算法 – 为什么SRP不是明文等价的？

关于SRP协议：
http://en.wikipedia.org/wiki/Secure_remote_password_protocol

我可以看到会话密钥(K)的生成是完全安全的,但在最后一步中,用户发送K(M)证明.如果网络不安全且midlle中的攻击者捕获了M,他就可以在没有K的情况下进行身份验证.

一点背景

众所周知的价值观(事先确定)：

n    A large prime number. All computations are performed modulo n.
  g    A primitive root modulo n (often called a generator).

用户密码建立为：

x = H(s, P)
v = g^x 

  H()  One-way hash function
  s    A random string used as the user's salt
  P    The user's password
  x    A private key derived from the password and salt
  v    The host's password verifier

身份验证：

+---+------------------------+--------------+----------------------+
|   | Alice                  | Public Wire  | Bob                  |
+---+------------------------+--------------+----------------------+
| 1 |                        |        C --> | (lookup s, v)        |
| 2 | x = H(s, P)            | <-- s        |                      |
| 3 | A = g^a                |        A --> |                      |
| 4 |                        | <-- B, u     | B = v + g^b          |
| 5 | S = (B - g^x)^(a + ux) |              | S = (A · v^u)^b      |
| 6 | K = H(S)               |              | K = H(S)             |
| 7 | M[1] = H(A, B, K)      |     M[1] --> | (verify M[1])        |
| 8 | (verify M[2])          | <-- M[2]     | M[2] = H(A, M[1], K) |
+---+------------------------+--------------+----------------------+

    u    Random scrambling parameter, publicly revealed
  a,b    Ephemeral private keys, generated randomly and not publicly revealed
  A,B    Corresponding public keys
  m,n    The two quantities (strings) m and n concatenated
    S    Calculated exponential value 
    K    Session key

你的问题的答案：

如您所见,双方根据每个人可用的值分别计算K(=会话密钥).
如果在步骤2中输入的Alice的密码P与她最初用于生成v的密码匹配,那么S的两个值将匹配.

然而,实际的会话密钥K从不通过网络发送,只是证明双方已成功计算出相同的会话密钥.因此,中间人可以重新发送证据,但由于他没有实际的会话密钥,他将无法对截获的数据做任何事情.