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-rw-r--r--crypto/bn/bn_kron.c246
1 files changed, 124 insertions, 122 deletions
diff --git a/crypto/bn/bn_kron.c b/crypto/bn/bn_kron.c
index 740359b7520d..88d731ac75cd 100644
--- a/crypto/bn/bn_kron.c
+++ b/crypto/bn/bn_kron.c
@@ -7,7 +7,7 @@
* are met:
*
* 1. Redistributions of source code must retain the above copyright
- * notice, this list of conditions and the following disclaimer.
+ * notice, this list of conditions and the following disclaimer.
*
* 2. Redistributions in binary form must reproduce the above copyright
* notice, this list of conditions and the following disclaimer in
@@ -61,124 +61,126 @@
/* Returns -2 for errors because both -1 and 0 are valid results. */
int BN_kronecker(const BIGNUM *a, const BIGNUM *b, BN_CTX *ctx)
- {
- int i;
- int ret = -2; /* avoid 'uninitialized' warning */
- int err = 0;
- BIGNUM *A, *B, *tmp;
- /* In 'tab', only odd-indexed entries are relevant:
- * For any odd BIGNUM n,
- * tab[BN_lsw(n) & 7]
- * is $(-1)^{(n^2-1)/8}$ (using TeX notation).
- * Note that the sign of n does not matter.
- */
- static const int tab[8] = {0, 1, 0, -1, 0, -1, 0, 1};
-
- bn_check_top(a);
- bn_check_top(b);
-
- BN_CTX_start(ctx);
- A = BN_CTX_get(ctx);
- B = BN_CTX_get(ctx);
- if (B == NULL) goto end;
-
- err = !BN_copy(A, a);
- if (err) goto end;
- err = !BN_copy(B, b);
- if (err) goto end;
-
- /*
- * Kronecker symbol, imlemented according to Henri Cohen,
- * "A Course in Computational Algebraic Number Theory"
- * (algorithm 1.4.10).
- */
-
- /* Cohen's step 1: */
-
- if (BN_is_zero(B))
- {
- ret = BN_abs_is_word(A, 1);
- goto end;
- }
-
- /* Cohen's step 2: */
-
- if (!BN_is_odd(A) && !BN_is_odd(B))
- {
- ret = 0;
- goto end;
- }
-
- /* now B is non-zero */
- i = 0;
- while (!BN_is_bit_set(B, i))
- i++;
- err = !BN_rshift(B, B, i);
- if (err) goto end;
- if (i & 1)
- {
- /* i is odd */
- /* (thus B was even, thus A must be odd!) */
-
- /* set 'ret' to $(-1)^{(A^2-1)/8}$ */
- ret = tab[BN_lsw(A) & 7];
- }
- else
- {
- /* i is even */
- ret = 1;
- }
-
- if (B->neg)
- {
- B->neg = 0;
- if (A->neg)
- ret = -ret;
- }
-
- /* now B is positive and odd, so what remains to be done is
- * to compute the Jacobi symbol (A/B) and multiply it by 'ret' */
-
- while (1)
- {
- /* Cohen's step 3: */
-
- /* B is positive and odd */
-
- if (BN_is_zero(A))
- {
- ret = BN_is_one(B) ? ret : 0;
- goto end;
- }
-
- /* now A is non-zero */
- i = 0;
- while (!BN_is_bit_set(A, i))
- i++;
- err = !BN_rshift(A, A, i);
- if (err) goto end;
- if (i & 1)
- {
- /* i is odd */
- /* multiply 'ret' by $(-1)^{(B^2-1)/8}$ */
- ret = ret * tab[BN_lsw(B) & 7];
- }
-
- /* Cohen's step 4: */
- /* multiply 'ret' by $(-1)^{(A-1)(B-1)/4}$ */
- if ((A->neg ? ~BN_lsw(A) : BN_lsw(A)) & BN_lsw(B) & 2)
- ret = -ret;
-
- /* (A, B) := (B mod |A|, |A|) */
- err = !BN_nnmod(B, B, A, ctx);
- if (err) goto end;
- tmp = A; A = B; B = tmp;
- tmp->neg = 0;
- }
-end:
- BN_CTX_end(ctx);
- if (err)
- return -2;
- else
- return ret;
- }
+{
+ int i;
+ int ret = -2; /* avoid 'uninitialized' warning */
+ int err = 0;
+ BIGNUM *A, *B, *tmp;
+ /*-
+ * In 'tab', only odd-indexed entries are relevant:
+ * For any odd BIGNUM n,
+ * tab[BN_lsw(n) & 7]
+ * is $(-1)^{(n^2-1)/8}$ (using TeX notation).
+ * Note that the sign of n does not matter.
+ */
+ static const int tab[8] = { 0, 1, 0, -1, 0, -1, 0, 1 };
+
+ bn_check_top(a);
+ bn_check_top(b);
+
+ BN_CTX_start(ctx);
+ A = BN_CTX_get(ctx);
+ B = BN_CTX_get(ctx);
+ if (B == NULL)
+ goto end;
+
+ err = !BN_copy(A, a);
+ if (err)
+ goto end;
+ err = !BN_copy(B, b);
+ if (err)
+ goto end;
+
+ /*
+ * Kronecker symbol, imlemented according to Henri Cohen,
+ * "A Course in Computational Algebraic Number Theory"
+ * (algorithm 1.4.10).
+ */
+
+ /* Cohen's step 1: */
+
+ if (BN_is_zero(B)) {
+ ret = BN_abs_is_word(A, 1);
+ goto end;
+ }
+
+ /* Cohen's step 2: */
+
+ if (!BN_is_odd(A) && !BN_is_odd(B)) {
+ ret = 0;
+ goto end;
+ }
+
+ /* now B is non-zero */
+ i = 0;
+ while (!BN_is_bit_set(B, i))
+ i++;
+ err = !BN_rshift(B, B, i);
+ if (err)
+ goto end;
+ if (i & 1) {
+ /* i is odd */
+ /* (thus B was even, thus A must be odd!) */
+
+ /* set 'ret' to $(-1)^{(A^2-1)/8}$ */
+ ret = tab[BN_lsw(A) & 7];
+ } else {
+ /* i is even */
+ ret = 1;
+ }
+
+ if (B->neg) {
+ B->neg = 0;
+ if (A->neg)
+ ret = -ret;
+ }
+
+ /*
+ * now B is positive and odd, so what remains to be done is to compute
+ * the Jacobi symbol (A/B) and multiply it by 'ret'
+ */
+
+ while (1) {
+ /* Cohen's step 3: */
+
+ /* B is positive and odd */
+
+ if (BN_is_zero(A)) {
+ ret = BN_is_one(B) ? ret : 0;
+ goto end;
+ }
+
+ /* now A is non-zero */
+ i = 0;
+ while (!BN_is_bit_set(A, i))
+ i++;
+ err = !BN_rshift(A, A, i);
+ if (err)
+ goto end;
+ if (i & 1) {
+ /* i is odd */
+ /* multiply 'ret' by $(-1)^{(B^2-1)/8}$ */
+ ret = ret * tab[BN_lsw(B) & 7];
+ }
+
+ /* Cohen's step 4: */
+ /* multiply 'ret' by $(-1)^{(A-1)(B-1)/4}$ */
+ if ((A->neg ? ~BN_lsw(A) : BN_lsw(A)) & BN_lsw(B) & 2)
+ ret = -ret;
+
+ /* (A, B) := (B mod |A|, |A|) */
+ err = !BN_nnmod(B, B, A, ctx);
+ if (err)
+ goto end;
+ tmp = A;
+ A = B;
+ B = tmp;
+ tmp->neg = 0;
+ }
+ end:
+ BN_CTX_end(ctx);
+ if (err)
+ return -2;
+ else
+ return ret;
+}