CSE 326: Data Structures Lecture #17 Priority Queues

1. CSE 326: Data StructuresLecture #17Priority Queues Alon Halevy Spring Quarter 2001

2. Heap-order property parent’s key is less than children’s keys result: minimum is always at the top Structure property complete tree with fringe nodes packed to the left result: depth is always O(log n); next open location always known 2 4 5 7 6 10 8 11 9 12 14 20 Binary Heap Priority Q Data Structure

3. 2 4 5 7 6 10 8 11 9 12 14 20 DeleteMin pqueue.deleteMin() 2 ? 4 5 7 6 10 8 11 9 12 14 20

4. Percolate Down ? 4 4 5 ? 5 7 6 10 8 7 6 10 8 11 9 12 14 20 11 9 12 14 20 4 4 6 5 6 5 7 ? 10 8 7 12 10 8 11 9 12 14 20 11 9 20 14 20

5. Finally… 4 6 5 7 12 10 8 11 9 20 14

6. DeleteMin Code int percolateDown(int hole, Object val) { while (2*hole <= size) { left = 2*hole; right = left + 1; if (right <= size && Heap[right] < Heap[left]) target = right; else target = left; if (Heap[target] < val) { Heap[hole] = Heap[target]; hole = target; } else break; } return hole; } Object deleteMin() { assert(!isEmpty()); returnVal = Heap[1]; size--; newPos = percolateDown(1, Heap[size+1]); Heap[newPos] = Heap[size + 1]; return returnVal; } runtime:

7. 2 4 5 7 6 10 8 11 9 12 14 20 Insert pqueue.insert(3) 2 4 5 7 6 10 8 11 9 12 14 20 ?

8. Percolate Up 2 2 4 5 4 5 3 7 6 10 8 7 6 ? 8 3 11 9 12 14 20 ? 11 9 12 14 20 10 2 2 3 4 ? 4 3 7 6 5 8 7 6 5 8 11 9 12 14 20 10 11 9 12 14 20 10

9. Insert Code int percolateUp(int hole, Object val) { while (hole > 1 && val < Heap[hole/2]) Heap[hole] = Heap[hole/2]; hole /= 2; } return hole; } void insert(Object o) { assert(!isFull()); size++; newPos = percolateUp(size,o); Heap[newPos] = o; } runtime:

10. Performance of Binary Heap • In practice: binary heaps much simpler to code, lower constant factor overhead

11. Changing Priorities • In many applications the priority of an object in a priority queue may change over time • if a job has been sitting in the printer queue for a long time increase its priority • unix “renice” • Alon’s story • Must have some (separate) way of find the position in the queue of the object to change (e.g. a hash table)

12. Other Priority Queue Operations • decreaseKey • given the position of an object in the queue, reduce its priority value • increaseKey • given the position of an an object in the queue, increase its priority value • remove • given the position of an object in the queue, remove it • buildHeap • given a set of items, build a heap

13. DecreaseKey, IncreaseKey, and Remove void decreaseKey(int pos, int delta){ temp = Heap[pos] - delta; newPos = percolateUp(pos, temp); Heap[newPos] = temp; } void increaseKey(int pos, int delta) { temp = Heap[pos] + delta; newPos = percolateDown(pos, temp); Heap[newPos] = temp; } void remove(int pos) { percolateUp(pos, NEG_INF_VAL); deleteMin(); }

14. BuildHeapFloyd’s Method. Thank you, Floyd. 12 5 11 3 10 6 9 4 8 1 7 2 pretend it’s a heap and fix the heap-order property! 12 Easy worst case bound: Easy average case bound: 5 11 3 10 6 9 4 8 1 7 2

15. Build(this)Heap 12 12 5 11 5 11 3 10 2 9 3 1 2 9 4 8 1 7 6 4 8 10 7 6 12 12 5 2 1 2 3 1 6 9 3 5 6 9 4 8 10 7 11 4 8 10 7 11

16. Finally… 1 3 2 4 5 6 9 12 8 10 7 11 runtime?

17. Complexity of Build Heap • Note: size of a perfect binary tree doubles (+1) with each additional layer • At most n/4 percolate down 1 levelat most n/8 percolate down 2 levelsat most n/16 percolate down 3 levels… O(n)

18. Thinking about Heaps • Observations • finding a child/parent index is a multiply/divide by two • operations jump widely through the heap • each operation looks at only two new nodes • inserts are at least as common as deleteMins • Realities • division and multiplication by powers of two are fast • looking at one new piece of data terrible in a cache line • with huge data sets, disk accesses dominate

19. Solution: d-Heaps 1 • Each node has d children • Still representable by array • Good choices for d: • optimize performance based on # of inserts/removes • choose a power of two for efficiency • fit one set of children on a memory page/disk block 3 7 2 4 8 5 12 11 10 6 9 12 1 3 7 2 4 8 5 12 11 10 6 9 What do d-heaps remind us of???

20. New Operation: Merge Given two heaps, merge them into one heap • first attempt: insert each element of the smaller heap into the larger. runtime: • second attempt: concatenate heaps’ arrays and run buildHeap. runtime: How about O(log n) time?

21. Idea: Hang a New Tree 2 1 + = 5 11 9 4 6 10 13 12 14 10 ? Note: we just gave up the nice structural property on binary heaps! Now, just percolate down! 2 1 5 11 9 4 6 10 13 12 14 10

22. Problem? 1 2 = + 4 Need some other kind of balance condition… 12 15 15 18 ? 1 2 4 12 15 15 18

23. Leftist Heaps • Idea: make it so that all the work you have to do in maintaining a heap is in one small part • Leftist heap: • almost all nodes are on the left • all the merging work is on the right

24. Random Definition:Null Path Length the null path length (npl) of a node is the number of nodes between it and a null in the tree • npl(null) = -1 • npl(leaf) = 0 • npl(single-child node) = 0 2 1 1 0 1 0 0 0 0 0

25. Leftist Heap Properties • Heap-order property • parent’s priority value is  to childrens’ priority values • result: minimum element is at the root • Leftist property • null path length of left subtree is  npl of right subtree • result: tree is at least as “heavy” on the left as the right Are leftist trees complete? Balanced?

26. Leftist tree examples NOT leftist leftist leftist 2 2 0 1 1 1 1 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 every subtree of a leftist tree is leftist, comrade! 0

27. Right Path in a Leftist Tree is Short 2 • If the right path has length at least r, the tree has at least 2r - 1 nodes • Proof by induction Basis: r = 1. Tree has at least one node: 21 - 1 = 1 Inductive step: assume true for r’ < r. The right subtree has a right path of at least r - 1 nodes, so it has at least 2r - 1 - 1 nodes. The left subtree must also have a right path of at least r - 1 (otherwise, there is a null path of r - 3, less than the right subtree). Again, the left has 2r - 1 - 1 nodes. All told then, there are at least: 2r - 1 - 1 + 2r - 1 - 1 + 1 = 2r - 1 • So, a leftist tree with at least n nodes has a right path of at most log n nodes 1 1 1 0 0 0 0 0 0

28. Merging Two Leftist Heaps • merge(T1,T2) returns one leftist heap containing all elements of the two (distinct) leftist heaps T1 and T2 merge T1 a a recursive merge L1 R1 L1 R1 a < b T2 b b L2 R2 L2 R2

29. Merge Continued a a npl(R’) > npl(L1) L1 R’ R’ L1 R’ = Merge(R1, T2) constant work at each merge; recursively traverse RIGHT right path of each tree; total work = O(log n)

30. Operations on Leftist Heaps • merge with two trees of total size n: O(log n) • insert with heap size n: O(log n) • pretend node is a size 1 leftist heap • insert by merging original heap with one node heap • deleteMin with heap size n: O(log n) • remove and return root • merge left and right subtrees merge merge

31. Example merge ? 1 3 5 merge 0 0 0 7 1 ? 10 12 5 5 0 1 merge 14 0 0 0 3 10 12 0 10 12 0 0 0 7 8 0 8 8 0 14 0 8 0 12

32. Sewing Up the Example ? ? 2 3 3 3 0 0 0 ? 7 1 1 7 7 5 5 5 0 0 0 0 0 0 0 0 14 14 14 8 10 0 10 8 10 8 0 0 12 0 12 12 Done?

33. Finally… 2 2 3 3 0 0 1 1 7 7 5 5 0 0 0 0 0 0 14 14 8 8 10 10 0 0 12 12

34. Skew Heaps • Problems with leftist heaps • extra storage for npl • two pass merge (with stack!) • extra complexity/logic to maintain and check npl • Solution: skew heaps • blind adjusting version of leftist heaps • amortized time for merge, insert, and deleteMin is O(log n) • worst case time for all three is O(n) • merge always switches children when fixing right path • iterative method has only one pass What do skew heaps remind us of?

35. Merging Two Skew Heaps merge T1 a a merge L1 R1 L1 R1 a < b T2 b b Notice the old switcheroo! L2 R2 L2 R2

36. Example merge 3 3 5 merge 7 7 5 10 12 5 merge 14 14 10 3 12 10 12 8 7 8 8 3 14 7 5 14 10 8 12

37. Skew Heap Code void merge(heap1, heap2) { case { heap1 == NULL: return heap2; heap2 == NULL: return heap1; heap1.findMin() < heap2.findMin(): temp = heap1.right; heap1.right = heap1.left; heap1.left = merge(heap2, temp); return heap1; otherwise: return merge(heap2, heap1); } }

38. Binary Heaps d-Heaps Binomial Queues Leftist Heaps Skew Heaps Comparing Heaps