License
c 2001-2016 T. Uyar, A. Yayımlı, E. Harmancı
Discrete Mathematics
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Trees
Under the following terms:
H. Turgut Uyar
Ayşegül Gençata Yayımlı
Emre Harmancı
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and indicate if changes were made.
I NonCommercial – You may not use the material for commercial purposes.
I ShareAlike – If you remix, transform, or build upon the material, you must
2001-2016
distribute your contributions under the same license as the original.
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Topics
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Tree
Definition
tree: connected graph with no cycle
Trees
Introduction
Rooted Trees
Binary Trees
Decision Trees
examples
Tree Problems
Minimum Spanning Tree
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Tree Theorems
Tree Theorems
Theorem
T is a tree (T is connected and contains no cycle).
⇔
There is one and only one path
between any two distinct nodes in T .
⇔
T is connected, but if any edge is removed
it will no longer be connected.
⇔
T contains no cycle, but if an edge is added
between any pair of nodes one and only one cycle will be formed.
Theorem
|E | = |V | − 1
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proof method: induction on the number of edges
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Tree Theorems
Tree Theorems
Proof: Induction step.
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|E | = k + 1
Proof: Base step.
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|E | = 0 ⇒ |V | = 1
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|E | = 1 ⇒ |V | = 2
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|E | = 2 ⇒ |V | = 3
I
assume that |E | = |V | − 1 for |E | ≤ k
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remove edge (y , z):
T1 = (V1 , E1 ), T2 = (V2 , E2 )
|V | = |V1 | + |V2 |
= |E1 | + 1 + |E2 | + 1
= (|E1 | + |E2 | + 1) + 1
= |E | + 1
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Tree Theorems
Tree Theorems
Theorem
In a tree, there are at least two nodes with degree 1.
Theorem
Proof.
T is a tree (T is connected and contains no cycle).
⇔
T is connected and |E | = |V | − 1.
⇔
T contains no cycle and |E | = |V | − 1.
P
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2|E | =
I
assume: only 1 node with degree 1:
⇒ 2|E | ≥ 2(|V | − 1) + 1
⇒ 2|E | ≥ 2|V | − 1
⇒ |E | ≥ |V | − 12 > |V | − 1
v ∈V
dv
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Rooted Tree
Node Level
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hierarchy between nodes
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level of node: distance from root
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creates implicit direction on edges: in and out degrees
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parent: adjacent node closer to root (only 1 such node)
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in-degree 0: root (only 1 such node)
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child: adjacent nodes further from root
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out-degree 0: leaf
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sibling: nodes with same parent
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not a leaf: internal node
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depth of tree: maximum level in tree
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Rooted Tree Example
Rooted Tree Example
Book
I C1
I
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leaves: x y z u v
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internal nodes: r p n t s q w
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parent of y : w
children of w : y and z
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root: r
I
I
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S1.1
S1.2
C2
C3
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y and z are siblings
S3.1
S3.2
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I
I
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Ordered Rooted Tree
S3.2.1
S3.2.2
S3.3
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Lexicographic Order
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siblings ordered from left to right
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address A comes before address B if one of:
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universal address system
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root: 0
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children of root: 1, 2, 3, . . .
A = x1 x2 . . . xi xj . . .
B = x1 x2 . . . xi xk . . .
xj comes before xk
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v : internal node with address a
children of v : a.1, a.2, a.3, . . .
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A = x1 x2 . . . xi
B = x1 x2 . . . xi xk . . .
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Lexicographic Order Example
Binary Trees
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0 - 1 - 1.1 - 1.2
- 1.2.1 - 1.2.2 - 1.2.3
- 1.2.3.1 - 1.2.3.2
- 1.3 - 1.4 - 2
- 2.1 - 2.2 - 2.2.1
- 3 - 3.1 - 3.2
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T = (V , E ) is a binary tree:
∀v ∈ V [dv o ∈ {0, 1, 2}]
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T = (V , E ) is a complete binary tree:
∀v ∈ V [dv o ∈ {0, 2}]
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Expression Tree
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binary operations can be represented as binary trees
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root: operator, children: operands
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mathematical expression can be represented as trees
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internal nodes: operators, leaves: variables and values
Expression Tree Examples
7−a
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a+b
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Expression Tree Examples
7−a
5
Expression Tree Examples
7−a
· (a + b)3
5
(a + b)3
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Expression Tree Examples
Expression Tree Traversals
1. inorder traversal:
traverse left subtree, visit root, traverse right subtree
t+
2. preorder traversal:
visit root, traverse left subtree, traverse right subtree
u∗v
w + x − yz
3. postorder traversal (reverse Polish notation):
traverse left subtree, traverse right subtree, visit root
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Inorder Traversal Example
Preorder Traversal Example
t + u ∗ v /w + x − y ↑ z
+t / ∗ uv + w − x ↑ y z
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Postorder Traversal Example
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Expression Tree Evaluation
t uv ∗ w x y z ↑ − + /+
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inorder traversal requires parantheses for precedence
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preorder and postorder traversals do not require parantheses
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Postorder Evaluation Example
Regular Trees
t uv ∗ w x y z ↑ − + /+
423 ∗ 1923 ↑ − + /+
4
4
4
4
4
4
7
∗
9 2 3
9 8 −
1 +
/
2 3
6 1
6 1
6 1
6 2
3 +
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T = (V , E ) is an m-ary tree:
∀v ∈ V [dv o ≤ m]
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T = (V , E ) is a complete m-ary tree:
∀v ∈ V [dv o ∈ {0, m}]
↑
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Regular Tree Theorem
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Regular Tree Examples
Theorem
T = (V , E ): complete m-ary tree
I
how many matches are played in a tennis tournament
with 27 players?
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every player is a leaf: l = 27
n =m·i +1
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every match is an internal node: m = 2
l = n − i = m · i + 1 − i = (m − 1) · i + 1
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number of matches: i =
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n: number of nodes
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l: number of leaves
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i: number of internal nodes
I
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i=
l−1
m−1
=
27−1
2−1
= 26
l −1
m−1
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Regular Tree Examples
I
Decision Trees
how many extension cords with 4 outlets are required
to connect 25 computers to a wall socket?
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every computer is a leaf: l = 25
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every extension cord is an internal node: m = 4
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number of cords: i =
l−1
m−1
=
25−1
4−1
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one of 8 coins is counterfeit (heavier)
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find counterfeit coin using a beam balance
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depth of tree: number of weighings
=8
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Decision Trees
Decision Trees
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Spanning Tree
I
I
Kruskal’s Algorithm
T = (V 0 , E 0 ) is a spanning tree of G (V , E ):
T is a subgraph of G
T is a tree
V0 = V
1. G 0 = (V 0 , E 0 ), V 0 = ∅, E 0 = ∅
minimum spanning tree:
total weight of edges in E 0 is minimal
4. if |E 0 | = |V | − 1: result is G 0
2. select e = (v1 , v2 ) ∈ E − E 0 such that:
E 0 ∪ {e} contains no cycle, and wt(e) is minimal
3. E 0 = E ∪ {e}, V 0 = V 0 ∪ {v1 , v2 }
5. go to step 2
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Kruskal’s Algorithm Example
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Kruskal’s Algorithm Example
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minimum weight: 1
(e, g )
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minimum weight: 2
(d, e), (d, f ), (f , g )
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E 0 = {(e, g )}
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E 0 = {(e, g ), (d, f )}
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|E 0 | = 1
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|E 0 | = 2
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Kruskal’s Algorithm Example
Kruskal’s Algorithm Example
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minimum weight: 2
(d, e), (f , g )
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E 0 = {(e, g ), (d, f ), (d, e)}
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|E 0 | = 3
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minimum weight: 2
(f , g ) forms a cycle
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minimum weight: 3
(c, e), (c, g ), (d, g )
(d, g ) forms a cycle
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E 0 = {(e, g ), (d, f ), (d, e), (c, e)}
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|E 0 | = 4
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Kruskal’s Algorithm Example
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Kruskal’s Algorithm Example
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E0 = {
(e, g ), (d, f ), (d, e),
(c, e), (b, e)
}
I
E0 = {
(e, g ), (d, f ), (d, e),
(c, e), (b, e), (a, b)
}
I
|E 0 | = 5
I
|E 0 | = 6
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Kruskal’s Algorithm Example
Prim’s Algorithm
1. T 0 = (V 0 , E 0 ), E 0 = ∅, v0 ∈ V , V 0 = {v0 }
2. select v ∈ V − V 0 such that for a node x ∈ V 0
e = (x, v ), E 0 ∪ {e} contains no cycle, and wt(e) is minimal
I
total weight: 17
3. E 0 = E ∪ {e}, V 0 = V 0 ∪ {x}
4. if |V 0 | = |V |: result is T 0
5. go to step 2
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Prim’s Algorithm Example
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Prim’s Algorithm Example
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E0 = ∅
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E 0 = {(a, b)}
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V 0 = {a}
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V 0 = {a, b}
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|V 0 | = 1
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|V 0 | = 2
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Prim’s Algorithm Example
Prim’s Algorithm Example
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E 0 = {(a, b), (b, e)}
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E 0 = {(a, b), (b, e), (e, g )}
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V 0 = {a, b, e}
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V 0 = {a, b, e, g }
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|V 0 | = 3
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|V 0 | = 4
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Prim’s Algorithm Example
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Prim’s Algorithm Example
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E 0 = {(a, b), (b, e), (e, g ), (d, e)}
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V 0 = {a, b, e, g , d}
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|V 0 | = 5
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E0 = {
(a, b), (b, e), (e, g ),
(d, e), (f , g )
}
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V 0 = {a, b, e, g , d, f }
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|V 0 | = 6
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Prim’s Algorithm Example
Prim’s Algorithm Example
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E0 = {
(a, b), (b, e), (e, g ),
(d, e), (f , g ), (c, g )
}
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V 0 = {a, b, e, g , d, f , c}
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|V 0 | = 7
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References
Required Reading: Grimaldi
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Chapter 12: Trees
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12.1. Definitions and Examples
12.2. Rooted Trees
Chapter 13: Optimization and Matching
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13.2. Minimal Spanning Trees:
The Algorithms of Kruskal and Prim
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total weight: 17
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