B-tree: shift key to sibling

General Information

Algorithmic problem: See B-tree

Type of algorithm: loop

Auxiliary data:

Abstract View

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

1. If [math]\displaystyle{ shiftRight }[/math]
   1. For [math]\displaystyle{ j \in \{ p.children[k].n-i+1, \dots ,p.children[k].n \{ }[/math], if [math]\displaystyle{ j\gt 0 }[/math], the original values of [math]\displaystyle{ p.children[k].key[j] }[/math] are now the values of [math]\displaystyle{ p.children[k].key[j+1] }[/math].
   2. For [math]\displaystyle{ j \in \{ p.children[k].n-i+1, \dots ,p.children[k].n \} }[/math], the original values of [math]\displaystyle{ p.children[k].children[j] }[/math] are now the values of [math]\displaystyle{ p.children[k].children[j+1] }[/math].
2. Otherwise:
   1. For [math]\displaystyle{ j \in \{2, \dots ,i+1 \} }[/math], id [math]\displaystyle{ j \le n }[/math], the original values of [math]\displaystyle{ p.children[k].key[j] }[/math] are now the values of [math]\displaystyle{ p.children[k].key[j-1] }[/math].
   2. For [math]\displaystyle{ j \in \{ 1, \dots ,i \} }[/math], the values of [math]\displaystyle{ p.children[k].children[j] }[/math] are now the values of [math]\displaystyle{ p.children[k].children[j-1] }[/math].
   3. A few moves have already been done (in the preprocessing, in fact):
      1. The former value of [math]\displaystyle{ p.keys[k] }[/math] is now the node [math]\displaystyle{ p.children[k-1] }[/math].
      2. The former value of [math]\displaystyle{ p.children[k].keys[1] }[/math] is now at [math]\displaystyle{ p.keys[k] }[/math].
      3. The children pointers of [math]\displaystyle{ p.children[k-1] }[/math] and [math]\displaystyle{ p.children[k]\lt /math are rearranged accordingly. '''Vairant:''' \lt math\gt i }[/math] increases by [math]\displaystyle{ 1 }[/math]

Break condition: [math]\displaystyle{ (shiftRight }[/math] ∧[math]\displaystyle{ i=p.children[k].n+1) }[/math]∨[math]\displaystyle{ ( }[/math]¬[math]\displaystyle{ shiftRight }[/math]∧[math]\displaystyle{ i=p.children[k].n) }[/math]

Before and after each iteration, [math]\displaystyle{ p }[/math] points to a node [math]\displaystyle{ N }[/math] such that the following holds:

1. In the case [math]\displaystyle{ p.n \lt 2M }[/math], all implementation invariants of B-trees are maintained if the following [math]\displaystyle{ K' }[/math] can be inserted at some position in [math]\displaystyle{ N }[/math] such that after insertion.
2. In the case [math]\displaystyle{ p.n = 2M }[/math], imagine that we could extend [math]\displaystyle{ p }[/math].keys by one more slot (and, consequently, extend .successors by one more slot). Then [math]\displaystyle{ K' }[/math] can be inserted in [math]\displaystyle{ N }[/math] at a position such that - except for the statement about the size of [math]\displaystyle{ N }[/math] - all implementation invariants of B-trees are again maintained after insertion.

Variant: The height level of the node to which [math]\displaystyle{ p }[/math] points is decreased by [math]\displaystyle{ 1 }[/math].

Break condition: [math]\displaystyle{ p.N \lt 2M }[/math] or (that is, inclusive-or) [math]\displaystyle{ p }[/math] points to the root.

Induction step

Abstract view:

1. Move the next key and children pointer.
2. If [math]\displaystyle{ shiftRight }[/math] and the break condition applies, the shift is to be done.

Implementation:

1. If [math]\displaystyle{ shiftRight }[/math]:
   1. If [math]\displaystyle{ i \le n }[/math], set [math]\displaystyle{ p.children[k]keys[p.children[k].n-i+2]:= p.children[k],keys[p.children[k]._n_-i_+1] }[/math]
   2. Set [math]\displaystyle{ p.children[k].children[p.children[k]._n-i+2]:= p.children[k].children[p.children[k]._n-i+1] }[/math]
   3. If [math]\displaystyle{ i=n+1 }[/math]:
      1. Set [math]\displaystyle{ p.children[k].keys[1]:= p.keys[k] ### Set \lt math\gt p.keys[k]:=p.children[k-1].keys[p.children[k-1]_n] ### Increment \lt math\gt p.children[k]_n by one ### Decrement \lt math\gt p.children[j-1]_n by one }[/math]
2. Otherwise:
   1. If [math]\displaystyle{ i\lt p.children[k]_n }[/math] set [math]\displaystyle{ p.children[k].keys[i]:= p.children[k].keys[i+1] }[/math]
   2. Set [math]\displaystyle{ p.children[k].children[i-1]:= p.children[k].children[i] }[/math]
   3. If [math]\displaystyle{ i=p.children[k]_n, }[/math]decrement [math]\displaystyle{ p.children[k]_n }[/math] by one

Correctness:

1. For the rearrangement of the children pointers, compare the induction basis.
2. Note that the increments and decrements of [math]\displaystyle{ p.children[k-1]_n }[/math] and [math]\displaystyle{ p.children[k]_n }[/math] are distributed over induction basis and induction step, but are correct in summary.

Induction Step

Abstract view:

1. If the node is not full, the key can be inserted, and we are done.
   1. If the new key is larger than all keys stored in the node, it can be appended.
   2. Otherwise, is must override one specific slot. For that, first the contents of this slot and of each slot to its right must be moved one position further right (the corresponding values and successor pointers have to be moved as well). Then the new (key,value)-pair may be safely inserted.
2. On the other hand, if the node is indeed full, the key cannot be inserted. In this case:
   1. The current node [math]\displaystyle{ N }[/math] will be split into two nodes, [math]\displaystyle{ N_1 }[/math] and [math]\displaystyle{ N_2 }[/math].
   2. [math]\displaystyle{ N_1 }[/math] will contain all keys strictly less than the median of { [math]\displaystyle{ p.keys[1], ... , p.keys[p.n], K' }[/math] }
   3. Analogously, [math]\displaystyle{ N_2 }[/math] will contain all keys strictly greater than the median of { [math]\displaystyle{ p.keys[1], ... , p.keys[p.n], K' }[/math] }.
   4. Next we try to
   5. If [math]\displaystyle{ K' }[/math] is that median itself, nothing is to be done on the current node.
   6. Otherwise, the median of [math]\displaystyle{ \{p.keys[1], \dots , p.keys[p.n], K'\} }[/math] is in one of the slots of the node. This slot is overridden (by moving the contents of each slot to the right one position further left). The previous contents of this slot is the new [math]\displaystyle{ K' }[/math] . If the current node is the root, which we next try to insert in the predecessor of

Implementation:

1. If [math]\displaystyle{ p.n = 2M }[/math]
   1. If [math]\displaystyle{ p.keys[M] \lt K' \lt p.keys[M + 1] }[/math], break the iteration and continue the loop.
   2. Otherwise: [math]\displaystyle{ x:= M }[/math] if [math]\displaystyle{ K' \lt p.keys[M] }[/math], and [math]\displaystyle{ x:= M+1 }[/math] otherwise.
   3. Set [math]\displaystyle{ K'' := p.keys[x] }[/math].
   4. For [math]\displaystyle{ j \in \{x+1, \dots, p.n \} }[/math] (in this order), set [math]\displaystyle{ p.keys [j-1]:= p.keys[j] }[/math] and [math]\displaystyle{ p. successor[j-1]:= p.successors[j] }[/math].
2. If [math]\displaystyle{ p.keys[p.n] \lt K' }[/math], insert the new (key, value)-pair at position [math]\displaystyle{ p.keys[p.n+1 }[/math]
3. Otherwise:
   1. Let [math]\displaystyle{ i \in \{ 1, \dots , p.n \} }[/math] be minimal subject to [math]\displaystyle{ K' \lt p.keys[i] }[/math]
   2. For [math]\displaystyle{ j \in \{ p.n, \dots , i \} }[/math] (in this order), set [math]\displaystyle{ p.keys[j+1]:= p.keys[j]. }[/math]
4. If [math]\displaystyle{ p.n \lt 2M }[/math], set [math]\displaystyle{ p.n:= p.n+1 }[/math] and terminate the algorithm.
5. Otherwise, set [math]\displaystyle{ K':= K'' }[/math].

Correctness: Easy to see.