4 ms·

I'm surprised nobody's given the obvious algorithm yet: while(i < BigInt(2).pow(10000)) ++i; Before anyone complains, this algorithm is correct and does not b

by IvyMike 14y ago

I'm surprised nobody's given the obvious algorithm yet: while(i < BigInt(2).pow(10000)) ++i; Before anyone complains, this algorithm is correct and does not break any of the rules as far as I can tell. :) I think the timing attack is probably what he's really looking for. Edit: as mattvanhorn pointed out, answer() is void, but that's ok... changed to a "constant time" algorithm. :)

mattvanhorn 14y ago
answer() has a void return, although I suppose you might be able to watch System.out to see if it worked.
- raylu 14y ago
  Nobody said you had to stop after reaching that line ;)
- nire 14y ago
  Yup, just set a new PrintStream(new ByteArrayOutputStream()) in System.setOut() and you're good.
lancefisher 14y ago
The BigInteger is really big. An i7 does about 109 gigaFLOPS, or 109 operations per nanosecond. [1] Suppose we can do 1 guess per operation. There are 2^10,000 possible roots. [2] 2^10,000 / 109 nanoseconds is 5.804×10^2991 years. [3] The stars will burn out before you brute force it. [1] http://en.wikipedia.org/wiki/FLOPS http://en.wikipedia.org/wiki/FLOPS [2] http://bit.ly/ZIWkkt http://bit.ly/ZIWkkt (parens in the original link) [3] http://www.wolframalpha.com/input/?i=2%5E10%2C000+%2F+109+nanoseconds+in+years http://www.wolframalpha.com/input/?i=2%5E10%2C000+%2F+109+na...
- IvyMike 14y ago
  I must dryly point out that this does not invalidate the correctness of the algorithm.
- IvyMike 14y ago
  As an optimist, I must point out that if Moore's law holds for the next 10000 doublings we'll have the answer in 20000 years, plus 9 picoseconds.
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