Max K-Vertex Cover

Introduction

The Max K-Vertex Cover problem [1] is a classical problem in graph theory and computer science, where we aim to find a set of vertices such that each edge of the graph is incident to at least one vertex in the set, and the size of the set does not exceed a given number $$k$$.

Mathematical Formulation

The Max-k Vertex Cover problem can be formulated as an Integer Linear Program (ILP):

Minimize: $$\sum_{(i,j) \in E} (1 - x_i)(1 - x_j)$$

Subject to: $$\sum_{i \in V} x_i = k$$

and $$x_i \in \{0, 1\} \quad \forall i \in V$$

Where:

• $$x_i$$ is a binary variable that equals 1 if node $$i$$ is in the cover and 0 otherwise

• $$E$$ is the set of edges in the graph

• $$V$$ is the set of vertices in the graph

• $$k$$ is the maximum number of vertices allowed in the cover

Solving with the Classiq platform

We go through the steps of solving the problem with the Classiq platform, using QAOA algorithm [2]. The solution is based on defining a pyomo model for the optimization problem we would like to solve.

from typing import cast

import networkx as nx
import numpy as np
import pyomo.core as pyo
from IPython.display import Markdown, display
from matplotlib import pyplot as plt


Building the Pyomo model from a graph input

We proceed by defining the pyomo model that will be used on the Classiq platform, using the mathematical formulation defined above:

import networkx as nx
import pyomo.core as pyo

def mvc(graph: nx.Graph, k: int) -> pyo.ConcreteModel:
model = pyo.ConcreteModel()
model.x = pyo.Var(graph.nodes, domain=pyo.Binary)
model.amount_constraint = pyo.Constraint(expr=sum(model.x.values()) == k)

def obj_expression(model):
# number of edges not covered
return sum((1 - model.x[i]) * (1 - model.x[j]) for i, j in graph.edges)

model.cost = pyo.Objective(rule=obj_expression, sense=pyo.minimize)

return model


The model contains:

• Index set declarations (model.Nodes, model.Arcs).

• Binary variable declaration for each node (model.x) indicating whether the variable is chosen for the set.

• Constraint rule – ensures that the set is of size k.

• Objective rule – counts the number of edges not covered; i.e., both related variables are zero.

import networkx as nx

K = 5
num_nodes = 10
p_edge = 0.5
graph = nx.erdos_renyi_graph(n=num_nodes, p=p_edge, seed=13)

mvc_model = mvc(graph, K)


mvc_model.pprint()

1 Set Declarations
x_index : Size=1, Index=None, Ordered=False
Key  : Dimen : Domain : Size : Members
None :     1 :    Any :   10 : {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

1 Var Declarations
x : Size=10, Index=x_index
Key : Lower : Value : Upper : Fixed : Stale : Domain
0 :     0 :  None :     1 : False :  True : Binary
1 :     0 :  None :     1 : False :  True : Binary
2 :     0 :  None :     1 : False :  True : Binary
3 :     0 :  None :     1 : False :  True : Binary
4 :     0 :  None :     1 : False :  True : Binary
5 :     0 :  None :     1 : False :  True : Binary
6 :     0 :  None :     1 : False :  True : Binary
7 :     0 :  None :     1 : False :  True : Binary
8 :     0 :  None :     1 : False :  True : Binary
9 :     0 :  None :     1 : False :  True : Binary

1 Objective Declarations
cost : Size=1, Index=None, Active=True
Key  : Active : Sense    : Expression
None :   True : minimize : (1 - x[0])*(1 - x[1]) + (1 - x[0])*(1 - x[5]) + (1 - x[0])*(1 - x[6]) + (1 - x[0])*(1 - x[7]) + (1 - x[0])*(1 - x[8]) + (1 - x[1])*(1 - x[2]) + (1 - x[1])*(1 - x[4]) + (1 - x[1])*(1 - x[5]) + (1 - x[1])*(1 - x[6]) + (1 - x[1])*(1 - x[9]) + (1 - x[2])*(1 - x[3]) + (1 - x[2])*(1 - x[4]) + (1 - x[3])*(1 - x[6]) + (1 - x[3])*(1 - x[8]) + (1 - x[4])*(1 - x[6]) + (1 - x[4])*(1 - x[7]) + (1 - x[4])*(1 - x[8]) + (1 - x[4])*(1 - x[9]) + (1 - x[5])*(1 - x[6]) + (1 - x[7])*(1 - x[8]) + (1 - x[8])*(1 - x[9])

1 Constraint Declarations
amount_constraint : Size=1, Index=None, Active=True
Key  : Lower : Body                                                                : Upper : Active
None :   5.0 : x[0] + x[1] + x[2] + x[3] + x[4] + x[5] + x[6] + x[7] + x[8] + x[9] :   5.0 :   True

4 Declarations: x_index x amount_constraint cost


Setting Up the Classiq Problem Instance

In order to solve the Pyomo model defined above, we use the Classiq combinatorial optimization engine. For the quantum part of the QAOA algorithm (QAOAConfig) - define the number of repetitions (num_layers):

from classiq import construct_combinatorial_optimization_model
from classiq.applications.combinatorial_optimization import OptimizerConfig, QAOAConfig

qaoa_config = QAOAConfig(num_layers=3)


For the classical optimization part of the QAOA algorithm we define the maximum number of classical iterations (max_iteration) and the $$\alpha$$-parameter (alpha_cvar) for running CVaR-QAOA, an improved variation of the QAOA algorithm [3]:

optimizer_config = OptimizerConfig(max_iteration=60, alpha_cvar=0.7)


Lastly, we load the model, based on the problem and algorithm parameters, which we can use to solve the problem:

qmod = construct_combinatorial_optimization_model(
pyo_model=mvc_model,
qaoa_config=qaoa_config,
optimizer_config=optimizer_config,
)


We also set the quantum backend we want to execute on:

from classiq import set_execution_preferences
from classiq.execution import ClassiqBackendPreferences, ExecutionPreferences

backend_preferences = ExecutionPreferences(
backend_preferences=ClassiqBackendPreferences(backend_name="simulator")
)

qmod = set_execution_preferences(qmod, backend_preferences)

from classiq import write_qmod

write_qmod(qmod, "max_k_vertex_cover")


Synthesizing the QAOA Circuit and Solving the Problem

We can now synthesize and view the QAOA circuit (ansatz) used to solve the optimization problem:

from classiq import show, synthesize

qprog = synthesize(qmod)
show(qprog)

Opening: https://platform.classiq.io/circuit/3d753f60-bd5e-40a9-9fcd-fd07cf4536e8?version=0.45.0.dev0%2Bcf1b9b7ccc


We now solve the problem by calling the execute function on the quantum program we have generated:

from classiq import execute

res = execute(qprog).result()


We can check the convergence of the run:

from classiq.execution import VQESolverResult

vqe_result = res[0].value
vqe_result.convergence_graph


Optimization Results

We can also examine the statistics of the algorithm:

import pandas as pd

from classiq.applications.combinatorial_optimization import (
get_optimization_solution_from_pyo,
)

solution = get_optimization_solution_from_pyo(
mvc_model, vqe_result=vqe_result, penalty_energy=qaoa_config.penalty_energy
)
optimization_result = pd.DataFrame.from_records(solution)

probability cost solution count
0 0.061 1.0 [1, 1, 0, 0, 1, 0, 1, 0, 1, 0] 61
3 0.022 1.0 [1, 1, 0, 1, 1, 0, 0, 0, 1, 0] 22
44 0.006 2.0 [1, 1, 0, 1, 1, 1, 0, 0, 1, 0] 6
32 0.007 2.0 [1, 1, 0, 0, 1, 1, 0, 0, 1, 0] 7
25 0.008 2.0 [1, 1, 0, 1, 1, 0, 0, 1, 0, 0] 8

And the histogram:

optimization_result.hist("cost", weights=optimization_result["probability"])

array([[<Axes: title={'center': 'cost'}>]], dtype=object)


Let us plot the solution:

best_solution = optimization_result.solution[optimization_result.cost.idxmin()]

best_solution

[1, 1, 0, 0, 1, 0, 1, 0, 1, 0]

def draw_solution(graph: nx.Graph, solution: list):
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 or v in solution_nodes
]
graph,
nodelist=solution_nodes,
edgelist=solution_edges,
node_color="r",
edge_color="y",
)

draw_solution(graph, best_solution)


Comparison to a classical solver

Lastly, we can compare to the classical solution of the problem:

from pyomo.opt import SolverFactory

solver = SolverFactory("couenne")
solver.solve(mvc_model)
classical_solution = [int(pyo.value(mvc_model.x[i])) for i in graph.nodes]

classical_solution

[1, 1, 0, 1, 1, 0, 0, 0, 1, 0]

draw_solution(graph, classical_solution)