Hot answers tagged

4

A key is derived from the password using a Password Based Key Derivation Function, in this case PBKDF2: Key = PBKDF2(HMAC−SHA1, passphrase, ssid, 4096, 256) PBKDF2 in turn is described by PKCS#5. These RSA cryptographic standards in turn are made available through RFC's nowadays, in this case RFC 2898: PKCS #5: Password-Based Cryptography Specification ...


2

Part of the problem you're having is that there are multiple distinct vulnerabilities in WEP, and you're getting confused by the sheer number. For example: I still don't have an understanding of how one might exploit [repeating IVs] to retrieve the key Answer: those are two separate vulnerabilities. The shortness of the IV space is an obvious ...


2

Yes. Both WPA-TKIP and WPA2-AES use the same 4-way handshake, and the same way of deriving the PTK from the PSK. AES does not affect this. To launch the brute-force attack on the PSK you only need to capture the 4-way handshake (including the nonces). This is also true for WPA2-AES-PSK.


1

The following answers are taken directly from the 802.11 standard (2012 version). The PTK is generated as follows: $$\mathrm{PTK} \gets \operatorname{\mathrm{PRF-X}(PMK, ``pairwise \ key \ expansion", \\ min(AA, SPA) || max(AA,SPA)|| \\ min(ANonce, SNonce) || max(ANonce, SNonce)) } $$ where $\mathrm{AA}$ and $\mathrm{SPA}$ are the MAC addresses of the ...


1

In pairwise key hierarchy there are two root keys: pre-shared key (PSK) and Authentication, Authorization, and Accounting Key (AAAK). From these two keys pairwise master key (PMK) is derived. In the case of PSK, PMK is equal to PSK. In other case, PMK is obtained by taking the first 256 bits of AAAK. Now, to the question about generating pairwise transient ...



Only top voted, non community-wiki answers of a minimum length are eligible