CS 3343: Analysis of Algorithms Quick sort and average running time analysis 2/25/2019

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 2/25/2019

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. 2/25/2019

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 2/25/2019

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) 2/25/2019

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 2/25/2019

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 2/25/2019

In-place Partition 3 6 2 10 5 6 8 3 13 8 3 2 10 11 2/25/2019

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. 2/25/2019

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 2/25/2019

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 2/25/2019

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 2/25/2019

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. 2/25/2019

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) 2/25/2019

Worst-case recursion tree T(n) = T(0) + T(n–1) + n 2/25/2019

Worst-case recursion tree T(n) = T(0) + T(n–1) + n T(n) 2/25/2019

Worst-case recursion tree T(n) = T(0) + T(n–1) + n n T(0) T(n–1) 2/25/2019

Worst-case recursion tree T(n) = T(0) + T(n–1) + n n T(0) (n–1) T(0) T(n–2) 2/25/2019

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) 2/25/2019

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) 2/25/2019

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) 2/25/2019

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) 2/25/2019

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? 2/25/2019

Analysis of “almost-best” case 2/25/2019

Analysis of “almost-best” case 2/25/2019

Analysis of “almost-best” case 2/25/2019

Analysis of “almost-best” case log10/9n … … … O(n) leaves Q(1) Q(1) 2/25/2019

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) 2/25/2019

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) 2/25/2019

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 2/25/2019

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) = The expected running time is an average of all cases Expectation 2/25/2019

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Solving recurrence Recursion tree (iteration) method - Good for guessing an answer Substitution method - Generic method, rigid, but may be hard Master method - Easy to learn, useful in limited cases only - Some tricks may help in other cases 2/25/2019

Substitution method The most general method to solve a recurrence (prove O and  separately): Guess the form of the solution: (e.g. using recursion trees, or expansion) Verify by induction (inductive step). 2/25/2019

Expected running time of Quicksort Guess We need to show that for some c and sufficiently large n Use T(n) instead of for convenience 2/25/2019

Need to show: T(n) ≤ c n log (n) Fact: Need to show: T(n) ≤ c n log (n) Assume: T(k) ≤ ck log (k) for 0 ≤ k ≤ n-1 Proof: using the fact that if c ≥ 4. Therefore, by defintion, T(n) =  (nlogn) 2/25/2019

Tightly Bounding The Key Summation Split the summation for a tighter bound What are we doing here? The lg k in the second term is bounded by lg n What are we doing here? Move the lg n outside the summation What are we doing here? 2/25/2019

Tightly Bounding The Key Summation The summation bound so far The lg k in the first term is bounded by lg n/2 What are we doing here? lg n/2 = lg n - 1 What are we doing here? Move (lg n - 1) outside the summation What are we doing here? 2/25/2019

Tightly Bounding The Key Summation The summation bound so far What are we doing here? Distribute the (lg n - 1) The summations overlap in range; combine them What are we doing here? The Guassian series What are we doing here? 2/25/2019

Tightly Bounding The Key Summation The summation bound so far Rearrange first term, place upper bound on second What are we doing here? What are we doing? Guassian series Multiply it all out What are we doing? 2/25/2019

Tightly Bounding The Key Summation 2/25/2019