Pseudo-Hadamard transform

Van Emde Boas tree
Type	Non-binary tree
Invented	1975
Invented by	Peter van Emde Boas
Asymptotic complexity in big O notation
Space	O(n)
Search	O(log log n)
Insert	O(log log n)
Delete	O(log log n)

A Van Emde Boas tree (or Van Emde Boas priority queue), also known as a vEB tree, is a tree data structure which implements an associative array with m-bit integer keys. It performs all operations in O(log m) time. Notice that m is the size of the keys — therefore O(log m) is O(log log n) in a tree where every key below n is set, exponentially better than a full self-balancing binary search tree. The vEB tree also has good space efficiency when it contains a large number of elements, as discussed below. It was invented by a team led by Peter van Emde Boas in 1975.

Supported operations

A vEB supports the operations of an ordered associative array, which includes the usual associative array operations along with two more order operations, FindNext and FindPrevious:

• Insert: insert a key/value pair with an m-bit key
• Delete: remove the key/value pair with a given key
• Lookup: find the value associated with a given key
• FindNext: find the key/value pair with the smallest key at least a given k
• FindPrevious: find the key/value pair with the largest key at most a given k

A vEB tree also supports the operations Minimum and Maximum, which return the minimum and maximum element stored in the tree respectively. These both run in O(1) time, since the minimum and maximum element are stored as attributes in each tree.

How it works

An example Van Emde Boas tree with dimension 5 and the root's aux structure after 1, 2, 3, 5, 8 and 10 have been inserted.

For the sake of simplicity, let log₂ m = k for some integer k. Define M=2^m. A vEB tree T over the universe {0,...,M-1} has a root node that stores an array T.children of length M^1/2. T.children[i] is a pointer to a vEB tree that is responsible for the values {iM^1/2,...,(i+1)M^1/2-1}. Additionally, T stores two values T.min and T.max as well as an auxiliary vEB tree T.aux.

Data is stored in a vEB tree as follows: The smallest value currently in the tree is stored in T.min and largest value is stored in T.max. Note that T.min is not stored anywhere else in the vEB tree, while T.max is. If T is empty then we use the convention that T.max=-1 and T.min=M. Any other value x is stored in the subtree T.children[i] where $i=\lfloor x/M^{1/2}\rfloor$. The auxiliary tree T.aux keeps track of which children are non-empty, so T.aux contains the value j if and only if T.children[j] is non-empty.

FindNext

The operation FindNext(T, x) that searches for the successor of an element x in a vEB tree proceeds as follows: If x≤T.min then the search is complete, and the answer is T.min. If x>T.max then the next element does not exist, return M. Otherwise, let i=x/M^1/2. If x≤T.children[i].max then the value being searched for is contained in T.children[i] so the search proceeds recursively in T.children[i]. Otherwise, We search for the value i in T.aux. This gives us the index j of the first subtree that contains an element larger than x. The algorithm then returns T.children[j].min. The element found on the children level needs to be composed with the high bits to form a complete next element.

function FindNext(T, x)
    if x ≤ T.min then
        return T.min
    if x > T.max then // no next element
        return M
    i = floor(x/$\sqrt{M}$)
    lo = x % $\sqrt{M}$
    hi = x - lo
    if lo ≤ T.children[i].max then
        return hi + FindNext(T.children[i], lo)
    return hi + T.children[FindNext(T.aux, i)].min
end

Note that, in any case, the algorithm performs O(1) work and then possibly recurses on a subtree over a universe of size M^1/2 (an m/2 bit universe). This gives a recurrence for the running time of T(m)=T(m/2) + O(1), which resolves to O(log m) = O(log log M).

Insert

The call insert(T, x) that inserts a value x into a vEB tree T operates as follows:

If T is empty then we set T.min = T.max = x and we are done.

Otherwise, if x<T.min then we insert T.min into the subtree i responsible for T.min and then set T.min = x. If T.children[i] was previously empty, then we also insert i into T.aux

Otherwise, if x>T.max then we insert x into the subtree i responsible for x and then set T.max = x. If T.children[i] was previously empty, then we also insert i into T.aux

Otherwise, T.min< x < T.max so we insert x into the subtree i responsible for x. If T.children[i] was previously empty, then we also insert i into T.aux.

In code:

function Insert(T, x)
    if T.min > T.max then // T is empty
        T.min = T.max = x;
        return
    if T.min == T.max then
        if x < T.min then
            T.min = x
            return
        if x > T.max then
            T.max = x
    if x < T.min then
        swap(x, T.min)
    if x > T.max then
        T.max = x
    i = floor(x / $\sqrt{M})$
    Insert(T.children[i], x % $\sqrt{M}$)
    if T.children[i].min == T.children[i].max then
        Insert(T.aux, i)
end

The key to the efficiency of this procedure is that inserting an element into an empty vEB tree takes O(1) time. So, even though the algorithm sometimes makes two recursive calls, this only occurs when the first recursive call was into an empty subtree. This gives the same running time recurrence of T(m)=T(m/2) + O(1) as before.

Delete

Deletion from vEB trees is the trickiest of the operations. The call Delete(T, x) that deletes a value x from a vEB tree T operates as follows:

If T.min = T.max = x then x is the only element stored in the tree and we set T.min = M and T.max = -1 to indicate that the tree is empty.

Otherwise, if x = T.min then we need to find the second-smallest value y in the vEB tree, delete it from its current location, and set T.min=y. The second-smallest value y is either T.max or T.children[T.aux.min].min, so it can be found in O(1) time. In the latter case we delete y from the subtree that contains it.

Similarly, if x = T.max then we need to find the second-largest value y in the vEB tree and set T.max=y. The second-largest value y is either T.min or T.children[T.aux.max].max, so it can be found in O(1) time. We alse delete x from the subtree that contains it.

In case where x is not T.min or T.max, and T has no other elements, we know x is not in T and return without further operations.

Otherwise, we have the typical case where x≠T.min and x≠T.max. In this case we delete x from the subtree T.children[i] that contains x.

In any of the above cases, if we delete the last element x or y from any subtree T.children[i] then we also delete i from T.aux

In code:

function Delete(T, x)
    if T.min == T.max == x then
        T.min = M
        T.max = -1
        return
    if x == T.min then
        if T.aux is empty then
            T.min = T.max
            return
        else
            x = T.children[T.aux.min].min
            T.min = x
    if x == T.max then
        if T.aux is empty then
            T.max = T.min
            return
        else
            T.max = T.children[T.aux.max].max
    if T.aux is empty then
        return
    i = floor(x / $\sqrt{M}$)
    Delete(T.children[i], x % $\sqrt{M}$)
    if T.children[i] is empty then
        Delete(T.aux, i)
end

Again, the efficiency of this procedure hinges on the fact that deleting from a vEB tree that contains only one element takes only constant time. In particular, the last line of code only executes if x was the only element in T.children[i] prior to the deletion.

Discussion

The assumption that log m is an integer is unnecessary. The operations x/$\sqrt{M}$ and x%$\sqrt{M}$ can be replaced by taking only higher-order ceil(m/2) and the lower-order floor(m/2) bits of x, respectively. On any existing machine, this is more efficient than division or remainder computations.

The implementation described above uses pointers and occupies a total space of $O(M)=O(2^m)$. This can be seen as follows. The recurrence is $S(M)=O(\sqrt{M})+(\sqrt{M}+1)\cdot S(\sqrt{M})$. Resolving that would lead to $S(M)\in (1+\sqrt{M}{)}^{\log \log M}+\log \log M\cdot O(\sqrt{M})$. One can, fortunately, also show that $S(M)=M-2$ by induction.

In practical implementations, especially on machines with shift-by-k and find first zero instructions, performance can further be improved by switching to a bit array once m equal to the word size (or a small multiple thereof) is reached. Since all operations on a single word are constant time, this does not affect the asymptotic performance, but it does avoid the majority of the pointer storage and several pointer dereferences, achieving a significant practical savings in time and space with this trick.

An obvious optimization of vEB trees is to discard empty subtrees. This makes vEB trees quite compact when they contain many elements, because no subtrees are created until something needs to be added to them. Initially, each element added creates about log(m) new trees containing about m/2 pointers all together. As the tree grows, more and more subtrees are reused, especially the larger ones. In a full tree of 2^m elements, only O(2^m) space is used. Moreover, unlike a binary search tree, most of this space is being used to store data: even for billions of elements, the pointers in a full vEB tree number in the thousands.

However, for small trees the overhead associated with vEB trees is enormous: on the order of 2^m/2. This is one reason why they are not popular in practice. One way of addressing this limitation is to use only a fixed number of bits per level, which results in a trie. Alternatively, each table may be replaced by a hash table, reducing the space to O(n) (where n is the number of elements stored in the data structure) at the expense of making the data structure randomized. Other structures, including y-fast tries and x-fast tries have been proposed that have comparable update and query times and also use randomized hash tables to reduce the space to O(n) or O(n log M).

References

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Further reading

• Erik Demaine, Shantonu Sen, and Jeff Lindy. Massachusetts Institute of Technology. 6.897: Advanced Data Structures (Spring 2003). Lecture 1 notes: Fixed-universe successor problem, van Emde Boas. Lecture 2 notes: More van Emde Boas, ....
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