# CS代考计算机代写 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 trees from 2-3-4 trees

2-3-4 trees are perfectly balanced ( height: ½ 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 BST by 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 trees from 2-3-4 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

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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

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Transformed Red-Black tree

bh = 3

black

height,

including

external

nodes

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54

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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.

8

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).

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Example operations

SEARCH 40

INSERT 40

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48

18

42

40

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Local Restructuring

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 INSERT part 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 INSERT part 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

13

Bottom-Up INSERT part 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 INSERT part 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 INSERT part 3a:4

Step 3a: Have reached a 3-node cluster.

a b

* K

a

b

g

a’

a”

b

g

2-3-4 tree

K a b

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 INSERT part 3b:4

Step 3b: Have reached a 3-node cluster.

a b

* K

a

b

g

a’

a”

b

g

2-3-4 tree

K a b

[ 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 INSERT part 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

a K b

TERMINAL CASE

b

K

a

b’

b”

*

b

a

g

LR Zig-Zag

D

x

p

g

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Bottom-Up INSERT part 3d:4

Step 3d: Have reached a 3-node cluster.

a b

* K

a

b

g

a

b’

b”

g

2-3-4 tree

a K 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.

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Bottom-Up INSERT part 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 INSERT part 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

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Bottom-Up INSERT summary

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] = nil then root[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

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90

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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.

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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. Adel’son-Vel’skiĭ, 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 of n 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 BST’s 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.] 27 RB-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). 28 Restricted 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 all possible 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 the subtree 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 a function of n? 29 END 30 . ) 1 n ( log bh ) 1 n ( log 2 1 + £ £ + . 1 ) 1 n ( log 2 1 bh 2 h height - + £ - £ /docProps/thumbnail.jpeg