Integer Linear Programming (ILP)

Integer Linear Programming (ILP) can be converted into a QUBO expression using PyQBPP. As an example, consider the following ILP:

\[\begin{aligned} \text{Maximize:} && 2x_0 +5x_1+5x_2\\ \text{Subject to:} && x_0 + 3 x_1 + x_2 &\leq 12 \\ && x_0 + 2x_2 &\leq 5\\ && x_1 + x_2 &\leq 4; \end{aligned}\]

PyQBPP program

import pyqbpp as qbpp

x = qbpp.between(qbpp.var_int("x", 3), 0, 5)
objective = 2 * x[0] + 5 * x[1] + 5 * x[2]
c1 = qbpp.between(x[0] + 3 * x[1] + x[2], 0, 12)
c2 = qbpp.between(x[0] + 2 * x[2], 0, 5)
c3 = qbpp.between(x[1] + x[2], 0, 4)

f = -objective + 100 * (c1 + c2 + c3)
f.simplify_as_binary()
solver = qbpp.EasySolver(f)
sol = solver.search({"time_limit": 1.0})
print(f"x0 = {sol(x[0])}, x1 = {sol(x[1])}, x2 = {sol(x[2])}")
print(f"objective = {sol(objective)}")
print(f"c1 = {sol(c1.body)}, c2 = {sol(c2.body)}, c3 = {sol(c3.body)}")

In this program, x is a vector of three integer variables, each taking a value in the range $[0, 5]$. The objective function and the three constraints are represented by objective, c1, c2, and c3.

The obtained solution is printed as follows:

x0 = 2, x1 = 3, x2 = 1
objective = 24
c1 = 12, c2 = 4, c3 = 4

Comparison with C++ QUBO++

| C++ QUBO++ | PyQBPP | |————————————|—————————————| | 0 <= x[0] + 3 * x[1] + x[2] <= 12 | between(x[0] + 3 * x[1] + x[2], 0, 12) | | sol(*c1) | sol(c1.body) |

整数線形計画法（ILP）

整数線形計画法（ILP） は、PyQBPPを使用してQUBO式に変換できます。 例として、以下のILPを考えます:

\[\begin{aligned} \text{Maximize:} && 2x_0 +5x_1+5x_2\\ \text{Subject to:} && x_0 + 3 x_1 + x_2 &\leq 12 \\ && x_0 + 2x_2 &\leq 5\\ && x_1 + x_2 &\leq 4; \end{aligned}\]

PyQBPPプログラム

import pyqbpp as qbpp

x = qbpp.between(qbpp.var_int("x", 3), 0, 5)
objective = 2 * x[0] + 5 * x[1] + 5 * x[2]
c1 = qbpp.between(x[0] + 3 * x[1] + x[2], 0, 12)
c2 = qbpp.between(x[0] + 2 * x[2], 0, 5)
c3 = qbpp.between(x[1] + x[2], 0, 4)

f = -objective + 100 * (c1 + c2 + c3)
f.simplify_as_binary()
solver = qbpp.EasySolver(f)
sol = solver.search({"time_limit": 1.0})
print(f"x0 = {sol(x[0])}, x1 = {sol(x[1])}, x2 = {sol(x[2])}")
print(f"objective = {sol(objective)}")
print(f"c1 = {sol(c1.body)}, c2 = {sol(c2.body)}, c3 = {sol(c3.body)}")

このプログラムでは、x は3つの整数変数のベクトルで、それぞれ $[0, 5]$ の範囲の値をとります。 目的関数と3つの制約は objective、c1、c2、c3 で表現されています。

得られた解は以下のように表示されます:

x0 = 2, x1 = 3, x2 = 1
objective = 24
c1 = 12, c2 = 4, c3 = 4

C++ QUBO++との比較

| C++ QUBO++ | PyQBPP | |————————————|—————————————| | 0 <= x[0] + 3 * x[1] + x[2] <= 12 | between(x[0] + 3 * x[1] + x[2], 0, 12) | | sol(*c1) | sol(c1.body) |