SORTING

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# SORTING - PowerPoint PPT Presentation

SORTING. Dan Barrish-Flood. heapsort. made file “3-Sorting-Intro-Heapsort.ppt”. Quicksort. Worst-case running time is Θ(n 2 ) on an input array of n numbers. Expected running time is Θ( n lg n ). Constants hidden by Θ are quite small. Sorts in place.

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SORTING
• Dan Barrish-Flood
heapsort
Quicksort
• Worst-case running time is Θ(n2) 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.
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!
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(n2) (worst-case)
• Under what circumstances do you suppose we get this worst-case behavior?
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...
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)
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.
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.
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! ≤ 2h
• Note the height is the worst-case number of comparisons that might be needed.
Show we can’t beat nlgn
• recall n! ≤ 2h ... now take logs
• lg(n!) ≤ lg(2h)
• 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.
Sorting in Linear Time !!!
• The Ω(nlgn) bound does not apply if we use info other than comparisons.
• Like what other info?
• Use the item as an array index.
• Examine the digits (or bits) of the item.
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.
• 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.
• Must use a stable sort.

1 fori← 1 to d

2 do use a stable sort to sort array A on digit i