TDDB56 DALGOPT-D DALG-C Lecture 8 – Sorting (part I)
TDDB56 DALGOPT-D DALG-C Lecture 8 – Sorting (part I)
The Sorting Problem
Sorting:
Input:
– Intro: aspects of sorting, different strategies
• A list L of data items with keys
Output:
• A list L’ of the same data items
in increasing order of keys,
– Insertion Sort, Selection Sort,
– Quick Sort
– Heap Sort,
Caution!
• Don’t over use sorting!
• Do you really need to have it sorted, or will a
dictionary do fine instead of a sorted array?
– Merge Sort
– Theoretical lower bound for comparison-based
sorting,
– Digital Sorting: BucketSort, RadixSort
Jan Maluszynski - HT 2006
8.1
Jan Maluszynski - HT 2006
TDDB56 DALGOPT-D DALG-C Lecture 8 – Sorting (part I)
TDDB56 DALGOPT-D DALG-C Lecture 8 – Sorting (part I)
Aspects of Sorting:
Strategies
8.2
• Insertion sorts:
•
•
•
•
For each new element to add to the sorted set, look for the right
place in that set to put the element...
Linear insertion, Shell sort, ...
Internal / External
In-place / With additional memory
Stable / unstable
Comparison-based / Digital
• Selection sorts:
In each iteration, search the unsorted set for the smallest (largest)
remaining item to add to the end of the sorted set
Straight selection, Heap sort, ...
• Exchange sorts:
Browse back and forth in some pattern, and whenever we are
looking at a pair with wrong relative order, swap them...
Quick sort, Merge sort...
Jan Maluszynski - HT 2006
8.3
Jan Maluszynski - HT 2006
8.4
1
TDDB56 DALGOPT-D DALG-C Lecture 8 – Sorting (part I)
TDDB56 DALGOPT-D DALG-C Lecture 8 – Sorting (part I)
(Linear) insertion sort
Analysis of Insertion Sort
”In each iteration, insert the first item from unsorted part
1
2
3
4
5
Its proper place in the sorted part”
In-place A[0..n-1] !
Data stored in A[0..n-1]
from i =1 to n-1:
– Sorted data in A[0.. i -1]
– Unsorted data in A[i..n-1]
• Scan sorted part for index s
for insertion of the selected item
• Increase i
i
•
•
•
•
•
•
•
s
i
Jan Maluszynski - HT 2006
8.5
TDDB56 DALGOPT-D DALG-C Lecture 8 – Sorting (part I)
TDDB56 DALGOPT-D DALG-C Lecture 8 – Sorting (part I)
(Straight) selection sort
Analysis of selection sort
remaining item to add to the end of the sorted set”
1
2
3
4
5
In-place A[0..n-1] !
from i =1 to n-1:
•
t1: n-1 passes
t2: n-1 passes...
t3: I = worst case no. of iterations in inner loop:
I = 1+2+…+n-1 = n(n-1)/2 thus O(n2)
t4: I passes
t5: n-1 passes
T: t1+t2+t3+t4+t5 thus O(n2) in worst case,
but …. good if file almost sorted
Jan Maluszynski - HT 2006
”In each iteration, search the unsorted set for the smallest
•
Procedure InsertionSort (table A[0..n-1]):
for i from 1 to n-1 do
j ← i; x ← A[i]
while j ≥1 and A[j-1]>x do
A[j] ← A[j-1] ; j ← j-1
A[j] ↔ x
i
– Sorted data in A[0.. i -1]
– Unsorted data in A[i..n-1]
Find index s of
smallest key
Swap places for A[i] and A[s]
i
Procedure Selectionsort (table A[0..n-1]):
for i from 0 to n-2 do
s←i
for j from i+1 to n-1 do
if A[j] < A[s] then s ← j
A[i] ↔ A[s]
• t1: n-1 passes
• t2: n-1 passes...
• t3: I = no. of iterations in inner loop:
I = n-2 + n-3 + n-4 +...+1 = (n-2)(n-1)/2 = (n2-3n+2)/2
• t4: I passes
• t5: n-1 passes
• T: t1+t2+t3+t4+t5 = 3*(n-1)+2*(n2-3n+2)/2
...thus O(n2) ...rather bad!
s
i i is is s
Jan Maluszynski - HT 2006
8.6
8.7
Jan Maluszynski - HT 2006
8.8
2