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