CE 221 Data Structures and Algorithms Chapter 7: Sorting (Insertion Sort, Shellsort) Text: Read Weiss, 7.1 7.4 1
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 2
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. 3
Insertion Sort - Algorithm N*N iterations, hence, time complexity = O(N 2 ) 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(N 2 ). If the input is sorted O(N). 4
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). 5
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 L R which is reversed in sequence. If L has x inversions, then L R 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. 6
Average Running Time for Simple Sorting - II Theorem: Any algorithm that sorts by exchanging adjacent elements requires Ω(N 2 ) time on average. Proof: Initially there exists N(N-1)/4 inversions on the average and each swap removes only one inversion, so Ω(N 2 ) 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(n 2 )), it must exchange elements that are far apart (eliminating more than just one inversion per exchange). 7
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 h 1, h 2,..., h t is called the increment sequence. h 1 = 1 always. After a phase, using h k, for every i, a[i] a[i+h k ]. The file is then said to be h k -sorted. 8
Shellsort - II An h k -sorted file that is then h k-1 -sorted remains h k -sorted. To h k -sort, for each i in h k,h k +1,...,N-1, place the element in the correct spot among i, i-h k, i-2h k. This is equivalent to performing an insertion sort on h k independent subarrays. 9
Shellsort - III Increment sequence by Shell: h t =floor(n/2), h k =floor(h k+1 /2) (poor) 10
Worst-Case Analysis of Shellsort - I Theorem: The worst case running time of Shellsort, using Shell s increments, is Θ(N 2 ). N/ 2 1 1 Proof: part I: prove Ω(N 2 )= i (i ) Why? smallest N/2 elements goes from position 2i-1 to i during the last pass. Previous passes all have even increments. 11
Worst-Case Analysis of Shellsort - II Proof: part II: prove O(N 2 ) A pass with increment h k consists of h k insertion sorts of about N/h k elements. One pass,hence, is O(h k (N/h k ) 2 ). Summing over all passes O 1 N 2 ( / h ) which is O(N 2 ). t i Shell s increments: pairs of increments are not relatively prime. Hibbard s increments: 1, 3, 7,..., 2 k -1 i 12
Worst-Case Analysis of Shellsort - III Theorem: The worst case running time of Shellsort, using Hibbard s increments, is Θ(N 3/2 ). Proof: (results from additive number theory) - for h k >N 1/2 use the bound O(N 2 /h k ) // h k =1, 3, 7,..., 2 t -1 - h k+2 done, h k+1 done, h k now. - a[p-i] < a[p] if i ah k 1 bhk 2 for a, b - but h k+2 =2h k+1 +1 hence gcd(h k+2, h k+1 ) = 1 i( h 1)( h 1) 8h 2 4h k1 k2 k k - Thus all can be expressed as such. - Therefore; innermost for loop executes O(h k ) times for each N-h k positions. This gives a bound of O(Nh k ) per pass. O( t / 2 k 1 Nh O( Nh k t / 2 t N k t / 21 N ) O( h 2 t / 2 2 / h k ) O( N ) O( N 3/ 2 ) t / 2 k 1 h k N Sinceh 2 t / 2 0 t k t / 21 1/ h ( k ) N ) 13
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. 14