19,339 reputation
23291
bio website
location Brooklyn, NY
age 33
visits member for 3 years, 2 months
seen 10 hours ago
Good Morning how are you, I'm dr jimbob
I'm interested in things.
I'm not a real dr,
But I am a real jim bob.

Have a PhD in Experimental High-Energy Physics, but left academia in mid-2010 to program professionally.

Mostly program/script in python, django, and jquery these days doing mostly web apps.

Also have experience programming in C, C++, java, haskell, php, and (bash) shell more in the past.

Linux as primary OS since 1999, ubuntu user since 2005 (Hoary).


May
14
answered Can I brute-force a password hash even if I don't know the hash-algorithm used?
May
12
answered Why was WPS not patched to make it secure?
May
12
answered Can a certification authority sign another CA's certs?
May
11
revised PRF, IKE and hash function
added 1901 characters in body
May
11
revised PRF, IKE and hash function
deleted 16 characters in body
May
11
revised PRF, IKE and hash function
added 169 characters in body
May
10
revised PRF, IKE and hash function
added 292 characters in body
May
10
answered PRF, IKE and hash function
May
9
answered On an “uncritical” site, is there any reason to NOT trust a self-signed certificate?
May
9
comment Can NSA generate all hashes for rainbow table to break md5?
"My definition of breaking would be obtaining a second preimage for a given hash" - Best second preimage attack on MD5 has complexity of 2^123. Collision attacks are much easier as you get to change both sides of the equation. Preimage attacks don't let you do that. With N=365 days, you need 41 ~ O(sqrt(N)) people in a room for two people to have 90% chance that at least one pair of two people have the same birthday (collision attack). But you need 329 ~ O(N) people in a room before its 90% likely that someone will have the same birthday as you (preimage / second preimage attack).
May
9
comment Can NSA generate all hashes for rainbow table to break md5?
@AntonGarciaDosil - Your statement "SHA-512 (equivalent of a 256 symmetric key)" is completely wrong. For the hash of a password, what matters is not a collision attack but preimage resistance, so for ultra-strong passwords sha512crypt has the strength of 512-bit (unless you fear your attacker has large scale fast quantum computers in which case Grover's algorithm reduces it to 256-bit strength). Granted, you only get 512-bits of security against a very secure passphrase with more than 512-bits of security. Most passwords are far weaker in the realm of 20-80 bits of entropy.
May
9
comment Can NSA generate all hashes for rainbow table to break md5?
@AntonGarciaDosil - The best collision attack for MD5 is only 2^18 time. Granted MD5 is still secure for preimage attacks on high-entropy information (best public attack is 2^123). MD5 shouldn't be used for passwords, mostly because users often choose weak low-entropy passwords and MD5 is too fast. Also note MD5crypt ($1$) and SHA512crypt ($6$) used in unix passwords are NOT simply an application of md5 or sha512. They are 1000 or 5000 (default) salted iterations of the simple hash. en.wikipedia.org/wiki/Crypt_(C)
May
9
comment Does OpenSSL leave any traces that a server uses it?
For example to get these headers: curl -I https://httpd.apache.org gives third line as Server: Apache/2.4.9 (Unix) mod_wsgi/3.4 Python/2.7.5 OpenSSL/1.0.1g
May
7
awarded  Announcer
May
7
comment SSH using pre-existing session (symmetric) keys
I'm curious why do you want to do this? Is the idea that if some group broke the common trapdoor functions (prime factorization, discrete log, elliptic curve discrete log) underlying Diffie Hellman and ECDH (as well as RSA, DSA), but symmetric ciphers are still strong then you want to have a protocol to rely on pre-shared symmetric ciphers?
May
7
revised Why is RSA decryption slow?
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May
7
revised Why is RSA decryption slow?
added 1100 characters in body
May
7
answered Why is RSA decryption slow?
May
7
comment What's the meaning of the term “ad-hoc MAC” (Message Authentication Code)?
@melostap Let's say I decided to make my own MAC as H(m+k) for some application I developed. This would be ad hoc versus using a widely used/analyzed MAC. By concatenating the key at the end of the message, this wouldn't be susceptible to the standard length extension attack that H(k+m) suffers from. But it is weaker than HMAC's construction, esp with say md5 where collision attacks exist. There are possibly other flaws and as this simpler scheme is not widely used makes it more questionable.
May
7
comment What's the meaning of the term “ad-hoc MAC” (Message Authentication Code)?
@melostap - But I think we agree that ad hoc security is frowned upon - instead of being able to use all the cryptanalysis of HMAC (as defined in RFC2104), you have to specifically analyze the custom HMAC used in just SSLv3. Today, HMAC is well used as an acceptable MAC. With MD hashes (MD5, SHA1, SHA2), the hash is vulnerable to length extension attacks. For that reason alone, it seems less secure to concatenate an ipad / opad to a secret key to create effectively an independent key. Using xor appears to be much more sensible under this threat.