15-211 fundamental data structures and algorithms

15-211Fundamental Data Structures and Algorithms

Ananda GunaFebruary 8, 2005

Splay Trees

Announcements

Homework 2 available!Due Monday, Feb.16, 11:59pm!More involved than hw1? Perhaps.

• Start early!Start early!

It is very important that you go to recitations:We will discuss splaying

Splay Trees (self adjusting trees)

Balanced BST’s

O(log n) performanceBest, average and worst cases

Need to store balance information per nodeexpensive

Difficult to implementInsertions and deletions are difficult

Even for easy inputs we still have O(log n) behaviorNo cost savings

Binary search trees

Simple binary search trees can have bad behavior for some insertion sequences.Average case O(log N), worst case O(N).

AVL trees maintain a balance invariant to prevent this bad behavior.Accomplished via rotations during insert.

Splay trees achieve amortized running time of O(log N).Accomplished via rotations during find.

Amortized running time

The analysis that allows us to conclude that we spend the same (or less) amount of time over a sequence of operations is called amortized analysis.

If we say that the amortized running time of a sequence of operations is O(f(N)):Some operations might be more than O(f(N)),

other less.But the average over the entire sequence is

O(f(N)).

Splay trees

Splay trees provide a guarantee that any sequence of M operations (starting from an empty tree) will require O(Mlog N) time.

Hence, each operation has amortized cost of O(log N).

It is possible that a single operation requires O(N) time.

But there are no bad sequences of operations on a splay tree.

A basic observation

Let’s suppose that a node requires O(N) time to find.

If and when this happens, we must move it somewhere closer to the root so that if we access it again, it will definitely require less than O(N) time.

If we don’t do this, then O(log N) amortized behavior is not possible.

Danny Sleator

The inventor of splay trees.

Winner of Kannelakis award.

See his splay tree demo at http://www.cs.cmu.edu/~sleator

(And definitely a procrastinator.)

The basic idea

Every time a node is accessed, move it to the root.

The move can be accomplished by performing AVL rotations.

Practical benefits:In practice, nodes are often accessed

multiple times.AVL rotations make trees more

balanced.

Using rotations

Suppose we perform a find operation, which accesses node n.

Then perform AVL rotations, starting from n, to move it up to the root.

Doing this requires O(d) time, where d is the depth of n.

But a subsequent access of n will require only O(1) time, and the tree will be better balanced, too.

A simple example

0 move 5 to root

\ 1 \ 2 \ 3 \ 4 \ 5

Example Ctd..

This is the "move-to-root heuristic".

Example 2 (move 2 to root)

6 / \ 1 7 / \0 3 / \ 2 5 / 4

Work sheet

Move to Root

Move-to-root() is an example of a "self-adjusting" heuristic

Every time we search for a node x, we do move-to-root(x)

Works well if we search for few different nodes

Idea is to try to maintain log(n) performance over a sequence of operations.

Important Concept - Amortized Time

Question: Is it possible that by applying move-to-root() on every search gives amortized time bounded by O(log n) for any access pattern?

Answer:

Bottom Up Splaying

Basic Cases

Zig-Zag:

z z x \ \ / \ y ===> x ===> z y / \ x y

Basic Cases

Zig-Zig: z y x / / \ \ y ===>x z ===> y / \ x z

Basic Cases

Zig y x / ===> \ x y

Example

6 / \ 1 7 / \0 2 \ 3 \ 5 / 4

Work area

Splaying, case 1

There are four cases to consider.

Case 1: The node is already the root.Nothing to do.

Splaying, case 2

Case 2: Accessed node’s parent is the root.Perform a single rotation.

Z

YX

ZYX

Splaying, cases 3 and 4

The next two cases cover the situation in which the accessed node has a grandparent.

Splaying, case 3

Case 3: Zig-zag (left).Perform an AVL double rotation.

a

Zb

X

Y1 Y2

a

Z

b

X Y1 Y2

Splaying, case 4

Case 4: Zig-zig (left).Special rotation.

a

Zb

Y

W X

a

Z

b

Y

W

X

Symmetry

And there are symmetric cases for zig-zag and zig-zig to the right.

Splay tree example

2

5

34

6

10

Insert {0, 1, 2, 3, 4, 5, 6}, then find 6

2

5

36

4

10

zig-zig right

Splay tree example, cont’d

2

5

36

4

10

6

5

32

4

10

zig-zag right

Splaying example, cont’d

zig-zag right

6

5

32

4

10

0

5

32

4

16

Result of splaying

Access of 6 required N nodes visited and modified.

But now accessing 5 requires only N/2 nodes visited and modified. Will also bring all nodes up

to N/4 of root. (Try it!)

And all nodes are shallower.

Every access will tend to improve the tree for future operations.

0

5

32

4

16

Operations on splay trees

Search – find the element and splay the resulting node

Insertion – insert the element and splay the node just inserted

Delete Operation

Deletion Splay the node we want to delete(say x)

to the root xA B

find the rightmost node of A (say y) and splay y to the root

Attach the parts

Splaying summary

Splaying has the effect of moving the accessed node to the root.

It also reduces the depth of almost all of the nodes along the access path.

Properties of Splaying

Theorem: A sequence of M splay operations on a tree of N nodes takes time O(M log N). (Assuming M > N)Proof: beyond the scope of this course.

Theorem: sequentially splaying all the nodes in the tree takes O(N) time. Proof: beyond the scope of this course.

Analysis of splay trees

The analysis of the running time of splay trees is quite difficult.

Any single find or insert might take O(N) time.

But any sequence of M operations, starting from an empty tree, will take only O(Mlog N) time.

In practice, splay trees work extremely well.

http://www.cs.technion.ac.il/~itai/ds2/framesplay/splay.html

BST’s vs Hash Tables

Question: Hash tables have O(1) performance for lookups, inserts and deletes, what is the use of search trees that can be O(log N) at best?

Answer:

Some nice things about BST’s

Find the maximum or minimum element Find the successor (predecessor) of a

given element in the set. Searching and computing the "rank" in

that ordering.The RANK of an element in an ordered list is

the number of elements before it in the list.

Splitting and joining (example from perl) Prefix matching

Thursday

Dynamic Programming Start HW2