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constraint-programming - 使用 Java 计算在区间中找到的变量的值

问题描述

我试图找到x和y可能采用的值，因此以下不等式成立：

1/24 < 1/15*y < 1/10*x < 2/24 < 2/15*y < 3/24

有没有办法在Java中制定这样的问题？约束编程可能会解决这样的问题，但有替代方法吗？

如果约束编程是唯一的方法，这看起来如何？

以下是我使用or-tools进行约束编程的尝试。如何制定严格的不等式？

MPSolver solver = new MPSolver(
            "SimpleMipProgram", MPSolver.OptimizationProblemType.CBC_MIXED_INTEGER_PROGRAMMING);
    // [END solver]

    // [START variables]
    double infinity = java.lang.Double.POSITIVE_INFINITY;
    // x and y are float/double variables.
    MPVariable x = solver.makeNumVar(0,1,"x"); //makeIntVar(0.0, infinity, "x");
    MPVariable y = solver.makeNumVar(0,1,"y"); //makeIntVar(0.0, infinity, "y");

    System.out.println("Number of variables = " + solver.numVariables());
    // [END variables]

    // [START constraints]
    // x + 7 * y <= 17.5.
    /*MPConstraint c0 = solver.makeConstraint(-1, 17.5, "c0");
    c0.setCoefficient(x, 1);
    c0.setCoefficient(y, 7);

    // x <= 3.5.
    MPConstraint c1 = solver.makeConstraint(-infinity, 3.5, "c1");
    c1.setCoefficient(x, 1);
    c1.setCoefficient(y, 0);*/

    // 1/24 < 1/15*y ---> -1/15 * y < -1/24
    MPConstraint c0 = solver.makeConstraint(-1000,-1/24.0,"c0");
    c0.setCoefficient(y,-1/15.0);

    //  1/15*y < 1/10*x ---> 1/15*y - 1/10*x < 0
    MPConstraint c1 = solver.makeConstraint(-1000,0,"c1");
    c1.setCoefficient(y,1/15.0);
    c1.setCoefficient(x,-1/10.0);

    // 1/10*x < 2/24 ---> 1/10*x < 2/24
    MPConstraint c2 = solver.makeConstraint(-1000,2/24.0,"c2");
    c2.setCoefficient(x,1/10.0);

    // 2/24 < 2/15*y ---> -2/15*y < -2/24
    MPConstraint c3 = solver.makeConstraint(-1000, -2/24.0);
    c3.setCoefficient(y,-2/15.0);

    // 2/15*y < 3/24 ---> 2/15*y < 3/24
    MPConstraint c4 = solver.makeConstraint(-1000,3/24.0);
    c4.setCoefficient(y,2/15.0);

标签： constraint-programmingor-toolsinequality

解决方案

这是使用整数求解器的工作代码

from __future__ import absolute_import
from __future__ import division
from __future__ import print_function

from ortools.sat.python import cp_model


model = cp_model.CpModel()
scale = 1000

x = model.NewIntVar(0, scale, 'x')
y = model.NewIntVar(0, scale, 'y')

# 1/24 < 1/15*y < 1/10*x < 2/24 < 2/15*y < 3/24

model.Add(5 * scale < 8 * y)
model.Add(8 * y < 12 * x)
model.Add(12 * x < 10 * scale)
model.Add(10 * scale < 16 * y)
model.Add(16 * y < 15 * scale)

solver = cp_model.CpSolver()
solver.parameters.log_search_progress = True
status = solver.Solve(model)

if status == cp_model.FEASIBLE:
    print('x =', solver.Value(x) * 1.0 / scale)
    print('y =', solver.Value(y) * 1.0 / scale)

当 scale = 1000 时，它输出：

x = 0.418
y = 0.626

使用 scale = 100，它输出：

x = 0.43
y = 0.63

scale = 10，它输出

x = 0.5
y = 0.7

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