1. Algorithmics - Lecture 8-91 LECTURE 8: Divide and conquer

2. Algorithmics - Lecture 8-92 In the previous lecture we saw … … how to analyze recursive algorithms –write a recurrence relation for the running time –solve the recurrence relation by forward or backward substitution … how to solve problems by decrease and conquer –decrease by a constant/variable –decrease by a constant/variable factor … sometimes decrease and conquer lead to more efficient algorithms than brute force techniques

3. Algorithmics - Lecture 8-93 Outline Basic idea of divide and conquer Examples Master theorem Mergesort Quicksort

4. Algorithmics - Lecture 8-94 Basic idea of divide and conquer T he problem is divided in several smaller instances of the same problem –The subproblems must be independent (each one will be solved at most once) –They should be of about the same size These subproblems are solved (by applying the same strategy or directly – if their size is small enough) –If the subproblem size is less than a given value (critical size) it is solved directly, otherwise it is solved recursively If necessary, the solutions obtained for the subproblems are combined

5. Algorithmics - Lecture 8-95 Basic idea of divide and conquer Divide&conquer (n) IF n ELSE FOR i:=1,k DO r i = Divide&conquer(n i ) ENDFOR ENDIF RETURN r

6. Algorithmics - Lecture 8-96 Example 1 Compute the maximum of an array x[1..n] 3 2 7 5 1 6 4 5 n=8, k=2 3 2 7 5 1 6 4 5 3 27 51 64 5 3 76 5 7 6 7 Divide Conquer Combine

7. Algorithmics - Lecture 8-97 Example 1 Algorithm: Maximum(x[left..right]) IF n=1 then RETURN x[left] ELSE m=(left+right) DIV 2 max1=maximum(x[left..m]) max2=maximum(x[m+1..right]) if max1>max2 THEN RETURN max1 ELSE RETURN max2 ENDIF Efficiency analysis Problem size: n Dominant operation: comparison Recurrence relation: 0, n=1 T(n)= T([n/2])+T(n-[n/2])+1, n>1

8. Algorithmics - Lecture 8-98 Example 1 Backward substitution: T(2 m ) = 2T(2 m-1 )+1 T(2 m-1 )=2T(2 m-2 )+1 |* 2 … T(2)=2T(1)+1 |* 2 m-1 T(1)=0 ---------------------------- T(n)=1+…+2 m-1 =2 m -1=n-1 0, n=1 T(n)= T([n/2])+T(n-[n/2])+1, n>1 Particular case: n=2 m 0, n=1 T(n)= 2T(n/2)+1, n>1

9. Algorithmics - Lecture 8-99 Example 1 General case. (a)Proof by complete mathematical induction First step. n=1 =>T(n)=0=n-1 Inductive step. Let us suppose that T(k)=k-1 for all k<n. Then T(n)=[n/2]-1+n-[n/2]-1+1=n-1 Thus T(n) =n-1 => T(n) belongs to Θ(n). 0, n=1 T(n)= T([n/2])+T(n-[n/2])+1, n>1 Particular case: n=2 m => T(n)=n-1

10. Algorithmics - Lecture 8-910 Example 1 General case. (b)Smoothness rule If T(n) belongs to  (f(n)) for n=b m T(n) is eventually nondecreasing (for n>n0 it is nondecreasing) f(n) is smooth (f(cn) belongs to  (f(n)) for any positive constant c) then T(n) belongs to  (f(n)) for all n Remarks. All functions that do not grow too fast are smooth (polynomial and logarithmic) For our example (“maximum” algorithm): T(n) is eventually nondecreasing, f(n)=n is smooth, thus T(n) is from  (n)

11. Algorithmics - Lecture 8-911 Example 2 – binary search Check if a given value, v, is an element of an increasingly sorted array, x[1..n] (x[i]<=x[i+1]) x 1 … x m-1 x m x m+1 … x n x 1 … x m’-1 x m’ x m’+1 … x m-1 x m+1 … x m’-1 x m’ x m’+1 … x n x m =v True v<x m v>x m True x m =v x left …..x right False left>right (empty array)

12. Algorithmics - Lecture 8-912 Example 2 – binary search Recursive variant: binsearch(x[left..right],v) IF left>right THEN RETURN False ELSE m:=(left+right) DIV 2 IF v=x[m] THEN RETURN True ELSE IF v<x[m] THEN RETURN binsearch(x[left..m-1],v) ELSE RETURN binsearch(x[m+1..right],v) ENDIF Remarks: n c =0 k=2 Only one of the two subproblems is solved This is rather a decrease & conquer approach

13. Algorithmics - Lecture 8-913 Example 2 – binary search First iterative variant: binsearch(x[1..n],v) left:=1 right:=n WHILE left<=right DO m:=(left+right) DIV 2 IF v=x[m] THEN RETURN True ELSE IF v<x[m] THEN right:=m-1 ELSE left:=m+1 ENDIF / ENDIF/ ENDWHILE RETURN False Second iterative variant: binsearch(x[1..n],v) left:=1 right:=n WHILE left<right DO m:=(left+right) DIV 2 IF v<=x[m] THEN right:=m ELSE left:=m+1 ENDIF / ENDWHILE IF x[left]=v THEN RETURN True ELSE RETURN False ENDIF

14. Algorithmics - Lecture 8-914 Example 2 – binary search Second iterative variant: binsearch(x[1..n],v) left:=1 right:=n WHILE left<right DO m:=(left+right) DIV 2 IF v<=x[m] THEN right:=m ELSE left:=m+1 ENDIF / ENDWHILE IF x[left]=v THEN RETURN True ELSE RETURN False ENDIF Correctness Precondition: n>=1 Postcondition: “returns True if v is in x[1..n] and False otherwise” Loop invariant: “if v is in x[1..n] then it is in x[left..right]” (i)left=1, right=n => the loop invariant is true (ii)It remains true after the execution of the loop body (iii)when right=left it implies the postcondition

15. Algorithmics - Lecture 8-915 Example 2 – binary search Second iterative variant: binsearch(x[1..n],v) left:=1 right:=n WHILE left<right DO m:=(left+right) DIV 2 IF v<=x[m] THEN right:=m ELSE left:=m+1 ENDIF / ENDWHILE IF x[left]=v THEN RETURN True ELSE RETURN False ENDIF Efficiency: Worst case analysis (n=2 m ) 1 n=1 T(n)= T(n/2)+1 n>1 T(n)=T(n/2)+1 T(n/2)=T(n/4)+1 … T(2)=T(1)+1 T(1)=1 T(n)=lg n+1 O(lg n)

16. Algorithmics - Lecture 8-916 Example 2 – binary search Remarks: By applying the smoothness rule one obtains that this result is true for arbitrary values of n The first iterative variant and the recursive one also belong to O(lg n)

17. Algorithmics - Lecture 8-917 Master theorem Let us consider the following recurrence relation: T 0 n<=n c T(n)= kT(n/m)+T DC (n) n>n c If T DC (n) belongs to  (n d ) (d>=0) then  (n d ) if k<m d T(n) belongs to  (n d lgn) if k=m d  (n lgk/lgm ) if k>m d A similar result holds for O and Ω notations

18. Algorithmics - Lecture 8-918 Master theorem Usefulness: It can be applied in the analysis of divide & conquer algorithms It avoids solving the recurrence relation In most practical applications the running time of the divide and combine steps has a polynomial order of growth It gives the efficiency class but it does not give the constants involved in the running time Example: maximum computation: k=2 (division in two subproblems, both should be solved) m=2 (each subproblem size is almost n/2) d=0 (the divide and combine steps are of constant cost) Since k>m d by applying the third case of the master theorem we obtain that T(n) belongs to  (n lg k/lg m )=  (n)

19. Algorithmics - Lecture 8-919 Master theorem Example 1: maximum computation: k=2 (division in two subproblems, both should be solved) m=2 (each subproblem size is almost n/2) d=0 (the divide and combine steps are of constant cost) Since k>m d by applying the third case of the master theorem we obtain that T(n) belongs to  (n lg k/lg m )=  (n) Example 2: binary search k=1 ( only one subproblem should be solved) m=2 (the subproblem size is almost n/2) d=0 (the divide and combine steps are of constant cost) Since k=m d by applying the second case of the master theorem we obtain that T(n) belongs to O(n d lg(n))=  (lg n)

20. Algorithmics - Lecture 8-920 Efficient sorting Elementary sorting methods belong to O(n 2 ) Idea to increase the efficiency of sorting process: –divide the sequence in contiguous subsequences (usually two) –sort each subsequence –combine the sorted subsequences to obtain the sorted sequence Divide Combine By position Merging Concatenation By value Merge sort Quicksort

21. Algorithmics - Lecture 8-921 Merge sort Basic idea: Divide x[1..n] in two subarrays x[1..[n/2]] and x[[n/2]+1..n] Sort each subarray Merge the elements of x[1..[n/2]] and x[[n/2]+1..n] and construct the sorted temporary array t[1..n]. Transfer the content of the temporary array in x[1..n] Remarks: Critical value: 1 (an array containing one element is already sorted) The critical value can be larger than 1 (e.g.10) and for the particular case one applies a basic sorting algorithm (e.g. insertion sort)

22. Algorithmics - Lecture 8-922 Merge sort

23. Algorithmics - Lecture 8-923 Mergesort Algorithm: mergesort(x[left..right]) IF left<right THEN m:=(left+right) DIV 2 x[left..m]:=mergesort(x[left..m]) x[m+1..right]:=mergesort(x[m+1..right]) x[left..right]:=merge(x[left..m],x[m+1..right]) ENDIF RETURN x[left..right] Remark: the algorithm will be called as mergesort(x[1..n])

24. Algorithmics - Lecture 8-924 Mergesort Merge step: merge (x[left..m],x[m+1..right]) i:=left; j:=m+1; k:=0; // scan simultaneously the arrays // and transfer the smallest element in t WHILE i<=m AND j<=right DO IF x[i]<=x[j] THEN k:=k+1 t[k]:=x[i] i:=i+1 ELSE k:=k+1 t[k]:=x[j] j:=j+1 ENDIF ENDWHILE // transfer the last elements of the first array (if it is the case) WHILE i<=m DO k:=k+1 t[k]:=x[i] i:=i+1 ENDWHILE // transfer the last elements of the second array (if it is the case) WHILE j<=right DO k:=k+1 t[k]:=x[j] j:=j+1 ENDWHILE RETURN t[1..k]

25. Algorithmics - Lecture 8-925 Mergesort The merge step can be used independently of the sorting process Variant of merge based on sentinels: –Merge two sorted arrays a[1..p] and b[1..q] –Add large values to the end of each array a[p+1]= , b[q+1]=  Merge(a[1..p],b[1..q]) a[p+1]:=  ; b[q+1]=  i:=1; j=1; FOR k:=1,p+q DO IF a[i]<=b[j] THEN c[k]:=a[i] i:=i+1 ELSE c[k]:=b[j] j:=j+1 ENDIF / ENDFOR RETURN c[1..p+q] Efficiency analysis of merge step Dominant operation: comparison T(p,q)=p+q In mergesort (p=[n/2], q=n-[n/2]): T(n)<=[n/2]+n-[n/2]=n Thus T(n) belongs to O(n)

26. Algorithmics - Lecture 8-926 Mergesort Efficiency analysis: 0 n=1 T(n)= T([n/2])+T(n-[n/2])+T M (n) n>1 Since k=2, m=2, d=1 (T M (n) is from O(n)) it follows (by the second case of the Master theorem) that T(n) belongs to O(nlgn). In fact T(n) is from  (nlgn) Remark. 1.The main disadvantage of merge sort is the fact that it uses an additional memory space of the array size 2.If in the merge step the comparison is <= then the mergesort is stable

27. Algorithmics - Lecture 8-927 Quicksort Idea: Divide the array x[1..n] in two subarrays x[1..q] and x[q+1..n] such that all elements of x[1..q] are smaller than the elements of x[q+1..n] Sort each subarray Concatenate the sorted subarrays

28. Algorithmics - Lecture 8-928 Quicksort Example 1 3 1 2 4 7 5 8 3 1 27 5 8 1 2 3 5 7 8 1 2 3 4 5 7 8 An element x[q] having the properties: (a)x[q]>=x[i], for all i<q (b)x[q] q is called pivot A pivot is placed on its final position A good pivot divides the array in two subarrays of almost the same size Sometimes –The pivot divides the array in an unbalanced manner –Does not exist a pivot => we must create one by swapping some elements Divide Combine

29. Algorithmics - Lecture 8-929 Quicksort Example 2 3 1 2 7 5 4 8 3 1 27 5 4 8 1 2 3 4 5 7 8 1 2 3 4 5 7 8 A position q having the property: (a)x[i]<=x[i], for all 1<=i<=q and all q+1<=j<=n is called partitioning position A good partitioning position divides the array in two subarrays of almost the same size Sometimes –The partitioning position divides the array in an unbalanced manner –Does not exist such a partitioning position => we must create one by swapping some elements Divide Combine

30. Algorithmics - Lecture 8-930 Quicksort The variant which uses a pivot: quicksort1(x[le..ri]) IF le<ri THEN q:=pivot(x[le..ri]) x[le..q-1]:=quicksort1(x[le..q-1]) x[q+1..ri]:=quicksort1(x[q+1..ri]) ENDIF RETURN x[le..ri] The variant which uses a partitioning position: quicksort2(x[le..ri]) IF le<ri THEN q:=partition(x[le..ri]) x[le..q]:=quicksort2(x[le..q]) x[q+1..ri]:=quicksort2(x[q+1..ri]) ENDIF RETURN x[le..ri]

31. Algorithmics - Lecture 8-931 Quicksort Constructing a pivot: Choose a value from the array (the first one, the last one or a random one) Rearrange the elements of the array such that all elements which are smaller than the pivot value are before the elements which are larger than the pivot value Place the pivot value on its final position (such that all elements on its left are smaller than it and all elements on its right are larger than it) Idea for rearranging the elements: use two pointers one starting from the first element and the other starting from the last element Increase/decrease the pointers until an inversion is found Repair the inversion Continue the process until the pointers cross each other

32. Algorithmics - Lecture 8-932 Quicksort How to construct a pivot 1 7 5 3 8 2 44 1 7 5 3 8 2 4 4 1 2 5 3 8 7 4 Choose the pivot value: 4 (the last one) Place a sentinel on the first position (only for the initial array) i=0, j=7 i=2, j=6 i=3, j=4 i=4, j=3 (the pointers crossed) The pivot is placed on its final position 0 1 2 3 4 5 6 7 4 1 2 3 5 8 7 4 4 1 2 3 4 8 7 5 Pivot value

33. Algorithmics - Lecture 8-933 Quicksort pivot(x[left..right]) v:=x[right] i:=left-1 j:=right WHILE i<j DO REPEAT i:=i+1 UNTIL x[i]>=v REPEAT j:=j-1 UNTIL x[j]<=v IF i<j THEN x[i]↔x[j] ENDIF ENDWHILE x[i] ↔ x[right] RETURN i Remarks: x[right] plays the role of a sentinel at the right At the left margin we can place explicitly a sentinel on x[0] (only for the initial array x[1..n]) The conditions x[i]>=v, x[j]<=v allow to stop the search when the sentinels are encountered. Also they allow obtaining a balanced splitting when the array contains equal elements. At the end of the while loop the pointers satisfy either i=j or i=j+1

34. Algorithmics - Lecture 8-934 Quicksort pivot(x[left..right]) v:=x[right] i:=left-1 j:=right WHILE i<j DO REPEAT i:=i+1 UNTIL x[i]>=v REPEAT j:=j-1 UNTIL x[j]<=v IF i<j THEN x[i] ↔x[j] ENDIF ENDWHILE x[i] ↔x[right] RETURN i Correctness: Loop invariant: If i<j then x[k]<=v for k=left..i x[k]>=v for k=j..right If i>=j then x[k]<=v for k=left..i x[k]>=v for k=j+1..right

35. Algorithmics - Lecture 8-935 Quicksort pivot(x[left..right]) v:=x[right] i:=left-1 j:=right WHILE i<j DO REPEAT i:=i+1 UNTIL x[i]>=v REPEAT j:=j-1 UNTIL x[j]<=v IF i<j THEN x[i] ↔x[j] ENDIF ENDWHILE x[i] ↔x[right] RETURN i Efficiency: Input size: n=right-left+1 Dominant operation: comparison T(n)=n+c, c=0 if i=j and c=1 if i=j+1 Thus T(n) belongs to  (n)

36. Algorithmics - Lecture 8-936 Quicksort Remark: the pivot position does not always divide the array in a balanced manner Balanced manner: the array is divided in two sequences of size almost n/2 If each partition is balanced then the algorithm executes few operations (this is the best case) Unbalanced manner: The array is divided in a subsequence of (n-1) elements the pivot and an empty subsequence If each partition is unbalanced then the algorithm executes much more operations (this is the worst case)

37. Algorithmics - Lecture 8-937 Quicksort Backward substitution: T(n)=T(n-1)+(n+1) T(n-1)=T(n-2)+n … T(2)=T(1)+3 T(1)=0 --------------------- T(n)=(n+1)(n+2)/2-3 Thus in the worst case quicksort belongs to  (n 2 ) Worst case analysis: 0 if n=1 T(n)= T(n-1)+n+1, if n>1

38. Algorithmics - Lecture 8-938 Quicksort By applying the second case of the master theorem (for k=2,m=2,d=1) one obtains that in the best case quicksort is  (nlgn) Best case analysis: 0, if n=1 T(n)= 2T(n/2)+n, if n>1 Thus quicksort belong to Ω(nlgn) and O(n 2 ) An average case analysis should be useful

39. Algorithmics - Lecture 8-939 Quicksort Average case analysis. Hypotheses: Each partitioning step needs at most (n+1) comparisons There are n possible positions for the pivot. Let us suppose that each position has the same probability to be selected (Prob(q)=1/n) If the pivot is on the position q then the number of comparisons satisfies T q (n)=T(q-1)+T(n-q)+(n+1)

40. Algorithmics - Lecture 8-940 Quicksort The average number of comparisons is T a (n)=(T 1 (n)+…+T n (n))/n =((T a (0)+T a (n-1))+(T a (1)+T a (n-2))+…+(T a (n-1)+T a (0)))/n + (n+1) =2(T a (0)+T a (1)+…+T a (n-1))/n+(n+1) Thus n T a (n) = 2(T a (0)+T a (1)+…+T a (n-1))+n(n+1) (n-1)T a (n-1)= 2(T a (0)+T a (1)+…+T a (n-2))+(n-1)n ----------------------------------------------------------------- By computing the difference between the last two equalities: nT a (n)=(n+1)T a (n-1)+2n T a (n)=(n+1)/n T a (n-1)+2

41. Algorithmics - Lecture 8-941 Quicksort Average case analysis. By backward substitution: T a (n) = (n+1)/n T a (n-1)+2 T a (n-1)= n/(n-1) T a (n-2)+2 |*(n+1)/n T a (n-2)= (n-1)/(n-2) T a (n-3)+2 |*(n+1)/(n-1) … T a (2) = 3/2 T a (1)+2 |*(n+1)/3 T a (1) = 0 |*(n+1)/2 ----------------------------------------------------- T a (n) = 2+2(n+1)(1/n+1/(n-1)+…+1/3) ≈ 2(n+1)(ln n-ln 3)+2 In the average case the complexity order is nlgn

42. Algorithmics - Lecture 8-942 Quicksort -variants Another way to construct a pivot pivot(x[left..right]) v:=x[left] i:=left FOR j=left+1,right IF x[j]<=v THEN i=i+1 x[i] ↔ x[j] ENDIF ENDFOR RETURN i Invariant: x[k]<=v for all left<=k<=i x[k]>v for all i<k<=j 3 7 5 2 1 4 8 v=3, i=1,j=2 3 7 5 2 1 4 8 i=2, j=4 3 2 5 7 1 4 8 i=3, j=5 3 2 1 7 5 4 8 i=3, j=8 Plasare pivot: 1 2 3 7 5 4 8 Pivot position: 3 Complexity order of partition: O(n)

43. Algorithmics - Lecture 8-943 Quicksort -variants Finding a partitioning position partition(x[left..right]) v:=x[left] i:=left j:=right+1 WHILE i<j DO REPEAT i:=i+1 UNTIL x[i]>=v REPEAT j:=j-1 UNTIL x[j]<=v IF i<j THEN x[i] ↔x[j] ENDIF ENDWHILE RETURN j 3 7 5 2 1 4 8 v=3 3 7 5 2 1 4 8 i=2, j=5 3 1 5 2 7 4 8 i=3, j=4 3 1 2 5 7 4 8 i=4, j=3 Partitioning position: 3 Complexity order of partition: O(n) Remark: The partition algorithm is to be used in the variant quicksort2

44. Algorithmics - Lecture 8-944 Next lecture will be on … … greedy strategy … and its applications