180 likes | 192 Views
An Inductive Chosen Plaintext Attack against WEP/WEP2. William A. Arbaugh University of Maryland, College Park waa@cs.umd.edu. Talk Outline. Introduction WEP/WEP2 IP Walker/Berkeley Attacks Attack Overview Attack Details Conclusions. 802.11 Hdr. ICV. Data. Encapsulate. Decapsulate.
E N D
An Inductive Chosen Plaintext Attack against WEP/WEP2 William A. Arbaugh University of Maryland, College Park waa@cs.umd.edu William Arbaugh, University of Maryland
Talk Outline • Introduction • WEP/WEP2 • IP • Walker/Berkeley Attacks • Attack Overview • Attack Details • Conclusions William Arbaugh, University of Maryland
802.11 Hdr ICV Data Encapsulate Decapsulate 802.11 Hdr IV Data WEP/WEP2 • Encryption Algorithm = RC4 • Per-packet encryption key = IV concatenated to a pre-shared key • WEP: 24 bit IV • WEP2: 128 bit IV • WEP allows IV to be reused with any frame • Data integrity provided by CRC-32 of the plaintext data (the “ICV”) • Data and ICV are encrypted under the per-packet encryption key William Arbaugh, University of Maryland
ICV 24 luxurious bits Encrypted under Key +IV using a Vernam Cipher 802.11 Hdr IV Data How to Read WEP Encrypted Traffic (1) • 50% chance of a collision exists already after only 4823 packets!!! • Pattern recognition can disentangle the XOR’d recovered plaintext. • Recovered ICV can tell you when you’ve disentangled plaintext correctly. • After only a few hours of observation, you can recover all 224 key streams. William Arbaugh, University of Maryland
How to Read WEP Encrypted Traffic (2) • Ways to accelerate the process: • Send spam into the network: no pattern recognition required! • Get the victim to send e-mail to you • The AP creates the plaintext for you! • Decrypt packets from one Station to another via an Access Point • If you know the plaintext on one leg of the journey, you can recover the key stream immediately on the other • Etc., etc., etc. William Arbaugh, University of Maryland
Observations • Walker/Berkeley attacks require either: • Depth and post analysis • Cooperating agent for known plain text • Can we do better? William Arbaugh, University of Maryland
Inductive Chosen Plain Text • Base Case: Recover an initial pseudo random stream of length n from known plain text. • Inductive step: Extend size of known pseudo random to n+1 by leveraging the redundant information in the CRC. William Arbaugh, University of Maryland
Base Case • Find initial pseudo random stream of size n. • Identify DHCP Discover messages from externals, e.g. size, and broadcast MAC address. • Known source (0.0.0.0), destination (255.255.255.255), header info • Allows the recovery of 24 bytes of pseudo random stream: Let n = 24 William Arbaugh, University of Maryland
Inductive Step • Create a datagram of size n-3 representing an ARP request, UDP open, ICMP etc. • Compute ICV and append only the first three bytes. • XOR with n bytes of pseudo random stream. • Append last byte as the n+1 byte William Arbaugh, University of Maryland
n-3 3 ICV-1 ICV 802.11 Hdr IV Data Data byte Iterate over the 255 possibilities Encrypted Data Pseudo Random Steam byte n+1 Inductive Step William Arbaugh, University of Maryland
Inductive Step 5. Now send datagram and wait for a response. 6. If no response, try another of the 254 remaining possibilities. 7. If there is a response, then we know: The n+1 byte was the last byte of the ICV, thus we have matching plaintext and ciphertext which gives us the n+1 byte of the pseudorandom stream. William Arbaugh, University of Maryland
ICV-1 ICV 802.11 Hdr IV Data Data n+1 ciphertext byte byte byte n+1 pseudo byte Encrypted Data Pseudo Random Steam After Response n-3 3 n+1 plaintext byte byte byte n+1 William Arbaugh, University of Maryland
Attack Cost • Assume moderately aggressive attacker: • ~100 attacker transmissions per second • NOTE: ICV failures will not be passed to OS and thus the attack is difficult to observe (failed ICV counter not withstanding) • 1.6 hours to recover 2300 byte MTU regardless of IV and key size in worst case • ~40 minutes in average case William Arbaugh, University of Maryland
WEP Costs • 46 hours to build full dictionary of <IV, pseudorandom> with one attacking host (~35GB) • But, the attack is embarrassingly parallel. • Four attacking hosts: 11.5 hours • Eight attacking hosts: 5.75 hours William Arbaugh, University of Maryland
WEP2 Costs • Prohibitive to build entire dictionary in terms of space and time, but we don’t need to do so. • Because, we can still find enough <IV,pseudorandom> pairs to find and attack a vulnerable host on the LAN and recover key actively, e.g. blind scans and blind attacks. William Arbaugh, University of Maryland
This Attack Works • Because of the redundant information provided by the CRC, and • Because of the lack of a keyed MIC William Arbaugh, University of Maryland
Stopping/Mitigating the Attack • Add a keyed MIC (stops attack) • Adding a replay window (mitigates attack) • Modifying the CRC such that it can’t be: • Easily determined by an attacker • Not linear (bit flipping attack) (mitigates attack) William Arbaugh, University of Maryland
Conclusions • Fundamental problem is that both WEP and WEP2 vulnerable to packet forgery. • It’s easy to dismiss this attack (and the Walker/Berkeley attacks) as “academic”. However, it’s only a matter of time before the attacks are implemented/scripted and released …What then? William Arbaugh, University of Maryland