Sorting
This presentation is the property of its rightful owner.
Sponsored Links
1 / 21

SORTING PowerPoint PPT Presentation


  • 67 Views
  • Uploaded on
  • Presentation posted in: General

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.

Download Presentation

SORTING

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Sorting

SORTING

  • Dan Barrish-Flood


Heapsort

heapsort

  • made file “3-Sorting-Intro-Heapsort.ppt”


Quicksort

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

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!


Quicksort1

Quicksort


Partition in action

Partition in action


Quicksort running time worse case

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

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

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

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

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.


Sorting

  • 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 n lg n

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

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

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.


Counting sort in action

Counting Sort in action


Radix sort

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 fori← 1 to d

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


Radix sort in action

Radix Sort in Action


Correctness of radix sort

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.


Running time of radix sort

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!


  • Login