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### CSE 326Heaps and the Priority Queue ADT

David Kaplan

Dept of Computer Science & Engineering

Autumn 2001

Back to Queues

- Some applications
- ordering CPU jobs
- simulating events
- picking the next search site
- Problems?
- short jobs should go first
- earliest (simulated time) events should go first
- most promising sites should be searched first

CSE 326 Autumn 2001 2

Priority Queue ADT

- Priority Queue operations
- create
- destroy
- insert
- deleteMin
- is_empty
- Priority Queue property: for two elements in the queue, x and y, if x has a lower priority value than y, x will be deleted before y

F(7) E(5)

D(100) A(4)

B(6)

deleteMin

insert

G(9)

C(3)

CSE 326 Autumn 2001 3

Applications of the Priority Q

- Hold jobs for a printer in order of length
- Store packets on network routers in order of urgency
- Simulate events
- Select symbols for compression
- Sort numbers
- Anything greedy

CSE 326 Autumn 2001 4

Naïve Priority Q Data Structures

- Unsorted list:
- insert:
- deleteMin:
- Sorted list:
- insert:
- deleteMin:

CSE 326 Autumn 2001 5

Binary Search TreePriority Q Data Structure (aka BST PQD :-)

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insert:

deleteMin:

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CSE 326 Autumn 2001 6

Heap-order property

parent’s key is less than children’s keys

result: minimum is always at the top

Structure property

complete tree: fringe nodes packed to the left

result: depth is always O(log n); next open location always known

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Binary HeapPriority Q Data StructureHow do we find the minimum?

CSE 326 Autumn 2001 7

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Nifty Storage Trick1

Calculations

- children:
- parent:
- root:
- next free:

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Note: Walking the array in index order gives us level-order traversal!!!

CSE 326 Autumn 2001 8

DeleteMin Code

Object 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:

CSE 326 Autumn 2001 11

Insert Code

void 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:

CSE 326 Autumn 2001 14

Performance of Binary Heap

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

CSE 326 Autumn 2001 15

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”
- Sysadmin may raise priority of a critical task

Must have some (separate) way to find the position in the queue of the object to change (e.g. a hash table)

- No log(N) find (as with BSTs) – why not?

CSE 326 Autumn 2001 16

Other Priority Queue Operations

decreaseKey

- given a pointer to an object in the queue, reduce its priority value

increaseKey

- given a pointer to an object in the queue, increase its priority value

remove

- given a pointer to an object in the queue, remove it

buildHeap

- given a set of items, build a heap

CSE 326 Autumn 2001 17

DecreaseKey, IncreaseKey, Remove

void decreaseKey(int obj) {

assert(size >= obj);

temp = Heap[obj];

newPos = percolateUp(obj, temp);

Heap[newPos] = temp;

}

void remove(int obj) {

assert(size >= obj);

percolateUp(obj,

NEG_INF_VAL);

deleteMin();

}

Note: changeKey functions assume that key value has already been changed!

void increaseKey(int obj) {

assert(size >= obj);

temp = Heap[obj];

newPos = percolateDown(obj, temp);

Heap[newPos] = temp;

}

CSE 326 Autumn 2001 18

BuildHeap (Floyd’s Method)

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

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Thank you, Floyd!

CSE 326 Autumn 2001 19

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)

CSE 326 Autumn 2001 22

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 sucks in a cache line
- with huge data sets, disk accesses dominate

CSE 326 Autumn 2001 23

Solution: d-Heaps

- Each node has d children
- Still representable by array
- Good choices for d:
- optimize performance based on # of inserts/removes
- d = 2k for efficiency (array index calcs)
- fit one set of children in a cache line
- fit one set of children on a memory page/disk block

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What do d-heaps remind us of???

CSE 326 Autumn 2001 24

Merging Heaps

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?

CSE 326 Autumn 2001 25

Solution: Leftist Heaps

Idea

Localize all maintenance work in one small part of the heap

Leftist heap:

- almost all nodes are on the left
- all the merging work is on the right

CSE 326 Autumn 2001 26

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

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another way of looking at it:

NPL is the height of complete subtree rooted at this node

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CSE 326 Autumn 2001 27

Leftist Heap Properties

Heap-order property

- parent’s priority value childrens’ priority values
- minimum element is at the root

Leftist property

- nodes, NPL(left subtree) NPL(right subtree)
- tree is at least as “heavy” on the left as the right

Are leftist trees complete? Balanced? Socialist?

CSE 326 Autumn 2001 28

Leftist Tree Examples

NOT leftist

leftist

leftist

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every subtree of a leftist

tree is leftist, comrade!

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CSE 326 Autumn 2001 29

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Right Path in a Leftist Tree is Short

Theorem:

If right path-length is 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.

Right subtree has a right path of at least r - 1 nodes, so it has at least 2(r-1) - 1 nodes.

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 2(r-1) - 1 nodes. All told then, there are at least:

2(r-1) - 1 + 2(r-1) - 1 + 1 = 2r - 1

Leftist tree with at least n nodes has right path of at most log n nodes

CSE 326 Autumn 2001 30

Merging Two Leftist Heaps

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

merge

a

T1

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merge

a < b

L1

R1

L1

R1

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T2

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L2

R2

L2

R2

CSE 326 Autumn 2001 31

Operations on Leftist Heaps

merge two trees of total size n: O(log n)

insert into 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

CSE 326 Autumn 2001 33

Iterative Leftist Merge

- Downward Pass
- Merge right paths

merge

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CSE 326 Autumn 2001 37

Iterative Leftist Merge (part deux)

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- Upward Pass
- Fix-up Leftist Heap Property

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What do we need to do leftist merge iteratively?

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(One More)Amortized Time

am·or·tize

To write off an expenditure for (office

equipment, for example) by prorating

over a certain period.

time

A nonspatial continuum in which

events occur in apparently

irreversible succession from the past

through the present to the future.

am·or·tized time

Running time limit resulting from writing off expensive

runs of an algorithm over multiple cheap runs of the

algorithm, usually resulting in a lower overall running time

than indicated by the worst possible case.

If M operations take total O(M log N) time, amortized time per operation is O(log N)

CSE 326 Autumn 2001 39

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

CSE 326 Autumn 2001 40

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);

}

}

CSE 326 Autumn 2001 43

Heaps o’ Heaps

CSE 326 Autumn 2001 44

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