

In [1]:

    
# %load ../../preconfig.py
%matplotlib inline
import matplotlib.pyplot as plt
import seaborn as sns
plt.rcParams['axes.grid'] = False

from __future__ import division

#import numpy as np
#import pandas as pd
#import itertools

import logging
logger = logging.getLogger()

18 B-Trees

B-Trees are balanced search trees, similar to red-black trees, but they are better at minizing disk I/O operations.

We use the number of pages read or written as a approximation of the total time spent accessing the disk.

A B-tree node is usually as large as a whole disk page, and this size limits the number of children a B-tree node can have.



In [2]:

    
plt.imshow(plt.imread('./res/fig18_3.png'))









    Out[2]:





<matplotlib.image.AxesImage at 0x109444c50>

18.1 Definition of B-trees

Properties

node attributes:
- $x.n$: the number of keys.
- $x.\text{keys}$: $x.\text{key}_1 \leq x.\text{key}_2 \leq \dotsb \leq x.\text{key}_n$.
- $x.leaf$: True if it is, False otherwise.
Each internal node has $x.n+1$ points $x.c_1, \dotsc, x.c_{x.n+1}$ to its children.
$x.c_i.keys \leq x.key_i \leq x.c_{i+1}.keys$
All leaves have the same depth $h$.
minimum degree of the B-tree $T$: a fixed integer $t \geq 2$.
- Every node other than the root must have at least $t-1$ keys.
- Every node may contain at most $2t-1$ keys. FULL.

The simplest B-tree occurs when $t=2$: 2-3-4 tree.

The worst-case height of B-tree

If $n \geq 1$, then for any $n$-key B-tree $T$ of height $h$ and minimum degree $t \geq 2$.
$$h \leq \log_t \frac{n+1}{2}$$

Proof: The root of a B-tree $T$ contains at least one key, and all other nodes contain at least $t-1$ keys.



In [3]:

    
plt.figure(figsize=(15,10))
plt.imshow(plt.imread('./res/fig18_4.png'))









    Out[3]:





<matplotlib.image.AxesImage at 0x10a750fd0>

the number $n$ of keys statisfies the inequality: $$ n \geq 1 + (t-1) \sum_{i=1}^h 2t^{i-1} = 2t^h -1$$

B-trees save a factor of about $\lg t$ over red-black trees in the number of nodes examined for most tree operations.

%maybe: exercise

18.2 Basic operations on B-trees

two conventions:

The root of the B-tree is always in main memory.
Any nodes that are passed as parameters must already have had a Disk-Read operation performed on them.

Searching a B-tree

We make a multiway branching decision according to the number of the node's children. $O(t lg_t n) = O(t) \times O(h)$.



In [4]:

    
#todo: code

Creating an empty B-tree

$O(1)$ time to allcate one disk page.



In [5]:

    
#todo: code

Inserting a key into a B-tree

As we trave down the tree, we split each full node we meet, in order to assure that the parent is always not full whenever we want to split a full node.

(a) B-TREE-SPLIT-CHILD(X,I)



In [6]:

    
plt.imshow(plt.imread('./res/fig18_5.png'))









    Out[6]:





<matplotlib.image.AxesImage at 0x10bfee110>

If $T$ is full tree, then create a new root node, and split them. (Height increase 1).

(b)

B-TREE-INSERT(T,K)
    if T is FULL:
        S = B-TREE-CREATE()
        S.children = T
        R = B-TREE-SPLIT-CHILD(S)
    else
        R = T

    B-TREE-INSERT-NONFULL(R,K)



In [7]:

    
plt.imshow(plt.imread('./res/fig18_6.png'))









    Out[7]:





<matplotlib.image.AxesImage at 0x10c1cfe90>

(c)

B-TREE-INSERT-NONFULL(T,K)
    if T is LEAF
        FIND_AND_INSERT(T,K)
    else
       C = FIND_CHILD(T,K)
       if C is FULL
           N = B-TREE-SPLIT-CHILD(C)
           R = FIND_CHILD(N,K)
       else
           R = T

       B-TREE-INSERT-NONFULL(R, K)



In [8]:

    
#todo: exercises

18.3 Deleting a key from a B-tree

When going down, if the node's child has only $t-1$ keys, move its key down to the child.
- if the node is root, then delete the root, use its child as new root. (Height -1).
If the node to delete is a leaf node, then directly remove it.
If the node to delete is a internal node, then
- If the key's left child has more than $t-1$ keys, move up the lastest key.
- If the key's right child has more than $t-1$ keys, move up the first key.
- otherwise, the sum of left and right child's keys is less than $2t-1$, merge them as a new node.

Base on the three priciples, we summary all possible cases as follows:



In [15]:

    
plt.figure(figsize=(10,20))
plt.subplot(2,1,1)
plt.imshow(plt.imread('./res/fig18_8.png'))
plt.subplot(2,1,2)
plt.imshow(plt.imread('./res/fig18_9.png'))









    Out[15]:





<matplotlib.image.AxesImage at 0x10ce6e610>



In [16]:

    
#todo: exercise