New attack on WPA using PMKID
hashcat.net
hashcat.net
The counterargument is keeping packet headers (i.e. remote IP addresses) and plaintext DNS queries private, but that's already the use case of a VPN. Even if it's just a "VPN" to your own home router. And then it protects you even against the operator of the access point (or someone impersonating it because, as usual, the passphrase is widely distributed).
Not to mention that most protocols in current use at minimum leak metadata. There would need to be a standard for an automatic authenticated VPN supported by hotspots and operating systems. Regular users shouldn't need to perform complex setup procedures.
And at that point, while I do like the seperation on concerns provided, why not just fix or replace WPA?
Meanwhile the guest users should have their own external VPN to protect them from you, which they should only have to set up once for all networks.
As long as you're legally responsible for the traffic coming out of your network, this is not a good thing to do. Unless people explicitly get the same protection an ISP gets, I'll keep advising them to not to share their connection openly.
That is obviously a jurisdiction-dependent legal question and anyone concerned about it should consult an attorney.
But if you're suggesting that, for example, the CDA or DMCA safe harbors only apply to Comcast and not book stores or auto shops or anyone else that provides public WiFi, I would be interested to see a citation for that.
But even with just DMCA to be a safe harbour you need to: have a service policy, show it to the users, have the possibility to prevent access for identified violations, and effectively keep some kind of connection record to be able to identify which users you need to terminate. I doubt anyone fulfills that at home. (I don't think shops and cafes do either)
It's even possible that not providing public access may increase certain risks. If you restrict access and someone guesses/cracks the password and does something terrible, that may make it harder to argue that it wasn't you.
I'm also not sure where you're reading the requirement to identify the users. There are many sites (e.g. Slashdot) where users can post anonymously (and via Tor or equivalent). Are you saying they don't qualify?
They have some info here:
https://openwireless.org/myths-legal.html
But notice that half the page is dedicated to extra-legal ISP shenanigans, which brings us back to routing your whole internet connection (guest net included) through a VPN. Which, again, you probably want even if you're the only one on your connection. It's not as if copyright trolls are renowned for their accuracy in targeting only people who are actually infringing something.
Not identify as in get their names. Just identify enough to know when they come back. Knowing which MAC to filter would probably be enough.
http://digital-law-online.info/lpdi1.0/treatise39.html
> First, the service provider is expected to adopt and reasonably implement a policy for the termination in appropriate circumstances of the accounts of subscribers of the provider’s service who are repeat online infringers of copyright.
You'd need to also identify which device was infringing by getting a connection time/destination.
I still don't see where it says you have to do that. Your link doesn't seem to say anything about it.
I question the value of MAC address blocking in general. Anyone can change their MAC address and popular systems are even using MAC address randomization by default now.
And in a physically local context like this, couldn't you just tell the person they're not allowed to use your wireless anymore, or remove them from the property?
The issue is who has to identify the user. If all they gave you was your own IP address with no accurate timestamp or ports, you wouldn't even be able to get the effectively-useless MAC address, even with the connection records most people don't keep. If they gave you the user's legal name (e.g. because the user signed up for the file sharing service with it) then you wouldn't need any connection records.
> couldn't you just tell the person they're not allowed to use your wireless anymore
The context we started with is wifi open to the public. You've never met your users and you may never see them (directional antenna from a distance), so the legal name is not useful either.
The situation where you know the users is much simpler.
You're thinking like a sysadmin. Think like an organization.
Compare the situation where you have a public space where everyone is welcome except Bob, because when Bob was there in the past he caused trouble and was asked never to come back.
You don't have to post guards checking ID because Bob knows he's not invited and the laws against trespassing deter him from showing up.
> The context we started with is wifi open to the public. You've never met your users and you may never see them (directional antenna from a distance), so the legal name is not useful either.
Seeing isn't required for telling. If you have the legal name, why can't you send a certified letter telling them they're not allowed to use your network anymore, then if they continue you call the police?
TLS for example is layered on top of Layer4(tcp/udp). It's not part of the tcp standard or something.
Wifi or wired,Segment level security should have it's own layer on top of or under Ethernet. Maybe 802.11ae falls under this?
Then I guess people would crack the VPN auth protocol.
in a corporate environment, use wpa2-enterprise, then password entropy doesnt matter quite as much.
The configuration space of WPA-EAP is huge and most combinations are horribly insecure, but as long as you stick with one of the "tunnel everything through TLS" EAPs (EAP-TTLS or PEAP) the result is safe against passive attackers even when you don't verify server certificates (obviously you should verify the certificates, because the active attack is trivial and does not have to interact with your network).
So you set up "eduroam" once on your phone, and then it works the same in a lecture theatre at Stanford, or in Nantes (France). So that's nice, and as dfox observes the AP isn't much involved, so the inevitable frailty of individual WiFi setups in less sophisticated institutions isn't a huge flaw in Eduroam or a grave risk for your home institution.
When working for an ISP it came up quite a few times that customers had extensive questions about security because they were genuinely worried about their ex-spouse spying on them. Even if they were all just "paranoid" in their specific cases (I wouldn't know), I think it's a fair concern. If all it takes is some googling and a bit of money to rent cloud GPU's, well, scorned lovers have done way more expensive and less effective things to cause damage or violate privacy.
If anyone knows of a WPA2 Enterprise setup guide that works well with minimal hassle (no CA/certificate installation hell, Linux+BSD+OSX+Windows as old as 8), I'd be eternally grateful.
Last time I ran the numbers an ISP default pattern password for ISPs around where I live (assuming perfect randomness within the ISP's pattern) was like $70 on Google Cloud GPUs (with half that on average).
And if your wifi has a default pattern SSID, then it probably has a default pattern password.
$70 is not atrociously high cost for a "last mile" security hop.
And if you have a botnet already then it's free.
These are people who would crack your wifi for the lulz (and have the stolen capacity to do it), get your house raided because they hack companies from behind your Internet connection...
... and then are stupid enough to when they hack and get access to a sensitive government database run a search for their own fucking name... and members of their family.
Anyway I'd suggest trying those first before digging into server setups or cloud offerings or the like.
Right now it appears they have an 30% success rate (if I'm reading it right).
Why say WPA/WPA instead of just WPA? Unless they meant WPA/WPA2.
Asking for a friend. :)
In other words, about 99% of the WiFi passwords out there are vulnerable to a brute force attack. But we knew that already, didn't we? Was it not already possible to brute force WPA2 before this new attack?
How much easier/faster is this new attack? It would have been nice if the article itself said something like: "This new attack increases brute force attack rate by a factor of 10x". Or whatever the right value is.
Do you ask people to download a QR code reader on their devices so they can access your WiFi router? That's a bit too crazy, don't you think?
edit: oh wow, as sibling commenter points out: the camera app detects QR codes now. neat!
No, no, no. The rest of your comment is spot-on, but I've done projects on passphrases and everyone gets this backwards. That is not what XKCD says. Let me quote the XKCD:
> four random common words
Random words, not a phrase like "may the force be with" and "my friend mark" (I'm sure you can find those two online). When cracking public hash dumps using phrases from public sources, I get hundreds of thousands of hits. If your phrase (partially) exists online, it's not a secret one.
Someone else mentioned diceware. Six random words from that dictionary (potentially generated using real dice) is pretty much unbreakable, though it has only a small security margin for when a protocol gets weakened but not broken.
Limit the dictionary to the top Xk most common English words, if leetspeak substitution is used you can also easily mask it.
We did this experiment and even randomly generating 6 word passwords using Wikipedia as dictionary resulted in passwords which are faster to crack than 16 character randomly generated passwords and that is because you can’t count the entropy as single characters sure if you brute force it char by char it will take longer but if you use words as your base unit then you only need to find 4-6 from a fairly limited pool of possibilities.
You can further limit it down by using grammar rules if your target is using passphrases those are even faster to crack and can be generated using markov chains or any basic grammar rule engine.
There is no fundamental difference. Either your pool of elements consists of 95 different symbols (ASCII printable characters and space) and you get passwords like 'F~iV3Bcv>\Q@' or your pool of elements consists of thousands of words and you get passwords like 'hubs exempted contend catchment others'.
As per Kerckhoff's principle, we should assume that the method of generating the password is known (which set of elements, i.e. your charset or dictionary, which RNG was used, and perhaps the length of the password).
The resulting strength in terms of bruteforceability of its hash is equal. Assuming one picks sane values, e.g. 6 elements when using a 7800 element set, or 12 elements when using a 95 element set, both are safe to use.
> randomly generating 6 word passwords using Wikipedia as dictionary
I don't understand what you mean by "using Wikipedia as dictionary". Wikipedia contains whole sentences. Did you download a dump of the English Wikipedia and split it on non-word characters and use that as dictionary? Did you remove duplicate words? Or did you take Wiktionary's words?
> if you use words as your base unit then you only need to find 4-6 from a fairly limited pool of possibilities.
I think you're wrong, but feel free to prove me wrong :). Here are some hashes:
f4cd51713a3ac5798b3a0b40fe61aa2f
f424f3432b7acbce2c2d3548490c9cc0
71fa597116bfc54ecc6c48162629d391
beb09edbc493bf826b61afa6905712df
42251a29996a78ef023bdb03a8c22ea9
3a80f15bd82c4f4c0a3338a1fa86adb9
They were generated using this script: function genphrase {
x=$1;
while [ $x -gt 0 ]; do
echo -n $(shuf /usr/share/dict/words | grep -v é | grep -v \' | tail -1)\ ;
let x=$x-1;
done;
}
for i in {1..6}; do
phrase="$(genphrase $i)";
sum=$(echo -n "$phrase" | md5sum | awk '{print $1}');
echo "$sum $phrase";
done;
Basically I take random words from /usr/share/dict/words so long as it does not contain an apostrophe or an accented e. The md5 hash of the result is generated, a very fast and well-supported algorithm (your favorite cracking program should support it without any trouble). As an example, from another run of the script, here is one of its output lines: 75e6881687661e09e404b517e7ee2ce3 goodliest villeins gymnast
You can use this sample output to verify that the hash is generated correctly and that your tool works. I expect the first two hashes should not be an issue, the third might be harder, the fourth would be impressive, and I expect that the fifth and sixth will remain uncracked.The dictionary version is 2018.04.16-1 (wamerican package in Debian), I uploaded it here: https://lucb1e.com/tmp/words (watch out clicking the link, it's a large file and your browser may just display it instead of downloading it)
> You can further limit it down by using grammar rules if your target is using passphrases those are even faster to crack and can be generated using markov chains or any basic grammar rule engine.
At least in my research, I've found that markov chains and n-grams are of much worse quality and slower than just using raw phrases from public sources. The downloading and processing of those sources takes more time, but that's a one-time action and not really part of the cracking process. And as I said in the comment you're replying to, if your 'phrase' is not random words but actually a logical 'phrase', then it's not suitable as passphrase. I think you're confusing random phrases with logical sentences.
Happened once then they stopped and just handed out the guest password.
ITYM 256 bits. :-)
More detail: https://stackoverflow.com/questions/18006390/why-is-the-wpa2...