2 * Copyright (c) 2018 Thomas Pornin <pornin@bolet.org>
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:
12 * The above copyright notice and this permission notice shall be
13 * included in all copies or substantial portions of the Software.
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
28 * In this file, we handle big integers with a custom format, i.e.
29 * without the usual one-word header. Value is split into 15-bit words,
30 * each stored in a 16-bit slot (top bit is zero) in little-endian
31 * order. The length (in words) is provided explicitly. In some cases,
32 * the value can be negative (using two's complement representation). In
33 * some cases, the top word is allowed to have a 16th bit.
37 * Negate big integer conditionally. The value consists of 'len' words,
38 * with 15 bits in each word (the top bit of each word should be 0,
39 * except possibly for the last word). If 'ctl' is 1, the negation is
40 * computed; otherwise, if 'ctl' is 0, then the value is unchanged.
43 cond_negate(uint16_t *a
, size_t len
, uint32_t ctl
)
50 for (k
= 0; k
< len
; k
++) {
61 * Finish modular reduction. Rules on input parameters:
63 * if neg = 1, then -m <= a < 0
64 * if neg = 0, then 0 <= a < 2*m
66 * If neg = 0, then the top word of a[] may use 16 bits.
68 * Also, modulus m must be odd.
71 finish_mod(uint16_t *a
, size_t len
, const uint16_t *m
, uint32_t neg
)
77 * First pass: compare a (assumed nonnegative) with m.
80 for (k
= 0; k
< len
; k
++) {
85 cc
= (aw
- mw
- cc
) >> 31;
90 * if neg = 1, then we must add m (regardless of cc)
91 * if neg = 0 and cc = 0, then we must subtract m
92 * if neg = 0 and cc = 1, then we must do nothing
95 ym
= -(neg
| (1 - cc
));
97 for (k
= 0; k
< len
; k
++) {
101 mw
= (m
[k
] ^ xm
) & ym
;
110 * a <- (a*pa+b*pb)/(2^15)
111 * b <- (a*qa+b*qb)/(2^15)
112 * The division is assumed to be exact (i.e. the low word is dropped).
113 * If the final a is negative, then it is negated. Similarly for b.
114 * Returned value is the combination of two bits:
115 * bit 0: 1 if a had to be negated, 0 otherwise
116 * bit 1: 1 if b had to be negated, 0 otherwise
118 * Factors pa, pb, qa and qb must be at most 2^15 in absolute value.
119 * Source integers a and b must be nonnegative; top word is not allowed
120 * to contain an extra 16th bit.
123 co_reduce(uint16_t *a
, uint16_t *b
, size_t len
,
124 int32_t pa
, int32_t pb
, int32_t qa
, int32_t qb
)
132 for (k
= 0; k
< len
; k
++) {
133 uint32_t wa
, wb
, za
, zb
;
140 * 0 <= wa <= 2^15 - 1
141 * 0 <= wb <= 2^15 - 1
144 * |za| <= (2^15-1)*(2^16) + (2^16-1) = 2^31 - 1
146 * Thus, the new value of cca is such that |cca| <= 2^16 - 1.
147 * The same applies to ccb.
151 za
= wa
* (uint32_t)pa
+ wb
* (uint32_t)pb
+ (uint32_t)cca
;
152 zb
= wa
* (uint32_t)qa
+ wb
* (uint32_t)qb
+ (uint32_t)ccb
;
154 a
[k
- 1] = za
& 0x7FFF;
155 b
[k
- 1] = zb
& 0x7FFF;
159 cca
= *(int16_t *)&tta
;
160 ccb
= *(int16_t *)&ttb
;
162 a
[len
- 1] = (uint16_t)cca
;
163 b
[len
- 1] = (uint16_t)ccb
;
164 nega
= (uint32_t)cca
>> 31;
165 negb
= (uint32_t)ccb
>> 31;
166 cond_negate(a
, len
, nega
);
167 cond_negate(b
, len
, negb
);
168 return nega
| (negb
<< 1);
173 * a <- (a*pa+b*pb)/(2^15) mod m
174 * b <- (a*qa+b*qb)/(2^15) mod m
176 * m0i is equal to -1/m[0] mod 2^15.
178 * Factors pa, pb, qa and qb must be at most 2^15 in absolute value.
179 * Source integers a and b must be nonnegative; top word is not allowed
180 * to contain an extra 16th bit.
183 co_reduce_mod(uint16_t *a
, uint16_t *b
, size_t len
,
184 int32_t pa
, int32_t pb
, int32_t qa
, int32_t qb
,
185 const uint16_t *m
, uint16_t m0i
)
188 int32_t cca
, ccb
, fa
, fb
;
192 fa
= ((a
[0] * (uint32_t)pa
+ b
[0] * (uint32_t)pb
) * m0i
) & 0x7FFF;
193 fb
= ((a
[0] * (uint32_t)qa
+ b
[0] * (uint32_t)qb
) * m0i
) & 0x7FFF;
194 for (k
= 0; k
< len
; k
++) {
195 uint32_t wa
, wb
, za
, zb
;
199 * In this loop, carries 'cca' and 'ccb' always fit on
200 * 17 bits (in absolute value).
204 za
= wa
* (uint32_t)pa
+ wb
* (uint32_t)pb
205 + m
[k
] * (uint32_t)fa
+ (uint32_t)cca
;
206 zb
= wa
* (uint32_t)qa
+ wb
* (uint32_t)qb
207 + m
[k
] * (uint32_t)fb
+ (uint32_t)ccb
;
209 a
[k
- 1] = za
& 0x7FFF;
210 b
[k
- 1] = zb
& 0x7FFF;
214 * The XOR-and-sub construction below does an arithmetic
215 * right shift in a portable way (technically, right-shifting
216 * a negative signed value is implementation-defined in C).
218 #define M ((uint32_t)1 << 16)
223 cca
= *(int32_t *)&tta
;
224 ccb
= *(int32_t *)&ttb
;
227 a
[len
- 1] = (uint32_t)cca
;
228 b
[len
- 1] = (uint32_t)ccb
;
234 * (this is a case of Montgomery reduction)
235 * The top word of 'a' and 'b' may have a 16-th bit set.
236 * We may have to add or subtract the modulus.
238 finish_mod(a
, len
, m
, (uint32_t)cca
>> 31);
239 finish_mod(b
, len
, m
, (uint32_t)ccb
>> 31);
244 br_i15_moddiv(uint16_t *x
, const uint16_t *y
, const uint16_t *m
, uint16_t m0i
,
248 * Algorithm is an extended binary GCD. We maintain four values
249 * a, b, u and v, with the following invariants:
251 * a * x = y * u mod m
252 * b * x = y * v mod m
254 * Starting values are:
261 * The formal definition of the algorithm is a sequence of steps:
263 * - If a is even, then a <- a/2 and u <- u/2 mod m.
264 * - Otherwise, if b is even, then b <- b/2 and v <- v/2 mod m.
265 * - Otherwise, if a > b, then a <- (a-b)/2 and u <- (u-v)/2 mod m.
266 * - Otherwise, b <- (b-a)/2 and v <- (v-u)/2 mod m.
268 * Algorithm stops when a = b. At that point, they both are equal
269 * to GCD(y,m); the modular division succeeds if that value is 1.
270 * The result of the modular division is then u (or v: both are
271 * equal at that point).
273 * Each step makes either a or b shrink by at least one bit; hence,
274 * if m has bit length k bits, then 2k-2 steps are sufficient.
277 * Though complexity is quadratic in the size of m, the bit-by-bit
278 * processing is not very efficient. We can speed up processing by
279 * remarking that the decisions are taken based only on observation
280 * of the top and low bits of a and b.
282 * In the loop below, at each iteration, we use the two top words
283 * of a and b, and the low words of a and b, to compute reduction
284 * parameters pa, pb, qa and qb such that the new values for a
287 * a' = (a*pa + b*pb) / (2^15)
288 * b' = (a*qa + b*qb) / (2^15)
290 * the division being exact.
292 * Since the choices are based on the top words, they may be slightly
293 * off, requiring an optional correction: if a' < 0, then we replace
294 * pa with -pa, and pb with -pb. The total length of a and b is
295 * thus reduced by at least 14 bits at each iteration.
297 * The stopping conditions are still the same, though: when a
298 * and b become equal, they must be both odd (since m is odd,
299 * the GCD cannot be even), therefore the next operation is a
300 * subtraction, and one of the values becomes 0. At that point,
301 * nothing else happens, i.e. one value is stuck at 0, and the
302 * other one is the GCD.
305 uint16_t *a
, *b
, *u
, *v
;
308 len
= (m
[0] + 15) >> 4;
313 memcpy(a
, y
+ 1, len
* sizeof *y
);
314 memcpy(b
, m
+ 1, len
* sizeof *m
);
315 memset(v
, 0, len
* sizeof *v
);
318 * Loop below ensures that a and b are reduced by some bits each,
319 * for a total of at least 14 bits.
321 for (num
= ((m
[0] - (m
[0] >> 4)) << 1) + 14; num
>= 14; num
-= 14) {
324 uint32_t a0
, a1
, b0
, b1
;
325 uint32_t a_hi
, b_hi
, a_lo
, b_lo
;
326 int32_t pa
, pb
, qa
, qb
;
330 * Extract top words of a and b. If j is the highest
331 * index >= 1 such that a[j] != 0 or b[j] != 0, then we want
332 * (a[j] << 15) + a[j - 1], and (b[j] << 15) + b[j - 1].
333 * If a and b are down to one word each, then we use a[0]
348 a0
^= (a0
^ aw
) & c0
;
349 a1
^= (a1
^ aw
) & c1
;
350 b0
^= (b0
^ bw
) & c0
;
351 b1
^= (b1
^ bw
) & c1
;
353 c0
&= (((aw
| bw
) + 0xFFFF) >> 16) - (uint32_t)1;
357 * If c1 = 0, then we grabbed two words for a and b.
358 * If c1 != 0 but c0 = 0, then we grabbed one word. It
359 * is not possible that c1 != 0 and c0 != 0, because that
360 * would mean that both integers are zero.
366 a_hi
= (a0
<< 15) + a1
;
367 b_hi
= (b0
<< 15) + b1
;
372 * Compute reduction factors:
377 * such that a' and b' are both multiple of 2^15, but are
378 * only marginally larger than a and b.
384 for (i
= 0; i
< 15; i
++) {
388 * a <- (a-b)/2 if: a is odd, b is odd, a_hi > b_hi
389 * b <- (b-a)/2 if: a is odd, b is odd, a_hi <= b_hi
390 * a <- a/2 if: a is even
391 * b <- b/2 if: a is odd, b is even
393 * We multiply a_lo and b_lo by 2 at each
394 * iteration, thus a division by 2 really is a
395 * non-multiplication by 2.
397 uint32_t r
, oa
, ob
, cAB
, cBA
, cA
;
400 * cAB = 1 if b must be subtracted from a
401 * cBA = 1 if a must be subtracted from b
402 * cA = 1 if a is divided by 2, 0 otherwise
406 * cAB and cBA cannot be both 1.
407 * if a is not divided by 2, b is.
410 oa
= (a_lo
>> i
) & 1;
411 ob
= (b_lo
>> i
) & 1;
413 cBA
= oa
& ob
& NOT(r
);
417 * Conditional subtractions.
421 pa
-= qa
& -(int32_t)cAB
;
422 pb
-= qb
& -(int32_t)cAB
;
425 qa
-= pa
& -(int32_t)cBA
;
426 qb
-= pb
& -(int32_t)cBA
;
431 a_lo
+= a_lo
& (cA
- 1);
432 pa
+= pa
& ((int32_t)cA
- 1);
433 pb
+= pb
& ((int32_t)cA
- 1);
434 a_hi
^= (a_hi
^ (a_hi
>> 1)) & -cA
;
436 qa
+= qa
& -(int32_t)cA
;
437 qb
+= qb
& -(int32_t)cA
;
438 b_hi
^= (b_hi
^ (b_hi
>> 1)) & (cA
- 1);
442 * Replace a and b with new values a' and b'.
444 r
= co_reduce(a
, b
, len
, pa
, pb
, qa
, qb
);
445 pa
-= pa
* ((r
& 1) << 1);
446 pb
-= pb
* ((r
& 1) << 1);
449 co_reduce_mod(u
, v
, len
, pa
, pb
, qa
, qb
, m
+ 1, m0i
);
453 * Now one of the arrays should be 0, and the other contains
454 * the GCD. If a is 0, then u is 0 as well, and v contains
455 * the division result.
456 * Result is correct if and only if GCD is 1.
458 r
= (a
[0] | b
[0]) ^ 1;
460 for (k
= 1; k
< len
; k
++) {