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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 ...


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 ...


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.


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 ...

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