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**Analysis of Algorithms**

Lecture Analysis of quick sort 4/8/2017

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Quick sort Another divide and conquer sorting algorithm – like merge sort Anyone remember the basic idea? The worst-case and average-case running time? Learn some new algorithm analysis tricks 4/8/2017

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**Quick sort Quicksort an n-element array:**

Divide: Partition the array into two subarrays around a pivot x such that elements in lower subarray £ x £ elements in upper subarray. Conquer: Recursively sort the two subarrays. Combine: Trivial. £ x x ≥ x Key: Linear-time partitioning subroutine. 4/8/2017

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**Partition All the action takes place in the partition() function £ x x**

Rearranges the subarray in place End result: two subarrays All values in first subarray all values in second Returns the index of the “pivot” element separating the two subarrays p q r £ x x ≥ x 4/8/2017

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**Pseudocode for quicksort**

QUICKSORT(A, p, r) if p < r then q PARTITION(A, p, r) QUICKSORT(A, p, q–1) QUICKSORT(A, q+1, r) Initial call: QUICKSORT(A, 1, n) 4/8/2017

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Idea of partition If we are allowed to use a second array, it would be easy 6 10 5 8 13 3 2 11 6 5 3 2 11 13 8 10 2 5 3 6 11 13 8 10 4/8/2017

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**Another idea Keep two iterators: one from head, one from tail 6 10 5 8**

13 3 2 11 6 2 5 3 13 8 10 11 3 2 5 6 13 8 10 11 4/8/2017

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In-place Partition 3 6 2 10 5 6 8 3 13 8 3 2 10 11 4/8/2017

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**Partition In Words Partition(A, p, r):**

Select an element to act as the “pivot” (which?) Grow two regions, A[p..i] and A[j..r] All elements in A[p..i] <= pivot All elements in A[j..r] >= pivot Increment i until A[i] > pivot Decrement j until A[j] < pivot Swap A[i] and A[j] Repeat until i >= j Swap A[j] and A[p] Return j Note: different from book’s partition(), which uses two iterators that both move forward. 4/8/2017

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**Partition Code What is the running time of partition()?**

Partition(A, p, r) x = A[p]; // pivot is the first element i = p; j = r + 1; while (TRUE) { repeat i++; until A[i] > x or i >= j; j--; until A[j] < x or j < i; if (i < j) Swap (A[i], A[j]); else break; } swap (A[p], A[j]); return j; What is the running time of partition()? partition() runs in (n) time 4/8/2017

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p r 6 10 5 8 13 3 2 11 x = 6 i j 6 10 5 8 13 3 2 11 scan i j 6 2 5 8 13 3 10 11 swap i j Partition example 6 2 5 8 13 3 10 11 scan i j 6 2 5 3 13 8 10 11 swap i j 6 2 5 3 13 8 10 11 scan j i p q r 3 2 5 6 13 8 10 11 final swap 4/8/2017

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6 10 5 8 11 3 2 13 Quick sort example 3 2 5 6 11 8 10 13 2 3 5 6 10 8 11 13 2 3 5 6 8 10 11 13 2 3 5 6 8 10 11 13 4/8/2017

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**Analysis of quicksort Assume all input elements are distinct.**

In practice, there are better partitioning algorithms for when duplicate input elements may exist. Let T(n) = worst-case running time on an array of n elements. 4/8/2017

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**Worst-case of quicksort**

Input sorted or reverse sorted. Partition around min or max element. One side of partition always has no elements. (arithmetic series) 4/8/2017

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**Worst-case recursion tree**

T(n) = T(0) + T(n–1) + n 4/8/2017

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**Worst-case recursion tree**

T(n) = T(0) + T(n–1) + n T(n) 4/8/2017

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**Worst-case recursion tree**

T(n) = T(0) + T(n–1) + n n T(0) T(n–1) 4/8/2017

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**Worst-case recursion tree**

T(n) = T(0) + T(n–1) + n n T(0) (n–1) T(0) T(n–2) 4/8/2017

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**Worst-case recursion tree**

T(n) = T(0) + T(n–1) + n n T(0) (n–1) T(0) (n–2) T(0) T(0) 4/8/2017

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**Worst-case recursion tree**

T(n) = T(0) + T(n–1) + n height n height = n T(0) (n–1) T(0) (n–2) T(0) T(0) 4/8/2017

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**Worst-case recursion tree**

T(n) = T(0) + T(n–1) + n n n height = n T(0) (n–1) T(0) (n–2) T(0) T(0) 4/8/2017

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**Worst-case recursion tree**

T(n) = T(0) + T(n–1) + n n n height = n Q(1) (n–1) Q(1) (n–2) T(n) = Q(n) + Q(n2) = Q(n2) Q(1) Q(1) 4/8/2017

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**Best-case analysis (For intuition only!)**

If we’re lucky, PARTITION splits the array evenly: T(n) = 2T(n/2) + Q(n) = Q(n log n) (same as merge sort) What if the split is always ? What is the solution to this recurrence? 4/8/2017

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**Analysis of “almost-best” case**

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**Analysis of “almost-best” case**

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**Analysis of “almost-best” case**

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**Analysis of “almost-best” case**

log10/9n … … … O(n) leaves Q(1) Q(1) 4/8/2017

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**Analysis of “almost-best” case**

log10n log10/9n … … O(n) leaves … Q(1) Q(n log n) Q(1) n log10n £ T(n) £ n log10/9n + O(n) 4/8/2017

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**Quicksort Runtimes Best-case runtime Tbest(n) (n log n)**

Worst-case runtime Tworst(n) (n2) Worse than mergesort? Why is it called quicksort then? Its average runtime Tavg(n) (n log n ) Better even, the expected runtime of randomized quicksort is (n log n) 4/8/2017

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**Randomized quicksort Randomly choose an element as pivot**

Every time need to do a partition, throw a die to decide which element to use as the pivot Each element has 1/n probability to be selected Rand-Partition(A, p, r) d = random(); // a random number between 0 and 1 index = p + floor((r-p+1) * d); // p<=index<=r swap(A[p], A[index]); Partition(A, p, r); // now do partition using A[p] as pivot 4/8/2017

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**Running time of randomized quicksort**

T(0) + T(n–1) + dn if 0 : n–1 split, T(1) + T(n–2) + dn if 1 : n–2 split, M T(n–1) + T(0) + dn if n–1 : 0 split, T(n) = 4/8/2017

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Divide-and-Conquer The most-well known algorithm design strategy: 1. Divide instance of problem into two or more smaller instances 2.Solve smaller instances.

Divide-and-Conquer The most-well known algorithm design strategy: 1. Divide instance of problem into two or more smaller instances 2.Solve smaller instances.

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