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Ask HN: SSH is still safe, after NSA?

14 pointsby hgfischeralmost 12 years ago
After all this news about NSA spying everything, is SSH still a safe tool?

4 comments

apialmost 12 years ago
Almost certainly, provided you&#x27;re running a recent-enough version.<p>A snooper at the line level would be able to see that you were SSH&#x27;ing to a given system and the amount of data transferred, but nothing more.<p>SSH has had very few vulnerabilities and has been really put through the ringer crypto-wise for quite some time. The protocol itself is likely quite solid. Of its common crypto algorithms, the only one I&#x27;d avoid is arcfour&#x2F;RC4. It&#x27;s an algorithm that&#x27;s known to be somewhat weaker than other common algos. Blowfish, AES, CAST, Salsa20, Twofish, etc. are not known to have any practical real-world-usable attacks against full-round versions.<p>Keep in mind that in the crypto world a &quot;break&quot; is anything that shortens the time to recover the key from that of a brute force search. So if I find a shortcut to crack a 2^128 key size symmetric cipher in &quot;only&quot; 2^112 iterations, that&#x27;s a break. But it&#x27;s not useful in the real world. To be useful in the real world, a break has to shorten things down to... well... depends on the adversary but probably &lt;2^64.<p>Of course you cannot rule out the possibility that the NSA has unpublished attacks against any of these, but most cryptographers I&#x27;ve read consider it somewhat unlikely that they have an unpublished attack good enough to <i>efficiently</i> crack them and read traffic in a real world scenario.
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keshyalmost 12 years ago
SSH is not safe if you ssh into a resource (directly or indirectly) controlled by NSA. The protocol as such is secure in the sense that for an &#x27;adversary&#x27; with average compute power, it is impossible to &#x27;break&#x27; the protocol.
ams0almost 12 years ago
As safe as your private key &#x2F; password.
bdunbaralmost 12 years ago
&gt; is SSH still a safe tool?<p>Is there something better?
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