RankBalanced Trees

1
Rank-Balanced Trees

• Siddhartha Sen, Princeton University
• WADS 2009
• Joint work with Bernhard Haeupler and
• Robert E. Tarjan

2
Observation

• Computer science is (still) a young field.
• We often settle for the first (good) solution.
• It may not be the best the design space is rich.

3
Research Agenda

• For fundamental problems, systematically explore
  the design space to find the best solutions,
  seeking
• elegance a quality of neatness and ingenious
  simplicity in the solution of a problem
  (especially in science or mathematics).
• ? wordnet.princeton.edu/perl/webwn
• Keep the design simple, allow complexity in the
  analysis.

4
Searching Dictionary Problem

• Maintain a set of items, so that
• Access find a given item
• Insert add a new item
• Delete remove an item
• are efficient.
• Assumption items are totally ordered, so that
  binary comparison is possible

5
Binary Search Tree
e
gt
lt
c
i
a
g
n
k

• Symmetric order

Why not hashing?
6
Binary Search Tree
e
gt
c
i
gt
a
g
n
lt
k

• Access k

7
Binary Search Tree
e
gt
c
i
lt
a
g
n
gt
k
h

• Insert h

8
Binary Search Tree
Find successor
Swap
e
Delete
c
i
k
k
a
g
n
k
h
i
m

• Delete i

9
Problem imbalance

• How to bound the height?
• Maintain local balance condition, rebalance after
  insert or delete balanced tree
• Restructure after each access self-adjusting
  tree
• Store balance information in nodes, guarantee
  O(log n) height
• After (during) insert/delete, restore balance
  bottom-up (top-down)
• Update balance information
• Restructure along access path

a
b
c
d
e
f
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Restructuring primitive(Single) Rotation
y
x
right left
x
y
C
A
A
B
B
C
Preserves symmetric order Changes heights Takes
O(1) time
11
Known Balanced Trees

• AVL trees (passé according to one author)
• weight balanced trees
• 2,3 trees
• B trees
• red-black trees
• etc.
• Goal small height, little rebalancing, simple
  algorithms

not binary
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Ranks

• Each node has an integer rank, a proxy for height
• Convention leaves have rank 0, missing nodes
  have rank -1
• rank of tree rank of root
• rank difference of a child
• rank of parent - rank of child
• i-child node of rank difference i
• i,j-node children have rank differences i and j

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Example of a rank-balanced tree
3
e
c
i
2
2
1
1
a
g
n
0
1
1
1
1
1
k
h
1
0
1
0
If all rank differences are positive, rank ?
height
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Rank Rules

• AVL trees every node is a 1,1- or 1,2-node
• Rank-balanced trees every node is a 1,1-, 1,2-,
  or 2,2-node (rank differences are 1 or 2)
• Red-black trees all rank differences are 0 or 1,
  no 0-child is the parent of another
• All need one balance bit per node.

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Height bounds

• nk minimum n for rank k
• AVL trees
• n0 1, n1 2, nk nk-1 nk-2 1, nk Fk3
  - 1
• nk Fk3 - 1, Fk2 lt ?k ? k ? log? n ? 1.44lg n
• Rank-balanced trees
• n0 1, n1 2, nk 2nk-2,
• nk 2??k/2? ? k ? 2lg n
• Same bound for red-black trees

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Insertion example
a
Demote a
Promote a
1
0
gt
b
Rotate left at b
Promote b
0
1
0
1
gt
c
1
0
0
Insert b
Insert c
Insert a
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Insertion example
b
Promote b
1
2
gt
Promote c
c
a
Demote c
2
1
0
1
0
1
0
gt
Rotate left at d
Promote d
d
0
1
0
1
gt
e
0
1
0
Insert e
Insert d
Insert c
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Insertion example
b
Demote b
2
gt
d
a
Rotate left at d
Promote d
1
2
1
0
0
2
gt
e
c
Promote e
0
1
2
1
0
0
1
gt
f
1
0
0
Insert e
Insert f
19
Insertion example
d
2
e
b
1
1
1
1
f
a
c
1
0
0
1
1
0
Insert f
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Rebalancing insertion

• 0, 1, or 2 rotations
• O(log n) rank changes
• No 2,2 nodes AVL trees
• ? log? n height!

21
Deletion example
e
Swap with successor
d
Double demote e
2
d
e
f
Delete
b
Demote b
1
2
1
0
1
1
f
a
c
1
0
0
1
1
0
Double rotate at c Double promote c
Delete a
Delete f
Delete d
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Deletion example
c
2
e
b
2
0
2
0
Delete f
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Rebalancing deletion

• 0, 1, or 2 rotations
• O(log n) rank changes

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Amortized (time-averaged) analysis

• If ti is the actual time of operation i and ?i is
  the potential of the data structure after
  operation i, the amortized time of operation i is

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Non-terminal cases
Must decrease potential!
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• Insertions 1,1-node ? 1,2-node
• Deletions 2,2-node ? 1,1- or 1,2-node
• ? 1,1-nodes 2 ? 2,2-nodes
• non-terminating steps are free,
• last insertion step ?? ? 2,
• last deletion step ?? ? 3
• If there are m inserts and d deletes (n m - d),
  the number of rebalancing steps is O(m d)

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• Rank-Balanced Trees
• height ? 2lg n
• 2 rotations per rebalancing
• O(1) amortized rebalancing time

• Red-Black Trees
• height ? 2lg n
• 3 rotations per rebalancing
• O(1) amortized rebalancing time

Yes.
No.
Are rank-balanced trees better?
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Better height bound?

• Sequential Insertions
• rank-balanced red-black
• height lg n (best) height 2lg n (worst)
• Theorem. The height of a rank-balanced tree is at
  most log? m.
• Degrades gracefully from AVL trees as d/m ? 1

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Proof

• Give a node a count of 1 when it is inserted
• Total amount of count in tree is m
• Potential of a node total count in its subtree
• When a node is deleted, its count is added to its
  parent if it has one
• Let ?k be the minimum potential of a node of rank
  k
• Claim ?k satisfies
• ?0 1, ?1 2, ?k 1 ?k-1 ?k-2 for k gt 1
• ? m ? Fk3 - 1 ? ?k

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Proof of Claim

• ?k 1 ?k-1 ?k-2 for k gt 1
• Easy for 1,1- and 1,2-nodes
• Harder for 2,2-nodes (created by deletions)
• But counts are inherited

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Rebalancing frequency

• How high does rebalancing propagate?
• O(m d) rebalancing steps total, which implies
• ? O((m d) / k) insertions/deletions at rank k
• Actually, we can show
• Theorem. There are O((m d) / 2k/3) rebalancing
  steps at rank k.

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Proof

• Use an exponential potential
• 1,1- and 2,2-nodes of rank i get potential bi
• 1,2-nodes of rank i get potential bi-2
• where b 21/3
• The potential change in the non-terminal steps
  telescopes. Combine this effect with
  initialization and terminal step.

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Telescoping potential
?? -bi3

bi3
bi2 -
0
1
bi2
bi1 -
0
1
2
1
bi1
bi -
0
0
1
2
1
bi
bi-1 -
0
1
1
2
1
2
1
0

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• Fix k. Cut off growth in potential at rank k
• 1,1- and 2,2-nodes of rank i bmini,k-3
• 1,2-nodes of rank i bmini-2,k-3
• Then a rebalancing that propagates to rank k or
  above decreases the potential by bk-3.
• The same idea works for red-black trees (we
  think).

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Preliminary evaluation
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Preliminary evaluation
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Preliminary evaluation
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Preliminary evaluation
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Conclusion

• Rank-balanced trees are a relaxation of AVL trees
  with behavior at least as good as red-black trees
  and better in important ways.
• Especially the height bound of min2lg n, log? m
• Exponential potential functions yield new
  insights into the efficiency of rebalancing. We
  anticipate more applications.

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Is rebalancing necessary?

• For insertions, yes. But what about deletions?
• Deletion rebalancing is complicated, ignored by
  textbooks, and many database systems do not do it
• So, can we avoid deletion rebalancing?
• Yes. Relaxation of AVL trees, ravl trees,
  achieves log? m access time using lglg m 1
  balance bits
• and no rebalancing during deletion!

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Thank you
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Extra slides