Van Emde Boas tree

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⌋. 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 successor of 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/√M)
    lo = x mod √M
    
    if lo < T.children[i].max then
        return (√M i) + FindNext(T.children[i], lo)
    j = FindNext(T.aux, i)
    return (√M j) + T.children[j].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 ${\displaystyle T(m)=T(m/2)+O(1)}$, which resolves to O(log m) = O(log log M).

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

1. If T is empty then we set T.min = T.max = x and we are done.
2. 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
3. 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
4. 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 x < T.min then
        swap(x, T.min)
    if x > T.max then
        T.max = x
    i = floor(x / √M)
    lo = x mod √M
    Insert(T.children[i], lo)
    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 ${\displaystyle T(m)=T(m/2)+O(1)}$ as before.

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:

1. 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.
2. 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 T.children[T.aux.min].min, so it can be found in O(1) time. We delete y from the subtree that contains it.
3. If x≠T.min and x≠T.max then we delete x from the subtree T.children[i] that contains x.
4. If x == T.max then we will need to find the second-largest value y in the vEB tree and set T.max=y. We start by deleting x as in previous case. Then value y is either T.min or T.children[T.aux.max].max, so it can be found in O(1) time.
5. 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
        hi = T.aux.min * √M
        j = T.aux.min
        T.min = x = hi + T.children[j].min
    i = floor(x / √M)
    lo = x mod √M
    Delete(T.children[i], lo)
    if T.children[i] is empty then
        Delete(T.aux, i)
    if x == T.max then
        if T.aux is empty then
            T.max = T.min
        else
            hi = T.aux.max * √M
            j = T.aux.max
            T.max = hi + T.children[j].max
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 second Delete call only executes if x was the only element in T.children[i] prior to the deletion.

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

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 M elements, only O(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.

The implementation described above uses pointers and occupies a total space of O(M) = O(2^m), proportional to the size of the key universe. This can be seen as follows. The recurrence is ${\displaystyle S(M)=O({\sqrt {M}})+({\sqrt {M}}+1)\cdot S({\sqrt {M}})}$. Resolving that would lead to ${\displaystyle S(M)\in (1+{\sqrt {M}})^{\log \log M}+\log \log M\cdot O({\sqrt {M}})}$. One can, fortunately, also show that S(M) = M−2 by induction.[4]