26. Divide and Conquer Algorithms

1. 26. Divide and Conquer Algorithms Binary Search Merge Sort Mesh Generation Recursion

2. An ordered (sorted) list The Manhattan phone book has 1,000,000+ entries. How is it possible to locate a name by examining just a tiny, tiny fraction of those entries?

3. Key idea of “phone book search”: repeated halving To find the page containing PatReed’s number… while (Phone book is longer than 1 page) Open to the middle page. if “Reed” comes before the first entry, Rip and throw away the 2nd half. else Rip and throw away the 1st half. end end

4. What happens to the phone book length? Original: 3000 pages After 1 rip: 1500 pages After 2 rips: 750 pages After 3 rips: 375 pages After 4 rips: 188 pages After 5 rips: 94 pages : After 12 rips: 1 page

5. Binary Search Repeatedly halving the size of the “search space” is the main idea behind the method of binary search. An item in a sorted array of length n can be located with just log2 n comparisons.

6. n log2(n) 100 7 1000 10 10000 13 Binary Search Repeatedly halving the size of the “search space” is the main idea behind the method of binary search. An item in a sorted array of length n can be located with just log2 n comparisons. “Savings” is significant!

7. 12 15 33 35 42 45 51 62 73 Binary search: target x = 70 1 2 3 4 5 6 7 8 9 10 11 12 v 75 86 98 L: 1 v(Mid) <= x So throw away the left half… Mid: 6 R: 12

8. 12 15 33 35 42 45 51 62 73 Binary search: target x = 70 1 2 3 4 5 6 7 8 9 10 11 12 v 75 86 98 L: 6 x < v(Mid) So throw away the right half… Mid: 9 R: 12

9. 12 15 33 35 42 45 51 62 73 Binary search: target x = 70 1 2 3 4 5 6 7 8 9 10 11 12 v 75 86 98 L: 6 v(Mid) <= x So throw away the left half… Mid: 7 R: 9

10. 12 15 33 35 42 45 51 62 73 Binary search: target x = 70 1 2 3 4 5 6 7 8 9 10 11 12 v 75 86 98 L: 7 v(Mid) <= x So throw away the left half… Mid: 8 R: 9

11. 12 15 33 35 42 45 51 62 73 Binary search: target x = 70 1 2 3 4 5 6 7 8 9 10 11 12 v 75 86 98 L: 8 Done because R-L = 1 Mid: 8 R: 9

12. function L = BinarySearch(a,x) % x is a row n-vector with x(1) < ... < x(n) % where x(1) <= a <= x(n) % L is the index such that x(L) <= a <= x(L+1) L = 1; R = length(x); % x(L) <= a <= x(R) while R-L > 1 mid = floor((L+R)/2); % Note that mid does not equal L or R. if a < x(mid) % x(L) <= a <= x(mid) R = mid; else % x(mid) <= a <== x(R) L = mid; end end

13. Binary search is efficient, but how do we sort a vector in the first place so that we can use binary search? • Many different algorithms out there... • Let’s look at merge sort • An example of the “divide and conquer” approach

14. Merge sort: Motivation • If I have two helpers, I’d… • Give each helper half the array to sort • Then I get back the sorted subarrays and merge them. What if those two helpers each had two sub-helpers? And the sub-helpers each had two sub-sub-helpers? And…

15. M H F M J A A R B H P P R B J F K N C D L Q G N L E E D C G K Q Subdivide the sorting task

16. A F R P P M J F R B M J B H A H C Q K G E N D D C L K N L Q E G Subdivide again

17. H E M G J R F P B A H M B C N G E K L D K Q And again A Q F L P D R C J N

18. And one last time H E M G B K A Q F L P D R C J N

19. E H G M B K Now merge A Q F L D P C R J N H E M G B K A Q F L P D R C J N

20. E H G M N C D L A K E H B J R M G B F P K Q And merge again A Q F L D P C R J N

21. K C C F L H P D L A E N H E M A N Q K G B R J P J D B R F Q G M And again

22. L C A M N E G H E J L F A C P B D F Q K G R H P M J D B K N And one last time Q R

23. E C A J G N L P H M F D K B Done! Q R

24. function y = mergeSort(x) % x is a vector. y is a vector % consisting of the values in x % sorted from smallest to largest. n = length(x); if n==1 y = x; else m = floor(n/2); yL = mergeSortL(x(1:m)); yR = mergeSortR(x(m+1:n)); y = merge(yL,yR); end

25. 12 33 35 45 15 12 42 15 33 55 65 35 75 42 45 55 65 75 The central sub-problem is the merging of two sorted arrays into one single sorted array

26. 12 33 35 45 15 42 55 65 75 Merge x: 1 ix: iy: 1 y: 1 iz: z: ix<=4 and iy<=5: x(ix) <= y(iy) ???

27. 12 33 35 45 15 12 42 55 65 75 Merge x: 1 ix: iy: 1 y: 1 iz: z: ix<=4 and iy<=5: x(ix) <= y(iy) YES

28. 12 33 35 45 15 12 42 55 65 75 Merge x: 2 ix: iy: 1 y: 2 iz: z: ix<=4 and iy<=5: x(ix) <= y(iy) ???

29. 12 33 35 45 15 12 15 42 55 65 75 Merge x: 2 ix: iy: 1 y: 2 iz: z: ix<=4 and iy<=5: x(ix) <= y(iy) NO

30. 12 33 35 45 15 12 15 42 55 65 75 Merge x: 2 ix: iy: 2 y: 3 iz: z: ix<=4 and iy<=5: x(ix) <= y(iy) ???

31. 12 33 35 45 15 12 42 15 33 55 65 75 Merge x: 2 ix: iy: 2 y: 3 iz: z: ix<=4 and iy<=5: x(ix) <= y(iy) YES

32. 12 33 35 45 15 12 42 15 33 55 65 75 Merge x: 3 ix: iy: 2 y: 4 iz: z: ix<=4 and iy<=5: x(ix) <= y(iy) ???

33. 12 33 35 45 15 12 42 15 33 55 65 35 75 Merge x: 3 ix: iy: 2 y: 4 iz: z: ix<=4 and iy<=5: x(ix) <= y(iy) YES

34. 12 33 35 45 15 12 42 15 33 55 65 35 75 Merge x: 4 ix: iy: 2 y: 5 iz: z: ix<=4 and iy<=5: x(ix) <= y(iy) ???

35. 12 33 35 45 15 12 42 15 33 55 65 35 75 42 Merge x: 4 ix: iy: 2 y: 5 iz: z: ix<=4 and iy<=5: x(ix) <= y(iy) NO

36. 12 33 35 45 15 12 42 15 33 55 65 35 75 42 Merge x: 4 ix: iy: 3 y: 5 iz: z: ix<=4 and iy<=5: x(ix) <= y(iy) ???

37. 12 33 35 45 15 12 42 15 33 55 65 35 75 42 45 Merge x: 4 ix: iy: 3 y: 5 iz: z: ix<=4 and iy<=5: x(ix) <= y(iy) YES

38. 12 33 35 45 15 12 42 15 33 55 65 35 75 42 45 Merge x: 5 ix: iy: 3 y: 6 iz: z: ix > 4

39. 12 33 35 45 15 12 42 15 55 33 65 35 75 42 45 55 Merge x: 5 ix: iy: 3 y: 6 iz: z: ix > 4: take y(iy)

40. 12 33 35 45 15 12 42 15 55 33 65 35 75 42 45 55 Merge x: 5 ix: iy: 4 y: 8 iz: z: iy <= 5

41. 12 33 35 45 15 12 42 15 55 33 65 35 75 42 45 55 65 Merge x: 5 ix: iy: 4 y: 8 iz: z: iy <= 5

42. 12 33 35 45 15 12 42 15 55 33 65 35 75 42 45 55 65 Merge x: 5 ix: iy: 5 y: 9 iz: z: iy <= 5

43. 12 33 35 45 15 12 42 15 55 33 65 35 75 42 45 55 65 75 Merge x: 5 ix: iy: 5 y: 9 iz: z: iy <= 5

44. function z = merge(x,y) nx = length(x); ny = length(y); z = zeros(1,nx+ny); ix = 1; iy = 1; iz = 1;

45. function z = merge(x,y) nx = length(x); ny = length(y); z = zeros(1, nx+ny); ix = 1; iy = 1; iz = 1; while ix<=nx && iy<=ny end % Deal with remaining values in x or y

46. function z = merge(x,y) nx = length(x); ny = length(y); z = zeros(1, nx+ny); ix = 1; iy = 1; iz = 1; while ix<=nx && iy<=ny ifx(ix) <= y(iy) z(iz)= x(ix); ix=ix+1; iz=iz+1; else z(iz)= y(iy); iy=iy+1; iz=iz+1; end end % Deal with remaining values in x or y

47. function z = merge(x,y) nx = length(x); ny = length(y); z = zeros(1, nx+ny); ix = 1; iy = 1; iz = 1; while ix<=nx && iy<=ny ifx(ix) <= y(iy) z(iz)= x(ix); ix=ix+1; iz=iz+1; else z(iz)= y(iy); iy=iy+1; iz=iz+1; end end while ix<=nx % copy remaining x-values z(iz)= x(ix); ix=ix+1; iz=iz+1; end while iy<=ny % copy remaining y-values z(iz)= y(iy); iy=iy+1; iz=iz+1; end

48. function y = mergeSort(x) % x is a vector. y is a vector % consisting of the values in x % sorted from smallest to largest. n = length(x); if n==1 y = x; else m = floor(n/2); yL = mergeSortL(x(1:m)); yR = mergeSortR(x(m+1:n)); y = merge(yL,yR); end

49. function y = mergeSortL(x) % x is a vector. y is a vector % consisting of the values in x % sorted from smallest to largest. n = length(x); if n==1 y = x; else m = floor(n/2); yL = mergeSortL_L(x(1:m)); yR = mergeSortL_R(x(m+1:n)); y = merge(yL,yR); end

50. function y = mergeSortL_L(x) % x is a vector. y is a vector % consisting of the values in x % sorted from smallest to largest. n = length(x); if n==1 y = x; else m = floor(n/2); yL = mergeSortL_L_L(x(1:m)); yR = mergeSortL_L_R(x(m+1:n)); y = merge(yL,yR); end There should be just one mergeSort function!