Introduction Fibonacci Heap Time Complexity...

IntroductionFibonacci Heap

Time Complexity Analysis

Fibonacci Heap

Group Minus One Second

December 6, 2016

Group Minus One Second Fibonacci Heap



Outline

IntroductionMotivationPerformance Goal

Fibonacci HeapStructureRankInsertDelete MinDecrease Key

Time Complexity AnalysisAmortized AnalysisInsert ComplexityDelete Min ComplexityDecrease Key Complexity




MotivationPerformance Goal

Motivation

I Design a data structure that supports the following operationsI Insert(x)I DeleteMin()I DecreaseKey(u, v)

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Motivation


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Insert(10)

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Motivation


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DeleteMin()

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Motivation


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DecreaseKey(21, 17)

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Performance Goal

I Our performance goals of this data structureI Insert(x): O(1)I DeleteMin(): O(log n)I DecreaseKey(u, v): O(1)




StructureRankInsertDelete MinDecrease Key

Structure

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

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roots

heap-ordered tree





Rank

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

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rank = 1 rank = 2 rank = 0 rank = 0 rank = 3





Insert

I To insert a new value x into Fibonacci Heap H, simply createa new tree that contains only x and add it to H.

I Example: Insert(21)

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Delete Min

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

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Delete Min

I We introduce consolidation, a process of merging trees of thesame rank.

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merge





Delete Min


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merge





Delete Min


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merge





Delete Min


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merge





Delete Min


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Delete Min

I After consolidation, we simply remove the minimum value andleave the remaining trees in the heap.

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Decrease Key

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

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Decrease Key

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

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Decrease Key

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

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Decrease Key


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Decrease Key


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Decrease Key

I But wait! Under this strategy, something undesirable mighthappen.

I The example below shows that this strategy can hurt theperformance of DeleteMin operation.

I Example:





Decrease Key with Marks

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






Case 1: no violation of heap property, finish.






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






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




Amortized AnalysisInsert ComplexityDelete Min ComplexityDecrease Key Complexity

Amortized Analysis

I We need a new technique to analyze the time complexity ofFibonacci heap

I Consider the following question:I Drop: Dropping a single coin onto a table costs 1 unit of timeI Clean: Collecting each coin into a bag costs 1 unit of time.

Emptying the bag costs 1 unit of time.I Question: What is the amortized time complexity of Clean

operation?





Amortized Analysis

I Observation: A coin that is cleaned must have been droppedbefore.

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

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

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

I Since no other operations are needed, each Insert operationstill takes $2 = O(1) time.

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$1 $1 $1 $1 $1 $1 $1





Delete Min Complexity

I Recall that DeleteMin opeartion consists of two parts:I Consolidate the heap: for each rank i 2 [0, k], merge two

trees of the same rank i .I Linearly search for the minimum: traverse all k + 1 roots.I k is the largest rank in the heap.

I We will make and prove a few claims.






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

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$1 $1

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$1

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






I Claim 2: Consolidation operations are of O(k) where k is thelargest rank.

I Proof: For each rank i 2 [0, k], consolidation calls merge togenerate new trees, which is free. So the only cost is the timeconsumed while travsering k + 1 ranks.






I Claim 3: Searching for minimum is of O(k) where k is thelargest rank.

I Proof: After consolidation there are at most k + 1 roots. Soa linear traversal will take O(k) time.

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I Combining all three claims, we can see that the complexity ofDeleteMin is O(k) where k is the largest rank.

I One last question: why k = O(log n)?I This is equivalent to ”what is the smallest number of nodes in

a tree of rank i?”






Some results for di↵erent ranks. Let Si be the smallest number ofnodes in a tree of rank i . We can see that

S0 = 1, S1 = 2, S2 = 3, S3 = 5, S4 = 8, ...

rank = 0







S0 = 1, S1 = 2, S2 = 3, S3 = 5, S4 = 8, ...

rank = 1







S0 = 1, S1 = 2, S2 = 3, S3 = 5, S4 = 8, ...

rank = 2







S0 = 1, S1 = 2, S2 = 3, S3 = 5, S4 = 8, ...

rank = 3







S0 = 1, S1 = 2, S2 = 3, S3 = 5, S4 = 8, ...

rank = 4






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’sname.

I Since Si � c i where c = 1+p5

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

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





Decrease Key Complexity

I By similar argument (using the amortized analysis), we canprove that the time complexity of DecreaseKey(u, v) isO(1).

I We will omit this part due to time limit.

I You can refer to the lecture note on the course website.


Thankyou

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