Thinking more…

今天繼續往下面想以後,發現我以前的想法錯誤了.
本想說,根據歷史資料去統計,以出現次數較少的數字取亂數,又,理論上每個數字出現的次數會趨近一致,所以應該會有較高的機率得到下期中獎的號碼.
回頭想想每期中獎號碼組合機率,應該是以 1 去除以 42 取 6 的所有組合數.
我以排列組合去Google找了很多資料:

• 組合
• 排列與組合
• 排列组合
• 排列组合
• 組合數學

要怎麼求組合數呢??我們將樂透的規則套入組合公式:
從 42 個數字中,取 6 個號碼,不論順序,組成一組,不同的組合數為: 排列數/42!
排列組的計算: 42 * (42-1) * (42-2) * (42-3) * (42-4) * (42-5)
42!: 42*41*40*….*2*1
所以寫好以後的程式就是這樣:
using System;
using System.Collections;

public class MyClass
{
    public static double calcP( double totalNumbers, double pickNumbers )
    {
        double possible = 1.0;
        
        for( int i=0; i<pickNumbers; i++ )
            possible = possible * (totalNumbers-i);
            
        return possible;
    }

    public static double calcC( double totalNumbers, double pickNumbers )
    {
        double mShock = 1.0; // m!
        
        for( long i=0; i<pickNumbers; i++ )
            mShock = mShock * (i+1);
        
        double p = calcP( totalNumbers, pickNumbers );
        
        return( p/mShock );
    }

    public static void Main()
    {
        Console.WriteLine( "組合數是: {0}", calcC( 42.0, 6.0 ) );
        
        RL();
    }
    
    #region Helper methods

    private static void RL()
    {
        Console.ReadLine();    
    }
    

    #endregion
}
最後算出來的結果是: 5,245,786
也就是說,如果有 5,245,786 人去買的話,至少會有一人中獎.
不過,還是覺得怪怪的…這數字會不會太小了呢??

之前騎車的時候想到的一個 idea,想要來預測下期樂透的號碼.
以亂數取某數字,理論上,當次數越趨近於無限的時候,每個數字出現的機率會趨近於相等.
所以我們寫了這麼一個程式來驗證其正確性:
using System;
using System.Collections;

public class MyClass
{
    public static void Main()
    {
        Random r = new Random();
        int[] statistics = new int[43];
        int i=0;
        int sum=0;
        
        // initialize
        for( i=0; i<statistics.Length; i++ )
            statistics[i]=0;
            
        for( i=0; i<100000; i++ )
        {
            // int index = Convert.ToInt32( 41*r.NextDouble()+1 );
            int index = r.Next(42)+1;
            statistics[ index ] = statistics[ index ] + 1;
        }
        
        for( i=0; i<statistics.Length; i++ )
        {
            Console.WriteLine( "statistics[{0}] = {1}", i, statistics[i] );
            sum+=statistics[i];
        }
        Console.WriteLine( "sum={0}", sum );
            
        RL();
    }
    
    #region Helper methods
    private static void RL()
    {
        Console.ReadLine();    
    }
    #endregion
}
這邊遇到一件很有趣的事情,如果你用 41*r.NextDouble()+1 來取的話
結果會是最前面數字與最後面數字出現的機率會比其他數字要少了約一倍.
所以這樣取會比較好些: r.Next(42)+1
另外我們捨棄 [0] 不用,純使用 [1] ~ [42], 以方便我們的統計.
參考資料:

• A Lottery Program in C#
• Generating Random Number and String in C#
• Random Numbers in C#
• 亂猜是猜中了，還是沒猜中？

…待續…

CheckSum 是一個很簡單的檢查方法,把內容拆解成一堆 byte (或word,dword), 如果把這些 bytes 相加起來得到的值一致,表示內容”應該”是沒問題.

using System;
using System.IO;
using System.Collections;

public class CheckSum
{
    private long sum=0;
    
    public CheckSum()
    {
    }

    public void Add( int i )
    {
        sum += i;
    }
    
    public void Add( byte b )
    {
        sum += b;
    }
    
    public long GetCheckSum()
    {
        return sum;
    }
}

public class MyClass
{
    public static void Main()
    {
        string filename;
        FileStream fs=null;
        CheckSum sum = new CheckSum();
        
        //Open the stream and read it back.
        Console.Write( "Please input filename:" );
        filename = Console.ReadLine();
        Console.WriteLine( "Trying to read {0}", filename );
        
        if( File.Exists( filename ) )
            Console.WriteLine( "{0} is existed.", filename );

        try
        {
            fs = File.Open( filename, FileMode.Open, FileAccess.Read );
            
            if( fs.CanRead )
            {
                int value=0;
                
                while( (value = fs.ReadByte()) >= 0)
                {
                    sum.Add( value );
                }
                
                Console.WriteLine( "Checksum = {0,8:x}", sum.GetCheckSum() );
            }
            else
                Console.WriteLine( "Cannot be read" );
        }
        catch( Exception ex )
        {
            Console.WriteLine( ex.Message );
        }
        finally
        {
            fs.Close();
        }
        
        RL();
    }
    
    #region Helper methods
    private static void RL()
    {
        Console.ReadLine();    
    }
    #endregion
}

在某論壇看到討論最大公因數的討論串,裡面有提到輾轉相除法.
老實說,當我看到最大公因數,我只想到暴力法.
從 1…n 開始 iterate, 能整除,表示是因數,把這些數字記起來,於是我們得到兩個集合.這兩個集合的交集,表示是共通因數,最大者則為最大公因數.
完全忘記有輾轉相除法這玩意兒,一時興起,port 到 c# 試試看.
using System;
using System.Collections;

public class MyClass
{
    public static int GCD( int a, int b )
    {
        if( a % b == 0 )
            return b;
        return GCD( b, a%b );
    }
    
    public static void Main()
    {
        int num1=1230, num2=460, num3=10;
        
        int result1 = GCD( num1, num2 );
        int result2 = GCD( result1, num3 );
        
        Console.WriteLine( "The GCD of {0} and {1} is {2}", num1, num2, result1 );
        Console.WriteLine( "The GCD of {0} and {1} is {2}", result1, num3, result2 );
        
        RL();
    }
    
    #region Helper methods
    private static void RL()
    {
        Console.ReadLine();    
    }
    #endregion
}