Strict Fibonacci heap

Strict Fibonacci heap
Type	Heap
Invented	2012
Invented by	Gerth S. Brodal, George Lagogiannis, and Robert E. Tarjan
Complexities in big O notation
Space complexity Time complexity Function Amortized Worst case Insert Θ(1) Find-min Θ(1) Delete-min O(log n) Decrease-key Θ(1) Merge Θ(1)

In computer science, a strict Fibonacci heap is a priority queue data structure with worst case time bounds equal to the amortized bounds of the Fibonacci heap. Along with Brodal queues, strict Fibonacci heaps belong to a class of asymptotically optimal data structures for priority queues. All operations on strict Fibonacci heaps run in worst case constant time except delete-min, which is necessarily logarithmic. This is optimal, because any priority queue can be used to sort a list of $n$ elements by performing $n$ insertions and $n$ delete-min operations. Strict Fibonacci heaps were invented in 2012 by Gerth Stølting Brodal, George Lagogiannis, and Robert E. Tarjan.

Structure

A strict Fibonacci heap. Nodes 5 and 2 are active roots.

A strict Fibonacci heap is a single tree satisfying the minimum-heap property.

The nodes in the heap have the following rules:

• All nodes are either active (colored white) or passive (colored red).
• The root is passive.
• For any node, the active children lie to the left of the passive children.

We also make the following definitions:

• An active root is an active node with a passive parent.
• A passive linkable node is a passive node where all its descendants are passive (a passive node with no children is considered to be linkable).
• The rank of an active node is the number of active children it has.
• The loss of an active node is the number of children it has lost, including active and passive children.
• An active root always has loss 0 (similar to how ordinary Fibonacci tree roots are always unmarked).

To facilitate root degree reduction (described below), the passive linkable children of the root always lie to the right of the passive non-linkable children.

Invariants

Intuitively, active nodes are the only nodes which have some kind of order, whilst passive nodes do not have any particular order. This is governed by the following structural invariant:

• For an active node, the $i$th rightmost active child $c_{i}$ (zero-indexed) satisfies $c_{i}.\mathrm {rank} +c_{i}.\mathrm {loss} \geq i$.

This invariant leads to similar structures found in binomial heaps and ordinary fibonacci heaps. For example, a subtree composed entirely of active nodes with no loss is a binomial tree.

The logarithmic bound of the delete-min operation is guaranteed by several additional invariants. A queue of all non-root nodes is also maintained

In the following, let the real number $R$ be defined as $2\lg n+6$, where $n$ is the number of nodes in the heap, and $\lg$ denotes the binary logarithm.

• The total loss in the heap is at most $R+1$.
• The total number of active nodes is at most $R+1$.
• The degree of the root is at most $R+3$.
• For an active node with loss 0, its degree is bounded by $2\lg(2n-p)+10$, where $p$ is its zero-indexed position in the queue (with 1 as the first element)
• For all other nodes, the degree is bounded by $2\lg(2n-p)+9$.
• (Corallary) The degree of any non-root node is at most $R+6$. (To see this, let $p$ = 0 in the previous invariant.)

Transformations

The following four transformations are intended to restore the above invariants after a priority queue operation has been performed. There are three main quantities we wish to minimize: the degree of the root, the total loss in the heap, and the number of active roots. All transformations can be performed in $O(1)$ time, which is possible by maintaining auxilliary data structures to track candidate nodes (described in the section on implementation).

Active root reduction

Let $x$ and $y$ be active roots with equal rank $r$, and assume $x.\mathrm {key} \leq y.\mathrm {key}$. Link $y$ to $x$ and increase the rank of $x$ by 1. If the rightmost child $z$ of $x$ is passive, link $z$ to the root.

As a result, $y$ is no longer an active root, so the number of active roots decreases by 1. However, the degree of the root node may increase by 1.

Root degree reduction

Let $x$, $y$, and $z$ be the three rightmost passive linkable children of the root. Detach them all from the root. Sort them such that $x.\mathrm {key} \leq y.\mathrm {key} \leq z.\mathrm {key}$. Change $x$ and $y$ to be active. Link $z$ to $y$, $y$ to $x$, and $x$ to the root. As a result, $x$ becomes an active root with rank 1 and loss 0. The rank and loss of $y$ is set to 0.

The net change of this transformation is that the degree of the root node decreases by 2. As a side effect, the number of active roots increases by 1.

One node loss reduction

Let $x$ be an active non-root with loss at least 2. Link $x$ to the root, thus turning it into an active root, and resetting its loss to 0. Let the original parent of $x$ be $y$. $y$ must be active, since otherwise $x$ would have previously been an active root, and thus could not have had positive loss. The rank of $y$ is decreased by 1. If $y$ is not an active root, increase its loss by 1.

Overall, the total loss decreases by 1 or 2. As a side effect, the root degree and number of active roots increase by 1, making it less preferrable to two node loss reduction, but still a necessary operation.

Two node loss reduction

Let $x$ and $y$ be active nodes with equal rank $r$ and loss equal to 1, and let $z$ be the parent of $y$. Without loss of generality, assume that $x.\mathrm {key} \leq y.\mathrm {key}$. Detach $y$ from $z$, and link $y$ to $x$. Increase the rank of $x$ by 1 and reset the loss of $x$ and $y$ from 1 to 0.

$z$ must be active, since $y$ had positive loss and could not have been an active root. Hence, the rank of $z$ is decreased by 1. If $z$ is not an active root, increase its loss by 1.

Overall, the total loss decreases by 1 or 2, with no side effects.

Summary

Effect of transformations

	Root degree	Total loss	Active roots
Active root reduction	$0{\text{ or }}{+}1$	$0$	$-1$
Root degree reduction	$-2$	$0$	$+1$
One node loss reduction	$+1$	${\text{at least }}{-}1$	$+1$
Two node loss reduction	$0$	$-1{\text{ or }}{-}2$	$0$

When deciding which transformations to perform, we consider only the worst case effect of these operations, for simplicity. The two types of loss reduction are considered to be the same operation. We define 'performing a loss reduction' to mean attempting each type of loss reduction in turn.

Worst case effect of transformations

	Root degree	Total loss	Active roots
Active root reduction	$+1$	$0$	$-1$
Root degree reduction	$-2$	$0$	$+1$
Loss reduction	$+1$	$-1$	$+1$

Implementation

Finding candidate nodes

The invariant restoring transformations rely on being able to find candidate nodes in $O(1)$ time. This means that we must keep track of active roots with the same rank, nodes with loss 1 of the same rank, and nodes with loss at least 2. There are several ways to do this.

The original paper by Brodal et al. described a fix-list and a rank-list.

Fix-list

The fix-list is divided into four parts:

1. Active roots ready for active root reduction – active roots with a partner of the same rank. Nodes with the same rank are kept adjacent.
2. Active roots not yet ready for active reduction – the only active roots for that rank.
3. Active nodes with loss 1 that are not yet ready for loss reduction – the only active nodes with loss 1 for that rank.
4. Active nodes that are ready for loss reduction – This includes active nodes with loss 1 that have a partner of the same rank, and active nodes with loss at least 2, which do not need partners to be reduced. Nodes with the same rank are kept adjacent.

To check if active root reduction is possible, we simply check if part 1 is non-empty. If it is non-empty, the first two nodes can be popped off and transformed.

Rank-list

The rank-list is a doubly linked list containing information about each rank, to allow nodes of the same rank to be partnered together in the fix-list.

For each node representing rank $r$ in the rank-list, we maintain:

• A pointer to the first active root in the fix-list with rank $r$. If such a node does not exist, this is NULL.
• A pointer to the first active node in the fix-list with rank $r$ and loss 1. If such a node does not exist, this is NULL.
• A pointer to the node representing rank $r+1$ and $r-1$, to facilitate the incrementation and decrementation of ranks.

The fix-list and rank-list require extensive bookkeeping, which must be done whenever a new active node arises, or when the rank or loss of a node is changed.

Shared flag

Using a shared flag to change make all nodes passive in $O(1)$ time

The merge operation changes all of the active nodes of the smaller heap into passive nodes. This can be done in $O(1)$ time by introducing a level of indirection. Instead of a boolean flag, each active node has a pointer towards an active flag object containing a boolean value. For passive nodes, it does not matter which active flag object they point to, as long as the flag object is set to passive, because it is not required to change many passive nodes into active nodes simultaneously.

Storing keys

The decrease-key operation requires a reference to the node of the item we wish to decrease the key of. However, the decrease-key operation itself sometimes swaps the items of nodes. To ensure the key of an item is always matched with the reference

Operations

Merge

Let $h_{1}$ and $h_{2}$ be strict Fibonacci heaps, where $|h_{1}|\leq |h_{2}|$. If either is empty, return the other. Henceforth, assume $h_{1}$ and $h_{2}$ are both non-empty. Since the size of the fix-list and rank-list of each heap is logarithmic with respect to the size of the heap, we cannot possibly merge these auxilliary structures in constant time. Instead, we throw away the structure of the smaller heap $h_{1}$ by discarding its fix-list and rank-list, and converting all of its nodes into passive nodes. This can be done in constant time, with a shared flag, as shown above. Link $h_{1}$ and $h_{2}$, letting the root with the smaller key become the parent of the other. Let $q_{1}$ and $q_{2}$ be the queues of $h_{1}$ and $h_{2}$ respectively. The combined queue is $q_{1}+\{r_{\mathrm {larger} }\}+q_{2}$, where $r_{\mathrm {larger} }$ is the root with the larger key.

The only possible structural violation is the root degree. This is solved by performing 1 active root reduction, and 1 root degree reduction, if each transformation is possible.

	Root degree	Total loss	Active roots
State after merge	$+1$	$0$	$0$
$1\times$Active root reduction	$+1$	$0$	$-1$
$1\times$Root degree reduction	$-2$	$0$	$+1$
Total	$0$	$0$	$0$

Insert

Insertion can be considered a special case of the merge operation. To insert a single key, create a new strict Fibonacci heap containing a single passive node and an empty queue. Merge it with the main heap.

Find-min

Due to the minimum-heap property, the node with the minimum key is always at the root, if it exists.

Delete-min

If the root is the only node in the heap, we are done by simply removing it. Otherwise, search the children of the root to find the node $x$ with minimum key, and set the new root to $x$. If $x$ is active, make it passive, causing all active children of $x$ to implicitly become active roots. Link the children of the old root to $x$. Since $x$ is now the root, move all of its passive linkable children to the right, and remove $x$ from the queue.

The degree of the root approximately doubles, because we have linked all the children of the old root to $x$. We perform the following restorative transformations:

1. Rotate the queue by moving the head $y$ of the queue to the back. If the two rightmost children of $y$ are passive, link them to the root.
2. Repeat step 1.
3. If a loss reduction is possible, perform it.
4. Perform active root reductions and root degree reductions until neither is possible.

To see how step 4 is bounded, consider the state after step 3:

	Root degree	Total loss	Active roots
State after delete-min	$+R+6$	$0$	$+R$
$2\times$Queue rotation	$+4$	$0$	$0$
$1\times$Loss reduction	$+1$	$-1$	$+1$
Total	$+R+11$	$-1$	$+R+1$

Observe that, 3 active root reductions and 2 root reductions decreases the root degree and active roots by 1:

	Root degree	Total loss	Active roots
$3\times$Active root reduction	$+3$	$0$	$-3$
$2\times$Root degree reduction	$-4$	$0$	$+2$
Total	$-1$	$0$	$-1$

Since $R=2\lg n+6$, step 4 never executes more than $O(\log n)$ times.

Decrease-key

Let $x$ be the node whose key has been decreased. If $x$ is the root, we are done. Otherwise, detach the subtree rooted at $x$, and link it to the root. If the key of $x$ is smaller than the key of the root, swap their keys.

Up to two structural violations may have occurred. Firstly, the degree of the root has increased (unless $x$ was already a child of the root). The second violation may have occured when $x$ was detached from its parent $y$.

• If $x$ was previously an active root, then moving $x$ from being a child of $y$ to a child of the root does not create any additional active roots, nor does it increase the loss of any node.
• If $x$ is passive and $y$ is passive, then there are no extra violations.
• If $x$ is passive and $y$ is active, then the loss of $y$ increases by 1, which may violate the total loss invariant.
• If both $x$ and $y$ are passive, then the loss of increases by 1, and an extra active root is created (by linking $x$ to the root).

In the worst case, all three quantities (root degree, total loss, active roots) increase by 1.

After performing 1 loss reduction, the worst case result is that the root degree and number of active roots have both increased by 2. Again, we use the fact that 3 active root reductions and 2 root reductions decreases both of these quantities by 1. Hence, applying these transformations 6 and 4 times respectively is sufficient to eliminate all violations.

	Root degree	Total loss	Active roots
State after decrease-key	$+1$	$+1$	$+1$
$1\times$Loss reduction	$+1$	$-1$	$+1$
$6\times$Active root reduction	$+6$	$0$	$-6$
$4\times$Root degree reduction	$-8$	$0$	$+4$
Total	$0$	$0$	$0$

Applications

Although they are theoretically optimal, strict Fibonacci heaps are not practical to use. Despite having simpler structure, experiments show that in practice the strict Fibonacci heap performs slower than more complicated Brodal queue and also slower than ordinary Fibonacci heaps. However, the Brodal queue makes extensive use of extendable arrays and redundant counters, whereas the strict Fibonacci heap is pointer based only.

Summary of running times

Here are time complexities of various heap data structures. The abbreviation am. indicates that the given complexity is amortized, otherwise it is a worst-case complexity. For the meaning of "O(f)" and "Θ(f)" see Big O notation. Names of operations assume a min-heap.

Operation	find-min	delete-min	decrease-key	insert	meld	make-heap
Binary	Θ(1)	Θ(log n)	Θ(log n)	Θ(log n)	Θ(n)	Θ(n)
Skew	Θ(1)	O(log n) am.	O(log n) am.	O(log n) am.	O(log n) am.	Θ(n) am.
Leftist	Θ(1)	Θ(log n)	Θ(log n)	Θ(log n)	Θ(log n)	Θ(n)
Binomial	Θ(1)	Θ(log n)	Θ(log n)	Θ(1) am.	Θ(log n)	Θ(n)
Skew binomial	Θ(1)	Θ(log n)	Θ(log n)	Θ(1)	Θ(log n)	Θ(n)
2–3 heap	Θ(1)	O(log n) am.	Θ(1)	Θ(1) am.	O(log n)	Θ(n)
Bottom-up skew	Θ(1)	O(log n) am.	O(log n) am.	Θ(1) am.	Θ(1) am.	Θ(n) am.
Pairing	Θ(1)	O(log n) am.	o(log n) am.	Θ(1)	Θ(1)	Θ(n)
Rank-pairing	Θ(1)	O(log n) am.	Θ(1) am.	Θ(1)	Θ(1)	Θ(n)
Fibonacci	Θ(1)	O(log n) am.	Θ(1) am.	Θ(1)	Θ(1)	Θ(n)
Strict Fibonacci	Θ(1)	Θ(log n)	Θ(1)	Θ(1)	Θ(1)	Θ(n)
Brodal	Θ(1)	Θ(log n)	Θ(1)	Θ(1)	Θ(1)	Θ(n)

External links

• JavaScript simulation of strict Fibonacci heap