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N cn2 therefore the worst case running time is n2 23

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Unformatted text preview: c(n - 1)2 + Θ(n) ≤ cn2 - 2c(n - 1) + 1 + Θ(n) ≤ cn2 Therefore, the worst case running time is Θ(n2) 23 Best Case Partition When the partitioning procedure produces two When regions of size n/2, we get the a balanced partition balanced with best case performance: best T(n) = 2T(n/2) + Θ(n) = Θ(n lg n) Average complexity is also Θ(n lg n) Average 24 Best Case Partitioning 25 Average Case Assuming random input, average-case running time Assuming is much closer to Θ(n lg n) than Θ(n2) First, a more intuitive explanation/example: First, Suppose that partition() always produces a 9-to-1 proportional split. This looks quite unbalanced! The recurrence is thus: T(n) = T(9n/10) + T(n/10) + Θ(n) = Θ(n lg n)! How deep will the recursion go? 26 Average Case T (n) = T (n /10) + T (9n /10) + Θ( n) = Θ( n log n)! For a split of proportionality α, where 0≤α≤½, the minimum depth of the tree is - lg n / lg α & the maximum depth is - lg n / lg(1-α). log2n = log10n/log102 27 Average Case Intuitively, a real-life run of quicksort will produce a Intuitively, mix of “bad” and “good” splits Randomly distributed among the recursion tree Pretend for intuition that they alternate between best-case (n/2:n/2) and worst-case (n-1:1) 28 Average Case What h...
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