1. 1 SORTING Dan Barrish-Flood

2. 2 heapsort made file “3-Sorting-Intro-Heapsort.ppt”

3. 3 Quicksort Worst-case running time is Θ(n 2 ) on an input array of n numbers. Expected running time is Θ(nlgn). Constants hidden by Θ are quite small. Sorts in place. Probably best sorting algorithm for large input arrays. Maybe.

4. 4 How does Quicksort work? based on “divide and conquer” paradigm (so is merge sort). Divide: Partition (re-arrange) the array A[p..r] into two (possibly empty) sub-arrays A[p.. q-1] and A[q+1.. r] such that each element of A[p.. q-1] is ≤ each element of A[q], which is, in turn, ≤ each element of A[q+1.. r]. Compute the index q as part of this partitioning procedure. Conquer: Sort the two sub-arrays A[p.. q-1] and A[q+1.. r] by recursive calls to quicksort. Combine: No combining needed; the entire array A[p.. r] is now sorted!

5. 5 Quicksort

6. 6 Partition in action

7. 7 Quicksort Running Time, worse-case worst-case occurs when partition yields one subproblem of size n-1 and one of size 0. Assume this “bad split” occurs at each recursive call. partition costs Θ(n). Recursive call to QS on array of size 0 just returns, so T(0) = 1, so we get: T(n) = T(n-1) + T(0) + Θ(n), same as... T(n) = T(n-1) + n just an arithmetic series! So... T(n) = Θ(n 2 ) (worst-case) Under what circumstances do you suppose we get this worst-case behavior?

8. 8 Quicksort, best-case In the most even possible split, PARTITION yields two subproblems each of size no more than n/2, since one is of size floor(n/2), and one is [ceiling(n/2)]-1. We get this recurrence, with some OK sloppiness: T(n) = 2T(n/2) + n (look familiar?) T(n) = O(nlgn) This is asymptotically superior to worst- case, but this ideal scenario is not likely...

9. 9 Quicksort, Average-case suppose the great and awful splits alternate levels in the tree. the running time for QS, when levels alternate between great and awful splits, is just the same as when all levels yield great splits! (with a slightly larger constant hidden by the big-oh notation). So, average case... T(n) = O(nlgn)

10. 10 A lower bound for sorting (Sorting, part 2) We will show that any sorting algorithm based only on comparison of the input values must run in Ω(nlgn) time.

11. 11 Decision Trees Tree of comparisons made by a sorting algorithm. Each comparison reduces the number of possible orderings. Eventually, only one must remain. A decision tree is a “full” (not “complete”) binary tree; each node is a leaf or has degree 2.

12. 12 Q. How many leaves does a decision tree have? A. There is one leaf for each permutation of n elements. There are n! permuatations. Q. What is the height of the tree? A. # of leaves = n! ≤ 2 h Note the height is the worst-case number of comparisons that might be needed.

13. 13 Show we can’t beat nlgn recall n! ≤ 2 h... now take logs lg(n!) ≤ lg(2 h ) lg(n!) ≤ h lg2 lg(n!) ≤ h... just flip it over h ≥ lg(n!) ( lg(n!) = Θ(nlgn) )...Stirling, CLRS p. 55 h = Ω(nlgn) QED In the worst case, Ω(nlgn) comparisons are needed to sort n items.

14. 14 Sorting in Linear Time !!! The Ω(nlgn) bound does not apply if we use info other than comparisons. Like what other info? 1.Use the item as an array index. 2.Examine the digits (or bits) of the item.

15. 15 Counting Sort Good for sorting integers in a narrow range Assume the input numbers (keys) are in the range 0..k Use an auxilliary array C[0..k] to hold the number of items less than i for 0 ≤ i ≤ k if k = O(n), then the running time is Θ(n). Counting sort is stable; it keeps records in their original order.

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17. 17 Counting Sort in action

18. 18 Radix Sort How IBM made its money, using punch card readers for census tabulation in early 1900’s. Card sorters worked on one column at a time. Sort each digit (or field) separately. Start with the least-significant digit. Must use a stable sort. RADIX-SORT(A, d) 1 for i ← 1 to d 2 do use a stable sort to sort array A on digit i

19. 19 Radix Sort in Action

20. 20 Correctness of Radix Sort induction on number of passes base case: low-order digit is sorted correctly inductive hypothesis: show that a stable sort on digit i leaves digits 1...i sorted –if 2 digits in position i are different, ordering by position i is correct, and positions 1.. i-1 are irrelevant –if 2 digits in position i are equal, numbers are already in the right order (by inductive hypotheis). The stable sort on digit i leaves them in the right order. Radix sort must invoke a stable sort.

21. 21 Running Time of Radix Sort use counting sort as the invoked stable sort, if the range of digits is not large if digit range is 1..k, then each pass takes Θ(n+k) time there are d passes, for a total of Θ(d(n+k)) if k = O(n), time is Θ(dn) when d is const, we have Θ(n), linear!