Fibonacci Heap Group Minus One Second December 6, 2016 Group Minus - - PowerPoint PPT Presentation

Fibonacci Heap

Group Minus One Second December 6, 2016

Outline

Introduction Motivation Performance Goal Fibonacci Heap Structure Rank Insert Delete Min Decrease Key Time Complexity Analysis Amortized Analysis Insert Complexity Delete Min Complexity Decrease Key Complexity

Motivation

I Design a data structure that supports the following operations

12 5 7 3 15 2 9 21 30

Motivation

I Design a data structure that supports the following operations

12 5 7 3 15 2 9 21 30 Insert(10) 10

Motivation

I Design a data structure that supports the following operations

12 5 7 3 15 2 9 21 30 DeleteMin() 10

Motivation

I Design a data structure that supports the following operations

12 5 7 3 15 9 21 30 DecreaseKey(21, 17) 10 17

Performance Goal

I Our performance goals of this data structure

Structure

I Fibonacci heap is essentially a set of heap-ordered trees.

17 24 30 26 46 35 23 7 3 18 52 41 44 39 roots heap-ordered tree

Rank

I The rank of a tree in a Fibonacci heap is the number of

children of the root.

17 24 30 26 46 35 23 7 3 18 52 41 44 39 rank = 1 rank = 2 rank = 0 rank = 0 rank = 3

Insert

I To insert a new value x into Fibonacci Heap H, simply create

a new tree that contains only x and add it to H.

I Example: Insert(21)

17 24 30 26 46 35 23 7 3 18 52 41 44 39 21

Insert

I To insert a new value x into Fibonacci Heap H, simply create

a new tree that contains only x and add it to H.

I Example: Insert(21)

17 24 30 26 46 35 23 7 3 18 52 41 44 39 21

Delete Min

I The minimum must be one of the roots. I But there can be too many(Ω(n)) roots! I Example: DeleteMin() for the heap below

17 24 23 7 3 21 46

Delete Min

I We introduce consolidation, a process of merging trees of the

same rank.

17 24 30 26 46 35 23 7 3 18 52 41 44 39 21 merge

Delete Min

I We introduce consolidation, a process of merging trees of the

same rank.

17 24 30 26 46 35 23 7 3 18 52 41 44 39 21 merge

Delete Min

I We introduce consolidation, a process of merging trees of the

same rank.

17 24 30 26 46 35 23 7 3 18 52 41 44 39 21 merge

Delete Min

I We introduce consolidation, a process of merging trees of the

same rank.

17 24 30 26 46 35 23 7 3 18 52 41 44 39 21 merge

Delete Min

I We introduce consolidation, a process of merging trees of the

same rank.

17 24 30 26 46 35 23 7 3 18 52 41 44 39 21

Delete Min

I After consolidation, we simply remove the minimum value and

leave the remaining trees in the heap.

17 24 30 26 46 35 23 7 18 52 41 44 39 21

Decrease Key

I Case 1: we can finish the operation immediately I Example: decreaseKey(46, 36)

17 24 30 26 46 35 23 7 18 52 41 44 39 21

Decrease Key

I Case 1: we can finish the operation immediately I Example: decreaseKey(46, 36)

17 24 30 26 36 35 23 7 18 52 41 44 39 21

Decrease Key

I Case 2: we have to cut the whole subtree rooted at u I Example: DecreaseKey(26, 16)

17 24 30 26 46 35 23 7 18 52 41 44 39 21

Decrease Key

I Case 2: we have to cut the whole subtree rooted at u I Example: DecreaseKey(26, 16)

17 24 30 16 46 35 23 7 18 52 41 44 39 21

Decrease Key

I Case 2: we have to cut the whole subtree rooted at u I Example: DecreaseKey(26, 16)

17 24 30 16 46 35 23 7 18 52 41 44 39 21

Decrease Key

I But wait! Under this strategy, something undesirable might

happen.

I The example below shows that this strategy can hurt the

performance of DeleteMin operation.

I Example:

Decrease Key with Marks

One node is marked if it has lost one of its children.

Decrease Key with Marks

Case 1: no violation of heap property, finish.

Decrease Key with Marks

Case 2: simply remove the subtree and mark the parent node.

Decrease Key with Marks

Case 2: simply remove the subtree and mark the parent node.

Decrease Key with Marks

Case 2: simply remove the subtree and mark the parent node.

Decrease Key with Marks

Case 3: recursively remove all subtrees whose parent is marked.

Decrease Key with Marks

Case 3: recursively remove all subtrees whose parent is marked.

Decrease Key with Marks

Case 3: recursively remove all subtrees whose parent is marked.

Decrease Key with Marks

Case 3: recursively remove all subtrees whose parent is marked.

Decrease Key with Marks

Case 3: recursively remove all subtrees whose parent is marked.

Amortized Analysis

I We need a new technique to analyze the time complexity of

Fibonacci heap

I Consider the following question:

Emptying the bag costs 1 unit of time.

• peration?

Amortized Analysis

I Observation: A coin that is cleaned must have been dropped

before.

I So why not prepay the cost of clean of each coin as the cost

• f drop?

I Drop: Dropping a single coin onto a table costs 2 units of

time

I Clean: Collecting each coin into a bag costs no time.

Emptying the bag costs 1 unit of time.

I The amortized time complexity of Clean operation is O(1).

Insert Complexity

I In addition to the cost of creating a tree($1), we also add a

merge credit (a prepaid constant cost,$1) to each new tree.

I Since no other operations are needed, each Insert operation

still takes $2 = O(1) time.

17 24 23 7 3 21 46 $1 $1 $1 $1 $1 $1 $1

Delete Min Complexity

I Recall that DeleteMin opeartion consists of two parts:

trees of the same rank i.

I We will make and prove a few claims.

Delete Min Complexity

I Claim 1: Merge operations are free. I Proof: Their costs are paid by merge credits.

17 46 $1 $1 17 46 $1 The other $ 1 is usd to pay for the cost of merge

Delete Min Complexity

I Claim 2: Consolidation operations are of O(k) where k is the

largest rank.

I Proof: For each rank i ∈ [0, k], consolidation calls merge to

generate new trees, which is free. So the only cost is the time consumed while travsering k + 1 ranks.

Delete Min Complexity

I Claim 3: Searching for minimum is of O(k) where k is the

largest rank.

I Proof: After consolidation there are at most k + 1 roots. So

a linear traversal will take O(k) time.

17 24 30 26 46 35 23 7 3 18 52 41 44 39 21

Delete Min Complexity

I Combining all three claims, we can see that the complexity of

DeleteMin is O(k) where k is the largest rank.

I One last question: why k = O(log n)?

a tree of rank i?”

Delete Min Complexity

Some results for different ranks. Let Si be the smallest number of nodes in a tree of rank i. We can see that S0 = 1, S1 = 2, S2 = 3, S3 = 5, S4 = 8, ...

rank = 0

Delete Min Complexity

Some results for different ranks. Let Si be the smallest number of nodes in a tree of rank i. We can see that S0 = 1, S1 = 2, S2 = 3, S3 = 5, S4 = 8, ...

rank = 1

Delete Min Complexity

Some results for different ranks. Let Si be the smallest number of nodes in a tree of rank i. We can see that S0 = 1, S1 = 2, S2 = 3, S3 = 5, S4 = 8, ...

rank = 2

Delete Min Complexity

Some results for different ranks. Let Si be the smallest number of nodes in a tree of rank i. We can see that S0 = 1, S1 = 2, S2 = 3, S3 = 5, S4 = 8, ...

rank = 3

Delete Min Complexity

Some results for different ranks. Let Si be the smallest number of nodes in a tree of rank i. We can see that S0 = 1, S1 = 2, S2 = 3, S3 = 5, S4 = 8, ...

rank = 4

Delete Min Complexity

I We can get a recurrence relation from the above observation:

Si = Si−1 + Si−2, S0 = 1, S1 = 2 which is exactly the Fibonacci sequence, and hence the heap’s name.

I Since Si ≥ ci where c = 1+ √ 5 2

, the number of nodes in each tree grows expoentially with the tree’s rank.

I Therefore, there can be at most O(log n) ranks in a Fibonacci

heap.

Decrease Key Complexity

I By similar argument (using the amortized analysis), we can

prove that the time complexity of DecreaseKey(u, v) is O(1).

I We will omit this part due to time limit. I You can refer to the lecture note on the course website.

Q& A

Thank you