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sshbn.c
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/*
* Bignum routines for RSA and DH and stuff.
*/
#include <stdio.h>
#include <assert.h>
#include <stdlib.h>
#include <string.h>
#include <limits.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 };
/*
* 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;
static Bignum newbn(int length)
{
Bignum b;
assert(length >= 0 && length < INT_MAX / BIGNUM_INT_BITS);
b = snewn(length + 1, BignumInt);
if (!b)
abort(); /* FIXME */
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 a BignumInt carried
* off the top.
*/
static BignumInt internal_add(const BignumInt *a, const BignumInt *b,
BignumInt *c, int len)
{
int i;
BignumDblInt carry = 0;
for (i = len-1; i >= 0; i--) {
carry += (BignumDblInt)a[i] + b[i];
c[i] = (BignumInt)carry;
carry >>= BIGNUM_INT_BITS;
}
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;
BignumDblInt carry = 1;
for (i = len-1; i >= 0; i--) {
carry += (BignumDblInt)a[i] + (b[i] ^ BIGNUM_INT_MASK);
c[i] = (BignumInt)carry;
carry >>= BIGNUM_INT_BITS;
}
}
/*
* 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;
BignumDblInt 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);
carry += c[i];
c[i] = (BignumInt)carry;
carry >>= BIGNUM_INT_BITS;
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;
BignumDblInt t;
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 ;) {
t = (MUL_WORD(*ap, *bp) + carry) + *cp;
*cp = (BignumInt) t;
carry = (BignumInt)(t >> BIGNUM_INT_BITS);
}
*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;
BignumDblInt t;
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 ;) {
t = (MUL_WORD(*ap, *bp) + carry) + *cp;
*cp = (BignumInt) t;
carry = (BignumInt)(t >> BIGNUM_INT_BITS);
}
}
}
}
/*
* 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;
BignumDblInt addend;
addend = (BignumDblInt)n << bshift;
while (addend) {
assert(word <= number[0]);
addend += number[word];
number[word] = (BignumInt) addend & BIGNUM_INT_MASK;
addend >>= BIGNUM_INT_BITS;
word++;
}
}
/*
* 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).
* The MSW of m MUST have its high bit set.
* Quotient is accumulated in the `quotient' array, which is a Bignum
* rather than the internal bigendian format. Quotient parts are shifted
* left by `qshift' before adding into quot.
*/
static void internal_mod(BignumInt *a, int alen,
BignumInt *m, int mlen,
BignumInt *quot, int qshift)
{
BignumInt m0, m1, h;
int i, k;
m0 = m[0];
assert(m0 >> (BIGNUM_INT_BITS-1) == 1);
if (mlen > 1)
m1 = m[1];
else
m1 = 0;
for (i = 0; i <= alen - mlen; i++) {
BignumDblInt t;
BignumInt q, r, c, ai1;
if (i == 0) {
h = 0;
} else {
h = a[i - 1];
a[i - 1] = 0;
}
if (i == alen - 1)
ai1 = 0;
else
ai1 = a[i + 1];
/* Find q = h:a[i] / m0 */
if (h >= m0) {
/*
* Special case.
*
* To illustrate it, suppose a BignumInt is 8 bits, and
* we are dividing (say) A1:23:45:67 by A1:B2:C3. Then
* our initial division will be 0xA123 / 0xA1, which
* will give a quotient of 0x100 and a divide overflow.
* However, the invariants in this division algorithm
* are not violated, since the full number A1:23:... is
* _less_ than the quotient prefix A1:B2:... and so the
* following correction loop would have sorted it out.
*
* In this situation we set q to be the largest
* quotient we _can_ stomach (0xFF, of course).
*/
q = BIGNUM_INT_MASK;
} else {
/* Macro doesn't want an array subscript expression passed
* into it (see definition), so use a temporary. */
BignumInt tmplo = a[i];
DIVMOD_WORD(q, r, h, tmplo, m0);
/* Refine our estimate of q by looking at
h:a[i]:a[i+1] / m0:m1 */
t = MUL_WORD(m1, q);
if (t > ((BignumDblInt) r << BIGNUM_INT_BITS) + ai1) {
q--;
t -= m1;
r = (r + m0) & BIGNUM_INT_MASK; /* overflow? */
if (r >= (BignumDblInt) m0 &&
t > ((BignumDblInt) r << BIGNUM_INT_BITS) + ai1) q--;
}
}
/* Subtract q * m from a[i...] */
c = 0;
for (k = mlen - 1; k >= 0; k--) {
t = MUL_WORD(q, m[k]);
t += c;
c = (BignumInt)(t >> BIGNUM_INT_BITS);
if ((BignumInt) t > a[i + k])
c++;
a[i + k] -= (BignumInt) t;
}
/* Add back m in case of borrow */
if (c != h) {
t = 0;
for (k = mlen - 1; k >= 0; k--) {
t += m[k];
t += a[i + k];
a[i + k] = (BignumInt) t;
t = t >> BIGNUM_INT_BITS;
}
q--;
}
if (quot)
internal_add_shifted(quot, q, qshift + BIGNUM_INT_BITS * (alen - mlen - i));
}
}
/*
* Compute (base ^ exp) % mod, the pedestrian way.
*/
Bignum modpow_simple(Bignum base_in, Bignum exp, Bignum mod)
{
BignumInt *a, *b, *n, *m, *scratch;
int mshift;
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];
/* Shift m left to make msb bit set */
for (mshift = 0; mshift < BIGNUM_INT_BITS-1; mshift++)
if ((m[0] << mshift) & BIGNUM_TOP_BIT)
break;
if (mshift) {
for (i = 0; i < mlen - 1; i++)
m[i] = (m[i] << mshift) | (m[i + 1] >> (BIGNUM_INT_BITS - mshift));
m[mlen - 1] = m[mlen - 1] << mshift;
}
/* 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;
}
}
/* 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, 0);
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, 0);
} else {
BignumInt *t;
t = a;
a = b;
b = t;
}
j--;
}
i++;
j = BIGNUM_INT_BITS-1;
}
/* Fixup result in case the modulus was shifted */
if (mshift) {
for (i = mlen - 1; i < 2 * mlen - 1; i++)
a[i] = (a[i] << mshift) | (a[i + 1] >> (BIGNUM_INT_BITS - mshift));
a[2 * mlen - 1] = a[2 * mlen - 1] << mshift;
internal_mod(a, mlen * 2, m, mlen, NULL, 0);
for (i = 2 * mlen - 1; i >= mlen; i--)
a[i] = (a[i] >> mshift) | (a[i - 1] << (BIGNUM_INT_BITS - mshift));
}
/* 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;
int mshift, 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];
/* Shift m left to make msb bit set */
for (mshift = 0; mshift < BIGNUM_INT_BITS-1; mshift++)
if ((m[0] << mshift) & BIGNUM_TOP_BIT)
break;
if (mshift) {
for (i = 0; i < mlen - 1; i++)
m[i] = (m[i] << mshift) | (m[i + 1] >> (BIGNUM_INT_BITS - mshift));
m[mlen - 1] = m[mlen - 1] << mshift;
}
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);
/* Main computation */
internal_mul(n, o, a, pqlen, scratch);
internal_mod(a, pqlen * 2, m, mlen, NULL, 0);
/* Fixup result in case the modulus was shifted */
if (mshift) {
for (i = 2 * pqlen - mlen - 1; i < 2 * pqlen - 1; i++)
a[i] = (a[i] << mshift) | (a[i + 1] >> (BIGNUM_INT_BITS - mshift));
a[2 * pqlen - 1] = a[2 * pqlen - 1] << mshift;
internal_mod(a, pqlen * 2, m, mlen, NULL, 0);
for (i = 2 * pqlen - 1; i >= 2 * pqlen - mlen; i--)
a[i] = (a[i] >> mshift) | (a[i - 1] << (BIGNUM_INT_BITS - mshift));
}
/* 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);