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Chapter 7: Sorting

(Insertion Sort, Shellsort)

CE 221

Data Structures and Algorithms

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Text: Read Weiss, § 7.1 – 7.4

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Preliminaries• Main memory sorting algorithms

• All algorithms are Interchangeable; an

array containing the N elements will be

passed.

• “<“, “>” (comparison) and “=“ (assignment)

are the only operations allowed on the

input data : comparison-based sorting

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Insertion Sort• One of the simplest sorting algorithms

• Consists of N-1 passes.

• for pass p = 1 to N-1 (0 thru p-1 already known to be sorted)

– elements in position 0 trough p (p+1

elements) are sorted by moving the element

left until a smaller element is encountered.

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Insertion Sort - Algorithm

N*N iterations, hence, time complexity = O(N2) in the worst case.This bound

is tight (input in the reverse order). Number of element comparisons in the

inner loop is p, summing up over all p = 1+2+...+N-1 = Θ(N2). If the input is

sorted O(N).Izmir University of Economics

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A Lower Bound for Simple Sorting

Algorithms

• An inversion in an array of numbers is

any ordered pair (i, j) such that a[i] > a[j].

• In the example; 9 inversions.

• Swapping two adjacent elements that are

out of place removes exactly one inversion

and a sorted array has no inversions.

• The running time of insertion sort O(I+N).

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Average Running Time for

Simple Sorting - I

• Theorem: The average number of inversions

for N distinct elements is N(N-1)/4.

• Proof: It is the sum of the number of inversions

in N! different permutations divided by N!. Each

permutation L has a corresponding permutation

LR which is reversed in sequence. If L has x

inversions, then LR has N(N-1)/2 – x inversions.

As a result ((N(N-1)/2) * (N!/2)) / N! = N(N-1)/4

is the number of inversions for an average list.

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• Theorem: Any algorithm that sorts by exchanging adjacent elements requires Ω(N2) time on average.

• Proof: Initially there exists N(N-1)/4 inversions on the average and each swap removes only one inversion, so Ω(N2) swaps are required.

• This is valid for all types of sorting algorithms (including those undiscovered) that perform only adjacent exchanges.

• Result: For a sorting algorithm to run subquadratic (o(N2)), it must exchange elements that are far apart (eliminating more than just one inversion per exchange).

Average Running Time for

Simple Sorting - II

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Shellsort - I

• Shellsort, invented by Donald Shell, works

by comparing elements that are distant;

the distance decreases as the algorithm

runs until the last phase (diminishing

increment sort)

• Sequence h1, h2, ..., ht is called the

increment sequence. h1 = 1 always. After

a phase, using hk, for every i, a[i] ≤ a[i+hk].

The file is then said to be hk-sorted.

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Shellsort - II• An hk-sorted file that is then hk-1-sorted

remains hk-sorted.

• To hk-sort, for each i in hk,hk+1,...,N-1,

place the element in the correct spot

among i, i-hk, i-2hk. This is equivalent to

performing an insertion sort on hk

independent subarrays.

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Shellsort - III

Increment sequence by Shell: ht=floor(N/2),

hk=floor(hk+1/2) (poor)

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Worst-Case Analysis of Shellsort - I

• Theorem: The worst case running time of

Shellsort, using Shell’s increments, is

Θ(N2).

• Proof: part I: prove Ω(N2)= Why?

smallest N/2 elements goes from position

2i-1 to i during the last pass. Previous

passes all have even increments.

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• Proof: part II: prove O(N2)

• A pass with increment hk consists of hk

insertion sorts of about N/hk elements.

One pass,hence, is O(hk(N/hk)2). Summing

over all passes which is O(N2).

• Shell’s increments: pairs of increments

are not relatively prime.

• Hibbard’s increments: 1, 3, 7,..., 2k-1

Worst-Case Analysis of Shellsort - II

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Worst-Case Analysis of Shellsort - III• Theorem: The worst case running time of Shellsort, using Hibbard’s

increments, is Θ(N3/2).

• Proof: (results from additive number theory)

- for hk>N1/2 use the bound O(N2/hk) // hk=1, 3, 7,..., 2t-1

- hk+2 done, hk+1done, hk now.

- a[p-i] < a[p] if

- but hk+2=2hk+1+1 hence gcd(hk+2, hk+1) = 1

- Thus all can be expressed as such .

- Therefore; innermost for loop executes O(hk) times for each N-hk

positions. This gives a bound of O(Nhk) per pass.

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Homework Assignments

• 7.1, 7.2, 7.3, 7.4

• You are requested to study and solve the

exercises. Note that these are for you to

practice only. You are not to deliver the

results to me.

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