/*
* It requires that all parameters be positive.
*
* @return {@code base<sup>exponent</sup> mod (2<sup>j</sup>)}.
* @see BigInteger#modPow(BigInteger, BigInteger)
*/
internal static BigInteger pow2ModPow(BigInteger baseJ, BigInteger exponent, int j)
{
// PRE: (base > 0), (exponent > 0) and (j > 0)
BigInteger res = BigInteger.ONE;
BigInteger e = exponent.copy();
BigInteger baseMod2toN = baseJ.copy();
BigInteger res2;
/*
* If 'base' is odd then it's coprime with 2^j and phi(2^j) = 2^(j-1);
* so we can reduce reduce the exponent (mod 2^(j-1)).
*/
if (baseJ.testBit(0))
{
inplaceModPow2(e, j - 1);
}
inplaceModPow2(baseMod2toN, j);
for (int i = e.bitLength() - 1; i >= 0; i--)
{
res2 = res.copy();
inplaceModPow2(res2, j);
res = res.multiply(res2);
if (BitLevel.testBit(e, i))
{
res = res.multiply(baseMod2toN);
inplaceModPow2(res, j);
}
}
inplaceModPow2(res, j);
return(res);
}
/*
* Performs a<sup>2</sup>
* @param a The number to square.
* @param aLen The length of the number to square.
*/
internal static int[] square(int[] a, int aLen, int[] res)
{
long carry;
for (int i = 0; i < aLen; i++)
{
carry = 0;
for (int j = i + 1; j < aLen; j++)
{
carry = unsignedMultAddAdd(a[i], a[j], res[i + j], (int)carry);
res[i + j] = (int)carry;
carry = java.dotnet.lang.Operator.shiftRightUnsignet(carry, 32);
}
res[i + aLen] = (int)carry;
}
BitLevel.shiftLeftOneBit(res, res, aLen << 1);
carry = 0;
for (int i = 0, index = 0; i < aLen; i++, index++)
{
carry = unsignedMultAddAdd(a[i], a[i], res[index], (int)carry);
res[index] = (int)carry;
carry = java.dotnet.lang.Operator.shiftRightUnsignet(carry, 32);
index++;
carry += res[index] & 0xFFFFFFFFL;
res[index] = (int)carry;
carry = java.dotnet.lang.Operator.shiftRightUnsignet(carry, 32);
}
return(res);
}
/*Implements the Montgomery modular exponentiation based in <i>The sliding windows algorithm and the Mongomery
* Reduction</i>.
*@ar.org.fitc.ref "A. Menezes,P. van Oorschot, S. Vanstone - Handbook of Applied Cryptography";
*@see #oddModPow(BigInteger, BigInteger,
* BigInteger)
*/
internal static BigInteger slidingWindow(BigInteger x2, BigInteger a2, BigInteger exponent, BigInteger modulus, int n2)
{
// fill odd low pows of a2
BigInteger [] pows = new BigInteger[8];
BigInteger res = x2;
int lowexp;
BigInteger x3;
int acc3;
pows[0] = a2;
x3 = monPro(a2, a2, modulus, n2);
for (int i = 1; i <= 7; i++)
{
pows[i] = monPro(pows[i - 1], x3, modulus, n2);
}
for (int i = exponent.bitLength() - 1; i >= 0; i--)
{
if (BitLevel.testBit(exponent, i))
{
lowexp = 1;
acc3 = i;
for (int j = java.lang.Math.max(i - 3, 0); j <= i - 1; j++)
{
if (BitLevel.testBit(exponent, j))
{
if (j < acc3)
{
acc3 = j;
lowexp = (lowexp << (i - j)) ^ 1;
}
else
{
lowexp = lowexp ^ (1 << (j - acc3));
}
}
}
for (int j = acc3; j <= i; j++)
{
res = monPro(res, res, modulus, n2);
}
res = monPro(pows[(lowexp - 1) >> 1], res, modulus, n2);
i = acc3;
}
else
{
res = monPro(res, res, modulus, n2);
}
}
return(res);
}
/** @see BigInteger#doubleValue() */
protected internal static double bigInteger2Double(BigInteger val)
{
// val.bitLength() < 64
if ((val.numberLength < 2) ||
((val.numberLength == 2) && (val.digits[1] > 0)))
{
return(val.longValue());
}
// val.bitLength() >= 33 * 32 > 1024
if (val.numberLength > 32)
{
return((val.sign > 0) ? java.lang.Double.POSITIVE_INFINITY
: java.lang.Double.NEGATIVE_INFINITY);
}
int bitLen = val.abs().bitLength();
long exponent = bitLen - 1;
int delta = bitLen - 54;
// We need 54 top bits from this, the 53th bit is always 1 in lVal.
long lVal = val.abs().shiftRight(delta).longValue();
/*
* Take 53 bits from lVal to mantissa. The least significant bit is
* needed for rounding.
*/
long mantissa = lVal & 0x1FFFFFFFFFFFFFL;
if (exponent == 1023)
{
if (mantissa == 0X1FFFFFFFFFFFFFL)
{
return((val.sign > 0) ? java.lang.Double.POSITIVE_INFINITY
: java.lang.Double.NEGATIVE_INFINITY);
}
if (mantissa == 0x1FFFFFFFFFFFFEL)
{
return((val.sign > 0) ? java.lang.Double.MAX_VALUE : -java.lang.Double.MAX_VALUE);
}
}
// Round the mantissa
if (((mantissa & 1) == 1) &&
(((mantissa & 2) == 2) || BitLevel.nonZeroDroppedBits(delta,
val.digits)))
{
mantissa += 2;
}
mantissa >>= 1; // drop the rounding bit
long resSign = (long)((val.sign < 0) ? 0x8000000000000000L : 0);
exponent = ((1023 + exponent) << 52) & 0x7FF0000000000000L;
long result = resSign | exponent | mantissa;
return(java.lang.Double.longBitsToDouble(result));
}
internal static BigInteger squareAndMultiply(BigInteger x2, BigInteger a2, BigInteger exponent, BigInteger modulus, int n2)
{
BigInteger res = x2;
for (int i = exponent.bitLength() - 1; i >= 0; i--)
{
res = monPro(res, res, modulus, n2);
if (BitLevel.testBit(exponent, i))
{
res = monPro(res, a2, modulus, n2);
}
}
return(res);
}
/*
* @param x an odd positive number.
* @param n the exponent by which 2 is raised.
* @return {@code x<sup>-1</sup> (mod 2<sup>n</sup>)}.
*/
internal static BigInteger modPow2Inverse(BigInteger x, int n)
{
// PRE: (x > 0), (x is odd), and (n > 0)
BigInteger y = new BigInteger(1, new int[1 << n]);
y.numberLength = 1;
y.digits[0] = 1;
y.sign = 1;
for (int i = 1; i < n; i++)
{
if (BitLevel.testBit(x.multiply(y), i))
{
// Adding 2^i to y (setting the i-th bit)
y.digits[i >> 5] |= (1 << (i & 31));
}
}
return(y);
}
/*
* Divides the array 'a' by the array 'b' and gets the quotient and the
* remainder. Implements the Knuth's division algorithm. See D. Knuth, The
* Art of Computer Programming, vol. 2. Steps D1-D8 correspond the steps in
* the algorithm description.
*
* @param quot the quotient
* @param quotLength the quotient's length
* @param a the dividend
* @param aLength the dividend's length
* @param b the divisor
* @param bLength the divisor's length
* @return the remainder
*/
internal static int[] divide(int[] quot, int quotLength, int[] a, int aLength,
int[] b, int bLength)
{
int[] normA = new int[aLength + 1]; // the normalized dividend
// an extra byte is needed for correct shift
int[] normB = new int[bLength + 1]; // the normalized divisor;
int normBLength = bLength;
/*
* Step D1: normalize a and b and put the results to a1 and b1 the
* normalized divisor's first digit must be >= 2^31
*/
int divisorShift = java.lang.Integer.numberOfLeadingZeros(b[bLength - 1]);
if (divisorShift != 0)
{
BitLevel.shiftLeft(normB, b, 0, divisorShift);
BitLevel.shiftLeft(normA, a, 0, divisorShift);
}
else
{
java.lang.SystemJ.arraycopy(a, 0, normA, 0, aLength);
java.lang.SystemJ.arraycopy(b, 0, normB, 0, bLength);
}
int firstDivisorDigit = normB[normBLength - 1];
// Step D2: set the quotient index
int i = quotLength - 1;
int j = aLength;
while (i >= 0)
{
// Step D3: calculate a guess digit guessDigit
int guessDigit = 0;
if (normA[j] == firstDivisorDigit)
{
// set guessDigit to the largest unsigned int value
guessDigit = -1;
}
else
{
long product = (((normA[j] & 0xffffffffL) << 32) + (normA[j - 1] & 0xffffffffL));
long res = Division.divideLongByInt(product, firstDivisorDigit);
guessDigit = (int)res; // the quotient of divideLongByInt
int rem = (int)(res >> 32); // the remainder of
// divideLongByInt
// decrease guessDigit by 1 while leftHand > rightHand
if (guessDigit != 0)
{
long leftHand = 0;
long rightHand = 0;
bool rOverflowed = false;
guessDigit++; // to have the proper value in the loop
// below
do
{
guessDigit--;
if (rOverflowed)
{
break;
}
// leftHand always fits in an unsigned long
leftHand = (guessDigit & 0xffffffffL)
* (normB[normBLength - 2] & 0xffffffffL);
/*
* rightHand can overflow; in this case the loop
* condition will be true in the next step of the loop
*/
rightHand = ((long)rem << 32)
+ (normA[j - 2] & 0xffffffffL);
long longR = (rem & 0xffffffffL)
+ (firstDivisorDigit & 0xffffffffL);
/*
* checks that longR does not fit in an unsigned int;
* this ensures that rightHand will overflow unsigned
* long in the next step
*/
if (java.lang.Integer.numberOfLeadingZeros((int)(java.dotnet.lang.Operator.shiftRightUnsignet(longR, 32))) < 32)
{
rOverflowed = true;
}
else
{
rem = (int)longR;
}
} while (((((ulong)leftHand) ^ 0x8000000000000000L) > (((ulong)rightHand) ^ 0x8000000000000000L)));
}
}
// Step D4: multiply normB by guessDigit and subtract the production
// from normA.
if (guessDigit != 0)
{
int borrow = Division.multiplyAndSubtract(normA, j
- normBLength, normB, normBLength,
guessDigit);
// Step D5: check the borrow
if (borrow != 0)
{
// Step D6: compensating addition
guessDigit--;
long carry = 0;
for (int k = 0; k < normBLength; k++)
{
carry += (normA[j - normBLength + k] & 0xffffffffL)
+ (normB[k] & 0xffffffffL);
normA[j - normBLength + k] = (int)carry;
carry = java.dotnet.lang.Operator.shiftRightUnsignet(carry, 32);
}
}
}
if (quot != null)
{
quot[i] = guessDigit;
}
// Step D7
j--;
i--;
}
/*
* Step D8: we got the remainder in normA. Denormalize it id needed
*/
if (divisorShift != 0)
{
// reuse normB
BitLevel.shiftRight(normB, normBLength, normA, 0, divisorShift);
return(normB);
}
java.lang.SystemJ.arraycopy(normA, 0, normB, 0, bLength);
return(normA);
}
/*
* 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);
}
/*
* @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));
}