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== md5 Collisions? == | == md5 Collisions? == |
Revision as of 17:39, 22 May 2008
This page serves as an archive for clearly Handled topics, as Talk:Main Page is getting a little cluttered.
md5 Collisions?
What about collisions in the md5 algorithm?
- What about them? In theory, two dates/stock market prices might end up with the same meetup point. Unlikely, and probably not a terribly big deal? Zigdon 07:39, 21 May 2008 (UTC)
- There are situations where cryptographic collisions are a problem, but this isn't one of them. md5 is just being used as a pseudorandom number generator here, and it has a very small-entropy seed (someone call Debian!). So its cryptographic strenth isn't particularly relevant. --Xkcd 07:43, 21 May 2008 (UTC)xkcd
OMG geohashing predicts future dow prices!
http://irc.peeron.com/xkcd/map/data/2009/05/27 --Ryan the leach 12:33, 21 May 2008 (UTC)
- heh yeh i saw that one too... this is deep magic SinJax
- Yeah, I totally wasn't testing stuff, <_< Zigdon 14:54, 21 May 2008 (UTC)
Wow, now you only need to break MD5 to get rich! 81.167.17.12 12:40, 21 May 2008 (UTC)
could be future meetup? --Ryan the leach 12:33, 21 May 2008 (UTC)
- Fairly inconsistent if so, what if the dow is different. WHAT THEN?!?
- It won't ;) --DarkRat
- well the tool wont say its different :P --Ryan the leach 12:33, 21 May 2008 (UTC)
- It won't ;) --DarkRat
geohash.org
There's already a thing called "geohash" which is a way to represent coordinates with arbitrary precision in a computer and query-friendly format. It's at http://geohash.org/ and it was designed and implemented by Gustavo Niemeyer.
- I don't see how that's relevant. We are talking about geohashing. It's a different word.
To decimal
...now, i'm likely just too thick to get it, but how does one convert back from the half md5 hash to decimals?
- It goes like this... each digit is below zero but not 10^-n but 16^-n....so the first hex digit is 16^-1, the next is 16^-2 and so on. And it goes from left to right so in the hex number "8d" you get:
f = (8 * 16^-1) + (13 * 16^-2) = 0.5507
i think anyway :) - SinJax
Or, interpret the bytes from md5 as two big-endian 64-bit integers, and divide both by 2^64.