Hey! I'm David, cofounder of zkSecurity and the author of the Real-World Cryptography book. I was previously a crypto architect at O(1) Labs (working on the Mina cryptocurrency), before that I was the security lead for Diem (formerly Libra) at Novi (Facebook), and a security consultant for the Cryptography Services of NCC Group. This is my blog about cryptography and security and other related topics that I find interesting.
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In this video I'm explaining what is that Galois Counter Mode that provides Authenticated Encryption with Associated Data (AEAD). You must have heard it combined with AES, and maybe used in TLS, ... This is just a small explanation, you can get more on the NIST specs.
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2 commentsMy coworker Tomek found a vulnerability in Rack, a Ruby thingy that is used pretty much in every web framework (Rails, Ramaze, Sinatra...) to translate queries (POST and GET) into Ruby objects.
Tomek: David you don't understand anything, go read this: https://www.omniref.com/ruby/gems/railties/4.2.0/symbols/Rails::Application#annotation=4084035&line=161
It all started when he read about how to_sym() used to work in Ruby versions prior to MRI 2.2. This function converts strings into symbols. A symbol is this thing: :symbol, that usually is tied to a value: :symbol => 'something' and that you must have seen written in a really nice way thanks to Ruby's lovely syntax sugar: symbol: 'something'. Developers use the to_sym() method a lot to transform the strings from GET/POST to symbols. Problem: the symbols created by to_sym() were stored in the memory and never freed. No garbage collection.
That is a problem if you let user input infinitely use that to_sym() method.
For example if you naively transformed the GET variables of a request to symbols through that function, then a malicious user could have queried that page with many ?stuff=something so that your code would have infinitely stocked symbols in memory until no memory was left.
Tomek was looking for the fastest method to fill up a server's memory. To do this, he tested various forms of parameters to see what would be the most effective. This included many small symbol names, large symbol names, and nested parameters. The most efficient way ended up being a lot of concurrent requests with huge symbol names, the server would go down in less than 2 minutes.
And it was through the nested parameter testing that he found CVE-2015-3225
https://groups.google.com/forum/#!topic/rubyonrails-security/gcUbICUmKMc
Potential Denial of Service Vulnerability in Rack
There is a potential denial of service vulnerability in Rack. This
vulnerability has been assigned the CVE identifier CVE-2015-3225.
Versions Affected: All.
Not affected: None.
Fixed Versions: 1.6.2, 1.5.4
Impact
------
Carefully crafted requests can cause a `SystemStackError` and potentially
cause a denial of service attack.
All users running an affected release should either upgrade or use one of the workarounds immediately. In Ruby on Rails, one of the nice things you can do is use associative arrays (called hashes in Ruby) in your forms.
For example, in your HTML
<input type="text" name="form[name]" value="david">would translates to Ruby as
{"form"=>{"name"=>"david"}}And that is thanks to Rack! But there is a small problem here, Rack parses these nested hashes by recursively calling itself:
def normalize_params(params, name, v = nil)
...
if params_hash_type?(params[k].last) && !params[k].last.key?(child_key)
normalize_params(params[k].last, child_key, v)
else
params[k] << normalize_params(params.class.new, child_key, v)Here, Ruby seems to push a bunch of info to the stack before calling the new function. A bit like an assembly prolog.
The thing is, after a while the stack gets full and Ruby throws an exception. And then? Rack catch the exception? Tries again? We don't really know, but the program hangs there.
So Tomek found out that when you would send a POST request with a deep enough hash you would cause the program to hang. This is bad because of DoS attacks.
Also, the problem cannot happen in json because json has a nest limit, and doesn't happen in GET requests either for the same reason.
CVE and Patches are talked about here: http://www.openwall.com/lists/oss-security/2015/06/16/14
Another thread there: https://hackerone.com/reports/42797
He's still trying to find out what is happening exactly, and he just opened a blog. So who knows, he might write something about it soon.
comment on this storyThis evening I was at Braintree's office in Chicago for the new edition of the Owasp meeting. The offices were amazing and huge!
They had beer pong tables, ping pong tables, a mini-arcade and a bar! (all that right next to the workspace)
Plenty of nice conference rooms
Some had whiteboard tables!
And in the largest building of Chicago!
Brandon Myers, a security researcher at Trustwave, was the first one to talk. He found out that when you executed gem fetch or gem install, the ruby package manager, it would allow a Man In The Middle to do a DNS poisoning attack to redirect you to his servers. Even though everything happens over TLS! This is because gem doesn't check for the domain in the certificate of rubygems.org: it just checks that the server has a valid certificate and that's all. You would then download the gem on his server and... game over.
He said that the same thing was happening when developers were pushing their gems to rubygems.org ...
That's a shame, and way worse than the downgrade https attack of go.
One way of mitigating the first MITM would be to just use curl or wget directly with https://www.rubygems.org and do whatever the gem fetch does to get the gems. Because curl or wget should have a correct implementation of TLS that dodge fake DNS responses (that's why DNSSEC is useless if you query a https webpage with a correctly signed certificate).
To mitigate the second attack Brandon talked about signed gems, and that it was far from being efficient since none of the top gems are signed.
Two other problem were that if a fix comes around, gem update --system is vulnerable to the attack (since gem is itself a gem) and not using the fully secure gem signing allows some dependencies of not being signed (and thus a MITM would be able to modify those).
Patrick Thomas and Alec Gleason followed by explaining how much they pwned their client with heartbleed, passwords in clear and github hooks. They then explained how you could get more information out of a pwned machine if there was git, vagrant, chef, docker and other non-crypto stuff installed on the machine.
All of this was facilitated by Devops,
a software development method that emphasizes communication, collaboration (information sharing and web service usage), integration, automation, and measurement of cooperation between software developers and other IT professionals.
https://en.wikipedia.org/wiki/DevOps
They said it was still a good thing but if done correctly.
After that I went to Chisec meet some people of the security community!
comment on this story
Basically, it's a way to get a quick x509 certificate for your server without knowing much about what is a x509 certificate:
You have a website. You want people to be able to log in on it from starbucks without the guy sitting at a near table reading the passwords in clear from the packets you're sending everyone around you. So you google a few random keywords like "secure website https" and you end up with a bunch of links and services (which may not be free) and you suddenly have to understand what are certificates, PKI, x509, public-key cryptography, RSA, FQDN, haaa.... Well, worry no more, Let's Encrypt was thought so that you wouldn't have to go through all that trouble and destroy your server on the way.
Oh and yeah. It's for free. But is hasn't been released yet.
You can learn more reading their technical overview, or some of their videos... or read my tl;dr:
sudo apt-get install lets-encryptlets-encrypt example.comAnd voila.
Behing the curtains this is what happens:
lets-encrypt will generate a pair of RSA private key/public key and contact the CA with your public key.
The CA will register your public key
The program will then ask the CA to verify your domain.
The CA will answer with a set of challenges. These are some tasks you can complete to prove to the CA you own that domain. One of the common one is to upload a certain file at a certain address of that domain.
The program you installed will then do that for you and poll the CA for a confirmation
The CA will tell you "OK man, all is good".
The program will then generate another long term pair of private key/public key, generate a CSR (Certificate Signing Request) with the new public key and send that CSR to the CA.
The CA will extract the information, create a beautiful x509 certificate, sign it and send it back to you.
lets-encrypt will install the certificate on your server and set certain options (or not) to force httpsBy the way, the certificate you will get will be a DV certificate, meaning that they only verified that you owned the domain, nothing more. If you want an EV certificate this is what you will have to go through (according to wikipedia):
So! The lets-encrypt program you run on your server is open sourced here: https://github.com/letsencrypt/lets-encrypt-preview. It is called lets-encrypt-preview I guess because it isn't done yet. It's written in Python and has to be run in sudo so that it can do most of the work for you. Note that it will install the certificates on your server only if you are using Apache or Nginx. Also, the address of the CA is hardcoded in the program so be sure to use the official lets-encrypt.
The program installed on the CA is also open sourced! So that anyone can publicly review and audit the code. It's called Boulder and it's here: https://github.com/letsencrypt/boulder and written in Go.
Lets-encrypt and Boulder both use the protocol ACME for Automated Certificate Management Environment specified here as a draft: https://letsencrypt.github.io/acme-spec/
ACME is written like a RFC. It actually wants to become an RFC eventually! So if you've read RFCs before you should feel like home.
The whole protocol is happening over TLS. As a result the exchanges are encrypted, you know that you are talking to the CA you want to talk to (eventhough you might not use DNSSEC) and replay attacks should be avoided.
The whole thing is actually a RESTful API. The client, you, can do GET or POST queries to certains URI on the CA webserver.
The first thing you want to do is register. Registration is actually done by generating a new pair of RSA keys and sending them the public key (along with your info).
ACME specifies that you should use JWS for the transport of data. Json Web Signature. It's basically Json with authentication (so that you can sign your messages). It actually uses a variant of JWS called Jose that doesn't use a normal base64 encoding but that's all you should know for now. If you really want to know more there is an RFC for it.
So here what a request should look like with JWS (your information are sent unencrypted in the payload field (but don't worry, everything is encrypted anyway because the exchange happens over TLS)):
POST /acme/new-registration HTTP/1.1
Host: example.com
{
"payload":"<payload contents>",
"signatures":[
{"protected":"<integrity-protected header 1 contents>",
"header":<non-integrity-protected header 1 contents>,
"signature":"<signature 1 contents>"}
}Boulder will check the signature with the public key you passed, verify the information you gave and eventually add you to its database.
Once you are registered, you can perform several actions:
The server will authenticate you because you will now send your public key along AND you will sign your requests. This all runs on top of TLS by the way. I know I already said that.
Boulder is separated in multiple components, this makes the code clearer and ensure that every piece of code does what it is supposed to do and nothing more.
One of the components is called the Web-Front-End (WFE) and is the only one accepting queries from the Client. It parses them, verifies them and passes them to the Registration Authority (RA) that combines the other authorities together to produce a response. The response is then passed back to the WFE and to the client over the ACME protocol. Are you following?
TWe'll see what other authorities the RA has access to in the next queries the client can do. But just to talk about the previous query, the new registration query, the RA talks to the Storage Authority that deals with the database (Which is currently SQLlite) to register your new account.
All the components can be run from a single machine (for the exception of the CA that runs on another library), but they can also be run seperately from different machines that will communicate on the same network via AMQP.
Now that you are registered, you have to validate a domain before you can request a certificate.
You make a request to a certain URI.
Well to be exact you make a POST request to /new-authz, and the response will be 201 if it works. It will also give you some information about where you can go next to do stuff (/authz)
Here's the current list of API calls you can do and the relevant answers.
The server will pass the info to the Policy Authority (PA) and ask it if it is willing to accept such a domain. If so, it will then answer with a list of challenges you can complete to prove you own the domain, along with combinations of accepted challenges to complete. For now they only have two challenges and you can complete either one:
SimpleHTTPS
If you choose SimpleHTTPS the lets-encrypt client will generate a random value and upload something at the address formed by a random value the CA sent you and the random value you generated.
If you choose DVSNI, the client will create a TLS certificate containing some of the info of the challenge and with the public key associated to his account.
The client then needs to query the CA again and the CA's Validation Authority (VA) will either check that the file has been uploaded or will perform a handshake with the client's server and verify that specific fields of the certificates have been correctly filled. If everything works out the VA will tell the RA that will tell the WFE that will tell you...
After that you are all good, you can now make a Certificate Signing Request :)
The agent will now generate a new pair of private key/public key. And create a CSR with it. So that your long term key used in your certificate is not the same as your let's encrypt account.
a CSR example, containing the identifier and the public key
After reception of it, the Web-Front-End of Boulder will pass it to the Registration Authority (RA) which will pass it to the Certificate Authority (CA) that will do all the work and will eventually sign it and send it back to the chain.
1: Client ---new-cert--> WFE
2: WFE ---NewCertificate--> RA
3: RA ---IssueCertificate--> CA
4: CA --> CFSSL
5: CA <-- CFSSL
6: RA <------return--------- CA
7: WFE <------return------- RA
8: Client <------------- WFEOh and also. the CA is talking to a CFSSL server which is CloudFlare's PKI Toolkit, a nice go library that you can use for many things and that is used here to act as the CA. CFSSL has recently pushed code to be compatible with the use of HSM which is a hardware device that you HAVE to use to sign keys when you are a CA.
After reception the lets-encrypt client will install the fresh certificate along with the chain to the root on your server and voila!
You can now revoke a certificate in the same way, but interestingly you won't need to sign the request with your account key, but with the private key associated to your certificate's public key. So that even if you lose your agent's key you can still revoke the certificate.
There are a bunch of stuff that you will be able to do with the lets-encrypt client but that haven't been implemented yet:
This post is a simplification of the protocol. If you want to know more and don't want to dig in the ACME specs right now you can also take a look at Boulder's flow diagrams.
If you've followed the news you should have seen that Let's Encrypt just generated its root certificate along with several other certificates: https://letsencrypt.org/2015/06/04/isrg-ca-certs.html
This is because when you are a CA you are suppose to keep the root certificate offline. So you sign a (few) certificate(s) with that root, you lock that root and you use the signed certificate(s) to sign other certificates. This is all very serious because if something goes wrong with the root certificate, you can't revoke anything and well... the internet goes wrong as a result (until vendors start removing these from their list of trusted roots).
So the keys for the certificate have to be generated during a "ceremony" where everything is filmed and everyone must authenticate oneself at least with two different documents, etc... Check Wikipedia's page on Key Ceremony it's "interesting".
Also, I received a note from Seth David Schoen and I thought that was an interesting anecdote to share :)
21 commentsthe name "Boulder" is a joke from the American children's cartoon Looney Tunes:
https://en.wikipedia.org/wiki/Wile_E._Coyote_and_The_Road_Runner
This is because the protocol that Boulder implements is called ACME, which was also the name of the company that made the products unsuccessfully used by the coyote to attempt to catch the roadrunner. https://en.wikipedia.org/wiki/Acme_Corporation
Two of those products were anvils (the original name of the CA software) and boulders.
The Hacking Week ended 2 weeks ago and EISTI got the victory.
I'm the proud creator of the crypto challenge number 4, still available here, which was solved 12 times.
I also wrote a Proof of Solvableness, reading it should teach you about a simple and elegant crypto attack on RSA: the Same Modulus Attack.
(Note that I wrote that back in January)
We are presentend with 4 files:
the irc.log reads like this:
Session Start: Thu Feb 05 20:49:04 2015
Session Ident: #mastercsi
[20:49] * Now talking in #mastercsi
[20:49] * Topic is 'http://www.math.u-bordeaux1.fr/CSI/ |||| http://www.youtube.com/watch?v=zuDtACzKGRs "das boot, ouh, ja" ||| http://www.koreus.com/video/chat-saut-balcon.html ||| http://blog.cryptographyengineering.com/ ||| http://www.youtube.com/watch?v=K1LZ60eMpiw ||| petit chat http://www.youtube.com/watch?v=eu2kVcWKvRo ||| sun : t'as le droit de boire quand même va'
[20:49] * Set by Jiss!~Jiss@2001:41d0:52:100::65d on Sat Nov 22 00:06:50
[20:49] <asdf> et donc j'ai chopé une vieille clé rsa qu'alice utilisait
[20:49] <qwer> alice la alice? tu te fous de moi ?
[20:49] <asdf> haha non
[20:49] <asdf> mais le truc est corrompu, ça a l'air de marcher pour chiffrer mais la moitié de la clé a disparu
[20:49] <qwer> attend j'ai sa clé publique qui traine quelque part, et même un fichier chiffré avec. me suis toujours demandé ce que c'était...
[20:50] <asdf> je t'ai envoyé le truc, mais ça m'étonnerait que ça soit la même clé non ?
[21:22] * Disconnected
Session Close: Thu Feb 05 21:22:11 2015so alice.pub seems to be alice public rsa key. secret seems to be the file encrypted under this key and mykey.pem should be the partial key which was found.
Private-Key: (1024 bit)
modulus:
00:c6:c8:35:29:a2:38:8f:14:63:65:c5:f5:fd:4b:
0d:88:89:61:b9:5d:e1:0f:fa:88:53:a3:c2:cb:ed:
75:0e:99:59:bd:0f:f8:72:c2:23:2f:6b:ad:32:62:
4f:35:6a:82:d0:62:75:5e:1e:4f:ed:ae:54:e8:ca:
24:71:fc:8d:13:ac:70:0e:e2:57:20:d4:d9:08:9f:
d6:fb:d4:2f:12:e6:a4:1e:1c:1d:e8:1f:57:8c:32:
13:2a:d0:85:94:e8:51:84:1d:02:39:cd:41:0d:ef:
11:d1:c1:5e:e7:5b:92:f8:6a:04:f7:c6:c7:f3:6b:
90:46:b8:fb:2f:e2:95:65:b1
publicExponent: 3 (0x3)
privateExponent:
00:84:85:78:c6:6c:25:b4:b8:42:43:d9:4e:a8:dc:
b3:b0:5b:96:7b:93:eb:5f:fc:5a:e2:6d:2c:87:f3:
a3:5f:10:e6:7e:0a:a5:a1:d6:c2:1f:9d:1e:21:96:
df:78:f1:ac:8a:ec:4e:3e:be:df:f3:c9:8d:f0:86:
c2:f6:a8:5e:0b:ef:c0:ca:19:c5:e2:49:55:49:fe:
e5:2e:51:3e:7b:e9:f2:22:07:d2:4b:84:7f:bb:0c:
b5:ba:b7:95:c6:90:05:3e:65:2d:11:53:9a:2d:96:
0f:ea:de:cb:9b:17:54:87:00:0f:78:12:ce:ac:f5:
db:83:30:16:06:cc:35:7d:a3
prime1: 245 (0xf5)
prime2: 207 (0xcf)
exponent1: 163 (0xa3)
exponent2: 138 (0x8a)
coefficient: 189 (0xbd)It looks like prime1, prime2 and some other stuff are pretty short. I guess this is what he meant by "half the key" is messed up.
By the way this is what a RSA PrivateKey should look like:
> RSAPrivateKey ::= SEQUENCE {
version Version,
modulus INTEGER, -- n
publicExponent INTEGER, -- e
privateExponent INTEGER, -- d
prime1 INTEGER, -- p
prime2 INTEGER, -- q
exponent1 INTEGER, -- d mod (p-1)
exponent2 INTEGER, -- d mod (q-1)
coefficient INTEGER, -- (inverse of q) mod p
otherPrimeInfos OtherPrimeInfos OPTIONAL
}So this is what exponent1, exponent2 and coefficient are. Just additional information so that computations are faster thanks to CRT.
Let's ignore that for the moment.
$ openssl rsa -pubin -in alice.pub -modulus -noout
Modulus=C6C83529A2388F146365C5F5FD4B0D888961B95DE10FFA8853A3C2CBED750E9959BD0FF872C2232F6BAD32624F356A82D062755E1E4FEDAE54E8CA2471FC8D13AC700EE25720D4D9089FD6FBD42F12E6A41E1C1DE81F578C32132AD08594E851841D0239CD410DEF11D1C15EE75B92F86A04F7C6C7F36B9046B8FB2FE29565B1
$ openssl rsa -in mykey.pem -modulus -noout
Modulus=C6C83529A2388F146365C5F5FD4B0D888961B95DE10FFA8853A3C2CBED750E9959BD0FF872C2232F6BAD32624F356A82D062755E1E4FEDAE54E8CA2471FC8D13AC700EE25720D4D9089FD6FBD42F12E6A41E1C1DE81F578C32132AD08594E851841D0239CD410DEF11D1C15EE75B92F86A04F7C6C7F36B9046B8FB2FE29565B1the partial key and alice public key seems to share the same modulus. this is vulnerable. If our public/private exponents are not messed up, this means we can factor the modulus and thus inverse Alice's public key.
Let's retrieve all the info we have and put them in a file:
openssl rsa -pubin -in alice.pub -modulus -noout | sed 's/Modulus=//' | xclip -selection cHere's the modulus. We know that our public key is 3, let's get the private key in the clipboard as well
openssl asn1parse -in mykey.pem | grep 129 | tail -n1 | awk '{ print $7}' | sed 's/://' | xclip -selection chere I parse mykey.pem with openssl. I select the lines I want with grep. It returns two results, the modulus and the private key. I select only the second line with tail. I select only the 7th column with awk. I remove the
:with sed. And now I have a beautiful integer in my clipboard.
Okay so let's do a bit of Sage now:
# let's write the info we have
modulus = int(0xC6C83529A2388F146365C5F5FD4B0D888961B95DE10FFA8853A3C2CBED750E9959BD0FF872C2232F6BAD32624F356A82D062755E1E4FEDAE54E8CA2471FC8D13AC700EE25720D4D9089FD6FBD42F12E6A41E1C1DE81F578C32132AD08594E851841D0239CD410DEF11D1C15EE75B92F86A04F7C6C7F36B9046B8FB2FE29565B1)
public = 3
private = int(0x848578C66C25B4B84243D94EA8DCB3B05B967B93EB5FFC5AE26D2C87F3A35F10E67E0AA5A1D6C21F9D1E2196DF78F1AC8AEC4E3EBEDFF3C98DF086C2F6A85E0BEFC0CA19C5E2495549FEE52E513E7BE9F22207D24B847FBB0CB5BAB795C690053E652D11539A2D960FEADECB9B175487000F7812CEACF5DB83301606CC357DA3)
# now let's factor the modulus
k = (private * public - 1)//2
carre = 1
g = 2
while carre == 1 or carre == modulus - 1:
g += 1
carre = power_mod(g, k, modulus)
p = gcd(carre - 1, modulus)
print(p)This does not work. This should work.
Let's re-do the maths:
We know that our private and public keys cancel out. This is RSA:
private * public = 1 mod phi(N)
so we have private * public - 1 = 0 mod phi(N)
So for any g in our ring, we should have g^(private * public - 1) = 1 mod N
This is how RSA works.
Let's write it like that: private * public - 1 = k with k a multiple of phi(n). And we know that phi(n) = (p-1)(q-1) is even. So it could be written like this: k = 2^t * r with r an odd number.
Now if we take a random g mod N and we do g^(k/2) it should be the square root of a 1.
The Chinese Remainder Theorem tells us that there are 4 square roots mod N:
1 mod p-1 mod p1 mod q-1 mod qand two of them should be 1 mod N and -1 mod N. The 2 others should be different from 1 and -1 mod N. That's what I was trying to find in my code.
Once we have found this x mod N which is a square root of 1 mod N, we know that it is either x = 1 mod p or x = -1 mod p.
If we are in the first case, we shoudl have x - 1 = 0 mod p which translates into x - 1 is a multiple of p.
Doing gcd(x - 1, N) should give us p the first prime. If you don't understand it maybe check Dan Boneh's explanation (proof end of page 3) which should be clearer than mine.
With p it's easy to get q the other prime.
But it doesn't work...
Ah! I forgot that g^(k/2) could equal 1 all the time if k/2 were to be a multiple of phi(n). So let's code a loop that divides k by 2 and tries any g^k until it is giving us something else than 1. Then we know how many times we have to divide k by 2 so it's not a multiple of phi(n) anymore.
It turns out we just have to do it 3 times. And then it magically works. A bit more of Sage gives us the primes:
# p and q our primes
p = gcd(carre - 1, modulus)
q = modulus // p
# now that we have factored N let's find alice decryption key
public = 65537
phi = (p - 1) * (q - 1)
private = inverse_mod(public, phi)Now that we have Alice's private key there are two ways to decrypt our secret:
Let's do the first one. We'll modify our mykey.pem for this:
openssl rsa -in mykey.pem -outform DER -out newkey.bin
xxd -p newkey.bin > newkey.hexwe get this:
3082012202010002818100c6c83529a2388f146365c5f5fd4b0d888961b9
5de10ffa8853a3c2cbed750e9959bd0ff872c2232f6bad32624f356a82d0
62755e1e4fedae54e8ca2471fc8d13ac700ee25720d4d9089fd6fbd42f12
e6a41e1c1de81f578c32132ad08594e851841d0239cd410def11d1c15ee7
5b92f86a04f7c6c7f36b9046b8fb2fe29565b102010302818100848578c6
6c25b4b84243d94ea8dcb3b05b967b93eb5ffc5ae26d2c87f3a35f10e67e
0aa5a1d6c21f9d1e2196df78f1ac8aec4e3ebedff3c98df086c2f6a85e0b
efc0ca19c5e2495549fee52e513e7be9f22207d24b847fbb0cb5bab795c6
90053e652d11539a2d960feadecb9b175487000f7812ceacf5db83301606
cc357da3020200f5020200cf020200a30202008a020200bdThis is a DER encoding. One particular encoding from the ASN.1 family. It is a TLV kind of encoding (Type Lenght Value).
For example in:
02 8181 00c6c83529a2388f146365c5f5fd4b0d888961b95de10ffa8853
a3c2cbed750e9959bd0ff872c2232f6bad32624f356a82d062755e1e4fed
ae54e8ca2471fc8d13ac700ee25720d4d9089fd6fbd42f12e6a41e1c1de8
1f578c32132ad08594e851841d0239cd410def11d1c15ee75b92f86a04f7
c6c7f36b9046b8fb2fe29565b1first is coded the type 02 (integer), then the length (81 repeated twice because the value block is bigger than 127bits, so we set the first byte to 81 (10000001, the first bit means it is a long way of defining the length, the 7 following bits are the number of byte it will take to define the length, in our case only one and it will be the next one) and the second byte to the actual size), then there is our modulo in hexadecimal. Note that the most significant bit of our value has to be zero if it is a positive integer, that's why we use 41 instead of 40 and lead the payload with 00.
So let's take the time and break this apart:
3082 // some header
0122 // the length of everything that follows (in byte)
0201 // integer of size 1
00
028181 // integer of size 0x81 (our modulus)
00c6c83529a2388f146365c5f5fd4b0d888961b95de10ffa8853a3c2cbed750e9959bd0ff872c2232f6bad32624f356a82d062755e1e4fedae54e8ca2471fc8d13ac700ee25720d4d9089fd6fbd42f12e6a41e1c1de81f578c32132ad08594e851841d0239cd410def11d1c15ee75b92f86a04f7c6c7f36b9046b8fb2fe29565b1
0201 // integer of size 1 (our public key)
03
028181 // integer of size 0x81 (our private key)
00848578c66c25b4b84243d94ea8dcb3b05b967b93eb5ffc5ae26d2c87f3a35f10e67e
0aa5a1d6c21f9d1e2196df78f1ac8aec4e3ebedff3c98df086c2f6a85e0b
efc0ca19c5e2495549fee52e513e7be9f22207d24b847fbb0cb5bab795c6
90053e652d11539a2d960feadecb9b175487000f7812ceacf5db83301606
cc357da3
0202 // integer of size 2 (prime 1)
00f5
0202 // integer of size 2 (prime 2)
00cf
0202 // integer of size 2 (exponent 1)
00a3
0202 // integer of size 2 (exponent 2)
008a
0202 // integer of size 2 (coefficient)
00bdNow let's remove everything which is after the modulus and let's refill the file with our own values. Let's go back in Sage to calculate them:
public = 65537
phi = (p - 1) * (q - 1)
private = inverse_mod(public, phi)
exponent1 = inverse_mod(private, p - 1)
exponent2 = inverse_mod(private, q - 1)
coefficient = inverse_mod(q, p)After filling and modifying the header's length accordingly, we obtain a nice hexadecimal file that we can transform back to binary:
xxd -r -p new_key.hex | openssl asn1parse -inform DERIt works! So now let's decrypt with is shall we?
xxd -r -p new_key.hex | openssl rsa -inform DER -outform PEM -out newkey.pem
openssl rsautl -decrypt -in secret -inkey newkey.pemWe have our secret :)!
3 commentsThe Hacking week just started, it's a CTF that happens over a week.
You'll find challenges about crypto, network, forensic, reverse and exploit.
And also, I have a challenge up there in the crypto challenge ^^
It's in french here: http://hackingweek.fr/challenges/ (click on "voir" next to crypto 4)
basically Alice encrypted the secret, you have to find what the secret is. What you have is a key that shares the same modulus as Alice.
comment on this storyDo you know what happens when you run this code in PHP?
<?php
var_dump(md5('240610708') == md5('QNKCDZO'));
var_dump(md5('aabg7XSs') == md5('aabC9RqS'));
var_dump(sha1('aaroZmOk') == sha1('aaK1STfY'));
var_dump(sha1('aaO8zKZF') == sha1('aa3OFF9m'));
var_dump('0010e2' == '1e3');
var_dump('0x1234Ab' == '1193131');
var_dump('0xABCdef' == ' 0xABCdef');
?>Check the answer here. That's right, everything is True.
This is because == doesn't check for type, if a string looks like an integer it will first try to convert it to an integer first and then compare it.
More about PHP == operator here
This is weird and you should use === instead.
Even better, you can use hash_equals (coupled with crypt)
Compares two strings using the same time whether they're equal or not.
This function should be used to mitigate timing attacks; for instance, when testing crypt() password hashes.
Here's the example from php.net:
<?php
$expected = crypt('12345', '$2a$07$usesomesillystringforsalt$');
$correct = crypt('12345', '$2a$07$usesomesillystringforsalt$');
$incorrect = crypt('apple', '$2a$07$usesomesillystringforsalt$');
hash_equals($expected, $correct);
?>Which will return True.
the hashed strings start with 0e, for example both strings are equals in php:
md5('240610708') = 0e462097431906509019562988736854
md5('QNKCDZO') = 0e830400451993494058024219903391because php understands them as both being zero to the power something big. So zero.
Now, if you're comparing unencrypted or unhashed strings and one of them is supposed to be secret, you might have potentialy created the setup for a timing-attack.
Always try to compare hashes instead of the plaintext!
4 commentsTo make it short, I did some research on the Boneh and Durfee bound, made some code and it worked. (The bound that allows you to find private keys if they are lesser than \(N^{0.292}\))
I noticed that many times, the lattice was imperfect as many vectors were unhelpful. I figured I could try to remove those and preserve a triangular basis, and I went even further, I removed some helpful vectors when they were annoying. The code is pretty straightforward (compare to the boneh and durfee algorithm here)
So what happens is that I make the lattice smaller, so when I feed it to the lattice reduction algorithm LLL it takes less time, and since the complexity of the whole attack is dominated by LLL, the whole attack takes less time.
It was all just theoric until I had to try the code on the plaid ctf challenge. There I used the normal code and solved it in ~3 minutes. Then I wondered, why not try running the same program but with the research branch?
That’s right, only 10 seconds. Because I removed some unhelpful vectors, I could use the value m=4 and it worked. The original algorithm needed m=5 and needed a lattice of dimension 27 when I successfully found a lattice of dimension 10 that worked out. I guess the same thing happened to the 59 triplets before that and that’s why the program ran way faster. 3 minutes to 10 seconds, I think we can call that a success!
The original code:
My book Real-World Cryptography is finished and shipping! You can purchase it here.
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