看到很多文章说ENCRYPTION_SEED3~ENCRYPTION_SEED4与私钥相关,得到了这两个key就可以搞定ecc,但通过初步分析,情况并没有这么乐观。
分析了一下,至少有两种情况:
1,同时定义了LM_SEED1~LM_SEED3和ENCRYPTION_SEED3~ENCRYPTION_SEED4,则私钥只与LM_SEED1~LM_SEED3有关,与ENCRYPTION_SEED3~ENCRYPTION_SEED4没有关系。
2,如果只定义了LM_SEED1~LM_SEED3,没有定义ENCRYPTION_SEED3~ENCRYPTION_SEED4,则ENCRYPTION_SEED3~ENCRYPTION_SEED4是由LM_SEED1~LM_SEED3产生且唯一。
对于情况2,私钥是通过genkeys产生的,函数如下
其中的seed为croseeds,croseeds来源于以下代码:
newseeds通过l_genrand产生,进而得到croseeds,seed3~seed4(ENCRYPTION_SEED3~ENCRYPTION_SEED4)。
从以上分析可以看出,私钥/公钥只与lmseed1~lmseed3有关,与seed3~seed4没有直接关联。
由此有如下结论:
1,seed3~seed4和私钥种子croseeds都是由LM_SEED1~LM_SEED3产生的,但这个过程不可逆,即不能由seed3~seed4和私钥croseeds反推出LM_SEED1~LM_SEED3;
2,即使找到seed3~seed4,同样无法获得私钥。
因此暴力求解flexlm ECC的计算量还是2^96,并没有减少到2^64.
以上个人的一些浅见,还望各位flexlm大神多多指正。
static
int
genkeys(unsigned
int
*
seed,
int
pubkey_strength,
sb_PrivateKey
*
privateKey,
sb_PublicKey
*
publicKey)
static
int
genkeys(unsigned
int
*
seed,
int
pubkey_strength,
sb_PrivateKey
*
privateKey,
sb_PublicKey
*
publicKey)
l_genrand(job, lmseed1, lmseed2, lmseed3, NEWSEEDSIZ, newseeds);
lc_free_job(job);
for
(i
=
0
;i <
4
; i
+
+
)
{
if
(lm_case
=
=
7
|| lm_case
=
=
8
)
/
*
new with v8.
1
+
*
/
{
seed1 |
=
newseeds[i] << (i
*
8
);
seed2 |
=
newseeds[i
+
4
] << (i
*
8
);
seed3 |
=
newseeds[i
+
8
] << (i
*
8
);
seed4 |
=
newseeds[i
+
12
] << (i
*
8
);
}
if
(lm_case
=
=
1
|| lm_case
=
=
2
)
/
*
Upgrade pre7.
2
to
8.1
+
*
/
{
seed3 |
=
newseeds[i
+
8
] << (i
*
8
);
seed4 |
=
newseeds[i
+
12
] << (i
*
8
);
}
croseeds[
0
][
0
] |
=
(newseeds[i
+
16
] << (i
*
8
));
croseeds[
0
][
1
] |
=
(newseeds[i
+
20
] << (i
*
8
));
croseeds[
0
][
2
] |
=
(newseeds[i
+
24
] << (i
*
8
));
}
........
l_gen_pkey_headers(croseeds, pubkey_strength, &public, vname);
l_genrand(job, lmseed1, lmseed2, lmseed3, NEWSEEDSIZ, newseeds);
lc_free_job(job);
for
(i
=
0
;i <
4
; i
+
+
)
{
if
(lm_case
=
=
7
|| lm_case
=
=
8
)
/
*
new with v8.
1
+
*
/
{
seed1 |
=
newseeds[i] << (i
*
8
);
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