Calculate Values of Variables Found in an Interval with Java - constraint-programming

Question

I am trying to find the values x and y may take so the following inequalities hold:
1/24 < 1/15*y < 1/10*x < 2/24 < 2/15*y < 3/24
Is there a way to formulate such a problem in Java?
Constraint Programming would probably solve such a problem but is there an alternative way?
If Constraint Programming is the only way, how does this look like?
The following is what I tried with constraint programming using or-tools. How to formulate strict inequalities?
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);

Answer 1

Here is a working code using the integer solver
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)
With scale = 1000, it outputs:
x = 0.418
y = 0.626
With scale = 100, it outputs:
x = 0.43
y = 0.63
With scale = 10, it outputs
x = 0.5
y = 0.7

Answer 2

I found the solution by writing down a loop that produces random values until all the statements are fulfilled.
Now I am interested in how wolfram alpha solves such problems so quickly.
public class inequalities {
private static double x;
private static double y;
private static double Ratio3 = 1/24.0;
private static double Ratio2 = 1/15.0;
private static double Ratio1 = 1/10.0;
public static void main(String[] args) {
x = Math.random();
y = Math.random();
boolean loop = true;
while (loop) {
loop = calculatingTheInequalities();
if (loop) {
x = Math.random();
y = Math.random();
}
}
System.out.println("x value: " + x);
System.out.println("y value: " + y);
}
public static boolean calculatingTheInequalities() {
if (Ratio3<Ratio2*y && Ratio2*y<Ratio1*x &&
Ratio1*x<2*Ratio3 && 2*Ratio3<2*Ratio2*y &&
2*Ratio2*y<3*Ratio3) {
return false;
} else {
return true;
}
/*if (Ratio3 < Ratio2 *y) {
if (Ratio2 *y < Ratio1 *x) {
if (Ratio1 *x<2* Ratio3) {
if (2* Ratio3 < 2* Ratio2 *y) {
if (2* Ratio2 *y < 3* Ratio3) {
return false;
} else {
return true;
}
} else {
return true;
}
} else {
return true;
}
} else {
return true;
}
} else {
return true;
}*/
}
}