- {
- BIGNUM *ret = in;
- int err = 1;
- int r;
- BIGNUM *b, *q, *t, *x, *y;
- int e, i, j;
-
- if (!BN_is_odd(p) || BN_abs_is_word(p, 1))
- {
- if (BN_abs_is_word(p, 2))
- {
- if (ret == NULL)
- ret = BN_new();
- if (ret == NULL)
- goto end;
- if (!BN_set_word(ret, BN_is_bit_set(a, 0)))
- {
- BN_free(ret);
- return NULL;
- }
- return ret;
- }
-
- BNerr(BN_F_BN_MOD_SQRT, BN_R_P_IS_NOT_PRIME);
- return(NULL);
- }
-
-#if 0 /* if BN_mod_sqrt is used with correct input, this just wastes time */
- r = BN_kronecker(a, p, ctx);
- if (r < -1) return NULL;
- if (r == -1)
- {
- BNerr(BN_F_BN_MOD_SQRT, BN_R_NOT_A_SQUARE);
- return(NULL);
- }
-#endif
-
- BN_CTX_start(ctx);
- b = BN_CTX_get(ctx);
- q = BN_CTX_get(ctx);
- t = BN_CTX_get(ctx);
- x = BN_CTX_get(ctx);
- y = BN_CTX_get(ctx);
- if (y == NULL) goto end;
-
- if (ret == NULL)
- ret = BN_new();
- if (ret == NULL) goto end;
-
- /* now write |p| - 1 as 2^e*q where q is odd */
- e = 1;
- while (!BN_is_bit_set(p, e))
- e++;
- if (!BN_rshift(q, p, e)) goto end;
- q->neg = 0;
-
- if (e == 1)
- {
- /* The easy case: (p-1)/2 is odd, so 2 has an inverse
- * modulo (p-1)/2, and square roots can be computed
- * directly by modular exponentiation.
- * We have
- * 2 * (p+1)/4 == 1 (mod (p-1)/2),
- * so we can use exponent (p+1)/4, i.e. (q+1)/2.
- */
- if (!BN_add_word(q,1)) goto end;
- if (!BN_rshift1(q,q)) goto end;
- if (!BN_mod_exp(ret, a, q, p, ctx)) goto end;
- err = 0;
- goto end;
- }
-
- /* e > 1, so we really have to use the Tonelli/Shanks algorithm.
- * First, find some y that is not a square. */
- i = 1;
- do
- {
- /* For efficiency, try small numbers first;
- * if this fails, try random numbers.
- */
- if (i < 20)
- {
- if (!BN_set_word(y, i)) goto end;
- }
- else
- {
- if (!BN_pseudo_rand(y, BN_num_bits(p), 0, 0)) goto end;
- if (BN_ucmp(y, p) >= 0)
- {
- if (!(p->neg ? BN_add : BN_sub)(y, y, p)) goto end;
- }
- /* now 0 <= y < |p| */
- if (BN_is_zero(y))
- if (!BN_set_word(y, i)) goto end;
- }
-
- r = BN_kronecker(y, p, ctx);
- if (r < -1) goto end;
- if (r == 0)
- {
- /* m divides p */
- BNerr(BN_F_BN_MOD_SQRT, BN_R_P_IS_NOT_PRIME);
- goto end;
- }
- }
- while (r == 1 && i++ < 80);
-
- if (r != -1)
- {
- /* Many rounds and still no non-square -- this is more likely
- * a bug than just bad luck.
- * Even if p is not prime, we should have found some y
- * such that r == -1.
- */
- BNerr(BN_F_BN_MOD_SQRT, BN_R_TOO_MANY_ITERATIONS);
- goto end;
- }
-
-
- /* Now that we have some non-square, we can find an element
- * of order 2^e by computing its q'th power. */
- if (!BN_mod_exp(y, y, q, p, ctx)) goto end;
- if (BN_is_one(y))
- {
- BNerr(BN_F_BN_MOD_SQRT, BN_R_P_IS_NOT_PRIME);
- goto end;
- }
-
- /* Now we know that (if p is indeed prime) there is an integer
- * k, 0 <= k < 2^e, such that
- *
- * a^q * y^k == 1 (mod p).
- *
- * As a^q is a square and y is not, k must be even.
- * q+1 is even, too, so there is an element
- *
- * X := a^((q+1)/2) * y^(k/2),
- *
- * and it satisfies
- *
- * X^2 = a^q * a * y^k
- * = a,
- *
- * so it is the square root that we are looking for.
- */
-
- /* t := (q-1)/2 (note that q is odd) */
- if (!BN_rshift1(t, q)) goto end;
-
- /* x := a^((q-1)/2) */
- if (BN_is_zero(t)) /* special case: p = 2^e + 1 */
- {
- if (!BN_nnmod(t, a, p, ctx)) goto end;
- if (BN_is_zero(t))
- {
- /* special case: a == 0 (mod p) */
- if (!BN_zero(ret)) goto end;
- err = 0;
- goto end;
- }
- else
- if (!BN_one(x)) goto end;
- }
- else
- {
- if (!BN_mod_exp(x, a, t, p, ctx)) goto end;
- if (BN_is_zero(x))
- {
- /* special case: a == 0 (mod p) */
- if (!BN_zero(ret)) goto end;
- err = 0;
- goto end;
- }
- }
-
- /* b := a*x^2 (= a^q) */
- if (!BN_mod_sqr(b, x, p, ctx)) goto end;
- if (!BN_mod_mul(b, b, a, p, ctx)) goto end;
-
- /* x := a*x (= a^((q+1)/2)) */
- if (!BN_mod_mul(x, x, a, p, ctx)) goto end;
-
- while (1)
- {
- /* Now b is a^q * y^k for some even k (0 <= k < 2^E
- * where E refers to the original value of e, which we
- * don't keep in a variable), and x is a^((q+1)/2) * y^(k/2).
- *
- * We have a*b = x^2,
- * y^2^(e-1) = -1,
- * b^2^(e-1) = 1.
- */
-
- if (BN_is_one(b))
- {
- if (!BN_copy(ret, x)) goto end;
- err = 0;
- goto end;
- }
-
-
- /* find smallest i such that b^(2^i) = 1 */
- i = 1;
- if (!BN_mod_sqr(t, b, p, ctx)) goto end;
- while (!BN_is_one(t))
- {
- i++;
- if (i == e)
- {
- BNerr(BN_F_BN_MOD_SQRT, BN_R_NOT_A_SQUARE);
- goto end;
- }
- if (!BN_mod_mul(t, t, t, p, ctx)) goto end;
- }
-
-
- /* t := y^2^(e - i - 1) */
- if (!BN_copy(t, y)) goto end;
- for (j = e - i - 1; j > 0; j--)
- {
- if (!BN_mod_sqr(t, t, p, ctx)) goto end;
- }
- if (!BN_mod_mul(y, t, t, p, ctx)) goto end;
- if (!BN_mod_mul(x, x, t, p, ctx)) goto end;
- if (!BN_mod_mul(b, b, y, p, ctx)) goto end;
- e = i;
- }
+{
+ BIGNUM *ret = in;
+ int err = 1;
+ int r;
+ BIGNUM *A, *b, *q, *t, *x, *y;
+ int e, i, j;
+
+ if (!BN_is_odd(p) || BN_abs_is_word(p, 1)) {
+ if (BN_abs_is_word(p, 2)) {
+ if (ret == NULL)
+ ret = BN_new();
+ if (ret == NULL)
+ goto end;
+ if (!BN_set_word(ret, BN_is_bit_set(a, 0))) {
+ if (ret != in)
+ BN_free(ret);
+ return NULL;
+ }
+ bn_check_top(ret);
+ return ret;
+ }
+
+ BNerr(BN_F_BN_MOD_SQRT, BN_R_P_IS_NOT_PRIME);
+ return (NULL);
+ }
+
+ if (BN_is_zero(a) || BN_is_one(a)) {
+ if (ret == NULL)
+ ret = BN_new();
+ if (ret == NULL)
+ goto end;
+ if (!BN_set_word(ret, BN_is_one(a))) {
+ if (ret != in)
+ BN_free(ret);
+ return NULL;
+ }
+ bn_check_top(ret);
+ return ret;
+ }
+
+ BN_CTX_start(ctx);
+ A = BN_CTX_get(ctx);
+ b = BN_CTX_get(ctx);
+ q = BN_CTX_get(ctx);
+ t = BN_CTX_get(ctx);
+ x = BN_CTX_get(ctx);
+ y = BN_CTX_get(ctx);
+ if (y == NULL)
+ goto end;
+
+ if (ret == NULL)
+ ret = BN_new();
+ if (ret == NULL)
+ goto end;
+
+ /* A = a mod p */
+ if (!BN_nnmod(A, a, p, ctx))
+ goto end;
+
+ /* now write |p| - 1 as 2^e*q where q is odd */
+ e = 1;
+ while (!BN_is_bit_set(p, e))
+ e++;
+ /* we'll set q later (if needed) */
+
+ if (e == 1) {
+ /*-
+ * The easy case: (|p|-1)/2 is odd, so 2 has an inverse
+ * modulo (|p|-1)/2, and square roots can be computed
+ * directly by modular exponentiation.
+ * We have
+ * 2 * (|p|+1)/4 == 1 (mod (|p|-1)/2),
+ * so we can use exponent (|p|+1)/4, i.e. (|p|-3)/4 + 1.
+ */
+ if (!BN_rshift(q, p, 2))
+ goto end;
+ q->neg = 0;
+ if (!BN_add_word(q, 1))
+ goto end;
+ if (!BN_mod_exp(ret, A, q, p, ctx))
+ goto end;
+ err = 0;
+ goto vrfy;
+ }
+
+ if (e == 2) {
+ /*-
+ * |p| == 5 (mod 8)
+ *
+ * In this case 2 is always a non-square since
+ * Legendre(2,p) = (-1)^((p^2-1)/8) for any odd prime.
+ * So if a really is a square, then 2*a is a non-square.
+ * Thus for
+ * b := (2*a)^((|p|-5)/8),
+ * i := (2*a)*b^2
+ * we have
+ * i^2 = (2*a)^((1 + (|p|-5)/4)*2)
+ * = (2*a)^((p-1)/2)
+ * = -1;
+ * so if we set
+ * x := a*b*(i-1),
+ * then
+ * x^2 = a^2 * b^2 * (i^2 - 2*i + 1)
+ * = a^2 * b^2 * (-2*i)
+ * = a*(-i)*(2*a*b^2)
+ * = a*(-i)*i
+ * = a.
+ *
+ * (This is due to A.O.L. Atkin,
+ * <URL: http://listserv.nodak.edu/scripts/wa.exe?A2=ind9211&L=nmbrthry&O=T&P=562>,
+ * November 1992.)
+ */
+
+ /* t := 2*a */
+ if (!BN_mod_lshift1_quick(t, A, p))
+ goto end;
+
+ /* b := (2*a)^((|p|-5)/8) */
+ if (!BN_rshift(q, p, 3))
+ goto end;
+ q->neg = 0;
+ if (!BN_mod_exp(b, t, q, p, ctx))
+ goto end;
+
+ /* y := b^2 */
+ if (!BN_mod_sqr(y, b, p, ctx))
+ goto end;
+
+ /* t := (2*a)*b^2 - 1 */
+ if (!BN_mod_mul(t, t, y, p, ctx))
+ goto end;
+ if (!BN_sub_word(t, 1))
+ goto end;
+
+ /* x = a*b*t */
+ if (!BN_mod_mul(x, A, b, p, ctx))
+ goto end;
+ if (!BN_mod_mul(x, x, t, p, ctx))
+ goto end;
+
+ if (!BN_copy(ret, x))
+ goto end;
+ err = 0;
+ goto vrfy;
+ }
+
+ /*
+ * e > 2, so we really have to use the Tonelli/Shanks algorithm. First,
+ * find some y that is not a square.
+ */
+ if (!BN_copy(q, p))
+ goto end; /* use 'q' as temp */
+ q->neg = 0;
+ i = 2;
+ do {
+ /*
+ * For efficiency, try small numbers first; if this fails, try random
+ * numbers.
+ */
+ if (i < 22) {
+ if (!BN_set_word(y, i))
+ goto end;
+ } else {
+ if (!BN_pseudo_rand(y, BN_num_bits(p), 0, 0))
+ goto end;
+ if (BN_ucmp(y, p) >= 0) {
+ if (!(p->neg ? BN_add : BN_sub) (y, y, p))
+ goto end;
+ }
+ /* now 0 <= y < |p| */
+ if (BN_is_zero(y))
+ if (!BN_set_word(y, i))
+ goto end;
+ }
+
+ r = BN_kronecker(y, q, ctx); /* here 'q' is |p| */
+ if (r < -1)
+ goto end;
+ if (r == 0) {
+ /* m divides p */
+ BNerr(BN_F_BN_MOD_SQRT, BN_R_P_IS_NOT_PRIME);
+ goto end;
+ }
+ }
+ while (r == 1 && ++i < 82);
+
+ if (r != -1) {
+ /*
+ * Many rounds and still no non-square -- this is more likely a bug
+ * than just bad luck. Even if p is not prime, we should have found
+ * some y such that r == -1.
+ */
+ BNerr(BN_F_BN_MOD_SQRT, BN_R_TOO_MANY_ITERATIONS);
+ goto end;
+ }
+
+ /* Here's our actual 'q': */
+ if (!BN_rshift(q, q, e))
+ goto end;
+
+ /*
+ * Now that we have some non-square, we can find an element of order 2^e
+ * by computing its q'th power.
+ */
+ if (!BN_mod_exp(y, y, q, p, ctx))
+ goto end;
+ if (BN_is_one(y)) {
+ BNerr(BN_F_BN_MOD_SQRT, BN_R_P_IS_NOT_PRIME);
+ goto end;
+ }
+
+ /*-
+ * Now we know that (if p is indeed prime) there is an integer
+ * k, 0 <= k < 2^e, such that
+ *
+ * a^q * y^k == 1 (mod p).
+ *
+ * As a^q is a square and y is not, k must be even.
+ * q+1 is even, too, so there is an element
+ *
+ * X := a^((q+1)/2) * y^(k/2),
+ *
+ * and it satisfies
+ *
+ * X^2 = a^q * a * y^k
+ * = a,
+ *
+ * so it is the square root that we are looking for.
+ */
+
+ /* t := (q-1)/2 (note that q is odd) */
+ if (!BN_rshift1(t, q))
+ goto end;
+
+ /* x := a^((q-1)/2) */
+ if (BN_is_zero(t)) { /* special case: p = 2^e + 1 */
+ if (!BN_nnmod(t, A, p, ctx))
+ goto end;
+ if (BN_is_zero(t)) {
+ /* special case: a == 0 (mod p) */
+ BN_zero(ret);
+ err = 0;
+ goto end;
+ } else if (!BN_one(x))
+ goto end;
+ } else {
+ if (!BN_mod_exp(x, A, t, p, ctx))
+ goto end;
+ if (BN_is_zero(x)) {
+ /* special case: a == 0 (mod p) */
+ BN_zero(ret);
+ err = 0;
+ goto end;
+ }
+ }
+
+ /* b := a*x^2 (= a^q) */
+ if (!BN_mod_sqr(b, x, p, ctx))
+ goto end;
+ if (!BN_mod_mul(b, b, A, p, ctx))
+ goto end;
+
+ /* x := a*x (= a^((q+1)/2)) */
+ if (!BN_mod_mul(x, x, A, p, ctx))
+ goto end;
+
+ while (1) {
+ /*-
+ * Now b is a^q * y^k for some even k (0 <= k < 2^E
+ * where E refers to the original value of e, which we
+ * don't keep in a variable), and x is a^((q+1)/2) * y^(k/2).
+ *
+ * We have a*b = x^2,
+ * y^2^(e-1) = -1,
+ * b^2^(e-1) = 1.
+ */
+
+ if (BN_is_one(b)) {
+ if (!BN_copy(ret, x))
+ goto end;
+ err = 0;
+ goto vrfy;
+ }
+
+ /* find smallest i such that b^(2^i) = 1 */
+ i = 1;
+ if (!BN_mod_sqr(t, b, p, ctx))
+ goto end;
+ while (!BN_is_one(t)) {
+ i++;
+ if (i == e) {
+ BNerr(BN_F_BN_MOD_SQRT, BN_R_NOT_A_SQUARE);
+ goto end;
+ }
+ if (!BN_mod_mul(t, t, t, p, ctx))
+ goto end;
+ }
+
+ /* t := y^2^(e - i - 1) */
+ if (!BN_copy(t, y))
+ goto end;
+ for (j = e - i - 1; j > 0; j--) {
+ if (!BN_mod_sqr(t, t, p, ctx))
+ goto end;
+ }
+ if (!BN_mod_mul(y, t, t, p, ctx))
+ goto end;
+ if (!BN_mod_mul(x, x, t, p, ctx))
+ goto end;
+ if (!BN_mod_mul(b, b, y, p, ctx))
+ goto end;
+ e = i;
+ }
+
+ vrfy:
+ if (!err) {
+ /*
+ * verify the result -- the input might have been not a square (test
+ * added in 0.9.8)
+ */
+
+ if (!BN_mod_sqr(x, ret, p, ctx))
+ err = 1;
+
+ if (!err && 0 != BN_cmp(x, A)) {
+ BNerr(BN_F_BN_MOD_SQRT, BN_R_NOT_A_SQUARE);
+ err = 1;
+ }
+ }