![Selection Sort
SelectionSort(A,n) // Sort A of size n
for (startpos = 0; startpos < n; startpos++)
// Scan segments A[0]..A[n-1], A[1]..A[n-1], …
// Locate position of minimum element in current segment
minpos = startpos;
for (i = minpos+1; i < n; i++)
if (A[i] < A[minpos])
minpos = i;
// Move minimum element to start of current segment
swap(A,startpos,minpos)](https://image.slidesharecdn.com/nptel-week2-module3-selectionsort-231011084609-0c361d65/85/selection-26-320.jpg)
![Recursive formulation
To sort A[i .. n-1]
Find minimum value in segment and move to A[i]
Apply Selection Sort to A[i+1..n-1]
Base case
Do nothing if i = n-1](https://image.slidesharecdn.com/nptel-week2-module3-selectionsort-231011084609-0c361d65/85/selection-28-320.jpg)
![Selection Sort, recursive
SelectionSort(A,start,n) // Sort A from start to n-1
if (start >= n-1)
return;
// Locate minimum element and move to start of segment
minpos = start;
for (i = start+1; i < n; i++)
if (A[i] < A[minpos])
minpos = i;
swap(A,start,minpos)
// Recursively sort the rest
SelectionSort(A,start+1,n)](https://image.slidesharecdn.com/nptel-week2-module3-selectionsort-231011084609-0c361d65/85/selection-29-320.jpg)
![Alternative calculation
t(n), time to run selection sort on length n
n steps to find minimum and move to position 0
t(n-1) time to run selection sort on A[1] to A[n-1]
Recurrence
t(n) = n + t(n-1)
t(1) = 1
t(n) = n + t(n-1) = n + ((n-1) + t(n-2)) = … =
n + (n-1) + (n-2) + … + 1 = n(n+1)/2 = O(n2)](https://image.slidesharecdn.com/nptel-week2-module3-selectionsort-231011084609-0c361d65/85/selection-30-320.jpg)
selection
- 1. DESIGN AND ANALYSIS OF ALGORITHMS Selection Sort MADHAVAN MUKUND, CHENNAI MATHEMATICAL INSTITUTE http://www.cmi.ac.in/~madhavan NPTEL MOOC,JAN-FEB 2015 Week 2, Module 3
- 2. Sorting Searching for a value Unsorted array — linear scan, O(n) Sorted array — binary search, O(log n) Other advantages of sorting Finding median value: midpoint of sorted list Checking for duplicates Building a frequency table of values
- 3. How to sort? You are a Teaching Assistant for a course The instructor gives you a stack of exam answer papers with marks, ordered randomly Your task is to arrange them in descending order
- 4. Strategy 1 Scan the entire stack and find the paper with minimum marks Move this paper to a new stack Repeat with remaining papers Each time, add next minimum mark paper on top of new stack Eventually, new stack is sorted in descending order
- 5. Strategy 1 … 74 32 89 55 21 64
- 6. Strategy 1 … 21 74 32 89 55 21 64
- 7. Strategy 1 … 21 32 74 32 89 55 21 64
- 8. Strategy 1 … 21 32 55 74 32 89 55 21 64
- 9. Strategy 1 … 21 32 55 64 74 32 89 55 21 64
- 10. Strategy 1 … 21 32 55 64 74 74 32 89 55 21 64
- 11. Strategy 1 … 21 32 55 64 74 89 74 32 89 55 21 64
- 12. Strategy 1 … Selection Sort Select the next element in sorted order Move it into its correct place in the final sorted list
- 13. Selection Sort Avoid using a second list Swap minimum element with value in first position Swap second minimum element to second position …
- 14. Selection Sort 74 32 89 55 21 64
- 15. Selection Sort 74 32 89 55 21 64
- 16. Selection Sort 21 32 89 55 74 64
- 17. Selection Sort 21 32 89 55 74 64
- 18. Selection Sort 21 32 89 55 74 64
- 19. Selection Sort 21 32 89 55 74 64
- 20. Selection Sort 21 32 55 89 74 64
- 21. Selection Sort 21 32 55 89 74 64
- 22. Selection Sort 21 32 55 64 74 89
- 23. Selection Sort 21 32 55 64 74 89
- 24. Selection Sort 21 32 55 64 74 89
- 25. Selection Sort 21 32 55 64 74 89
- 26. Selection Sort SelectionSort(A,n) // Sort A of size n for (startpos = 0; startpos < n; startpos++) // Scan segments A[0]..A[n-1], A[1]..A[n-1], … // Locate position of minimum element in current segment minpos = startpos; for (i = minpos+1; i < n; i++) if (A[i] < A[minpos]) minpos = i; // Move minimum element to start of current segment swap(A,startpos,minpos)
- 27. Analysis of Selection Sort Finding minimum in unsorted segment of length k requires one scan, k steps In each iteration, segment to be scanned reduces by 1 t(n) = n + (n-1) + (n-2) + … + 1 = n(n+1)/2 = O(n2)
- 28. Recursive formulation To sort A[i .. n-1] Find minimum value in segment and move to A[i] Apply Selection Sort to A[i+1..n-1] Base case Do nothing if i = n-1
- 29. Selection Sort, recursive SelectionSort(A,start,n) // Sort A from start to n-1 if (start >= n-1) return; // Locate minimum element and move to start of segment minpos = start; for (i = start+1; i < n; i++) if (A[i] < A[minpos]) minpos = i; swap(A,start,minpos) // Recursively sort the rest SelectionSort(A,start+1,n)
- 30. Alternative calculation t(n), time to run selection sort on length n n steps to find minimum and move to position 0 t(n-1) time to run selection sort on A[1] to A[n-1] Recurrence t(n) = n + t(n-1) t(1) = 1 t(n) = n + t(n-1) = n + ((n-1) + t(n-2)) = … = n + (n-1) + (n-2) + … + 1 = n(n+1)/2 = O(n2)