/*                        ESc101N Laboratory Assignment Solution
                     		 Wednesday 18/8/04
PROBLEM: Write a program (with one class and that too having only the 
"main" method) to compute the solution of system of linear eqautions 
with 3 variables using Gaussian Elimination. You may assume that a 
unique solution exists

*/


import javabook.*;
import java.awt.*;
import java.lang.*;

class gaussian
{
	public static void main(String[] args)
	{
		int a11,a12,a13,a21,a22,a23,a31,a32,a33,c1,c2,c3;    
		MainWindow mainwindow=new MainWindow();
		
		InputBox inputbox=new InputBox(mainwindow);    
		OutputBox outputbox=new OutputBox(mainwindow);  
		
		
		// Take each coefficient as input one by one from the user 
		// by invoking the getInteger() method of InputBox class. 
		// Error message will be displayed if the value entered 
		// cannot be converted to an int.
		
		a11=inputbox.getInteger("Please Enter a11");
		a12=inputbox.getInteger("Please Enter a12");
		a13=inputbox.getInteger("Please Enter a13");
		c1=inputbox.getInteger("Please Enter c1");
		a21=inputbox.getInteger("Please Enter a21");
		a22=inputbox.getInteger("Please Enter a22");
		a23=inputbox.getInteger("Please Enter a23");
		c2=inputbox.getInteger("Please Enter c2");
		a31=inputbox.getInteger("Please Enter a31");
		a32=inputbox.getInteger("Please Enter a32");
		a33=inputbox.getInteger("Please Enter a33");
		c3=inputbox.getInteger("Please Enter c3");
		
		/*The three equations in x,y and z are
		(a11)x + (a12)y + (a13)z = c1  ... (1)
		(a21)x + (a22)y + (a23)z = c2  ... (2)
		(a31)x + (a32)y + (a33)z = c3  ... (3)


		Reduce eqns (1)and (2)to linear equations in 2 variables using the following steps :-
		a) Multipy eqn(1) by (a21)
			(a21) * [(a11)x + (a12)y + (a13)z] = (a21) * c1  ... (4)

		b) Multipy eqn(2) by (a21)
			(a11) * [(a21)x + (a22)y + (a23)z] = (a11) * c2  ... (5)

		c) Subtract eqn(5) from eqn(4)
			[(a21)*(a12) - (a11)*(a22)]y + [(a21)*(a13)-(a11)*(a23)]z = (a21)*c1 - (a11)*c2  ... (6)
			

		Similarly reduce, eqns (2) and (3) to linear equations in 2 variables to get the following eqn. :
			[(a31)*(a22) - (a21)*(a32)]y + [(a31)*(a23)-(a21)*(a33)]z = (a31)*c2 - (a21)*c2 ... (7)
			
		Equation (6) can be written as:
			(l1)y + (m1)z = n1   .....(8)
			where l1=(a21)*(a12) - (a11)*(a22)
			      m1=(a21)*(a13)-(a11)*(a23)
			      n1=(a21)*c1 - (a11)*c2
		Similiarly (7) can be written as:
			(l2)y + (m2)z = n2   .....(9)
			where l2=(a31)*(a22) - (a21)*(a32)
			      m2=(a31)*(a23)-(a21)*(a33)
			      n2=(a31)*c2 - (a21)*c3
			     
		The variables l1,l2,m1,m2,n1,n2 can be taken as float to avoid truncation while division.
		*/
		
		float l1,l2,m1,m2,n1,n2;
		l1=(a21)*(a12) - (a11)*(a22);
		l2=(a31)*(a22) - (a21)*(a32);
		m1=(a21)*(a13)-(a11)*(a23);
		m2=(a31)*(a23)-(a21)*(a33);
		n1=(a21)*c1 - (a11)*c2;
		n2=(a31)*c2 - (a21)*c3;
		
		/*
		z can be calculated by reducing equations (8) and (9) to linear equations in one variable z by following steps:
		a)  Multiplying equation (8) by l2 and equation (9) by l1
			l2[(l1)y + (m1)z] = n1*l2 ....  (10)
			l1[(l2)y + (m2)z] = n2*l1 ....  (11)
			
		b)  Substracting (10) from (11) and solving for z:
			z[l2*m1-l1*m2] = n1*l2 - n2*l1
			
		  So, z= (n1*l2 - n2*l1)/(l2*m1-l1*m2)
		*/
			
		float x,y,z;
		z=(n1*l2-n2*l1)/(l2*m1-l1*m2);
		
		//by putting the value of z in equation (8) y can be calculated.
		
		y=(n1-m1*z)/l1;
		
		//by putting the values of y and z in equation (1) x can be calculated.
		x=(c1-a12*y-a13*z)/a11;
		
		
		//To show the output invoke setVisible() method of OutputBox with an argument true.
		outputbox.setVisible(true);
		
		//Print the equations entered by the user 
		outputbox.printLine("You have entered the following equations:\n");
		outputbox.printLine(a11+"x + "+a12+"y + "+a13+"z = "+c1);
		outputbox.printLine(a21+"x + "+a22+"y + "+a23+"z = "+c2);
		outputbox.printLine(a31+"x + "+a32+"y + "+a33+"z = "+c3);
		
		//print the solution i.e. values of x,y and z
		outputbox.printLine("\nThe solution of the equations is:\n");
		outputbox.printLine("x = "+x);
		outputbox.printLine("y = "+y);
		outputbox.printLine("z = "+z);
	}
}