github
/
putty
зеркало из https://github.com/github/putty.git
1
0
Форкнуть 0
Код Задачи Пакеты Проекты Релизы Вики Активность
putty/sshbn.c

2184 строки
59 KiB
C
Исходник Ответственный История

/*
* Bignum routines for RSA and DH and stuff.
*/
#include <stdio.h>
#include <assert.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <ctype.h>
#include "misc.h"
#include "sshbn.h"
#define BIGNUM_INTERNAL
typedef BignumInt *Bignum;
#include "ssh.h"
BignumInt bnZero[1] = { 0 };
BignumInt bnOne[2] = { 1, 1 };
BignumInt bnTen[2] = { 1, 10 };
/*
* The Bignum format is an array of `BignumInt'. The first
* element of the array counts the remaining elements. The
* remaining elements express the actual number, base 2^BIGNUM_INT_BITS, _least_
* significant digit first. (So it's trivial to extract the bit
* with value 2^n for any n.)
*
* All Bignums in this module are positive. Negative numbers must
* be dealt with outside it.
*
* INVARIANT: the most significant word of any Bignum must be
* nonzero.
*/
Bignum Zero = bnZero, One = bnOne, Ten = bnTen;
static Bignum newbn(int length)
{
Bignum b;
assert(length >= 0 && length < INT_MAX / BIGNUM_INT_BITS);
b = snewn(length + 1, BignumInt);
memset(b, 0, (length + 1) * sizeof(*b));
b[0] = length;
return b;
}
void bn_restore_invariant(Bignum b)
{
while (b[0] > 1 && b[b[0]] == 0)
b[0]--;
}
Bignum copybn(Bignum orig)
{
Bignum b = snewn(orig[0] + 1, BignumInt);
if (!b)
abort(); /* FIXME */
memcpy(b, orig, (orig[0] + 1) * sizeof(*b));
return b;
}
void freebn(Bignum b)
{
/*
* Burn the evidence, just in case.
*/
smemclr(b, sizeof(b[0]) * (b[0] + 1));
sfree(b);
}
Bignum bn_power_2(int n)
{
Bignum ret;
assert(n >= 0);
ret = newbn(n / BIGNUM_INT_BITS + 1);
bignum_set_bit(ret, n, 1);
return ret;
}
/*
* Internal addition. Sets c = a - b, where 'a', 'b' and 'c' are all
* big-endian arrays of 'len' BignumInts. Returns the carry off the
* top.
*/
static BignumCarry internal_add(const BignumInt *a, const BignumInt *b,
BignumInt *c, int len)
{
int i;
BignumCarry carry = 0;
for (i = len-1; i >= 0; i--)
BignumADC(c[i], carry, a[i], b[i], carry);
return (BignumInt)carry;
}
/*
* Internal subtraction. Sets c = a - b, where 'a', 'b' and 'c' are
* all big-endian arrays of 'len' BignumInts. Any borrow from the top
* is ignored.
*/
static void internal_sub(const BignumInt *a, const BignumInt *b,
BignumInt *c, int len)
{
int i;
BignumCarry carry = 1;
for (i = len-1; i >= 0; i--)
BignumADC(c[i], carry, a[i], ~b[i], carry);
}
/*
* Compute c = a * b.
* Input is in the first len words of a and b.
* Result is returned in the first 2*len words of c.
*
* 'scratch' must point to an array of BignumInt of size at least
* mul_compute_scratch(len). (This covers the needs of internal_mul
* and all its recursive calls to itself.)
*/
#define KARATSUBA_THRESHOLD 50
static int mul_compute_scratch(int len)
{
int ret = 0;
while (len > KARATSUBA_THRESHOLD) {
int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */
int midlen = botlen + 1;
ret += 4*midlen;
len = midlen;
}
return ret;
}
static void internal_mul(const BignumInt *a, const BignumInt *b,
BignumInt *c, int len, BignumInt *scratch)
{
if (len > KARATSUBA_THRESHOLD) {
int i;
/*
* Karatsuba divide-and-conquer algorithm. Cut each input in
* half, so that it's expressed as two big 'digits' in a giant
* base D:
*
* a = a_1 D + a_0
* b = b_1 D + b_0
*
* Then the product is of course
*
* ab = a_1 b_1 D^2 + (a_1 b_0 + a_0 b_1) D + a_0 b_0
*
* and we compute the three coefficients by recursively
* calling ourself to do half-length multiplications.
*
* The clever bit that makes this worth doing is that we only
* need _one_ half-length multiplication for the central
* coefficient rather than the two that it obviouly looks
* like, because we can use a single multiplication to compute
*
* (a_1 + a_0) (b_1 + b_0) = a_1 b_1 + a_1 b_0 + a_0 b_1 + a_0 b_0
*
* and then we subtract the other two coefficients (a_1 b_1
* and a_0 b_0) which we were computing anyway.
*
* Hence we get to multiply two numbers of length N in about
* three times as much work as it takes to multiply numbers of
* length N/2, which is obviously better than the four times
* as much work it would take if we just did a long
* conventional multiply.
*/
int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */
int midlen = botlen + 1;
BignumCarry carry;
#ifdef KARA_DEBUG
int i;
#endif
/*
* The coefficients a_1 b_1 and a_0 b_0 just avoid overlapping
* in the output array, so we can compute them immediately in
* place.
*/
#ifdef KARA_DEBUG
printf("a1,a0 = 0x");
for (i = 0; i < len; i++) {
if (i == toplen) printf(", 0x");
printf("%0*x", BIGNUM_INT_BITS/4, a[i]);
}
printf("\n");
printf("b1,b0 = 0x");
for (i = 0; i < len; i++) {
if (i == toplen) printf(", 0x");
printf("%0*x", BIGNUM_INT_BITS/4, b[i]);
}
printf("\n");
#endif
/* a_1 b_1 */
internal_mul(a, b, c, toplen, scratch);
#ifdef KARA_DEBUG
printf("a1b1 = 0x");
for (i = 0; i < 2*toplen; i++) {
printf("%0*x", BIGNUM_INT_BITS/4, c[i]);
}
printf("\n");
#endif
/* a_0 b_0 */
internal_mul(a + toplen, b + toplen, c + 2*toplen, botlen, scratch);
#ifdef KARA_DEBUG
printf("a0b0 = 0x");
for (i = 0; i < 2*botlen; i++) {
printf("%0*x", BIGNUM_INT_BITS/4, c[2*toplen+i]);
}
printf("\n");
#endif
/* Zero padding. midlen exceeds toplen by at most 2, so just
* zero the first two words of each input and the rest will be
* copied over. */
scratch[0] = scratch[1] = scratch[midlen] = scratch[midlen+1] = 0;
for (i = 0; i < toplen; i++) {
scratch[midlen - toplen + i] = a[i]; /* a_1 */
scratch[2*midlen - toplen + i] = b[i]; /* b_1 */
}
/* compute a_1 + a_0 */
scratch[0] = internal_add(scratch+1, a+toplen, scratch+1, botlen);
#ifdef KARA_DEBUG
printf("a1plusa0 = 0x");
for (i = 0; i < midlen; i++) {
printf("%0*x", BIGNUM_INT_BITS/4, scratch[i]);
}
printf("\n");
#endif
/* compute b_1 + b_0 */
scratch[midlen] = internal_add(scratch+midlen+1, b+toplen,
scratch+midlen+1, botlen);
#ifdef KARA_DEBUG
printf("b1plusb0 = 0x");
for (i = 0; i < midlen; i++) {
printf("%0*x", BIGNUM_INT_BITS/4, scratch[midlen+i]);
}
printf("\n");
#endif
/*
* Now we can do the third multiplication.
*/
internal_mul(scratch, scratch + midlen, scratch + 2*midlen, midlen,
scratch + 4*midlen);
#ifdef KARA_DEBUG
printf("a1plusa0timesb1plusb0 = 0x");
for (i = 0; i < 2*midlen; i++) {
printf("%0*x", BIGNUM_INT_BITS/4, scratch[2*midlen+i]);
}
printf("\n");
#endif
/*
* Now we can reuse the first half of 'scratch' to compute the
* sum of the outer two coefficients, to subtract from that
* product to obtain the middle one.
*/
scratch[0] = scratch[1] = scratch[2] = scratch[3] = 0;
for (i = 0; i < 2*toplen; i++)
scratch[2*midlen - 2*toplen + i] = c[i];
scratch[1] = internal_add(scratch+2, c + 2*toplen,
scratch+2, 2*botlen);
#ifdef KARA_DEBUG
printf("a1b1plusa0b0 = 0x");
for (i = 0; i < 2*midlen; i++) {
printf("%0*x", BIGNUM_INT_BITS/4, scratch[i]);
}
printf("\n");
#endif
internal_sub(scratch + 2*midlen, scratch,
scratch + 2*midlen, 2*midlen);
#ifdef KARA_DEBUG
printf("a1b0plusa0b1 = 0x");
for (i = 0; i < 2*midlen; i++) {
printf("%0*x", BIGNUM_INT_BITS/4, scratch[2*midlen+i]);
}
printf("\n");
#endif
/*
* And now all we need to do is to add that middle coefficient
* back into the output. We may have to propagate a carry
* further up the output, but we can be sure it won't
* propagate right the way off the top.
*/
carry = internal_add(c + 2*len - botlen - 2*midlen,
scratch + 2*midlen,
c + 2*len - botlen - 2*midlen, 2*midlen);
i = 2*len - botlen - 2*midlen - 1;
while (carry) {
assert(i >= 0);
BignumADC(c[i], carry, c[i], 0, carry);
i--;
}
#ifdef KARA_DEBUG
printf("ab = 0x");
for (i = 0; i < 2*len; i++) {
printf("%0*x", BIGNUM_INT_BITS/4, c[i]);
}
printf("\n");
#endif
} else {
int i;
BignumInt carry;
const BignumInt *ap, *bp;
BignumInt *cp, *cps;
/*
* Multiply in the ordinary O(N^2) way.
*/
for (i = 0; i < 2 * len; i++)
c[i] = 0;
for (cps = c + 2*len, ap = a + len; ap-- > a; cps--) {
carry = 0;
for (cp = cps, bp = b + len; cp--, bp-- > b ;)
BignumMULADD2(carry, *cp, *ap, *bp, *cp, carry);
*cp = carry;
}
}
}
/*
* Variant form of internal_mul used for the initial step of
* Montgomery reduction. Only bothers outputting 'len' words
* (everything above that is thrown away).
*/
static void internal_mul_low(const BignumInt *a, const BignumInt *b,
BignumInt *c, int len, BignumInt *scratch)
{
if (len > KARATSUBA_THRESHOLD) {
int i;
/*
* Karatsuba-aware version of internal_mul_low. As before, we
* express each input value as a shifted combination of two
* halves:
*
* a = a_1 D + a_0
* b = b_1 D + b_0
*
* Then the full product is, as before,
*
* ab = a_1 b_1 D^2 + (a_1 b_0 + a_0 b_1) D + a_0 b_0
*
* Provided we choose D on the large side (so that a_0 and b_0
* are _at least_ as long as a_1 and b_1), we don't need the
* topmost term at all, and we only need half of the middle
* term. So there's no point in doing the proper Karatsuba
* optimisation which computes the middle term using the top
* one, because we'd take as long computing the top one as
* just computing the middle one directly.
*
* So instead, we do a much more obvious thing: we call the
* fully optimised internal_mul to compute a_0 b_0, and we
* recursively call ourself to compute the _bottom halves_ of
* a_1 b_0 and a_0 b_1, each of which we add into the result
* in the obvious way.
*
* In other words, there's no actual Karatsuba _optimisation_
* in this function; the only benefit in doing it this way is
* that we call internal_mul proper for a large part of the
* work, and _that_ can optimise its operation.
*/
int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */
/*
* Scratch space for the various bits and pieces we're going
* to be adding together: we need botlen*2 words for a_0 b_0
* (though we may end up throwing away its topmost word), and
* toplen words for each of a_1 b_0 and a_0 b_1. That adds up
* to exactly 2*len.
*/
/* a_0 b_0 */
internal_mul(a + toplen, b + toplen, scratch + 2*toplen, botlen,
scratch + 2*len);
/* a_1 b_0 */
internal_mul_low(a, b + len - toplen, scratch + toplen, toplen,
scratch + 2*len);
/* a_0 b_1 */
internal_mul_low(a + len - toplen, b, scratch, toplen,
scratch + 2*len);
/* Copy the bottom half of the big coefficient into place */
for (i = 0; i < botlen; i++)
c[toplen + i] = scratch[2*toplen + botlen + i];
/* Add the two small coefficients, throwing away the returned carry */
internal_add(scratch, scratch + toplen, scratch, toplen);
/* And add that to the large coefficient, leaving the result in c. */
internal_add(scratch, scratch + 2*toplen + botlen - toplen,
c, toplen);
} else {
int i;
BignumInt carry;
const BignumInt *ap, *bp;
BignumInt *cp, *cps;
/*
* Multiply in the ordinary O(N^2) way.
*/
for (i = 0; i < len; i++)
c[i] = 0;
for (cps = c + len, ap = a + len; ap-- > a; cps--) {
carry = 0;
for (cp = cps, bp = b + len; bp--, cp-- > c ;)
BignumMULADD2(carry, *cp, *ap, *bp, *cp, carry);
}
}
}
/*
* Montgomery reduction. Expects x to be a big-endian array of 2*len
* BignumInts whose value satisfies 0 <= x < rn (where r = 2^(len *
* BIGNUM_INT_BITS) is the Montgomery base). Returns in the same array
* a value x' which is congruent to xr^{-1} mod n, and satisfies 0 <=
* x' < n.
*
* 'n' and 'mninv' should be big-endian arrays of 'len' BignumInts
* each, containing respectively n and the multiplicative inverse of
* -n mod r.
*
* 'tmp' is an array of BignumInt used as scratch space, of length at
* least 3*len + mul_compute_scratch(len).
*/
static void monty_reduce(BignumInt *x, const BignumInt *n,
const BignumInt *mninv, BignumInt *tmp, int len)
{
int i;
BignumInt carry;
/*
* Multiply x by (-n)^{-1} mod r. This gives us a value m such
* that mn is congruent to -x mod r. Hence, mn+x is an exact
* multiple of r, and is also (obviously) congruent to x mod n.
*/
internal_mul_low(x + len, mninv, tmp, len, tmp + 3*len);
/*
* Compute t = (mn+x)/r in ordinary, non-modular, integer
* arithmetic. By construction this is exact, and is congruent mod
* n to x * r^{-1}, i.e. the answer we want.
*
* The following multiply leaves that answer in the _most_
* significant half of the 'x' array, so then we must shift it
* down.
*/
internal_mul(tmp, n, tmp+len, len, tmp + 3*len);
carry = internal_add(x, tmp+len, x, 2*len);
for (i = 0; i < len; i++)
x[len + i] = x[i], x[i] = 0;
/*
* Reduce t mod n. This doesn't require a full-on division by n,
* but merely a test and single optional subtraction, since we can
* show that 0 <= t < 2n.
*
* Proof:
* + we computed m mod r, so 0 <= m < r.
* + so 0 <= mn < rn, obviously
* + hence we only need 0 <= x < rn to guarantee that 0 <= mn+x < 2rn
* + yielding 0 <= (mn+x)/r < 2n as required.
*/
if (!carry) {
for (i = 0; i < len; i++)
if (x[len + i] != n[i])
break;
}
if (carry || i >= len || x[len + i] > n[i])
internal_sub(x+len, n, x+len, len);
}
static void internal_add_shifted(BignumInt *number,
BignumInt n, int shift)
{
int word = 1 + (shift / BIGNUM_INT_BITS);
int bshift = shift % BIGNUM_INT_BITS;
BignumInt addendh, addendl;
BignumCarry carry;
addendl = n << bshift;
addendh = (bshift == 0 ? 0 : n >> (BIGNUM_INT_BITS - bshift));
assert(word <= number[0]);
BignumADC(number[word], carry, number[word], addendl, 0);
word++;
if (!addendh && !carry)
return;
assert(word <= number[0]);
BignumADC(number[word], carry, number[word], addendh, carry);
word++;
while (carry) {
assert(word <= number[0]);
BignumADC(number[word], carry, number[word], 0, carry);
word++;
}
}
static int bn_clz(BignumInt x)
{
/*
* Count the leading zero bits in x. Equivalently, how far left
* would we need to shift x to make its top bit set?
*
* Precondition: x != 0.
*/
/* FIXME: would be nice to put in some compiler intrinsics under
* ifdef here */
int i, ret = 0;
for (i = BIGNUM_INT_BITS / 2; i != 0; i >>= 1) {
if ((x >> (BIGNUM_INT_BITS-i)) == 0) {
x <<= i;
ret += i;
}
}
return ret;
}
static BignumInt reciprocal_word(BignumInt d)
{
BignumInt dshort, recip, prodh, prodl;
int corrections;
/*
* Input: a BignumInt value d, with its top bit set.
*/
assert(d >> (BIGNUM_INT_BITS-1) == 1);
/*
* Output: a value, shifted to fill a BignumInt, which is strictly
* less than 1/(d+1), i.e. is an *under*-estimate (but by as
* little as possible within the constraints) of the reciprocal of
* any number whose first BIGNUM_INT_BITS bits match d.
*
* Ideally we'd like to _totally_ fill BignumInt, i.e. always
* return a value with the top bit set. Unfortunately we can't
* quite guarantee that for all inputs and also return a fixed
* exponent. So instead we take our reciprocal to be
* 2^(BIGNUM_INT_BITS*2-1) / d, so that it has the top bit clear
* only in the exceptional case where d takes exactly the maximum
* value BIGNUM_INT_MASK; in that case, the top bit is clear and
* the next bit down is set.
*/
/*
* Start by computing a half-length version of the answer, by
* straightforward division within a BignumInt.
*/
dshort = (d >> (BIGNUM_INT_BITS/2)) + 1;
recip = (BIGNUM_TOP_BIT + dshort - 1) / dshort;
recip <<= BIGNUM_INT_BITS - BIGNUM_INT_BITS/2;
/*
* Newton-Raphson iteration to improve that starting reciprocal
* estimate: take f(x) = d - 1/x, and then the N-R formula gives
* x_new = x - f(x)/f'(x) = x - (d-1/x)/(1/x^2) = x(2-d*x). Or,
* taking our fixed-point representation into account, take f(x)
* to be d - K/x (where K = 2^(BIGNUM_INT_BITS*2-1) as discussed
* above) and then we get (2K - d*x) * x/K.
*
* Newton-Raphson doubles the number of correct bits at every
* iteration, and the initial division above already gave us half
* the output word, so it's only worth doing one iteration.
*/
BignumMULADD(prodh, prodl, recip, d, recip);
prodl = ~prodl;
prodh = ~prodh;
{
BignumCarry c;
BignumADC(prodl, c, prodl, 1, 0);
prodh += c;
}
BignumMUL(prodh, prodl, prodh, recip);
recip = (prodh << 1) | (prodl >> (BIGNUM_INT_BITS-1));
/*
* Now make sure we have the best possible reciprocal estimate,
* before we return it. We might have been off by a handful either
* way - not enough to bother with any better-thought-out kind of
* correction loop.
*/
BignumMULADD(prodh, prodl, recip, d, recip);
corrections = 0;
if (prodh >= BIGNUM_TOP_BIT) {
do {
BignumCarry c = 1;
BignumADC(prodl, c, prodl, ~d, c); prodh += BIGNUM_INT_MASK + c;
recip--;
corrections++;
} while (prodh >= ((BignumInt)1 << (BIGNUM_INT_BITS-1)));
} else {
while (1) {
BignumInt newprodh, newprodl;
BignumCarry c = 0;
BignumADC(newprodl, c, prodl, d, c); newprodh = prodh + c;
if (newprodh >= BIGNUM_TOP_BIT)
break;
prodh = newprodh;
prodl = newprodl;
recip++;
corrections++;
}
}
return recip;
}
/*
* Compute a = a % m.
* Input in first alen words of a and first mlen words of m.
* Output in first alen words of a
* (of which first alen-mlen words will be zero).
* Quotient is accumulated in the `quotient' array, which is a Bignum
* rather than the internal bigendian format.
*
* 'recip' must be the result of calling reciprocal_word() on the top
* BIGNUM_INT_BITS of the modulus (denoted m0 in comments below), with
* the topmost set bit normalised to the MSB of the input to
* reciprocal_word. 'rshift' is how far left the top nonzero word of
* the modulus had to be shifted to set that top bit.
*/
static void internal_mod(BignumInt *a, int alen,
BignumInt *m, int mlen,
BignumInt *quot, BignumInt recip, int rshift)
{
int i, k;
#ifdef DIVISION_DEBUG
{
int d;
printf("start division, m=0x");
for (d = 0; d < mlen; d++)
printf("%0*llx", BIGNUM_INT_BITS/4, (unsigned long long)m[d]);
printf(", recip=%#0*llx, rshift=%d\n",
BIGNUM_INT_BITS/4, (unsigned long long)recip, rshift);
}
#endif
/*
* Repeatedly use that reciprocal estimate to get a decent number
* of quotient bits, and subtract off the resulting multiple of m.
*
* Normally we expect to terminate this loop by means of finding
* out q=0 part way through, but one way in which we might not get
* that far in the first place is if the input a is actually zero,
* in which case we'll discard zero words from the front of a
* until we reach the termination condition in the for statement
* here.
*/
for (i = 0; i <= alen - mlen ;) {
BignumInt product;
BignumInt aword, q;
int shift, full_bitoffset, bitoffset, wordoffset;
#ifdef DIVISION_DEBUG
{
int d;
printf("main loop, a=0x");
for (d = 0; d < alen; d++)
printf("%0*llx", BIGNUM_INT_BITS/4, (unsigned long long)a[d]);
printf("\n");
}
#endif
if (a[i] == 0) {
#ifdef DIVISION_DEBUG
printf("zero word at i=%d\n", i);
#endif
i++;
continue;
}
aword = a[i];
shift = bn_clz(aword);
aword <<= shift;
if (shift > 0 && i+1 < alen)
aword |= a[i+1] >> (BIGNUM_INT_BITS - shift);
{
BignumInt unused;
BignumMUL(q, unused, recip, aword);
(void)unused;
}
#ifdef DIVISION_DEBUG
printf("i=%d, aword=%#0*llx, shift=%d, q=%#0*llx\n",
i, BIGNUM_INT_BITS/4, (unsigned long long)aword,
shift, BIGNUM_INT_BITS/4, (unsigned long long)q);
#endif
/*
* Work out the right bit and word offsets to use when
* subtracting q*m from a.
*
* aword was taken from a[i], which means its LSB was at bit
* position (alen-1-i) * BIGNUM_INT_BITS. But then we shifted
* it left by 'shift', so now the low bit of aword corresponds
* to bit position (alen-1-i) * BIGNUM_INT_BITS - shift, i.e.
* aword is approximately equal to a / 2^(that).
*
* m0 comes from the top word of mod, so its LSB is at bit
* position (mlen-1) * BIGNUM_INT_BITS - rshift, i.e. it can
* be considered to be m / 2^(that power). 'recip' is the
* reciprocal of m0, times 2^(BIGNUM_INT_BITS*2-1), i.e. it's
* about 2^((mlen+1) * BIGNUM_INT_BITS - rshift - 1) / m.
*
* Hence, recip * aword is approximately equal to the product
* of those, which simplifies to
*
* a/m * 2^((mlen+2+i-alen)*BIGNUM_INT_BITS + shift - rshift - 1)
*
* But we've also shifted recip*aword down by BIGNUM_INT_BITS
* to form q, so we have
*
* q ~= a/m * 2^((mlen+1+i-alen)*BIGNUM_INT_BITS + shift - rshift - 1)
*
* and hence, when we now compute q*m, it will be about
* a*2^(all that lot), i.e. the negation of that expression is
* how far left we have to shift the product q*m to make it
* approximately equal to a.
*/
full_bitoffset = -((mlen+1+i-alen)*BIGNUM_INT_BITS + shift-rshift-1);
#ifdef DIVISION_DEBUG
printf("full_bitoffset=%d\n", full_bitoffset);
#endif
if (full_bitoffset < 0) {
/*
* If we find ourselves needing to shift q*m _right_, that
* means we've reached the bottom of the quotient. Clip q
* so that its right shift becomes zero, and if that means
* q becomes _actually_ zero, this loop is done.
*/
if (full_bitoffset <= -BIGNUM_INT_BITS)
break;
q >>= -full_bitoffset;
full_bitoffset = 0;
if (!q)
break;
#ifdef DIVISION_DEBUG
printf("now full_bitoffset=%d, q=%#0*llx\n",
full_bitoffset, BIGNUM_INT_BITS/4, (unsigned long long)q);
#endif
}
wordoffset = full_bitoffset / BIGNUM_INT_BITS;
bitoffset = full_bitoffset % BIGNUM_INT_BITS;
#ifdef DIVISION_DEBUG
printf("wordoffset=%d, bitoffset=%d\n", wordoffset, bitoffset);
#endif
/* wordoffset as computed above is the offset between the LSWs
* of m and a. But in fact m and a are stored MSW-first, so we
* need to adjust it to be the offset between the actual array
* indices, and flip the sign too. */
wordoffset = alen - mlen - wordoffset;
if (bitoffset == 0) {
BignumCarry c = 1;
BignumInt prev_hi_word = 0;
for (k = mlen - 1; wordoffset+k >= i; k--) {
BignumInt mword = k<0 ? 0 : m[k];
BignumMULADD(prev_hi_word, product, q, mword, prev_hi_word);
#ifdef DIVISION_DEBUG
printf(" aligned sub: product word for m[%d] = %#0*llx\n",
k, BIGNUM_INT_BITS/4,
(unsigned long long)product);
#endif
#ifdef DIVISION_DEBUG
printf(" aligned sub: subtrahend for a[%d] = %#0*llx\n",
wordoffset+k, BIGNUM_INT_BITS/4,
(unsigned long long)product);
#endif
BignumADC(a[wordoffset+k], c, a[wordoffset+k], ~product, c);
}
} else {
BignumInt add_word = 0;
BignumInt c = 1;
BignumInt prev_hi_word = 0;
for (k = mlen - 1; wordoffset+k >= i; k--) {
BignumInt mword = k<0 ? 0 : m[k];
BignumMULADD(prev_hi_word, product, q, mword, prev_hi_word);
#ifdef DIVISION_DEBUG
printf(" unaligned sub: product word for m[%d] = %#0*llx\n",
k, BIGNUM_INT_BITS/4,
(unsigned long long)product);
#endif
add_word |= product << bitoffset;
#ifdef DIVISION_DEBUG
printf(" unaligned sub: subtrahend for a[%d] = %#0*llx\n",
wordoffset+k,
BIGNUM_INT_BITS/4, (unsigned long long)add_word);
#endif
BignumADC(a[wordoffset+k], c, a[wordoffset+k], ~add_word, c);
add_word = product >> (BIGNUM_INT_BITS - bitoffset);
}
}
if (quot) {
#ifdef DIVISION_DEBUG
printf("adding quotient word %#0*llx << %d\n",
BIGNUM_INT_BITS/4, (unsigned long long)q, full_bitoffset);
#endif
internal_add_shifted(quot, q, full_bitoffset);
#ifdef DIVISION_DEBUG
{
int d;
printf("now quot=0x");
for (d = quot[0]; d > 0; d--)
printf("%0*llx", BIGNUM_INT_BITS/4,
(unsigned long long)quot[d]);
printf("\n");
}
#endif
}
}
#ifdef DIVISION_DEBUG
{
int d;
printf("end main loop, a=0x");
for (d = 0; d < alen; d++)
printf("%0*llx", BIGNUM_INT_BITS/4, (unsigned long long)a[d]);
if (quot) {
printf(", quot=0x");
for (d = quot[0]; d > 0; d--)
printf("%0*llx", BIGNUM_INT_BITS/4,
(unsigned long long)quot[d]);
}
printf("\n");
}
#endif
/*
* The above loop should terminate with the remaining value in a
* being strictly less than 2*m (if a >= 2*m then we should always
* have managed to get a nonzero q word), but we can't guarantee
* that it will be strictly less than m: consider a case where the
* remainder is 1, and another where the remainder is m-1. By the
* time a contains a value that's _about m_, you clearly can't
* distinguish those cases by looking at only the top word of a -
* you have to go all the way down to the bottom before you find
* out whether it's just less or just more than m.
*
* Hence, we now do a final fixup in which we subtract one last
* copy of m, or don't, accordingly. We should never have to
* subtract more than one copy of m here.
*/
for (i = 0; i < alen; i++) {
/* Compare a with m, word by word, from the MSW down. As soon
* as we encounter a difference, we know whether we need the
* fixup. */
int mindex = mlen-alen+i;
BignumInt mword = mindex < 0 ? 0 : m[mindex];
if (a[i] < mword) {
#ifdef DIVISION_DEBUG
printf("final fixup not needed, a < m\n");
#endif
return;
} else if (a[i] > mword) {
#ifdef DIVISION_DEBUG
printf("final fixup is needed, a > m\n");
#endif
break;
}
/* If neither of those cases happened, the words are the same,
* so keep going and look at the next one. */
}
#ifdef DIVISION_DEBUG
if (i == mlen) /* if we printed neither of the above diagnostics */
printf("final fixup is needed, a == m\n");
#endif
/*
* If we got here without returning, then a >= m, so we must
* subtract m, and increment the quotient.
*/
{
BignumCarry c = 1;
for (i = alen - 1; i >= 0; i--) {
int mindex = mlen-alen+i;
BignumInt mword = mindex < 0 ? 0 : m[mindex];
BignumADC(a[i], c, a[i], ~mword, c);
}
}
if (quot)
internal_add_shifted(quot, 1, 0);
#ifdef DIVISION_DEBUG
{
int d;
printf("after final fixup, a=0x");
for (d = 0; d < alen; d++)
printf("%0*llx", BIGNUM_INT_BITS/4, (unsigned long long)a[d]);
if (quot) {
printf(", quot=0x");
for (d = quot[0]; d > 0; d--)
printf("%0*llx", BIGNUM_INT_BITS/4,
(unsigned long long)quot[d]);
}
printf("\n");
}
#endif
}
/*
* Compute (base ^ exp) % mod, the pedestrian way.
*/
Bignum modpow_simple(Bignum base_in, Bignum exp, Bignum mod)
{
BignumInt *a, *b, *n, *m, *scratch;
BignumInt recip;
int rshift;
int mlen, scratchlen, i, j;
Bignum base, result;
/*
* The most significant word of mod needs to be non-zero. It
* should already be, but let's make sure.
*/
assert(mod[mod[0]] != 0);
/*
* Make sure the base is smaller than the modulus, by reducing
* it modulo the modulus if not.
*/
base = bigmod(base_in, mod);
/* Allocate m of size mlen, copy mod to m */
/* We use big endian internally */
mlen = mod[0];
m = snewn(mlen, BignumInt);
for (j = 0; j < mlen; j++)
m[j] = mod[mod[0] - j];
/* Allocate n of size mlen, copy base to n */
n = snewn(mlen, BignumInt);
i = mlen - base[0];
for (j = 0; j < i; j++)
n[j] = 0;
for (j = 0; j < (int)base[0]; j++)
n[i + j] = base[base[0] - j];
/* Allocate a and b of size 2*mlen. Set a = 1 */
a = snewn(2 * mlen, BignumInt);
b = snewn(2 * mlen, BignumInt);
for (i = 0; i < 2 * mlen; i++)
a[i] = 0;
a[2 * mlen - 1] = 1;
/* Scratch space for multiplies */
scratchlen = mul_compute_scratch(mlen);
scratch = snewn(scratchlen, BignumInt);
/* Skip leading zero bits of exp. */
i = 0;
j = BIGNUM_INT_BITS-1;
while (i < (int)exp[0] && (exp[exp[0] - i] & ((BignumInt)1 << j)) == 0) {
j--;
if (j < 0) {
i++;
j = BIGNUM_INT_BITS-1;
}
}
/* Compute reciprocal of the top full word of the modulus */
{
BignumInt m0 = m[0];
rshift = bn_clz(m0);
if (rshift) {
m0 <<= rshift;
if (mlen > 1)
m0 |= m[1] >> (BIGNUM_INT_BITS - rshift);
}
recip = reciprocal_word(m0);
}
/* Main computation */
while (i < (int)exp[0]) {
while (j >= 0) {
internal_mul(a + mlen, a + mlen, b, mlen, scratch);
internal_mod(b, mlen * 2, m, mlen, NULL, recip, rshift);
if ((exp[exp[0] - i] & ((BignumInt)1 << j)) != 0) {
internal_mul(b + mlen, n, a, mlen, scratch);
internal_mod(a, mlen * 2, m, mlen, NULL, recip, rshift);
} else {
BignumInt *t;
t = a;
a = b;
b = t;
}
j--;
}
i++;
j = BIGNUM_INT_BITS-1;
}
/* Copy result to buffer */
result = newbn(mod[0]);
for (i = 0; i < mlen; i++)
result[result[0] - i] = a[i + mlen];
while (result[0] > 1 && result[result[0]] == 0)
result[0]--;
/* Free temporary arrays */
smemclr(a, 2 * mlen * sizeof(*a));
sfree(a);
smemclr(scratch, scratchlen * sizeof(*scratch));
sfree(scratch);
smemclr(b, 2 * mlen * sizeof(*b));
sfree(b);
smemclr(m, mlen * sizeof(*m));
sfree(m);
smemclr(n, mlen * sizeof(*n));
sfree(n);
freebn(base);
return result;
}
/*
* Compute (base ^ exp) % mod. Uses the Montgomery multiplication
* technique where possible, falling back to modpow_simple otherwise.
*/
Bignum modpow(Bignum base_in, Bignum exp, Bignum mod)
{
BignumInt *a, *b, *x, *n, *mninv, *scratch;
int len, scratchlen, i, j;
Bignum base, base2, r, rn, inv, result;
/*
* The most significant word of mod needs to be non-zero. It
* should already be, but let's make sure.
*/
assert(mod[mod[0]] != 0);
/*
* mod had better be odd, or we can't do Montgomery multiplication
* using a power of two at all.
*/
if (!(mod[1] & 1))
return modpow_simple(base_in, exp, mod);
/*
* Make sure the base is smaller than the modulus, by reducing
* it modulo the modulus if not.
*/
base = bigmod(base_in, mod);
/*
* Compute the inverse of n mod r, for monty_reduce. (In fact we
* want the inverse of _minus_ n mod r, but we'll sort that out
* below.)
*/
len = mod[0];
r = bn_power_2(BIGNUM_INT_BITS * len);
inv = modinv(mod, r);
assert(inv); /* cannot fail, since mod is odd and r is a power of 2 */
/*
* Multiply the base by r mod n, to get it into Montgomery
* representation.
*/
base2 = modmul(base, r, mod);
freebn(base);
base = base2;
rn = bigmod(r, mod); /* r mod n, i.e. Montgomerified 1 */
freebn(r); /* won't need this any more */
/*
* Set up internal arrays of the right lengths, in big-endian
* format, containing the base, the modulus, and the modulus's
* inverse.
*/
n = snewn(len, BignumInt);
for (j = 0; j < len; j++)
n[len - 1 - j] = mod[j + 1];
mninv = snewn(len, BignumInt);
for (j = 0; j < len; j++)
mninv[len - 1 - j] = (j < (int)inv[0] ? inv[j + 1] : 0);
freebn(inv); /* we don't need this copy of it any more */
/* Now negate mninv mod r, so it's the inverse of -n rather than +n. */
x = snewn(len, BignumInt);
for (j = 0; j < len; j++)
x[j] = 0;
internal_sub(x, mninv, mninv, len);
/* x = snewn(len, BignumInt); */ /* already done above */
for (j = 0; j < len; j++)
x[len - 1 - j] = (j < (int)base[0] ? base[j + 1] : 0);
freebn(base); /* we don't need this copy of it any more */
a = snewn(2*len, BignumInt);
b = snewn(2*len, BignumInt);
for (j = 0; j < len; j++)
a[2*len - 1 - j] = (j < (int)rn[0] ? rn[j + 1] : 0);
freebn(rn);
/* Scratch space for multiplies */
scratchlen = 3*len + mul_compute_scratch(len);
scratch = snewn(scratchlen, BignumInt);
/* Skip leading zero bits of exp. */
i = 0;
j = BIGNUM_INT_BITS-1;
while (i < (int)exp[0] && (exp[exp[0] - i] & ((BignumInt)1 << j)) == 0) {
j--;
if (j < 0) {
i++;
j = BIGNUM_INT_BITS-1;
}
}
/* Main computation */
while (i < (int)exp[0]) {
while (j >= 0) {
internal_mul(a + len, a + len, b, len, scratch);
monty_reduce(b, n, mninv, scratch, len);
if ((exp[exp[0] - i] & ((BignumInt)1 << j)) != 0) {
internal_mul(b + len, x, a, len, scratch);
monty_reduce(a, n, mninv, scratch, len);
} else {
BignumInt *t;
t = a;
a = b;
b = t;
}
j--;
}
i++;
j = BIGNUM_INT_BITS-1;
}
/*
* Final monty_reduce to get back from the adjusted Montgomery
* representation.
*/
monty_reduce(a, n, mninv, scratch, len);
/* Copy result to buffer */
result = newbn(mod[0]);
for (i = 0; i < len; i++)
result[result[0] - i] = a[i + len];
while (result[0] > 1 && result[result[0]] == 0)
result[0]--;
/* Free temporary arrays */
smemclr(scratch, scratchlen * sizeof(*scratch));
sfree(scratch);
smemclr(a, 2 * len * sizeof(*a));
sfree(a);
smemclr(b, 2 * len * sizeof(*b));
sfree(b);
smemclr(mninv, len * sizeof(*mninv));
sfree(mninv);
smemclr(n, len * sizeof(*n));
sfree(n);
smemclr(x, len * sizeof(*x));
sfree(x);
return result;
}
/*
* Compute (p * q) % mod.
* The most significant word of mod MUST be non-zero.
* We assume that the result array is the same size as the mod array.
*/
Bignum modmul(Bignum p, Bignum q, Bignum mod)
{
BignumInt *a, *n, *m, *o, *scratch;
BignumInt recip;
int rshift, scratchlen;
int pqlen, mlen, rlen, i, j;
Bignum result;
/*
* The most significant word of mod needs to be non-zero. It
* should already be, but let's make sure.
*/
assert(mod[mod[0]] != 0);
/* Allocate m of size mlen, copy mod to m */
/* We use big endian internally */
mlen = mod[0];
m = snewn(mlen, BignumInt);
for (j = 0; j < mlen; j++)
m[j] = mod[mod[0] - j];
pqlen = (p[0] > q[0] ? p[0] : q[0]);
/*
* Make sure that we're allowing enough space. The shifting below
* will underflow the vectors we allocate if pqlen is too small.
*/
if (2*pqlen <= mlen)
pqlen = mlen/2 + 1;
/* Allocate n of size pqlen, copy p to n */
n = snewn(pqlen, BignumInt);
i = pqlen - p[0];
for (j = 0; j < i; j++)
n[j] = 0;
for (j = 0; j < (int)p[0]; j++)
n[i + j] = p[p[0] - j];
/* Allocate o of size pqlen, copy q to o */
o = snewn(pqlen, BignumInt);
i = pqlen - q[0];
for (j = 0; j < i; j++)
o[j] = 0;
for (j = 0; j < (int)q[0]; j++)
o[i + j] = q[q[0] - j];
/* Allocate a of size 2*pqlen for result */
a = snewn(2 * pqlen, BignumInt);
/* Scratch space for multiplies */
scratchlen = mul_compute_scratch(pqlen);
scratch = snewn(scratchlen, BignumInt);
/* Compute reciprocal of the top full word of the modulus */
{
BignumInt m0 = m[0];
rshift = bn_clz(m0);
if (rshift) {
m0 <<= rshift;
if (mlen > 1)
m0 |= m[1] >> (BIGNUM_INT_BITS - rshift);
}
recip = reciprocal_word(m0);
}
/* Main computation */
internal_mul(n, o, a, pqlen, scratch);
internal_mod(a, pqlen * 2, m, mlen, NULL, recip, rshift);
/* Copy result to buffer */
rlen = (mlen < pqlen * 2 ? mlen : pqlen * 2);
result = newbn(rlen);
for (i = 0; i < rlen; i++)
result[result[0] - i] = a[i + 2 * pqlen - rlen];
while (result[0] > 1 && result[result[0]] == 0)
result[0]--;
/* Free temporary arrays */
smemclr(scratch, scratchlen * sizeof(*scratch));
sfree(scratch);
smemclr(a, 2 * pqlen * sizeof(*a));
sfree(a);
smemclr(m, mlen * sizeof(*m));
sfree(m);
smemclr(n, pqlen * sizeof(*n));
sfree(n);
smemclr(o, pqlen * sizeof(*o));
sfree(o);
return result;
}
Bignum modsub(const Bignum a, const Bignum b, const Bignum n)
{
Bignum a1, b1, ret;
if (bignum_cmp(a, n) >= 0) a1 = bigmod(a, n);
else a1 = a;
if (bignum_cmp(b, n) >= 0) b1 = bigmod(b, n);
else b1 = b;
if (bignum_cmp(a1, b1) >= 0) /* a >= b */
{
ret = bigsub(a1, b1);
}
else
{
/* Handle going round the corner of the modulus without having
* negative support in Bignum */
Bignum tmp = bigsub(n, b1);
assert(tmp);
ret = bigadd(tmp, a1);
freebn(tmp);
}
if (a != a1) freebn(a1);
if (b != b1) freebn(b1);
return ret;
}
/*
* Compute p % mod.
* The most significant word of mod MUST be non-zero.
* We assume that the result array is the same size as the mod array.
* We optionally write out a quotient if `quotient' is non-NULL.
* We can avoid writing out the result if `result' is NULL.
*/
static void bigdivmod(Bignum p, Bignum mod, Bignum result, Bignum quotient)
{
BignumInt *n, *m;
BignumInt recip;
int rshift;
int plen, mlen, i, j;
/*
* The most significant word of mod needs to be non-zero. It
* should already be, but let's make sure.
*/
assert(mod[mod[0]] != 0);
/* Allocate m of size mlen, copy mod to m */
/* We use big endian internally */
mlen = mod[0];
m = snewn(mlen, BignumInt);
for (j = 0; j < mlen; j++)
m[j] = mod[mod[0] - j];
plen = p[0];
/* Ensure plen > mlen */
if (plen <= mlen)
plen = mlen + 1;
/* Allocate n of size plen, copy p to n */
n = snewn(plen, BignumInt);
for (j = 0; j < plen; j++)
n[j] = 0;
for (j = 1; j <= (int)p[0]; j++)
n[plen - j] = p[j];
/* Compute reciprocal of the top full word of the modulus */
{
BignumInt m0 = m[0];
rshift = bn_clz(m0);
if (rshift) {
m0 <<= rshift;
if (mlen > 1)
m0 |= m[1] >> (BIGNUM_INT_BITS - rshift);
}
recip = reciprocal_word(m0);
}
/* Main computation */
internal_mod(n, plen, m, mlen, quotient, recip, rshift);
/* Copy result to buffer */
if (result) {
for (i = 1; i <= (int)result[0]; i++) {
int j = plen - i;
result[i] = j >= 0 ? n[j] : 0;
}
}
/* Free temporary arrays */
smemclr(m, mlen * sizeof(*m));
sfree(m);
smemclr(n, plen * sizeof(*n));
sfree(n);
}
/*
* Decrement a number.
*/
void decbn(Bignum bn)
{
int i = 1;
while (i < (int)bn[0] && bn[i] == 0)
bn[i++] = BIGNUM_INT_MASK;
bn[i]--;
}
Bignum bignum_from_bytes(const unsigned char *data, int nbytes)
{
Bignum result;
int w, i;
assert(nbytes >= 0 && nbytes < INT_MAX/8);
w = (nbytes + BIGNUM_INT_BYTES - 1) / BIGNUM_INT_BYTES; /* bytes->words */
result = newbn(w);
for (i = 1; i <= w; i++)
result[i] = 0;
for (i = nbytes; i--;) {
unsigned char byte = *data++;
result[1 + i / BIGNUM_INT_BYTES] |=
(BignumInt)byte << (8*i % BIGNUM_INT_BITS);
}
bn_restore_invariant(result);
return result;
}
Bignum bignum_from_bytes_le(const unsigned char *data, int nbytes)
{
Bignum result;
int w, i;
assert(nbytes >= 0 && nbytes < INT_MAX/8);
w = (nbytes + BIGNUM_INT_BYTES - 1) / BIGNUM_INT_BYTES; /* bytes->words */
result = newbn(w);
for (i = 1; i <= w; i++)
result[i] = 0;
for (i = 0; i < nbytes; ++i) {
unsigned char byte = *data++;
result[1 + i / BIGNUM_INT_BYTES] |=
(BignumInt)byte << (8*i % BIGNUM_INT_BITS);
}
bn_restore_invariant(result);
return result;
}
Bignum bignum_from_decimal(const char *decimal)
{
Bignum result = copybn(Zero);
while (*decimal) {
Bignum tmp, tmp2;
if (!isdigit((unsigned char)*decimal)) {
freebn(result);
return 0;
}
tmp = bigmul(result, Ten);
tmp2 = bignum_from_long(*decimal - '0');
freebn(result);
result = bigadd(tmp, tmp2);
freebn(tmp);
freebn(tmp2);
decimal++;
}
return result;
}
Bignum bignum_random_in_range(const Bignum lower, const Bignum upper)
{
Bignum ret = NULL;
unsigned char *bytes;
int upper_len = bignum_bitcount(upper);
int upper_bytes = upper_len / 8;
int upper_bits = upper_len % 8;
if (upper_bits) ++upper_bytes;
bytes = snewn(upper_bytes, unsigned char);
do {
int i;
if (ret) freebn(ret);
for (i = 0; i < upper_bytes; ++i)
{
bytes[i] = (unsigned char)random_byte();
}
/* Mask the top to reduce failure rate to 50/50 */
if (upper_bits)
{
bytes[i - 1] &= 0xFF >> (8 - upper_bits);
}
ret = bignum_from_bytes(bytes, upper_bytes);
} while (bignum_cmp(ret, lower) < 0 || bignum_cmp(ret, upper) > 0);
smemclr(bytes, upper_bytes);
sfree(bytes);
return ret;
}
/*
* Read an SSH-1-format bignum from a data buffer. Return the number
* of bytes consumed, or -1 if there wasn't enough data.
*/
int ssh1_read_bignum(const unsigned char *data, int len, Bignum * result)
{
const unsigned char *p = data;
int i;
int w, b;
if (len < 2)
return -1;
w = 0;
for (i = 0; i < 2; i++)
w = (w << 8) + *p++;
b = (w + 7) / 8; /* bits -> bytes */
if (len < b+2)
return -1;
if (!result) /* just return length */
return b + 2;
*result = bignum_from_bytes(p, b);
return p + b - data;
}
/*
* Return the bit count of a bignum, for SSH-1 encoding.
*/
int bignum_bitcount(Bignum bn)
{
int bitcount = bn[0] * BIGNUM_INT_BITS - 1;
while (bitcount >= 0
&& (bn[bitcount / BIGNUM_INT_BITS + 1] >> (bitcount % BIGNUM_INT_BITS)) == 0) bitcount--;
return bitcount + 1;
}
/*
* Return the byte length of a bignum when SSH-1 encoded.
*/
int ssh1_bignum_length(Bignum bn)
{
return 2 + (bignum_bitcount(bn) + 7) / 8;
}
/*
* Return the byte length of a bignum when SSH-2 encoded.
*/
int ssh2_bignum_length(Bignum bn)
{
return 4 + (bignum_bitcount(bn) + 8) / 8;
}
/*
* Return a byte from a bignum; 0 is least significant, etc.
*/
int bignum_byte(Bignum bn, int i)
{
if (i < 0 || i >= (int)(BIGNUM_INT_BYTES * bn[0]))
return 0; /* beyond the end */
else
return (bn[i / BIGNUM_INT_BYTES + 1] >>
((i % BIGNUM_INT_BYTES)*8)) & 0xFF;
}
/*
* Return a bit from a bignum; 0 is least significant, etc.
*/
int bignum_bit(Bignum bn, int i)
{
if (i < 0 || i >= (int)(BIGNUM_INT_BITS * bn[0]))
return 0; /* beyond the end */
else
return (bn[i / BIGNUM_INT_BITS + 1] >> (i % BIGNUM_INT_BITS)) & 1;
}
/*
* Set a bit in a bignum; 0 is least significant, etc.
*/
void bignum_set_bit(Bignum bn, int bitnum, int value)
{
if (bitnum < 0 || bitnum >= (int)(BIGNUM_INT_BITS * bn[0])) {
if (value) abort(); /* beyond the end */
} else {
int v = bitnum / BIGNUM_INT_BITS + 1;
BignumInt mask = (BignumInt)1 << (bitnum % BIGNUM_INT_BITS);
if (value)
bn[v] |= mask;
else
bn[v] &= ~mask;
}
}
/*
* Write a SSH-1-format bignum into a buffer. It is assumed the
* buffer is big enough. Returns the number of bytes used.
*/
int ssh1_write_bignum(void *data, Bignum bn)
{
unsigned char *p = data;
int len = ssh1_bignum_length(bn);
int i;
int bitc = bignum_bitcount(bn);
*p++ = (bitc >> 8) & 0xFF;
*p++ = (bitc) & 0xFF;
for (i = len - 2; i--;)
*p++ = bignum_byte(bn, i);
return len;
}
/*
* Compare two bignums. Returns like strcmp.
*/
int bignum_cmp(Bignum a, Bignum b)
{
int amax = a[0], bmax = b[0];
int i;
/* Annoyingly we have two representations of zero */
if (amax == 1 && a[amax] == 0)
amax = 0;
if (bmax == 1 && b[bmax] == 0)
bmax = 0;
assert(amax == 0 || a[amax] != 0);
assert(bmax == 0 || b[bmax] != 0);
i = (amax > bmax ? amax : bmax);
while (i) {
BignumInt aval = (i > amax ? 0 : a[i]);
BignumInt bval = (i > bmax ? 0 : b[i]);
if (aval < bval)
return -1;
if (aval > bval)
return +1;
i--;
}
return 0;
}
/*
* Right-shift one bignum to form another.
*/
Bignum bignum_rshift(Bignum a, int shift)
{
Bignum ret;
int i, shiftw, shiftb, shiftbb, bits;
BignumInt ai, ai1;
assert(shift >= 0);
bits = bignum_bitcount(a) - shift;
ret = newbn((bits + BIGNUM_INT_BITS - 1) / BIGNUM_INT_BITS);
if (ret) {
shiftw = shift / BIGNUM_INT_BITS;
shiftb = shift % BIGNUM_INT_BITS;
shiftbb = BIGNUM_INT_BITS - shiftb;
ai1 = a[shiftw + 1];
for (i = 1; i <= (int)ret[0]; i++) {
ai = ai1;
ai1 = (i + shiftw + 1 <= (int)a[0] ? a[i + shiftw + 1] : 0);
ret[i] = ((ai >> shiftb) | (ai1 << shiftbb)) & BIGNUM_INT_MASK;
}
}
return ret;
}
/*
* Left-shift one bignum to form another.
*/
Bignum bignum_lshift(Bignum a, int shift)
{
Bignum ret;
int bits, shiftWords, shiftBits;
assert(shift >= 0);
bits = bignum_bitcount(a) + shift;
ret = newbn((bits + BIGNUM_INT_BITS - 1) / BIGNUM_INT_BITS);
shiftWords = shift / BIGNUM_INT_BITS;
shiftBits = shift % BIGNUM_INT_BITS;
if (shiftBits == 0)
{
memcpy(&ret[1 + shiftWords], &a[1], sizeof(BignumInt) * a[0]);
}
else
{
int i;
BignumInt carry = 0;
/* Remember that Bignum[0] is length, so add 1 */
for (i = shiftWords + 1; i < ((int)a[0]) + shiftWords + 1; ++i)
{
BignumInt from = a[i - shiftWords];
ret[i] = (from << shiftBits) | carry;
carry = from >> (BIGNUM_INT_BITS - shiftBits);
}
if (carry) ret[i] = carry;
}
return ret;
}
/*
* Non-modular multiplication and addition.
*/
Bignum bigmuladd(Bignum a, Bignum b, Bignum addend)
{
int alen = a[0], blen = b[0];
int mlen = (alen > blen ? alen : blen);
int rlen, i, maxspot;
int wslen;
BignumInt *workspace;
Bignum ret;
/* mlen space for a, mlen space for b, 2*mlen for result,
* plus scratch space for multiplication */
wslen = mlen * 4 + mul_compute_scratch(mlen);
workspace = snewn(wslen, BignumInt);
for (i = 0; i < mlen; i++) {
workspace[0 * mlen + i] = (mlen - i <= (int)a[0] ? a[mlen - i] : 0);
workspace[1 * mlen + i] = (mlen - i <= (int)b[0] ? b[mlen - i] : 0);
}
internal_mul(workspace + 0 * mlen, workspace + 1 * mlen,
workspace + 2 * mlen, mlen, workspace + 4 * mlen);
/* now just copy the result back */
rlen = alen + blen + 1;
if (addend && rlen <= (int)addend[0])
rlen = addend[0] + 1;
ret = newbn(rlen);
maxspot = 0;
for (i = 1; i <= (int)ret[0]; i++) {
ret[i] = (i <= 2 * mlen ? workspace[4 * mlen - i] : 0);
if (ret[i] != 0)
maxspot = i;
}
ret[0] = maxspot;
/* now add in the addend, if any */
if (addend) {
BignumCarry carry = 0;
for (i = 1; i <= rlen; i++) {
BignumInt retword = (i <= (int)ret[0] ? ret[i] : 0);
BignumInt addword = (i <= (int)addend[0] ? addend[i] : 0);
BignumADC(ret[i], carry, retword, addword, carry);
if (ret[i] != 0 && i > maxspot)
maxspot = i;
}
}
ret[0] = maxspot;
smemclr(workspace, wslen * sizeof(*workspace));
sfree(workspace);
return ret;
}
/*
* Non-modular multiplication.
*/
Bignum bigmul(Bignum a, Bignum b)
{
return bigmuladd(a, b, NULL);
}
/*
* Simple addition.
*/
Bignum bigadd(Bignum a, Bignum b)
{
int alen = a[0], blen = b[0];
int rlen = (alen > blen ? alen : blen) + 1;
int i, maxspot;
Bignum ret;
BignumCarry carry;
ret = newbn(rlen);
carry = 0;
maxspot = 0;
for (i = 1; i <= rlen; i++) {
BignumInt aword = (i <= (int)a[0] ? a[i] : 0);
BignumInt bword = (i <= (int)b[0] ? b[i] : 0);
BignumADC(ret[i], carry, aword, bword, carry);
if (ret[i] != 0 && i > maxspot)
maxspot = i;
}
ret[0] = maxspot;
return ret;
}
/*
* Subtraction. Returns a-b, or NULL if the result would come out
* negative (recall that this entire bignum module only handles
* positive numbers).
*/
Bignum bigsub(Bignum a, Bignum b)
{
int alen = a[0], blen = b[0];
int rlen = (alen > blen ? alen : blen);
int i, maxspot;
Bignum ret;
BignumCarry carry;
ret = newbn(rlen);
carry = 1;
maxspot = 0;
for (i = 1; i <= rlen; i++) {
BignumInt aword = (i <= (int)a[0] ? a[i] : 0);
BignumInt bword = (i <= (int)b[0] ? b[i] : 0);
BignumADC(ret[i], carry, aword, ~bword, carry);
if (ret[i] != 0 && i > maxspot)
maxspot = i;
}
ret[0] = maxspot;
if (!carry) {
freebn(ret);
return NULL;
}
return ret;
}
/*
* Create a bignum which is the bitmask covering another one. That
* is, the smallest integer which is >= N and is also one less than
* a power of two.
*/
Bignum bignum_bitmask(Bignum n)
{
Bignum ret = copybn(n);
int i;
BignumInt j;
i = ret[0];
while (n[i] == 0 && i > 0)
i--;
if (i <= 0)
return ret; /* input was zero */
j = 1;
while (j < n[i])
j = 2 * j + 1;
ret[i] = j;
while (--i > 0)
ret[i] = BIGNUM_INT_MASK;
return ret;
}
/*
* Convert an unsigned long into a bignum.
*/
Bignum bignum_from_long(unsigned long n)
{
const int maxwords =
(sizeof(unsigned long) + sizeof(BignumInt) - 1) / sizeof(BignumInt);
Bignum ret;
int i;
ret = newbn(maxwords);
ret[0] = 0;
for (i = 0; i < maxwords; i++) {
ret[i+1] = n >> (i * BIGNUM_INT_BITS);
if (ret[i+1] != 0)
ret[0] = i+1;
}
return ret;
}
/*
* Add a long to a bignum.
*/
Bignum bignum_add_long(Bignum number, unsigned long n)
{
const int maxwords =
(sizeof(unsigned long) + sizeof(BignumInt) - 1) / sizeof(BignumInt);
Bignum ret;
int words, i;
BignumCarry carry;
words = number[0];
if (words < maxwords)
words = maxwords;
words++;
ret = newbn(words);
carry = 0;
ret[0] = 0;
for (i = 0; i < words; i++) {
BignumInt nword = (i < maxwords ? n >> (i * BIGNUM_INT_BITS) : 0);
BignumInt numword = (i < number[0] ? number[i+1] : 0);
BignumADC(ret[i+1], carry, numword, nword, carry);
if (ret[i+1] != 0)
ret[0] = i+1;
}
return ret;
}
/*
* Compute the residue of a bignum, modulo a (max 16-bit) short.
*/
unsigned short bignum_mod_short(Bignum number, unsigned short modulus)
{
unsigned long mod = modulus, r = 0;
/* Precompute (BIGNUM_INT_MASK+1) % mod */
unsigned long base_r = (BIGNUM_INT_MASK - modulus + 1) % mod;
int i;
for (i = number[0]; i > 0; i--) {
/*
* Conceptually, ((r << BIGNUM_INT_BITS) + number[i]) % mod
*/
r = ((r * base_r) + (number[i] % mod)) % mod;
}
return (unsigned short) r;
}
#ifdef DEBUG
void diagbn(char *prefix, Bignum md)
{
int i, nibbles, morenibbles;
static const char hex[] = "0123456789ABCDEF";
debug(("%s0x", prefix ? prefix : ""));
nibbles = (3 + bignum_bitcount(md)) / 4;
if (nibbles < 1)
nibbles = 1;
morenibbles = 4 * md[0] - nibbles;
for (i = 0; i < morenibbles; i++)
debug(("-"));
for (i = nibbles; i--;)
debug(("%c",
hex[(bignum_byte(md, i / 2) >> (4 * (i % 2))) & 0xF]));
if (prefix)
debug(("\n"));
}
#endif
/*
* Simple division.
*/
Bignum bigdiv(Bignum a, Bignum b)
{
Bignum q = newbn(a[0]);
bigdivmod(a, b, NULL, q);
while (q[0] > 1 && q[q[0]] == 0)
q[0]--;
return q;
}
/*
* Simple remainder.
*/
Bignum bigmod(Bignum a, Bignum b)
{
Bignum r = newbn(b[0]);
bigdivmod(a, b, r, NULL);
while (r[0] > 1 && r[r[0]] == 0)
r[0]--;
return r;
}
/*
* Greatest common divisor.
*/
Bignum biggcd(Bignum av, Bignum bv)
{
Bignum a = copybn(av);
Bignum b = copybn(bv);
while (bignum_cmp(b, Zero) != 0) {
Bignum t = newbn(b[0]);
bigdivmod(a, b, t, NULL);
while (t[0] > 1 && t[t[0]] == 0)
t[0]--;
freebn(a);
a = b;
b = t;
}
freebn(b);
return a;
}
/*
* Modular inverse, using Euclid's extended algorithm.
*/
Bignum modinv(Bignum number, Bignum modulus)
{
Bignum a = copybn(modulus);
Bignum b = copybn(number);
Bignum xp = copybn(Zero);
Bignum x = copybn(One);
int sign = +1;
assert(number[number[0]] != 0);
assert(modulus[modulus[0]] != 0);
while (bignum_cmp(b, One) != 0) {
Bignum t, q;
if (bignum_cmp(b, Zero) == 0) {
/*
* Found a common factor between the inputs, so we cannot
* return a modular inverse at all.
*/
freebn(b);
freebn(a);
freebn(xp);
freebn(x);
return NULL;
}
t = newbn(b[0]);
q = newbn(a[0]);
bigdivmod(a, b, t, q);
while (t[0] > 1 && t[t[0]] == 0)
t[0]--;
while (q[0] > 1 && q[q[0]] == 0)
q[0]--;
freebn(a);
a = b;
b = t;
t = xp;
xp = x;
x = bigmuladd(q, xp, t);
sign = -sign;
freebn(t);
freebn(q);
}
freebn(b);
freebn(a);
freebn(xp);
/* now we know that sign * x == 1, and that x < modulus */
if (sign < 0) {
/* set a new x to be modulus - x */
Bignum newx = newbn(modulus[0]);
BignumInt carry = 0;
int maxspot = 1;
int i;
for (i = 1; i <= (int)newx[0]; i++) {
BignumInt aword = (i <= (int)modulus[0] ? modulus[i] : 0);
BignumInt bword = (i <= (int)x[0] ? x[i] : 0);
newx[i] = aword - bword - carry;
bword = ~bword;
carry = carry ? (newx[i] >= bword) : (newx[i] > bword);
if (newx[i] != 0)
maxspot = i;
}
newx[0] = maxspot;
freebn(x);
x = newx;
}
/* and return. */
return x;
}
/*
* Render a bignum into decimal. Return a malloced string holding
* the decimal representation.
*/
char *bignum_decimal(Bignum x)
{
int ndigits, ndigit;
int i, iszero;
BignumInt carry;
char *ret;
BignumInt *workspace;
/*
* First, estimate the number of digits. Since log(10)/log(2)
* is just greater than 93/28 (the joys of continued fraction
* approximations...) we know that for every 93 bits, we need
* at most 28 digits. This will tell us how much to malloc.
*
* Formally: if x has i bits, that means x is strictly less
* than 2^i. Since 2 is less than 10^(28/93), this is less than
* 10^(28i/93). We need an integer power of ten, so we must
* round up (rounding down might make it less than x again).
* Therefore if we multiply the bit count by 28/93, rounding
* up, we will have enough digits.
*
* i=0 (i.e., x=0) is an irritating special case.
*/
i = bignum_bitcount(x);
if (!i)
ndigits = 1; /* x = 0 */
else
ndigits = (28 * i + 92) / 93; /* multiply by 28/93 and round up */
ndigits++; /* allow for trailing \0 */
ret = snewn(ndigits, char);
/*
* Now allocate some workspace to hold the binary form as we
* repeatedly divide it by ten. Initialise this to the
* big-endian form of the number.
*/
workspace = snewn(x[0], BignumInt);
for (i = 0; i < (int)x[0]; i++)
workspace[i] = x[x[0] - i];
/*
* Next, write the decimal number starting with the last digit.
* We use ordinary short division, dividing 10 into the
* workspace.
*/
ndigit = ndigits - 1;
ret[ndigit] = '\0';
do {
iszero = 1;
carry = 0;
for (i = 0; i < (int)x[0]; i++) {
/*
* Conceptually, we want to compute
*
* (carry << BIGNUM_INT_BITS) + workspace[i]
* -----------------------------------------
* 10
*
* but we don't have an integer type longer than BignumInt
* to work with. So we have to do it in pieces.
*/
BignumInt q, r;
q = workspace[i] / 10;
r = workspace[i] % 10;
/* I want (BIGNUM_INT_MASK+1)/10 but can't say so directly! */
q += carry * ((BIGNUM_INT_MASK-9) / 10 + 1);
r += carry * ((BIGNUM_INT_MASK-9) % 10);
q += r / 10;
r %= 10;
workspace[i] = q;
carry = r;
if (workspace[i])
iszero = 0;
}
ret[--ndigit] = (char) (carry + '0');
} while (!iszero);
/*
* There's a chance we've fallen short of the start of the
* string. Correct if so.
*/
if (ndigit > 0)
memmove(ret, ret + ndigit, ndigits - ndigit);
/*
* Done.
*/
smemclr(workspace, x[0] * sizeof(*workspace));
sfree(workspace);
return ret;
}
Ссылка в новой задаче Показать git blame Копировать постоянную ссылку