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package symjava.examples;

import static symjava.math.SymMath.PI;

import static symjava.math.SymMath.exp;

import static symjava.math.SymMath.log;

import static symjava.math.SymMath.pow;

import static symjava.math.SymMath.sin;

import static symjava.math.SymMath.sqrt;

import static symjava.symbolic.Symbol.a;

import static symjava.symbolic.Symbol.b;

import static symjava.symbolic.Symbol.c;

import static symjava.symbolic.Symbol.d;

import static symjava.symbolic.Symbol.t;

import static symjava.symbolic.Symbol.x;

import static symjava.symbolic.Symbol.y;

import static symjava.symbolic.Symbol.z;

import symjava.bytecode.BytecodeFunc;

import symjava.domains.Domain;

import symjava.domains.Domain2D;

import symjava.domains.Domain3D;

import symjava.domains.DomainND;

import symjava.domains.Interval;

import symjava.relational.Ge;

import symjava.relational.Le;

import symjava.symbolic.Expr;

import symjava.symbolic.Integrate;

import symjava.symbolic.Symbol;

import symjava.symbolic.utils.JIT;

public class NumericalIntegration {

public static void main(String[] args) {

test_1D();

test_2D();

test_ND();

//Expr i = Integrate.apply(exp(pow(x,2)), Interval.apply(a, b).setStepSize(0.001));

//BytecodeFunc fi = JIT.compile(new Expr[]{a,b}, i);

//System.out.println(fi.apply(1,2));

test_paper_example1();

test_paper_example2();

}

public static void test_1D() {

//Define the interal

Domain I = Interval.apply(-10, 0).setStepSize(0.01);

//Define the integral: cumulative distribution function for standard normal distribution

Expr cdf = Integrate.apply(exp(-0.5*pow(x,2))/sqrt(2*PI), I);

System.out.println(cdf); //\int_{-10.0}^{10.0}{1/\sqrt{2*\pi}*e^{-0.5*x^2}}dx

//Compile cdf to perform numerical integration

BytecodeFunc f = JIT.compile(cdf);

System.out.println(f.apply()); //0.5

//Define the integral: cumulative distribution function for a general normal distribution

Symbol mu = new Symbol("\\mu");

Symbol sigma = new Symbol("\\sigma");

Expr cdf2 = Integrate.apply(exp(-pow(x-mu,2)/(2*sigma*sigma))/(sigma*sqrt(2*PI)), I);

System.out.println(cdf2); //\int_{-10.0}^{0.0}{1/(\sigma*\sqrt{2*\pi})*e^{-(-\mu + x)^2/(2*\sigma*\sigma)}}dx

//Compile cdf2 to perform numerical integration

BytecodeFunc f2 = JIT.compile(new Expr[]{mu, sigma}, cdf2);

System.out.println(f2.apply(-5.0, 1.0)); //~1.0

}

public static void test_2D() {

//http://turing.une.edu.au/~amth142/Lectures/Lecture_14.pdf

/*

Example:

We will evaluate the integral

I = \int_{\Omega} sin(sqrt(log(x+y+1))) dxdy

where \Omega is the disk

(x-1/2)^2 + (y-1/2)^2 <= 1/4

Since the disk Ω is contained within the square [0, 1] × [0, 1], we can

generate x and y as uniform [0, 1] random numbers, and keep those which

lie in the disk Ω.

Matlab code:

function ii = monte2da(n)

k = 0 // count no. of points in disk

sumf = 0 // keep running sum of function values

while (k < n) // keep going until we get n points

x = rand(1,1)

y = rand(1,1)

if ((x-0.5)^2 + (y-0.5)^2 <= 0.25) then // (x,y) is in disk

k = k + 1 // increment count

sumf = sumf + sin(sqrt(log(x+y+1))) // increment sumf

end

end

ii = (%pi/4)*(sumf/n) // %pi/4 = volume of disk

endfunction

-->monte2da(100000)

ans =

0.5679196

*/

Domain omega = new Domain2D("\\Omega", x, y)

.setConstraint( Le.apply((x-0.5)*(x-0.5) + (y-0.5)*(y-0.5), 0.25) )

.setBound(x, 0, 1)

.setBound(y, 0, 1);

Expr I = Integrate.apply(sin(sqrt(log(x+y+1))), omega);

System.out.println(I);

BytecodeFunc fI = JIT.compile(I);

System.out.println(fI.apply());

test_2D_verifiy();

}

public static void test_2D_verifiy() {

double xMin=0, xMax=1, xStep=0.001;

double yMin=0, yMax=1, yStep=0.001;

double sum = 0.0;

for(double x=xMin; x<=xMax; x+=xStep) {

for(double y=yMin; y<=yMax; y+=yStep) {

if((x-0.5)*(x-0.5) + (y-0.5)*(y-0.5) < 0.5*0.5)

sum += Math.sin(Math.sqrt(Math.log(x+y+1)))*xStep*yStep;

}

}

System.out.println("verify="+sum);

}

public static void test_ND() {

double r = 1.0;

Domain omega = new Domain3D("\\Omega", x, y, z)

.setConstraint( Le.apply(x*x + y*y + z*z, r*r) )

.setBound(x, -1, 1)

.setBound(y, -1, 1)

.setBound(z, -1, 1);

Expr ii = Integrate.apply(1, omega);

System.out.println(ii);

BytecodeFunc f = JIT.compile(ii);

System.out.println(f.apply());

System.out.println(4.0*Math.PI/3.0*Math.pow(r, 3));

Domain omega2 = new DomainND("\\Omega", x, y, z, t)

.setConstraint( Le.apply(x*x + y*y + z*z + t*t, r*r) )

.setBound(x, -1, 1)

.setBound(y, -1, 1)

.setBound(z, -1, 1)

.setBound(t, -1, 1);

Expr ii2 = Integrate.apply(1, omega2);

System.out.println(ii);

BytecodeFunc f2 = JIT.compile(ii2);

System.out.println(f2.apply());

System.out.println(0.5*Math.PI*Math.PI*Math.pow(r, 4));

}

/**

* I = \int_{\Omega} sin(sqrt(log(x+y+1))) dxdy

* where

* \Omega={ (x,y), where (x-1/2)^2 + (y-1/2)^2 <= 0.25 }

*/

public static void test_paper_example1() {

Domain omega = new Domain2D("\\Omega", x, y)

.setBound(x, 0.5-sqrt(0.25-(y-0.5)*(y-0.5)), 0.5+sqrt(0.25-(y-0.5)*(y-0.5)))

.setBound(y, 0, 1)

.setStepSize(0.001);

Expr I = Integrate.apply( sin(sqrt(log(x+y+1)) ), omega);

System.out.println(I);

BytecodeFunc fI = JIT.compile(I);

System.out.println(fI.apply());

}

/**

* Monte Carlo integration on an annulus.

* I = \int_{\Omega} sin(sqrt(log(x+y+1))) dxdy

* where

* \Omega= { (x,y), where

* a^2 <= (x-1/2)^2 + (y-1/2)^2 <= b^2 or

* c^2 <= (x-1/2)^2 + (y-1/2)^2 <= d^2

* }

* we choose

* a=0.25, b=0.5, c=0.75, d=1.0

*/

public static void test_paper_example2() {

Expr eq = (x-0.5)*(x-0.5) + (y-0.5)*(y-0.5);

Domain omega = new Domain2D("\\Omega", x, y)

.setConstraint(

( Ge.apply(eq, a*a) & Le.apply(eq, b*b)) |

( Ge.apply(eq, c*c) & Le.apply(eq, d*d) ))

.setBound(x, 0, 1)

.setBound(y, 0, 1);

Expr I = Integrate.apply(sin(sqrt(log(x+y+1)) ), omega);

System.out.println(I);

BytecodeFunc fI = JIT.compile(new Expr[]{a, b, c, d}, I);

System.out.println(fI.apply(0.25, 0.5, 0.75, 1.0));

test_paper_example_verifiy();

}

public static void test_paper_example_verifiy() {

double xMin=0, xMax=1, xStep=0.001;

double yMin=0, yMax=1, yStep=0.001;

double sum = 0.0;

double a=0.25, b=0.5, c=0.75, d=1.0;

for(double x=xMin; x<=xMax; x+=xStep) {

for(double y=yMin; y<=yMax; y+=yStep) {

double disk = (x-0.5)*(x-0.5) + (y-0.5)*(y-0.5);

if((a*a <= disk && disk <= b*b) ||

(c*c <= disk && disk <= d*d )

) {

sum += Math.sin(Math.sqrt(Math.log(x+y+1)))*xStep*yStep;

}

}

}

System.out.println("verify="+sum);

}

}