B-tree: find

General Information

Algorithmic problem: Sorted sequence: find

Type of algorithm: loop

Auxiliary data: A pointer [math]\displaystyle{ p }[/math] of type "pointer to a B-tree node of key type [math]\displaystyle{ \mathcal{K} }[/math]".

Abstract View

Invariant: After [math]\displaystyle{ i\geq 0 }[/math] iterations:

1. pointer [math]\displaystyle{ p }[/math] points to some node of the B-tree on height level [math]\displaystyle{ i }[/math] and
2. the searched key is in the range of that node.

Variant: [math]\displaystyle{ i }[/math] is increased by [math]\displaystyle{ 1 }[/math].

Break condition:

1. [math]\displaystyle{ p }[/math] points to a leaf of the B-tree or (that is, inclusive-or)
2. the searched key is in the node to which [math]\displaystyle{ p }[/math] points.

Induction Basis

Abstract view: p is initialized so as to point to the root of the B-tree.

Implementation: Obvious.

Proof: Obvious.

Induction Step

Abstract view:

1. Let [math]\displaystyle{ N }[/math] denote the node to which [math]\displaystyle{ p }[/math] currently points.
2. If the searched key is in [math]\displaystyle{ N }[/math], terminate the algorithm and return true.
3. Otherwise, if [math]\displaystyle{ N }[/math] is a leaf, terminate the algorithm and return false.
4. Otherwise, let [math]\displaystyle{ p }[/math] point the child of [math]\displaystyle{ N }[/math] such that the searched key is in the range of that child.

Implementation:

1. If [math]\displaystyle{ K }[/math] is one of the values [math]\displaystyle{ p }[/math].keys[math]\displaystyle{ [1],\dots,p }[/math].keys[math]\displaystyle{ [p.n] }[/math], terminate the algorithm and return true.
2. If [math]\displaystyle{ p }[/math].children[math]\displaystyle{ [0] = }[/math] void (that is, the current node is a leaf), terminate the algorithm and return false.
3. If [math]\displaystyle{ K \lt p }[/math].keys[math]\displaystyle{ [1] }[/math] set [math]\displaystyle{ p := p }[/math].children[math]\displaystyle{ [0] }[/math].
4. Otherwise, if [math]\displaystyle{ K \gt p }[/math].keys[math]\displaystyle{ [p.n] }[/math] set [math]\displaystyle{ p := p }[/math].children[math]\displaystyle{ [p.n] }[/math].
5. Otherwise, there is exactly one [math]\displaystyle{ i \in \{1,\dots,p.n-1\} }[/math] such that [math]\displaystyle{ p }[/math].keys[math]\displaystyle{ [i] \lt K \lt p }[/math].keys[math]\displaystyle{ [i+1] }[/math].
6. Set [math]\displaystyle{ p := p }[/math].children[math]\displaystyle{ [i] }[/math].

Correctness: Obvious.

Complexity

Statement: The asymptotic complexity is in [math]\displaystyle{ \Theta(T\cdot\log n) }[/math] in the worst case, where [math]\displaystyle{ T }[/math] is the complexity of the comparison and [math]\displaystyle{ n }[/math] the total number of keys in the B-tree.

Proof: Follows immediately from the the maximum height of the B-tree.

Pseudocode

B-TREE-FIND(x,k)
1 i = 1
2 while i ≤ x.n and k > x.keyi 
3        i = i + 1
4 if i ≤ x.n and k == x.keyi 
5        return (x.i)
6 elseif  x.leaf
7        return NIL
8 else DISK-READ(x.ci) 
9        return B-TREE-FIND(x.ci,k)