[SOLVED] CS scheme data structure chain AVL algorithm PowerPoint Presentation

PowerPoint Presentation

EECS 4101/5101
Red-Black
Tree
Prof. Andy Mirzaian

Lists
Move-to-Front
Search Trees
Binary Search Trees
Multi-Way Search Trees
B-trees
Splay Trees
2-3-4 Trees
Red-Black Trees
SELF ADJUSTING
WORST-CASE EFFICIENT
competitive
competitive?
Linear Lists
Multi-Lists
Hash Tables
DICTIONARIES
2

References:

[CLRS] chapter 13
AAW animation

3

Binary Search treesfrom2-3-4 trees
2-3-4 trees areperfectly balanced( height: log(n+1) h log(n+1) )
search trees that use 2-nodes, 3-nodes, and 4-nodes.
We can transform a 2-3-4 tree to an O(log n) height BSTby replacing each
3-node and 4-node by a small-clustered BST with 2 or 3 binary nodes.
Dilemma: How do we distinguish node clusters?

Disallowed
4-node
clusters:

2-node

right slant
left slant
OR
3-node

4-node
4

Red-Black treesfrom2-3-4 trees
Although the resulting BST has height O(log n), it loses the node cluster information and renders the 2-3-4 tree algorithms obsolete. We use a node cluster colour-coding scheme to resolve this.
Convention:each cluster top level is black (including external nodes),
lower levels within each cluster are red.
So, any red node is clustered within its parent cluster.

Disallowed
4-node
clusters:

2-node

right slant
left slant
OR
3-node

4-node

5

Example:2-3-4 tree

36
12 18 26
42 48
6 8 10
14 16
20 22 24
28 30 32
38
44
50 52 54

bh = 3

black
height,
including
external
nodes
6

Transformed Red-Black tree

bh = 3

black
height,
including
external
nodes

36

10
6
14
20
24
32
28
54
50
8
16
22
30
38
44
52
12
26
48
18
42

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Definition: Red-Black tree
DEFINITION:T is a Red-Black tree if it satisfies the following:

T is a Binary Search Tree with a red/black colour bit per node.

Every red node has a black parent. ( root is black.)

By convention we assume all external nodes are black.

All external nodes have the same number (namely, bh) of black
proper ancestors.
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Implementing Operations
Access operations:

SEARCH

MINIMUM
MAXIMUM
PREDECESSOR
SUCCESSOR

Update Operations:

INSERT
DELETE

Use the BST algorithms without change.
(Simply ignore the node colours.)

Worst-case time: O(h) = O(log n).
Simulate the 2-3-4 tree algorithms
as shown next.

Worst-case time: O(h) = O(log n).
9

Example operations
SEARCH40
INSERT40

36

10
6
14
20
24
32
28
54
50
8
16
22
30
38
44
52
12
26
48
18
42

40

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LocalRestructuring
INSERT and DELETE need the following local operations that take O(1) time each:

Node Colour Switch:red black

Link Rotation

a
b
a
b
g
D
p

b
a
a
b
g
D
p

Right Rotate D
Left Rotate D
FACT:Rotation preserves INORDER sequence of the tree nodes
and changes slant of link D.
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Bottom-Up INSERTpart 0:4
Bottom-Up INSERT (K,T)
Simulate 2-3-4 tree bottom up insertion.

When inserting K at the bottom of the tree, repeatedly split 4-nodes upwards until you either split the root, or reach a 2-node or a 3-node.

Simulate this on the Red-Black tree.

a1 a2 a3

b1 b2 b3

K

c1 c2 c3
12

Bottom-Up INSERTpart 1:4
Bottom-Up INSERT (K,T)
Step 1:Follow the search path of K down Red-Black tree T.
If K is found, return. Otherwise, we reach a black external
node x. Convert x into a red leaf and store K in it.
At the end of the algorithm we will re-colour the root black even if it might have temporarily become red.
K is red means x splits into x & x, & K is inserted into the parent cluster.
Now T satisfies all 4 properties of RB-trees except property 2: red x might have a red parent p. We need to climb up the tree to fix up.
If p is black,
we are done.

If p is red, more work
follows.
x

a

p
root[T]

a
a
K
x

p
x
x

root[T]
a

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Bottom-Up INSERTpart 2a:4
Step 2:While parent is part of a 4-node cluster, keep splitting it and promote its middle key up. (May repeat many times.)

a b c
* K

K a
c
a
b
g
d
p

* b
p
p

a
a
b
g
d
2-3-4 tree

Red-Black tree

b
a
c
a

a
K
a
b
g
d
p
*
Colour Switch
p, u, g.

[ K up from d is symmetric ]

b
a
c
a

a
*
K
a
b
g
d
p

x
p
u
g
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Bottom-Up INSERTpart 2b:4
Step 2:While parent is part of a 4-node cluster, keep splitting it and promote its middle key up. (May repeat many times.)

a b c
* K

a
b
g
d
p

* b
p
p

a
b
b
g
d
2-3-4 tree
a K
c

Red-Black tree

b
a
c
b

b
K
a
b
g
d
p
*
Colour Switch
p, u, g.

[ K up from g is symmetric ]

b
a
c
b

b
*
K
a
b
g
d
p

x
p
u
g
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Bottom-Up INSERTpart 3a:4
Step 3a:Have reached a 3-node cluster.

a b
* K

a
b
g

a
a
b
g
2-3-4 tree
Kab
TERMINAL CASE

Red-Black tree
No Change.

a
K
b
a
a
a
b
g

[ K up from g is symmetric ]

a
K
b
a
a
*
a
b
g
Opposite Slant

x
p
s
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Bottom-Up INSERTpart 3b:4
Step 3b:Have reached a 3-node cluster.

a b
* K

a
b
g

a
a
b
g
2-3-4 tree
Kab
[ K up from g is symmetric ]

Red-Black tree
Colour Switch p, g.

Rotate D.

a
K
b
a
a
a
b
g
D

TERMINAL CASE

b
K
a
a
a
*
b
a
g
Same Slant
Zig-Zig

D
x
p
g
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Bottom-Up INSERTpart 3c:4
Step 3c:Have reached a 3-node cluster.

Red-Black tree
Colour Switch x, g.
DoubleRotate(x,p,g):
i.e.,Rotate b
then Rotate D.

K
a
b
a
b
b
g
D
b

a b
* K

a
b
g

a
b
b
g
2-3-4 tree
aK b

TERMINAL CASE

b
K
a
b
b
*
b
a
g
LR Zig-Zag

D
x
p
g
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Bottom-Up INSERTpart 3d:4
Step 3d:Have reached a 3-node cluster.

a b
* K

a
b
g

a
b
b
g
2-3-4 tree

aK b
TERMINAL CASE
RL Zig-Zag

a
K
b
b
b
*
a
b
g

D
x
p
g

Red-Black tree

K
a
b
a
b
D
g
b
b

Colour Switch x, g.
DoubleRotate(x,p,g):
i.e.,Rotate b
then Rotate D.
19

Bottom-Up INSERTpart 3e:4
Step 3e:Have reached a 2-node cluster.
[ K up from b is symmetric ]

Red-Black tree
No change.

a
a
b

K
a
a

a
* K

a
b

a
a
b
2-3-4 tree

K a
TERMINAL CASE

a
a
b

K
a
a
*
x
p
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Bottom-Up INSERTpart 4:4
Step 4:Have reached nil (parent of root).
END

Red-Black tree
colour[root[T]] black

K
a
a
a = nil
This last step always resets root colour to black.
Black-height increases by one if root was temporarily red.

* K (root level)

a
a
2-3-4 tree
K
a = nil

a
a
*
a = nil
x

K
21

Bottom-Up INSERTsummary
Bottom-Up INSERT (K,T)
Step 1:Follow the search path of K down Red-Black tree T.
If K is found then return
Otherwise, we have reached a black external node x.
Convert x into a red leaf and store K in it.

Step 2:while (p, u)=red(* x=red, parent in 4-node cluster *)
do SwitchColour(p,u,g);x g end-while

Step 3:if p=red &u=black (u possibly external)(* x=red*)
then case:(* 3-node turns into unbalanced 4-node*)
[Zig-Zig (x,p,g)]: SwitchColour(p,g); Rotate(p,g)
[Zig-Zag(x,p,g)]: SwitchColour(x,g); DoubleRotate(x,p,g)
end-case

Step 4:if root[T] = nilthenroot[T] x
colour[root[T]] black
end
x = current node, p = parent of x, u = uncle of x, g = grand parent of x.
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Example:Bottom-Up Insert 45

80
30

90
10
20
50
60
40
70

A
B
J
C
H
I
D
E
F
G

add 45 @
E &
split
4-nodes
up

90
20
50
60
40

J
C
H
I
D
E
F
G

45
E

A
B
80
30
10
70

balance up
unbalanced
4-node

80
30

90
10
20
50
60
40
70

A
B
J
C
H
I
D
E
F
G

45
E
*
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INSERT & DELETE
FACT 1: Bottom-Up Insert makes at most 2 rotations.

This occurs when after the 4-node splitting loop a 3-node turns into a zig-zag 4-node. It needs 2 rotations to turn it into a balanced 4-node cluster.
FACT 2: Bottom-Up Delete makes at most 3 rotations.
Top-Down Insert (exercise).
Bottom-Up Delete (see [CLRS]).
Top-Down Delete:
While going down the search path, maintain LI:
[LI: current node x is either root[T] or x is red or x has a red child.]
With LI, splice-out can finish the task in O(1) time.
FACT 3: Top-Down Insert & Delete may make up to Q(log n) rotations.Explain why.

This makes them unsuitable for persistent data structures.
24

Bibliography:
Red-Black Trees:[R. Bayer, E.M. McCreight, Symmetric binary B-trees: Data Structure and maintenance algorithms, Acta Informatica, 1:290-306, 1972.]
Colour coded Red-Black Trees:[L.J. Guibas, R. Sedgewick, A dichromatic framework for balanced trees,in Proceedings of the 19th Annual Symposium on Foundations of Computer Science (FOCS), pp: 8-21, IEEE Computer Society, 1978.]
AA-trees:[A. Andersson, Balanced search trees made simple, in Proceedings 3rd Workshop on Algorithms and Data Structures (WADS), in Lecture Notes in Computer Science LNCS709, pp: 60-71, Springer-Verlag, 1993.]
Scapegoat trees:[I. Galperin, R.L. Rivest, Scapegoat trees, in Proc. 4th ACM-SIAM Symp. on Discrete Algorithms (SODA), pp: 165-174, 1993.]
Treaps:[R. Seidel, C.R. Aragon, Randomized search trees, Algorithmica, 16:464-497, 1996.]
AVL trees:[G.M. Adelson-Velski, E.M. Landis, An algorithm for the organization of information, Soviet Mathematics Doklady, 3:1259-1263, 1962.]
Weight balanced trees:[J. Nievergelt, E.M. Reingold, Binary search trees of bounded balance, SIAM J. of Computing, 2(1):33-43, 1973.]
K-neighbor trees:[H.A. Mauer, Th. Ottmann, H.-W. Six, Implementing dictionaries using binary trees of very small height, Information Processing Letters, 5(1):11-14, 1976.]
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Exercises
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[CLRS, Exercise 13.2-4, p. 314]Rotation Sequence:
Show that any BST holding a set ofn keys can be transformed into any other BST holding the same set of keys using O(n) rotations. [Hint: first show that at most n-1 right rotations suffice to transform the tree into a right-going chain.]

[CLRS, Exercise 13.2-5, p. 314]Right Rotation Sequence:
We say that a BST T1 can be right-converted to BST T2 if it is possible to obtain T2 from T1 via a series of right rotations. Give an example of two trees T1 and T2, holding the same set of keys, such that T1 cannot be right-converted to T2. Then show that if a tree T1 can be right-converted to T2, it can be done so using O(n2) right rotations, where n is the number of keys in T1.

Top-Down Insert & Delete: Design and analyze O(log n) time top-down Delete and Insert on red-black trees. [Hint: simulate top-down 2-3-4 tree procedures & use the hints given in these slides.]

[CLRS, Exercise 13.4-7, p. 330] Suppose we use bottom-up insert and delete on red-black tree T. Suppose that a key K is inserted into T and then immediately deleted from it. Is the resulting tree always the same as the initial tree? Justify your answer.

Split and Join on Red-Black trees:These are cut and paste operations on dictionaries. The Split operation takes as input a dictionary (a set of keys) A and a key value K (not necessarily in A), and splits A into two disjoint dictionaries B = { xA | key[x] K } and C = { xA | key[x] > K }. (Dictionary A is destroyed as a result of this operation.) The Join operation is essentially the reverse; it takes two input dictionaries A and B such that every key in A < every key in B, and replaces them with their union dictionary C = AB. (A and B are destroyed as a result of this operation.) Design and analyze efficient Split and Join on red-black trees. [Note: Definition of Split and Join here is the same we gave on BSTs and 2-3-4 trees, and slightly different than the one in Problem 13-2, pages 332-333 of [CLRS].]Red-Black tree Insertion sequence: Bottom-Up Insert integers 1..n one at a time in increasing order into an initially empty red-black tree in O(n) time total.[Hint: This is related to exercise 4 in the Slide on B-trees. Keep a pointer to the largest key node.]27RB-Balance:We say a binary tree T is RB-balanced if it satisfies the following property:RB-balance:for every node x in tree T, the length of a longest path from x to any of its descendant external nodes is at most twice the length of a shortest path from x to any of its descendant external nodes.(a) Show that every red-black tree is an RB-balanced Binary Search Tree.(b) Now we want to prove that the converse also holds. Let T be an arbitrary RB-balanced BST with n nodes. We want to show (algorithmically) that it is always possible to colour each node of T red or black to make it a red-black tree. [Note that we make no structural change to T other than colouring its nodes.] Design and analyze an O(n) time algorithm to do the node colouring to turn T into a red-black tree.[You may assume there is space for a colour-bit in each node of T.](c) Carefully prove the correctness of your algorithm in part (b).BST to RB-tree Conversion: Given an n-node arbitrary BST, design and analyze an O(n) time algorithm to construct an equivalent red-black tree (i.e., one that contains the same set of keys).Range-Search Reporting in RB-tree:Let T be a given red-black tree. We are also given a pair of key values a and b, a < b (not necessarily in T). We want to report every item x in T such that a key[x] b. Design an algorithm that solves this problem and takes O(R+log n) time in the worst case, where n is the number of items in T and R is the number of reported items (i.e., the output size). 28Restricted Red-Black Tree:We define a Restricted Red-Black Tree (RRB-tree) to be a standard Red-Black Tree with the added structural constraint that every red node must be the right child of its parent. So, every left child is black. (In comparison with 2-3-4 trees, this indicates that we have no 4-node clusters, and every 3-node cluster is right slanted.) We want to show that it is possible to maintain such a structure while performing dictionary operations efficiently. Let T be an arbitrary n-node RRB-tree.(a)Obtain tight lower and upper bounds on height of T as a function of n.(b)Show how to perform the dictionary insert operation on T efficiently. Make sure you consider allpossible cases in the algorithm. What is its worst-case running time as a function of n?(c)Consider a leaf node x in T. (Note x is not an external node.) What are possible structures of thesubtree rooted at the sibling of x?(d)Using your answer to part (c), show how to perform the dictionary delete operation on T efficiently.Make sure you consider all possible cases in the algorithm. What is its worst-case running time as afunction of n?29END30.)1n(logbh)1n(log21++.1)1n(log21bh2hheight -+-/docProps/thumbnail.jpeg