/*
* Copyright (c) 2016 Thomas Pornin
*
* Permission is hereby granted, free of charge, to any person obtaining
* a copy of this software and associated documentation files (the
* "Software"), to deal in the Software without restriction, including
* without limitation the rights to use, copy, modify, merge, publish,
* distribute, sublicense, and/or sell copies of the Software, and to
* permit persons to whom the Software is furnished to do so, subject to
* the following conditions:
*
* The above copyright notice and this permission notice shall be
* included in all copies or substantial portions of the Software.
*
* THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND,
* EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF
* MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND
* NONINFRINGEMENT. IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS
* BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN
* ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN
* CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
* SOFTWARE.
*/
#include "inner.h"
/* see inner.h */
void
br_i31_muladd_small(uint32_t *x, uint32_t z, const uint32_t *m)
{
uint32_t m_bitlen;
unsigned mblr;
size_t u, mlen;
uint32_t a0, a1, b0, hi, g, q, tb;
uint32_t under, over;
uint32_t cc;
/*
* We can test on the modulus bit length since we accept to
* leak that length.
*/
m_bitlen = m[0];
if (m_bitlen == 0) {
return;
}
if (m_bitlen <= 31) {
uint32_t lo;
hi = x[1] >> 1;
lo = (x[1] << 31) | z;
x[1] = br_rem(hi, lo, m[1]);
return;
}
mlen = (m_bitlen + 31) >> 5;
mblr = (unsigned)m_bitlen & 31;
/*
* Principle: we estimate the quotient (x*2^31+z)/m by
* doing a 64/32 division with the high words.
*
* Let:
* w = 2^31
* a = (w*a0 + a1) * w^N + a2
* b = b0 * w^N + b2
* such that:
* 0 <= a0 < w
* 0 <= a1 < w
* 0 <= a2 < w^N
* w/2 <= b0 < w
* 0 <= b2 < w^N
* a < w*b
* I.e. the two top words of a are a0:a1, the top word of b is
* b0, we ensured that b0 is "full" (high bit set), and a is
* such that the quotient q = a/b fits on one word (0 <= q < w).
*
* If a = b*q + r (with 0 <= r < q), we can estimate q by
* doing an Euclidean division on the top words:
* a0*w+a1 = b0*u + v (with 0 <= v < w)
* Then the following holds:
* 0 <= u <= w
* u-2 <= q <= u
*/
hi = x[mlen];
if (mblr == 0) {
a0 = x[mlen];
memmove(x + 2, x + 1, (mlen - 1) * sizeof *x);
x[1] = z;
a1 = x[mlen];
b0 = m[mlen];
} else {
a0 = ((x[mlen] << (31 - mblr)) | (x[mlen - 1] >> mblr))
& 0x7FFFFFFF;
memmove(x + 2, x + 1, (mlen - 1) * sizeof *x);
x[1] = z;
a1 = ((x[mlen] << (31 - mblr)) | (x[mlen - 1] >> mblr))
& 0x7FFFFFFF;
b0 = ((m[mlen] << (31 - mblr)) | (m[mlen - 1] >> mblr))
& 0x7FFFFFFF;
}
/*
* We estimate a divisor q. If the quotient returned by br_div()
* is g:
* -- If a0 == b0 then g == 0; we want q = 0x7FFFFFFF.
* -- Otherwise:
* -- if g == 0 then we set q = 0;
* -- otherwise, we set q = g - 1.
* The properties described above then ensure that the true
* quotient is q-1, q or q+1.
*
* Take care that a0, a1 and b0 are 31-bit words, not 32-bit. We
* must adjust the parameters to br_div() accordingly.
*/
g = br_div(a0 >> 1, a1 | (a0 << 31), b0);
q = MUX(EQ(a0, b0), 0x7FFFFFFF, MUX(EQ(g, 0), 0, g - 1));
/*
* We subtract q*m from x (with the extra high word of value 'hi').
* Since q may be off by 1 (in either direction), we may have to
* add or subtract m afterwards.
*
* The 'tb' flag will be true (1) at the end of the loop if the
* result is greater than or equal to the modulus (not counting
* 'hi' or the carry).
*/
cc = 0;
tb = 1;
for (u = 1; u <= mlen; u ++) {
uint32_t mw, zw, xw, nxw;
uint64_t zl;
mw = m[u];
zl = MUL31(mw, q) + cc;
cc = (uint32_t)(zl >> 31);
zw = (uint32_t)zl & (uint32_t)0x7FFFFFFF;
xw = x[u];
nxw = xw - zw;
cc += nxw >> 31;
nxw &= 0x7FFFFFFF;
x[u] = nxw;
tb = MUX(EQ(nxw, mw), tb, GT(nxw, mw));
}
/*
* If we underestimated q, then either cc < hi (one extra bit
* beyond the top array word), or cc == hi and tb is true (no
* extra bit, but the result is not lower than the modulus). In
* these cases we must subtract m once.
*
* Otherwise, we may have overestimated, which will show as
* cc > hi (thus a negative result). Correction is adding m once.
*/
over = GT(cc, hi);
under = ~over & (tb | LT(cc, hi));
br_i31_add(x, m, over);
br_i31_sub(x, m, under);
}