Reputation: 321
I have three equations like the following ones:
How can I find the values of x, y, and z with Java?
String equation1="x+y+z=100;";
String equation2="x+y-z=50;";
String equation3="x-y-z=10;";
int[] SolveEquations(equation1,equation2,equation3) {
// to do
// how to do?
}
Do you have any possible solutions or other common frameworks?
Upvotes: 15
Views: 107690
Reputation: 672
try this, please:
import org.apache.commons.math3.linear.*;
import org.junit.Assert;
import org.junit.Test;
/**
* Author: Andrea Ciccotta
*/
public class LinearSystemTest extends Assert {
/**
* Ax = B
* 2x + 3y - 2z = 1
* -x + 7y + 6x = -2
* 4x - 3y - 5z = 1
* <p>
* it will use the LUDecomposition:
* LU decomposition:
* 1. find A = LU where LUx = B
* 2. solve Ly = B
* 4. solve Ux = y
*/
@Test
public void linearSystem3x3Test() {
final RealMatrix coefficients = new Array2DRowRealMatrix(new double[][]{{2, 3, -2}, {-1, 7, 6}, {4, -3, -5}});
final DecompositionSolver solver = new LUDecomposition(coefficients).getSolver();
final RealVector constants = new ArrayRealVector(new double[]{1, -2, 1}, false);
final RealVector solution = solver.solve(constants);
final double[] arraySolution = solution.toArray();
assertEquals(arraySolution[0], -0.36986301369863006, 0);
assertEquals(arraySolution[1], 0.1780821917808219, 0);
assertEquals(arraySolution[2], -0.6027397260273972, 0);
}
}
Upvotes: 2
Reputation: 11
There are many ways to solve linear system equations. There is a simplest way to perform this. In the example the java code solve for two variables USING Matrix method but you can modify to perform 3 variables calculations.
import java.util.Scanner; //OBJETO SCANNER
public class SYS2 {
public static void main (String args[]) {
//VARIABLE DECLARATION SPACE
int i=0,j = 0;
float x,y;
Scanner S = new Scanner (System.in);
int EC3[][]= new int [2][3]; //ARRAY TO STORE EQUATION 2X3
float DET1=0;
float DET2 =0;
float DETA=0;
float DETB=0;
//END VARIABLE DECLARATIONS
System.out.println("Enter Equation System : ");
for (i=0; i< 2; i++) {
for (j=0; j< 3; j++)
EC3[i][j] = S.nextInt();
}
System.out.println("SISTEMA DE ECUACION LINEAL: "); //THIS SENTENCE ONLY PRINT THE CATCHED VALUES OF EQUATION
for (i=0; i< 2; i++) {
for (j=0; j< 3; j++)
System.out.print(EC3[i][j] + " ");
System.out.println();
}
// System.out.print("Determinante A de la Matriz: ");
// System.out.print((EC3[0][2] * EC3[1][1]) - (EC3[0][1]*EC3[1][2]) );
for (i=0;i<2;i++) {
for (j=0; j<2;j++)
DET1= ((EC3[0][2] * EC3[1][1]) -( EC3[0][1]*EC3[1][2]));
}
// System.out.print(DET1 );
// System.out.println();
for (i=0;i<2;i++) {
for (j=0; j<2;j++)
DET2= ((EC3[0][0] * EC3[1][1]) - (EC3[0][1]*EC3[1][0]));
}
// System.out.print("Determinante B de la Matriz: ");
// System.out.println(DET2 );
x = (DET1 / DET2);
System.out.println();
System.out.println("X = " + x);
System.out.print("=======================");
//FIN PARA VALOR DE X
//COMIENZO DE VALOR DE Y
// System.out.print("Determinante A de la Matriz Y: ");
for (i=0;i<2;i++) {
for (j=0; j<2;j++)
DETA= EC3[0][0] * EC3[1][2] - EC3[0][2]*EC3[1][0];
// System.out.print(DETA );
// System.out.println();
}
for (i=0;i<2;i++) {
for (j=0; j<2;j++)
DETB= EC3[0][0] * EC3[1][1] - EC3[0][1]*EC3[1][0];
}
// System.out.print("Determinante B de la Matriz Y: ");
// System.out.println(DETB );
y = DETA / DETB;
System.out.print("=======================");
System.out.println();
System.out.println("Y = " + y);
System.out.print("=======================");
}
}
Upvotes: 1
Reputation: 309008
Since you're writing Java, you can use the JAMA package to solve this. I'd recommend a good LU decomposition method.
It's a simple linear algebra problem. You should be able to solve it by hand or using something like Excel pretty easily. Once you have that you can use the solution to test your program.
There's no guarantee, of course, that there is a solution. If your matrix is singular, that means there is no intersection of those three lines in 3D space.
Upvotes: 12
Reputation: 101
you can use the java matrix package JAMA. See the full page of this example below here
/*
*Solving three variable linear equation system
* 3x + 2y - z = 1 ---> Eqn(1)
* 2x - 2y + 4z = -2 ---> Eqn(2)
* -x + y/2- z = 0 ---> Eqn(3)
*/
import Jama.Matrix;
import java.lang.Math.*;
public class Main {
public Main() {
//Creating Arrays Representing Equations
double[][] lhsArray = {{3, 2, -1}, {2, -2, 4}, {-1, 0.5, -1}};
double[] rhsArray = {1, -2, 0};
//Creating Matrix Objects with arrays
Matrix lhs = new Matrix(lhsArray);
Matrix rhs = new Matrix(rhsArray, 3);
//Calculate Solved Matrix
Matrix ans = lhs.solve(rhs);
//Printing Answers
System.out.println("x = " + Math.round(ans.get(0, 0)));
System.out.println("y = " + Math.round(ans.get(1, 0)));
System.out.println("z = " + Math.round(ans.get(2, 0)));
}
public static void main(String[] args) {
new Main();
}
}
Upvotes: 10
Reputation: 34421
Create a parser using ANTLR. Then evaluate the AST using Gaussian elimination.
Upvotes: 4
Reputation: 11669
You can also use Commons Math. They have a section of this in their userguide (see 3.4)
Upvotes: 5
Reputation: 4913
Use Gaussian_elimination it's incredibly easy, but there are some values you may have hard life calculating.
Upvotes: 2
Reputation: 30573
You can use determinant to calculate values of x y and z. Logic can be found out here http://www.intmath.com/Matrices-determinants/1_Determinants.php
And then you need to implement it in java using 3 dimensional arrays.
Upvotes: 12