Ejemplo n.º 1
0
        /**
         * Calculates a.modInverse(p) Based on: Savas, E; Koc, C "The Montgomery Modular
         * Inverse - Revised"
         */
        internal static BigInteger modInverseMontgomery(BigInteger a, BigInteger p)
        {
            if (a.sign == 0)
            {
                // ZERO hasn't inverse
                // math.19: BigInteger not invertible
                throw new ArithmeticException("BigInteger not invertible");
            }


            if (!p.testBit(0))
            {
                // montgomery inverse require even modulo
                return(modInverseHars(a, p));
            }

            int m = p.numberLength * 32;
            // PRE: a \in [1, p - 1]
            BigInteger u, v, r, s;

            u = p.copy();  // make copy to use inplace method
            v = a.copy();
            int max = java.lang.Math.max(v.numberLength, u.numberLength);

            r           = new BigInteger(1, 1, new int[max + 1]);
            s           = new BigInteger(1, 1, new int[max + 1]);
            s.digits[0] = 1;
            // s == 1 && v == 0

            int k = 0;

            int lsbu = u.getLowestSetBit();
            int lsbv = v.getLowestSetBit();
            int toShift;

            if (lsbu > lsbv)
            {
                BitLevel.inplaceShiftRight(u, lsbu);
                BitLevel.inplaceShiftRight(v, lsbv);
                BitLevel.inplaceShiftLeft(r, lsbv);
                k += lsbu - lsbv;
            }
            else
            {
                BitLevel.inplaceShiftRight(u, lsbu);
                BitLevel.inplaceShiftRight(v, lsbv);
                BitLevel.inplaceShiftLeft(s, lsbu);
                k += lsbv - lsbu;
            }

            r.sign = 1;
            while (v.signum() > 0)
            {
                // INV v >= 0, u >= 0, v odd, u odd (except last iteration when v is even (0))

                while (u.compareTo(v) > BigInteger.EQUALS)
                {
                    Elementary.inplaceSubtract(u, v);
                    toShift = u.getLowestSetBit();
                    BitLevel.inplaceShiftRight(u, toShift);
                    Elementary.inplaceAdd(r, s);
                    BitLevel.inplaceShiftLeft(s, toShift);
                    k += toShift;
                }

                while (u.compareTo(v) <= BigInteger.EQUALS)
                {
                    Elementary.inplaceSubtract(v, u);
                    if (v.signum() == 0)
                    {
                        break;
                    }
                    toShift = v.getLowestSetBit();
                    BitLevel.inplaceShiftRight(v, toShift);
                    Elementary.inplaceAdd(s, r);
                    BitLevel.inplaceShiftLeft(r, toShift);
                    k += toShift;
                }
            }
            if (!u.isOne())
            {
                // in u is stored the gcd
                // math.19: BigInteger not invertible.
                throw new ArithmeticException("BigInteger not invertible");
            }
            if (r.compareTo(p) >= BigInteger.EQUALS)
            {
                Elementary.inplaceSubtract(r, p);
            }

            r = p.subtract(r);

            // Have pair: ((BigInteger)r, (Integer)k) where r == a^(-1) * 2^k mod (module)
            int n1 = calcN(p);

            if (k > m)
            {
                r = monPro(r, BigInteger.ONE, p, n1);
                k = k - m;
            }

            r = monPro(r, BigInteger.getPowerOfTwo(m - k), p, n1);
            return(r);
        }
Ejemplo n.º 2
0
        /**
         * @param m a positive modulus
         * Return the greatest common divisor of op1 and op2,
         *
         * @param op1
         *            must be greater than zero
         * @param op2
         *            must be greater than zero
         * @see BigInteger#gcd(BigInteger)
         * @return {@code GCD(op1, op2)}
         */
        internal static BigInteger gcdBinary(BigInteger op1, BigInteger op2)
        {
            // PRE: (op1 > 0) and (op2 > 0)

            /*
             * Divide both number the maximal possible times by 2 without rounding
             * gcd(2*a, 2*b) = 2 * gcd(a,b)
             */
            int lsb1      = op1.getLowestSetBit();
            int lsb2      = op2.getLowestSetBit();
            int pow2Count = java.lang.Math.min(lsb1, lsb2);

            BitLevel.inplaceShiftRight(op1, lsb1);
            BitLevel.inplaceShiftRight(op2, lsb2);

            BigInteger swap;

            // I want op2 > op1
            if (op1.compareTo(op2) == BigInteger.GREATER)
            {
                swap = op1;
                op1  = op2;
                op2  = swap;
            }

            do   // INV: op2 >= op1 && both are odd unless op1 = 0

            // Optimization for small operands
            // (op2.bitLength() < 64) implies by INV (op1.bitLength() < 64)
            {
                if ((op2.numberLength == 1) ||
                    ((op2.numberLength == 2) && (op2.digits[1] > 0)))
                {
                    op2 = BigInteger.valueOf(Division.gcdBinary(op1.longValue(),
                                                                op2.longValue()));
                    break;
                }

                // Implements one step of the Euclidean algorithm
                // To reduce one operand if it's much smaller than the other one
                if (op2.numberLength > op1.numberLength * 1.2)
                {
                    op2 = op2.remainder(op1);
                    if (op2.signum() != 0)
                    {
                        BitLevel.inplaceShiftRight(op2, op2.getLowestSetBit());
                    }
                }
                else
                {
                    // Use Knuth's algorithm of successive subtract and shifting
                    do
                    {
                        Elementary.inplaceSubtract(op2, op1);                   // both are odd
                        BitLevel.inplaceShiftRight(op2, op2.getLowestSetBit()); // op2 is even
                    } while (op2.compareTo(op1) >= BigInteger.EQUALS);
                }
                // now op1 >= op2
                swap = op2;
                op2  = op1;
                op1  = swap;
            } while (op1.sign != 0);
            return(op2.shiftLeft(pow2Count));
        }