B-tree
Why B-Trees
• When the data is too big, we will have to use
disk storage instead of putting all the data in
main memory
• In such case, we have to take into account the
fact that disk access takes much longer time
than many other instructions
Disk Access Time
• Access time = seek time + rotational delay (latency) +
transfer time
– Seek time is slow, it depends on the mechanical movement of
the disk head to position the head at the correct track of the
disk
– Latency is the time required to position the head above the
correct block and on average, it is one-half of a revolution.
• Ex. to transfer 5kb from a disk requires 40ms to locate a
track, making 7200 RPM and with a data transfer rate of
1000kb per second, access time = 40ms + 4ms + 5ms = 49
ms
Why B-Trees
• Assume that we use a Balanced Binary tree to
store all 20 million recordslog2 20,000,000 is
about 25, end up with a very deep tree
• it will take more than 1 second to transfer a
record
• cannot improve on the log n for a binary tree
• The solution is to use more branches and thus
less height, as branching increases, depth
decreases
Definition of a B-tree
• A B-tree of order m is an m-way tree (i.e., a
tree where each node may have up to m
children) in which:
1.the root has at least two subtree unless it is a leaf
2. each non-leaf and nonroot nodes have k-1 keys
and k pointers where m / 2 <= k <= m
3. Each leaf node holds k-1 keys where m / 2 <= k
<= m
4. All leaves are on the same level.
An example B-Tree
A B-tree of order 5
3
1
2
6
7
8
9
12 14 16 19
B-Trees Example: 2-3 Tree
If we take m = 3, we get a 2-3 tree, in which non-leaf
nodes have two or three children (i.e., one or two
keys), and it is balanced as B-Trees are always
balanced (since the leaves are all at the same level),
Search B-Tree
B-Tree: Insertion
• insert the new key into a leaf
• If the resulting leaf becoming too big, split the leaf into
two, promoting the middle key to the leaf’s parent
• If the promotion results in the parent becoming too
big, split the parent into two, promoting the middle key
• This strategy might have to be repeated all the way to
the top
• If necessary, the root is split in two and the middle key
is promoted to a new root, making the tree one level
higher
B-tree insertion example (order 3)
From
Wiki
B-tree: Deletion
During insertion, the key always goes into a leaf.
For deletion we wish to remove from a leaf. There
are three possible ways we can do this:
1. If the key is already in a leaf node, and removing
it doesn’t cause that leaf node to have too few keys,
then simply remove the key to be deleted.
2. If the key is in a non-leaf node, then delete the
key and promote the predecessor or successor key
to the non-leaf deleted key’s position
B-tree: Deletion
If 1 or 2 cause a leaf node containing less than the minimum
number of keys then we either get help from sibling or merge
nodes.
3. Check if one of the siblings immediately adjacent to the leaf
in question has more than the minimum number of keys, if
yes, then promote one of its keys to the parent and take the
parent key into the lacking leaf
4. if neither of them has more than the minimum number of
keys then merge the lacking leaf and one of its neighbours
with their shared parent (the opposite of promoting a key)
if the merge step causes the parent with too few keys, then
we repeat the process up to the root, if required
Analysis of B-Trees
• The maximum number of items in a B-tree of order m and height h:
root
level 1
level 2
. . .
level h
m–1
m(m – 1)
m2(m – 1)
mh(m – 1)
• the total number of items is
(1 + m + m2 + m3 + … + mh)(m – 1) =
[(mh+1 – 1)/ (m – 1)] (m – 1) = mh+1 – 1
• When m = 5 and h = 2 this gives 53 – 1 = 124
• Demo
Revisit: B-Trees Motivation
• When searching tables held on disc, the cost
of each disc transfer is high
– If we use a B-tree of order 101, and assume that
we can transfer each node in one disc read
operation
– A B-tree of order 101 and height 3 can hold 1014 –
1 items (approximately 100 million) and any item
can be accessed with 3 disc reads (assuming we
hold the root in memory)