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An attempt to backdoor the kernelAn attempt to backdoor the kernelPosted Nov 7, 2003 18:55 UTC (Fri) by zooko (subscriber, #2589)In reply to: An attempt to backdoor the kernel by lm Parent article: An attempt to backdoor the kernel If you can find collisions in SHA-1, you can probably use that to forge digital signatures and gain remote authorizations to any system that uses cryptography for authentication. (This includes, among others, any system which uses SSH, TLS, or a cryptographically authenticated VPN.) Is it a lawsuit waiting to happen to run sshd?
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An attempt to backdoor the kernel Posted Nov 8, 2003 18:54 UTC (Sat) by Stephen_Beynon (✭ supporter ✭, #4090) [Link] You could only do that if you manage to find a way of generating a blockof data with a pre-determined hash. The problem of finding 2 blocks of data which generate the same hash (any hash) is a far smaller problem.
An attempt to backdoor the kernel Posted Nov 13, 2003 9:59 UTC (Thu) by ekj (subscriber, #1524) [Link] Sure. More spesifically, the magnitude of the problem is proportiaonal to the square root ofthe other problem. So finding two strings with the same 160 bit hash requires you to generate and hash on the order of 2^80 strings. I wouldn't loose much sleep over this, but if you do, there's no problem with going to a bigger hash. If you could generate and hash 2^30 strings a second (and store all the hashes you already have created...) you'd still need 2^50 seconds before you'd on the average get lucky. Everyone designing cryptographic hashes is aware of this issue, which is why the hashes are so big in the first place.
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