1 CSE 326: Data Structures Priority Queues (Heaps) Lecture 9: Monday, Jan 27, 2003.

1

CSE 326: Data Structures Priority Queues (Heaps)

Lecture 9: Monday, Jan 27, 2003

2

Not Quite Queues

• Consider applications– ordering CPU jobs– searching for the exit in a maze– emergency room admission processing

• Problems?– short jobs should go first– most promising nodes should be searched first– most urgent cases should go first

3

Priority Queue ADT

• Priority Queue operations

• 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)

insert deleteMin

G(9) C(3)

create : heapinsert : heap value heapfindMin : heap valuedeleteMin : heap heapis_empty : heap boolean

4

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• Anything greedy

5

Discrete Event Simulation

• An event is a pair (x,t) where x describes the event and t is time it should occur

• A discrete event simulator (DES) maintains a set S of events which it intends to simulate in time order

repeat { Find and remove (x0,t0) from S such that t0 is minimum; Do whatever x0 says to do; in the process new events (x1,t1)…(xk,tk) may be generated; Insert the new events into S; }

6

Emergency Room Simulation • Patient arrive at time t with injury of criticality C

– If no patients waiting and a free doctor, assign them to doctor and create a future departure event; else put patient in the Criticality priority queue

• Patient departs at time t– If someone in Criticality queue, pull out most critical and

assign to doctor; create a future departure event

timequeue

arrive(t,c)criticality(triage)queueassign

patient todoctor

patientgenerator depart(t)

depart(t)

arrive(t,c)

7

Naïve Priority Queue Data Structures

• Unsorted list:– insert:– findMin:– deleteMin:

• Sorted list:– insert:– findMin:– deleteMin:

8

BST Tree Priority Queue Data Structure

4

121062

115

8

14137 9

•Regular BST:–insert:–findMin:–deleteMin:

•AVL Tree:–insert:–findMin:–deleteMin:

Can we do better?

9

Binary Heap Priority Q Data Structure

201412911

81067

54

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

How do we find the minimum?

10

201412911

81067

54

2

2 4 5 7 6 10 8 11 9 12 14 20121 2 3 4 5 6 7 8 9 10 11 12

1

2 3

4 5 6 7

8 9

10 11 12

Nifty Storage Trick• Calculations:

– child:

– parent:

– root:

– next free:

0

11

201412911

81067

54

2

2 4 5 7 6 10 8 11 9 12 14 20121 2 3 4 5 6 7 8 9 10 11 12

1

2 3

4 5 6 7

8 9

10 11 12

Nifty Storage Trick• Calculations:

– child: left = 2*node right=2*node+1– parent: floor(node/2)

– root: 1

– next free: length+1

0

12

findMin

201412911

81067

54

2

pqueue.findMin()

2

Time = O(1)

13

DeleteMin

201412911

81067

54

202

201412911

81067

54

2

pqueue.deleteMin()

14

Percolate Down

1412911

81067

54

20

1412911

81067

520

4

1412911

810207

56

4

1420911

810127

56

4

15

DeleteMin CodeComparable deleteMin(){ x = A[1]; A[1]=A[size--]; percolateDown(1); return x;}

percolateDown(int hole) { tmp=A[hole]; while (2*hole <= size) { left = 2*hole; right = left + 1; if (right <= size && A[right] < A[left]) target = right; else target = left; if (A[target] < tmp) { A[hole] = A[target]; hole = target; } else break; } A[hole] = tmp;}

Trick to avoid repeatedly copying the value at A[1]

Move downTime = O(log n)(why ?)

16

Insert

201412911

81067

54

2

201412911

81067

54

2

pqueue.insert(3)

3

17

Percolate Up

201412911

81067

54

2

3 201412911

8367

54

2

10

201412911

8567

34

2

10

18

Insert Codevoid insert(Comparable x) { // Efficiency hack: we won’t actually put x // into the heap until we’ve located the position // it goes in. This avoids having to copy it // repeatedly during the percolate up. int hole = ++size; // Percolate up for( ; hole>1 && x < A[hole/2] ; hole = hole/2) A[hole] = A[hole/2]; A[hole] = x;}

Time = O(log n)(why ?)

19

Performance of Binary Heap

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

Binary heapworst case

Binary heap avg case

AVL tree worst case

BST 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)

20

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”

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

21

Other Priority Queue Operations

• decreaseKey – Given the position of an object in the queue, increase

its priority (lower its key). Fix heap property by:• increaseKey

– given the position of an an object in the queue, decrease its priority (increase its key). Fix heap property by:

• remove– given the position of an an object in the queue, remove

it. Do increaseKey to infinity then …

22

BuildHeap

• Task: Given a set of n keys, build a heap all at once

• Approach 1: Repeatedly perform Insert(key)

• Complexity:

23

BuildHeapFloyd’s Method

5 11 3 10 6 9 4 8 1 7 212

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

27184

96103

115

12buildHeap(){

for (i=size/2; i>0; i--)

percolateDown(i);

}

24

Build(this)Heap

67184

92103

115

12

671084

9213

115

12

1171084

9613

25

12

1171084

9653

21

12

25

Finally…

11710812

9654

23

1

26

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…

O(n)

22

4242...

163

82

41

log

1

log

12

nninninnn n

ii

n

ii

27

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

28

4

9654

23

1

8 1012

7

11

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 in a cache line

– fit one set of children on a memory page/disk block

3 7 2 8 5 12 11 10 6 9112

29

New Operation: Merge

Merge(H1,H2): Merge two heaps H1 and H2 of size O(N). – E.g. Combine queues from two different sources to run on one

CPU.

1. Can do O(N) Insert operations: O(N log N) time2. Better: Copy H2 at the end of H1 (assuming array

implementation) and use Floyd’s Method for BuildHeap.Running Time: O(N)

Can we do even better? (i.e. Merge in O(log N) time?)

30

Binomial Queues

• All in O(log n) time• Recursive Definition of Binomial Tree Bk of height k:

– B0 = single root node– Bk = Attach Bk-1 to root of another Bk-1

• Idea: a binomial heap H is a forest of binomial trees:H = B0 B1 B2 ... Bk

where each Bi may be present, or may be empty

insert : heap value heapfindMin : heap valuedeleteMin : heap heapmerge : heap heap heap

31

Building a Binomial Tree• To construct a binomial tree Bk of height k:

1. Take the binomial tree Bk-1 of height k-1

2. Place another copy of Bk-1 one level below the first

3. Attach the root nodes• Binomial tree of height k has exactly 2k nodes (by induction)

B0 B1 B2 B3

32





B0 B1 B2 B3

33





B0 B1 B2 B3

34





B0 B1 B2 B3

35





B0 B1 B2 B3

36





B0 B1 B2 B3

37





B0 B1 B2 B3

Recall: a binomial heap may have any subset of these trees

38

Why Binomial?• Why are these trees called binomial?

– Hint: how many nodes at depth d?

B0 B1 B2 B3

39

Why Binomial?• Why are these trees called binomial?

– Hint: how many nodes at depth d?Number of nodes at different depths d for Bk =

[1], [1 1], [1 2 1], [1 3 3 1], …Binomial coefficients of (a + b)k : k!/((k-d)!d!)

B0 B1 B2 B3

40

Binomial Queue PropertiesSuppose you are given a binomial queue of N nodes1. There is a unique set of binomial trees for N nodes2. What is the maximum number of trees that can be in

an N-node queue?– 1 node 1 tree B0; 2 nodes 1 tree B1; 3 nodes 2 trees

B0 and B1; 7 nodes 3 trees B0, B1 and B2 …

– Trees B0, B1, …, Bk can store up to 20 + 21 + … + 2k = 2k+1 – 1 nodes = N.

– Maximum is when all trees are used. So, solve for (k+1).– Number of trees is log(N+1) = O(log N)

41

Definition of Binomial Queues

3

Binomial Queue = “forest” of heap-ordered binomial trees

1

7

-1

2 1 3

8 11 5

6

5

9 6

7

21

B0 B2 B0 B1 B3

Binomial queue H1

5 elements = 101 base 2 B2 B0

Binomial queue H2

11 elements = 1011 base 2 B3 B1 B0

42

findMin()• In each Bi, the minimum key is at the root• So scan sequentially B1, B2, ..., Bk, compute the

smallest of their keys:• Time: O(log n) (why ?)

B0 B1 B2 B3

43

Binomial Queues: Merge• Main Idea: Merge two binomial queues by merging

individual binomial trees– Since Bk+1 is just two Bk’s attached together, merging trees

is easy• Steps for creating new queue by merging:

1. Start with Bk for smallest k in either queue.

2. If only one Bk, add Bk to new queue and go to next k.

3. Merge two Bk’s to get new Bk+1 by making larger root the child of smaller root. Go to step 2 with k = k + 1.

44

Example: Binomial Queue Merge

31

7

-1

2 1 3

8 11 5

6

5

9 6

7

21

H1: H2:

45


31

7

-1

2 1 3

8 11 5

6

5

9 6

7

21

H1: H2:

46


3

1

7

-1

2 1 3

8 11 5

6

5

9 6

721

H1: H2:

47


3

1

7

-1

2 1 3

8 11 5

6

5

9 6

7

21

H1: H2:

48


3

1

7

-1

2 1 3

8 11 5

6

5

9 6

7

21

H1: H2:

49


3

1

7

-1

2 1 3

8 11 5

6

5

9 6

7

21

H1: H2:

50

Binomial Queues: Merge and Insert

• What is the run time for Merge of two O(N) queues?

• How would you insert a new item into the queue?

51

Binomial Queues: Merge and Insert

• What is the run time for Merge of two O(N) queues?– O(number of trees) = O(log N)

• How would you insert a new item into the queue?– Create a single node queue B0 with new item and

merge with existing queue– Again, O(log N) time

• Example: Insert 1, 2, 3, …,7 into an empty binomial queue

52

Insert 1,2,…,7

1

53

Insert 1,2,…,7

1

2

54

Insert 1,2,…,7

1

2

3

55

Insert 1,2,…,7

1

2

3

4

56

Insert 1,2,…,7

1

2 3

4

57

Insert 1,2,…,7

1

2 3

4

5

58

Insert 1,2,…,7

1

2 3

4

5

6

59

Insert 1,2,…,7

1

2 3

4

5

6

7

60

Binomial Queues: DeleteMin• Steps:

1. Find tree Bk with the smallest root

2. Remove Bk from the queue

3. Delete root of Bk (return this value); You now have a new queue made up of the forest B0, B1, …, Bk-1

4. Merge this queue with remainder of the original (from step 2)

• Run time analysis: Step 1 is O(log N), step 2 and 3 are O(1), and step 4 is O(log N). Total time = O(log N)

• Example: Insert 1, 2, …, 7 into empty queue and DeleteMin

61

Insert 1,2,…,7

1

2 3

4

5

6

7

62

DeleteMin

2 3

4

5

6

7

63

Merge

2 3

4

5

6

7

64

Merge

2 3

4

5

6

7

65

Merge

2

3

4

5

6

7

66

Merge

2

3

4

5

6

7

DONE!

67

Implementation of Binomial Queues• Need to be able to scan through all trees, and given

two binomial queues find trees that are same size– Use array of pointers to root nodes, sorted by size– Since is only of length log(N), don’t have to worry

about cost of copying this array– At each node, keep track of the size of the (sub) tree

rooted at that node• Want to merge by just setting pointers

– Need pointer-based implementation of heaps • DeleteMin requires fast access to all subtrees of root

– Use First-Child/Next-Sibling representation of trees

68

Efficient BuildHeap for Binomial Queues

• Insert one at a time - O(n log n)• Better algorithm:

– Start with each element as a singleton tree– Merge trees of size 1– Merge trees of size 2– Merge trees of size 4

• Complexity:

log log

1 1

log

1

(1 log1) (1 log 2) (1 log 4) ...2 4 8

(1 ) ( )2 2

because 22

n n

i ii i

n

ii

n n n

n i iO n O n

i

1 CSE 326: Data Structures Priority Queues (Heaps) Lecture 9: Monday, Jan 27, 2003.

Documents

Transcript of 1 CSE 326: Data Structures Priority Queues (Heaps) Lecture 9: Monday, Jan 27, 2003.