# Difference between revisions of "TADM2E 2.45"

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− | The expected number of times the assignment to | + | The expected number of times the assignment to <i>tmp</i> is made is the sum of the probabilities that the <math>n^{th}</math> element is the <i>minimum</i>. If we assume the minimum is distributed uniformly in our sequence then the probability any given element is the minimum is <math>1/n</math>. |

Expected time is E(n) = E(n-1) + 1/n, E[1] = 0 | Expected time is E(n) = E(n-1) + 1/n, E[1] = 0 | ||

− | To compute expected value we sum this quantity for | + | To compute expected value we sum this quantity for <math>n</math>: |

− | + | <math>\sum_{i=1}^{n} \frac{1}{i}</math> | |

− | and recognize this as the definition of the | + | and recognize this as the definition of the <math>n^{th}</math> <i>Harmonic number</i> |

− | + | <math>H(n) = \sum_{i=1}^{n} \frac{1}{i} \sim \ln n</math> | |

− | so our expected value approaches | + | so our expected value approaches <math>\ln n</math> as <math>n</math> grows large. |

''Return to [[Algo-analysis-TADM2E]]'' | ''Return to [[Algo-analysis-TADM2E]]'' |

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

The expected number of times the assignment to *tmp* is made is the sum of the probabilities that the $ n^{th} $ element is the *minimum*. If we assume the minimum is distributed uniformly in our sequence then the probability any given element is the minimum is $ 1/n $.

Expected time is E(n) = E(n-1) + 1/n, E[1] = 0

To compute expected value we sum this quantity for $ n $:

$ \sum_{i=1}^{n} \frac{1}{i} $

and recognize this as the definition of the $ n^{th} $ *Harmonic number*

$ H(n) = \sum_{i=1}^{n} \frac{1}{i} \sim \ln n $

so our expected value approaches $ \ln n $ as $ n $ grows large.

*Return to Algo-analysis-TADM2E*