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3/12/2015
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Chapter 11 Balanced Trees
Why care about advanced implementations?
Same entries, different insertion sequence:
Not good! Would like to keep tree balanced.
Balanced binary tree
• The disadvantage of a binary search tree is that its height can be as large as N‐1
• This means that the time needed to perform insertion and deletion and many other operations can be O(N) in the worst case
• We want a tree with small height
• A binary tree with N node has height at least (log N) • Thus, our goal is to keep the height of a binary search tree
O(log N)
• Such trees are called balanced binary search trees. Examples are AVL tree, red‐black tree.
AVL Trees
• A sorted binary tree
• The heights of two subtrees at any given node differ by at most 1
New Height Definition
For convenience, we redefine Height of a node
• The height of a leaf is 1. The height of a null node is zero.
• The height of an internal node is the maximum height of its children plus 1
Note that this definition of height is different from the one we
defined previously (we defined the height of a leaf as zero previously).
AVL Tree
Class BinaryNode
KeyType: Key
integer: Height
BinaryNode: LeftChild
BinaryNode: RightChild
BinaryNode: parent // optional
Constructor(KeyType: key)
Key = key
Height = 1
End Constructor
End Class
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AVL tree
• An AVL tree is a binary search tree in which
– for every node in the tree, the height of the left and right subtrees differ by at most 1.
AVL property violated here
The left tree is AVL; the right tree is not AVL.
AVL tree• Let x be the root of an AVL tree of height h• Let Nh denote the minimum number of nodes in an AVL
tree of height h• We have
• By repeated substitution, we obtain the general form
• The boundary conditions are: N1=1 and N2 =2. This implies that h = O(log Nh).
• Thus, many operations (searching, insertion, deletion) on an AVL tree will take O(log N) time.
2
2
21
2
12
1
h
h
hhh
N
N
NNN
ihi
h NN 22
Insertion
• First, insert the new key as a new leaf just as in an ordinary binary search tree
• Then trace the path from the new leaf towards the root. For each node x encountered, check if heights of x.LeftChild and x.RightChild differ by at most 1.
• If yes, proceed to parent(x). If not, restructure by doing either a single rotation or a double rotation.
• For insertion, once we perform a rotation at a node x, we won’t need to perform any rotation at any ancestor of x.
Rotations
• Since an insertion/deletion involves adding/deleting a single node, this can only increase/decrease the height of some subtree by 1
• Thus, if the AVL tree property is violated at a node x, it means that the heights of left(x) ad right(x) differ by exactly 2.
• Rotations will be applied to x to restore the AVL tree property.
Rotations
• When the tree structure changes (e.g., insertion or deletion), we need to transform the tree to restore the AVL tree property.
• This is done using single rotations or double rotations.
x
y
AB
C
y
x
AB C
Before RotationAfter Rotation
e.g. Single Rotation
Insertion & Rotation
• Let x be the node at which left(x) and right(x) differ by 2
• Assume that the height of x is h+3
• There are 4 cases
– Height of left(x) is h+2 (i.e. height of right(x) is h)
• Height of left(left(x)) is h+1 single rotate with left child
• Height of right(left(x)) is h+1 double rotate with left child
– Height of right(x) is h+2 (i.e. height of left(x) is h)
• Height of right(right(x)) is h+1 single rotate with right child
• Height of left(right(x)) is h+1 double rotate with right child
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Single Right Rotation
The new key is inserted in the subtree A. The AVL-property is violated at xheight of left(x) is h+2height of right(x) is h.
Single Left Rotation
Single rotation takes O(1) time. Insertion takes O(log N) time.
The new key is inserted in the subtree C. The AVL-property is violated at x.
keys(A) < x < keys(B) < y < keys(C)
5
3
1 4
Insert 0.8
AVL Tree
8
0.8
5
3
1 4
8
x
y
A
B
C
3
51
0.84 8
After rotation
rightRotate( BinaryNode: x ){
BinaryNode y, z;
y = x.LeftChild;x.LeftChild = y.RightChild;(y.RightChild).parent = x;y.RightChild = x;z = y.parent = x.parent;x.parent = y;
if (z == Nil )Root = y;
else if ( x == z.LeftChild )z.LeftChild = y;
elsez.RightChild = y;
}
z z
leftRotate( BinaryNode: x ){
BinaryNode y, z;
y = x.RightChild;x.RightChild = y.LeftChild;(y.LeftChild).parent = x;y.LeftChild = x;z = y.parent = x.parent;x.parent = y;
if (z == Nil )Root = y;
else if ( x == z.LeftChild )z.LeftChild = y;
elsez.RightChild = y;
}
z z
Double rotationThe new key is inserted in the subtree B1 or B2. The AVL-property is violated at x.x-y-z forms a zig-zag shape
also called left-right rotate
Keys(A) < y < keys(B1) < z < keys(B2) < x < keys(C)
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Double rotation
The new key is inserted in the subtree B1 or B2. The AVL-property is violated at x.
also called right-left rotate
keys(A) < x < keys(B1) < z < keys(B2) < y < keys(C)
5
3
1 4
Insert 3.5
AVL Tree
8
3.5
5
3
1 4
8
4
5
1
3
3.5 After Rotation
x
y
A z
B
C
8
leftRightRotate( BinaryNode: x ){
leftRotate(x.LeftChild);rightRotate(x);
}
rightLeftRotate( BinaryNode: x ){
rightRotate(x.RightChild);leftRotate(x);
}
An Extended Example
Insert 3,2,1,4,5,6,7, 16,15,14
3
Fig 1
3
2
Fig 2
3
2
1
Fig 3
2
1 3Fig 4
2
1 3
4Fig 5
2
1 3
4
5
Fig 6
Single rotation
Single rotation
2
1 4
53
Fig 7 6
2
1 4
53
Fig 8
4
2 5
61 3
Fig 9
4
2 5
61 3
7Fig 10
4
2 6
71 3
5 Fig 11
Single rotation
Single rotation
4
2 6
71 3
5 16
Fig 12
4
2 6
71 3
5 16
15Fig 13
4
2 6
151 3 5
167Fig 14
Double rotation
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5
4
2 7
151 3 6
1614
Fig 16
4
2 6
151 3 5
167
14
Fig 15
Double rotation
Deleting Nodes
• Use the same rotations
Deletion
• Delete a node x as in ordinary binary search tree. Note that the last node deleted is a node with 0 or 1 child.
• Then trace the path from that node towards the root.
• For each node x encountered, check if heights of x.LeftChild and x.RightChild differ by at most 1. If yes, proceed to x.parent. If not, perform an appropriate rotation at x. There are 4 cases as in the case of insertion.
• For deletion, after we perform a rotation at x, we may have to perform a rotation at some ancestor of x. Thus, we must continue to trace the path until we reach the root.
Deletion
• On closer examination: the single rotations for deletion can be divided into 4 cases (instead of 2 cases)
– Two cases for rotate with left child
– Two cases for rotate with right child
Single rotations in deletion
rotate with left child
In both figures, a node is deleted in subtree C, causing the height to drop to h. The height of y is h+2. When the height of subtree A is h+1, the height of B can be h or h+1. Fortunately, the same single rotation can correct both cases.
Single rotations in deletion
rotate with right child
In both figures, a node is deleted in subtree A, causing the height to drop to h. The height of y is h+2. When the height of subtree C is h+1, the height of B can be h or h+1. A single rotation can correct both cases.
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Rotations in deletion
• There are 4 cases for single rotations, but we do not need to distinguish among them.
• There are exactly two cases for double rotations (as in the case of insertion)
• Therefore, we can reuse exactly the same procedure for insertion to determine which rotation to perform
2‐3 Trees
• A node contains one or two key values (called 2‐nodes or 3‐nodes, respectively)
• Every internal 2‐node has two children
• Every internal 3‐node has three children
• All leaves are at the same level.
2‐3 Trees
each internal node has either 2 or 3 children all leaves are at the same level
Features
Example of 2‐3 Tree
2‐3 Trees with Ordered Nodes
2-node 3-node
• leaf node can be either a 2-node or a 3-node
Traversing a 2‐3 Treeinorder(TwoThreeTree: ttt)
if (ttt is a leaf)visit the data item of ttt
else if (ttt has two data items){
inorder(left subtree of ttt)visit the first data iteminorder(middle subtree of ttt)visit the second data iteminorder(right subtree of ttt)
}else{
inorder(left subtree of ttt)visit the data iteminorder(right subtree of ttt)
}
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Searching a 2‐3 Tree
TreeItemType: retrieveItem(TwoThreeTree: ttt, KeyType: key)
if(key is inside ttt’s node)return the data portion of key
else if (ttt is a leaf)return NIL
else {let sttt be the appropriate subtreereturn retrieveItem(sttt, key)
}
What did we gain?
What is the time efficiency of searching for an item?
Because every internal node has at least two children, a tree containing N nodes can have a height of at most log2(N)
Gain: Ease of Keeping the Tree Balanced
Binary SearchTree
2-3 Tree
both trees afterinserting items39, 38, ... 32
Inserting ItemsInsert 39
Inserting Items
Insert 38
1) insert 38 in leaf2) divide leaf
and move middlevalue up to parent
3) result
Inserting Items
Insert 37
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Inserting ItemsInsert 36
insert in leaf
divide leafand move middlevalue up to parent
overcrowdednode
Inserting Items... still inserting 36
divide overcrowded node,move middle value up to parent,
attach children to smallest and largest
result
Inserting Items
After Insertion of 35, 34, 33
Inserting so far
Inserting so far Inserting ItemsHow do we insert 32?
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Inserting Items creating a new root if necessary tree grows at the root
Inserting ItemsFinal Result
70
Deleting ItemsDelete 70
80
Deleting ItemsDeleting 70: swap 70 with inorder successor (80)
Deleting Items
Deleting 70: ... get rid of 70
Deleting ItemsResult
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Deleting ItemsDelete 100
Deleting ItemsDeleting 100
Deleting ItemsResult
Deleting ItemsDelete 80
Deleting ItemsDeleting 80 ...
Deleting ItemsDeleting 80 ...
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Deleting ItemsDeleting 80 ...
Deleting ItemsFinal Result
comparison withbinary search tree
Deletion Algorithm I
1. Locate node n, which contains item I
2. If node n is not a leaf swap I with inorder successor
deletion always begins at a leaf
3. If leaf node n contains another item, just delete item Ielse
try to redistribute nodes from siblings (see next slide)if not possible, merge node (see next slide)
Deleting item I:
Deletion Algorithm II
A sibling has 2 items: redistribute item
between siblings andparent
No sibling has 2 items: merge node move item from parent
to sibling
Redistribution
Merging
Deletion Algorithm III
Internal node n has no item left redistribute
Redistribution not possible: merge node move item from parent
to sibling adopt child of n
If n's parent ends up without item, apply process recursively
Redistribution
Merging
Deletion Algorithm IV
If merging process reaches the root and root is without item delete root
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Operations of 2‐3 Trees
all operations have time complexity of log n
2‐3‐4 Trees• similar to 2-3 trees• 4-nodes can have 3 items and 4 children
4-node
2‐3‐4 Tree Example 2‐3‐4 Tree: Insertion
Insertion procedure:• similar to insertion in 2-3 trees• items are inserted at the leafs• since a 4-node cannot take another item,
4-nodes are split up during insertion process
Strategy• on the way from the root down to the leaf:
split up all 4-nodes "on the way" insertion can be done in one pass
(remember: in 2-3 trees, a reverse pass might be necessary)
2‐3‐4 Tree: Insertion
Inserting 60, 30, 10, 20, 50, 40, 70, 80, 15, 90, 100
2‐3‐4 Tree: Insertion
Inserting 60, 30, 10, 20 ...
... 50, 40 ...
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2‐3‐4 Tree: Insertion
Inserting 50, 40 ...
... 70, ...
2‐3‐4 Tree: Insertion
Inserting 70 ...
... 80, 15 ...
2‐3‐4 Tree: Insertion
Inserting 80, 15 ...
... 90 ...
2‐3‐4 Tree: Insertion
Inserting 90 ...
... 100 ...
(b)
2‐3‐4 Tree: Insertion
Inserting 100 ...
2‐3‐4 Tree: Insertion Procedure
Splitting 4-nodes during Insertion
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2‐3‐4 Tree: Insertion Procedure
Splitting a 4-node whose parent is a 2-node during insertion
2‐3‐4 Tree: Insertion Procedure
Splitting a 4-node whose parent is a 3-node during insertion
2‐3‐4 Tree: DeletionDeletion procedure:
• similar to deletion in 2-3 trees• items are deleted at the leafs swap item of internal node with inorder successor
• note: a 2-node leaf creates a problem
Strategy (different strategies possible)
• on the way from the root down to the leaf:turn 2-nodes (except root) into 3-nodes
deletion can be done in one pass(remember: in 2-3 trees, a reverse pass might be necessary)
Red-Black Tree
• A ref-black tree is a binarysearch such that each nodehas a color of either red or black.
• The root is black.
• Every path from a node to a leaf contains the samenumber of black nodes.
• If a node is red then itsparent must be black.
Class BinaryNodeKeyType: KeyBoolean: isRedBinaryNode: LeftChildBinaryNode: RightChildBinaryNode: parent
Constructor(KeyType: key)Key = keyisRed = true
End ConstructorEnd Class
ExampleThe root is black.
The parent of any red node must be
black.
Maintaining the Red Black Properties in a Tree
• Insertions
• Must maintain rules of Red Black Tree.
• New Node always a leaf
– can't be black or we will violate rule of the same # of blacks along any path
– therefore the new leaf must be red
– If parent is black, done (trivial case)
– if parent red, things get interesting because a red leaf with a red parent violates no double red rule.
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The parent of a red node must be black.
Algorithm: InsertionA red-black tree is a particular binary search tree, so create a new node as red and insert it as in normal search tree.
What property may be violated?
579
Violation!
7
Algorithm: InsertionWe have detected a need for balance when z is red and its parent, too.
• If z has a red uncle: colour the parent and uncle black, and grandparent red. Then replace z by grandparent to see if new z’s parent is red.
z
Algorithm: InsertionWe have detected a need for balance when z is red and his parent too.
• If z has a red uncle: colour the parent and uncle black, and grandparent red. Then replace z by grandparent to see if new z’s parent is red.
• If z is a left child and has a black uncle: colour the parent black and the grandparent red, then rightRotate(z.parent.parent)
z
rotateRight(G)G
P S
EDX C
A B
Relative to G, X is at left-left positions. rotateRight(G) will exchange of roles between G and P, so P becomes G's parent. Also must recolor P and G.
89
After rotateRight(G)P
X G
SCA B
EDApparent rule violation?
rotateLeft(G) will handle the case when X is at right right position relative to G.
Algorithm: InsertionWe have detected a need for balance when z is red and his parent too.
• If z has a red uncle: colour the parent and uncle black, and grandparent red. Then replace z by grandparent to see if z’s parent is red.
• If z is a left child and has a black uncle: colour the parent black and the grandparent red, then rotateRight(z.parent.parent)• If z is a right child and has a black uncle, then rotateLeft(z.parent) and
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Double Rotation• What if X is at left right relative to G?
– a single rotation will not work
• Must perform a double rotation
– rotate X and P
– rotate X and GG
P S
EDXA
B C
After Double Rotati
X
P G
SCA B
EDDouble rotation is also needed when X is at right left position relative to G.
G
X S
EDP C
A B
Example of Inserting Sorted Numbers
• 1 2 3 4 5 6 7 8 9 10
1
Insert 1. A leaf so red. Realize it isroot so recolorto black.
1
Insert 2
1
2
make 2 red. Parentis black so done.
Insert 3
1
2
3
Insert 3. Parent is red. Parent's sibling is black(null) 3 is outside relative to grandparent. Rotateparent and grandparent
2
1 3
Insert 4
2
1 3
On way down see2 with 2 red children.Recolor 2 red andchildren black.Realize 2 is rootso color back to black
2
1 3
4
When adding 4parent is blackso done.
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Insert 52
1 3
4
5
5's parent is red.Parent's sibling isblack (null). 5 isoutside relative tograndparent (3) so rotateparent and grandparent thenrecolor
Finish insert of 52
1 4
3 5
Insert 62
1 4
3 5
On way down see4 with 2 redchildren. Make4 red and childrenblack. 4's parent isblack so no problem.
Finishing insert of 62
1 4
3 5
6
6's parent is blackso done.
Insert 72
1 4
3 5
6
7
7's parent is red.Parent's sibling isblack (null). 7 isoutside relative tograndparent (5) so rotate parent and grandparent then recolor
Finish insert of 72
1 4
3 6
5 7
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Insert 82
1 4
3 6
5 7
On way down see 6with 2 red children.Make 6 red andchildren black. Thiscreates a problembecause 6's parent, 4, isalso red. Must performrotation.
Still Inserting 82
1 4
3 6
5 7
Recolored nowneed torotate
Finish inserting 84
2
3
6
5 71
8
Recolored nowneed torotate
Insert 94
2
3
6
5 71
8
9
On way down see 4 has two red childrenso recolor 4 red and children black. Realize 4 is the root so recolor black
Finish Inserting 94
2
3
6
5 81
7 9After rotations and recoloring
Insert 104
2
3
6
5 81
7 9On way down see 8 has twored children so change 8 tored and children black
10
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Insert 114
2
3
6
5 81
7 9
10
11
Again a rotation isneeded.
110
Finish inserting 114
2
3
6
5 81
7 10
9 11
Properties of Red Black Trees
• If a Red node has any children, it must have two children and they must be Black. (Why?)
• If a Black node has only one child that child must be a Red leaf. (Why?)
• Due to the rules there are limits on how unbalanced a Red Black tree may become.
Red‐Black Tree vs 2‐3‐4 Tree
• binary-search-tree representation of 2-3-4 tree
• 3- and 4-nodes are represented by equivalent binary trees
• Each 2-3-4 node generates exactly one black node (on the top), and zero red node for 2-nodes, one red for 3-nodes, and two red ones for 4-nodes.
Red‐Black Representation of 4‐node Red‐Black Representation of 3‐node
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Red‐Black Tree Example Red‐Black Tree Example
117
Multiway Search Trees
A multiway search tree of order m, or an m‐way search tree, is an m‐ary tree in which:
1. Each node has up to m children and m‐1 keys
2. The keys in each node are in ascending order
3. The keys in the first i children are smaller than the ith key
4. The keys in the last m‐i children are larger than the ith key
118
A 5‐Way Search Tree
393533
14
52
55402216
1918
159 25
14131110
B‐tree• B‐tree is a generalization of 2‐3‐4 tree with a large number of
branches.
• A B‐tree of order m is an m‐way search tree (i.e., a tree where each node may have up to m children) in which:
1. the number of keys in each non‐leaf node is one less than the number of its children.
2. all leaves are on the same level
3. all non‐leaf nodes except the root have at least m / 2children
4. the root is either a leaf node, or it has from two to mchildren
5. a leaf node contains no more than m – 1 keys
• The number m is always odd120
An example B‐Tree
51 6242
6 12
26
55 60 7064 9045
1 2 4 7 8 13 15 18 25
27 29 46 48 53
A B-tree of order 5containing 26 items
Note that all the leaves are at the same level
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121
A Typical Disk Drive
122
Disk Access
Disk Access Time =
Seek Time (moving disk head to correct track)
+ Rotational Delay (rotating disk to correct block in track)
+ Transfer Time (time to transfer block of
data to main memory)
Motivation for B‐Trees
• Index structures for large datasets cannot be stored in main memory
• Storing it on disk requires different approach to efficiency
• Assuming that a disk spins at 3600 RPM, one revolution occurs in 1/60 of a second, or 16.7ms
• Crudely speaking, one disk access takes about the same time as 200,000 instructions
Motivation (cont.)
• Assume that we use an AVL tree to store about 20 million records
• We end up with a very deep binary tree with lots of different disk accesses; log2 20,000,000 is about 24, so this takes about 0.2 seconds
• We know we can’t improve on the log n lower bound on search for a binary tree
• But, the solution is to use more branches and thus reduce the height of the tree!
– As branching increases, depth decreases
125
A B‐Tree of Order 5
To find the location of a key, traverse the keys at the root sequentially until at a pointer where any key before it is less than the search key and any key after it is greater than or equal to the search key.
Follow that pointer and proceed in the same way with the keys at that node until the search key is found, or are at a leaf and the search key is not in the leaf.
2216
393533191852
126
A B‐Tree of Order 1001
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127
B‐Tree Insertion Case 1:A key is placed in a leaf that still has some room
39353395
2216
1918
Shift keys to preserve ordering & insert new key.
Insert 7
393533975
2216
1918
128
B‐Tree Insertion Case 2: A key is placed in a leaf that is full
3935339752
2216
1918
Split the leaf, creating a new leaf, and move half the keys from full leaf to new leaf.
Insert 8
39353352
2216
191897
129
B‐Tree Insertion: Case 2
39353352
2216
1918
Move median key to parent, and add pointer to new leaf in parent.
Insert 8
97
39353352
22167
191898
130
B‐Tree Insertion: Case 3 The root is full and must be split
39353352
4022167
1918
In this case, a new node must be created at each level, plus a new root. This split results in an increase in the height of the tree.
Insert 15
141298 595543
131
B‐Tree Insertion: Case 3 The root is full and must be split
39353352
4022167
1918
Insert 15
98 1412
39353352
127
191898 1514
4022Move 12 & 16 up
132
B‐Tree Insertion: Case 3
16
This is the only case in which the height of the
B-tree increases.
39353352
127
191898 1514
4022
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133
B+‐Tree
39353352
33188
1918
A B+-Tree has all keys, with attached records, at the leaf level. Search keys, without attached records, are duplicated at upper levels. A B+-tree also has links between the leaves.
141298
42
134
Application: Web Search Engine
A web crawler program gathers information about web pages and stores it in a database for later retrieval by keyword by a search engine such as Google.
• Search Engine Task: Given a keyword, return the list of web pages containing the keyword.
• Assumptions:– The list of keywords can fit in internal memory, but the list of webpages (urls) for each keyword (potentially millions) cannot.
– Query could be for single or multiple keywords, in which pages contain all of the keywords, but pages are not ranked.
What data structures should be used?
Summary
• AVL trees, 2-3 trees, 2-3-4 tress, red-black trees, and B-tree are all balanced trees, with O(log(N)) heights.
• 2-3-4 trees and red-black trees have one-to-one correspondence.
• AVL trees and red-black trees are special binary search trees with simple data structure.
• B-trees with large of number of branches have small height and are suitable for storing large data sets on slow disks.
136
External Sorting
Problem: If a list is too large to fit in main memory, the time required to access a data value on a disk dominates any efficiency analysis.
1 disk access ≡ Several millionmachine instructions
Solution: Develop external sorting algorithms that minimize disk accesses
137
Basic External Sorting Algorithm
• Assume unsorted data is on disk at start
• Let M = maximum number of records that can be stored & sorted in internal memory at one time
AlgorithmRepeat:
1. Read M records into main memory & sort internally.
2. Write this sorted sub‐list onto disk. (This is one “run”).
Until all data is processed into runs
Repeat:
1. Merge two runs into one sorted run twice as long
2. Write this single run back onto disk
Until all runs processed into runs twice as long
Merge runs again as often as needed until only one large run: the sorted list
138
Basic External Sorting
11 96 12 35 17 99 28 58 41 75 159481
Unsorted Data on Disk
Assume M = 3 (M would actually be much larger, of course.) First step is to read 3 data items at a time into main memory, sort them and write them back to disk as runs of length 3.
11 9481
9612 35
17 9928
5841 75
15
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139
Basic External Sorting
Next step is to merge the runs of length 3 into runs of length 6.
11 9481 9612 35
17 9928 5841 75
1511 9481
9612 35
17 9928
5841 75
15140
Basic External Sorting
Next step is to merge the runs of length 6 into runs of length 12.
11 9481 9612 3517 9928 5841 75
15
15
11 9481 9612 35
17 9928 5841 75
141
Basic External Sorting
Next step is to merge the runs of length 12 into runs of length 24. Here we have less than 24, so we’re finished.
11 9481 9612 3517 9928 5841 7515
11 9481 9612 3517 9928 5841 75
15
2‐Way Sort: Requires 3 Buffers
• Pass 1: Read one page, sort it, write it.
– one buffer page used
• Pass 2, 3, …, etc.:
– three buffer pages used.
Main memory buffers
INPUT 1
INPUT 2
OUTPUT
DiskDisk
Two‐Way External Merge Sort
• Each pass we read + write each page in file.
• N pages in the file => the number of passes
• So toal cost is:
• Idea: Divide and conquer: sort subfiles and merge
log2 1N
2 12N Nlog
Input file
1-page runs
2-page runs
4-page runs
8-page runs
PASS 0
PASS 1
PASS 2
PASS 3
9
3,4 6,2 9,4 8,7 5,6 3,1 2
3,4 5,62,6 4,9 7,8 1,3 2
2,34,6
4,7
8,91,35,6 2
2,3
4,46,7
8,9
1,23,56
1,2
2,3
3,4
4,56,6
7,8Assuming one page has 2 records
General External Merge Sort
• To sort a file with N pages using B buffer pages:
– Pass 0: use B buffer pages. Produce sorted runs of B pages each.
– Pass 1, pass 2, …, etc.: merge B‐1 runs.
N B/
B Main memory buffers
INPUT 1
INPUT B-1
OUTPUT
DiskDisk
INPUT 2
. . . . . .. . .
More than 3 buffer pages. How can we utilize them?
3/12/2015
25
Cost of External Merge Sort
• Number of passes:
• Cost = 2N * (# of passes)
• E.g., with 5 buffer pages, to sort 108 page file:
– Pass 0: = 22 sorted runs of 5 pages each (last run is only 3 pages)
– Pass 1: = 6 sorted runs of 20 pages each (last run is only 8 pages)
– Pass 2: 2 sorted runs, 80 pages and 28 pages
– Pass 3: Sorted file of 108 pages
1 1 log /B N B
108 5/
22 4/
Number of Passes of External Sort
N B=3 B=5 B=9 B=17 B=129 B=257100 7 4 3 2 1 11,000 10 5 4 3 2 210,000 13 7 5 4 2 2100,000 17 9 6 5 3 31,000,000 20 10 7 5 3 310,000,000 23 12 8 6 4 3100,000,000 26 14 9 7 4 41,000,000,000 30 15 10 8 5 4
