public LongArray ModMultiply(LongArray other, int m, int[] ks)
{
/*
* Find out the degree of each argument and handle the zero cases
*/
int aDeg = Degree();
if (aDeg == 0)
{
return this;
}
int bDeg = other.Degree();
if (bDeg == 0)
{
return other;
}
/*
* Swap if necessary so that A is the smaller argument
*/
LongArray A = this, B = other;
if (aDeg > bDeg)
{
A = other; B = this;
int tmp = aDeg; aDeg = bDeg; bDeg = tmp;
}
/*
* Establish the word lengths of the arguments and result
*/
int aLen = (int)((uint)(aDeg + 63) >> 6);
int bLen = (int)((uint)(bDeg + 63) >> 6);
int cLen = (int)((uint)(aDeg + bDeg + 62) >> 6);
if (aLen == 1)
{
long a0 = A.m_ints[0];
if (a0 == 1L)
{
return B;
}
/*
* Fast path for small A, with performance dependent only on the number of set bits
*/
long[] c0 = new long[cLen];
MultiplyWord(a0, B.m_ints, bLen, c0, 0);
/*
* Reduce the raw answer against the reduction coefficients
*/
return ReduceResult(c0, 0, cLen, m, ks);
}
/*
* Determine if B will get bigger during shifting
*/
int bMax = (int)((uint)(bDeg + 7 + 63) >> 6);
/*
* Lookup table for the offset of each B in the tables
*/
int[] ti = new int[16];
/*
* Precompute table of all 4-bit products of B
*/
long[] T0 = new long[bMax << 4];
int tOff = bMax;
ti[1] = tOff;
Array.Copy(B.m_ints, 0, T0, tOff, bLen);
for (int i = 2; i < 16; ++i)
{
ti[i] = (tOff += bMax);
if ((i & 1) == 0)
{
ShiftUp(T0, (int)((uint)tOff >> 1), T0, tOff, bMax, 1);
}
else
{
Add(T0, bMax, T0, tOff - bMax, T0, tOff, bMax);
}
}
/*
* Second table with all 4-bit products of B shifted 4 bits
*/
long[] T1 = new long[T0.Length];
ShiftUp(T0, 0, T1, 0, T0.Length, 4);
// ShiftUp(T0, bMax, T1, bMax, tOff, 4);
long[] a = A.m_ints;
long[] c = new long[cLen << 3];
int MASK = 0xF;
/*
* Lopez-Dahab (Modified) algorithm
*/
for (int aPos = 0; aPos < aLen; ++aPos)
{
long aVal = a[aPos];
int cOff = aPos;
for (;;)
{
int u = (int)aVal & MASK;
aVal = (long)((ulong)aVal >> 4);
int v = (int)aVal & MASK;
AddBoth(c, cOff, T0, ti[u], T1, ti[v], bMax);
aVal = (long)((ulong)aVal >> 4);
if (aVal == 0L)
{
break;
}
cOff += cLen;
}
}
{
int cOff = c.Length;
while ((cOff -= cLen) != 0)
{
AddShiftedUp(c, cOff - cLen, c, cOff, cLen, 8);
}
}
/*
* Finally the raw answer is collected, reduce it against the reduction coefficients
*/
return ReduceResult(c, 0, cLen, m, ks);
}
public LongArray ModMultiplyAlt(LongArray other, int m, int[] ks)
{
/*
* Find out the degree of each argument and handle the zero cases
*/
int aDeg = Degree();
if (aDeg == 0)
{
return this;
}
int bDeg = other.Degree();
if (bDeg == 0)
{
return other;
}
/*
* Swap if necessary so that A is the smaller argument
*/
LongArray A = this, B = other;
if (aDeg > bDeg)
{
A = other; B = this;
int tmp = aDeg; aDeg = bDeg; bDeg = tmp;
}
/*
* Establish the word lengths of the arguments and result
*/
int aLen = (int)((uint)(aDeg + 63) >> 6);
int bLen = (int)((uint)(bDeg + 63) >> 6);
int cLen = (int)((uint)(aDeg + bDeg + 62) >> 6);
if (aLen == 1)
{
long a0 = A.m_ints[0];
if (a0 == 1L)
{
return B;
}
/*
* Fast path for small A, with performance dependent only on the number of set bits
*/
long[] c0 = new long[cLen];
MultiplyWord(a0, B.m_ints, bLen, c0, 0);
/*
* Reduce the raw answer against the reduction coefficients
*/
return ReduceResult(c0, 0, cLen, m, ks);
}
// NOTE: This works, but is slower than width 4 processing
// if (aLen == 2)
// {
// /*
// * Use common-multiplicand optimization to save ~1/4 of the adds
// */
// long a1 = A.m_ints[0], a2 = A.m_ints[1];
// long aa = a1 & a2; a1 ^= aa; a2 ^= aa;
//
// long[] b = B.m_ints;
// long[] c = new long[cLen];
// multiplyWord(aa, b, bLen, c, 1);
// add(c, 0, c, 1, cLen - 1);
// multiplyWord(a1, b, bLen, c, 0);
// multiplyWord(a2, b, bLen, c, 1);
//
// /*
// * Reduce the raw answer against the reduction coefficients
// */
// return ReduceResult(c, 0, cLen, m, ks);
// }
/*
* Determine the parameters of the Interleaved window algorithm: the 'width' in bits to
* process together, the number of evaluation 'positions' implied by that width, and the
* 'top' position at which the regular window algorithm stops.
*/
int width, positions, top, banks;
// NOTE: width 4 is the fastest over the entire range of sizes used in current crypto
// width = 1; positions = 64; top = 64; banks = 4;
// width = 2; positions = 32; top = 64; banks = 4;
// width = 3; positions = 21; top = 63; banks = 3;
width = 4; positions = 16; top = 64; banks = 8;
// width = 5; positions = 13; top = 65; banks = 7;
// width = 7; positions = 9; top = 63; banks = 9;
// width = 8; positions = 8; top = 64; banks = 8;
/*
* Determine if B will get bigger during shifting
*/
int shifts = top < 64 ? positions : positions - 1;
int bMax = (int)((uint)(bDeg + shifts + 63) >> 6);
int bTotal = bMax * banks, stride = width * banks;
/*
* Create a single temporary buffer, with an offset table to find the positions of things in it
*/
int[] ci = new int[1 << width];
int cTotal = aLen;
{
ci[0] = cTotal;
cTotal += bTotal;
ci[1] = cTotal;
for (int i = 2; i < ci.Length; ++i)
{
cTotal += cLen;
ci[i] = cTotal;
}
cTotal += cLen;
}
// NOTE: Provide a safe dump for "high zeroes" since we are adding 'bMax' and not 'bLen'
++cTotal;
long[] c = new long[cTotal];
// Prepare A in Interleaved form, according to the chosen width
Interleave(A.m_ints, 0, c, 0, aLen, width);
// Make a working copy of B, since we will be shifting it
{
int bOff = aLen;
Array.Copy(B.m_ints, 0, c, bOff, bLen);
for (int bank = 1; bank < banks; ++bank)
{
ShiftUp(c, aLen, c, bOff += bMax, bMax, bank);
}
}
/*
* The main loop analyzes the Interleaved windows in A, and for each non-zero window
* a single word-array XOR is performed to a carefully selected slice of 'c'. The loop is
* breadth-first, checking the lowest window in each word, then looping again for the
* next higher window position.
*/
int MASK = (1 << width) - 1;
int k = 0;
for (;;)
{
int aPos = 0;
do
{
long aVal = (long)((ulong)c[aPos] >> k);
int bank = 0, bOff = aLen;
for (;;)
{
int index = (int)(aVal) & MASK;
if (index != 0)
{
/*
* Add to a 'c' buffer based on the bit-pattern of 'index'. Since A is in
* Interleaved form, the bits represent the current B shifted by 0, 'positions',
* 'positions' * 2, ..., 'positions' * ('width' - 1)
*/
Add(c, aPos + ci[index], c, bOff, bMax);
}
if (++bank == banks)
{
break;
}
bOff += bMax;
aVal = (long)((ulong)aVal >> width);
}
}
while (++aPos < aLen);
if ((k += stride) >= top)
{
if (k >= 64)
{
break;
}
/*
* Adjustment for window setups with top == 63, the final bit (if any) is processed
* as the top-bit of a window
*/
k = 64 - width;
MASK &= MASK << (top - k);
}
/*
* After each position has been checked for all words of A, B is shifted up 1 place
*/
ShiftUp(c, aLen, bTotal, banks);
}
int ciPos = ci.Length;
while (--ciPos > 1)
{
if ((ciPos & 1L) == 0L)
{
/*
* For even numbers, shift contents and add to the half-position
*/
AddShiftedUp(c, ci[(uint)ciPos >> 1], c, ci[ciPos], cLen, positions);
}
else
{
/*
* For odd numbers, 'distribute' contents to the result and the next-lowest position
*/
Distribute(c, ci[ciPos], ci[ciPos - 1], ci[1], cLen);
}
}
/*
* Finally the raw answer is collected, reduce it against the reduction coefficients
*/
return ReduceResult(c, ci[1], cLen, m, ks);
}
