1. CSE 326: Data Structures Trees
Lecture 8: Friday, Jan 24, 2003

2. Today: Splay Trees
Fast both in worst-case amortized analysis and in practice
Are used in the kernel of NT for keep track of process information!
Invented by Sleator and Tarjan (1985)
Details:
Weiss 4.5 (basic splay trees)
11.5 (amortized analysis)
12.1 (better “top down” implementation)
We’ll start by introducing AVL trees.
Then, I’d like to spend some time talking about double-tailed distributions and means.
Next, we’ll gind out what AVL stands for.
Finally, you’ll receive a special bonus if we get to it!
(Unfortunately, the bonus is AVL tree deletion)

3. Basic Idea “Blind” rebalancing – no height info kept!
Worst-case time per operation is O(n)
Worst-case amortized time is O(log n)
Insert/find always rotates node to the root!
Good locality:
Most commonly accessed keys move high in tree – become easier and easier to find

4. Idea
move n to root by series of zig-zag and zig-zig rotations, followed by a final single rotation (zig) if necessary 10
You’re forced to make
a really deep access: 17
Since you’re down there anyway,
fix up a lot of deep nodes! 5 2 9 3

5. Zig-Zag* n n Helped Unchanged Hurt g X p g p W X Y Z W Y Z
up 2 X p g
down 1 p
down 1
up 1 n W
This is just a double rotation. X Y Z W Y Z
*This is just a double rotation

6. Zig-Zig n n g W p p Z X g Y Y Z W X
Can anyone tell me how to implement this with two rotations?
There are two possibilities:
Start with rotate n
or rotate p?
Rotate p!
Rotate n makes p n’s left child and then we’re hosed.
Then, rotate n.
This helps all the nodes in blue and hurts the ones in red. So, in some sense, it helps and hurts the same number of nodes on one rotation.
Question: what if we keep rotating? What happens to this whole subtree?
It gets helped! Y Z W X

7. Why Splaying Helps Node n and its children are always helped (raised)
Except for last step, nodes that are hurt by a zig-zag or zig-zig are later helped by a rotation higher up the tree!
Result:
shallow nodes may increase depth by one or two
helped nodes decrease depth by a large amount
If a node n on the access path is at depth d before the splay, it’s at about depth d/2 after the splay
Exceptions are the root, the child of the root, and the node splayed
Alright, remember what we did on Monday.
We learned how to splay a node to the root of a search tree.
We decided it would help because we’d go a lot of fixing up if we had an expensive access.
That means we have to fix up the tree on every expensive access.

8. Splaying Example 1 1 2 2
zig-zig 3 3
Find(6) 4 6 5 4 5 6

9. Still Splaying 6 1 1 2 6
zig-zig 3 3 6 5 4 2 5 4

10. Almost There, Stay on Target1 6 1 6
zig 3 3 2 5 2 5 4 4

11. Splay Again 6 1 6 1
zig-zag 3 4
Find(4) 2 5 3 5 4 2

12. Example Splayed Out 6 1 4 1 6
zig-zag 3 5 4 2 3 5 2

13. Locality
“Locality” – if an item is accessed, it is likely to be accessed again soon
Why?
Assume m  n access in a tree of size n
Total worst case time is O(m log n)
O(log n) per access amortized time
Suppose only k distinct items are accessed in the m accesses.
Time is O(n log n + m log k )
Compare with O( m log n ) for AVL tree
those k items are all at the top of the tree
getting those k items near root

14. Splay Operations: Insert
To insert, could do an ordinary BST insert
but would not fix up tree
A BST insert followed by a find (splay)?
Better idea: do the splay before the insert!
How?
What about insert?
Ideas?
Can we just do BST insert?
NO. Because then we could do an expensive operation without fixing up the tree.

15. Split Split(T, x) creates two BST’s L and R:
All elements of T are in either L or R
All elements in L are  x
All elements in R are  x
L and R share no elements
Then how do we do the insert?
What about insert?
Ideas?
Can we just do BST insert?
NO. Because then we could do an expensive operation without fixing up the tree.

16. Split Split(T, x) creates two BST’s L and R:
All elements of T are in either L or R
All elements in L are  x
All elements in R are > x
L and R share no elements
Then how do we do the insert?
Insert as root, with children L and R
What about insert?
Ideas?
Can we just do BST insert?
NO. Because then we could do an expensive operation without fixing up the tree.

17. Splitting in Splay Trees
How can we split?
We have the splay operation
We can find x or the parent of where x would be if we were to insert it as an ordinary BST
We can splay x or the parent to the root
Then break one of the links from the root to a child
How can we implement this?
We can splay.
We can find x or where x ought to be.
We can splay that spot to the root.
Now, what do we have?
The left subtree is all <= x
The right is all >= x

18. could be x, or what would have been the parent of x
Split
could be x, or what would have been the parent of x
split(x)
splay T L R
if root is > x
if root is  x
So, a split just splays x’s spot to the root then hacks off one subtree.
This code is _very_ pseudo. You should only use it as a general guideline. OR L R L R
 x
> x
< x
> x

19. Back to Insert x split(x) L R L R  x > x Insert(x): Split on x
Now, If we can split on x and produce one subtree smaller and one larger than x, insert is easy!
Just split on x.
Then, hang the left (smaller) subtree on the left of x.
Hang the right (larger) subtree on the right of x.
Pretty simple, huh?
Are we fixing up deep paths?
Insert(x):
Split on x
Join subtrees using x as root

20. Insert Example Insert(5) 6 4 4 6 1 9 split(5) 1 6 1 9 9 4 7 2 2 7 7 2
Let’s do some examples. 4 6 1 9 2 7

21. Splay Operations: Deletex
find(x)
delete x L R L R
< x
> x
OK, we’ll do something similar for delete.
We know x is in the tree.
Find it and bring it to the root.
Remove it.
Now, we have to split subtrees.
How do we put them back together?
Now what?

22. Join Join(L, R): given two trees such that L < R, merge them
Splay on the maximum element in L then attach R R L
splay L R
The join operation puts two subtrees together as long as one has smaller keys to begin with.
First, splay the max element of L to the root.
Now, that’s gauranteed to have no right child, right?
Just snap R onto that NULL right side of the max.

23. Delete Completed x T find(x) delete x L R L R < x > x Join(L,R)
So, we just join the two subtrees for delete.
T - x

24. Delete Example Delete(4) 6 4 6 1 9 find(4) 1 6 1 9 9 4 7 2 2 7
Find max 7 2 2 2 1 6 1 6 9 9 7 7

25. Splay Trees, Summary
Splay trees are arguably the most practical kind of self-balancing trees
If number of finds is much larger than n, then locality is crucial!
Example: word-counting
Also supports efficient Split and Join operations – useful for other tasks
E.g., range queries

26. Dictionary & Search ADTs
Dictionary ADT (aka map ADT) Stores values associated with user-specified keys
keys may be any (homogenous) comparable type
values may be any (homogenous) type
Search ADT: (aka Set ADT) stores keys only
Dictionaries associate some key with a value, just like a real dictionary (where the key is a word and the value is its definition).
In this example, I’ve stored user-IDs associated with descriptions of their coolness level.
This is probably the most valuable and widely used ADT we’ll hit.
I’ll give you an example in a minute that should firmly entrench this concept.

27. Dictionary & Search ADTs
create :  dictionary
insert : dictionary  key  values  dictionary
find : dictionary  key  values
delete : dictionary  key  dictionary
kim chi
spicy cabbage
Kreplach
tasty stuffed dough
Kiwi
Australian fruit
insert(kohlrabi, upscale tuber)
Dictionaries associate some key with a value, just like a real dictionary (where the key is a word and the value is its definition).
In this example, I’ve stored user-IDs associated with descriptions of their coolness level.
This is probably the most valuable and widely used ADT we’ll hit.
find(kreplach)
kreplach: tasty stuffed dough

28. Dictionary Implementations
Arrays:
Unsorted
Sorted
Linked lists
BST
Random
AVL
Splay

29. Dictionary Implementations
Arrays
Lists
Binary Search Trees
unsorted
sorted
AVL
splay
insert
O(1)
O(n)
O(log n)
amortized
find
delete
find + O(1)

30. The last dictionary we discuss: B-Trees
Suppose we want to store the data on disk
A disk access is a lot more expensive than one CPU operation
Example
1,000,000 entries in the dictionary
An AVL tree requires log(1,000,000)  20 disk accesses – this is expensive
Idea in B Trees:
Increase the fan-out, decrease the hight
Make 1 node = 1 block

31. B-Trees Basics All keys are stored at leaves
Nonleaf nodes have guidance keys, to help the search
Parameter d = the degree
book uses the order M = 2d+1)
Rules for Keys:
The root is either a leaf, or has between 1 and 2d keys
All other nodes (except the root) have between d and 2d keys
Rule for number of children:
Each node (except leaves) has one more children than keys
Balance rule:
The tree is perfectly balanced !

32. B-Trees Basics A non-leaf node: A leaf node: Then called a B+ tree30
120
240
Keys k < 30
30<=k<120
120<=k<240
Keys 240<=k
Then called
a B+ tree 40 50 60
Next leaf
Record with key 40
Record with key 50
Record with key 60

33. B+Tree Example
d = 2 (M = 5)
Find the key 40 80
40  80 20 60
100
120
140
20 < 40  60 10 15 18 20 30 40 50 60 65 80 85 90
30 < 40  40 10 15 18 20 30 40 50 60 65 80 85 90

34. B+Tree Design How large d ? Example: 2d x 4 + (2d+1)  8 <= 4096
Key size = 4 bytes
Pointer size = 8 bytes
Block size = 4096 byes
2d x 4 + (2d+1)  8 <= 4096
d = 170

35. B+ Trees Depth Assume d = 170 How deep is the B-tree ?
Depth = 0 (just the root)  at least 170 keys
Depth = 1  at least 171  30103 keys
Depth = 2   1712  5106 keys
Depth = 3    860 106 keys
Depth = 4    147 109 keys
Nobody has more keys !
With a B tree we can find any data item with at most 5 disk accesses !

36. Insertion in a B+ Tree Insert (K, P) Find leaf where K belongs, insert
If no overflow (2d keys or less), halt
If overflow (2d+1 keys), split node, insert in parent:
If leaf, keep K3 too in right node
When root splits, new root has 1 key only
parent
parent K3 K1 K2 K3 K4 K5 P0 P1 P2 P3 P4 p5 K1 K2 P0 P1 P2 K4 K5 P3 P4 p5

37. Insertion in a B+ Tree Insert K=19 80 20 60 100 120 140 10 15 18 20 3050 60 65 80 85 90 10 15 18 20 30 40 50 60 65 80 85 90

38. Insertion in a B+ Tree After insertion 80 20 60 100 120 140 10 15 1819 20 30 40 50 60 65 80 85 90 10 15 18 19 20 30 40 50 60 65 80 85 90

39. Insertion in a B+ Tree Now insert 25 80 20 60 100 120 140 10 15 18 1930 40 50 60 65 80 85 90 10 15 18 19 20 30 40 50 60 65 80 85 90

40. Insertion in a B+ Tree After insertion 80 20 60 100 120 140 10 15 1819 20 25 30 40 50 60 65 80 85 90 10 15 18 19 20 25 30 40 50 60 65 80 85 90

41. Insertion in a B+ Tree But now have to split ! 80 20 60 100 120 140 1015 18 19 20 25 30 40 50 60 65 80 85 90 10 15 18 19 20 25 30 40 50 60 65 80 85 90

42. Insertion in a B+ Tree After the split 80 20 30 60 100 120 140 10 1518 19 20 25 30 40 50 60 65 80 85 90 10 15 18 19 20 25 30 40 50 60 65 80 85 90

43. Deletion from a B+ Tree Delete 30 80 20 30 60 100 120 140 10 15 18 1925 30 40 50 60 65 80 85 90 10 15 18 19 20 25 30 40 50 60 65 80 85 90

44. Deletion from a B+ Tree After deleting 30 May change to 40, or not 8020 30 60
100
120
140 10 15 18 19 20 25 40 50 60 65 80 85 90 10 15 18 19 20 25 40 50 60 65 80 85 90

45. Deletion from a B+ Tree Now delete 25 80 20 30 60 100 120 140 10 15 1819 20 25 40 50 60 65 80 85 90 10 15 18 19 20 25 40 50 60 65 80 85 90

46. Deletion from a B+ Tree After deleting 25 Need to rebalance Rotate 8020 30 60
100
120
140 10 15 18 19 20 40 50 60 65 80 85 90 10 15 18 19 20 40 50 60 65 80 85 90

47. Deletion from a B+ Tree Now delete 40 80 19 30 60 100 120 140 10 15 1850 60 65 80 85 90 10 15 18 19 20 40 50 60 65 80 85 90

48. Deletion from a B+ Tree After deleting 40 Rotation not possible
Need to merge nodes 80 19 30 60
100
120
140 10 15 18 19 20 50 60 65 80 85 90 10 15 18 19 20 50 60 65 80 85 90

49. Deletion from a B+ Tree Final tree 80 19 60 100 120 140 10 15 18 19 2050 60 65 80 85 90 10 15 18 19 20 50 60 65 80 85 90