Skip to content

Number Partition Problem

Introduction

In the Number Partitioning Problem [1] we need to find how to partition a set of integers into two subsets of equal sums. In case such a partition does not exist, we can ask for a partition where the difference between the sums is minimal.

Mathematical formulation

Given a set of numbers \(S=\{s_1,s_2,...,s_n\}\), a partition is defined as \(P_1,P_2 \subset \{1,...,n\}\), with \(P_1\cup P_2=\{1,...,n\}\) and \(P_1\cap P_2=\emptyset\). In the Number Partitioning Problem we need to determine a partition such that \(|\sum_{j\in P_1}s_j-\sum_{j\in P_2}s_j|\) is minimal. A partition can be represented by a binary vector \(x\) of size \(n\), where we assign 0 or 1 for being in \(P_1\) or \(P_2\), respectively. The quantity we ask to minimize is \(|\vec{x}\cdot \vec{s}-(1-\vec{x})\cdot\vec{s}|=|(2\vec{x}-1)\cdot\vec{s}|\). In practice we will minimize the square of this expression.

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:

# we define a matrix which gets a set of integers s and returns a pyomo model for the partitioning problem


def partite(s) -> pyo.ConcreteModel:
    model = pyo.ConcreteModel()
    SetSize = len(s)  # the set size
    model.x = pyo.Var(
        range(SetSize), domain=pyo.Binary
    )  # our variable is a binary vector

    # we define a cost function
    model.cost = pyo.Objective(
        expr=sum(((2 * model.x[i] - 1) * s[i]) for i in range(SetSize)) ** 2,
        sense=pyo.minimize,
    )

    return model
Myset = np.random.randint(1, 12, 10)
mylist = [int(x) for x in Myset]
print("This is my list: ", mylist)
set_partition_model = partite(mylist)
This is my list:  [4, 8, 11, 7, 1, 8, 8, 5, 7, 11]
set_partition_model.pprint()
1 Set Declarations
    x_index : Size=1, Index=None, Ordered=Insertion
        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 : ((2*x[0] - 1)*4 + (2*x[1] - 1)*8 + (2*x[2] - 1)*11 + (2*x[3] - 1)*7 + 2*x[4] - 1 + (2*x[5] - 1)*8 + (2*x[6] - 1)*8 + (2*x[7] - 1)*5 + (2*x[8] - 1)*7 + (2*x[9] - 1)*11)**2

3 Declarations: x_index x 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=set_partition_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="aer_simulator")
)

qmod = set_execution_preferences(qmod, backend_preferences)
from classiq import write_qmod

write_qmod(qmod, "set_partition")

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/0ff868cf-2ccd-47e7-b5f7-4500845af8c3?version=0.41.0.dev39%2B79c8fd0855

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

png

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(
    set_partition_model,
    vqe_result=vqe_result,
    penalty_energy=qaoa_config.penalty_energy,
)
optimization_result = pd.DataFrame.from_records(solution)
optimization_result.sort_values(by="cost", ascending=True).head(5)
probability cost solution count
525 0.001 0.0 [0, 1, 1, 1, 1, 0, 1, 0, 0, 0] 1
29 0.004 0.0 [1, 1, 1, 0, 0, 0, 0, 1, 1, 0] 4
511 0.001 0.0 [1, 1, 0, 0, 0, 1, 1, 0, 1, 0] 1
494 0.001 0.0 [0, 1, 0, 0, 0, 1, 1, 0, 0, 1] 1
424 0.001 0.0 [0, 0, 0, 1, 1, 1, 1, 0, 0, 1] 1

And the histogram:

optimization_result.hist("cost", weights=optimization_result["probability"])
array([[<Axes: title={'center': 'cost'}>]], dtype=object)

png

Let us plot the solution:

best_solution = optimization_result.solution[optimization_result.cost.idxmin()]
p1 = [mylist[i] for i in range(len(mylist)) if best_solution[i] == 0]
p2 = [mylist[i] for i in range(len(mylist)) if best_solution[i] == 1]
print("P1=", p1, ", total sum: ", sum(p1))
print("P2=", p2, ", total sum: ", sum(p2))
print("difference= ", abs(sum(p1) - sum(p2)))
P1= [7, 1, 8, 8, 11] , total sum:  35
P2= [4, 8, 11, 5, 7] , total sum:  35
difference=  0

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

from pyomo.opt import SolverFactory

solver = SolverFactory("couenne")
solver.solve(set_partition_model)

set_partition_model.display()
Model unknown

  Variables:
    x : Size=10, Index=x_index
        Key : Lower : Value : Upper : Fixed : Stale : Domain
          0 :     0 :   1.0 :     1 : False : False : Binary
          1 :     0 :   1.0 :     1 : False : False : Binary
          2 :     0 :   1.0 :     1 : False : False : Binary
          3 :     0 :   0.0 :     1 : False : False : Binary
          4 :     0 :   1.0 :     1 : False : False : Binary
          5 :     0 :   0.0 :     1 : False : False : Binary
          6 :     0 :   0.0 :     1 : False : False : Binary
          7 :     0 :   0.0 :     1 : False : False : Binary
          8 :     0 :   0.0 :     1 : False : False : Binary
          9 :     0 :   1.0 :     1 : False : False : Binary

  Objectives:
    cost : Size=1, Index=None, Active=True
        Key  : Active : Value
        None :   True :   0.0

  Constraints:
    None
classical_solution = [pyo.value(set_partition_model.x[i]) for i in range(len(mylist))]
p1 = [mylist[i] for i in range(len(mylist)) if classical_solution[i] == 0]
p2 = [mylist[i] for i in range(len(mylist)) if classical_solution[i] == 1]
print("P1=", p1, ", total sum: ", sum(p1))
print("P2=", p2, ", total sum: ", sum(p2))
print("difference= ", abs(sum(p1) - sum(p2)))
P1= [7, 8, 8, 5, 7] , total sum:  35
P2= [4, 8, 11, 1, 11] , total sum:  35
difference=  0

References

[1]: Number Partitioning Problem (Wikipedia)

[2]: Farhi, Edward, Jeffrey Goldstone, and Sam Gutmann. "A quantum approximate optimization algorithm." arXiv preprint arXiv:1411.4028 (2014).

[3]: Barkoutsos, Panagiotis Kl, et al. "Improving variational quantum optimization using CVaR." Quantum 4 (2020): 256.