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SORTING

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- Dan Barrish-Flood

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

- 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.

- 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!

- 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?

- 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...

- 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)

- We will show that any sorting algorithm based only on comparison of the input values must run in Ω(nlgn) time.

- 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.

- 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.

- 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.

- 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.
- 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

- 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.

- 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!