1 files changed, 140 insertions, 0 deletions
diff --git a/openssl-1.1.0h/crypto/bn/bn_kron.c b/openssl-1.1.0h/crypto/bn/bn_kron.c
new file mode 100644
index 0000000..b9bc6cc
--- /dev/null
+++ b/openssl-1.1.0h/crypto/bn/bn_kron.c
@@ -0,0 +1,140 @@
+/*
+ * Copyright 2000-2016 The OpenSSL Project Authors. All Rights Reserved.
+ *
+ * Licensed under the OpenSSL license (the "License").  You may not use
+ * this file except in compliance with the License.  You can obtain a copy
+ * in the file LICENSE in the source distribution or at
+ * https://www.openssl.org/source/license.html
+ */
+
+#include "internal/cryptlib.h"
+#include "bn_lcl.h"
+
+/* least significant word */
+#define BN_lsw(n) (((n)->top == 0) ? (BN_ULONG) 0 : (n)->d[0])
+
+/* 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, implemented 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;
+}