Question

It is said that in B-trees with realistically large minimum degrees, almost all keys are stored in the leaf nodes. Justify this statement mathematically. Numerically demonstrate your mathematical justification with examples of t = 100 and 200.

Answer 1

ANSWER:-

Insertion
1) Initialize x as root.
2) While x is not leaf, do following
..a) Find the child of x that is going to to be traversed next. Let the child be y.
..b) If y is not full, change x to point to y.
..c) If y is full, split it and change x to point to one of the two parts of y. If k is smaller than mid key in y, then set x as first part of y. Else second part of y. When we split y, we move a key from y to its parent x.
3) The loop in step 2 stops when x is leaf. x must have space for 1 extra key as we have been splitting all nodes in advance. So simply insert k to x.

Note that the algorithm follows the Cormen book. It is actually a proactive insertion algorithm where before going down to a node, we split it if it is full.

The advantage of splitting before is, we never traverse a node twice. If we don’t split a node before going down to it and split it only if new key is inserted (reactive), we may end up traversing all nodes again from leaf to root.

This happens in cases when all nodes on the path from root to leaf are full. So when we come to the leaf node, we split it and move a key up. Moving a key up will cause a split in parent node (because parent was already full).

This cascading effect never happens in this proactive insertion algorithm. There is a disadvantage of this proactive insertion though, we may do unnecessary splits.

Let us understand the algorithm with an example tree of minimum degree ‘t’ as 3 and a sequence of integers 10, 20, 30, 40, 50, 60, 70, 80 and 90 in an initially empty B-Tree.

Initially root is NULL. Let us first insert 10.

Btree1
Let us now insert 20, 30, 40 and 50. They all will be inserted in root because maximum number of keys a node can accommodate is 2*t – 1 which is 5.

BTree2Ins
Let us now insert 60. Since root node is full, it will first split into two, then 60 will be inserted into the appropriate child.

BTreeIns3
Let us now insert 70 and 80. These new keys will be inserted into the appropriate leaf without any split.

BTreeIns4
Let us now insert 90. This insertion will cause a split. The middle key will go up to the parent.

BTreeIns6

/ C++ program for B-Tree insertion

#include<iostream>

using namespace std;

// A BTree node

class BTreeNode

{

int *keys; // An array of keys

int t; // Minimum degree (defines the range for number of keys)

BTreeNode **C; // An array of child pointers

int n; // Current number of keys

bool leaf; // Is true when node is leaf. Otherwise false

public:

BTreeNode(int t, bool leaf); // Constructor

// A utility function to insert a new key in the subtree rooted with

// this node. The assumption is, the node must be non-full when this

// function is called

void insertNonFull(int k);

void splitChild(int i, BTreeNode *y);

// A function to traverse all nodes in a subtree rooted with this node

void traverse();

// A function to search a key in subtree rooted with this node.

BTreeNode *search(int k); // returns NULL if k is not present.

// Make BTree friend of this so that we can access private members of this

// class in BTree functions

friend class BTree;

};

// A BTree

class BTree

{

BTreeNode *root; // Pointer to root node

int t; // Minimum degree

public:

// Constructor (Initializes tree as empty)

BTree(int _t)

{ root = NULL; t = _t; }

// function to traverse the tree

void traverse()

{ if (root != NULL) root->traverse(); }

// function to search a key in this tree

BTreeNode* search(int k)

{ return (root == NULL)? NULL : root->search(k); }

// The main function that inserts a new key in this B-Tree

void insert(int k);

};

BTreeNode::BTreeNode(int t1, bool leaf1)

{

// Copy the given minimum degree and leaf property

t = t1;

leaf = leaf1;

// Allocate memory for maximum number of possible keys

// and child pointers

keys = new int[2*t-1];

C = new BTreeNode *[2*t];

// Initialize the number of keys as 0

n = 0;

}

// Function to traverse all nodes in a subtree rooted with this node

void BTreeNode::traverse()

{

// There are n keys and n+1 children, travers through n keys

// and first n children

int i;

for (i = 0; i < n; i++)

{

// If this is not leaf, then before printing key[i],

// traverse the subtree rooted with child C[i].

if (leaf == false)

C[i]->traverse();

cout << " " << keys[i];

}

// Print the subtree rooted with last child

if (leaf == false)

C[i]->traverse();

}

// Function to search key k in subtree rooted with this node

BTreeNode *BTreeNode::search(int k)

{

// Find the first key greater than or equal to k

int i = 0;

while (i < n && k > keys[i])

i++;

// If the found key is equal to k, return this node

if (keys[i] == k)

return this;

// If key is not found here and this is a leaf node

if (leaf == true)

return NULL;

// Go to the appropriate child

return C[i]->search(k);

}

void BTree::insert(int k)

{

// If tree is empty

if (root == NULL)

{

// Allocate memory for root

root = new BTreeNode(t, true);

root->keys[0] = k; // Insert key

root->n = 1; // Update number of keys in root

}

else // If tree is not empty

{

// If root is full, then tree grows in height

if (root->n == 2*t-1)

{

// Allocate memory for new root

BTreeNode *s = new BTreeNode(t, false);

// Make old root as child of new root

s->C[0] = root;

// Split the old root and move 1 key to the new root

s->splitChild(0, root);

// New root has two children now. Decide which of the

// two children is going to have new key

int i = 0;

if (s->keys[0] < k)

i++;

s->C[i]->insertNonFull(k);

// Change root

root = s;

}

else // If root is not full, call insertNonFull for root

root->insertNonFull(k);

}

}

// A utility function to insert a new key in this node

// The assumption is, the node must be non-full when this

// function is called

void BTreeNode::insertNonFull(int k)

{

// Initialize index as index of rightmost element

int i = n-1;

// If this is a leaf node

if (leaf == true)

{

// The following loop does two things

// a) Finds the location of new key to be inserted

// b) Moves all greater keys to one place ahead

while (i >= 0 && keys[i] > k)

{

keys[i+1] = keys[i];

i--;

}

// Insert the new key at found location

keys[i+1] = k;

n = n+1;

}

else // If this node is not leaf

{

// Find the child which is going to have the new key

while (i >= 0 && keys[i] > k)

i--;

// See if the found child is full

if (C[i+1]->n == 2*t-1)

{

// If the child is full, then split it

splitChild(i+1, C[i+1]);

// After split, the middle key of C[i] goes up and

// C[i] is splitted into two. See which of the two

// is going to have the new key

if (keys[i+1] < k)

i++;

}

C[i+1]->insertNonFull(k);

}

}

// A utility function to split the child y of this node

// Note that y must be full when this function is called

void BTreeNode::splitChild(int i, BTreeNode *y)

{

// Create a new node which is going to store (t-1) keys

// of y

BTreeNode *z = new BTreeNode(y->t, y->leaf);

z->n = t - 1;

// Copy the last (t-1) keys of y to z

for (int j = 0; j < t-1; j++)

z->keys[j] = y->keys[j+t];

// Copy the last t children of y to z

if (y->leaf == false)

{

for (int j = 0; j < t; j++)

z->C[j] = y->C[j+t];

}

// Reduce the number of keys in y

y->n = t - 1;

// Since this node is going to have a new child,

// create space of new child

for (int j = n; j >= i+1; j--)

C[j+1] = C[j];

// Link the new child to this node

C[i+1] = z;

// A key of y will move to this node. Find location of

// new key and move all greater keys one space ahead

for (int j = n-1; j >= i; j--)

keys[j+1] = keys[j];

// Copy the middle key of y to this node

keys[i] = y->keys[t-1];

// Increment count of keys in this node

n = n + 1;

}

// Driver program to test above functions

int main()

{

BTree t(3); // A B-Tree with minium degree 3

t.insert(10);

t.insert(20);

t.insert(5);

t.insert(6);

t.insert(12);

t.insert(30);

t.insert(7);

t.insert(17);

cout << "Traversal of the constucted tree is ";

t.traverse();

int k = 6;

(t.search(k) != NULL)? cout << "\nPresent" : cout << "\nNot Present";

k = 15;

(t.search(k) != NULL)? cout << "\nPresent" : cout << "\nNot Present";

return 0;

}