3 ms·
From my educated guess, A single 4090 can crack it around ~140, since there will be some scaling losses. Also, this is an optimistic take since we don't expect
by bayindirh 2y ago
From my educated guess, A single 4090 can crack it around ~140, since there will be some scaling losses. Also, this is an optimistic take since we don't expect the VRAM won't have any bit-flip events during this time, under load.
If you can have 10 Tesla cards, the number will be a bit shorter (around 14, I guess), since NVLink is much more efficient and creates a mesh between cards without hitting the PCIe.
- patates 2y agoRemembering the good old days when you could use it to connect consumer cards...
- rtkwe 2y agoThe advantages from SLI were always well below doubling and usually more like 50%. As frame rates and everything else has gotten higher and more complex the overhead of SLI got more and more onerous to the point where it barely gave much of an advantage to support at all.
- fwip 2y agoDoesn't have any scaling losses - it's a very parallel problem. Divide the keyspace N ways, run N brute-force searches. Similarly, it doesn't benefit from a faster connection to other nodes or main memory. Most bit-flips won't matter - either you get a false positive which is ruled out trivially, or the 1-in-$SearchSpace chance you get the false negative.
- jerf 2y agoFor education's sake and better internet search terms, there's some terms for this, the most popular of which is "embarassingly parallel": https://en.wikipedia.org/wiki/Embarrassingly_parallel https://en.wikipedia.org/wiki/Embarrassingly_parallel And agreed; coordination costs are negligible next to the cost of the calculation, so it should be effectively linear, and dominated by the luck of the draw on when the correct key is selected.