170 likes | 294 Views
Priority Queues. And the amazing binary heap Chapter 20 in DS&PS Chapter 6 in DS&AA. Definition. Priority queue has interface: insert, findMin, and deleteMin or: insert, findMax, deleteMax Priority queue is not a search tree Implementation: Binary heap Heap property:
E N D
Priority Queues And the amazing binary heap Chapter 20 in DS&PS Chapter 6 in DS&AA
Definition • Priority queue has interface: • insert, findMin, and deleteMin • or: insert, findMax, deleteMax • Priority queue is not a search tree • Implementation: Binary heap • Heap property: • every node is smaller than its sons. • Note: X is larger than Y means key of X is larger than key for Y.
Representation • array A of objects: start at A[1]. • Growable, if necessary. • Better: bound on number of elements • Recall: • root is at A[1] • father of A[i] is A[i/2] (integer division) • Left son of A[i] is A[2*i] • Right son of A[i] is A[2*i+1] • Efficient in storage and manipulation • Machines like to multiple and divide by 2. • Just a shift.
Insertion Example • Data: 8, 5, 12, 7, 4 • Algorithm: (see example in text) • 1. Insert at end • 2. bubble up I.e. repeat: if father larger, swap • Easier if you draw the tree 8 -> [8] 5 -> [8,5] whoops: swap: [5,8] 12 -> [5,8,12] fine 7 -> [5,8,12,7] swap with father: [5,7,12,8] 4 --> [5,7,12,8,4] --> [5,4,12,8,7] -->[4,5,12,8,7]
Insert • Insert(object) … bubble up A[firstFree] = object int i = firstFree while ( A[i] < A[i/2] & i>0) swap A[i] and A[i/2] i = i/2 … recall i/2 holds father node end while • O(d) = O(log n) • Why is heap property maintained?
Deletion Example • Heap = [4,5,12,7,11,15,18,10] • Idea: 1. put last element into first position 2. fix by swapping downwards (percolate) • Easier with picture • delete min: [-,5,12,7,11,15,18,10] [10,5,12,7,11,15,18] … not a heap [5,10,12,7,11,15,18] … again [5,7,12,10,11,15,18] … done • You should be able to do this.
DeleteMin • Idea: put last element at root and swap down as needed • int i = 1 • repeat // Percolate left = 2*i right = 2*i+1 if ( A[left] < A[i]) swap A[left] and A[i]; i = left else if (A[right] < A[i]) swap A[right] and A[i]; i = right • until i > size/2 why? • O(log n) • Why is heap property maintained?
HeapSort • Simple verions 1. Put the n elements in Heap • time: log n per element, or O(n log n) 2. Pull out min element n times • time: log n per element or O(nlogn) • Yields O(n log n) algorithm • But • faster to use a buildHeap routine • Some authors use heap for best elements • Bounded heap to keep track of top n • top 100 applicants from a pool of 10000+ • here better to use minHeap. Why?
Example of Buildheap • Idea: for each non-leaf in deepest to shallowest order percolate (swap down) • Data: [5,4,7,3,8,2,1] -> [5,4,1,3,8,2,7] -> [5,3,1,4,8,2,7] -> [5,3,1,4,8,2,7] -> [1,3,5,4,8,2,7] -> [1,3,2,4,8,5,7] Done. (easier to see with trees) Easier to do than to analyze
BuildHeap (FixHeap) • Faster HeapSort: replace step 1. • O(N) algorithm for making a heap • Starts with all elements. • Algorithm: for (int i = size/2; i>0; i--) percolateDown(i) • That’s it! • When done, a heap. Why? • How much work? • Bounded by sum of heights of all nodes
Event Simulation Application • Wherever there are multiple servers • Know service time (probability distribution) and arrival time (probability distribution). • Questions: • how long will lines get • expected wait time • each teller modeled as a queue • waiting line for customers modeled as queue • departure times modeled as priority queue
Analysis I • Theorem: A perfect binary tree of height H has 2^(H+1)-1 nodes and the sum of heights of all nodes is N-H-1, i.e. O(N). • Proof: use induction on H • recall: 1+2+4+…+2^k = 2^(k+1)-1 if H=0 then N = 1 and height sum = 0. Check. • Assume for H = k, ie. • Binary tree of height k has sum = 2^(K+1)-1 -k -1. • Consider Binary tree of height k+1. • Heights of all non-leafs is 1 more than Binary tree of height k. • Number of non-leafs = number of nodes in binary tree of size k, ie. 2^(k+1) -1.
Induction continued HeightSum(k+1) = HeightSum(k) +2^(k+1)-1 = 2^(k+1) -k-1 -1 +2^(k+1)-1 = 2^(k+2) - k-2 -1 = N - (k+2) - 1 And that’s the inductive hypothesis. Still not sure: write a (recursive) program to compute Note: height(node) = 1+max(height(node.left),height(node.right))
Analysis II: tree marking • N is size (number of nodes) of tree • for each node n of height h • Darken edges as follows: • go down left one edge, then go right to leaf • darken all edges traversed • darkened edges “counts” height of node • Note: • N-1 edges in tree (all nodes have a father except root) • no edge darkened twice • rightmost path from root to leaf is not darkened (H edges) • sum of darkened = N-1-H = sum of heights!
D-heaps • Generalization of 2-heap or binary heap, but d-arity tree • same representation trick works • e.g. 3-arity tree can be stored in locations: • 0: root • 1,2,3 first level • 4,5,6 for sons of 1; 7, 8 ,9 for sons of 2 etc. • Access time is now: O(d *log( d *N)) • Representation trick useful for storing trees of fixed arity.
Leftist Heaps • Merge is expensive in ordinary Heaps • Define null path length (npl) of a node as the length of the shortest path from the node to a descendent without 2 children. • A node has the null path length property if the nlp of its left son is greater than or equal to the nlp of its right son. • If every node has the null path length property, then it is a leftist heap. • Leftist heaps tend to be left leaning. • Value? A step toward Skew Heaps.
Skew Heaps • Splaying again. What is the goal? • Skew heaps are self-adjusting leftist heaps. • Merge easier to implement. • Worst case run time is O(N) • Worse case run time for M operations is O(M log N) • Amortized time complexity is O( log N). • Merge algo: see text