Revision as of 18:14, 11 September 2014

2-23.

If I prove that an algorithm takes <math>O(n^2)</math> worst-case time, is it possible that it takes <math>O(n)</math> on some inputs?

Answer: Yes.

Explanation:

<math>O(n^2)</math> worst-case means that the worst-case is bound from above by <math>O(n^2)</math>; it does not necessarily mean that all cases must follow that complexity. Thus, there could be some inputs that are <math>O(n)</math>.

If I prove that an algorithm takes <math>O(n^2)</math> worst-case time, is it possible that it takes <math>O(n)</math> on all inputs?

Answer: Yes.

Explanation:

<math>O(n^2)</math> worst-case is only an upper bound on the worst-case. It is possible that all inputs can be done in <math>O(n)</math>, which still follows this upper bound.

If I prove that an algorithm takes <math>\Theta(n^2)</math> worst-case time, is it possible that it takes <math>O(n)</math> on some inputs?

Answer: Yes.

Explanation:

Although the worst case is <math>\Theta(n^2)</math>, this does not mean all cases are <math>\Theta(n^2)</math>.

If I prove that an algorithm takes <math>\Theta(n^2)</math> worst-case time, is it possible that it takes <math>O(n)</math> on all inputs?

Answer: No.

Explanation:

The worst-case input must follow <math>\Theta(n^2)</math>, so it can't be <math>O(n)</math>. Therefore, all cases are not <math>O(n)</math>

Is the function <math>f(n) = \Theta(n^2)</math>, where <math>f(n) = 100n^2</math> for even <math>n</math> and <math>f(n) = 20n^{2} - n * log_2 n</math> for odd <math>n</math>?

Answer: Yes.

Explanation: Both even and odd functions are <math>\Theta(n^2)</math>.