_ Git - BearSSL/blob - src/int/i31_muladd.c

   1 /*
   2  * Copyright (c) 2016 Thomas Pornin <pornin@bolet.org>
   3  *
   4  * Permission is hereby granted, free of charge, to any person obtaining
   5  * a copy of this software and associated documentation files (the
   6  * "Software"), to deal in the Software without restriction, including
   7  * without limitation the rights to use, copy, modify, merge, publish,
   8  * distribute, sublicense, and/or sell copies of the Software, and to
   9  * permit persons to whom the Software is furnished to do so, subject to
  10  * the following conditions:
  11  *
  12  * The above copyright notice and this permission notice shall be
  13  * included in all copies or substantial portions of the Software.
  14  *
  15  * THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND,
  16  * EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF
  17  * MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND
  18  * NONINFRINGEMENT. IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS
  19  * BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN
  20  * ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN
  21  * CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
  22  * SOFTWARE.
  23  */
  24
  25 #include "inner.h"
  26
  27 /* see inner.h */
  28 void
  29 br_i31_muladd_small(uint32_t *x, uint32_t z, const uint32_t *m)
  30 {
  31         uint32_t m_bitlen;
  32         unsigned mblr;
  33         size_t u, mlen;
  34         uint32_t a0, a1, b0, hi, g, q, tb;
  35         uint32_t under, over;
  36         uint32_t cc;
  37
  38         /*
  39          * We can test on the modulus bit length since we accept to
  40          * leak that length.
  41          */
  42         m_bitlen = m[0];
  43         if (m_bitlen == 0) {
  44                 return;
  45         }
  46         if (m_bitlen <= 31) {
  47                 uint32_t lo;
  48
  49                 hi = x[1] >> 1;
  50                 lo = (x[1] << 31) | z;
  51                 x[1] = br_rem(hi, lo, m[1]);
  52                 return;
  53         }
  54         mlen = (m_bitlen + 31) >> 5;
  55         mblr = (unsigned)m_bitlen & 31;
  56
  57         /*
  58          * Principle: we estimate the quotient (x*2^31+z)/m by
  59          * doing a 64/32 division with the high words.
  60          *
  61          * Let:
  62          *   w = 2^31
  63          *   a = (w*a0 + a1) * w^N + a2
  64          *   b = b0 * w^N + b2
  65          * such that:
  66          *   0 <= a0 < w
  67          *   0 <= a1 < w
  68          *   0 <= a2 < w^N
  69          *   w/2 <= b0 < w
  70          *   0 <= b2 < w^N
  71          *   a < w*b
  72          * I.e. the two top words of a are a0:a1, the top word of b is
  73          * b0, we ensured that b0 is "full" (high bit set), and a is
  74          * such that the quotient q = a/b fits on one word (0 <= q < w).
  75          *
  76          * If a = b*q + r (with 0 <= r < q), we can estimate q by
  77          * doing an Euclidean division on the top words:
  78          *   a0*w+a1 = b0*u + v  (with 0 <= v < b0)
  79          * Then the following holds:
  80          *   0 <= u <= w
  81          *   u-2 <= q <= u
  82          */
  83         hi = x[mlen];
  84         if (mblr == 0) {
  85                 a0 = x[mlen];
  86                 memmove(x + 2, x + 1, (mlen - 1) * sizeof *x);
  87                 x[1] = z;
  88                 a1 = x[mlen];
  89                 b0 = m[mlen];
  90         } else {
  91                 a0 = ((x[mlen] << (31 - mblr)) | (x[mlen - 1] >> mblr))
  92                         & 0x7FFFFFFF;
  93                 memmove(x + 2, x + 1, (mlen - 1) * sizeof *x);
  94                 x[1] = z;
  95                 a1 = ((x[mlen] << (31 - mblr)) | (x[mlen - 1] >> mblr))
  96                         & 0x7FFFFFFF;
  97                 b0 = ((m[mlen] << (31 - mblr)) | (m[mlen - 1] >> mblr))
  98                         & 0x7FFFFFFF;
  99         }
 100
 101         /*
 102          * We estimate a divisor q. If the quotient returned by br_div()
 103          * is g:
 104          * -- If a0 == b0 then g == 0; we want q = 0x7FFFFFFF.
 105          * -- Otherwise:
 106          *    -- if g == 0 then we set q = 0;
 107          *    -- otherwise, we set q = g - 1.
 108          * The properties described above then ensure that the true
 109          * quotient is q-1, q or q+1.
 110          *
 111          * Take care that a0, a1 and b0 are 31-bit words, not 32-bit. We
 112          * must adjust the parameters to br_div() accordingly.
 113          */
 114         g = br_div(a0 >> 1, a1 | (a0 << 31), b0);
 115         q = MUX(EQ(a0, b0), 0x7FFFFFFF, MUX(EQ(g, 0), 0, g - 1));
 116
 117         /*
 118          * We subtract q*m from x (with the extra high word of value 'hi').
 119          * Since q may be off by 1 (in either direction), we may have to
 120          * add or subtract m afterwards.
 121          *
 122          * The 'tb' flag will be true (1) at the end of the loop if the
 123          * result is greater than or equal to the modulus (not counting
 124          * 'hi' or the carry).
 125          */
 126         cc = 0;
 127         tb = 1;
 128         for (u = 1; u <= mlen; u ++) {
 129                 uint32_t mw, zw, xw, nxw;
 130                 uint64_t zl;
 131
 132                 mw = m[u];
 133                 zl = MUL31(mw, q) + cc;
 134                 cc = (uint32_t)(zl >> 31);
 135                 zw = (uint32_t)zl & (uint32_t)0x7FFFFFFF;
 136                 xw = x[u];
 137                 nxw = xw - zw;
 138                 cc += nxw >> 31;
 139                 nxw &= 0x7FFFFFFF;
 140                 x[u] = nxw;
 141                 tb = MUX(EQ(nxw, mw), tb, GT(nxw, mw));
 142         }
 143
 144         /*
 145          * If we underestimated q, then either cc < hi (one extra bit
 146          * beyond the top array word), or cc == hi and tb is true (no
 147          * extra bit, but the result is not lower than the modulus). In
 148          * these cases we must subtract m once.
 149          *
 150          * Otherwise, we may have overestimated, which will show as
 151          * cc > hi (thus a negative result). Correction is adding m once.
 152          */
 153         over = GT(cc, hi);
 154         under = ~over & (tb | LT(cc, hi));
 155         br_i31_add(x, m, over);
 156         br_i31_sub(x, m, under);
 157 }