Fibonacci Heap Group Paradox December 21, 2016 Contents 1. - - PowerPoint PPT Presentation

Fibonacci Heap

Group Paradox December 21, 2016

Contents

• 1. Introduction
• 2. Operations
• 3. Why it called Fibonacci (the proof)

Priority Queue in Dijkstra

Fibonacci Heap

Operation Binary Heap Fibonacci Heap Find-Min O(1) O(1) Insert O(lg(n)) O(1) Extract-Min O(lg(n)) O(lg(n)) Decrease-Key O(lg(n)) O(1) Merge O(n) O(1)

Structure

root list link a bunch of trees minimum pointer

Insert

Just add them into the root list

Extract-Min

• 1. extract the minimum X
• 2. add the children of X to the root list
• 3. consolidate (make the degree of nodes unique)
• 4. find the new minimum

Amortized Analysis

◮ we pay one coin for every re-link operation. ◮ a coin takes O(1) time ◮ requirement : every root has a coin ◮ deposit a coin on new node when we insert it to meet the

requirement

◮ we can use these coins during the consolidation,

so consolidation costs no new coin

◮ Extract-Min only needs at most O(max degree) coins for

re-linking children and finding new minimum

Amortized Analysis

◮ we pay one coin for every re-link operation. ◮ a coin takes O(1) time ◮ requirement : every root has a coin ◮ deposit a coin on new node when we insert it to meet the

requirement

◮ we can use these coins during the consolidation,

so consolidation costs no new coin

◮ Extract-Min only needs at most O(max degree) coins for

re-linking children and finding new minimum

Amortized Analysis

◮ we pay one coin for every re-link operation. ◮ a coin takes O(1) time ◮ requirement : every root has a coin ◮ deposit a coin on new node when we insert it to meet the

requirement

◮ we can use these coins during the consolidation,

so consolidation costs no new coin

◮ Extract-Min only needs at most O(max degree) coins for

re-linking children and finding new minimum

Amortized Analysis

◮ we pay one coin for every re-link operation. ◮ a coin takes O(1) time ◮ requirement : every root has a coin ◮ deposit a coin on new node when we insert it to meet the

requirement

◮ we can use these coins during the consolidation,

so consolidation costs no new coin

◮ Extract-Min only needs at most O(max degree) coins for

re-linking children and finding new minimum

Amortized Analysis

◮ we pay one coin for every re-link operation. ◮ a coin takes O(1) time ◮ requirement : every root has a coin ◮ deposit a coin on new node when we insert it to meet the

requirement

◮ we can use these coins during the consolidation,

so consolidation costs no new coin

◮ Extract-Min only needs at most O(max degree) coins for

re-linking children and finding new minimum

Amortized Analysis

◮ we pay one coin for every re-link operation. ◮ a coin takes O(1) time ◮ requirement : every root has a coin ◮ deposit a coin on new node when we insert it to meet the

requirement

◮ we can use these coins during the consolidation,

so consolidation costs no new coin

◮ Extract-Min only needs at most O(max degree) coins for

re-linking children and finding new minimum

Note that one coin takes O(1) amortized cost Operation Coin Needed at most Amortized Cost Insert 2 O(1) Extract-Min (2 + 1) · maxdegree O(lg(n))

Lemma

The maximum degree of nodes in a Fibonacci heap is O(lg(n))

The Idea of Amortized Analysis

... ... ... ...

insert extract-min

Running Time Running Time

insert insert insert insert extract-min insert insert insert

Decrease-Key

• 1. cut it to the root list
• 2. make the cascading cut

Coin Analysis

Cut a node to root costs 2 coin, mark a node costs 2 coins. Operation Coin Needed at most Amortized Cost Insert 2 O(1) Extract-Min (2 + 1) · maxdegree O(lg(n)) Decrease-Key 4 O(1)

Delete

• 1. decrease it to −∞
• 2. extract the minimum

◮ amortized cost is O(1) + O(lg(n)) = O(lg(n))

Why it called Fibonacci

Lemma : Bounding the Maximum Degree

Let x be any node in a Fibonacci heap, and let k = x.degree. Then size(x) ≥ Fk+2

Proof

recall two lemmas about the Fibonacci number Fk+2 ≤ φk and Fk = k−2 Fi + 1 Let Sk be the possible minimum size of any nodes of degree k, we can derive Sk ≥

k−2

• Si + 1

thus, Sk ≥ Fk+2

Why it called Fibonacci

Lemma : Bounding the Maximum Degree

Let x be any node in a Fibonacci heap, and let k = x.degree. Then size(x) ≥ Fk+2

Proof

recall two lemmas about the Fibonacci number Fk+2 ≤ φk and Fk = k−2 Fi + 1 Let Sk be the possible minimum size of any nodes of degree k, we can derive Sk ≥

k−2

• Si + 1

thus, Sk ≥ Fk+2

Why it called Fibonacci

Lemma : Bounding the Maximum Degree

Let x be any node in a Fibonacci heap, and let k = x.degree. Then size(x) ≥ Fk+2

Proof

recall two lemmas about the Fibonacci number Fk+2 ≤ φk and Fk = k−2 Fi + 1 Let Sk be the possible minimum size of any nodes of degree k, we can derive Sk ≥

k−2

• Si + 1

thus, Sk ≥ Fk+2