call the sample method: 
optimization_result = combi.sample(combi.optimized_params)
optimization_result.sort_values(by="cost").head(5)
solutionprobabilitycost
37{‘x’: [1, 1, 1, 1, 0, 0, 0]}0.006836-4
52{‘x’: [0, 1, 1, 1, 0, 1, 0]}0.004395-4
44{‘x’: [1, 0, 0, 1, 1, 0, 0]}0.005371-3
34{‘x’: [0, 0, 0, 1, 1, 1, 0]}0.007812-3
50{‘x’: [1, 1, 0, 0, 0, 0, 1]}0.004883-3
Compare the optimized results to uniformly sampled results: 
uniform_result = combi.sample_uniform()
And compare the histograms: 
optimization_result["cost"].plot(
    kind="hist",
    bins=40,
    edgecolor="black",
    weights=optimization_result["probability"],
    alpha=0.6,
    label="optimized",
)
uniform_result["cost"].plot(
    kind="hist",
    bins=40,
    edgecolor="black",
    weights=uniform_result["probability"],
    alpha=0.6,
    label="uniform",
)
plt.legend()
plt.ylabel("Probability", fontsize=16)
plt.xlabel("cost", fontsize=16)
plt.tick_params(axis="both", labelsize=14)
output Plot the solution: 
best_solution = optimization_result.solution[optimization_result.cost.idxmin()]
best_solution
Output:
{'x': [1, 1, 1, 1, 0, 0, 0]}
  


solution_nodes = [v for v in graph.nodes if best_solution["x"][v]]
solution_edges = [
    (u, v) for u, v in graph.edges if u in solution_nodes and v in solution_nodes
]
nx.draw_kamada_kawai(graph, with_labels=True)
nx.draw_kamada_kawai(
    graph,
    with_labels=True,
    nodelist=solution_nodes,
    edgelist=solution_edges,
    node_color="r",
    edge_color="r",
)
output

​
Comparing to a Classical Solver

Lastly, compare to the classical solution of the problem: 
from pyomo.opt import SolverFactory

solver = SolverFactory("couenne")
solver.solve(max_clique_model)
classical_solution = [
    int(pyo.value(max_clique_model.x[i])) for i in range(len(max_clique_model.x))
]
print("Classical solution:", classical_solution)
Output:

Classical solution: [1, 1, 1, 1, 0, 0, 0]
  


solution = [int(pyo.value(max_clique_model.x[i])) for i in graph.nodes]
solution_nodes = [v for v in graph.nodes if solution[v]]
solution_edges = [
    (u, v) for u, v in graph.edges if u in solution_nodes and v in solution_nodes
]
nx.draw_kamada_kawai(graph, with_labels=True)
nx.draw_kamada_kawai(
    graph,
    with_labels=True,
    nodelist=solution_nodes,
    edgelist=solution_edges,
    node_color="r",
    edge_color="r",
)
output

​
References

[1] Farhi, Edward, Jeffrey Goldstone, and Sam Gutmann. (2014). A quantum approximate optimization algorithm. arXiv preprint arXiv:1411.4028. [2] Barkoutsos, Panagiotis Kl, et al. (2020). Improving variational quantum optimization using CVaR. Quantum 4: 256.