Niclas Alexandersson edited d) Quicksort average complexity.tex  about 10 years ago

Commit id: 0cdc53c6ed0520bfff1fd14f0ce3406fa5357d85

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\Updownarrow\\  \sum_{m=0}^{n-1}\left[\frac{C(m+1)}{\prod_{k=0}^{m}\left[1+\frac{1}{k+1}\right]} - \frac{C(m)}{\prod_{k=0}^{m-1}\left[1+\frac{1}{k+1}\right]}\right] = \sum_{m=0}^{n-1}\left[\frac{2 - \frac{1}{m+1}}{\prod_{k=0}^{m}\left[1+\frac{1}{k+1}\right]}\right]\\  \Updownarrow\\  \frac{C(n)}{\prod_{k=0}^{n-1}\left[1+\frac{1}{k+1}\right]} - \frac{C(0)}{\prod_{k=0}^{-1}\left[1+\frac{1}{k+1}\right]} = \sum_{m=0}^{n-1}\left[\frac{2 - \frac{1}{m+1}}{\prod_{k=0}^{m}\left[1+\frac{1}{k+1}\right]}\right] \frac{1}{m+1}}{\prod_{k=0}^{m}\left[1+\frac{1}{k+1}\right]}\right]\\  \Updownarrow\\  C(n) = \left(\prod_{k=0}^{n-1}\left[1+\frac{1}{k+1}\right]\right)\left(C(0) + \sum_{m=0}^{n-1}\left[\frac{2 - \frac{1}{m+1}}{\prod_{k=0}^{m}\left[1+\frac{1}{k+1}\right]}\right]\right)  \]