TDDB57 DALG – Lecture 9: Sorting I: Comparison-based sorting.
Page 2
L j if i
Page 4
J. Maluszynski, IDA, Linköpings Universitet, 2004.
Page 1
TDDB57 DALG – Lecture 9: Sorting I: Comparison-based sorting.
The Sorting Problem
Sorting: Overview
Input:
A list L of data items with keys.
L i
j
Output:
A list L of the same items placed in order of keys.
[L/D 11.1]
[L/D 11.2]
[L/D 11.3]
Lecture 9: Comparison-based sorting
Kinds of sorting
Insertion sort
Selection Sort and Heap Sort
Labintroduction heapsort
[L/D 11.4]
Lecture 10: QuickSort
QuickSort, Labintroduction
Labintroduction quicksort
0
0
J. Maluszynski, IDA, Linköpings Universitet, 2004.
j
i
i
n-1
n-1
x
J. Maluszynski, IDA, Linköpings Universitet, 2004.
If the items are much larger than the keys (e.g., database records):
do not move the items themselves,
but sort an array of pointers to them.
[L/D p. 29]
[L/D 11.5]
[L/D 11.6]
Lecture 11: Comparison-based sorting 2; digital sorting
Mergesort
Lower Bound of Comparison-based Sorting
Bucket sort and Radix sort
TDDB57 DALG – Lecture 9: Sorting I: Comparison-based sorting.
J. Maluszynski, IDA, Linköpings Universitet, 2004.
Page 3
TDDB57 DALG – Lecture 9: Sorting I: Comparison-based sorting.
Insertion Sort
Kinds of Sorting
In place sorting algorithm
using a table A 0 n 1
Internal sorting: all data kept in main memory
External sorting: not all data can be kept in main memory.
Sorting in place:
the original data items are rearranged within their positions in memory
Sorting with auxiliary data structure: extra memory is used
At the beginning of each step i,
for 1 i n, the table consists of:
left, sorted part A 0 i 1 ,
marked element x A i ,
right, unsorted part A i 1 n 1 .
The marked element x is inserted
in the sorted part; the rest of the
sorted part is shifted to the right:
j i
while j 1 and A j 1 x
A j
A j 1
j
j 1
x
A j
Sorting by comparison:
the (keys of) data items are compared to establish their relative order.
Digital sorting:
the binary representation of (the key of) a data item is used
to determine its resulting position.
Stable sorting algorithm:
items with equal keys never change their relative position
Unstable sorting algorithm: otherwise
1
x
1
J. Maluszynski, IDA, Linköpings Universitet, 2004.
Page 6
TDDB57 DALG – Lecture 9: Sorting I: Comparison-based sorting.
in place,
using a table A 0 n
min
min
swap
min
swap
J. Maluszynski, IDA, Linköpings Universitet, 2004.
Motivation:
reduce the maximal shifting
distance in each phase.
Improvement:
iterate on k, decreasing hk
InsertionSort the result
InsertionSort only elements
at positions i i h i 2h
Repeat for i 0 1 h 1
0
0
10
J. Maluszynski, IDA, Linköpings Universitet, 2004.
1
1
11
11 10
10 11
Shell Sort Idea
7 6 5 4 3 2
6
5
4
3
7
2
6
5
9
8
2
7
2 6 5 9 8 7
4 5 6 7 8 9
5)
Page 8
O n2
A 11 10 9 8
A0 11
A1
10
A2
9
A3
8
A4
A0 1
A1
0
A2
4
A3
3
A4
A 1 0 4 3
A 0 1 2 3
(Example for h
TDDB57 DALG – Lecture 9: Sorting I: Comparison-based sorting.
Iterative in-place sorting
Sorting by insertion:
Place next unsorted element
in its place in the sorted part.
Insertion Sort, Shell Sort
Sorting by selection:
Find the least unsorted element
attach it to the sorted list.
Selection Sort, Heap Sort
Selection sort
Page 5
1
2
J. Maluszynski, IDA, Linköpings Universitet, 2004.
1
TDDB57 DALG – Lecture 9: Sorting I: Comparison-based sorting.
Insertion Sort Complexity
What is the worst case?
n
Page 7
Consider the number of inversions in the unsorted array
= number of pairs A i A j , for all i j, with A i A j
TDDB57 DALG – Lecture 9: Sorting I: Comparison-based sorting.
Each inversion requires a comparison A j
2
Choosing Shell Sort Increments
1, Shell Sort is InsertionSort.
For h
How many increments?
Empirical studies show good results with:
h1 1
hi 1 3hi 1 while hi 1 n
h
1 4 13 40 121
taken in decreasing order.
Number of comparisons in the worst case:
nn
1
For which data does it work well? Estimation?
TDDB57 DALG – Lecture 9: Sorting I: Comparison-based sorting.
2
2
2
2
0
5
5
5
5
5
1
3
3
3
3
3
2
2
2
4
1
1
1
1
3
1
1
4
4
4
4
4
Page 9
Page 11
1
3
3 4
1 5
3
1
1 5
n
4
2
1 4
3 2
3 2
1 5
3 1
1 5
3
1n
2
3 2
4
2 0
2 0
2 0
1 5
J. Maluszynski, IDA, Linköpings Universitet, 2004.
O n2
2
J. Maluszynski, IDA, Linköpings Universitet, 2004.
4
2 0
4
3 2
2 0
3 2
1 4
4 0
TDDB57 DALG – Lecture 9: Sorting I: Comparison-based sorting.
Heap Sort
Improving Selection Sort:
Page 10
Put data items on heap in O n log n time
How to do this in place?
TDDB57 DALG – Lecture 9: Sorting I: Comparison-based sorting.
1
1
1
1
4
0
2
2
2
5
5
5
1
3
5
3
3
3
3
2
4
4
4
4
2
2
3
Heap to Sorted: In Place
1
1
4
4
2
5
3
5
Page 12
3 2
1 5
3 4
1 5
3 4
1
3 2
1 4
4 0
3 2
2 4
0
5 2
3 4
0
1
3 4
1
3 4
1 5
3 2
Delete minimal data items from heap until empty
Complexity of Selection Sort
n
Unsorted to Heap: In Place
2
3
1
5
Is there a “worst case”?
Heap Sort
4
Number of comparisons:
n
How to improve it?
TDDB57 DALG – Lecture 9: Sorting I: Comparison-based sorting.
J. Maluszynski, IDA, Linköpings Universitet, 2004.
J. Maluszynski, IDA, Linköpings Universitet, 2004.
3 2
4 4
0
2
3 2
5 4
0
5 4
0
TDDB57 DALG – Lecture 9: Sorting I: Comparison-based sorting.
Page 13
J. Maluszynski, IDA, Linköpings Universitet, 2004.
TDDB57 DALG – Lecture 9: Sorting I: Comparison-based sorting.
Lab 5 Introduction
Page 14
Complexity of Heap Sort
Yet another heap, and heap sort
The main idea
Partially ordered trees with reverse order:
parent not smaller than the children
The worst time of HeapSort is
Heapification:
usual ( Lecture 8) heap with reverse order:
heapification in T with root at T 0
maximal key at T 0
J. Maluszynski, IDA, Linköpings Universitet, 2004.
Sorting:
Residual heap kept at the beginning of T
deleteMax exchanges T 0 with the last element of the heap
shi f tdown: you have to implement re-heapification,
elements beyond heap are sorted.
O n log n
Explain!