lecture 8
- 1. CS 332: Algorithms Quicksort
- 2. Review: Analyzing Quicksort What will be the worst case for the algorithm? Partition is always unbalanced What will be the best case for the algorithm? Partition is balanced Which is more likely? The latter, by far, except... Will any particular input elicit the worst case? Yes: Already-sorted input
- 3. Review: Analyzing Quicksort In the worst case: T(1) = (1) T(n) = T(n - 1) + (n) Works out to T(n) = (n 2 )
- 4. Review: Analyzing Quicksort In the best case: T(n) = 2T(n/2) + (n) What does this work out to? T(n) = (n lg n)
- 5. Review: Analyzing Quicksort (Average Case) Intuitively, a real-life run of quicksort will produce a 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) What happens if we bad-split root node, then good-split the resulting size (n-1) node?
- 6. Review: Analyzing Quicksort (Average Case) Intuitively, a real-life run of quicksort will produce a 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) What happens if we bad-split root node, then good-split the resulting size (n-1) node? We end up with three subarrays, size 1, (n-1)/2, (n-1)/2 Combined cost of splits = n + n -1 = 2n -1 = O(n) No worse than if we had good-split the root node!
- 7. Review: Analyzing Quicksort (Average Case) Intuitively, the O(n) cost of a bad split (or 2 or 3 bad splits) can be absorbed into the O(n) cost of each good split Thus running time of alternating bad and good splits is still O(n lg n), with slightly higher constants How can we be more rigorous?
- 8. Analyzing Quicksort: Average Case For simplicity, assume: All inputs distinct (no repeats) Slightly different partition() procedure partition around a random element, which is not included in subarrays all splits (0:n-1, 1:n-2, 2:n-3, … , n-1:0) equally likely What is the probability of a particular split happening? Answer: 1/n
- 9. Analyzing Quicksort: Average Case So partition generates splits (0:n-1, 1:n-2, 2:n-3, … , n-2:1, n-1:0) each with probability 1/n If T(n) is the expected running time, What is each term under the summation for? What is the (n) term for?
- 10. Analyzing Quicksort: Average Case So… Write it on the board
- 11. Analyzing Quicksort: Average Case We can solve this recurrence using the dreaded substitution method Guess the answer Assume that the inductive hypothesis holds Substitute it in for some value < n Prove that it follows for n
- 12. Analyzing Quicksort: Average Case We can solve this recurrence using the dreaded substitution method Guess the answer What’s the answer? Assume that the inductive hypothesis holds Substitute it in for some value < n Prove that it follows for n
- 13. Analyzing Quicksort: Average Case We can solve this recurrence using the dreaded substitution method Guess the answer T(n) = O(n lg n) Assume that the inductive hypothesis holds Substitute it in for some value < n Prove that it follows for n
- 14. Analyzing Quicksort: Average Case We can solve this recurrence using the dreaded substitution method Guess the answer T(n) = O(n lg n) Assume that the inductive hypothesis holds What’s the inductive hypothesis? Substitute it in for some value < n Prove that it follows for n
- 15. Analyzing Quicksort: Average Case We can solve this recurrence using the dreaded substitution method Guess the answer T( n ) = O( n lg n ) Assume that the inductive hypothesis holds T( n ) an lg n + b for some constants a and b Substitute it in for some value < n Prove that it follows for n
- 16. Analyzing Quicksort: Average Case We can solve this recurrence using the dreaded substitution method Guess the answer T( n ) = O( n lg n ) Assume that the inductive hypothesis holds T( n ) an lg n + b for some constants a and b Substitute it in for some value < n What value? Prove that it follows for n
- 17. Analyzing Quicksort: Average Case We can solve this recurrence using the dreaded substitution method Guess the answer T( n ) = O( n lg n ) Assume that the inductive hypothesis holds T( n ) an lg n + b for some constants a and b Substitute it in for some value < n The value k in the recurrence Prove that it follows for n
- 18. Analyzing Quicksort: Average Case We can solve this recurrence using the dreaded substitution method Guess the answer T( n ) = O( n lg n ) Assume that the inductive hypothesis holds T( n ) an lg n + b for some constants a and b Substitute it in for some value < n The value k in the recurrence Prove that it follows for n Grind through it…
- 19. Analyzing Quicksort: Average Case Note: leaving the same recurrence as the book What are we doing here? The recurrence to be solved What are we doing here? What are we doing here? Plug in inductive hypothesis Expand out the k=0 case 2b/n is just a constant, so fold it into (n)
- 20. Analyzing Quicksort: Average Case What are we doing here? What are we doing here? Evaluate the summation: b+b+…+b = b (n-1) The recurrence to be solved Since n-1<n, 2b(n-1)/n < 2b What are we doing here? Distribute the summation This summation gets its own set of slides later
- 21. Analyzing Quicksort: Average Case How did we do this? Pick a large enough that an/4 dominates (n)+b What are we doing here? Remember, our goal is to get T(n) an lg n + b What the hell? We’ll prove this later What are we doing here? Distribute the (2a/n) term The recurrence to be solved
- 22. Analyzing Quicksort: Average Case So T( n ) an lg n + b for certain a and b Thus the induction holds Thus T(n) = O(n lg n) Thus quicksort runs in O(n lg n) time on average (phew!) Oh yeah, the summation…
- 23. Tightly Bounding The Key Summation 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? Split the summation for a tighter bound
- 24. Tightly Bounding The Key Summation The summation bound so far What are we doing here? 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
- 25. Tightly Bounding The Key Summation The summation bound so far What are we doing here? Distribute the ( lg n - 1 ) What are we doing here? The summations overlap in range; combine them What are we doing here? The Guassian series
- 26. Tightly Bounding The Key Summation The summation bound so far What are we doing here? Rearrange first term, place upper bound on second What are we doing? X Guassian series What are we doing? Multiply it all out
- 27. Tightly Bounding The Key Summation