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Priority Queues (Heaps)

Priority Queues (Heaps). Sections 6.1 to 6.5. The Priority Queue ADT. DeleteMin log N time Insert log N time Other operations FindMin Constant time Initialize N time. 2. Applications of Priority Queues. Any event/job management that assign priority to events/jobs

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Priority Queues (Heaps)

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  1. Priority Queues (Heaps) Sections 6.1 to 6.5

  2. The Priority Queue ADT DeleteMin log N time Insert log N time Other operations FindMin Constant time Initialize N time 2

  3. Applications of Priority Queues • Any event/job management that assign priority to events/jobs • In Operating Systems • Scheduling jobs • In Simulators • Scheduling the next event (smallest event time)

  4. Priority Queue Implementation • Implemented as adaptor class around • Linked lists • O(N) worst-case time on either insert() or deleteMin() • Binary Search Trees • O(log(N)) average time on insert() and delete() • Overkill: all elements are sorted • However, we only need the minimum element • Heaps • This is what we’ll study and use to implement Priority Queues • O(logN) worst case for both insertion and delete operations

  5. Partially Ordered Trees • A partially ordered tree (POT) is a tree T such that: • There is an order relation <= defined for the vertices of T • For any vertex p and any child c of p, p <= c • Consequences: • The smallest element in a POT is the root • No conclusion can be drawn about the order of children

  6. Binary Heaps • A binary heap is a partially ordered completebinary tree • The tree is completely filled on all levels except possibly the lowest. • In a more general d-Heap • A parent node can have d children • We simply refer to binary heaps as heaps 0 root 3 2 4 5

  7. 1 2 3 4 5 6 7 Vector Representation of Complete Binary Tree • Storing elements in vector in level-order • Parent of v[k] = v[k/2] • Left child of v[k] = v[2*k] • Right child of v[k] = v[2*k + 1] R root l r ll lr rl rr R l r ll lr rl rr

  8. Heap example • Parent of v[k] = v[k/2] • Left child of v[k] = v[2*k] • Right child of v[k] = v[2*k + 1]

  9. Examples Which one is a heap?

  10. Implementation of Priority Queue (heap)

  11. Insertion Example: insert(14) 14 14 14

  12. Basic Heap Operations: insert(x) • Maintain the complete binary tree property and fix any problem with the partially ordered tree property • Create a leaf at the end • Repeat • Locate parent • if POT not satisfied • Swap with parent • else • Stop • Insert x into its final location

  13. Implementation of insert

  14. deleteMin() example 31

  15. deleteMin() Example (Cont’d) 31 31 31

  16. Basic Heap Operations: deleteMin() • Replace root with the last leaf ( last element in the array representation • This maintains the complete binary tree property but may violate the partially ordered tree property • Repeat • Find the smaller child of the “hole” • If POT not satisfied • Swap hole and smaller child • else • Stop

  17. Implementation of deleteMin()

  18. Implementation of deleteMin()

  19. Constructor • Construct heap from a collection of item • How to? • Naïve methods • Insert() each element • Worst-case time: O(N(logN)) • We show an approach taking O(N) worst-case • Basic idea • First insert all elements into the tree without worrying about POT • Then, adjust the tree to satisfy POT, starting from the bottom

  20. Constructor

  21. Example percolateDown(7) percolateDown(6) percolateDown(5)

  22. percolateDown(4) percolateDown(3) percolateDown(2) percolateDown(1)

  23. Complexity Analysis Consider a tree of height h with 2h-1 nodes Time = 1•(h) + 2•(h-1) + 4•(h-2) + ... + 2h-1•1 = i=1h 2h-i i = 2hi=1h i/2i = 2h O(1) = O(2h) = O(N) Proof for i=1h i/2i = O(1) i/2i ≤ ∫i-1i (x+1)/2x dx i=1h i/2i ≤ i=1∞ i/2i ≤ ∫0∞ (x+1)/2x dx Note: ∫u dv = uv - ∫v du, with dv = 2-x and u = x and ∫2-x dx = -2-x/ln 2 ∫0∞ (x+1)/2x dx = -x 2-x/ln 2|0∞ +1/ln 2∫0∞ 2-x dx + ∫0∞ 2-x dx = -2-x/ln2 2|0∞ - 2-x/ln 2|0∞ = (1 + 1/ln 2)/ln2 = O(1) 23

  24. Alternate Proof Prove i=1h i/2i = O(1) i=0∞ xi = 1/(1-x) when |x| < 1 Differentiating both sides with respect to x we get i=0∞ i xi-1 = 1/(1-x)2 So, i=0∞ i xi = x/(1-x)2 Substituting x = 1/2 above gives i=0∞ i 2-i = 0.5/(1-0.5)2 = 2 = O(1) 24

  25. C++ STL Priority Queues • priority_queue class template • Implements deleteMax instead of deleteMin in default • MaxHeap instead of MinHeap • Template • Item type • container type (default vector) • comparator (default less) • Associative queue operations • Void push(t) • void pop() • T& top() • void clear() • bool empty()

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