1、MergeSort,Algorithm : Design & Analysis 6,In the last class,General pattern of divide-and conquer Quicksort: the Strategy Quicksort: the Algorithm Analysis of Quicksort Improvements of the Algorithm,Mergesort,Mergesort Worst Case Analysis of Mergesort Lower Bounds for Sorting by Comparison of Keys W
2、orst Case Average Behavior,MergeSort: the Strategy,Easy division No comparison is done during the division Minimizing the size difference between the divided subproblems Merging two sorted subranges Using Merge,Merging Sorted Arrays,A0,Ak-1,B0,Bm-1,Merge: the Specification,Input: Array A with k elem
3、ents and B with m elements, each in nondecreasing order of their key. Output: C, an array containing n=k+m elements from A and B in nondecreasing order. C is passed in and the algorithm fills it.,Merge: the Recursive Version,merge(A,B,C)if (A is empty)rest of C = rest of Belse if (B is empty)rest of
4、 C = rest of Aelseif (first of A first of B)first of C =first of Amerge(rest of A, B, rest of C)elsefirst of C =first of Bmerge(A, rest of B, rest of C)return,Base cases,Worst Case Complexity of Merge,Observations: After each comparison, one element is inserted into Array C, at least. After entering
5、 Array C, an element will never be compared again After the last comparison, at least two elements have not yet been moved to Array C. So at most n-1 comparisons are done. Worst case is that the last comparison is conducted between Ak-1 and Bm-1 In worst case, n-1 comparisons are done, where n=k+m,O
6、ptimality of Merge,Any algorithm to merge two sorted arrays, each containing k=m=n/2 entries, by comparison of keys, does at least n-1 comparisons in the worst case. Choose keys so that: b0a0b1 a1.biaibi+1,.,bm-1ak-1 Then the algorithm must compare ai with bi for every i in 0,m-1, and must compare a
7、i with bi+1 for every i in 0, m-2, so, there are n-1 comparisons.,Valid for |k-m|1, as well.,Space Complexity of Merge,A algorithm is “in space”, if the extra space it has to use is in (1) Merge is not a algorithm “in space”, since it need enough extra space to store the merged sequence during the m
8、erging process.,Overlapping Arrays for Merge,0,k-1,m-1,0,Before the merge,k+m-1,A,B,Merge from the right,0,k-1,m-1,0,k+m-1,0,k-1,m-1,0,k+m-1,A/C,Finished,Merged,extra space,Partly finished,MergeSort,Input: Array E and indexes first, and last, such that the elements of Ei are defined for firstilast.
9、Output: Efirst,Elast is a sorted rearrangement of the same elements. Procedure void mergeSort(Element E, int first, int last)if (firstlast)int mid=(first+last)/2;mergeSort(E, first, mid);mergeSort(E, mid+1, last);merge(E, first, mid, last)return,Analysis of Mergesort,The recurrence equation for Merg
10、esort W(n)=W(n/2)+W(n/2)+n-1 W(1)=0 Where n=last-first+1, the size of range to be sorted The Master Theorem applies for the equation, so: W(n)(nlogn),k/2 may be k/2 or k/2,Base cases occur at depth lg(n+1)-1 and lg(n+1),Recursion Tree for Mergesort,T(n),n-1,T(n/2),n/2-1,T(n/8),n/8-1,T(n/4),n/4-1,n-1
11、,n-2,n-4,n-8,Level 0,Level 1,Level 2,Level 3,Note: nonrecursive costs on level k is n-2k for all level without basecase node,Non-complete Recursive Tree,B base-case nodes on the second lowest level,n-B base-case nodes No nonbase-case nodes at this depth,2D-1 nodes,Since each nonbase-case node has 2
12、children, there are (n-B)/2 nonbase-case nodes at depth D-1,Example: n=11,Number of Comparison of Mergesort,The maximum depth D of the recursive tree is lg(n+1). Let B base case nodes on depth D-1, and n-B on depth D, (Note: base case node has nonrecursive cost 0). (n-B)/2 nonbase case nodes at dept
13、h D-1, each has nonrecursive cost 1. So:nlg(n)-n+1 number of comparison nlg(n)-0.914n,Decision Tree for Sorting,A example for n=3Decision tree is a 2-tree.(Assuming no same keys) The action of Sort on a particular input corresponds to following on path in its decision tree from the root to a leaf as
14、sociated to the specific output,Internal node,External node,Characteristics of the Decision Tree,For a sequence of n distinct elements, there are n! different permutation, so, the decision tree has at least n! leaves, and exactly n! leaves can be reached from the root. So, for the purpose of lower b
15、ounds evaluation, we use trees with exactly n! leaves. The number of comparison done in the worst case is the height of the tree. The average number of comparison done is the average of the lengths of all paths from the root to a leaf.,Lower Bound for Worst Case,Theorem: Any algorithm to sort n item
16、s by comparisons of keys must do at least lgn!, or approximately nlgn-1.443n, key comparisons in the worst case. Note: Let L=n!, which is the number of leaves, then L2h, where h is the height of the tree, that is h lgL=lgn! For the asymptotic behavior:derived using:,Lower Bound for Average Behavior,
17、Since a decision tree with L leaves is a 2-tree, the average path length from the root to a leaf is .The trees that minimize epl are as balanced as possible. Recall that epl Llg(L). Theorem: The average number of comparison done by an algorithm to sort n items by comparison of keys is at least lg(n!
18、), which is about nlgn-1.443n.,to be proved,Reducing External Path Length,Assuming that h-k1, when calculating epl, h+h+k is replaced by (h-1)+2(k+1). The net change in epl is k-h+10, that is, the epl decreases. So, more balanced 2-tree has smaller epl.,X,Y,X,Y,level,k,level,h,-,1,level,h,level,k,+1
19、,Mergesort Has Optimal Average Performance,We have proved that the average number of comparisons done by an algorithm to sort n items by comparison of keys is at least about nlgn-1.443n The worst complexity of mergesort is in (nlgn) But, the average performance can not be worse the the worst case performance. So, mergesort is optimal as for its average performance.,Home Assignment,pp.212- 4.24 4.25 4.27 4.29 4.30 4.32,
