mergesort, quicksort Oct. 6, 2017 1 Time complexity 2 2 - PowerPoint PPT Presentation

COMP 250 Lecture 13 mergesort, quicksort Oct. 6, 2017 1

Time complexity 𝑃 𝑜 2 𝑃 𝑚𝑝𝑕 2 𝑜 𝑃 𝑜 • convert to binary • List operations: • insertion/selection/ findMax, remove bubble sort • grade school addition • grade school • binary search or subtraction multiplication • ….. • …… • …… 2

Computers perform ~10 9 operations per second. 2 10 ≈ 10 3 2 20 ≈ 10 6 2 30 ≈ 10 9

Computers perform ~10 9 operations per second. 𝑜 2 𝑚𝑝𝑕 2 𝑜 𝑜 2 10 ≈ 10 3 10 10 6 2 20 ≈ 10 6 20 10 12 minutes…hours 2 30 ≈ 10 9 30 10 18 centuries second

Better sorting algorithms ? 𝑃( 𝑜 2 ) 𝑃 𝑜 < ? < 5

Mergesort Given a list, partition it into two halves (1 st & 2 nd ). Sort each half (recursively). Merge the two halves. This turns out to be much faster than the other list sorting algorithms we have seen. 6

8 1 8 1 10 3 10 2 mergesort 3 6 3 3 11 8 11 4 6 10 6 5 1 11 1 6 partition merge 7 7 2 7 16 8 4 16 2 10 2 5 mergesort 5 11 7 5 4 13 13 4 13 16 13 16 7

mergesort(list){ if list.length == 1 return list else{ mid = (list.size - 1) / 2 list1 = list. getElements (0,mid) list2 = list. getElements (mid+1, list.size-1) list1 = mergesort(list1) list2 = mergesort(list2) return merge( list1, list2 ) } } 8

mergesort(list){ if list.length == 1 return list else{ mid = (list.size - 1) / 2 list1 = list.getElements(0,mid) list2 = list.getElements(mid+1, list.size-1) list1 = mergesort(list1) list2 = mergesort(list2) return merge( list1, list2 ) } } 9

mergesort(list){ if list.length == 1 return list else{ mid = (list.size - 1) / 2 list1 = list.getElements(0,mid) list2 = list.getElements(mid+1, list.size-1) list1 = mergesort(list1) list2 = mergesort(list2) return merge( list1, list2 ) } } 10

1 8 8 1 3 10 10 2 mergesort 6 3 3 3 8 11 11 4 10 6 6 5 11 1 1 6 partition merge 7 7 2 7 16 8 4 16 2 10 5 2 mergesort 5 11 7 5 4 13 13 4 13 16 16 13 11

head1 1 3 6 8 10 11 merge 2 head2 4 5 7 13 16 12

1 1 3 head1 6 8 10 11 merge 2 head2 4 5 7 13 16 13

1 1 3 head1 2 6 8 10 11 merge 2 4 head2 5 7 13 16 14

1 1 3 2 6 head1 3 8 10 11 merge 2 4 head2 5 7 13 16 15

1 1 3 2 6 head1 3 8 4 10 11 merge 2 4 5 head2 7 13 16 16

1 1 3 2 6 head1 3 8 4 10 5 11 merge 2 4 5 7 head2 13 16 17

1 1 3 2 6 3 8 head1 4 10 5 11 6 merge 2 4 5 7 head2 13 16 18

1 1 3 2 6 3 8 head1 4 10 5 11 6 merge 7 2 4 5 7 13 head2 16 19

1 1 3 2 ….and so on until 6 3 one list is empty. 8 4 10 5 11 6 merge 7 2 8 4 10 5 11 7 13 head2 16 20

1 1 3 2 Then, copy the 6 3 remaining 8 4 elements. 10 5 11 6 merge 7 2 8 4 10 5 11 7 13 13 16 16 21

merge( list1, list2){ initialize list to be empty while (list1 is not empty) & (list2 is not empty){ if (list1.first < list2.first) list.addlast( list1.removeFirst(list1) ) else list.addlast( list2.removeFirst(list2) ) } while list1 is not empty list.addlast( list1.removeFirst(list1) ) while list2 is not empty list.addlast( list2.removeFirst(list2) ) return list } 22

merge( list1, list2){ initialize list to be empty while (list1 is not empty) & (list2 is not empty){ if (list1.first < list2.first) list.addlast( list1.removeFirst(list1) ) else list.addlast( list2.removeFirst(list2) ) } while list1 is not empty list.addlast( list1.removeFirst(list1) ) while list2 is not empty list.addlast( list2.removeFirst(list2) ) return list } 23

8 mergesort(list){ 10 if list.length == 1 3 return list 11 else{ 6 mid = (list.size - 1) / 2 1 list1 = list.getElements(0,mid) list2 = list.getElements(mid+1, list.size-1) 7 list1 = mergesort(list1) 16 list2 = mergesort(list2) 2 return merge( list1, list2 ) 5 } 4 } 13

8 8 10 10 3 list1 3 11 11 6 6 1 1 7 7 16 16 2 2 list2 5 5 4 4 13 13

8 8 8 10 list1 10 10 3 3 3 11 11 11 6 list2 6 6 1 1 1 7 7 16 16 2 2 5 5 4 4 13 13

8 list1 8 10 8 8 10 3 list2 10 10 3 3 3 11 11 11 6 6 6 1 1 1 7 7 16 16 2 2 5 5 4 4 13 13

8 8 8 list1 8 10 10 10 8 8 10 3 list2 10 10 3 3 3 11 11 11 6 6 6 1 1 1 7 7 16 16 2 2 5 5 4 4 13 13 28

8 8 8 8 10 3 10 10 8 8 10 8 list1 3 10 10 3 10 3 3 11 11 11 6 list2 6 6 1 1 1 7 7 16 16 2 2 5 5 4 4 13 13

8 8 8 8 10 3 10 10 8 8 10 8 list1 3 10 10 3 10 3 3 11 11 6 11 1 11 11 6 11 6 6 6 6 list2 6 1 1 11 1 1 7 7 16 16 2 2 5 5 4 4 13 13

8 8 8 8 10 3 10 10 1 8 8 10 8 3 3 10 10 3 10 6 3 list1 3 11 11 6 11 8 1 11 11 6 11 6 6 10 6 6 6 1 11 1 11 1 1 7 7 16 16 2 2 list2 5 5 4 4 13 13

8 8 8 8 10 3 10 10 1 8 8 10 8 3 3 10 10 3 10 6 3 list1 3 11 11 6 11 8 1 11 11 6 11 6 6 10 6 6 6 1 11 1 11 1 1 7 7 7 7 7 7 2 2 16 16 16 16 16 16 4 7 2 2 2 2 list2 5 16 5 5 5 7 5 5 4 4 4 4 4 13 4 13 5 4 13 5 16 13 13 13 32

8 8 8 8 10 3 10 10 1 8 8 10 1 8 3 3 10 10 3 2 10 6 3 3 3 11 11 6 11 8 1 11 11 4 6 11 6 6 10 6 6 6 5 1 11 1 11 1 1 6 7 7 7 7 7 7 7 2 2 16 16 16 8 16 16 16 4 7 2 2 10 2 2 5 16 5 11 5 5 7 5 5 4 4 13 4 4 4 13 4 13 5 16 4 13 5 16 13 13 13 33

Q: How many operations are required to mergesort a list of size 𝑜 ? 𝑜 𝑜 A: 𝑃( 𝑜 𝑚𝑝𝑕 2 𝑜 ) This will become more clear a few lectures from now when we discuss recurrences. 𝑚𝑝𝑕 2 𝑜 𝑚𝑝𝑕 2 𝑜 34

𝑜 𝑚𝑝𝑕 2 𝑜 is much closer to 𝑜 than to 𝑜 2 𝑜 2 𝑚𝑝𝑕 2 𝑜 𝑜 𝒐 𝒎𝒑𝒉 𝟑 𝒐 2 10 ≈ 10 3 𝟐𝟏 𝟓 10 10 6 2 20 ≈ 10 6 ~𝟐𝟏 𝟖 20 10 12 2 30 ≈ 10 9 ~𝟐𝟏 𝟐𝟏 30 10 18

Computers perform ~10 9 operations per second. 𝑜 2 𝑚𝑝𝑕 2 𝑜 𝑜 𝒐 𝒎𝒑𝒉 𝟑 𝒐 2 10 ≈ 10 3 𝟐𝟏 𝟓 10 10 6 2 20 ≈ 10 6 ~𝟐𝟏 𝟖 20 10 12 2 30 ≈ 10 9 ~𝟐𝟏 𝟐𝟏 30 10 18 milliseconds minutes hours centuries

𝑃( 𝑜 2 ) 𝑃 𝑜 < 𝑃( 𝑜 𝑚𝑝𝑕 2 𝑜) ≪ mergesort bubble sort quicksort selection sort insertion sort 37

Quicksort quicksort(list){ if list.length <= 1 return list else{ pivot = list.removeFirst() // or some other element list1 = list.getElementsLessThan(pivot) list2 = list.getElementsGreaterOrEqual(pivot) list1 = quicksort(list1) list2 = quicksort(list2) } return concatenate( list1, e, list2 ) } 38

Quicksort quicksort(list){ if list.length <= 1 return list else{ pivot = list.removeFirst() // or some other element list1 = list.getElementsLessThan(pivot) list2 = list.getElementsGreaterOrEqual(pivot) list1 = quicksort(list1) list2 = quicksort(list2) } return concatenate( list1, e, list2 ) } 39

1 3 pivot 1 8 2 6 2 quicksort 10 3 1 3 3 4 7 4 11 5 2 5 6 6 5 6 partition 1 7 4 7 concatenate 7 8 16 10 10 10 quicksort 2 11 11 11 5 13 16 13 4 16 13 16 13 40

Quicksort can be done “in place” quicksort( low, high ){ if low > high return else{ pivot = ___; // select index in {low, …, high} partitionIndex = makePartition (low, high, pivot) quicksort(low, partitionIndex - 1) quicksort(partionIndex + 1, high) } } Quicksort partitioning can be done ‘in place’ using a clever swapping and scanning technique. (See web for details, if interested.) 41

pivot 3 1 8 6 2 10 quicksort 1 3 3 7 4 11 2 5 6 partition 5 6 1 4 7 7 8 8 16 10 10 2 11 11 5 quicksort 16 13 4 13 16 13 42

Mergesort vs. Quicksort • Mergesort typically uses an extra list. More space can hurt performance for big lists. • We will discuss worst case performance of quicksort later in the course. • See stackoverflow if you want opinions on which is better. 43