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			140 lines
		
	
	
	
		
			3.2 KiB
		
	
	
	
		
			C
		
	
	
	
	
	
			
		
		
	
	
			140 lines
		
	
	
	
		
			3.2 KiB
		
	
	
	
		
			C
		
	
	
	
	
	
| /*
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|  * Copyright 2000-2016 The OpenSSL Project Authors. All Rights Reserved.
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|  *
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|  * Licensed under the OpenSSL license (the "License").  You may not use
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|  * this file except in compliance with the License.  You can obtain a copy
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|  * in the file LICENSE in the source distribution or at
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|  * https://www.openssl.org/source/license.html
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|  */
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| 
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| #include "internal/cryptlib.h"
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| #include "bn_local.h"
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| 
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| /* least significant word */
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| #define BN_lsw(n) (((n)->top == 0) ? (BN_ULONG) 0 : (n)->d[0])
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| 
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| /* Returns -2 for errors because both -1 and 0 are valid results. */
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| int BN_kronecker(const BIGNUM *a, const BIGNUM *b, BN_CTX *ctx)
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| {
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|     int i;
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|     int ret = -2;               /* avoid 'uninitialized' warning */
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|     int err = 0;
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|     BIGNUM *A, *B, *tmp;
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|     /*-
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|      * In 'tab', only odd-indexed entries are relevant:
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|      * For any odd BIGNUM n,
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|      *     tab[BN_lsw(n) & 7]
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|      * is $(-1)^{(n^2-1)/8}$ (using TeX notation).
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|      * Note that the sign of n does not matter.
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|      */
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|     static const int tab[8] = { 0, 1, 0, -1, 0, -1, 0, 1 };
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| 
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|     bn_check_top(a);
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|     bn_check_top(b);
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| 
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|     BN_CTX_start(ctx);
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|     A = BN_CTX_get(ctx);
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|     B = BN_CTX_get(ctx);
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|     if (B == NULL)
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|         goto end;
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| 
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|     err = !BN_copy(A, a);
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|     if (err)
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|         goto end;
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|     err = !BN_copy(B, b);
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|     if (err)
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|         goto end;
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| 
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|     /*
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|      * Kronecker symbol, implemented according to Henri Cohen,
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|      * "A Course in Computational Algebraic Number Theory"
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|      * (algorithm 1.4.10).
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|      */
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| 
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|     /* Cohen's step 1: */
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| 
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|     if (BN_is_zero(B)) {
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|         ret = BN_abs_is_word(A, 1);
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|         goto end;
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|     }
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| 
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|     /* Cohen's step 2: */
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| 
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|     if (!BN_is_odd(A) && !BN_is_odd(B)) {
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|         ret = 0;
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|         goto end;
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|     }
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| 
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|     /* now  B  is non-zero */
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|     i = 0;
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|     while (!BN_is_bit_set(B, i))
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|         i++;
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|     err = !BN_rshift(B, B, i);
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|     if (err)
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|         goto end;
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|     if (i & 1) {
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|         /* i is odd */
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|         /* (thus  B  was even, thus  A  must be odd!)  */
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| 
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|         /* set 'ret' to $(-1)^{(A^2-1)/8}$ */
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|         ret = tab[BN_lsw(A) & 7];
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|     } else {
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|         /* i is even */
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|         ret = 1;
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|     }
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| 
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|     if (B->neg) {
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|         B->neg = 0;
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|         if (A->neg)
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|             ret = -ret;
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|     }
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| 
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|     /*
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|      * now B is positive and odd, so what remains to be done is to compute
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|      * the Jacobi symbol (A/B) and multiply it by 'ret'
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|      */
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| 
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|     while (1) {
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|         /* Cohen's step 3: */
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| 
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|         /*  B  is positive and odd */
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| 
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|         if (BN_is_zero(A)) {
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|             ret = BN_is_one(B) ? ret : 0;
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|             goto end;
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|         }
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| 
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|         /* now  A  is non-zero */
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|         i = 0;
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|         while (!BN_is_bit_set(A, i))
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|             i++;
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|         err = !BN_rshift(A, A, i);
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|         if (err)
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|             goto end;
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|         if (i & 1) {
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|             /* i is odd */
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|             /* multiply 'ret' by  $(-1)^{(B^2-1)/8}$ */
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|             ret = ret * tab[BN_lsw(B) & 7];
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|         }
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| 
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|         /* Cohen's step 4: */
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|         /* multiply 'ret' by  $(-1)^{(A-1)(B-1)/4}$ */
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|         if ((A->neg ? ~BN_lsw(A) : BN_lsw(A)) & BN_lsw(B) & 2)
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|             ret = -ret;
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| 
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|         /* (A, B) := (B mod |A|, |A|) */
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|         err = !BN_nnmod(B, B, A, ctx);
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|         if (err)
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|             goto end;
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|         tmp = A;
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|         A = B;
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|         B = tmp;
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|         tmp->neg = 0;
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|     }
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|  end:
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|     BN_CTX_end(ctx);
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|     if (err)
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|         return -2;
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|     else
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|         return ret;
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| }
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