使用PLINQ加速计算pi

Question

我需要使用PLINQ - 库通过Monte-Carlo方法计算Pi数，但是当我的并行程序运行时，它计算的Pi数比它的不平行模拟长得多。如何解决？并行计算类及其不可比拟的模拟如下：

class CalcPiPLINQ
    {
        int n;
        double aPi;
        int counter;
        public StringBuilder Msg; // diagnostic message

        public void Init(int aN)
        {
            //stopWatch.Start();
            n = aN; // save total calculate-iterations amount
            aPi = -1; // flag, if no any calculate-iteration has been completed
            Msg = new StringBuilder("No any calculate-iteration has been completed");
        }
        public void Run()
        {
            if (n < 1)
            {
                Msg = new StringBuilder("Invalid N-value");
                return;
            }
            Random rnd = new Random(); // create random-numbers generator

            Stopwatch stopWatch = Stopwatch.StartNew();
            var numbers = ParallelEnumerable.Range(0, n);
            var result = Enumerable
            .Range(0, n)
            .Select(i => new { X = rnd.NextDouble(), Y = rnd.NextDouble() })
            .AsParallel()
            .Select(p => Check(p.X, p.Y))
            .Count(b => b);

            aPi = 4.0 * ((double)result / (double)n); // calculate approximate Pi-value
            //stopWatch.Stop();
            TimeSpan ts = stopWatch.Elapsed;
            string elapsedTime = String.Format("{0:00}:{1:00}:{2:00}.{3:00}",
            ts.Hours, ts.Minutes, ts.Seconds,
            ts.Milliseconds / 10);
            Console.WriteLine("RunTime " + elapsedTime);
        }
        public double Done()
        {
            if (aPi > 0)
            {
                Msg = new StringBuilder("Calculations are completed successful!");
                return aPi; // return calculated value
            }
            else
            {
                return 0; // no result
            }
        }
        public bool Check(double x, double y)
        {
            return ((x - 0.5) * (x - 0.5) + (y - 0.5) * (y - 0.5)) < 0.25;
        }

        public void CheckWithCounter(double x, double y)
        {
            if(((x - 0.5) * (x - 0.5) + (y - 0.5) * (y - 0.5)) < 0.25)
            {
                Interlocked.Increment(ref counter);
            }
        }
    }

不平行的模拟：

class TCalcPi//unparallel calculating method
    {
        int N;
        int N_0;
        double aPi;
        public StringBuilder Msg; // diagnostic message
        double x, y;
        Stopwatch stopWatch = new Stopwatch();

        public void Init(int aN)
        {
            stopWatch.Start();
            N = aN; // save total calculate-iterations amount
            aPi = -1; // flag, if no any calculate-iteration has been completed
            Msg = new StringBuilder("No any calculate-iteration has been completed");
        }
        public void Run()
        {
            if (N < 1)
            {
                Msg = new StringBuilder("Invalid N - value");
                return;
            }

            int i;
            Random rnd = new Random(); // to create randomizer
            for (i = 1; i <= N; i++)
            {
                x = rnd.NextDouble(); // to generate coordinates
                y = rnd.NextDouble(); // 
                if (((x -  0.5) * (x -  0.5) + (y -  0.5) * (y -  0.5)) <  0.25)
                {
                    N_0++; // coordinate in a circle! mark it by incrementing N_0
                }
            }
            aPi = 4.0 * ((double)N_0 / (double)N); // to calculate approximate Pi - value
            stopWatch.Stop();
            TimeSpan ts = stopWatch.Elapsed;
            string elapsedTime = String.Format("{0:00}:{1:00}:{2:00}.{3:00}",
            ts.Hours, ts.Minutes, ts.Seconds,
            ts.Milliseconds / 10);
            Console.WriteLine("RunTime " + elapsedTime);
        }
        public double Done()
        {
            if (aPi > 0)
            {
                Msg = new StringBuilder("Calculates has been completed successful");
                return aPi; // return gotten value
            }
            else
            {
                return 0; // no result
            }
        }
    }