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Understanding Fibonacci Heaps: Structure, Operations, and Performance

Fibonacci heaps are advanced data structures composed of a collection of heap-ordered trees that support efficient priority queue operations. This guide covers the structure of Fibonacci heaps, including the parent-child relationships and how nodes are linked. Key operations such as Insert, Minimum, Extract-Min, and Union are explained, showcasing their amortized costs. The importance of potential functions and their role in analyzing performance is also discussed. This resource is ideal for computer science students and professionals interested in data structure optimization.

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Understanding Fibonacci Heaps: Structure, Operations, and Performance

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  1. Fibonacci Heap

  2. H.min H.min (b) (a) 23 23 7 7 3 3 17 17 24 24 26 26 46 30 30 52 52 18 18 38 38 46 35 35 39 39 41 41

  3. H.min 23 7 3 17 24 (a) 26 30 52 18 38 46 21 (b) 23 7 3 17 24 35 39 41 26 30 52 18 38 46 35 39 41 Insert key 21 H.min

  4. Fibonacci Heaps: A collection of heap-ordered trees. trees: rooted but unordered Each node x: x.p points to its parent x.child points to any one of its children children of x are linked together in a circular doubly linked list x.Left, x.right: points to its left and right siblings. x.degree: number of children in the child list of x x.mark: indicate whether node x has lost a child since the last time x was mode the child of another node H.min: points to the root of the tree containing a minimum key H.n: number of nodes in H

  5. Potential function: # of marked nodes in H  (H) = t(H) + 2m(H) # of trees in the rooted list of H Fibonacci heap D(n): upper bound on the max degree of any node in an n-node Fibonacci heap

  6. Unordered binomial tree: U0: a single node Uk: consists of 2 unordered binomial trees Uk-1 for which the root of one is made into any child of the root of the other 若只用 Make-Heap, Insert, Minimum, Extract-Min & Union. Fibonacci heap 只含 unordered binomial trees. Make-Heap, Make-Fib-Heap(H): Allocate and return the Fibonacci heap object H with H.n=0 and H.min=nil t(H)=0 , m(H)=0 so  (H)=0  The amortized cost of Make-Fib-Heap is equal to its O(1) actual cost.

  7. Fib-Heap-Insert(H, x) • x.degree = 0 • x.p = NIL • x.child= NIL • x.mark= FALSE • if H.min == NIL • create a root list for H containing just x • H.min = x • else insert x into H’s root list • if x.key < H.min.key • H.min = x • H.n = H.n +1 Actual cost: O(1); Amortized cost: O(1) + 1

  8. Finding the minimum node: min[H]  O(1) Amortized cost O(1)  is not changed • Uniting 2 Fibonacci heaps: Fib-Heap-Union(H1, H2) • H =Make-Fib-Heap( ) • H.min = H1.min • concatenate the root list of H2 with the root list of H • if (H1.min == NIL) or (H2.min  NIL and H2.min.key<H1.min.key) • H.min = H2.min • H.n = H1.n + H2.n • return H

  9. t(H) = t(H1) + t(H2) , m(H) = m(H1) + m(H2) • (H) - ((H1)+(H2)) = (t(H)+2m(H)) – ((t(H1)+2m(H1)) + (t(H2)+2m(H2))) = 0 Thus the amortized cost of Fib-Heap-Union is therefore O(1) actual cost

  10. Extracting the minimum node: Fib-Heap-Extract-Min(H) • z = H.min • if z  NIL • for each child x of z • do { add x to the root list of H • x.p = NIL } • remove z from the root list of H • if z == z.right • H. min = NIL • else H.min = z.right • Consolidate(H) • H.n = H.n – 1 • return z

  11. Fib-Heap-Link(H, y, x) {1. remove y from the root list of H; 2. make y a child of x; x.degree =x.degree+1; 3. y.mark = FALSE; } Consolidate(H) • let A[0..D(H.n)] be a new array • for i = 0 to D(n[H]) do A[i]=NIL • for each node w in the root list of H • do { x = w ; d = x.degree; • while A[d]  NIL • do { y = A[d] • if x.key > y.key exchange xy • Fib-Heap-Link(H, y, x) • A[d] = NIL ; d = d+1; } • A[d] = x; } • H.min = NIL • for i = 0 to D(H.n) do • if A[i]  NIL • if H.min==NIL • create a root list for H containing just A[i]; H.min =A[i]; • else insert A[i] into H’s root list • if A[i].key < H.min.key H.min = A[i]

  12. 21 (a) 23 7 3 17 24 21 (b) 23 7 17 24 52 26 30 52 18 38 46 26 30 18 38 46 35 39 41 35 39 41 H.min H.min

  13. w,x 0 1 2 3 4 0 1 2 3 4 A A w,x 21 21 (d) (c) 23 23 7 7 17 17 24 24 52 52 26 26 30 30 18 18 38 38 46 46 35 35 39 39 41 41

  14. w,x 0 1 2 3 4 0 1 2 3 4 A A 21 (e) 23 7 17 17 24 52 x 7 (f) 24 21 52 26 30 30 18 18 38 38 46 w 23 26 46 35 39 39 41 41 35

  15. x 7 (g) 24 21 52 w 23 26 46 0 1 2 3 4 A 35 0 1 2 3 4 A 17 x 7 21 52 (h) 30 18 18 38 38 w 23 24 17 39 39 41 41 26 30 46 35

  16. 0 1 2 3 4 A w, x 7 21 52 (i) 23 24 17 26 30 46 0 1 2 3 4 35 A w, x 7 21 52 (j) 23 24 18 18 38 38 17 26 30 46 39 39 41 41 35

  17. 0 1 2 3 4 A w, x 7 18 (k) 23 24 21 39 17 26 30 46 52 35 0 1 2 3 4 A w, x 7 18 (l) 38 38 23 24 21 39 17 26 30 46 41 41 52 35

  18. H.min 7 18 (m) 23 24 21 39 17 26 30 46 52 35 38 41

  19. Decreasing a key and deleting a node: do not preserve the property that all trees in the Fibonacci heap are unordered binomial trees. Fib-Heap-Decrease-key(H, x, k) • if k>x.key • error “new key is greater than current key” • x.key = k • y x.p • if yNIL and x.key< y.key • { CUT(H, x, y) • CASCADING-CUT(H, y) } • if x.key< H.min.key • H.min = x

  20. CUT(H, x, y) 1. remove x from the child list of y, decrease y.degree 2. add x to the root list of H 3. x.p= NIL 4. x.mark = FALSE CASCADING-CUT(H, y) • zy.p • if zNIL • if y.mark == FALSE • y.mark= TRUE • else CUT(H, y, z) • CASCADING-CUT(H, z) Fib-Heap-Delete(H, x) { Fib-Heap-Decrease-key(H, x, -) Fib-Heap-Extract-Min(H) }

  21. H.min 7 18 (a) 23 24 21 39 17 26 30 46 52 35 H.min 15 7 18 (b) 23 24 21 39 38 38 17 26 30 52 41 41 35

  22. H.min 15 5 7 (c) 23 24 17 30 26 H.min 15 5 7 18 18 26 (d) 23 24 21 21 39 39 38 38 17 30 52 52 41 41

  23. H.min 15 5 7 24 26 (e) 23 17 30 18 21 39 38 52 41

  24. Analysis of Decrease-key: Actual cost : O(c) suppose CASCADING-CUT is called c times • Each recursive call of CASCADING-CUT except for the last one, cuts a marked node and clears the mark bit. • After Decrease-key, there are at most t(H)+c trees, and at most m(H)-c+2 marked nodes. Last call of CASCADING-CUT may have marked a node Thus; the potential change is : [t(H)+c+2(m(H)-c+2)] - [t(H)+2m(H)] = 4-c Amortized cost: O(c)+4-c = O(1) By scaling up the units of potential to dominate the constant hidden in O(c)

  25. At most D(n)+1 nodes remain on the list and no nodes become marked • Analysis of Fib-Heap-Extract-Min: H : n-node Fib-Heap Actual cost : O(D(n)) : for-loop in Fib-Heap-Extract-Min D(n)+t(H)-1 : size of the root list Total actual cost: O(D(n))+t(H) Potential before extracting : t(H)+2m(H) Potential after extracting :  D(n)+1+2m(H) Thus the amortized cost is at most: O(D(n))+t(H)+[(D(n)+1+2m(H)) – (t(H)+2m(H))] = O(D(n)+t(H)-t(H)) = O(D(n))

  26. Bounding the maximum degree: Goal : D(n)  logn ,  = Let size(x) be the number of nodes, including x itself, in the subtree rooted at x

  27. x yk y1 y2 • Lemma 1 x : any node in a Fibonacci heap and x.degree=k y1, …, yk : children of x in the order in which they were linked to x. (from the earliest to the latest) Then, y1.degree  0 and yi.degree  i-2 for i=2,3,…,k Pf: Clearly, y1.degree  0 For i  2, note that when yi was linked to x, all of y1, …, yi-1 were children of x, so we MUST have had x.degree  i-1. Node yi is linked to x only if x.degree = yi.degree, thus yi.degree  i-1 at that time. Since then, node yi has lost at most ONE child, since it would have been cut from x if it had lost two children. We conclude that yi.degree  i-2

  28. Fibonacci number: • Lemma 2: For all integer k0, pf: By induction on k k=0, F2=F1+F0=1 = 1+F0 Suppose

  29. x yk y1 y2  Sk-2  S0 • Lemma 3: x: any node in a Fibonacci heap, and let k=x.degree Then, size(x)  Fk+2 k pf: Sk : denote the min possible value of size(z) over all nodes z such that z.degree=k. Trivially, S0=1, S1=2, and S2=3 Sk size(x) , size(y1)  1 size(x)  Sk  2+i=2,…,k Si-2 By induction on k that SkFk+2 Clearly for k=0 and 1 Assume that k2 and that SiFi+2 for i=0,…,k-1 We have Sk 2+i=2,…,k Si-2  2+i=2,…,k Fi = 1+ i=0,…,k Fi = Fk+2 Thus, size(x)  Sk Fk+2 k

  30. Corollary 4: The max degree D(n) of any node in an n-node Fibonacci heap is O(lg n) pf: x: any node in an n-node Fibonacci heap k=degree[x] n  size(x) k logn  k Thus the max degree D(n) of any node is O(lg n)

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