Using Java random numbers to calculate pi π

I. Theory Article 1. Mathematical formulas

Circle Area formula: Π*r*r, where π is pi, R is the circle radius;

Square area formula: S*s, where S is the edge length;

Pythagorean Theorem: a*a + b*b = c*c, where A/b is right triangle two right-angled edges, and C is the hypotenuse.

2. Calculation method

Consider that the square with the side length r is embedded with a 1/4 circle with a radius of R.

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Insidecircle area = Circle area with radius of R/4 =Π*R*R/4

Square area = R*r

So, insidecircle area/Square area = (Π*R*R/4)/(R*R) =Π/4

The polygons consist of lines, and the lines consist of dots. Therefore, if there are several points evenly falling into the square, then the number of points falling into the insidecircle will be Π/4, thus the value of π will be counted.

Second, the actual combat Chapter 1. Shotgun calculation

The concrete way is: make a like the same square plank, with the shotgun against it a random sweep, and finally count the number of bullet holes and landing point, so as to obtain the value of π.

In fact, there are really researchers who have done experiments with the brain hole larger than black holes, and they got 3.13 in 30,857 samples, which is a good result.

See: Http://www.techug.com/compute-pi-with-gun

2. Java Random number calculation

As a yard farmer, of course, can not afford to play shotgun this high-grade toys, then the next code to play a piece.

The specific idea is this: R directly takes a value of 1.0, also need to define a drop in the square of all points of the number pointnumber, each point has a coordinate (x, y), x. Y takes a value of 0.0-1.0, using the Java random number to generate each point, and then use the Pythagorean theorem to determine whether the point falls within the circle or the circle, and the number of points falling within the circle Insidecirclenumber, then π=insidecirclenumber/ Pointnumber*4. Of course, if the calculation is only one time, the error may be larger, you can increase the number of calculations Calctimes, and then the average.

The following results are calculated as follows:

Pointnumber	Calctimes	Pi	The closest value to π
10000	10000	3.141617279999959	3.1416
100000	10000	3.1415569599999684	3.1416
1000000	10000	3.1415845499999953	3.141592
10000000	10000	3.1415924761886806	3.1415928

Attached source code:

Package Com.test.pai; Import Org.apache.commons.lang.math.RandomUtils; public class Calcpai { public static Boolean incircle (Double x, double y) { Return (y <= math.sqrt (1-x * x)); } public static double calcpaibypointnumber (long num) { Double incirclenum = 0.0; for (long i = 0; i < num; i++) { if (Calcpai.incircle (randomutils.nextdouble (), randomutils.nextdouble ())) { incirclenum++; } } Double pai = incirclenum * 4/num; Return pai; } public static void Main (string[] args) { Double Realpai = 3.14159265; Currresult Currresult = new Currresult (0.0, Realpai, 0.0); Long times = 10000; Long num = 1000000; for (Long i = 1; I <= times; i++) { Double pai = calcpai.calcpaibypointnumber (num); Currresult.settotalpai (Currresult.gettotalpai () + pai); Double diff = Math.Abs (realpai-pai); if (diff < currresult.getdifference ()) { Currresult.setcurrpai (PAI); Currresult.setdifference (diff); } System.out.println ("No." + i + "/" + times + "\ T" + pai + "\ T" + currresult.getcurrpai () + "\ T" + Currresult.gettotalpai ()/i); } } }

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Using Java random numbers to calculate pi π