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