bitCount() public method

public bitCount ( ) : int
return int
Exemplo n.º 1
0
 /**
  * Create an Erlang integer from the given value.
  *
  * @param val
  *                the long value to use.
  */
 public OtpErlangLong(BigInteger v)
 {
     if (v == null)
     {
         throw new NullReferenceException();
     }
     if (v.bitCount() < 64)
     {
         val = v.LongValue();
     }
     else
     {
         bigVal = v;
     }
 }
Exemplo n.º 2
0
        //***********************************************************************
        // Performs the calculation of the kth term in the Lucas Sequence.
        // For details of the algorithm, see reference [9].
        //
        // k must be odd.  i.e LSB == 1
        //***********************************************************************

        private static BigInteger[] LucasSequenceHelper(BigInteger P, BigInteger Q,
                                                        BigInteger k, BigInteger n,
                                                        BigInteger constant, int s)
        {
                BigInteger[] result = new BigInteger[3];

                if((k.data[0] & 0x00000001) == 0)
                        throw (new ArgumentException("Argument k must be odd."));

                int numbits = k.bitCount();
                uint mask = (uint)0x1 << ((numbits & 0x1F) - 1);

                // v = v0, v1 = v1, u1 = u1, Q_k = Q^0

                BigInteger v = 2 % n, Q_k = 1 % n,
                           v1 = P % n, u1 = Q_k;
                bool flag = true;

                for(int i = k.dataLength - 1; i >= 0 ; i--)     // iterate on the binary expansion of k
                {
                        //Console.WriteLine("round");
                        while(mask != 0)
                        {
                                if(i == 0 && mask == 0x00000001)        // last bit
                                        break;

                                if((k.data[i] & mask) != 0)             // bit is set
                                {
                                        // index doubling with addition

                                        u1 = (u1 * v1) % n;

                                        v = ((v * v1) - (P * Q_k)) % n;
                                        v1 = n.BarrettReduction(v1 * v1, n, constant);
                                        v1 = (v1 - ((Q_k * Q) << 1)) % n;

                                        if(flag)
                                                flag = false;
                                        else
                                                Q_k = n.BarrettReduction(Q_k * Q_k, n, constant);

                                        Q_k = (Q_k * Q) % n;
                                }
                                else
                                {
                                        // index doubling
                                        u1 = ((u1 * v) - Q_k) % n;

                                        v1 = ((v * v1) - (P * Q_k)) % n;
                                        v = n.BarrettReduction(v * v, n, constant);
                                        v = (v - (Q_k << 1)) % n;

                                        if(flag)
                                        {
                                                Q_k = Q % n;
                                                flag = false;
                                        }
                                        else
                                                Q_k = n.BarrettReduction(Q_k * Q_k, n, constant);
                               }

                               mask >>= 1;
                        }
                        mask = 0x80000000;
                }

                // at this point u1 = u(n+1) and v = v(n)
                // since the last bit always 1, we need to transform u1 to u(2n+1) and v to v(2n+1)

                u1 = ((u1 * v) - Q_k) % n;
                v = ((v * v1) - (P * Q_k)) % n;
                if(flag)
                        flag = false;
                else
                        Q_k = n.BarrettReduction(Q_k * Q_k, n, constant);

                Q_k = (Q_k * Q) % n;


                for(int i = 0; i < s; i++)
                {
                        // index doubling
                        u1 = (u1 * v) % n;
                        v = ((v * v) - (Q_k << 1)) % n;

                        if(flag)
                        {
                                Q_k = Q % n;
                                flag = false;
                        }
                        else
                                Q_k = n.BarrettReduction(Q_k * Q_k, n, constant);
                }

                result[0] = u1;
                result[1] = v;
                result[2] = Q_k;

                return result;
        }
Exemplo n.º 3
0
        //***********************************************************************
        // Modulo Exponentiation
        //***********************************************************************

        public BigInteger modPow(BigInteger exp, BigInteger n)
        {
                if((exp.data[maxLength-1] & 0x80000000) != 0)
                        throw (new ArithmeticException("Positive exponents only."));

                BigInteger resultNum = 1;
	        BigInteger tempNum;
	        bool thisNegative = false;

	        if((this.data[maxLength-1] & 0x80000000) != 0)   // negative this
	        {
	                tempNum = -this % n;
	                thisNegative = true;
	        }
	        else
	                tempNum = this % n;  // ensures (tempNum * tempNum) < b^(2k)

	        if((n.data[maxLength-1] & 0x80000000) != 0)   // negative n
	                n = -n;

                // calculate constant = b^(2k) / m
                BigInteger constant = new BigInteger();

                int i = n.dataLength << 1;
                constant.data[i] = 0x00000001;
                constant.dataLength = i + 1;

                constant = constant / n;
                int totalBits = exp.bitCount();
                int count = 0;

                // perform squaring and multiply exponentiation
                for(int pos = 0; pos < exp.dataLength; pos++)
                {
                        uint mask = 0x01;
                        //Console.WriteLine("pos = " + pos);

                        for(int index = 0; index < 32; index++)
                        {
                                if((exp.data[pos] & mask) != 0)
                                        resultNum = BarrettReduction(resultNum * tempNum, n, constant);

                                mask <<= 1;

                                tempNum = BarrettReduction(tempNum * tempNum, n, constant);


                                if(tempNum.dataLength == 1 && tempNum.data[0] == 1)
                                {
                                        if(thisNegative && (exp.data[0] & 0x1) != 0)    //odd exp
                                                return -resultNum;
                                        return resultNum;
                                }
                                count++;
                                if(count == totalBits)
                                        break;
                        }
                }

                if(thisNegative && (exp.data[0] & 0x1) != 0)    //odd exp
                        return -resultNum;

	        return resultNum;
        }
Exemplo n.º 4
0
 public void write_big_integer(BigInteger v)
 {
     if (v.bitCount() < 64)
     {
         this.write_long(v.LongValue(), true);
         return;
     }
     int signum = (v > (BigInteger)0) ? 1 : (v < (BigInteger)0) ? -1 : 0;
     if (signum < 0)
     {
         v = -v;
     }
     byte[] magnitude = v.getBytes();
     int n = magnitude.Length;
     // Reverse the array to make it little endian.
     for (int i = 0, j = n; i < j--; i++)
     {
         // Swap [i] with [j]
         byte b = magnitude[i];
         magnitude[i] = magnitude[j];
         magnitude[j] = b;
     }
     if ((n & 0xFF) == n)
     {
         write1(OtpExternal.smallBigTag);
         write1(n); // length
     }
     else
     {
         write1(OtpExternal.largeBigTag);
         write4BE(n); // length
     }
     write1(signum < 0 ? 1 : 0); // sign
     // Write the array
     writeN(magnitude);
 }