# Difference between revisions of "TADM2E 2.23"

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− | == If I prove that an algorithm takes | + | == 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. | Answer: Yes. | ||

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Explanation: | 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 | + | == 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. | Answer: Yes. | ||

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Explanation: | 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 | + | == 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. | Answer: Yes. | ||

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Explanation: | Explanation: | ||

− | Although the worst case is | + | 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 | + | == 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. | Answer: No. | ||

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Explanation: | Explanation: | ||

− | The worst-case input must follow | + | 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 | + | == 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. | Answer: Yes. | ||

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

## Latest revision as of 18:23, 11 September 2014

## Contents

- 1 2-23.
- 1.1 If I prove that an algorithm takes $ O(n^2) $ worst-case time, is it possible that it takes $ O(n) $ on
*some*inputs? - 1.2 If I prove that an algorithm takes $ O(n^2) $ worst-case time, is it possible that it takes $ O(n) $ on
*all*inputs? - 1.3 If I prove that an algorithm takes $ \Theta(n^2) $ worst-case time, is it possible that it takes $ O(n) $ on
*some*inputs? - 1.4 If I prove that an algorithm takes $ \Theta(n^2) $ worst-case time, is it possible that it takes $ O(n) $ on
*all*inputs? - 1.5 Is the function $ f(n) = \Theta(n^2) $, where $ f(n) = 100n^2 $ for even $ n $ and $ f(n) = 20n^{2} - n * log_2 n $ for odd $ n $?

- 1.1 If I prove that an algorithm takes $ O(n^2) $ worst-case time, is it possible that it takes $ O(n) $ on

# 2-23.

## If I prove that an algorithm takes $ O(n^2) $ worst-case time, is it possible that it takes $ O(n) $ on *some* inputs?

Answer: Yes.

Explanation:

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

## If I prove that an algorithm takes $ O(n^2) $ worst-case time, is it possible that it takes $ O(n) $ on *all* inputs?

Answer: Yes.

Explanation:

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

## If I prove that an algorithm takes $ \Theta(n^2) $ worst-case time, is it possible that it takes $ O(n) $ on *some* inputs?

Answer: Yes.

Explanation:

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

## If I prove that an algorithm takes $ \Theta(n^2) $ worst-case time, is it possible that it takes $ O(n) $ on *all* inputs?

Answer: No.

Explanation:

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

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

Answer: Yes.

Explanation: Both even and odd functions are $ \Theta(n^2) $.