TTIT33 Algorithms and Optimization – DALG Lecture 6
TTIT33 Algorithms and Optimization – DALG Lecture 6
Content:
• Balanced Search Trees
TTIT33- Algorithms and optimization
– AVL Trees
[GT 10.2, LD 7.1]
– Multi-way search trees
[GT 10.4, LD 7.2, CL 18.1-2]
– Information on B-trees
Algorithms Lecture 5
• Piority Queues – Heaps
[GT 14.3, LD 7.2, CL 18.1-2]
[GT 8.1,8.3, LD 9.1, CL 6.1-5]
Balanced Search Trees,
Heaps
Jan Maluszynski - HT 2006
6.1
Jan Maluszynski - HT 2006
TTIT33 Algorithms and Optimization – DALG Lecture 6
TTIT33 Algorithms and Optimization – DALG Lecture 6
AVL Trees
Worst AVL Tree
6.2
• A worst AVL tree is a maximally unbalanced AVL
tree…
…or a tree of a given height with a minimum
number of internal nodes
LeftHeight(v)-RightHeight(v)
(Example: insert 44, 17, 78, 32, 50, 88, 62)
A BST
Jan Maluszynski - HT 2006
6.3
Jan Maluszynski - HT 2006
6.4
TTIT33 Algorithms and Optimization – DALG Lecture 6
TTIT33 Algorithms and Optimization – DALG Lecture 6
AVL-Tree Insertion – Rebalance…
AVL-Tree Insertion… (Goodrich/Tamassia)
How to insert an item with key 50?
• Perform a LookUp(50)
• While searching for 50,
keep track of last passed
node with balance ≠ 0
Îcritical node
• If not found, insert at the
leaf where search ended
• Recompute balance on
the way back
• Check critical node!
If balance ∉ [-1..1], rebalance!
Jan Maluszynski - HT 2006
6.5
Jan Maluszynski - HT 2006
30
b:
b: -1
0
20
b: 1
10
b: 0
60
b: 0
1
40
b:
b: -1
0
70
b: 0
50
b: 0
6.6
1
TTIT33 Algorithms and Optimization – DALG Lecture 6
TTIT33 Algorithms and Optimization – DALG Lecture 6
AVL-Tree Insertion & Re-balance…
Time to re-balance…
Now try an item with key 15...
• Perform a LookUp(15)
• While searching for 15,
keep track of last passed
node with balance ≠ 0
Îcritical node
• If not found, insert at the
leaf where search ended
• Recompute balance on
the way back
• Check critical node!
If balance ∉ [-1..1], rebalance!
• Label the critical node
and its 2 descendants on the
30
b: ?
path to ”15” as a, b, c,
such that a < b < c, in
c
20
60
an in-order traversal
b: 2
b: 1
• Re-structure the nodes
a
10
a, b and c so that
n b:40-1 b:700
b: -1
b has a and c as
children!
b 15
50
30
b:
b: -1
?
20
b: 12
10
b:
b: -1
0
60
b: 1
40
b: -1
15
b: 0
70
b: 0
50
b: 0
k
l
Jan Maluszynski - HT 2006
6.7
TTIT33 Algorithms and Optimization – DALG Lecture 6
Time to re-balance…
Removal of a node...
•
•
•
Jan Maluszynski - HT 2006
TTIT33 Algorithms and Optimization – DALG Lecture 6
TTIT33 Algorithms and Optimization – DALG Lecture 6
Tri-node restructuring = rotations....
Double rotations...
b
c
l
Jan Maluszynski - HT 2006
c
c
l
a
m
b
b
a
a
c
k
6.10
Two rotations are needed when the nodes to
re-balance are placed in a zig-zag pattern...
1. Rotate up b over a
Note: Labeling of a, b
and c same as before!
2. Rotate up b over c
c
b
j
j
6.8
Label the critical node, the child on the deepest side,
and its descendants on the deepest side as a, b, c,
such that a < b < c, in an in-order traversal
Re-structure as previous
Go to #2 and continue update and check towards the
root (we may have to re-balance more than once!)
6.9
Other authors use left and right rotations:
Single left rotation:
a
• left part of the subb
tree (a and j) is lowered
j
• We have ”rotated (up)
k
b over a”...
m
Same thing, but in reverse!
1. Perform a LookUp and Remove as in
an ordinary binary tree
2. Update the balance on the way back
to the root
3. If too unbalanced: Re-structure! ...but:
• Label the critical node
and its 2 descendants on the
30
b:
b: -1
?
path to ”15” as a, b, c,
such that a < b < c, in
60
b 15 20 c
an in-order traversal
b: 1
b: 0 b: 2
c
20
• Re-structure the nodes ab 15
a
10b: 0 b: 2
70
c 40
a, b and c so that
10 b
20
b: -115
n b: -1 b: 0
b:
b: -1
0
b: 0
2
b has a and c as
b: 0b 15
n
b: 0
children!
50
k
n
b: 0
k
• Update balance!
l
m
Jan Maluszynski - HT 2006
b: 0
Jan Maluszynski - HT 2006
TTIT33 Algorithms and Optimization – DALG Lecture 6
a
b: 0
m
m
k
6.11
j
k
l
m
j
k
l
m
l
Jan Maluszynski - HT 2006
6.12
2
TTIT33 Algorithms and Optimization – DALG Lecture 6
TTIT33 Algorithms and Optimization – DALG Lecture 6
New approach: relax some condition...
(2,3) trees [or (2,4) trees, or (a,b)-trees…]
• AVL-Tree: binary tree, accepts a small unbalance
• Recall:
Previously:
• A single ”pivot element”
• If larger we search to the right
• If smaller we search to the left
Full binary tree: nonempty; degree is either 0 or 2 for each
node
Perfect binary tree: full, all leaves have the same depth
• Can we build and maintain a perfect tree (if we skip
”binary”) ??
6.13
8
5 10
2
8
12
Jan Maluszynski - HT 2006
6.14
TTIT33 Algorithms and Optimization – DALG Lecture 6
TTIT33 Algorithms and Optimization – DALG Lecture 6
(2,3) trees, (2,4) trees, or (a,b) trees...
Insert in (a,b)-tree where a=2 and b=3
• Each node is either a leaf,
or has c children where a ≤ c ≤ b
• As long as there is room in
the child we find, add element
to that child...
5
Insert(10)
• LookUp works aproximately as before
5 10
...may happen repeatedly
5 10 15
• Delete must check that a node does not
become empty (then we transfer or merge
nodes)
Jan Maluszynski - HT 2006
• If full, split and push the selected
pivot element upwards...
Insert(15)
• Insert must check that a node does not
overflow (then we split the node)
10
5
Insert(18)
15
10
5
6.15
10
Insert(17)
15 18
5
TTIT33 Algorithms and Optimization – DALG Lecture 6
TTIT33 Algorithms and Optimization – DALG Lecture 6
Delete in a (2,3)-tree
Delete in a (2,3)-tree
30
Delete(25)
?
30
18
25
18
6.16
Delete(18)
5
15
20
?
20
10 17
5
35 40
5
30
15
35
10
10
35
35
?
15
15
35 40
10 17
10 17
5
No pivot ??
Too many children!
10 17
20
Transfer of
30 and 35
15 17 18
2. A leaf is removed (becomes empty)
Î transfer some other key to that leaf, or
Î merge (fuse) it with a neighbor
…ok if we have a sibling with 2+ elements
20
10 17
Jan Maluszynski - HT 2006
Three cases:
1. No constraints are violated by removal
2. A leaf is removed (becomes empty)
Î transfer some other key to that leaf,
5
2
Now:
• Multiple pivot elements
• No. of children =
no. of pivot elements + 1
Î we would always know the worst
search time exactly!
Jan Maluszynski - HT 2006
5
18
30
15
18
30
40
5
15
17
5
15 17
30
40
40
35 40
Jan Maluszynski - HT 2006
6.17
Jan Maluszynski - HT 2006
6.18
3
TTIT33 Algorithms and Optimization – DALG Lecture 6
TTIT33 Algorithms and Optimization – DALG Lecture 6
Delete in a (2,3)-tree
Properties of a (2,3) tree
3. An internal node becomes empty
Root: replace with in-order pred. or succ.
Î repair inconsistencies with suitable
merge and transfer operations…
Internal underflow...
...merge nodes
Replace...
...merge leafs
Delete(20)
?
10
5
35
17
30
40
5
?
35
17
?
∑
17
20
10
• Always a perfect tree
h +1
• A minimal tree of height h will have n = 2 − 1
nodes (it’s a full binary tree with full set of
nodes at all levels)
• A maximal (2,3)-tree will have a branching
factor of 3, thus n = h 3i =(3h +1 − 1) / 2
30
40
5 10
i =0
…2 keys in every node Î k = 3
35
30
h +1
−1
17 35
40
5 10 30
Jan Maluszynski - HT 2006
40
• Thus the height h = log 3 k 
6.19
Jan Maluszynski - HT 2006
TTIT33 Algorithms and Optimization – DALG Lecture 6
TTIT33 Algorithms and Optimization – DALG Lecture 6
B-Tree
ADT Priority Queue:
• Used to keep an index over external data
(e.g. content of a disc)
No of keys
in a tree
6.20
– Linearly ordered set K of keys
– We store pairs <k,I> (as in Dictionary), multiple pairs
with same key are allowed. The key denotes priority
– Items retrieved by priority: i.e. by the minimal key.
• It’s only an (a,b)-tree where a = b/2
• We may now choose b so that b-1
references to children (other disc blocks)
fit into a single disc block
• By defining a = b/2 we will always fill up
a disc block when two blocks are merged!
Jan Maluszynski - HT 2006
6.21
Jan Maluszynski - HT 2006
TTIT33 Algorithms and Optimization – DALG Lecture 6
TTIT33 Algorithms and Optimization – DALG Lecture 6
Implementing Priority Queues
Updates on a Heap Structure
6.22
• DeleteMin = deletion of the root
– Replace root by last leaf
– Restore partial order by swaping nodes
downwards
”down-heap bubbling”
This is a
complete
binary tree!
• Insert
– Insert new node after last leaf
– Restore partial ordering by ”up-heap bubbling”
”last” leaf
This Jan
is Maluszynski
also called
a HEAP
- HT 2006
6.23
Jan Maluszynski - HT 2006
6.24
4
TTIT33 Algorithms and Optimization – DALG Lecture 6
TTIT33 Algorithms and Optimization – DALG Lecture 6
HEAP Properties:
HEAP example – bubble-up after insert(4)
We store pairs of <key, data> in
the tree.
Recall tree representation in a
table:
• getLeftChild(i) = 2*i+1
• getRightChild(i) = 2*i+2
...where i is the table index, not
the key!
In [G/T] the root
starts at index 1
and representation
is:
-getLeft(i) = 2*I
-getRight(i) = 2*i+1
• size(), minElemenent():
O(1)
• minElement(), minKey(): O(1)
• insertItem(), removeMin(): O(log n)
>> HeapSort O(n log n) sorting algorithm
...in [L/D] the root
starts at 0...
Recall vector representation of BST!
A complete binary tree...
• Compact vector representation
• Bubble-up and bubble-down have fast
implementations
Jan Maluszynski - HT 2006
6.25
Jan Maluszynski - HT 2006
6.26
TTIT33 Algorithms and Optimization – DALG Lecture 6
Heap Sort
Kinds of heaps :
• minKey in the root => (Min)Heap
• maxKey in the root => MaxHeap
Heap Sort: input D[0..n]
• Heapify D => Max Heap
worst case O(n log n)
• For i from 0 to n-1
DeleteMax => put in D[n-i]
(worst case O(log n) for restoring heap D[0..i]
Jan Maluszynski - HT 2006
6.27
5