CSC 2300 Data Structures & Algorithms

CSC 2300Data Structures & Algorithms March 23, 2007 Chapter 7. Sorting

Today – Sorting • Quicksort – • Algorithm • Pivot • Analysis • Worst Case • Best Case • Average Case

Quicksort – Algorithm • If the number of elements in S is 0 or 1, then return. • Pick any element v in S. This is called the pivot. • Partition S – {v} into two disjoint groups: S1 = { x ε S – {v} | x ≤ v} and S2 = { x ε S – {v} | x ≥ v}. • Return { quicksort(S1) followed by v followed by quicksort(S2)}.

Quicksort – Example

Quicksort – Partition Strategy • Example. Input: 8, 1, 4, 9, 6, 3, 5, 2, 7, 0. Say 6 is chosen as pivot. • 8 1 4 9 0 3 5 2 7 6 i j pivot • 8 1 4 9 0 3 5 2 7 6 i j • 2 1 4 9 0 3 5 8 7 6 i j • 2 1 4 9 0 3 5 8 7 6 i j • 2 1 4 5 0 3 9 8 7 6 i j • 2 1 4 5 0 3 9 8 7 6 j i pivot • 2 1 4 5 0 3 6 8 7 9 pivot

Choices of Pivot • Four suggestions: • First element of array; • Larger of first two distinct elements of array; • Middle element of array; • Randomly. • What do you think about these choices? • All bad choices. Why?

Good Choice of Pivot • Best choice: median of array. • Disadvantage? • Practical choice: Median of Three. • What is it? • Median of left, right, and center elements. • Example: 8, 1, 4, 9, 6, 3, 5, 2, 7, 0. • Median of 8, 6, and 0.

Example • Example: 8, 1, 4, 9, 6, 3, 5, 2, 7, 0. • Pivot = Median of 8, 6, and 0. • What should new array look like? • Recall what we have done: 8 1 4 9 0 3 5 2 7 6 i j pivot • Can we do better? 0 1 4 9 6 3 5 2 7 8 i pivot j • Where should we move pivot? 0 1 4 9 7 3 5 2 6 8 i j pivot

Median-of-Three Code

Quicksort – Analysis • Quicksort is recursive. • We thus get a recurrence formula: T(0) = T(1) = 1, T(N) = T(i) + T(N – i – 1) + cN, where i denotes the number of elements in S1. • What value of i gives worst case? • What value of i gives best case?

Worst Case Analysis • We have i = 0, always. • What does that say about the pivot? • Always the smallest element. • Recurrence becomes T(N) = T(0) + T(N – 1) + cN. • Ignore T(0), and get T(N) = T(N – 1) + cN. • Hence T(N – 1) = T(N – 2) + c(N – 1), T(N – 2) = T(N – 3) + c(N – 2), … T(2) = T(1) + c(2). • We get T(N) = T(1) + c ∑ i = 1 + c [ N(N+1)/2 – 1] = O(N2).

Best Case Analysis • We have i = N/2, always. • What does that say about the pivot? • Always the median. • Recurrence becomes T(N) = T(N/2) + T(N/2) + cN = 2 T(N/2) + cN. • Do you remember how to solve this recurrence? • Divide by N to get T(N)/N = T(N/2)/(N/2) + c. • Thus, T(N/2)/(N/2) = T(N/4)/(N/4) + c, T(N/4)/(N/4) = T(N/8)/(N/8) + c, … T(2)/2 = T(1)/1 + c. • We get T(N)/N = T(1)/1 + c logN, and so T(N) = N + c N logN = O(N log N).

Average Case Analysis • Always much harder than worst and best cases. • What can we assume about the pivot? • Assume that each of the sizes for S1 is equally likely and thus has probability 1/N. • The average value of T(i) is thus (1/N) ∑ T(j). • What can we say about the value of T(N – i – 1)? • Recurrence becomes T(N) = (2/N) ∑ T(j) + cN. • Does this recurrence look familiar? • When we did an internal path length analysis in Chapter 4 (Trees).

Average Case Analysis • Recurrence: T(N) = (2/N) ∑ T(j) + cN. • How can we solve this recurrence? • Divide by N? • No, multiply by N! • We get this recurrence: N T(N) = 2 ∑ T(j) + cN2. • How do we get rid of the ∑ T(j) ? • We use this recurrence: (N – 1)T(N – 1) = 2 ∑ T(j) + c(N – 1)2. • Subtracting one recurrence from the other, we get NT(N) – (N – 1)T(N – 1) = 2 T(N – 1) + c(2N – 1). • Simplifying and dropping the c term, we get NT(N) = (N+1) T(N – 1) + 2cN.

Recurrence • Recurrence: NT(N) = (N+1) T(N – 1) + 2cN. • How can we solve this recurrence? • Divide by N? • Divide by N+1? • No, divide by N(N+1)! • We get this recurrence: T(N)/(N+1) = T(N – 1)/N + 2c/(N+1). • What to do now? • We can telescope: T(N – 1)/N = T(N – 2)/(N – 1) + 2c/N, T(N – 2)/(N – 1) = T(N – 3)/(N – 2) + 2c/(N – 1), … T(2)/3 = T(1)/2 + 2c/3. • We get this solution: T(N)/(N+1) = T(1)/2 + 2c ∑ (1/i). • What does ∑ (1/i) equal? • We get T(N) = O(N log N).

CSC 2300 Data Structures & Algorithms