Sorting

Sorting Sorting is important! Things that would be much more difficult without sorting: –finding a telephone number –looking up a word in the dictionary –finding a book in the library –buying a cd –renting a movie –reading a book –reading new

How to sort? Think about sorting a hand of cards

Comparison-based sorting Sort by making comparisons Ordering decision based on comparison of elements Requires that elements are comparable –have a natural order numerical –numbers lexicographic (alphabetical) –characters chronological –dates

Insertion sort Input: n elements in arbitrary order For k = 2, …, n –compare elements in position k and position k-1 if elements are unordered, swap them –compare elements in position k-1 and position k-2 if elements are unordered, swap them –continue until you compare ordered elements or all elements have been swapped once elements are sorted, they remain sorted relative to each other Output: n elements in sorted order

Insertion sort k=18167 First element is sorted k=28167 Compare 1 and Swap 1 and First two elements are sorted k=31867 Compare 6 and Swap 6 and Compare 6 and First three elements are sorted k=41687 Compare 7 and Swap 7 and Compare 7 and All elements are sorted

Worst case behavior At each step k –compare element in position k with previous elements until first k elements are in sorted order In the worst case, we compare each element with all of the previous elements –at step 2 we make 1 comparison –at step 3 we make 2 comparisons  –at step n we make n-1 comparisons Worst case behavior occurs when the array is in reverse sorted order …

Insertion sort – worst case k=18632 First element is sorted k=28632 Compare 6 and Swap 6 and 8 k=36832 Compare 3 and Swap 3 and Compare 3 and Swap 3 and 6 k=43682 Compare 2 and Swap 2 and Compare 2 and Swap 2 and Compare 2 and Swap 2 and All elements are sorted

Best case behavior What if the elements are already sorted? How many comparisons per element will be made by insertion sort?

Insertion sort – best case k= First element is sorted k= Compare 3 and First two elements are sorted k= Compare 5 and First three elements are sorted k= Compare 6 and First four elements are sorted k= Compare 7 and First five elements are sorted k= Compare 9 and All elements are sorted

Comparisons in insertion sort Worst case –element k requires (k-1) comparisons –total number of comparisons: … + (n-1) = ½ (n)(n-1) = ½ (n 2 -n) Best case –elements 2 through n each require one comparison –total number of comparisons: … + 1 = n-1 (n-1) times

The divide-and-conquer approach Insertion sort uses an incremental approach –puts elements in correct place one at a time We can design more efficient sorting algorithms using the divide-and-conquer approach: –Divide the problem into a number of subproblems –Conquer the subproblems by solving them recursively until the subproblems are small enough to solve directly –Merge the solutions for the subproblems into the solution for the original problem

Mergesort Divide-and-conquer sorting algorithm Given a list of n elements in arbitrary order –Divide the list of n elements into two sub-lists of n/2 elements –Conquer each sub-list by sorting it recursively –Merge the solutions to all sub-lists into a single sorted list of n elements

Mergesort divide merge

Merging Idea: –suppose you have an unsorted deck of cards –you split the deck in half, give half to your friend to sort and sort the other half yourself –now you have two half-decks, each sorted –how do you combine them into a single sorted deck?

Merging two sorted lists result of mergesecond listfirst list

Comparisons in mergesort Dividing a list of elements requires no comparisons –just compute midpoint Merging two lists of size m requires at least m and at most 2m - 1 comparisons –one comparison at each step of the merge –elements in result list are not compared again –add at least one element to result at each step Total number of comparisons is the sum of the number of comparisons made at each merge