Heapsortokkay

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Heapsortokkay

  1. 1. Heapsort
  2. 2. Why we study Heapsort? • It is a well-known, traditional sorting algorithm you will be expected to know • Heapsort is always O(n log n) – Quicksort is usually O(n log n) but in the worst case slows to O(n2 ) – Quicksort is generally faster, but Heapsort is better in time-critical applications • Heapsort have really cool algorithm!
  3. 3. What is a “heap”? • Definitions of heap: 1. A large area of memory from which the programmer can allocate blocks as needed, and deallocate them (for ex:“new and delete”) when no longer needed 2. A balanced, left-justified binary tree in which no node has a value greater than the value in its parent(heap property) • These two definitions have little in common • Heapsort uses the second definition
  4. 4. Binary Tree Typical example for binary tree Remember: The depth of a node is its distance from the root The depth of a tree is the depth of the deepest node Depth,of tree , h=log2(n) no: of nodes, n=2h+1 -1 n=15,h=3 0 3 2 1 root back
  5. 5. Balanced binary trees • A binary tree of depth n is balanced if all the nodes at depths 0 through n-2 have two children Balanced Balanced Not balanced n-2 n-1 n back
  6. 6. Left-justified binary trees • A balanced binary tree is left-justified if: – all the leaves are at the same depth, or – all the leaves at depth n+1 are to the left of all the nodes at depth n (nodes must be filled from left) Left-justified Not left-justified left-justified
  7. 7. The heap property • A node has the heap property if the value in the node is as large as or larger than the values in its children • A binary tree is a heap if all nodes in it have the heap property 12 8 3 Blue node has heap property 12 8 12 Blue node has heap property 12 8 14 Blue node does not have heap property
  8. 8. SiftUp • If a node does not have heap property, you can give it the heap property by exchanging its value with the value of the larger child • This is sometimes called sifting up • Notice that the child may have lost the heap property 14 8 12 Blue node has heap property 12 8 14 Blue node does not have heap property
  9. 9. Constructing a heap • A tree consisting of a single node is automatically a heap • We construct a heap by adding nodes one at a time: – Add the node just to the left of the leftmost node in the deepest level – If the deepest level is full, start a new level • Examples: Add a new node here Add a new node here
  10. 10. Before turn on to sorting Remember Sift up=Trickle down Array Binary tree Binary tree Heapifying Heap Heap must be always balanced & left justified For a heap largest value will be in the root
  11. 11. Sorting • What do heaps have to do with sorting an array? • The answer is: – Because the binary tree is balanced and left justified, it can be represented as an array – All our operations on binary trees can be represented as operations on arrays – To sort: heapify the array; while the array isn’t empty { remove and replace the root; re-heap the new root node; }
  12. 12. 21 15 25 3 5 12 7 19 45 2 9 0 1 2 3 4 5 6 7 8 9 10 Mapping an array into Binary tree • Notice: – The left child of index i is at index 2*i+1 – The right child of index i is at index 2*i+2 – Example: the children of node 3 (3) are 7 (19) and 8 (45) 3 4519 5 2 12 9 7 21 2515 0 1 2 3 4 5 6 7 8 9 10
  13. 13. Heapify the Binary tree 3 4519 5 2 12 9 7 21 2515 21 15 25 3 5 12 7 19 45 2 9 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10
  14. 14. Heapify the Binary tree 3 4519 5 2 12 9 7 21 2515 21 15 25 3 5 12 7 19 45 2 9 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10
  15. 15. Heapify the Binary tree 9 519 3 2 12 45 7 21 2515 21 15 25 3 9 12 7 19 45 2 5 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10
  16. 16. Heapify the Binary tree 9 519 15 2 1245 7 21 25 3 21 15 25 45 9 12 7 19 3 2 5 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10
  17. 17. Heapify the Binary tree 9 5 45 15 2 12 19 7 21 25 3 21 45 25 15 9 12 7 19 3 2 5 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10
  18. 18. Heapify the Binary tree 9 515 21 2 12 45 719 25 3 21 45 25 19 9 12 7 15 3 2 5 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10
  19. 19. We Heapified the Binary tree 9 515 45 2 12 21 719 25 3 45 21 25 19 9 12 7 15 3 2 5 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 largest
  20. 20. step 2: Removing the root • Recall that the largest number is now in the root • Suppose we discard the root: • How can we fix the binary tree so it is once again balanced and left-justified? • Solution: remove the rightmost leaf at the deepest level and use it for the new root 19 315 9 2 12 5 7 45 2521 0 1 2 3 4 5 6 7 8 9 10
  21. 21. Removing and replacing the root • The “root” is the first element in the array • The “rightmost node at the deepest level” is the last element • Swap them... • ...And pretend that the last element in the array no longer exists—that is, the “last index” is 9 (2) 45 21 25 19 9 12 7 15 3 2 5 0 1 2 3 4 5 6 7 8 9 10 5 21 25 19 9 12 7 15 3 2 45 0 1 2 3 4 5 6 7 8 9 10
  22. 22. 9 15 5 2 12 25 719 21 3 0 1 2 3 4 5 6 7 8 9 The reHeap method I • Our tree is balanced and left-justified, but no longer a heap • However, only the root lacks the heap property • We can siftUp() the root • After doing this, one and only one of its children may have lost the heap property 5 21 25 19 9 12 7 15 3 2 45 0 1 2 3 4 5 6 7 8 9 10
  23. 23. The reHeap method II • Now the left child of the root (still the number 5) lacks the heap property 9 15 12 2 25 5 719 21 3 0 1 2 3 4 5 6 7 8 9 25 21 5 19 9 12 7 15 3 2 45 0 1 2 3 4 5 6 7 8 9 10
  24. 24. Heap again….. • Our tree is once again a heap, because every node in it has the heap property • Once again, the largest (or a largest) value is in the root • We can repeat this process until the tree becomes empty • This produces a sequence of values in order largest to smallest 9 15 12 2 25 5 719 21 3 0 1 2 3 4 5 6 7 8 9
  25. 25. reHeap method • Now the left child of the root (still the number 2) lacks the heap property 9 15 21 5 2 719 12 3 0 1 2 3 4 5 6 7 8 2 21 5 19 9 12 7 15 3 25 45 0 1 2 3 4 5 6 7 8 9 10
  26. 26. 9 15 2 519 7 21 12 3 0 1 2 3 4 5 6 7 8 21 2 12 19 9 5 7 15 3 25 45 0 1 2 3 4 5 6 7 8 9 10
  27. 27. 9 19 15 52 7 21 12 3 0 1 2 3 4 5 6 7 8 21 2 12 19 9 5 7 15 3 25 45 0 1 2 3 4 5 6 7 8 9 10
  28. 28. 9 2 19 515 7 21 12 0 1 2 3 4 5 6 7 8 21 19 12 15 9 5 7 2 3 25 45 0 1 2 3 4 5 6 7 8 9 10 Once again we heapified 3
  29. 29. 9 2 19 5 3 7 12 15 0 1 2 3 4 5 6 7 8 3 19 12 15 9 12 7 2 0 1 2 3 4 5 6 7 8 9 10 21 25 45
  30. 30. 9 2 15 5 3 7 12 19 0 1 2 3 4 5 6 7 8 19 3 12 15 9 12 7 2 0 1 2 3 4 5 6 7 8 9 10 21 25 45
  31. 31. 9 2 3 5 15 7 12 19 0 1 2 3 4 5 6 7 8 21 25 45 Once again we heapified 19 15 12 3 9 12 7 2 0 1 2 3 4 5 6 7 8 9 10
  32. 32. 93 15 5 2 7 12 0 1 2 3 4 5 6 2 15 12 3 9 12 7 19 0 1 2 3 4 5 6 7 8 9 10 21 25 45
  33. 33. 3 15 9 5 2 7 12 0 1 2 3 4 5 6 15 2 12 3 9 12 7 19 0 1 2 3 4 5 6 7 8 9 10 21 25 45
  34. 34. 23 5 9 7 12 15 0 1 2 3 4 5 6 Once again we heapified 15 9 12 3 2 5 7 19 0 1 2 3 4 5 6 7 8 9 10 21 25 45
  35. 35. 2 7 5 12 3 9 0 1 2 3 4 5 7 9 12 3 2 5 15 19 0 1 2 3 4 5 6 7 8 9 10 21 25 45
  36. 36. 2 12 5 9 3 7 0 1 2 3 4 5 12 9 7 3 2 5 15 19 0 1 2 3 4 5 6 7 8 9 10 21 25 45 Once again we heapified
  37. 37. 2 5 79 3 0 1 2 3 4 5 9 12 3 7 12 15 19 0 1 2 3 4 5 6 7 8 9 10 21 25 45
  38. 38. 2 5 7 9 3 0 1 2 3 4 9 5 12 3 7 12 15 19 0 1 2 3 4 5 6 7 8 9 10 21 25 45 Once again we heapified
  39. 39. 2 5 7 3 0 1 2 3 2 5 7 3 9 12 15 19 0 1 2 3 4 5 6 7 8 9 10 21 25 45
  40. 40. 7 5 2 3 0 1 2 3 7 5 2 3 9 12 15 19 0 1 2 3 4 5 6 7 8 9 10 21 25 45 Once again we heapified
  41. 41. 3 25 0 1 2 3 5 2 7 9 12 15 19 0 1 2 3 4 5 6 7 8 9 10 21 25 45
  42. 42. 5 3 2 0 1 2 5 3 2 7 9 12 15 19 0 1 2 3 4 5 6 7 8 9 10 21 25 45 Once again we heapified
  43. 43. 2 3 0 1 2 3 5 7 9 12 15 19 0 1 2 3 4 5 6 7 8 9 10 21 25 45
  44. 44. 3 2 0 1 3 2 5 7 9 12 15 19 0 1 2 3 4 5 6 7 8 9 10 21 25 45 Once again we heapified
  45. 45. 2 2 0 3 5 7 9 12 15 19 1 2 3 4 5 6 7 8 9 10 21 25 45 0 2 3 5 7 9 12 15 19 21 25 45 0 1 2 3 4 5 6 7 8 9 10
  46. 46. Heapify (Heapify (AA,, ii)) 1. l ← left [i] 2. r ← right [i] 3. if l ≤ heap-size [A] and A[l] > A[i] 4. then largest ← l 5. else largest ← i 6. if r ≤ heap-size [A] and A[i] > A[largest] 7. then largest ← r 8. if largest ≠ i 9. then exchange A[i] ↔ A[largest] 10. Heapify (A, largest)
  47. 47. Void CreateHeap(int arr[],int size) { int i,father,son; for(i=1;i<size;i++) { int element=arr[i]; son=i; father=floor(son/2); while((son>0)&&(arr[father]<element)) { arr[son]=arr[father]; son=father; father=floor(son/2); } arr[son]=element; } CODE 4 HEAPIFYING
  48. 48. Heap sort make 2 standard heap operations: insertion & root deletion. We delete the maximum,We delete the maximum, Swap it with the last elementSwap it with the last element Pretend that the last element in the array no longer existsPretend that the last element in the array no longer exists ThenThen Sort root of the heap.Sort root of the heap. Heapify the remaining binary tree.Heapify the remaining binary tree. Repeat it until the tree became emptyRepeat it until the tree became empty
  49. 49. void root_deletion( ) { for ( int i = count - 1 ; i > 0 ; i-- ) { int temp = arr[i] ;last value arr[i] = arr[0] ;passing the root value arr[0]=temp; makeheap(i); } } Root deletion program implementation
  50. 50. Analysis I • Here’s how the algorithm starts: heapify the array; • Heapifying the array: we compare each of n nodes – Each node has to be sifted up, possibly as far as the root • Since the binary tree is perfectly balanced, sifting up a single node takes O(log n) time – Since we do this n times, heapifying takes n*O(log n) time, that is, O(n log n) time
  51. 51. Analysis II • Here’s the rest of the algorithm: while the array isn’t empty { remove and replace the root; reheap the new root node; } • We do the while loop n times (actually, n-1 times), because we remove one of the n nodes each time • Removing and replacing the root takes O(1) time • Therefore, the total time is n times however long it takes the reheap method
  52. 52. Analysis III • To reheap the root node, we have to follow one path from the root to a leaf node (and we might stop before we reach a leaf) • The binary tree is perfectly balanced • Therefore, this path is O(log n) long – And we only do O(1) operations at each node – Therefore, reheaping takes O(log n) times • Since we reheap inside a while loop that we do n times, the total time for the while loop is n*O(log n), or O(n log n)
  53. 53. Analysis IV • Here’s the algorithm again: heapify the array; while the array isn’t empty { remove and replace the root; reheap the new root node; } • We have seen that heapifying takes O(n log n) time • The while loop takes O(n log n) time • The total time is therefore O(n log n) + O(n log n) • This is the same as O(n log n) time
  54. 54. The End
  55. 55. Jinsha Jinsha M idhun M idhun Navya Navya Remesha Remesha Shybin Shybin
  56. 56. WelcomeWelcome JinshaJinsha MidhunMidhun NavyaNavya Remesha Remesha ShybinShybin

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