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02-01-2012
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#1 |
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Senior Member
 
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Kakaroto talks about ECDSA algorithm
PS3 developer Kakaroto, has been hard at work on jailbreaking 4.00 firmware. He has taken the time t... [Read More]
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 TrueBlue, You Lose. |
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02-01-2012
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#3 | |
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Homebrew Developer
 
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Fix: http://kakaroto.homelinux.net/2012/01/how-the-ecdsa-algorithm-works/
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If you are going to promote TB at least do it right!!!, or better yet DON'T!!!
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02-01-2012
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#4 |
  
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First I need that aspirin KaKaRoTo is talking about, second I think I have to go back to school after reading that (very interesting) piece of information
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Sony just lost the PS3's chastity belt keys, secret fun spots are open to explore ...
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02-01-2012
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#5 |
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Apprentice
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kakaroto is really working hard on the alogrithm thing.i still wonder if how true blue people are able to bend the things to play the new games except that ACID team is correct in saying that they dont have the keys as according to kakaroto its nearly impossible to get the keys except that if we had a hacker like in the movie named 'THE CORE'...just kidding friends
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02-01-2012
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#6 | |
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after reading it I would need an aspirin to |
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02-01-2012
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#7 |
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Senior Member
 
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Mathematically is not to hard to understand, i deal with similar **** on university everyday, although this is harder, kakaroto's explanation is better than any of my teachers explanation, whats hard is the actual implementation of it...
Now that its explained, i hope you all understand that it is impossible to get the NPDRM keys. Edit1 Although i have been thinking, about a possible timing attack, but i don't think its possible. 
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<eussNL> judge: ΄so why did you torrent 5 million AVI of women moaning΄ <eussNL> <TizzyT> i dont judge if it sounds good i listen |
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02-01-2012
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#8 |
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now i know how to get the keys i only must study mathematic for the algho then study developing and last but not least study laws because sony black man...so no prob ill be back in 15 years with keys
************* [ - Post Merged - ] ************* S S = k^-1 (z + dA*R) k^-1 (z + da*R) = k^-1 (z + da*R z -dA*R) = k^-1 (z z) So : k = (z z) / (S S) is it an alien speech ?
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02-01-2012
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#9 | |
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The S and S' are to be thought as "first S" and "second S", whereas they share some properties but have their own variables. Hence when you put them together, the properties that are the same are eliminated and leave only their own variables, which are known. Thus the k = (z - z') / (S - S'); you're supposed to know the values of z, z', S, and S' to obtain k. This is just simple multivariate calculus, I'll ask my teachers in university what the know about the ECDSA (in March). Doubt they'll know much, but I think I could at least get some more clarifications. |
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