CSE 326: Data StructuresLecture #17

Priority Queues

Alon Halevy

Spring Quarter 2001

Binary Heap Priority Q Data Structure

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• 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

DeleteMin

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pqueue.deleteMin()

Percolate Down

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Finally…

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DeleteMin CodeObject deleteMin() {

assert(!isEmpty());

returnVal = Heap[1];

size--;

newPos =

percolateDown(1,

Heap[size+1]);

Heap[newPos] =

Heap[size + 1];

return returnVal;

}

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;}

runtime:

Insert

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pqueue.insert(3)

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Percolate Up

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Insert Codevoid insert(Object o) {

assert(!isFull());

size++;

newPos =

percolateUp(size,o);

Heap[newPos] = o;

}

int percolateUp(int hole, Object val) { while (hole > 1 && val < Heap[hole/2]) Heap[hole] = Heap[hole/2]; hole /= 2; } return hole;}

runtime:

Performance of Binary Heap

• In practice: binary heaps much simpler to code, lower constant factor overhead

Binary heap

worst case

Binary heap avg case

AVL tree worst case

AVL tree avg case

Insert O(log n) O(1)

percolates 1.6 levels

O(log n) O(log n)

Delete Min

O(log n) O(log n) O(log n) O(log n)

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)

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

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();}

BuildHeapFloyd’s Method. Thank you, Floyd.

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pretend it’s a heap and fix the heap-order property!

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Easy average case bound:

Build(this)Heap

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Finally…

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runtime?

Complexity of Build Heap• Note: size of a perfect binary tree doubles (+1)

with each additional layer• At most n/4 percolate down 1 level

at most n/8 percolate down 2 levelsat most n/16 percolate down 3 levels…

log

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1

1 2 3 ...4 8 16 2

(2)2 2 2

n

ii

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n n n ni

n i nn

O(n)

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

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Solution: d-Heaps

• 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 8 5 12 11 10 6 9112

What do d-heaps remind us of???

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?

Idea: Hang a New Tree

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Now, just percolate down!

Note: we just gave up the nice structural property on binary heaps!

Problem?

Need some other kind of balance condition…

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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

the null path length (npl) of a node is the number of nodes between it and a null in the tree

Random Definition:Null Path Length

• npl(null) = -1• npl(leaf) = 0• npl(single-child node) = 0

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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?

Leftist tree examples

NOT leftist leftist

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every subtree of a leftist tree is leftist, comrade!

Right Path in a Leftist Tree is Short• If the right path has length

at least r, the tree has at least 2r - 1 nodes

• Proof by inductionBasis: 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

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Merging Two Leftist Heaps

• merge(T1,T2) returns one leftist heap containing all elements of the two (distinct) leftist heaps T1 and T2

a

L1 R1

b

L2 R2

mergeT1

T2

a < b

a

L1

recursive merge

b

L2 R2

R1

Merge Continued

a

L1 R’

R’ = Merge(R1, T2)

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R’ L1

npl(R’) > npl(L1)

constant work at each merge; recursively traverse RIGHT right path of each tree; total work = O(log n)

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

Example

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Sewing Up the Example

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Done?

Finally…

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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?

Merging Two Skew Heaps

a

L1 R1

b

L2 R2

mergeT1

T2

a < b

a

L1

merge

b

L2 R2

R1

Notice the old switcheroo!

Example

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Skew Heap Codevoid 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);

}}

Comparing Heaps

• Binary Heaps

• d-Heaps

• Binomial Queues

• Leftist Heaps

• Skew Heaps