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| + | MATT!!! VS FUCKYOUMATT!!!!!!!!!!!!!!!!! |
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| − | == 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? ==
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| − | Answer: Yes.
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| − | Explanation:
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| − | <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>.
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| − | == 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? ==
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| − | Answer: Yes.
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| − | Explanation:
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| − | <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.
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| − | == 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? ==
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| − | Answer: Yes.
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| − | Explanation:
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| − | Although the worst case is <math>\Theta(n^2)</math>, this does not mean all cases are <math>\Theta(n^2)</math>.
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| − | == 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? ==
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| − | Answer: No.
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| − | Explanation:
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| − | 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>
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| − | == 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>? ==
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| − | Answer: Yes.
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| − | Explanation: Both even and odd functions are <math>\Theta(n^2)</math>.
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Revision as of 01:33, 4 August 2020
MATT!!! VS FUCKYOUMATT!!!!!!!!!!!!!!!!!
