LEC 8-DS ALGO(heaps).pdf
- 1. Data Structures & Algorithm CS-102 Lecture 8 Binary Heaps A priority Queue Data Structures Lecturer: Syeda Nazia Ashraf 1
- 2. 2 Heap • A heap is a complete binary tree that conforms to the heap order. • The heap order property: in a (min) heap, for every node X, the key in the parent is smaller than (or equal to) the key in X. • Or, the parent node has key smaller than or equal to both of its children nodes.
- 3. 3 Heap • Analogously, we can define a max-heap, where the parent has a key larger than the its two children. • Thus the largest key would be in the root.
- 4. 4
- 5. 5 Heap 13 16 21 31 26 24 65 32 68 19 This is a min heap
- 6. 6 Heap Not a heap: (min)heap property violated 13 19 21 31 26 6 65 32 68 16 19>16 6<21
- 9. 9 Inserting into a Heap 13 16 21 31 26 24 65 32 68 19 13 21 16 24 31 19 68 65 26 32 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0 1 2 3 7 6 5 4 8 9 10 This is an existing heap
- 10. 10 Inserting into a Heap insert(14) 13 16 21 31 26 24 65 32 68 19 13 21 16 24 31 19 68 65 26 32 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0 1 2 3 7 6 5 4 8 9 10 14 14 11
- 11. 11 Inserting into a Heap insert(14) 13 16 21 31 26 24 65 32 68 19 13 21 16 24 31 19 68 65 26 32 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0 1 2 3 7 6 5 4 8 9 10 11
- 12. 12 Inserting into a Heap insert(14) 13 16 21 31 26 24 65 32 68 19 13 21 16 24 31 19 68 65 26 32 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0 1 2 3 7 6 5 4 8 9 10 11
- 13. 13 Inserting into a Heap insert(14) 13 16 21 31 26 24 65 32 68 19 13 21 16 24 31 19 68 65 26 32 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0 1 2 3 7 6 5 4 8 9 10 11
- 14. 14 Inserting into a Heap insert(14) 13 16 21 31 26 24 65 32 68 19 13 21 16 24 31 19 68 65 26 32 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0 1 2 3 7 6 5 4 8 9 10 11 14 14
- 15. 15 Inserting into a Heap insert(14) with exchange 13 16 21 31 26 24 65 32 68 19 13 21 16 24 31 19 68 65 26 32 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0 1 2 3 7 6 5 4 8 9 10 14 14 11
- 16. 16 Inserting into a Heap insert(14) with exchange 13 16 21 31 26 24 65 32 68 19 13 21 16 24 31 19 68 65 26 32 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0 1 2 3 7 6 5 4 8 9 10 14 14 11
- 17. 17 Inserting into a Heap insert(14) with exchange 13 16 21 31 26 24 65 32 68 19 13 21 16 24 31 19 68 65 26 32 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0 1 2 3 7 6 5 4 8 9 10 14 14 11
- 18. 18 Inserting into a Heap insert(15) with exchange 13 16 21 31 26 24 65 32 68 19 13 21 16 24 31 19 68 65 26 32 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0 1 2 3 7 6 5 4 8 9 10 14 14 11 15 15 12
- 19. 19 Inserting into a Heap insert(15) with exchange 13 16 21 31 26 24 65 32 68 19 13 21 16 24 31 19 68 65 26 32 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0 1 2 3 7 6 5 4 8 9 10 14 14 11 15 15 12
- 20. 20 Inserting into a Heap insert(15) with exchange 13 16 21 31 26 24 65 32 68 19 13 21 16 24 31 19 68 65 26 32 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0 1 2 3 7 6 5 4 8 9 10 14 14 11 15 15 12
- 21. 21 Inserting into a Heap insert(15) with exchange 13 16 21 31 26 24 65 32 68 19 13 21 16 24 31 19 68 65 26 32 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0 1 2 3 7 6 5 4 8 9 10 14 14 11 15 15 12
- 22. Inserting into a Heap 22 Max Heap To add an item to a heap, we follow the reverse procedure. Place it in the next leaf position and move it up. Again, we require O(h) or O(logn) exchanges.
- 23. 23 P<X N<X
- 24. 24 T<X
- 25. 25 Insert in Min Heap
- 26. 26 Insert in Max Heap
- 27. 27 DeleteMin • Deletion in Max (or Min) Heap always happens at the root to remove the Maximum (or minimum) value. • Deleting it (or removing it) from root causes a hole which needs to be filled. 13 16 21 31 26 24 65 32 68 19 1 2 3 7 6 5 4 8 9 10 14 11
- 32. 32 DeleteMin deleteMin(): heap size is reduced by 1. 14 16 31 26 24 65 32 68 19 1 2 3 7 6 5 4 8 9 10 21
- 33. DeleteMin 33
- 34. 34
- 35. 35 DelMin
- 36. Let's start with this heap. A deletion will remove the T at the root. To work out how we're going to maintain the heap property, use the fact that a complete tree is filled from the left. So that the position which must become empty is the one occupied by the M. Put it in the vacant root position. DeleteMax 36
- 37. This has violated the condition that the root must be greater than each of its children. So interchange the M with the larger of its children. The left subtree has now lost the heap property. So again interchange the M with the larger of its children. This tree is now a heap again, so we're finished. DeleteMax We need to make at most h interchanges of a root of a subtree with one of its children to fully restore the heap property. Thus deletion from a heap is O(h) or O(logn). 37
- 38. 38 BuildHeap • Suppose we are given as input N keys (or items) and we want to build a heap of the keys. • Obviously, this can be done with N successive inserts. • Each call to insert will either take unit time (leaf node) or log2N (if new key percolates all the way up to the root).
- 39. 39 BuildHeap • The worst time for building a heap of N keys could be Nlog2N. • It turns out that we can build a heap in linear time.
- 40. 40 BuildHeap • Suppose we have a method percolateDown(p) which moves down the key in node p downwards. • This is what was happening in deleteMin.
- 41. 41 BuildHeap Initial data (N=15) 65 21 19 26 14 16 68 13 24 15 1 2 3 5 6 7 8 9 10 11 12 13 14 0 31 32 15 4 5 70 12
- 42. 42 BuildHeap Initial data (N=15) 65 19 21 14 24 26 13 15 68 16 65 21 19 26 14 16 68 13 24 15 1 2 3 5 6 7 8 9 10 11 12 13 14 0 1 2 3 7 6 5 4 8 9 10 31 31 11 32 32 12 15 4 5 13 70 14 12 15 5 70 12
- 43. 43 BuildHeap • The general algorithm is to place the N keys in an array and consider it to be an unordered binary tree. • The following algorithm will build a heap out of N keys. for( i = N/2; i > 0; i-- ) percolateDown(i);
- 44. 44 BuildHeap i = 15/2 = 7 65 19 21 14 24 26 13 15 68 16 65 21 19 26 14 16 68 13 24 15 1 2 3 5 6 7 8 9 10 11 12 13 14 0 1 2 3 7 6 5 4 8 9 10 31 31 11 32 32 12 15 4 5 13 70 14 12 15 5 70 12 i i i=n/2 Minimum Heap
- 45. 45 BuildHeap i = 15/2 = 7 65 19 21 14 24 26 13 15 12 16 65 21 19 26 14 16 12 13 24 15 1 2 3 5 6 7 8 9 10 11 12 13 14 0 1 2 3 7 6 5 4 8 9 10 31 31 11 32 32 12 15 4 5 13 70 14 68 15 5 70 68 i i
- 46. 46 BuildHeap i = 6 65 19 21 14 24 26 13 15 12 16 65 21 19 26 14 16 12 13 24 15 1 2 3 5 6 7 8 9 10 11 12 13 14 0 1 2 3 7 6 5 4 8 9 10 31 31 11 32 32 12 15 4 5 13 70 14 68 15 5 70 68 i i
- 47. BuildHeap i = 5 65 5 21 14 24 26 13 15 12 16 65 21 5 26 14 16 12 13 24 15 1 2 3 5 6 7 8 9 10 11 12 13 14 0 1 2 3 7 6 5 4 8 9 10 31 31 11 32 32 12 15 4 19 13 70 14 68 15 19 70 68 i i 47
- 48. BuildHeap i = 4 65 5 14 21 24 26 13 15 12 16 65 14 5 26 21 16 12 13 24 15 1 2 3 5 6 7 8 9 10 11 12 13 14 0 1 2 3 7 6 5 4 8 9 10 31 31 11 32 32 12 15 4 19 13 70 14 68 15 19 70 68 i i 48
- 49. BuildHeap i = 3 65 5 14 21 24 13 26 15 12 16 65 14 5 13 21 16 12 26 24 15 1 2 3 5 6 7 8 9 10 11 12 13 14 0 1 2 3 7 6 5 4 8 9 10 31 31 11 32 32 12 15 4 19 13 70 14 68 15 19 70 68 i i 49
- 50. BuildHeap i = 2 65 16 14 21 24 13 26 15 12 32 65 14 16 13 21 32 12 26 24 15 1 2 3 5 6 7 8 9 10 11 12 13 14 0 1 2 3 7 6 5 4 8 9 10 31 31 11 5 5 12 15 4 19 13 70 14 68 15 19 70 68 i i 50
- 51. BuildHeap i = 1 65 16 14 21 31 24 26 15 12 32 65 14 16 24 21 32 12 26 31 15 1 2 3 5 6 7 8 9 10 11 12 13 14 0 1 2 3 7 6 5 4 8 9 10 13 13 11 5 5 12 15 4 19 13 70 14 68 15 19 70 68 i i 51
- 52. BuildHeap Min heap 5 16 14 21 31 24 26 15 65 32 5 14 16 24 21 32 65 26 31 15 1 2 3 5 6 7 8 9 10 11 12 13 14 0 1 2 3 7 6 5 4 8 9 10 13 13 11 12 12 12 15 4 19 13 70 14 68 15 19 70 68 52
- 53. 53 Build Min Heap i i i i=n/2 i = 5/2 = 2
- 54. 54 Build Max Heap Insert sequence: RPOFESSIRKAL Letter which comes after, it is greater
- 55. 55 Max Heap
- 56. 56
- 57. Other Heap Operations • decreaseKey(p, delta): lowers the value of the key at position ‘p’ by the amount ‘delta’. Since this might violate the heap order, the heap must be reorganized with percolate up (in min heap) or down (in max heap). • increaseKey(p, delta): opposite of decreaseKey. • remove(p): removes the node at position p from the heap. This is done by first decreaseKey(p, delta) and then performing deleteMin(). 57