B-treesds162/wiki.files/06-B-tree.pdf1000 1000 1000 1000 1000 1000 1000 1000 1001·1000 1001 2·1000...

B-trees

B-trees

Motivation

When the size of a data-structure is very big, it is kept onsecondary memory (hard-disk/SSD).

Secondary memory is slow compared to RAM.

The time for reading a block in secondary memoryconsists of seek time (the delay between the request andthe transfer of the data) and transfer time.

The seek time is ∼ 10 milliseconds in hard-disks, and∼ 0.1 milliseconds in SSDs.

We want a data-structure that minimizes the diskaccesses.

B-trees

Motivation

Suppose we need to store a dynamic set with 109

elements on secondary memory.

The height of a binary search tree is at leastblog 109c = 29. Thus, a search operation in an optimalbinary search tree requires 30 disk accesses in the worstcase.

1

1 1

1 1 1 1

1

2

22

23

1000

1000 1000 1000

1000 1000 1000

1000

1001·1000

10012·1000

B-trees

Motivation

A B-tree is a generalization of binary search tree, that canstore many elements in one node.

If for example we store 1000 elements in a node (andeach internal node has 1001 children), a tree of height 2suffices. Thus, a search operation in an optimal B-treerequires 3 disk accesses in the worst case.

In fact, we can store the top two levels in RAM, and thena search operation requires only 1 disk access.

1

1 1

1 1 1 1

1

2

22

23

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1000 1000 1000

1000 1000 1000

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1001·1000

10012·1000

B-trees

B-tree

A B-tree is a rooted tree with the following properties.

1 A node x has the following fields.1 x .n, the number of keys stored at x .2 The keys of the elements, x .key1, x .key2, . . . , x .keyx .n in

increasing order (x .key1 ≤ x .key2 ≤ · · · ).3 Pointers to the satellite data of the keys.4 A boolean value x .leaf which is TRUE if and only if x is

a leaf.

t=2

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8 9 10 15 16 17 19 20 22 234 5 61 2

3 7 11

14

12 13

B-trees

B-tree

2 An internal nodes x has x .n + 1 children. The node xstores pointers to it children x .c1, x .c2, . . . , x .cx .n+1.

3 The keys x .key1, . . . , x .keyx .n separate the ranges of keysin the subtrees of x :If ki is some key in the subtree of x .ci thenk1 ≤ x .key1 ≤ k2 ≤ x .key2 ≤ · · · ≤ x .keyx .n ≤ kx .n+1.

t=2

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8 9 10 15 16 17 19 20 22 234 5 61 2

3 7 11

14

12 13

B-trees

B-tree

4 All leaves have the same depth.

5 The tree has a fixed parameter t, called minimum degree,that defines upper and lower bounds on the number ofkeys stored at each node:

1 Every node except the root must have at least t− 1 keys.2 If the tree is not empty, the root must have at least one

key.3 Every node can have at most 2t − 1 keys.

t=2

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3 7 11

14

12 13

B-trees

The height of a B-tree

Theorem

For n ≥ 1, the height of a B-tree with minimum degree tcontaining n keys is at most logt

n+12.

Consider a B-tree of height h with minimal number ofkeys: 1 key at the root and t − 1 keys at each other node.

Number of keys in nodes of depth i ≥ 1 is 2t i−1(t − 1).

t-1

0

1

2t-1

t-1

t-1

t-1 t-1 t-1 t-1

t-1

t-1

t-1

t-1 t-1 t-1

1

3

1

2

2t 2t(t-1)

2t2 2t2(t-1)

2(t-1)

1depth nodes keys

B-trees

The height of a B-tree

The number of keys in a minimal B-tree is

1 +h∑

i=1

2t i−1(t − 1) = 1 + 2(t − 1)h∑

i=1

t i−1

= 1 + 2(t − 1)

(th − 1

t − 1

)= 2th − 1.

For any B-tree we have n ≥ 2th − 1.

Therefore, h ≤ logtn+12

.

B-trees

Searching

Search(x , k)

(1) i ← 1(2) while i ≤ x .n AND k > x .keyi(3) i ← i + 1(4) if i ≤ x .n AND k = x .keyi(5) return (x , i)(6) else if x .leaf(7) return NULL(8) else(9) Disk-Read(x .ci)(10) return Search(x .ci , k)

t=2Search(T.root,8)

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3 7 11

14

12 13

xi=1

We assume that the root is stored in RAM.

CPU time: Θ(th) = Θ(t logt n).

Disk accesses: h ≤ logt n.

B-trees

Searching

Search(x , k)


t=2Search(T.root,8)

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8 9 10 15 16 17 19 20 22 234 5 61 2

3 7 11

14

12 13

xi=2




B-trees

Searching

Search(x , k)


t=2Search(T.root,8)

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8 9 10 15 16 17 19 20 22 234 5 61 2

3 7 11

14

12 13

xi=3




B-trees

Searching

Search(x , k)


t=2Search(T.root,8)

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8 9 10 15 16 17 19 20 22 234 5 61 2

3 7 11

14

12 13

xi=1




B-trees

Creating an empty B-tree

Create(T )

(1) x ← Allocate-Node()(2) x .leaf ← TRUE(3) x .n← 0(4) diskWrite(x)(5) T .root← x

CPU time: Θ(1).

Disk accesses: 1.

B-trees

Insertion

The goal is to insert the new element into an existing leaf.

This is easy if the leaf has less than 2t − 1 elements.

10 30 50

2 4 8

t=2

12 20 30 60 65

Insert(T,15)

15 can be inserted to this leaf

B-trees

Insertion

The hard case is when the leaf has 2t − 1 elements.

The solution is to perform the following step during thedescend in the tree: Before entering a node, if the nodehas 2t − 1 elements, perform node splitting on the nodein order to reduce the number of elements.

10 30 50

2 4 8

t=2

12 20 30 60 65

Insert(T,9)

B-trees

Node splitting

Let v be a node with 2t − 1 elements. The node splittingoperation on v replaces v with two nodes:one containing the t − 1 smallest elements of v , andanother containing the t − 1 largest elements.

The median element of v is moved to the parent of v .If v is the root, a new root is created.

10 40 60

15 20 25 30 35

10 40 60

15 20 30 35

25

B C D E A CA F B D FE

t=3

B-trees

Node splitting

If v is not the root, the parent of v has less than 2t − 1elements, so it is legal to move the median to the parent.

10 40 60

15 20 25 30 35

10 40 60

15 20 30 35

25

B C D E A CA F B D FE

t=3

B-trees

Example

10 30 50

2 4 8

t=2

12 20 30 60 65

Insert(T,9)

B-trees

Example

10 30 50

2 4 8

t=2

12 20 30 60 65

Insert(T,9)

Split node

B-trees

Example

10

2 4 8

t=2

12 20 30 60 65

Insert(T,9)

30

50

B-trees

Example

10

2 4 8

t=2

12 20 30 60 65

Insert(T,9)

30

50Split node

B-trees

Example

4,10

t=2

12 20 30 60 65

Insert(T,9)

30

50

2 8

B-trees

Example

4,10

t=2

12 20 40 60 65

Insert(T,9)

30

50

2 8 9

9 is addedto the leaf

B-trees

SplitChild(x , i)

(1) y ← x .ci(2) z ← AllocateNode()(3) z .leaf ← y .leaf(4) z .n← t − 1(5) for j = 1 to t − 1(6) z .keyj ← y .keyj+t

(7) if not y .leaf(8) for j = 1 to t(9) z .cj ← y .cj+t

(10) for j = x .n + 1 downtoi + 1

(11) x .cj+1 ← x .cj(12) x .ci+1 ← z(13) for j = x .n downto i(14) x .keyj+1 ← x .keyj(15) x .keyi ← y .keyt(16) x .n← x .n + 1(17) y .n← t − 1(18) DiskWrite(y)(19) DiskWrite(z)(20) DiskWrite(x)

B CA

t=310key

c

1520key

c

xi=2

yn: 5 leaf: FALSE

D E F

303525

4060n: 3 leaf: FALSE

B-trees

SplitChild(x , i)



(10) for j = x .n + 1 downtoi + 1


B CA

t=310key

c

1520key

c

xi=2

yn: 5 leaf: FALSE

D E F

303525 keyn: 2 leaf: FALSE

c

z


B-trees

SplitChild(x , i)



(10) for j = x .n + 1 downtoi + 1


B CA

t=310key

c

1520key

c

xi=2

yn: 5 leaf: FALSE

D E F


c

z3035


B-trees

SplitChild(x , i)



(10) for j = x .n + 1 downtoi + 1


B CA

t=310key

c

1520key

c

xi=2

yn: 5 leaf: FALSE

D E F


c

z3035

D E F


B-trees

SplitChild(x , i)



(10) for j = x .n + 1 downtoi + 1


B CA

t=310key

c

1520key

c

xi=2

yn: 2 leaf: FALSE

keyn: 2 leaf: FALSE

c

z3035

D E F


B-trees

Insert — Pseudo-code

Insert(T , k)

(1) r ← T .Root(2) if r .n = 2t − 1(3) s ← AllocateNode()(4) T .root← s(5) s.leaf ←FALSE(6) s.n← 0(7) s.c1 ← r(8) SplitChild(s, 1)(9) InsertNonfull(s, k)(10) else(11) InsertNonfull(r , k)

B-trees

Insert — Pseudo-code

InsertNonfull(x , k)

(1) i ← x .n(2) if x .leaf(3) while i ≥ 1 AND

k < x .keyi(4) x .keyi+1 ← x .keyi(5) i ← i − 1(6) x .keyi+1 ← k(7) x .n← x .n + 1(8) DiskWrite(x)(9) else

(10) while i ≥ 1 ANDk < x .keyi

(11) i ← i − 1(12) i ← i + 1(13) DiskRead(x .ci)(14) if x .ci .n = 2t − 1(15) SplitChild(x , i)

if k > x .keyi(16) i ← i + 1(17) InsertNonfull(x .ci , k)

B-trees

Deletion

The easy case is deleting an element from a leaf that hasmore than t − 1 elements.

The hard cases are deleting an item from a leaf that hast − 1 elements, and deleting an item from an internalnode.

The following step is performed during the descend in thetree: Before entering a node v , if v has t − 1 elements,perform shifting or merging in order to increase thenumber of elements.

B-trees

Shifting

If v has an immediate left sibling u and u has at least telements:

1 Move the element in the parent of v that “separates” vand u to v .

2 Move the maximum element of u to the parent of v .

10 40

E FA

5020 25 30

A C

10

B

20 25 5040

30t=2

B C D D FE

vu

B-trees

Shifting

Otherwise, if v has an immediate right sibling w , and whas at least t elements then

1 Move the element of the parent of v that “separates” vand w to v .

2 Move the minimum element of w to the parent of v .

10 40

A B C E

30 50 60

D D EA C

10

60

B

30 40

50t=2

v w

5 5

B-trees

Merging

If v has no immediate sibling with at least t elements:1 Merge v with an immediate sibling

(merge with the immediate left sibling if it exists, andotherwise merge with the right immediate right sibling).

2 Move the “separating element” from the parent of v tothe new node.

10 40

A B

30

A CB

30 40 50

t=2

50

C D

10

D

v

B-trees

Deleting a key — Case 1

When reaching the node x that contains the key k we want todelete. there are several cases.

Case 1: If x is a leaf, delete the key.

20 25 30 20 30

t=2

x

B-trees


Suppose that x is an internal node, and k = x .keyi .Let y = x .ci and z = x .ci+1.

Case 2: If y has at least t elements, recursively delete thepredecessor of k (which is in the subtree of y) and put it in xinstead of k .

t=2

zy

x Delete(x,40)10 40

20 30

23 26 35

predecessor

B-trees




t=2

zy

x Delete(x,40)10 40

20 26

23 30 35

B-trees




t=2

zy

x Delete(x,40)

20 26

23 30

10 35

B-trees


Case 3: If y has t − 1 elements and z has at least t elements,recursively delete the successor of k (which is in the subtree ofz) and put it in x instead of k .

t=2

zy

x Delete(x,40)

30 50 60

43 46

successor

10 40

B-trees


Case 3: If y has t − 1 elements and z has at least t elements,recursively delete the successor of k (which is in the subtree ofz) and put it in x instead of k .

t=2

zy

x Delete(x,40)

30 50 60

10 43

46

B-trees


Case 4: If both y and z have t − 1 elements, merge y and z ,and recursively delete k from the merged node.

t=2

zy

x Delete(x,40)

30

10 40

50

25 35 45 55

B-trees



t=2

zy

x

Delete(x,40)

30 40 50

25 35 45 55

10

B-trees



t=2

x

Delete(x,40)

30 50

25 55

10

35 40 45

B-trees



t=2

x

Delete(x,40)

30 50

25 55

10

35 45

B-trees

Example

t=3

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14

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delete(T,6)

B-trees

Example

t=3

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14

12 13

delete(T,6)

6 < 14 so the next node is the 1st child of the current node.This child has 3 > t−1 elements, so no modification is needed.

B-trees

Example

t=3

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3 7 11

14

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delete(T,6)

3 < 6 < 7 so the next node is the 2nd child of the current node.This child has 3 > t−1 elements, so no modification is needed.

B-trees

Example

t=3

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3 7 11

14

12 13

delete(T,6)

B-trees

Example

t=3

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3 7 11

14 delete(T,6)

The element 6 is deleted from the current node.

B-trees

Example

t=3

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3 7 11

14 delete(T,11)

B-trees

Example

t=3

18 21

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3 7 11

14 delete(T,11)


B-trees

Example

t=3

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3 7 11

14 delete(T,11)

11 is the 3rd item of the node.The 4th child has 2 = t − 1 elements, so we don’t replace 11with its successor.The 3rd child has 3 > t − 1 so we replace 11 by its predecessorand delete the predecessor.

B-trees

Example

t=3

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3 7 11

14 delete(T,11)

B-trees

Example

t=3

18 21

8 9 15 16 17 19 20 22 2312 13 4 51 2

3 7 11

14 delete(T,11)


B-trees

Example

t=3

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3 7 10

14 delete(T,11)

The element 11 is replaced by 10.

B-trees

Example

t=3

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3 7 10

14 delete(T,7)

B-trees

Example

t=3

18 21

8 9 15 16 17 19 20 22 2312 13 4 51 2

3 7 10

14 delete(T,7)


B-trees

Example

t=3

18 21

8 9 15 16 17 19 20 22 2312 13 4 51 2

3 7 10

14 delete(T,7)

7 is the 2nd item of the node.Both the 2nd and 3rd children of the node have 2 = t − 1elements. Therefore we merge these children and move 7 tothe new node.

B-trees

Example

t=3

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15 16 17 19 20 22 2312 13 4 5 7 8 91 2

3 10

14 delete(T,7)

B-trees

Example

t=3

18 21

15 16 17 19 20 22 2312 13 4 5 8 91 2

3 10

14 delete(T,7)


B-trees

Example

t=3

18 21

15 16 17 19 20 22 2312 13 4 5 8 91 2

3 10

14 delete(T,4)

B-trees

Example

t=3

18 21

15 16 17 19 20 22 2312 13 4 5 8 91 2

3 10

14 delete(T,4)

4 < 14 so the next node is the 1st child of the current node.This child has 2 > t − 1 elements, so a modification is needed.This child has no immediate sibling with more than 2 = t − 1elements, so perform merging on this child and its sibling.The height of the tree decreases by 1.

B-trees

Example

t=3

15 16 17 19 20 22 2312 13 4 5 8 91 2

3 10 14 18 21

delete(T,4)

3 < 4 < 10 so the next node is the 2nd child of the currentnode.This child has 4 > t−1 elements, so no modification is needed.

B-trees

Example

t=3

15 16 17 19 20 22 2312 13 4 5 8 91 2

3 10 14 18 21

delete(T,4)

B-trees

Example

t=3

15 16 17 19 20 22 2312 13 5 8 91 2

3 10 14 18 21

delete(T,4)


B-trees

Example

t=3

15 16 17 19 20 22 2312 13 5 8 91 2

3 10 14 18 21

delete(T,2)

B-trees

Example

t=3

15 16 17 19 20 22 2312 13 5 8 91 2

3 10 14 18 21

delete(T,2)

2 < 3 so the next node is the 1st child of the current node.This child has 2 = t − 1 elements, so a modification is needed.The immediate sibling of this child has 3 > t − 1 elements, sowe perform shifting.

B-trees

Example

t=3

15 16 17 19 20 22 2312 13 8 91 2 3

5 10 14 18 21

delete(T,2)

B-trees

Example

t=3

15 16 17 19 20 22 2312 13 8 91 3

5 10 14 18 21

delete(T,2)


B-trees

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