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sshbn.c
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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 <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];