Download MSc Computer Science ICS 801 Design and Analysis of Algorithms

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Randomized Binary Search Trees
 All the basic operations in a binary search tree run in
O(h) time
MSc Computer Science
 Height of the tree varies on how the items are inserted
and deleted
ICS 801
 If the items are inserted in a strictly increasing order,
the tree will be a chain with height n-1
Design and Analysis of Algorithms
 If we assume that the n keys are inserted in the tree in
random order such that any of the n! permutations are
equally likely
 The expected height of the tree is O(log n)
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Randomized Binary Search Trees
Randomized Binary Search Trees
 Inserting at the root requires use of two operations
 The randomized algorithm is based on the probability of the
1. Rotate left
value being inserted being the root of the tree
 Used to exchange a node with its right child hence the
right child goes one level up
 This can be given by 1/(n+1) where n is the number of nodes
currently in the tree
2. Rotate right
 This means that we will make the node we are currently
 Used to exchange a node with its left child hence the left
child goes one level up
inserting into the tree with probability 1/(n+1)
 The diagrams in the next slide show examples of rotating
left and rotating right
 If we cannot make it the root of the tree, then we will make it the
root of either the left subtree or the right subtree
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Randomized Binary Search Trees
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Randomized Binary Search Tree Algorithm
Node insert(Node h, Value val)
{
if (h == null)
return new Node(val)
if (Math.random()*(h.N + 1) <1)
return rootInsert(h, val)
if (val< h.val)
h.l = insert(h.l, val)
else
h.r = insert(h.r, val)
h.N++;
return h;
}
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Randomized Binary Search Tree Algorithm
Randomized Binary Search Tree Algorithm
Node rotL(Node h) {
Node x = h.r;
h.r = x.l;
x.l = h;
return x;
}
Node rotR(Node h) {
Node x = h.l;
h.l = x.r;
x.r = h;
return x;
}
Node rootInsert(Node h, Value val)
{
if (h == null)
return new Node(val);
if (val< h.val)) {
h.l = rootInsert(h.l, val);
h = rotR(h);
}
else {
h.r = rootInsert(h.r, val);
h = rotL(h);
}
return h;
}
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Example
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Red-black tree
 A type of self balancing binary search tree
 Given the following tree show how to insert the value
of 10 at the root node
 Typically used to implement associative arrays
 It can search, insert and delete in guaranteed O(log n) time
where n is the number of elements in the tree
 It has one extra bit of storage per node,
 The extra bit is used to store color information
 The color is either red or black
 By constraining the way the nodes can be colored in any path from
the root to the leaf,
 The red-black trees ensure that no path is more than twice as
long as any other
 So the tree is approximately balanced
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Red-black tree
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Red-black tree Example

If a node has a null child, the null child is considered to be a leaf node
and its color is black

A red-black tree thus satisfies the following properties
1.
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Every node is either red or black
2. The root is black
3. Every leaf (null) is black
4. If a node is red, then both its children are black
5. For each node, all paths from the node to descendant leaves
contain the same number of black nodes

A red-black tree with n internal nodes has height at most
2 lg(n+1)

The subtree rooted at any node x contains at least 2bh(x)-1 internal
nodes. Where bh(x) is the black height of node x without counting x
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Red-black tree insertion
Red-black tree insertion
 The following operations are needed to perform an insertion into a
red-black tree
 In insertion, a node is inserted just in the same way it
could be inserted in a normal search tree
1. Left rotation
 After it is inserted, it is colored red
A node becomes the left child of its right child
2. Right rotation
 There is a possibility that the parent of the inserted node
A node becomes the right child of its left child
is colored red
3. Uncle
 This then leads to a violation of property 4
Get the uncle of a given node
4. Grandparent
 Adjustments have to be carried out in the tree to ensure
Get the grandparent of a given node
that it satisfies the violated property
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 The operations performed on the tree to return the tree to the
required state are described below. Assume the inserted node is z
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Red-black tree insertion
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Red-black tree insertion
1. If the parent of the inserted node (z=14) is black,
2. If the parent of z (27) is red,
 There is no violation, end
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 There is a violation of property 4
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Red-black tree insertion
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Red-black tree insertion
2.1 The uncle of z is red
Recolor the parent (28 and 30 to black and the grandparent (29)
to red as shown below. 29 becomes the new z
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Red-black tree insertion
Red-black tree insertion
2.2 The uncle of z is black,
 The black height of the paths from 30 does not change because
2.2.1 z (27) is the right child of its parent, and z’s parent is the left
child
the number of black nodes along those paths has not increased.
The uncle turned black and the grand parent red, so the number
of black nodes in those paths has remained the same
 The number of black nodes in the paths via 28 have not changed
either,
One extra black node was added by turning 28 black
An extra black node removed by making 29 (grandparent red)
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Red-black tree insertion
2.2.1
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Red-black tree insertion
left rotate the parent of z, 26 becomes the new z
 The right children of 27 have one less node above them.
This node was red. So the number of black nodes along those
paths has not changed
 The left children of 27 become the right children of 26.
The have one extra red node above them
So the number of black nodes along those paths has not
changed
 The left children of 26 have one extra red node above them
So the number of black nodes along those paths has not
changed
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Red-black tree insertion
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Red-black tree insertion
2.2.2
z (26) is the left child of its parent which is the
left child of its grandparent
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2.2.2 Right rotate the grand parent (29) and color 29 red
and 27 black
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Red-black tree insertion
Red-black tree insertion
 The right children of 29 have an extra node above them (27)
 In case z is the root, just color it black
This node is black, but since 29 witch originally was black
becomes red, the number of black nodes in the paths remain the
same
All paths in the tree get one extra black node.
So if they originally had an equal number of black nodes, they
still have an equal number of black nodes
 The right children of 27 become the left children of 29. the
number of black nodes above them has not changed
Exercise
 The left children of 27 have one less node above them (29) which
was black.
How do you delete an item from a red-black tree?
But since 27 which was red becomes black, the number of black
nodes remain the same
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