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Analysis of Algorithms. Trees-II. Prof. Muhammad Saeed. Tree. B-Trees. Tree. Motivation for B-Trees …. We can store an entire data structure in main memory What if we have so much data that it won’t fit? We will have to use disk storage but when this happens our time complexity fails

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Trees-II

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Trees ii

Analysis of Algorithms

Trees-II

Prof. Muhammad Saeed

Analysis Of Algorithms Trees-II


Trees ii

Tree

B-Trees

Analysis Of Algorithms Trees-II


Motivation for b trees

Tree

Motivation for B-Trees …..

  • We can store an entire data structure in main memory

  • What if we have so much data that it won’t fit?

  • We will have to use disk storage but when this happens our time complexity fails

  • The problem is that Big-Oh analysis assumes that all operations take roughly equal time

  • This is not the case when disk access is involved

Analysis Of Algorithms Trees-II


Motivation

Tree

….. Motivation…….

Assume that a disk spins at 3600 RPM

In 1 minute it makes 3600 revolutions, hence one revolution occurs in 1/60 of a second, or 16.7ms

On average what we want is half way round this disk – it will take 8ms

This sounds good until you realize that we get 120 disk accesses a second – the same time as 25 million instructions

In other words, one disk access takes about the same time as 200,000 instructions

It is worth executing lots of instructions to avoid a disk access

Analysis Of Algorithms Trees-II


Motivation1

Tree

….. Motivation

  • Assume that we use an Binary tree to store all the details of people in Canada (about 32 million records)

  • We still end up with a very deep tree with lots of different disk accesses; log2 20,000,000 is about 25, so this takes about 0.21 seconds (if there is only one user of the program)

  • We know we can’t improve on the log n for a binary tree

  • But, the solution is to use more branches and thus less height!

  • As branching increases, depth decreases

Analysis Of Algorithms Trees-II


Trees ii

Tree

B-Tree

  • A B-tree of order m is a tree where each node may have up to m children in which:

  • the number of keys in each non-leaf node is one less than the number of its children and these keys partition the keys in the children in the fashion of a search tree

  • all leaves are on the same level

  • all non-leaf nodes except the root have at least m / 2 children

  • the root is either a leaf node, or it has from two to m children

  • a leaf node contains no more than m – 1 keys

  • The number m should always be odd

Analysis Of Algorithms Trees-II


Trees ii

Tree

A B-tree of order 5 containing 26 items

26

6

12

42

51

62

1

2

4

7

8

13

15

18

25

27

29

45

46

48

53

55

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90

Analysis Of Algorithms Trees-II


Trees ii

Tree

  • Constructing a B-Tree …..

  • Suppose we start with an empty B-tree and keys arrive in the following order:1 12 8 2 25 6 14 28 17 7 52 16 48 68 3 26 29 53 55 45

  • We want to construct a B-tree of order 5

  • The first four items go into the root:

  • To put the fifth item in the root would violate condition 5

  • Therefore, when 25 arrives, pick the middle key to make a new root

1

2

8

12

Analysis Of Algorithms Trees-II


Constructing a b tree

Tree

….. Constructing a B-tree …..

Add 25 to the tree

Exceeds Order. Promote middle and split.

1 12 8 2 25 6 14 28 17 7 52 16 48 68 3 26 29 53 55 45

1

2

8

12

25

Analysis Of Algorithms Trees-II


Constructing a b tree1

Tree

….. Constructing a B-tree …..

8

1

2

12

25

1 12 8 225 6 14 28 17 7 52 16 48 68 3 26 29 53 55 45

6, 14, 28 get added to the leaf nodes:

8

14

28

1

2

6

1

2

12

25

Analysis Of Algorithms Trees-II


Constructing a b tree2

Tree

….. Constructing a B-tree …..

Adding 17 to the right leaf node would over-fill it, so we take the middle key, promote it (to the root) and split the leaf

8

1 12 8 2256 14 28 17 7 52 16 48 68 3 26 29 53 55 45

1 12 8 2256 14 28 17 7 52 16 48 68 3 26 29 53 55 45

25

28

28

17

1

6

2

2

12

14

Analysis Of Algorithms Trees-II


Trees ii

Tree

….. Constructing a B-tree …..

7, 52, 16, 48 get added to the leaf nodes

8

17

1 12 8 2256 14 28 17 7 52 16 48 68 3 26 29 53 55 45

7

25

28

48

52

6

1

2

12

14

16

Analysis Of Algorithms Trees-II


Constructing a b tree3

Tree

….. Constructing a B-tree …..

Adding 68 causes us to split the right most leaf, promoting 48 to the root

1 12 8 2256 14 28 17 7 52 16 48 68 3 26 29 53 55 45

8

17

52

7

48

6

28

2

25

68

1

16

14

12

Analysis Of Algorithms Trees-II


Constructing a b tree4

Tree

….. Constructing a B-tree …..

Adding 3 causes us to split the left most leaf

48

17

8

1 12 8 2256 14 28 17 7 52 16 48 68 3 26 29 53 55 45

7

7

16

14

12

6

2

3

1

25

28

52

68

Analysis Of Algorithms Trees-II


Constructing a b tree5

Tree

….. Constructing a B-tree …..

Add 26, 29, 53, 55 then go into the leaves

48

17

8

3

1 12 8 2256 14 28 17 7 52 16 48 68 3 26 29 53 55 45

16

14

12

1

2

6

7

25

28

52

68

55

26

53

29

Analysis Of Algorithms Trees-II


Constructing a b tree6

Tree

….. Constructing a B-tree …..

Add 45 increases the trees level

Exceeds Order. Promote middle and split.

1 12 8 2256 14 28 17 7 52 16 48 68 3 26 29 53 55 45

Exceeds Order. Promote middle and split.

48

17

8

3

16

29

14

28

45

68

12

26

55

25

53

52

1

2

6

7

Analysis Of Algorithms Trees-II


Inserting into a b tree

Tree

Inserting into a B-Tree

  • Attempt to insert the new key into a leaf

  • If this would result in that leaf becoming too big, split the leaf into two, promoting the middle key to the leaf’s parent

  • If this would result 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

Analysis Of Algorithms Trees-II


Exercise in inserting a b tree

Tree

Exercise in Inserting a B-Tree

  • Insert the following keys to a 5-way B-tree:

  • 3, 7, 9, 23, 45, 1, 5, 14, 25, 24, 13, 11, 8, 19, 4, 31, 35, 56

Analysis Of Algorithms Trees-II


Answer to exercise

Tree

Answer to Exercise

Analysis Of Algorithms Trees-II


Removal from a b tree

Tree

Removal from a B-tree …..

  • 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 not in a leaf then it is guaranteed (by the nature of a B-tree) that its predecessor or successor will be in a leaf -- in this case can we delete the key and promote the predecessor or successor key to the non-leaf deleted key’s position.

Analysis Of Algorithms Trees-II


Removal from a b tree1

Tree

….. Removal from a B-tree

  • If (1) or (2) lead to a leaf node containing less than the minimum number of keys then we have to look at the siblings immediately adjacent to the leaf in question:

  • 3: if one of them has more than the min’ number of keys then we can promote one of its keys to the parent and take the parent key into our lacking leaf

  • 4: if neither of them has more than the min’ number of keys then the lacking leaf and one of its neighbours can be combined with their shared parent (the opposite of promoting a key) and the new leaf will have the correct number of keys; if this step leave the parent with too few keys then we repeat the process up to the root itself, if required

Analysis Of Algorithms Trees-II


Type 1 simple leaf deletion

12

2

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29

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69

9

72

52

15

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22

43

Tree

Type #1: Simple leaf deletion

Assuming a 5-way

B-Tree, as before...

Delete 2: Since there are enough

keys in the node, just delete it

Analysis Of Algorithms Trees-II


Type 2 simple non leaf deletion

56

12

29

69

52

72

7

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31

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Tree

Type #2: Simple non-leaf deletion

56

Delete 52

Borrow the predecessor

or (in this case) successor

Analysis Of Algorithms Trees-II


Type 4 too few keys in node and its siblings

12

29

56

Join back together

7

9

15

31

43

22

69

72

Tree

Type #4: Too few keys in node and its siblings

Too few keys!

Delete 72

Analysis Of Algorithms Trees-II


Type 4 too few keys in node and its siblings1

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29

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69

Tree

Type #4: Too few keys in node and its siblings

Analysis Of Algorithms Trees-II


Type 3 enough siblings

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29

Demote root key and

promote leaf key

7

9

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22

31

43

56

69

Tree

Type #3: Enough siblings

Delete 22

Analysis Of Algorithms Trees-II


Type 3 enough siblings1

43

56

69

Tree

Type #3: Enough siblings

12

31

7

9

15

29

Analysis Of Algorithms Trees-II


Exercise in removal from a b tree

Tree

Exercise in Removal from a B-Tree

  • Given 5-way B-tree created by these data

  • 3, 7, 9, 23, 45, 1, 5, 14, 25, 24, 13, 11, 8, 19, 4, 31, 35, 56

    Add these further keys: 2, 6,12

  • Delete these keys: 4, 5, 7, 3, 14

  • Analysis Of Algorithms Trees-II


    Answer to exercise1

    Tree

    Answer to Exercise

    Analysis Of Algorithms Trees-II


    Analysis of b trees

    Tree

    Analysis of B-Trees

    • The maximum number of items in a B-tree of order m and height h:

      rootm – 1

      level 1m(m – 1)

      level 2m2(m – 1)

      . . .

      level hmh(m – 1)

    • So, 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

    Analysis Of Algorithms Trees-II


    Reasons for using b trees

    Tree

    Reasons for using B-Trees

    • When searching tables held on disc, the cost of each disc transfer is high but doesn't depend much on the amount of data transferred, especially if consecutive items are transferred

    • If we use a B-tree of order 101, say, 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)

    • 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)

    • B-Trees are always balanced (since the leaves are all at the same level), so 2-3 trees make a good type of balanced tree

    Analysis Of Algorithms Trees-II


    Trees ii

    Tree

    • B-Trees:widely used for file systems and databases

      • Windows: HPFS.

      • Mac: HFS, HFS+.

      • Linux: ReiserFS, XFS, Ext3FS, JFS.

      • Databases: ORACLE, DB2, INGRES, SQL, PostgreSQL

    Analysis Of Algorithms Trees-II


    Trees ii

    Tree

    Binomial-Trees

    Analysis Of Algorithms Trees-II


    Trees ii

    Bo

    Bo

    B1

    B1

    Tree

    Binomial Trees …..

    The binomial tree Bk is an ordered tree defined recursively.

    B0

    B1

    B2

    Analysis Of Algorithms Trees-II


    Trees ii

    B2

    B3

    B2

    B3

    Tree

    ….. Binomial Trees …..

    B3

    B4

    Analysis Of Algorithms Trees-II


    Trees ii

    Tree

    ….. Binomial Trees

    • Properties for tree Bk:

    • There are 2k nodes

    • The height of the tree is k

    • The number of nodes at depth i for i = 0…k is

    • The root has degree k which is greater than any other node

    Analysis Of Algorithms Trees-II


    Trees ii

    Tree

    Binomial Heaps …..

    • A binomial heap H is a set of binomials trees that satisfies the following binomial-heap properties:

    • Each binomial tree in H obeys the min-heap property.

    • For any nonnegative integer k, there is at most one binomial tree in H whose root has degree k.

    • Binomial trees will be joined by a linked list of the roots

    Analysis Of Algorithms Trees-II


    Trees ii

    Tree

    ….. Binomial Heaps …..

    An n node binomial heap consists of at most Floor(log n) + 1 binomial trees.

    Analysis Of Algorithms Trees-II


    Trees ii

    Tree

    ….. Binomial Heaps …..

    How many binary bits are needed to count the nodes in any given Binomial Tree? Answer: k for Bk, where k is the degree of the root.

    B3 3 bits

    B2 2 bits

    B0 0 bits

    B1 1 bits

    11

    111

    1

    1

    1

    4

    2

    101

    10

    01

    3

    2

    100

    0

    6

    3

    2

    110

    4

    011

    00

    4

    001

    7

    8

    4

    010

    000

    9

    Analysis Of Algorithms Trees-II


    Trees ii

    Tree

    ….. Binomial Heaps …..

    Head

    2

    1

    1

    <1110>

    4

    3

    2

    6

    3

    9

    4

    8

    7

    4

    9

    • There are 14 nodes, which is 1110 in binary. This also can be written as <1110>, which means there is no B0, one B1, one B2 and one B3.

    • There is a corresponding set of trees for a heap of any size!

    Analysis Of Algorithms Trees-II


    Trees ii

    Tree

    ….. Binomial Heaps …..

    • Node Representation

    Analysis Of Algorithms Trees-II


    Trees ii

    Tree

    ….. Binomial Heaps …..

    • Binomial Min-Heap

    • Walk across roots, find minimum

    • O(log n) since at most log n + 1 trees

    Head

    2

    1

    3

    4

    3

    2

    6

    4

    9

    4

    8

    7

    6

    9

    Analysis Of Algorithms Trees-II


    Trees ii

    z

    y

    z

    y

    z

    y

    z

    y

    Tree

    ….. Binomial Heaps …..

    Binomial-Link(y,z)

    1.p[y]  z

    2.sibling[y]  child[z]

    3.child[z]  y

    4.degree[z]  degree[z] + 1

    Link binomial trees with the same degree. Note that z, the second argument to BL(), becomes the parent, and y becomes the child.

    (1) Runtime

    Analysis Of Algorithms Trees-II


    Trees ii

    Tree

    ….. Binomial Heaps …..

    Binomial-Heap-Union(H1, H2)

    • H  Binomial-Heap-Merge(H1, H2)

    • This merges the root lists of H1 and H2 in increasing order of root degree

    • Walk across the merged root list, merging binomial trees of equal degree. If there are three such trees in a row only merge the last two together (to maintain property of increasing order of root degree as we walk the roots)

    • Runtime: Merge time plus Walk Time: O(log n)

    Analysis Of Algorithms Trees-II


    Trees ii

    2

    60

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    18

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    Tree

    ….. Binomial Heaps …..

    Starting with the following two binomial heaps:

    Merge root lists, but now we have two trees of same degree

    Combine trees of same degree using binomial link, make smaller key the root of the combined tree

    Analysis Of Algorithms Trees-II


    Trees ii

    6

    head[H1]

    head[H2]

    3

    18

    7

    15

    12

    37

    29

    10

    44

    8

    25

    28

    33

    48

    22

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    23

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    50

    24

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    45

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    Tree

    ….. Binomial Heaps …..

    Merge Heap H1 and H2

    Analysis Of Algorithms Trees-II


    Trees ii

    Tree

    ….. Binomial Heaps …..

    Binomial-Heap-Insert(H)

    • To insert a new node, simple create a new Binomial Heap with one node (the one to insert) and then Union it with the heap

    • H’  Make-Binomial-Heap()

    • p[x]  NIL

    • child[x]  NIL

    • sibling[x]  NIL

    • degree[x]  0

    • head[H’]  x

    • H  Binomial-Heap-Union(H, H’)

    • Runtime: O(log n)

    Analysis Of Algorithms Trees-II


    Trees ii

    Broken into separate Binomial Trees after root’s extraction

    Part of original heap showing binomial tree with minimal root.

    H

    Reversal of roots, and combining into new heap

    Tree

    ….. Binomial Heaps …..

    Binomial-Heap-Extract-Min(H)

    With a min-heap, the root has the least value in the heap.

    Notice that if we remove the root from the figure below, we are left with four heaps, and they are in decreasing order of degree. So to extract the min we create a root list of the children of the node being extracted, but do so in reverse order. Then call Binomial-Heap-Union(..)

    • Runtime: Θ(log n)

    Analysis Of Algorithms Trees-II


    Trees ii

    Tree

    ….. Binomial Heaps …..

    head[H]

    1

    37

    10

    16

    12

    25

    6

    41

    28

    13

    26

    29

    18

    23

    8

    14

    77

    42

    38

    17

    11

    27

    Extract-Min(H)

    Analysis Of Algorithms Trees-II


    Trees ii

    Tree

    ….. Binomial Heaps …..

    Heap Decrease Key

    • Same as decrease-key for a binary heap

    • Move the element upward, swapping values, until we reach a position where the value is  key of its parent

    • Runtime: Θ(log n)

    Heap Delete Key

    • Set key of node to delete to –infinity and extract it:

    • Decrease-key(H, p, -∞)

    • Extract-min(H)

    • Runtime: Θ(log n)

    Analysis Of Algorithms Trees-II


    Trees ii

    12

    18

    16

    6

    26

    29

    23

    8

    14

    37

    10

    42

    38

    17

    11

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    41

    28

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    head[H]

    77

    25

    Tree

    ….. Binomial Heaps

    Decrease key 26 to 7

    Analysis Of Algorithms Trees-II


    Trees ii

    Analysis Of Algorithms Trees-II


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