Operations on b-trees, Data Structure & Algorithms

Assignment Help:

Operations on B-Trees

Given are various operations which can be performed on B-Trees:

  • Search
  • Create
  • Insert

B-Tree does effort to minimize disk access and the nodes are usually stored on disk

All the nodes are supposed to be stored into secondary storage instead of primary storage. All references to a given node are preceded through a read operation. Likewise, once a node is changed and it is no longer required, it has to be written out to secondary storage with write operation.

Given is the algorithm for searching a B-tree:

B-Tree Search (x, k)

i < - 1

while i < = n [x] and k > keyi[x]

do i ← i + 1

if i < = n [x] and k = key1 [x]

then return (x, i)

if leaf [x]

then return NIL

else Disk - Read (ci[x])

return B - Tree Search (Ci[x], k)

The search operation is alike to binary tree. Instead of selecting between a left and right child as in binary tree, a B-tree search have to make an n-way choice.

The right child is selected by performing a linear search of the values into the node. After determining the value greater than or equal to desired value, the child pointer to the instantaneous left to that value is followed.

The exact running time of search operation based upon the height of the tree. Given is the algorithm for the creation of a B-tree:

B-Tree Create (T)

x ← Allocate-Node ( )

 Leaf [x] ← True

n [x] ← 0

Disk-write (x)

root [T] ← x

 

The above denoted algorithm creates an empty B-tree through allocating a new root which has no keys and is a leaf node.

Given is the algorithm for insertion into a B-tree:

B-Tree Insert (T,K)

r ← root (T)

if n[r] = 2t - 1

then S ← Allocate-Node ( )

root[T] ← S

leaf [S] ← FALSE

n[S] ← 0

C1 ← r

B-Tree-Split-Child (s, I, r)

B-Tree-Insert-Non full (s, k)

else

B - Tree-Insert-Non full (r, k)

To carry on an insertion on B-tree, the proper node for the key has to be located. Next, the key has to be inserted into the node.

If the node is not full prior to the insertion, then no special action is needed.

If node is full, then the node has to be split to make room for the new key. As splitting the node results in moving one key to the parent node, the parent node ha not be full. Else, another split operation is required.

This procedure may repeat all the way up to the root and may need splitting the root node.


Related Discussions:- Operations on b-trees

State about the pre- and post conditions, State about the pre- and post con...

State about the pre- and post conditions Programmers can easily document other pre- and post conditions and class invariants, though, and insert code to check most value preco

Find the optimal control, 1. Use the Weierstrass condition, find the (Stron...

1. Use the Weierstrass condition, find the (Strongly) minimizing curve and the value of J min for the cases where x(1) = 0, x(2) = 3. 2. The system = x 1 + 2u; where

Need help with working out. I dont really get it, Suppose there are exactly...

Suppose there are exactly five packet switches (Figure 4) between a sending host and a receiving host connected by a virtual circuit line (shown as dotted line in figure 4). The tr

Demonstration of polynomial using linked list, Demonstration of Polynomial ...

Demonstration of Polynomial using Linked List # include # include Struct link { Char sign; intcoef; int expo; struct link *next; }; Typedefstruct link

Program segment for deletion of any element from the queue, Program segment...

Program segment for deletion of any element from the queue delete() { int delvalue = 0; if (front == NULL) printf("Queue Empty"); { delvalue = front->value;

Algorithm for determining strongly connected components, Algorithm for dete...

Algorithm for determining strongly connected components of a Graph: Strongly Connected Components (G) where d[u] = discovery time of the vertex u throughout DFS , f[u] = f

Algorithams example, any simple algoritham questions with answers

any simple algoritham questions with answers

Method to add an element in circular queue, Q. Let us consider a queue is h...

Q. Let us consider a queue is housed in an array in circular fashion or trend. It is required to add new items to the queue. Write down a method ENQ to achieve this also check whet

Tradeoff between space and time complexity, We might sometimes seek a trade...

We might sometimes seek a tradeoff among space & time complexity. For instance, we may have to select a data structure which requires a lot of storage to reduce the computation tim

The space - time trade off, The best algorithm to solve a given problem is ...

The best algorithm to solve a given problem is one that requires less space in memory and takes less time to complete its execution. But in practice it is not always possible to

Write Your Message!

Captcha
Free Assignment Quote

Assured A++ Grade

Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!

All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd