158 lines
3.8 KiB
C
158 lines
3.8 KiB
C
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/* TomsFastMath, a fast ISO C bignum library.
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*
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* This project is meant to fill in where LibTomMath
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* falls short. That is speed ;-)
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*
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* This project is public domain and free for all purposes.
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*
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* Tom St Denis, tomstdenis@gmail.com
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*/
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#include "bignum_fast.h"
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/* a/b => cb + d == a */
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int fp_div(fp_int *a, fp_int *b, fp_int *c, fp_int *d)
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{
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fp_int q, x, y, t1, t2;
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int n, t, i, norm, neg;
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/* is divisor zero ? */
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if (fp_iszero (b) == 1) {
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return FP_VAL;
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}
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/* if a < b then q=0, r = a */
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if (fp_cmp_mag (a, b) == FP_LT) {
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if (d != NULL) {
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fp_copy (a, d);
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}
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if (c != NULL) {
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fp_zero (c);
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}
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return FP_OKAY;
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}
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fp_init(&q);
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q.used = a->used + 2;
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fp_init(&t1);
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fp_init(&t2);
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fp_init_copy(&x, a);
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fp_init_copy(&y, b);
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/* fix the sign */
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neg = (a->sign == b->sign) ? FP_ZPOS : FP_NEG;
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x.sign = y.sign = FP_ZPOS;
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/* normalize both x and y, ensure that y >= b/2, [b == 2**DIGIT_BIT] */
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norm = fp_count_bits(&y) % DIGIT_BIT;
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if (norm < (int)(DIGIT_BIT-1)) {
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norm = (DIGIT_BIT-1) - norm;
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fp_mul_2d (&x, norm, &x);
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fp_mul_2d (&y, norm, &y);
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} else {
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norm = 0;
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}
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/* note hac does 0 based, so if used==5 then its 0,1,2,3,4, e.g. use 4 */
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n = x.used - 1;
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t = y.used - 1;
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/* while (x >= y*b**n-t) do { q[n-t] += 1; x -= y*b**{n-t} } */
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fp_lshd (&y, n - t); /* y = y*b**{n-t} */
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while (fp_cmp (&x, &y) != FP_LT) {
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++(q.dp[n - t]);
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fp_sub (&x, &y, &x);
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}
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/* reset y by shifting it back down */
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fp_rshd (&y, n - t);
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/* step 3. for i from n down to (t + 1) */
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for (i = n; i >= (t + 1); i--) {
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if (i > x.used) {
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continue;
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}
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/* step 3.1 if xi == yt then set q{i-t-1} to b-1,
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* otherwise set q{i-t-1} to (xi*b + x{i-1})/yt */
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if (x.dp[i] == y.dp[t]) {
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q.dp[i - t - 1] = ((((fp_word)1) << DIGIT_BIT) - 1);
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} else {
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fp_word tmp;
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tmp = ((fp_word) x.dp[i]) << ((fp_word) DIGIT_BIT);
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tmp |= ((fp_word) x.dp[i - 1]);
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tmp /= ((fp_word) y.dp[t]);
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q.dp[i - t - 1] = (fp_digit) (tmp);
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}
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/* while (q{i-t-1} * (yt * b + y{t-1})) >
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xi * b**2 + xi-1 * b + xi-2
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do q{i-t-1} -= 1;
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*/
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q.dp[i - t - 1] = (q.dp[i - t - 1] + 1);
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do {
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q.dp[i - t - 1] = (q.dp[i - t - 1] - 1);
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/* find left hand */
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fp_zero (&t1);
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t1.dp[0] = (t - 1 < 0) ? 0 : y.dp[t - 1];
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t1.dp[1] = y.dp[t];
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t1.used = 2;
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fp_mul_d (&t1, q.dp[i - t - 1], &t1);
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/* find right hand */
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t2.dp[0] = (i - 2 < 0) ? 0 : x.dp[i - 2];
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t2.dp[1] = (i - 1 < 0) ? 0 : x.dp[i - 1];
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t2.dp[2] = x.dp[i];
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t2.used = 3;
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} while (fp_cmp_mag(&t1, &t2) == FP_GT);
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/* step 3.3 x = x - q{i-t-1} * y * b**{i-t-1} */
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fp_mul_d (&y, q.dp[i - t - 1], &t1);
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fp_lshd (&t1, i - t - 1);
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fp_sub (&x, &t1, &x);
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/* if x < 0 then { x = x + y*b**{i-t-1}; q{i-t-1} -= 1; } */
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if (x.sign == FP_NEG) {
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fp_copy (&y, &t1);
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fp_lshd (&t1, i - t - 1);
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fp_add (&x, &t1, &x);
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q.dp[i - t - 1] = q.dp[i - t - 1] - 1;
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}
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}
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/* now q is the quotient and x is the remainder
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* [which we have to normalize]
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*/
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/* get sign before writing to c */
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x.sign = x.used == 0 ? FP_ZPOS : a->sign;
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if (c != NULL) {
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fp_clamp (&q);
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fp_copy (&q, c);
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c->sign = neg;
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}
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if (d != NULL) {
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fp_div_2d (&x, norm, &x, NULL);
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/* the following is a kludge, essentially we were seeing the right remainder but
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with excess digits that should have been zero
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*/
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for (i = b->used; i < x.used; i++) {
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x.dp[i] = 0;
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}
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fp_clamp(&x);
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fp_copy (&x, d);
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}
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return FP_OKAY;
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}
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/* $Source: /cvs/libtom/tomsfastmath/src/divide/fp_div.c,v $ */
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/* $Revision: 1.1 $ */
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/* $Date: 2006/12/31 21:25:53 $ */
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