RegexThoughts

Equations to be Identified

y=...

A*sin(x...) + D

sin(x...)/A + D

(1/A)


Solve the problem by adding a check between the ampersands for a A` so we can look for

(A/78787)*sin(B*C)&/d& + D



.

 Coefficient part one
\*  connects the coefficient 
sin\(([^)\n]*)\) --sin and the middle gets us B and C
--second part after sin


It could be possible that we could try a different framework and reverse engineer the graph itself to get the features


	public static void main(String[] args){
		//TODO USE EXPRESSION THE EVALUATE A AND D
		
		String getCoeff = "y=(-?\\d*[\\./]?\\d*)\\*sin\\([^)\\n]*\\)([\\+-]\\d*[\\./]?\\d*)?";
		//String insideSIN = "sin\\(([^)\\n]*)\\)";
		//String getOffset = "sin\\([^)\\n]*\\)([\\+\\-]\\d*[\\./]?\\d*)";
		//String all = "(-?\\d*[\\./]?\\d*)\\*sin\\(([^)\\n]*)\\)";
		String[] test = {"y= sin( x)","y=-sin(4 * x)", "y= sin (x)", "y=-3 * sin(4*x+20)", "y=-4.3*sin(4*x+432)+9"," y=5 /3 * sin( x/3)-4"};
		
		for(int i= 0; i<test.length; i++){
			test[i] = test[i].replaceAll(" ", "");
			test[i] = test[i].replaceAll("-sin", "-1*sin");
			System.out.println("\nTest case "+i);
			System.out.println("Equation: " + test[i]);
			System.out.println("   Coeff: " + test[i].replaceAll(getCoeff, "____ $1 ____"));
			//System.out.println("    Sine: " + test[i].replaceAll(insideSIN, "____ $1 ____"));
			//System.out.println("  Offset: " + test[i].replaceAll(getOffset, "____ $1 ____"));
			//System.out.println("     All: " + test[i].replaceAll(all, "____ $1 ____ $2 ____"));
		}