B*-tree of order m is a search tree that is either empty or that satisfies three properties:
- The root node has minimum two and maximum 2 floor ((2m-2)/3) +1 children
- Other internal nodes have the minimum floor ((2m-1)/3) and maximum m children
- All external nodes are on the same level.
The advantage of using B* trees over B-trees is a unique feature called the ‘two-to-three’ split. By this, the minimum number of keys in each node is not half the maximum number, but two-thirds of it, making data far more compact. However, the disadvantage of this is a complex deletion operation.
The difficulties in practically implementing a B-star algorithm contribute to why it’s not as regularly used as its B and B+ counterparts.
Below is a basic implementation of the B-star insertion function – just to demonstrate its contrast from B (full implementation would be far more lengthy and complex).
The unique parts of the algorithm for B* Tree insertion are as follows:
Two-Three Split
1. If inserting into a full leaf node (which is not the root) and which has a full right sibling (and whose parent has at least one free key):
- Take an array (‘marray’) consisting of ‘m-1’ keys of the full leaf-node, the parent key of this node, the new key to be inserted, and the ‘m-1’ keys of its right sibling (Totally m-1 + 1 + 1 + m-1 = 2m keys)
- Sort these keys
- Create three new nodes:
- p – whose keys are the first (2m – 2)/3 elements of ‘marray’
The element at index (2m – 2)/3 is stored as ‘parent1’ - q – whose keys are the next (2m – 1)/3 elements of ‘marray’ after parent1
The element at index (4m)/3 is stored as ‘parent2’ - r – whose keys are the last (2m)/3 elements of ‘marray’ after parent2
- p – whose keys are the first (2m – 2)/3 elements of ‘marray’
- The key in the leaf’s parent which points to this leaf should have its value replaced as ‘parent1’
- If the parent key in iv) has any adjacent keys, they should be shifted to the right. In the space that remains, place ‘parent2’.
- p, q and r must be made child keys of parent1 and parent2 (if ‘parent1’ and ‘parent2’ are the first two keys in the parent node), else p, q, r must be made the child keys of the key before parent 1, parent 1, and parent 2 respectively.
Before Insertion :
After insertion:
2. If inserting into a full leaf node (which is not the root) with empty/non-full right sibling.
3. The other cases are the same as for B-Trees.
Examples:
Input: Add 4 to 1 2 3 L 5 R 7 8 9
Output: 1 2 L 3 7 R 4 5 R 8 9
3 and 7 become the parent keys by the two-three splitInput : Add 5 to 2 3 4 L 6 R 8 9 11
Output : 2 3 L 4 8 R 5 6 R 9 11
3 and 6 become the parent keys by the two-three split
Below is the implementation of the above approach :
// CPP program to implement B* tree #include <bits/stdc++.h> using namespace std;
// This can be changed to any value - // it is the order of the B* Tree #define N 4 struct node {
// key of N-1 nodes
int key[N - 1];
// Child array of 'N' length
struct node* child[N];
// To state whether a leaf or not; if node
// is a leaf, isleaf=1 else isleaf=0
int isleaf;
// Counts the number of filled keys in a node
int n;
// Keeps track of the parent node
struct node* parent;
}; // This function searches for the leaf // into which to insert element 'k' struct node* searchforleaf( struct node* root, int k,
struct node* parent, int chindex)
{ if (root) {
// If the passed root is a leaf node, then
// k can be inserted in this node itself
if (root->isleaf == 1)
return root;
// If the passed root is not a leaf node,
// implying there are one or more children
else {
int i;
/*If passed root's initial key is itself g
reater than the element to be inserted,
we need to insert to a new leaf left of the root*/
if (k < root->key[0])
root = searchforleaf(root->child[0], k, root, 0);
else {
// Find the first key whose value is greater
// than the insertion value
// and insert into child of that key
for (i = 0; i < root->n; i++)
if (root->key[i] > k)
root = searchforleaf(root->child[i], k, root, i);
// If all the keys are less than the insertion
// key value, insert to the right of last key
if (root->key[i - 1] < k)
root = searchforleaf(root->child[i], k, root, i);
}
}
}
else {
// If the passed root is NULL (there is no such
// child node to search), then create a new leaf
// node in that location
struct node* newleaf = new struct node;
newleaf->isleaf = 1;
newleaf->n = 0;
parent->child[chindex] = newleaf;
newleaf->parent = parent;
return newleaf;
}
} struct node* insert( struct node* root, int k)
{ if (root) {
struct node* p = searchforleaf(root, k, NULL, 0);
struct node* q = NULL;
int e = k;
// If the leaf node is empty, simply
// add the element and return
for ( int e = k; p; p = p->parent) {
if (p->n == 0) {
p->key[0] = e;
p->n = 1;
return root;
}
// If number of filled keys is less than maximum
if (p->n < N - 1) {
int i;
for (i = 0; i < p->n; i++) {
if (p->key[i] > e) {
for ( int j = p->n - 1; j >= i; j--)
p->key[j + 1] = p->key[j];
break ;
}
}
p->key[i] = e;
p->n = p->n + 1;
return root;
}
// If number of filled keys is equal to maximum
// and it's not root and there is space in the parent
if (p->n == N - 1 && p->parent && p->parent->n < N) {
int m;
for ( int i = 0; i < p->parent->n; i++)
if (p->parent->child[i] == p) {
m = i;
break ;
}
// If right sibling is possible
if (m + 1 <= N - 1)
{
// q is the right sibling
q = p->parent->child[m + 1];
if (q) {
// If right sibling is full
if (q->n == N - 1) {
struct node* r = new struct node;
int * z = new int [((2 * N) / 3)];
int parent1, parent2;
int * marray = new int [2 * N];
int i;
for (i = 0; i < p->n; i++)
marray[i] = p->key[i];
int fege = i;
marray[i] = e;
marray[i + 1] = p->parent->key[m];
for ( int j = i + 2; j < ((i + 2) + (q->n)); j++)
marray[j] = q->key[j - (i + 2)];
// marray=bubblesort(marray, 2*N)
// a more rigorous implementation will
// sort these elements
// Put first (2*N-2)/3 elements into keys of p
for ( int i = 0; i < (2 * N - 2) / 3; i++)
p->key[i] = marray[i];
parent1 = marray[(2 * N - 2) / 3];
// Put next (2*N-1)/3 elements into keys of q
for ( int j = ((2 * N - 2) / 3) + 1; j < (4 * N) / 3; j++)
q->key[j - ((2 * N - 2) / 3 + 1)] = marray[j];
parent2 = marray[(4 * N) / 3];
// Put last (2*N)/3 elements into keys of r
for ( int f = ((4 * N) / 3 + 1); f < 2 * N; f++)
r->key[f - ((4 * N) / 3 + 1)] = marray[f];
// Because m=0 and m=1 are children of the same key,
// a special case is made for them
if (m == 0 || m == 1) {
p->parent->key[0] = parent1;
p->parent->key[1] = parent2;
p->parent->child[0] = p;
p->parent->child[1] = q;
p->parent->child[2] = r;
return root;
}
else {
p->parent->key[m - 1] = parent1;
p->parent->key[m] = parent2;
p->parent->child[m - 1] = p;
p->parent->child[m] = q;
p->parent->child[m + 1] = r;
return root;
}
}
}
else // If right sibling is not full
{
int put;
if (m == 0 || m == 1)
put = p->parent->key[0];
else
put = p->parent->key[m - 1];
for ( int j = (q->n) - 1; j >= 1; j--)
q->key[j + 1] = q->key[j];
q->key[0] = put;
p->parent->key[m == 0 ? m : m - 1] = p->key[p->n - 1];
}
}
}
}
/*Cases of root splitting, etc. are omitted
as this implementation is just to demonstrate
the two-three split operation*/
}
else {
// Create new node if root is NULL
struct node* root = new struct node;
root->key[0] = k;
root->isleaf = 1;
root->n = 1;
root->parent = NULL;
}
} // Driver code int main()
{ /* Consider the following tree that has been obtained
from some root split:
6
/ \
1 2 4 7 8 9
We wish to add 5. This makes the B*-tree:
4 7
/ \ \
1 2 5 6 8 9
Contrast this with the equivalent B-tree, in which
some nodes are less than half full
4 6
/ \ \
1 2 5 7 8 9
*/
// Start with an empty root
struct node* root = NULL;
// Insert 6
root = insert(root, 6);
// Insert 1, 2, 4 to the left of 6
root->child[0] = insert(root->child[0], 1);
root->child[0] = insert(root->child[0], 2);
root->child[0] = insert(root->child[0], 4);
root->child[0]->parent = root;
// Insert 7, 8, 9 to the right of 6
root->child[1] = insert(root->child[1], 7);
root->child[1] = insert(root->child[1], 8);
root->child[1] = insert(root->child[1], 9);
root->child[1]->parent = root;
cout << "Original tree: " << endl;
for ( int i = 0; i < root->n; i++)
cout << root->key[i] << " " ;
cout << endl;
for ( int i = 0; i < 2; i++) {
cout << root->child[i]->key[0] << " " ;
cout << root->child[i]->key[1] << " " ;
cout << root->child[i]->key[2] << " " ;
}
cout << endl;
cout << "After adding 5: " << endl;
// Inserting element '5':
root->child[0] = insert(root->child[0], 5);
// Printing nodes
for ( int i = 0; i <= root->n; i++)
cout << root->key[i] << " " ;
cout << endl;
for ( int i = 0; i < N - 1; i++) {
cout << root->child[i]->key[0] << " " ;
cout << root->child[i]->key[1] << " " ;
}
return 0;
} |
Output:
Original Tree: 6 1 2 4 7 8 9 After adding 5: 4 7 1 2 5 6 8 9