Graph Theory

1
Graph Theory

• Trees
• Algorithms

2
Graphs Basic Definitions 4

• Let n be the number of nodes (stations) and e be
  the number of edges (links).
• A graph is connected if there is a sequence of
  nodes and edges between every pair of nodes.
• A cycle is a sequence of nodes and edges visiting
  none more than once, except for the first/last
  one
• Degree of edges incident on node

3
Trees 6

• A tree is a simple graph having the property that
  there is a unique path between every pair of
  nodes.
• A rooted tree is one in which a particular node
  is designated as the root
• A node in a rooted tree is a leaf if it is at
  root i (igt0) and not adjacent to any nodes at
  level i1
• A tree which is a subgraph of a graph G and
  contains every node of G is called a spanning
  tree of G
• A node which is not a leaf is an internal node
• Chartrand p. 87/ 1,2

4
Trees (cont.)
   
 
5
Spanning and Rooted trees

• Petersen Graph (find spanning treesMaple
  counttrees)
• Rooted tree
• Rooted tree 2

 
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Theorem Let T be a simple graph with n nodes.
TFAE

• (a) T is a tree
• (b) T is connected and acyclic
• (c)T is connected and has n-1 edges
• (d) T has n-1 edges and is acyclic

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Proof (partial)

• (a)?(b)
• There is a path between every pair of nodes
  (definition of tree)
• Hence, T is connected
• If T had a cycle, then there would exist at least
  two paths from node a to node b
• But the path from a to b is unique (by
  definition)
• Hence, T is acyclic.
• ?
• (b)?(c),(c)?(d)

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Proof (partial--continued)

• To show (d)?(a)
• Assume acyclic and n-1 edges
• RTStree(unique path betw. every node pair)
• WTST connected
• Assume not then T1,T2,Tk are components
• (disjoint connected subgraphs)
• Let Ti have ni nodes no cycles, so each must
  have ni-1 edges
• n-1(n1-1)(n2-1)(n3-1)(nk-1) lt
  (n1n2n3nk)-1n-1
• ??Contradiction T connected, so T is a tree

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Uses of trees 9

• Ex/ Huffman codes
• ASCII-each character encoded in a 7-bit string
• A 100 0000
• B100 0001
• C 100 0010
• 1 011 0001
• 2011 0010
• ! 010 0001
• 010 1010

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Uses of trees (cont)

• See board (BNF 15.3,15.4)

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Depth first search

• Suppose we wish to search the nodes of a graph,
  beginning at a specified node. Two types of
  strategy could be used
• forge ahead, moving on to a new node whenever one
  is available
• Spread outchecking all nodes at a each level
  before moving on
• First strategy depth-first search

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DFS algorithm

• This algorithm will assign labels to nodes and
  select edges of a graph G. ? is the set of nodes
  with labels, ? is the set of edges selected, and
  the predecessor of a node Y is a node in ? that
  is used in labeling Y.

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DFS algorithm

• STEP 1 (Start) Pick a node u and assign it label
  1. Let k1, ? u, and ? ? and Xu.
• STEP 2(check for completion) If ? contains each
  node of G, then StopG connected
• STEP 3(Find next node and edge) Find node Y not
  in ? that is adjacent to X if no such node, go
  to Step 4. Otherwise, putX,Y in ?, increment k
  to k1, assign the label k to Y, and put Y in ?.
  Replace X by Y, and go to Step 2
• (Back up) If Xu, then Stop. G is not connected.
  Else, replace X by the predecessor of X and go to
  Step 3

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BFS algorithm

• This algorithm will find a spanning tree, if it
  exists, for a graph on n nodes. In the algorithm,
  ? is the set of nodes with labels and ? is the
  set of edges connecting nodes in ?.

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BFS algorithm

• STEP 1 (Start) Pick a node u and assign it label
  0. Let ? u, and ? ? and Xu.
• STEP 2(check for completion) If ?ltn, go to
  Step 3. Otherwise, if ?n, stop the edges in ?
  and the nodes in ? form a spanning tree for G.
• STEP 3 (label the nodes) Find the nodes not in
  ? that are adjacent to nodes in ? having the
  largest label number call it k. If there are no
  such nodes, then the graph has no spanning tree.
  Otherwise, assign the newly found nodes the label
  k1, and put them in ?. For each new node with
  label k1, place in ? one edge joining this node
  to a node with label k. If there is more than one
  such edge, choose one arbitrarily. Go to Step 2.

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Minimum Weight Spanning Tree algorithm (Kruskal)

• This algorithm will find a minimal spanning tree,
  if one exists, for a weighted graph G with n
  nodes, e edges.

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MWST algorithm--Kruskal

• STEP 1 (Start) If no edges, G is not connected
  else, pick an edge with the smallest weight (ties
  can be broken arbitrarily). Place edge in E and
  node in N.
• STEP 2(check for completion) If E contains n-1
  edges, Stop have MWST else, go to Step 3.
• STEP 3 (pick next edge) Find the edges of
  smallest weight which do not form a cycle with
  any of the edges of E. If no such edges, G is not
  connected and there is no spanning tree. Else,
  choose one such edge and place it in E and the
  nodes in N. Go to Step 2.

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Binary trees and traversals 6

• Binary Tree a rooted tree in which each node has
  at most two children and each child is designated
  left child or right child.
• Thus in a binary tree, each node may have 0,1, or
  2 children
• Left childleft and below parent
• Right childright and below parent
• The left subtree of a node N in a binary tree is
  the graph formed by the left child L, the
  descendants of L, and the edges connecting these
  nodes
• Right subtreedefined similarly

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Binary trees and traversals 6

• A is the root, A has two children, left child B
  and right child C
• Node B has one child, left child D
• Node C has right child E but no left child.

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Rooted Tree
    D

B A
C
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Expression trees

• 4.6.
• Polish notationPolish mathematician
  Lukasiewiczno parens needed
• RT4

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Traversal 4

• Traversal visit each node of a graph exactly
  once
• BFS/DFStraversal of a connected graphnodes are
  visited , i.e., labeled, exactly once
• A preorder traversal of a binary tree is
  characterized by visiting the parent before the
  children and the left child before the right
  child
• Listing the nodes in the order they are visited
  is called a preorder listing

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Preorder Traversal (i.e., DFS w/ choosing left
before right) 4

• STEP 1 (Visit) Visit the root
• STEP 2(go left) Go to the left subtree, if one
  exists, and do a preorder traversal
• STEP 3 Go to the right subtree, if one exists,
  and do a preorder traversal.
• 4.6.4,4.6.5

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Postorder Traversal 6

• RPNReverse Polish Notation
• Operation sign is followed by the operands (HP
  calculator)
• (2-34)(48/2)-2
• Pre -2344/82 Preorderlook for operation
  sign followed by two numbers
• Post 234-482/
• Postorderlook for two condecutive numbers
  followed by an operation
• By using a traversal called postorder, can obtain
  the RPN for an expression

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Postorder Traversal (child before parent, left
before right) 4

• STEP 1 (Start) Go to the root
• STEP 2(go left) Go to the left subtree, if one
  exists, and do a postorder traversal
• STEP 3 (go right) Go to the right subtree, if
  one exists, and do a postorder traversal.
• Step 4 (Visit) Visit the root
• 4.6.7, 4.6.8

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Binary Search Tree

• (to determine if an element a is in a binary
  search tree)
• Step 1(Start) Let V be the root of the binary
  search tree.
• Step 2 (compare) If a V, then A is in the tree
  STOP. Else, go to Step 3
• Step 3 (if smaller, go left) If Vlta, go to Step
  4. Otherwise, altV
• If no left child of V, a is not in tree. STOP
• Else, V has left child L let VL and go to Step
  2
• Step 4 (if larger, go right)