java解多元一次方程组 结果值是区间 返回多种结果 所有结果都大于0 demo 代码

在多元一次方程组中，如果需要返回的是结果值的区间，可以将其转换为线性规划问题，然后使用线性规划的方法来求解。以下是一个简单的Java代码演示如何使用线性规划来解多元一次方程组，并返回所有结果都大于0的多种结果。

import org.apache.commons.math3.optim.linear.LinearConstraint;
import org.apache.commons.math3.optim.linear.LinearObjectiveFunction;
import org.apache.commons.math3.optim.linear.Relationship;
import org.apache.commons.math3.optim.linear.SimplexSolver;

import java.util.ArrayList;
import java.util.List;

public class LinearProgrammingSolver {
    public static List<double[]> solve(double[][] A, double[] b) {
        int n = A[0].length;
        LinearObjectiveFunction f = new LinearObjectiveFunction(new double[n], 0);
        List<LinearConstraint> constraints = new ArrayList<>();
        for (int i = 0; i < A.length; i++) {
            double[] coefficients = new double[n];
            System.arraycopy(A[i], 0, coefficients, 0, n);
            constraints.add(new LinearConstraint(coefficients, Relationship.LEQ, b[i]));
        }
        SimplexSolver solver = new SimplexSolver();
        double[] x0 = new double[n];
        Arrays.fill(x0, 1);
        double minValue = Double.NEGATIVE_INFINITY;
        for (int i = 0; i < n; i++) {
            double[] c = new double[n];
            c[i] = -1;
            LinearObjectiveFunction f1 = new LinearObjectiveFunction(c, 0);
            double[] x = solver.optimize(f1, constraints).getPoint().toArray();
            if (x != null && isValidSolution(x)) {
                double value = computeValue(A, b, x);
                if (value > minValue) {
                    minValue = value;
                    x0 = x;
                }
            }
        }
        List<double[]> solutions = new ArrayList<>();
        solutions.add(x0);
        for (int i = 0; i < n; i++) {
            double[] c = new double[n];
            c[i] = -1;
            LinearObjectiveFunction f1 = new LinearObjectiveFunction(c, 0);
            double[] x = solver.optimize(f1, constraints).getPoint().toArray();
            if (x != null && isValidSolution(x) && !isEqual(x, x0)) {
                solutions.add(x);
            }
        }
        return solutions;
    }

    private static boolean isValidSolution(double[] x) {
        for (double value : x) {
            if (value < 0) {
                return false;
            }
        }
        return true;
    }

    private static boolean isEqual(double[] x, double[] y) {
        for (int i = 0; i < x.length; i++) {
            if (Math.abs(x[i] - y[i]) > 1e-6) {
                return false;
            }
        }
        return true;
    }

    private static double computeValue(double[][] A, double[] b, double[] x) {
        double value = 0;
        for (int i = 0; i < A.length; i++) {
            double sum = 0;
            for (int j = 0; j < A[i].length; j++) {
                sum += A[i][j] * x[j];
            }
            value += Math.min(sum, b[i]);
        }
        return value;
    }

    public static void main(String[] args) {
        double[][] A = {{1, 2, 3}, {4, 5, 6}, {7, 8, 10}};
        double[] b = {6, 15, 27};
        List<double[]> solutions = solve(A, b);
        if (solutions.isEmpty()) {
            System.out.println("No solution");
        } else {
            for (double[] x : solutions) {
                System.out.println(Arrays.toString(x));
            }
        }
    }
}

在上面的代码中，我们首先将多元一次方程组转换为线性规划问题，然后使用线性规划的方法来求解。我们首先求解目标函数为每个变量的系数都为-1的线性规划问题，得到一个最小值。然后，我们依次求解每个变量的系数为-1的线性规划问题，并找到所有最小值等于原问题最小值的解。最后，我们将所有的解返回，并确保它们都大于0。

需要注意的是，在线性规划问题中，通常会存在多个最优解，因此返回的结果可能不是唯一的。