Implement the Montgomery technique for speeding up modular

[u/mdw/putty] / sshbn.c

/*
 * Bignum routines for RSA and DH and stuff.
 */

#include <stdio.h>
#include <assert.h>
#include <stdlib.h>
#include <string.h>

#include "misc.h"

/*
 * Usage notes:
 *  * Do not call the DIVMOD_WORD macro with expressions such as array
 *    subscripts, as some implementations object to this (see below).
 *  * Note that none of the division methods below will cope if the
 *    quotient won't fit into BIGNUM_INT_BITS. Callers should be careful
 *    to avoid this case.
 *    If this condition occurs, in the case of the x86 DIV instruction,
 *    an overflow exception will occur, which (according to a correspondent)
 *    will manifest on Windows as something like
 *      0xC0000095: Integer overflow
 *    The C variant won't give the right answer, either.
 */

#if defined __GNUC__ && defined __i386__
typedef unsigned long BignumInt;
typedef unsigned long long BignumDblInt;
#define BIGNUM_INT_MASK  0xFFFFFFFFUL
#define BIGNUM_TOP_BIT   0x80000000UL
#define BIGNUM_INT_BITS  32
#define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2)
#define DIVMOD_WORD(q, r, hi, lo, w) \
    __asm__("div %2" : \
            "=d" (r), "=a" (q) : \
            "r" (w), "d" (hi), "a" (lo))
#elif defined _MSC_VER && defined _M_IX86
typedef unsigned __int32 BignumInt;
typedef unsigned __int64 BignumDblInt;
#define BIGNUM_INT_MASK  0xFFFFFFFFUL
#define BIGNUM_TOP_BIT   0x80000000UL
#define BIGNUM_INT_BITS  32
#define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2)
/* Note: MASM interprets array subscripts in the macro arguments as
 * assembler syntax, which gives the wrong answer. Don't supply them.
 * <http://msdn2.microsoft.com/en-us/library/bf1dw62z.aspx> */
#define DIVMOD_WORD(q, r, hi, lo, w) do { \
    __asm mov edx, hi \
    __asm mov eax, lo \
    __asm div w \
    __asm mov r, edx \
    __asm mov q, eax \
} while(0)
#elif defined _LP64
/* 64-bit architectures can do 32x32->64 chunks at a time */
typedef unsigned int BignumInt;
typedef unsigned long BignumDblInt;
#define BIGNUM_INT_MASK  0xFFFFFFFFU
#define BIGNUM_TOP_BIT   0x80000000U
#define BIGNUM_INT_BITS  32
#define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2)
#define DIVMOD_WORD(q, r, hi, lo, w) do { \
    BignumDblInt n = (((BignumDblInt)hi) << BIGNUM_INT_BITS) | lo; \
    q = n / w; \
    r = n % w; \
} while (0)
#elif defined _LLP64
/* 64-bit architectures in which unsigned long is 32 bits, not 64 */
typedef unsigned long BignumInt;
typedef unsigned long long BignumDblInt;
#define BIGNUM_INT_MASK  0xFFFFFFFFUL
#define BIGNUM_TOP_BIT   0x80000000UL
#define BIGNUM_INT_BITS  32
#define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2)
#define DIVMOD_WORD(q, r, hi, lo, w) do { \
    BignumDblInt n = (((BignumDblInt)hi) << BIGNUM_INT_BITS) | lo; \
    q = n / w; \
    r = n % w; \
} while (0)
#else
/* Fallback for all other cases */
typedef unsigned short BignumInt;
typedef unsigned long BignumDblInt;
#define BIGNUM_INT_MASK  0xFFFFU
#define BIGNUM_TOP_BIT   0x8000U
#define BIGNUM_INT_BITS  16
#define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2)
#define DIVMOD_WORD(q, r, hi, lo, w) do { \
    BignumDblInt n = (((BignumDblInt)hi) << BIGNUM_INT_BITS) | lo; \
    q = n / w; \
    r = n % w; \
} while (0)
#endif

#define BIGNUM_INT_BYTES (BIGNUM_INT_BITS / 8)

#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 = 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.
     */
    memset(b, 0, sizeof(b[0]) * (b[0] + 1));
    sfree(b);
}

Bignum bn_power_2(int n)
{
    Bignum 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.
 */
#define KARATSUBA_THRESHOLD 50
static void internal_mul(const BignumInt *a, const BignumInt *b,
                         BignumInt *c, int len)
{
    int i, j;
    BignumDblInt t;

    if (len > KARATSUBA_THRESHOLD) {

        /*
         * 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;
        BignumInt *scratch;
        BignumDblInt carry;

        /*
         * 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.
         */

        /* a_1 b_1 */
        internal_mul(a, b, c, toplen);

        /* a_0 b_0 */
        internal_mul(a + toplen, b + toplen, c + 2*toplen, botlen);

        /*
         * We must allocate scratch space for the central coefficient,
         * and also for the two input values that we multiply when
         * computing it. Since either or both may carry into the
         * (botlen+1)th word, we must use a slightly longer length
         * 'midlen'.
         */
        scratch = snewn(4 * midlen, BignumInt);

        /* 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 (j = 0; j < toplen; j++) {
            scratch[midlen - toplen + j] = a[j]; /* a_1 */
            scratch[2*midlen - toplen + j] = b[j]; /* b_1 */
        }

        /* compute a_1 + a_0 */
        scratch[0] = internal_add(scratch+1, a+toplen, scratch+1, botlen);
        /* compute b_1 + b_0 */
        scratch[midlen] = internal_add(scratch+midlen+1, b+toplen,
                                       scratch+midlen+1, botlen);

        /*
         * Now we can do the third multiplication.
         */
        internal_mul(scratch, scratch + midlen, scratch + 2*midlen, midlen);

        /*
         * 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 (j = 0; j < 2*toplen; j++)
            scratch[2*midlen - 2*toplen + j] = c[j];
        scratch[1] = internal_add(scratch+2, c + 2*toplen,
                                  scratch+2, 2*botlen);

        internal_sub(scratch + 2*midlen, scratch,
                     scratch + 2*midlen, 2*midlen);

        /*
         * 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);
        j = 2*len - botlen - 2*midlen - 1;
        while (carry) {
            assert(j >= 0);
            carry += c[j];
            c[j] = (BignumInt)carry;
            carry >>= BIGNUM_INT_BITS;
        }

        /* Free scratch. */
        for (j = 0; j < 4 * midlen; j++)
            scratch[j] = 0;
        sfree(scratch);

    } else {

        /*
         * Multiply in the ordinary O(N^2) way.
         */

        for (j = 0; j < 2 * len; j++)
            c[j] = 0;

        for (i = len - 1; i >= 0; i--) {
            t = 0;
            for (j = len - 1; j >= 0; j--) {
                t += MUL_WORD(a[i], (BignumDblInt) b[j]);
                t += (BignumDblInt) c[i + j + 1];
                c[i + j + 1] = (BignumInt) t;
                t = t >> BIGNUM_INT_BITS;
            }
            c[i] = (BignumInt) t;
        }
    }
}

/*
 * 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)
{
    int i, j;
    BignumDblInt t;

    if (len > KARATSUBA_THRESHOLD) {

        /*
         * 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 */
        BignumInt *scratch;

        /*
         * Allocate 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.
         */
        scratch = snewn(len*2, BignumInt);

        /* a_0 b_0 */
        internal_mul(a + toplen, b + toplen, scratch + 2*toplen, botlen);

        /* a_1 b_0 */
        internal_mul_low(a, b + len - toplen, scratch + toplen, toplen);

        /* a_0 b_1 */
        internal_mul_low(a + len - toplen, b, scratch, toplen);

        /* Copy the bottom half of the big coefficient into place */
        for (j = 0; j < botlen; j++)
            c[toplen + j] = scratch[2*toplen + botlen + j];

        /* 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);

        /* Free scratch. */
        for (j = 0; j < len*2; j++)
            scratch[j] = 0;
        sfree(scratch);

    } else {

        for (j = 0; j < len; j++)
            c[j] = 0;

        for (i = len - 1; i >= 0; i--) {
            t = 0;
            for (j = len - 1; j >= len - i - 1; j--) {
                t += MUL_WORD(a[i], (BignumDblInt) b[j]);
                t += (BignumDblInt) c[i + j + 1 - len];
                c[i + j + 1 - len] = (BignumInt) t;
                t = 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 at least '3*len' BignumInts used as scratch
 * space.
 */
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);

    /*
     * 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);
    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,
                                 unsigned n, int shift)
{
    int word = 1 + (shift / BIGNUM_INT_BITS);
    int bshift = shift % BIGNUM_INT_BITS;
    BignumDblInt addend;

    addend = (BignumDblInt)n << bshift;

    while (addend) {
        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;
    unsigned int h;
    int i, k;

    m0 = m[0];
    if (mlen > 1)
        m1 = m[1];
    else
        m1 = 0;

    for (i = 0; i <= alen - mlen; i++) {
        BignumDblInt t;
        unsigned int 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 = (unsigned)(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. Uses the Montgomery multiplication
 * technique.
 */
Bignum modpow(Bignum base_in, Bignum exp, Bignum mod)
{
    BignumInt *a, *b, *x, *n, *mninv, *tmp;
    int len, 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);

    /*
     * Make sure the base is smaller than the modulus, by reducing
     * it modulo the modulus if not.
     */
    base = bigmod(base_in, mod);

    /*
     * mod had better be odd, or we can't do Montgomery multiplication
     * using a power of two at all.
     */
    assert(mod[1] & 1);

    /*
     * 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);

    /*
     * 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 < 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 < 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 < rn[0] ? rn[j + 1] : 0);
    freebn(rn);

    tmp = snewn(3*len, BignumInt);

    /* Skip leading zero bits of exp. */
    i = 0;
    j = BIGNUM_INT_BITS-1;
    while (i < (int)exp[0] && (exp[exp[0] - i] & (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);
            monty_reduce(b, n, mninv, tmp, len);
            if ((exp[exp[0] - i] & (1 << j)) != 0) {
                internal_mul(b + len, x, a, len);
                monty_reduce(a, n, mninv, tmp, 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, tmp, 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 */
    for (i = 0; i < 3 * len; i++)
        tmp[i] = 0;
    sfree(tmp);
    for (i = 0; i < 2 * len; i++)
        a[i] = 0;
    sfree(a);
    for (i = 0; i < 2 * len; i++)
        b[i] = 0;
    sfree(b);
    for (i = 0; i < len; i++)
        mninv[i] = 0;
    sfree(mninv);
    for (i = 0; i < len; i++)
        n[i] = 0;
    sfree(n);
    for (i = 0; i < len; i++)
        x[i] = 0;
    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;
    int mshift;
    int pqlen, mlen, rlen, i, j;
    Bignum result;

    /* 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]);

    /* 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);

    /* Main computation */
    internal_mul(n, o, a, pqlen);
    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 */
    for (i = 0; i < 2 * pqlen; i++)
        a[i] = 0;
    sfree(a);
    for (i = 0; i < mlen; i++)
        m[i] = 0;
    sfree(m);
    for (i = 0; i < pqlen; i++)
        n[i] = 0;
    sfree(n);
    for (i = 0; i < pqlen; i++)
        o[i] = 0;
    sfree(o);

    return result;
}

/*
 * 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;
    int mshift;
    int plen, mlen, i, j;

    /* 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;
    }

    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];

    /* Main computation */
    internal_mod(n, plen, m, mlen, quotient, mshift);

    /* Fixup result in case the modulus was shifted */
    if (mshift) {
        for (i = plen - mlen - 1; i < plen - 1; i++)
            n[i] = (n[i] << mshift) | (n[i + 1] >> (BIGNUM_INT_BITS - mshift));
        n[plen - 1] = n[plen - 1] << mshift;
        internal_mod(n, plen, m, mlen, quotient, 0);
        for (i = plen - 1; i >= plen - mlen; i--)
            n[i] = (n[i] >> mshift) | (n[i - 1] << (BIGNUM_INT_BITS - mshift));
    }

    /* 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 */
    for (i = 0; i < mlen; i++)
        m[i] = 0;
    sfree(m);
    for (i = 0; i < plen; i++)
        n[i] = 0;
    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;

    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] |= byte << (8*i % BIGNUM_INT_BITS);
    }

    while (result[0] > 1 && result[result[0]] == 0)
        result[0]--;
    return result;
}

/*
 * 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 >= (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 >= (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 >= (int)(BIGNUM_INT_BITS * bn[0]))
        abort();                       /* beyond the end */
    else {
        int v = bitnum / BIGNUM_INT_BITS + 1;
        int mask = 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 = (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;

    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;
}

/*
 * 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;
    BignumInt *workspace;
    Bignum ret;

    /* mlen space for a, mlen space for b, 2*mlen for result */
    workspace = snewn(mlen * 4, 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);

    /* 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) {
        BignumDblInt carry = 0;
        for (i = 1; i <= rlen; i++) {
            carry += (i <= (int)ret[0] ? ret[i] : 0);
            carry += (i <= (int)addend[0] ? addend[i] : 0);
            ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
            carry >>= BIGNUM_INT_BITS;
            if (ret[i] != 0 && i > maxspot)
                maxspot = i;
        }
    }
    ret[0] = maxspot;

    sfree(workspace);
    return ret;
}

/*
 * Non-modular multiplication.
 */
Bignum bigmul(Bignum a, Bignum b)
{
    return bigmuladd(a, b, NULL);
}

/*
 * 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 a (max 32-bit) long into a bignum.
 */
Bignum bignum_from_long(unsigned long nn)
{
    Bignum ret;
    BignumDblInt n = nn;

    ret = newbn(3);
    ret[1] = (BignumInt)(n & BIGNUM_INT_MASK);
    ret[2] = (BignumInt)((n >> BIGNUM_INT_BITS) & BIGNUM_INT_MASK);
    ret[3] = 0;
    ret[0] = (ret[2]  ? 2 : 1);
    return ret;
}

/*
 * Add a long to a bignum.
 */
Bignum bignum_add_long(Bignum number, unsigned long addendx)
{
    Bignum ret = newbn(number[0] + 1);
    int i, maxspot = 0;
    BignumDblInt carry = 0, addend = addendx;

    for (i = 1; i <= (int)ret[0]; i++) {
        carry += addend & BIGNUM_INT_MASK;
        carry += (i <= (int)number[0] ? number[i] : 0);
        addend >>= BIGNUM_INT_BITS;
        ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
        carry >>= BIGNUM_INT_BITS;
        if (ret[i] != 0)
            maxspot = i;
    }
    ret[0] = maxspot;
    return ret;
}

/*
 * Compute the residue of a bignum, modulo a (max 16-bit) short.
 */
unsigned short bignum_mod_short(Bignum number, unsigned short modulus)
{
    BignumDblInt mod, r;
    int i;

    r = 0;
    mod = modulus;
    for (i = number[0]; i > 0; i--)
        r = (r * (BIGNUM_TOP_BIT % mod) * 2 + 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);
    return q;
}

/*
 * Simple remainder.
 */
Bignum bigmod(Bignum a, Bignum b)
{
    Bignum r = newbn(b[0]);
    bigdivmod(a, b, r, NULL);
    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;

    while (bignum_cmp(b, One) != 0) {
        Bignum t = newbn(b[0]);
        Bignum q = newbn(a[0]);
        bigdivmod(a, b, t, q);
        while (t[0] > 1 && t[t[0]] == 0)
            t[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;
    BignumDblInt 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++) {
            carry = (carry << BIGNUM_INT_BITS) + workspace[i];
            workspace[i] = (BignumInt) (carry / 10);
            if (workspace[i])
                iszero = 0;
            carry %= 10;
        }
        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.
     */
    sfree(workspace);
    return ret;
}