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Oct 15, 2009 at 6:04 comment added Darsh Ranjan When I originally asked this question (in the seminar mentioned in the posted question), I did ask for an efficient algorithm, but I probably didn't emphasize it enough. :-) Solving the 1D version by continued fractions, we can achieve complexity proportional to the number of terms in the continued fraction of the answer, which is at worst O(log(d)) (where d is the smallest denominator). That may be too optimistic for 2D, but could polylog(d) be achievable?
Oct 5, 2009 at 0:13 history edited Kim Morrison CC BY-SA 2.5
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Oct 5, 2009 at 0:08 vote accept Kim Morrison
Oct 4, 2009 at 22:13 history answered Ilya Nikokoshev CC BY-SA 2.5