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Consider an algorithm A with running time n√n. Assume that A can solve instances of size n on a machine that takes 10−12 seconds per operation. What size instances (in terms of n) can A solve in the same time on a machine that takes 10−15 seconds per operation?

Question

Consider an algorithm A with running time n√n. Assume that A can solve instances of size n on a machine that takes 10−12 seconds per operation. What size instances (in terms of n) can A solve in the same time on a machine that takes 10−15 seconds per operation?

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Solution

The running time of the algorithm A is proportional to n√n. This means that if we increase the speed of the machine by a factor of 1000 (from 10^-12 seconds per operation to 10^-15 seconds per operation), we can also increase the size of the instances that the algorithm can solve in the same time by the cube root of 1000, because n√n = n^(3/2), and the cube root of 1000 is 10.

So, if the algorithm can solve instances of size n on the slower machine, it can solve instances of size 10n on the faster machine.

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