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bn_kron.c
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1 /* crypto/bn/bn_kron.c */
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50  * This product includes cryptographic software written by Eric Young
51  * ([email protected]). This product includes software written by Tim
52  * Hudson ([email protected]).
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54  */
55 
56 #include "cryptlib.h"
57 #include "bn_lcl.h"
58 
59 /* least significant word */
60 #define BN_lsw(n) (((n)->top == 0) ? (BN_ULONG) 0 : (n)->d[0])
61 
62 /* Returns -2 for errors because both -1 and 0 are valid results. */
63 int BN_kronecker(const BIGNUM *a, const BIGNUM *b, BN_CTX *ctx)
64  {
65  int i;
66  int ret = -2; /* avoid 'uninitialized' warning */
67  int err = 0;
68  BIGNUM *A, *B, *tmp;
69  /* In 'tab', only odd-indexed entries are relevant:
70  * For any odd BIGNUM n,
71  * tab[BN_lsw(n) & 7]
72  * is $(-1)^{(n^2-1)/8}$ (using TeX notation).
73  * Note that the sign of n does not matter.
74  */
75  static const int tab[8] = {0, 1, 0, -1, 0, -1, 0, 1};
76 
77  bn_check_top(a);
78  bn_check_top(b);
79 
80  BN_CTX_start(ctx);
81  A = BN_CTX_get(ctx);
82  B = BN_CTX_get(ctx);
83  if (B == NULL) goto end;
84 
85  err = !BN_copy(A, a);
86  if (err) goto end;
87  err = !BN_copy(B, b);
88  if (err) goto end;
89 
90  /*
91  * Kronecker symbol, imlemented according to Henri Cohen,
92  * "A Course in Computational Algebraic Number Theory"
93  * (algorithm 1.4.10).
94  */
95 
96  /* Cohen's step 1: */
97 
98  if (BN_is_zero(B))
99  {
100  ret = BN_abs_is_word(A, 1);
101  goto end;
102  }
103 
104  /* Cohen's step 2: */
105 
106  if (!BN_is_odd(A) && !BN_is_odd(B))
107  {
108  ret = 0;
109  goto end;
110  }
111 
112  /* now B is non-zero */
113  i = 0;
114  while (!BN_is_bit_set(B, i))
115  i++;
116  err = !BN_rshift(B, B, i);
117  if (err) goto end;
118  if (i & 1)
119  {
120  /* i is odd */
121  /* (thus B was even, thus A must be odd!) */
122 
123  /* set 'ret' to $(-1)^{(A^2-1)/8}$ */
124  ret = tab[BN_lsw(A) & 7];
125  }
126  else
127  {
128  /* i is even */
129  ret = 1;
130  }
131 
132  if (B->neg)
133  {
134  B->neg = 0;
135  if (A->neg)
136  ret = -ret;
137  }
138 
139  /* now B is positive and odd, so what remains to be done is
140  * to compute the Jacobi symbol (A/B) and multiply it by 'ret' */
141 
142  while (1)
143  {
144  /* Cohen's step 3: */
145 
146  /* B is positive and odd */
147 
148  if (BN_is_zero(A))
149  {
150  ret = BN_is_one(B) ? ret : 0;
151  goto end;
152  }
153 
154  /* now A is non-zero */
155  i = 0;
156  while (!BN_is_bit_set(A, i))
157  i++;
158  err = !BN_rshift(A, A, i);
159  if (err) goto end;
160  if (i & 1)
161  {
162  /* i is odd */
163  /* multiply 'ret' by $(-1)^{(B^2-1)/8}$ */
164  ret = ret * tab[BN_lsw(B) & 7];
165  }
166 
167  /* Cohen's step 4: */
168  /* multiply 'ret' by $(-1)^{(A-1)(B-1)/4}$ */
169  if ((A->neg ? ~BN_lsw(A) : BN_lsw(A)) & BN_lsw(B) & 2)
170  ret = -ret;
171 
172  /* (A, B) := (B mod |A|, |A|) */
173  err = !BN_nnmod(B, B, A, ctx);
174  if (err) goto end;
175  tmp = A; A = B; B = tmp;
176  tmp->neg = 0;
177  }
178 end:
179  BN_CTX_end(ctx);
180  if (err)
181  return -2;
182  else
183  return ret;
184  }