Trees Dr B Prabhakaran prabautdallas edu Application Examples

Trees Dr. B. Prabhakaran {praba@utdallas. edu}

Application Examples n n Useful for locating items in a list Used in Huffman coding algorithm Study games like checkers and chess to determine winning strategies Weighted trees used to model various categories of problems

Trees Definition: A connected undirected graph with no simple circuits Characteristics - No multiple edges - No loops

Example UT Dallas School of Engineering CS EE TE School of Management Cohort MBA School of Arts School of Social Sciences

Which of the following graphs are trees? a e b b d c a a b f e a c c f f f e A d c B e d C d b D A & B : Trees C: Not a Tree (Cycle abdcfa) D: Not a Tree (Not connected). However it is a forest i. e. unconnected graphs

Tree Theorem 1 : An undirected Graph is a tree if and only if there is a unique simple path between any two of its vertices Proof (=>) The graph is a Tree and it does not have cycles. If there were cycles then there will be more than one simple path between two vertices. Hence, there is a unique simple path between every pair of vertices Proof (<=) There exists a unique simple path between any pair of vertices. Therefore the graph is connected and there are no cycles in the graph. Hence, the graph is a tree

Tree terminology Rooted tree: one vertex designated as root and every edge directed away Parent: u is the parent of v iff (if and only if) there is an edge directed from u to v Child: v is called the child of u Every non-root node has a unique parent (? )

Tree terminology Siblings: vertices with same parent Ancestors: all vertices on path from the root to this vertex, excluding the vertex Descendants: Descendants of vertex v are all vertices that have v as an ancestor Internal Nodes: Nodes that have children External or Leaf Nodes: Nodes that have no children

Given : Tree rooted at A Find: Descendants (B), Ancestor (A), Siblings (G) ? A Root Node = A B E D G H Internal Nodes = B, C, E and F C F I J K External Nodes = D, G, H , I, J and K

Definition - Level n The level of vertex v in a rooted tree is the length of the unique path from the root to v

What is the level of Ted? Lou Hal Ed Max Ken Joe Sue Ted

Definition - Height n The height of a rooted tree is the maximum of the levels of its vertices

What is the height? Lou Hal Ed Max Ken Joe Sue Ted

Definition: m-ary trees n n n Rooted tree where every vertex has no more than ‘m’ children Full m-ary if every internal vertex has exactly ‘m’ children (i. e. , except leaf/external vertices). m=2 gives a binary tree

Example: 3 -ary tree

Definition: Binary Tree n n n Every internal vertex has a maximum of 2 children An ordered rooted tree is a rooted tree where the children of each internal vertex are ordered. In an ordered binary tree, the two possible children of a vertex are called the left child and the right child, if they exist.

An Ordered Binary Tree Lou Hal Ed Max Ken Joe Sue Ted

Definition: Balanced n A rooted binary tree of height h is called balanced if all its leaves are at levels h or h-1

Is this tree balanced? Lou Hal Ed Max Ken Joe Sue Ted

Tree Property 1 Theorem 2 : A tree with n vertices has n-1 edges Proof: By induction Basis Step: for n = 1 there are (1 – 1) 0 edges Inductive Step: We assume that n vertices has n – 1 edges. We need to prove that n + 1 vertices have n edges. A tree with n + 1 vertices can be achieved by adding an edge from a vertex of the tree to the new vertex (why 1 edge ? ). The total number of edges is (n – 1 + 1 = n).

Tree Property 2 Theorem 3: A full m-ary tree with i internal vertices contains n = mi +1 vertices Proof: i internal vertices have m children. Therefore, we have mi vertices. Since the root is not a child we have n = mi + 1 vertices.

Tree Property 3 Theorem 4: A full m-ary tree with 1) n vertices has i = (n-1)/m internal vertices and l = [(m-1)n+1]/m leaves 2) i internal vertices has n = mi + 1 vertices and l = (m-1)i+1 leaves 3) l leaves has n = (ml-1)/(m-1) vertices and i = (l-1)/(m-1) internal vertices

Tree Property 4 Theorem 5 – Proof for 1) l: Number of leaves i: Number of internal vertices n: total number of vertices. By theorem 3, n = mi + 1, and we have n = l + i (why ? ) n = mi + 1 => i = (n-1)/m By subsitution in n = l + i Þ l = n – (n-1)/m Þ [(m-1)n+1]/m

Tree Property 4 Theorem 5 – Proof for 2) l: Number of leaves i: Number of internal vertices n: total number of vertices. By theorem 3, n = mi + 1, and we have n = l + i By subsitution in n = l + i Þ l + i = mi + 1 Þ l = (m-1) i + 1

Tree Property 4 Theorem 5 – Proof for 3) l: Number of leaves i: Number of internal vertices n: total number of vertices. By theorem 3, n = mi + 1, and we have n = l + i (i = n – l) By subsitution of i in n = mi + 1 Þ n = m (n –l) + 1 Þ n (m -1) = (ml -1) Þ n = (ml -1)/ (m -1) By subsitution of n in i = n – l we have i = (l – 1)/(m – 1)

Tree Property 5 Theorem 6: There at most mh leaves in a m-ary tree of height h Proof: By Induction Basic Step: m-ary tree of height 1 => leaves = m which is true since the root has m-ary children

Tree Property 5 Inductive Step: n Assume that the result is true for tree of height less than h n Let Tree T be m-ary of height h n n To build T we can add m such sub-tree of height ≤ h -1 such that the root node acts as the child to the root of T T has ≤ m. mh -1 ≤ mh

Applications of trees n Binary Search Trees n Decision Trees n Prefix Codes n Game Trees

Binary Search Trees (BST) Applications n n A Binary Search Tree can be used to store items in its vertices It enables efficient searches

Definition: BST A special kind of binary tree in which: n n n Each vertex contains a distinct key value, The key values in the tree can be compared using “greater than” and “less than”, and The key value of each vertex in the tree is less than every key value in its right subtree, and greater than every key value in its left subtree.

Shape of a binary search tree Depends on its key values and their order of insertion. Example: Insert the elements 6, 3, 4, 2, 10 in that order. The first value to be inserted is put into the root 6

Shape of a binary search tree Inserting 3 into the BST 6 3

Shape of a binary search tree Inserting 4 into the BST 6 3 4

Shape of a binary search tree Inserting 4 into the BST 6 3 2 4

Shape of a binary search tree Inserting 10 into the BST 6 10 3 2 4 12

Binary search tree algorithm n n n n procedure insertion (T: binary search tree, X: item) v: = root of T {a vertex not present in T has the value null} while v != null and label(v) != x begin if x < label(v) then if left child of v != null then v: =left child of v else add new vertex as left child of v and set v: = null else if right child of v != null then v: =right child of v else add new vertex as right child of v to T and set v : = null end if root of T = null then add a vertex v to the tree and label it with x else if v is null or label(v) != x then label new vertex with x and let v be this new vertex {v = location of x}

Codes Text Encoding Our next goal is to develop a code that represents a given text as compactly as possible. A standard encoding is ASCII, which represents every character using 7 bits: “An English sentence” = 133 bits ≈ 17 bytes 1000001 (A) 1101110 (n) 0100000 ( ) 1000101 (E) 1101110 (n) 1100111 (g)1101100 (l) 1101001 (i) 1110011 (s) 1101000 (h) 0100000 ( ) 1110011 (s) 1100101 (e) 1101110 (n) 1110100 (t) 1100101 (e) 1101110 (n) 1100011 (c) 1100101 (e)

Codes Text Encoding Of course, this is wasteful because we can encode 12 characters in 4 bits: ‹space› = 0000 A = 0001 E = 0010 c = 0011 e = 0100 g = 0101 h = 0110 i = 0111 l = 1000 n = 1001 s = 1010 t = 1011 Then we encode the phrase as 0001 (A) 1001 (n) 0000 ( ) 0010 (E) 1001 (n) 0101 (g) 1000 (l) 0111 (i) 1010 (s) 0110 (h) 0000 ( ) 1010 (s) 0100 (e) 1001 (n) 1011 (t) 0100 (e) 1001 (n) 0011 (c) 0100 (e) This requires 76 bits ≈ 10 bytes

Codes Text Encoding An even better code is given by the following encoding: ‹space› = 000 A = 0010 E = 0011 s = 010 c = 0110 g = 0111 h = 1000 i = 1001 l = 1010 t = 1011 e = 110 n = 111 Then we encode the phrase as 0010 (A) 111 (n) 000 ( ) 0011 (E) 111 (n) 0111 (g) 1010 (l) 1001 (i) 010 (s) 1000 (h) 000 ( ) 010 (s) 110 (e) 111 (n) 1011 (t) 110 (e) 111 (n) 0110 (c) 110 (e) This requires 65 bits ≈ 9 bytes

Codes That Can Be Decoded Fixed-length codes: n Every character is encoded using the same number of bits. To determine the boundaries between characters, we form groups of w bits, where w is the length of a character. n Examples: ■ ASCII ■ Our first improved code Prefix codes: n n No character is the prefix of another character. Examples: ■ Fixed-length codes ■ Huffman codes n

Why Prefix Codes ? n Consider a code that is not a prefix code: n a = 01 m = 10 n = 111 o = 0 r = 11 s = 1 t = 0011 Now you send a fan-letter to your favorite movie star. One of the sentences is “You are a star. ” n You encode “star” as “ 1 0011 01 11”. Your idol receives the letter and decodes the text using your coding table: 100110111 = 10 0 111 = “moron”

Representing a Prefix-Code Dictionary Our example: ‹space› = 000 A = 0010 E = 0011 s = 010 c = 0110 g = 0111 h = 1000 i = 1001 l = 1010 t = 1011 e = 110 n = 111 0 0 0 1 1 ‹spc› 1 0 1 A E 0 s 0 1 1 1 0 0 1 0 1 c g h i l t 0 e 1 n

Huffman’s Algorithm Huffman(C) 1 n ← | C| 2 Q←C 3 for i = 1. . n – 1 4 do allocate a new node z 5 left[z] ← x ← Delete-Min(Q) 6 right[z] ← y ← Delete-Min(Q) 7 f [ z ] ← f [ x ] + f [ y] 8 Insert(Q, z) 9 return Delete-Min(Q) f: 5 e: 9 c: 12 b: 13 d: 16 a: 45

Huffman’s Algorithm 14 Huffman(C) 1 n ← | C| 2 Q←C 3 for i = 1. . n – 1 4 do allocate a new node z 5 left[z] ← x ← Delete-Min(Q) 6 right[z] ← y ← Delete-Min(Q) 7 f [ z ] ← f [ x ] + f [ y] 8 Insert(Q, z) 9 return Delete-Min(Q) f: 5 e: 9 c: 12 b: 13 d: 16 a: 45

Huffman’s Algorithm Huffman(C) 1 n ← | C| 2 Q←C 3 for i = 1. . n – 1 4 do allocate a new node z 5 left[z] ← x ← Delete-Min(Q) 6 right[z] ← y ← Delete-Min(Q) 7 f [ z ] ← f [ x ] + f [ y] 8 Insert(Q, z) 9 return Delete-Min(Q) c: 12 b: 13 f: 5 14 d: 16 e: 9 a: 45

Huffman’s Algorithm Huffman(C) 1 n ← | C| 2 Q←C 3 for i = 1. . n – 1 4 do allocate a new node z 5 left[z] ← x ← Delete-Min(Q) 6 right[z] ← y ← Delete-Min(Q) 7 f [ z ] ← f [ x ] + f [ y] 8 Insert(Q, z) 9 return Delete-Min(Q) 25 c: 12 b: 13 f: 5 14 d: 16 e: 9 a: 45

Huffman’s Algorithm Huffman(C) 1 n ← | C| 2 Q←C 3 for i = 1. . n – 1 4 do allocate a new node z 5 left[z] ← x ← Delete-Min(Q) 6 right[z] ← y ← Delete-Min(Q) 7 f [ z ] ← f [ x ] + f [ y] 8 Insert(Q, z) 9 return Delete-Min(Q) 14 f: 5 d: 16 e: 9 c: 12 25 a: 45 b: 13

Huffman’s Algorithm Huffman(C) 1 n ← | C| 2 Q←C 3 for i = 1. . n – 1 4 do allocate a new node z 5 left[z] ← x ← Delete-Min(Q) 6 right[z] ← y ← Delete-Min(Q) 7 f [ z ] ← f [ x ] + f [ y] 8 Insert(Q, z) 9 return Delete-Min(Q) 30 14 f: 5 d: 16 e: 9 c: 12 25 a: 45 b: 13

Huffman’s Algorithm Huffman(C) 1 n ← | C| 2 Q←C 3 for i = 1. . n – 1 4 do allocate a new node z 5 left[z] ← x ← Delete-Min(Q) 6 right[z] ← y ← Delete-Min(Q) 7 f [ z ] ← f [ x ] + f [ y] 8 Insert(Q, z) 9 return Delete-Min(Q) 25 c: 12 30 14 b: 13 f: 5 a: 45 d: 16 e: 9

Huffman’s Algorithm Huffman(C) 1 n ← | C| 2 Q←C 3 for i = 1. . n – 1 4 do allocate a new node z 5 left[z] ← x ← Delete-Min(Q) 6 right[z] ← y ← Delete-Min(Q) 7 f [ z ] ← f [ x ] + f [ y] 8 Insert(Q, z) 9 return Delete-Min(Q) 55 25 c: 12 30 14 b: 13 f: 5 a: 45 d: 16 e: 9

Huffman’s Algorithm Huffman(C) 1 n ← | C| 2 Q←C 3 for i = 1. . n – 1 4 do allocate a new node z 5 left[z] ← x ← Delete-Min(Q) 6 right[z] ← y ← Delete-Min(Q) 7 f [ z ] ← f [ x ] + f [ y] 8 Insert(Q, z) 9 return Delete-Min(Q) 55 a: 45 25 c: 12 30 14 b: 13 f: 5 d: 16 e: 9

Huffman’s Algorithm Huffman(C) 1 n ← | C| 2 Q←C 3 for i = 1. . n – 1 4 do allocate a new node z 5 left[z] ← x ← Delete-Min(Q) 6 right[z] ← y ← Delete-Min(Q) 7 f [ z ] ← f [ x ] + f [ y] 8 Insert(Q, z) 9 return Delete-Min(Q) 100 0 1 a: 45 0 55 0 1 25 0 c: 12 100 30 1 14 b: 13 d: 16 101 0 1 111 f: 5 e: 9 1100 1101

Game Trees n n n Writing down all the possible moves by both players down onto a tree Root node represents the current status Internal nodes at even levels correspond to positions in which the first player is to move (MAX nodes) Internal nodes at odd levels correspond to positions in which the second player is to move (Min nodes) Leaves correspond to positions at which the game has ended. One player or the other has won, or perhaps the game is drawn.

Game Trees: Partial Game Tree for Tic-Tac-Toe Here's a little piece of the game tree for Tic-Tac-Toe, starting from an empty board. Note that even for this trivial game, the search tree is quite big.

Game Trees n In case of TIC TAC TOE, there are 9 choices for the first player n Next, the opponent have 8 choices, …, and so on n Number of nodes = n Exponential order of nodes n n 1 + 9*8*7 + … +9*8*7*6*5*4*3*2*1 = ? ? ? If the depth of the tree is m and there are b moves at each node, the time complexity of evaluating all the nodes are O(bm) Even in case of TICTACTOE, there are so many nodes

Game Trees: Min-Max Procedure n n So instead of expanding the whole tree, we use a min-max strategy Starting from the current game position as the node, expand the tree to some depth Apply the evaluation function at each of the leaf nodes “Back up” values for each of the non-leaf nodes until a value is computed for the root node n n n At MIN nodes, the backed-up value is the minimum of the values associated with its children. At MAX nodes, the backed-up value is the maximum of the values associated with its children. Pick the edge at the root associated with the child node whose backed-up value determined the value at the root

Game Trees: Minimax Search 2 1 2 2 7 1 Static evaluator value 8 2 7 1 8 2 1 2 7 1 8 2 This is the move selected by minimax 2 1 MAX MIN 2 7 1 8

Tree Traversal n n n Information is stored in Ordered Rooted tree Methods are needed for visiting each vertex of the tree in order to access data Three most commonly used traversal algorithms are: 1. Preorder Traversal 2. Inorder Traversal 3. Postorder Traversal

Preorder Traversal Let T be an ordered binary tree with root r If T has only r, then r is the preorder traversal. Otherwise, suppose T 1, T 2 are the left and right subtrees at r. The preorder traversal begins by visiting r. Then traverses T 1 in preorder, then traverses T 2 in preorder.

Preorder Traversal Example: J E A H T M Y ‘J’ ‘T’ ‘E’ ‘A’ ‘H’ ‘M’ ‘Y’

Preorder Traversal Procedure preorder (T: ordered rooted tree) r : = root of T List r; for each child c of r from left to right T(c) : = subtree with c as its root preorder(T(c)); End.

Inorder Traversal Let T be an ordered binary tree with root r. If T has only r, then r is the inorder traversal. Otherwise, suppose T 1, T 2 are the left and right subtrees at r. The inorder traversal begins by traversing T 1 in inorder. Then visits r, then traverses T 2 in inorder.

Inorder Traversal: A E H J M T Y ‘J’ ‘T’ ‘E’ ‘A’ ‘H’ ‘M’ ‘Y’

Inorder Traversal Procedure inorder (T: ordered rooted tree) r : = root of T if r is a leaf then list r else begin l : = first child of r from left to right; T(l) : = subtree with l as its root inorder (T(l)); list r; for each child c of r except for l from left to right T(c) : = subtree with c as its root inorder(T(c)); End.

Postorder Traversal Let T be an ordered binary tree with root r. If T has only r, then r is the postorder traversal. Otherwise, suppose T 1, T 2 are the left and right subtrees at r. The postorder traversal begins by traversing T 1 in postorder. Then traverses T 2 in postorder, then ends by visiting r.

Postorder Traversal: A H E M Y T J ‘J’ ‘T’ ‘E’ ‘A’ ‘H’ ‘M’ ‘Y’

Postorder Traversal Procedure postorder (T: ordered rooted tree) r : = root of T for each child c of r from left to right T(c) : = subtree with c as its root postorder(T(c)); End. List r.

Binary Expression Tree n n Ordered rooted tree representing complicated expression such compound propositions, combinations of sets, and arithmetic expression are called Binary Expression Tree. In this special kind of tree, 1. Each leaf node contains a single operand, 2. Each nonleaf node contains a single binary operator, and 3. The left and right subtrees of an operator node represent subexpressions that must be evaluated before applying the operator at the root of the subtree.

Infix, Prefix, and Postfix Expressions ‘*’ ‘-’ ‘ 8’ ‘/’ ‘ 5’ ‘ 3’ ‘+’ ‘ 4’ ‘ 2’ What infix, prefix, postfix expressions does it represent?

Infix, Prefix, and Postfix Expressions ‘*’ ‘/’ ‘-’ ‘ 8’ ‘ 5’ ‘ 4’ Infix: ‘ 3’ ‘+’ ‘ 2’ ((8 -5)*((4+2)/3)) Prefix: *-85 /+423 Postfix: 85 - 42+3/*

Spanning Tree A spanning tree of a (connected) graph G is a connected, acyclic subgraph of G that contains all of the vertices of G

Spanning Tree Example Below is an example of a spanning tree T, where the edges in T are drawn as solid lines and the edges in G but not in T are drawn as dotted lines. B A H C F D G E I J

Algorithms to find spanning tree in a graph n Depth-First-Search (DFS) n Breadth-First-Search (BFS)

Depth First Search (DFS) • The algorithm starts from a node, selects a node connected to it, then selects a node connected to this new node and so on, till no new nodes remain. • Then, it backtracks to the node next to the last one and discovers any new nodes connected to it. • Move back 2 vertices in the path and try again. • Repeat this move back operation until no more edges to be added. • A different ordering of the vertices can lead to a different DFS spanning tree • Stacks are used to implement DFS algorithms

DFS Example 1 2 3 5 4

DFS Example 1 2 3 5 4

DFS Example 1 2 3 5 4

DFS Example 1 2 3 5 4

DFS Example 1 2 3 5 4

Algorithm for DFS

Breadth First Search (BFS) n n Compared to DFS breadth first search is a simple algorithm Timidly tries one edge. Totally exhaust neighbors of a vertex then goes to next neighbors. It radiates in waves in balanced manner. Implemented using queues Whatever is in queues tells u what to explore next. Once the queue is empty algorithm comes to an end

Algorithm for BFS

BFS Example 1 2 3 5 4

BFS Example 1 2 3 5 4

BFS Example 1 2 3 5 4

BFS Example 1 2 3 5 4

BFS Example 1 2 3 5 4

Minimum Spanning Tree (Min SPT) Given: Graph G=(V, E), |V|=n Cost function c: E R. Goal: Find a minimum-cost spanning tree for V i. e. , find a subset of arcs E* E which connects any two nodes of V with minimum possible cost Min. span. tree: G*=(V, E*) Example: G=(V, E) 3 3 b d 2 2 8 8 a 3 3 5 a 5 7 7 4 4 e c 4 4 Red bold arcs are in E*

Algorithms for Min SPT n Prim’s Algorithm n Kruskal’s Algorithm

Prims Algorithm Procedure Prims (G: weighted connected undirected graph with n vertices) T : = a minimum-weight edge For i : = 1 to n-2 Begin e: = an edge of minimum weight incident to a vertex in T and not forming a simple circuit in T if added to T T : = T with e added End {T is a minimum spanning tree of G}

Prim’s Example n Initially 3 b 2 d 8 a 3 4 c 5 7 4 e T = {}

Prim’s Example n Initial step, chose min weight edge 3 b 2 d 8 a 3 4 c 5 7 4 e T = {{a, b}}

Prim’s Example n Iteration 1 3 b 2 d 8 a 3 4 c 5 7 4 e T = {{a, b}, {b, c}}

Prim’s Example n Iteration 2 3 b 2 d 8 a 3 4 c 5 7 4 e T = {{a, b}, {b, c}, {b, d}}

Prim’s Example n Iteration 3 3 b 2 d 8 a 3 4 c 5 7 4 e T = {{a, b}, {b, c}, {b, d} , {c, e}}

Kruskal’s algorithm Procedure Kruskal (G: weighted connected undirected graph with n vertices) T : = empty graph For i : = 1 to n-1 Begin e : = any edge in G with smallest weight that does not form simple circuit when added to T T : = T with e added End { T is the minimum spanning tree of G}

Kruskal’s Example n Initially 3 b 2 d 8 a 3 4 c 5 7 4 e T = {}

Kruskal’s Example n Iteration 1 3 b 2 d 8 a 3 4 c 5 7 4 e T = {{a, b}}

Kruskal’s Example n Iteration 2 3 b 2 d 8 a 3 4 c 5 7 4 e T = {{a, b}, {b, c}}

Kruskal’s Example n Iteration 3 3 b 2 d 8 a 3 4 c 5 7 4 e T = {{a, b}, {b, c}, {b, d}}

Kruskal’s Example n Iteration 4 3 b 2 d 8 a 3 4 c 5 7 4 e T = {{a, b}, {b, c}, {b, d} , {c, e}}

Prim’s and Kruskal’s n Prim’s n n Chosen edges: minimum weight & incident to a vertex already in the tree. Kruskal’s n Chosen edge: minimum weight & not necessarily incident to a vertex in the tree.