HandOut and Work Sheet CS 309 GRAPH THEORY AND COMBINATORICS Semester: Five Academic Year: 2018-2019 Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics HandOut and Work Sheet CS 309 GRAPH THEORY AND COMBINATORICS Semester: Five FACULTY NAME DESIGNATION DEPARTMENT : KEERTHI A S PILLAI : ASSISTANT PROFESSOR : COMPUTER SCIENCE AND ENGINEERING VERIFIED BY, 2 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE 3 HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE 4 HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE 5 HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics MODULE I Graph A linear graph (graph) G=(V,E) consists of a set of objects V={v1,v2,...} called vertices, and another set E={e1,e2,...} called edges, such that each edge ek is identified with an unordered pair (vi,vj) of vertices. The vertices vi,vj associated with edge ek are called the end vertices of ek. Vertex is also referred to as a node, a junction, a point, 0- cell, or an 0-simplex. Edges are also referred to as a branch, a line, an element, a 1-cell , an arc and a 1simplex. Graph can be represented by means of diagrams in which vertices are represented as points and each edge as a line segment joining its end vertices Self loop: An edge having the same vertex as both its end vertices is called a self loop(loop). Such an edge is associated with vertex pair (vi,vi). In the above figure, edge e1 is a self loop associated with vertex pair(v2,v2) Parallel edges: More than one edge associated with a given pair of vertices are called parallel edges. In the above figure e4 and e5 are parallel edges. Simple Graph: A graph that has neither self loops nor parallel edges is called simple graph. A graph is also called a linear complex, a 1-complex, or a one dimensional complex. Examples of simple graph are shown below: 6 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Applications of Graph a) Konigsberg Bridge Problem: Two islands C and D were connected to each other and to the banks A and B with seven bridges as shown in figure. The problem was to start at any land areas A, B, C or D , walk over each of the seven bridges exactly once, and return to the starting point Euler represented this problem by means of a graph. Vertices represent the land areas and the edges represents the bridges Euler proved that a solution for this problem does not exists b) Utilities Problem: There are three houses H1 , H2 and H3, each to be connected to each of the three utilities water(W), gas(G) and electricity(E) by means of conduits. Is it possible to make such connections without any crossovers of conduits? This problem also can be represented by a graph where edges represented the conduits and verticesrepresented the houses and utility centres 7 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics From the graph it is clear that no solution is exists for ulitilty problem c) Electrical Network Problem: Topology of a electrical network is studied by means of graphs. Vertices represented the electrical network junctions and the edges represented the branches Electrical Network and its graph d) Seating problem: Problem: Problem: Nine members of a new club meet each day for lunch at a round table. They decide to sit such that every member has different neighbours at ach lunch. How many days can this arrangement last? Solution: Seating arrangement of nine members in a round table can be represented by means of a graph. Each vertex represents a member and an edge joining two vertices represents the relationship of sitting next to each other 8 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Finite and Infinite Graph A graph with finite number of vertices and finite number of edges is called a finite graph, otherwise it is an infinite graph Finite Graph Infinite Graph Incidence and Degree When a vertex vi is an end vertex of some ej, vi and ej are said to incident with each other.Two non parallel edges said to be adjacent if they are incident on a common vertex. For example: a) e3,e4,e5 incident on the vertex v1 b) e3,e1,e2 incident on the vertex v2 c) e4,e5,e6 incident on the vertex v3 d) e2,e6,e7 incident on the vertex v4 e) e7 incident on the vertex v5 Degree or valency of a vertex The number of edges incident on a vertex vi, with self loop counted twice, is called the degree , d(vi), of vertex vi. For example: a) d(v1)= d(v3)= d(v4)=3 9 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics b) d(v2)=4 c) d(v5)=1 Sum of the degrees of all vertices in G is twice the number of edges in G 𝒏 𝒅 𝒗𝒊 = 𝟐𝒆 𝒊=𝟏 For example: sum of degrees of vertices=d(v1)+d(v2)+d(v3)+ d(v4)+d(v5) = 3+4+3+3+1 =14 = twice the number of edges Theorem 1.1 :The number of vertices of odd degree in a graph is always even Proof: Sum of the degrees of all vertices in G = 𝑛𝑖=1 𝑑(𝑣𝑖 ) If we consider the vertices with odd and even degree separately, the above equation can be expressed as the sum of vertices of even and odd degree as follows: 𝑛 𝑑(𝑣𝑖 ) = 𝑖=1 𝑑 𝑣𝑗 + 𝑒𝑣𝑒𝑛 𝑑 𝑣𝑘 𝑜𝑑𝑑 Since the left hand side of the equation is even, and the first expression on the right hand side is even (being a sum of even numbers), the second expression must also be even: 𝑑 𝑣𝑘 = 𝑎𝑛 𝑒𝑣𝑒𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑑𝑑 Because each 𝑑 𝑣𝑘 is odd, the total number of terms in the sum must be even to make the sum an even number. Hence the theorem Steps for checking the given degree sequence form a graph Step 1: Sort the degrees in decending order Step 2: If the first degree is 0, then graph can be formed Step 3: If the first degree has value d ,the following d degrees must be greater than 0. If not, no such graph can be formed Step 4: Take away(remove) the first degree (value of d) and reduce the following d degrees by one Step 5: Repeat the steps until all degrees are zero Steps for drawing a graph, if the degree sequence is given Step 1: Sort the degrees in decending order Step 2: Connect the highest degree d to the next d vertices 10 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Step 3: Take away(remove) the first degree (value of d) and reduce the following d degrees by one Step 4: Repeat the steps until all degrees are zero 11 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Regular Graph A graph in which all vertices are of equal degree is called a regular graph Isolated Vertex A vertex having no incident edge is called isolated vertex. Isolated vertices are vertices with zero degree Pendent Vertex or End vertex A vertex of degree one is called a pendant vertex or an end vertex Example Isolated vertex: v4,v7 Pendant vertex: v3 Null Graph A graph without any edges is called null graph Every vertex in null graph is an isolated vertex 12 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Isomorphism Two graphs G and G’ are said to be isomorphic if there is a one to one correspondence between their vertices and between their edges such that the incidence relationship is preserved The two isomorphic graph must have a) same number of vertices b) same number of edges c) equal number of vertices with a given degree QN: Check whether the following pair of graphs are isomorphic or not 1. G1 G2 Vertices and 𝒂, 𝒃, 𝒄, 𝒅, 𝒆 𝒗𝟏, 𝒗𝟐, 𝒗𝟑, 𝒗𝟒, 𝒗𝟓 Total number of vertices 5 5 Vertices with degree (arranged in decending order) a, c, d, b, e v1, v3, v4, v2, v5 Total number of edges= sum of degrees of all vertices/2 (3+3+3+2+1)/2=6 (3+3+3+2+1)/2=6 Since G1 and G2 have same number of vertices and same number of edges, G1 and G2 are isomorphic Graphs 2. 13 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics G1 𝑨𝟏, 𝑨𝟐, 𝑨𝟑, 𝑨𝟒, 𝑨𝟓, 𝑨𝟔, 𝑨𝟕, 𝑨𝟖 G2 𝑩𝟏, 𝑩𝟐, 𝑩𝟑, 𝑩𝟒, 𝑩𝟓, 𝑩𝟔, 𝑩𝟕, 𝑩𝟖 Vertices and Total number of vertices 8 8 Vertices with degree (arranged in 𝑨𝟏, 𝑨𝟐, 𝑨𝟑, 𝑨𝟒, 𝑨𝟓, 𝑨𝟔, 𝑨𝟕, 𝑨𝟖 𝑩𝟏, 𝑩𝟐, 𝑩𝟑, 𝑩𝟒, 𝑩𝟓, 𝑩𝟔, 𝑩𝟕, 𝑩𝟖 decending order) Total number of edges= sum of (3+3+3+3+3+3+3+3)/2=12 (3+3+3+3+3+3+3+3)/2=12 degrees of all vertices/2 Since G1 and G2 have same number of vertices and same number of edges, G1 and G2 are isomorphic Graphs 3. Vertices and G1 G2 𝑨𝟏, 𝑨𝟐, 𝑨𝟑, 𝑨𝟒, 𝑨𝟓 𝑩𝟏, 𝑩𝟐, 𝑩𝟑, 𝑩𝟒, 𝑩𝟓, , 𝑨𝟔, 𝑨𝟕, 𝑨𝟖, 𝑨𝟗, 𝑨𝟏𝟎 𝑩𝟔, 𝑩𝟕, 𝑩𝟖, 𝑩𝟗, 𝑩𝟏𝟎 G3 𝑪𝟏, 𝑪𝟐, 𝑪𝟑, 𝑪𝟒, 𝑪𝟓, 𝑪𝟔, 𝑪𝟕, 𝑪𝟖, 𝑪𝟗, 𝑪𝟏𝟎 Total number of vertices 10 10 10 Vertices with degree (arranged in 𝑨𝟏, 𝑨𝟐, 𝑨𝟑, 𝑨𝟒, 𝑨𝟓, 𝑩𝟏, 𝑩𝟐, 𝑩𝟑, 𝑩𝟒, 𝑩𝟓, 𝑪𝟏, 𝑪𝟐, 𝑪𝟑, 𝑪𝟒, 𝑪𝟓, decending order) 𝑨𝟔, 𝑨𝟕, 𝑨𝟖, 𝑨𝟗, 𝑨𝟏𝟎 𝑩𝟔, 𝑩𝟕, 𝑩𝟖, 𝑩𝟗,B10 𝑪𝟔, 𝑪𝟕, 𝑪𝟖, 𝑪𝟗,C10 Total number of edges= sum of (3+3+3+3+3+3+3+3 (3+3+3+3+3+3+3+3+ (3+3+3+3+3+3+3+3+ degrees of all +3+3)/2=15 3+3)/2=15 3+3)/2=15 vertices/2 Since G1,G2 and G3 have same number of vertices and same number of edges, G1, G2 and G3 are isomorphic Graphs 14 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics 4) G1 𝑨𝟏, 𝑨𝟐, 𝑨𝟑, 𝑨𝟒, 𝑨𝟓, 𝑨𝟔 G2 𝑩𝟏, 𝑩𝟐, 𝑩𝟑, 𝑩𝟒, 𝑩𝟓, 𝑩𝟔 Vertices and Total number of vertices 6 6 Vertices with degree (arranged in 𝑨𝟒, 𝑨𝟐, 𝑨𝟑, 𝑨𝟏, 𝑨𝟓, 𝑨𝟔 𝑩𝟑, 𝑩𝟐, 𝑩𝟓, 𝑩𝟏, 𝑩𝟒, 𝑩𝟔 decending order) Total number of edges= sum of (3+2+2+1+1+1)/2=5 (3+2+2+1+1+1)/2=5 degrees of all vertices/2 Since G1 and G2 have same number of vertices and same number of edges, G1 and G2 are isomorphic Graphs Sub graphs A graph g is said to be a sub graph of a graph G, if all the vertices and all the edges of g are in G, and each edge of g has the same end vertices in g as in G. And it is denoted by g G graph G graph g Properties of sub graph Every graph is its own sub graph A sub graph of a sub graph of G is a sub graph of G A single vertex in a graph G is a sub graph of G A single edge in G, together with its end vertices , is also a sub graph of G 15 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Edge Disjoint Sub Graphs Two sub graphs g1 and g2 of a graph G are said to be edge disjoint if g1 and g2 do not have any edge in common A Puzzle with multi coloured cubes Problem: We are given four cubes. The six faces of of every cube are various coloured blue, green, red or white. Is it possible to stack the cubes one on top of another to form a column such that no color appears twice on any of the four sides of theis column. Solution: Step 1: Draw a graph with four vertices B, G, R and W one for each color. Three opposite faces of the cube are represented by three edges. ie., if a blue face in cube 1 has a white face opposite to it, draw an edge between vertices B and W in the graph. Put label 1 on all the three edges resulting from cube 1. Repeat the procedure for all the three cubes one by one 16 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics on the same graph until we have a graph with four vertices and 12 edges. A particular set of four coloured cubes and their graphs are shown below. Step 2: Consider the graph resulting from the representation. The degree of each vertex is the total number of faces with the corresponding color. Four cubes can be arranged (to form a coloumn such that no color appears more than once on any side) if and only if there exists two edge disjoint sub graphs, each with four edges, each of the edges labelled differently, and such that each vertex is of degree two. For the above set of cubes, this condition is satisfied, and the two sub graphs are shown below. Walks A walk is defined as a finite alternating sequence of vertices and edges, beginning and ending with vertices, such that each edge is incident with the preceding and following it No edge appeared more than once in a walk. A vertex may appear more than once A walk is also referred to as an edge train or chain. 17 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Vertices with which a walk begins and ends are called its terminal vertices Closed Walk: If a walk begins and ends with same vertex, then it is called closed walk Open Walk: If a walk is not closed is called Open Walk Path An open walk in which no vertex appears more than once is called a path The number of edges in a path is called the length of a path A self loop can be included in a walk but not in a path The terminal vertices of a path are of degree one and the rest of the vertices are of degree two. Circuit A closed walk in which no vertex appears more than one is called a circuit Circuit is a closed non intersecting walk 18 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics If the circuit is a sub graph of another graph , one must count degrees contributed by the edges in the circuit only A circuit is also called a cycle, elementary cycle, circular path and polygon Every self loop is a circuit, but not every circuit is a self loop Connected and Disconnected Graph A graph G is said to be connected if there is atleast one path between every pair of vertices in G. Otherwise G is disconnected A null graph of more than one vertex is disconnected A disconnected graph consists of two or more connected graph. Each of these connected subgraph is called a component Theorem 1.2: A graph G is disconnected if and only if its vertex set V can be partitioned into two non empty, disjoint subsets V1 and V2 such that there exists no edge in G whose one end vertex vertex is in subset V1 and other in subset V2 Proof: Suppose V can be partitioned into two disjoint subsets V1 and V2 Consider two arbitrary vertices a and b of G, such that 𝑎 ∈ 𝑉1 and 𝑏 ∈ 𝑉2 19 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics No path can exist between vertices a and b; otherwise, there would be atleast one edge whose one end vertex would be in V1 and the other is in V2 Hence if a partition exists, G is not connected Conversely, let G be a disconnected graph. Consider a vertex a in G. Let V1 be the set of all vertices that are joined by paths to a. Since G is disconnected, V1 does not include all vertices of G. The remaining vertices will form a set V2 No vertex in V1 is joined to any in V2 by an edge. Hence the partition. Theorem 1.3: If a graph (connected or disconnected) has exactly two vertices of odd degree, there must be a path joining these two vertices Proof: Let G be a graph with all even vertices(vertice with even degree) except vertices v1 and v2, which are odd. For every component of a disconnected graph , no graph can have an odd number of odd vertices. Therfore in graph G, v1 and v2 must belong to the same component , and hence must have a path between them. Theorem 1.4: A simple graph (ie., a graph without parallel edges or self loops) with n vertices and k components can have at most 𝒏 − 𝒌 (𝒏 − 𝒌 + 𝟏)/𝟐 edges Proof: Let the number of vertices in each of the k components of a graph G be n1,n2,..., nk. Thus we have 𝑛1 + 𝑛2 + ⋯ + 𝑛𝑘 = 𝑛 𝑛𝑖 ≥ 1 Maximum number of edges in the ith component of G is2 𝑛𝑖 (𝑛𝑖 − 1) Therfore maximum number of edges in G is 1 1 2 𝑘 𝑖=1 1 𝑛𝑖 − 1 𝑛𝑖 = 2 = 𝑘 𝑖=1 𝑘 1 2 1 𝑛𝑖 2 − 2 𝑛𝑖 2 − 𝑖=1 𝑘 𝑛𝑖 𝑖=1 𝑛 2 1 ≤ 𝑛2 − 𝑘 − 1 2𝑛 − 𝑘 2 1 = 𝑛−𝑘 𝑛−𝑘−1 2 20 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor − 𝑛 2 Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Points to Remember A linear graph (graph) G=(V,E) consists of a set of vertices and set of edges Applcation of Graphs: Konigsberge Bridge Problem, Utility Problem, Electrical Network Problem, Seating Problem A graph with finite number of vertices and finite number of edges is called a finite graph, otherwise it is an infinite graph The number of edges incident on a vertex vi, with self loop counted twice, is called the degree , d(vi), of vertex vi. Sum of the degrees of all vertices in G is twice the number of edges in G 𝒏 𝒅 𝒗𝒊 = 𝟐𝒆 𝒊=𝟏 The number of vertices of odd degree in a graph is always even A graph in which all vertices are of equal degree is called a regular graph Isolated vertices are vertices with zero degree A vertex of degree one is called a pendant vertex or an end vertex A graph without any edges is called null graph The two isomorphic graph must have a) same number of vertices b) same number of edges c) equal number of vertices with a given degree A graph g is said to be a sub graph of a graph G, if all the vertices and all the edges of g are in G, and each edge of g has the same end vertices in g as in G Two sub graphs g1 and g2 of a graph G are said to be edge disjoint if g1 and g2 do not have any edge in common No edge appeared more than once in a walk. A vertex may appear more than once Closed Walk: If a walk begins and ends with same vertex, then it is called closed walk Open Walk: If a walk is not closed is called Open Walk An open walk in which no vertex appears more than once is called a path A closed walk in which no vertex appears more than one is called a circuit A graph G is said to be connected if there is atleast one path between every pair of vertices in G. Otherwise G is disconnected A simple graph (ie., a graph without parallel edges or self loops) with n vertices and k components can have at most 𝒏 − 𝒌 (𝒏 − 𝒌 + 𝟏)/𝟐 edges PROBLEMS 1. Prove that any two simple connected graph with n vertices , all of degree two are isomorphic 2. Are the two graphs in the figure are isomorphic? Why? 21 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics 3. Show that two graphs in the figure are isomorphic. 4. Prove that a simple graph with n vertices must be connected if it has more than [ 𝑛 − 1 𝑛 − 2 ]/2 edges (Hint: Use Theorem1.4) 5. Prove that if a connected graph G is decomposed into two sub graphs g 1 and g2, there must be atleast one vertex common between g1 and g2 6. Prove that a connected graph G remains connected after removing an edge e i from G, if and only if ei is in some circuit G 7. Draw a connected graph that becomes disconnected when any edge is removed from it 8. Prove that a graph with n vertices satisfying the condition –“ a connected graph that becomes disconnected when any edge is removed from it “ is a)simple b) has exactly n-1 edges 9. What is the length of the path from the entrance to the center of the maze in the following figure. 10. List all different paths between vertices 5 and 6 in the figure. Give the length of each of these paths? 22 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics PREVIOUS YEAR UNIVERSITY QUESTIONS 1. Consider a graph G with 4 vertices: v1, v2, v3 and v4 and the degrees of vertices are 3, 5, 2 and 1 respectively. Is it possible to construct such a graph G? If not, why? 2. Draw a disconnected simple graph G1 with 10 vertices and 4 components and also calculate the maximum number of edges possible in G1. 3. Write any two applications of graphs with sufficient explanation 4. Discuss the Konigsberg Bridge problem.Is there any solution to the problem? Justify your answer. 5. Prove that a simple graph with n vertices must be connected, if it has more than (n-1)(n2)/2 edges. 6. Explain any two applications of graphs 7. Define the term a)simple graph b)Complete graph c) Null graph 8. Define Simple Graph, Multi Graph and Pseudo Graph with examples 9. Explain the terms: a) Null Graph b) Complete Graph and c) Pendant Vertex with examples 10. Prove that the number of vertices of odd degree in a graph is always even OR Prove that in any undirected graph, the number of vertices of odd degree should be even 11. Differentiate a Walk, Path & Circuit in a Graph 12. What are the steps involved in proving that two graphs are isomorphic or not? 13. Prove that, if a graph has exactly two vertices of odd degree, then there must be a path joining these two vertices 14. Prove that “ a simple graph with n vertices and k components can have atmost 𝑛 − 𝑘(𝑛−𝑘+1)/2 edges 15. Prove the following: “A graph G is disconnected if and only if its vertex set V can be partitioned into two non empty disjoint subsets V1 and V2 such that there exists no edge in G whose one end vertex is in V1 and other in V2” 16. State four cube problem and find the solution to the four cube problem for the set of cubes drawn below 23 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics WORKSHEET 1. Draw graphs representing problems of a)two houses and three utilities; b)four houses and four utilities. say, water, gas, electricity, and telephone 2. Consider a graph G with 4 vertices: v1, v2, v3 and v4 and the degrees of vertices are 3, 5, 2 and 1 respectively. Is it possible to construct such a graph G? If not, why? 3. Draw a disconnected simple graph G1 with 10 vertices and 4 components and also calculate the maximum number of edges possible in G1. 4. What are the basic conditions to be satisfied for two graphs to be isomorphic? Are the two graphs below isomorphic? Explain with valid reasons 5. Define isomorphism between two graphs. Are the following graphs are isomorphic to each other? Justify your answer. 24 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics 6. Draw all simple graph of one, two, three and four vertices 7. Prove that the maximum number of edges in a simple graph with n vertices is 𝑛∗ 𝑛−1 2 8. Determine whether the given pair of graph is isomorphic or not 9. Draw two isomorphic graphs with 6 vertices and 9 edges 10. State Seating problem and find all its solutions 11. Prove that a graph with n vertices, n-1 edges and no circuits is connected 12. Check whether the two graphs given below are isomorphic? 13. Prove that the ring sum of two circuits in a graph G is either a circuit or an edge disjoint union of circuits 14. What is the largest possible number of vertices in a graph with 19 edges and all vertices are of degree at least 3 15. Prove that the ring sum of two circuits in a graph G is either a circuit or an edge disjoint union of circuits 16. Is it possible to have simple graph with the following degree sequences? If yes, draw the graphs. a. 2,3,3,3,3,3,4,5 b. 1,3,3,4,5,6,6 c. 1,2,3,3,4,5,6 17. Show that there is no graph G with 𝑉(𝐺) = 12 and 𝐸(𝐺) = 28 in which each vertex is of degree 3 or 6 18. Does there exists a graph with 28 edges and 12 vertices each of degree 3 or 4? 25 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics MODULE II Euler Graph If some closed walk in a graph contains all the edges of the graph, then the walk is called an Euler Line and a graph that contain Euler line is called Euler graph Euler graph is always connected Theorem 2.1: A given connected graph G is an Euler graph if and only if all vertices of G are of even degree Proof: Suppose that G is and Euler graph. It therefore contains an Euler line( which is a closed walk). In tracing this walk, observe that every time the walk meets a vertex v it goes through two “new” edges incident on v –with one we entered v and with the other “exited”. This is true not only of all intermediate vertices of the walk but also of the terminal vertex, because we “exited” and entered the same vertex at the beginning and end of the walk, respectively. Thus if G is an Euler graph, the degree of every vertex is even Conversely, let us assume that each vertex of G has even degree. We need to show that G is Eulerian. Let us start with a vertex v0 ∈ 𝑉(𝐺). Assume G is connected, there exists a vertex v1 ∈ 𝑉(𝐺) that is adjacent to v0. Since G is a simple graph and𝑑 𝑣𝑖 ≥ 2, for each vertex vi ∈ 𝑉(𝐺), there exists a vertex v2 ∈ 𝑉(𝐺), that is adjacent that is adjacent to v1 with 𝑣2 ≠ 𝑣0 .Similarly, there exists a vertex v3 ∈ 𝑉(𝐺), that is adjacent that is adjacent to v2 with 𝑣3 ≠ 𝑣1 As the number of vertices is finite, the process of getting a new vertex will finally end with a vertex 𝑣𝑖 being adjacent to a vertex 𝑣𝑘 for some 𝑖, 0 ≤ 𝑖 ≤ 𝑘 − 2. Hence, 𝑣𝑖 − 𝑣𝑖+1 − 𝑣𝑖+2 − ⋯ − 𝑣𝑘 − 𝑣𝑖 forms a circuit, say 𝐶, in 𝐺. If 𝐶 contains every edge of 𝐺, then 𝐶 gives rise to a closed Eulerian trail and we are done. Let us assume that 𝐸 𝐶 is a proper subset of 𝐸 𝐺 . Now consider the graph 𝐺1 that is obtained by removing all the edges in 𝐶 from 𝐺. Then, 𝐺1 may be a disconnected graph but each vertex of 𝐺1 still has even degree. Hence, we can do the same process explained above to 𝐺1 also to get a closed Eulerian trail, say 𝐶1 . If 𝑪𝟏 contains all edges of 𝑮𝟏 , then 𝑪 ∪ 𝑪𝟏 is a closed Euler trail in G. If not, let 𝐺2 be the graph obtained by removing the edges of 𝐶1 from 𝐺1 . 26 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Since G is a finite graph, we can proceed to find out a finite number of cycles only. Le the process of finding cycles, as explained above, ends after a finite number of steps say, 𝑟. Then the reduced graph 𝐺𝑟 = 𝐺𝑟−1 − 𝐸 𝐶𝑟−1 = 𝐺 − 𝐸(𝐶 ∪ 𝐶1 ∪ 𝐶𝑟−1 ) will be an empty graph(null graph). Then 𝐶 ∪ 𝐶1 ∪ 𝐶2 … ∪ 𝐶𝑟−1 is a closed Euler trail in G Therefore, G is Eulerian, This completes the proof. Konigsberg Bridge Problem: Two islands C and D were connected to each other and to the banks A and B with seven bridges as shown in figure. The problem was to start at any land areas A, B, C or D , walk over each of the seven bridges exactly once, and return to the starting point It can be represented by means of the following graph and the problem can be solved by checking whether the following graph is an Euler or not. If it is Euler graph then we can solve the problem. Otherwise not. From the figure we find that not all its vertices are of even degree . Hence it is not an Euler graph. Thus it is not possible to walk over each of the seven bridges exactly once and return to the starting point Unicursal line or open Euler line: 27 An open walk that includes all edges of a graph without retracing any edge is a an open Euler line or Unicursal line. A graph that has a unicursal line is called a unicursal graph Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics A connected graph is unicursal if and only if it has exactly two vertices of odd degree Theorem 2.2: In a connected graph G with exactly 2k odd vertices, there exist k edge-disjoint subgraphs such that they together contain all edges of G and that each is a unicursal graph Proof: Let the odd vertices of the given graph G be named as v1,v2,...,vk; w1,w2,...,wk in any arbitrary order. Add k edges to G between the vertex pairs (v1,w1),( v2,w2),...,( wk,wk) to form a new graph G’. Since every vertex of G’ is of even degree, G’ consists of anEuler line 𝜌 . Now if we remove 𝜌 the k edges we just added, 𝜌 will split into k walks, each of which is a unicursal line The first removal will leave a single unicursal line; the second removal will split that into two unicursal line; and each successive removal will split a unicursal line into two, until there are k of them. Thus the theorem. Theorem2.3: A connected graph G is an decomposed into circuits Euler graph if and only if it can be Proof: Suppose graph G can be decomposed into circuits; that is , G is a union of edge disjoint circuits. Since the degree of every vertex in a circuit is two , the degree of every vertex in G is even. Hence G is an Euler graph Arbitrarily Traceable Graphs A vertex v in an Euler graph have a property that an Euler line is always obtained when one follows any walk from vertex v. Such a graph is called an arbitrarily traceable graph from vertex v. 28 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Hamiltonian Circuits A circuit in a connected graph G is said to be Hamiltonian if it includes every vertex of G In the figure a, starting at vertex v, if one traverse along the edges shown in heavy lines- passing through each vertex exactly once- one gets a Hamiltonian circuit Hamiltonian circuit in a graph of n vertices consists of exactly n edges Hamiltonian Path If we remove any one edge from a Hamiltonian circuit , we are left with a path. This path is called a Hamiltonian Path The length of a Hamiltonian path in a connected graph of n vertices is n-1 29 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Since Hamiltonian circuit traverses every vertex exactly once, it cannot include a self loop or a set of parallel edges. Complete Graph A simple graph in which there exists an edge between every pair of vertices is called complete graph. A complete graph is also called universal graph or clique Since every vertex is joined with every other vertex through one edge , the degree of every vertex is n-1 in a completed graph G of n vertices Total number of edges in G is 𝒏(𝒏 − 𝟏)/𝟐 Note: It is easy to construct a Hamiltonian circuit in a complete graph of n vertices. Let the vertices be numbered v1, v2,..., vn. Since an edge exists between any two vertices, we can start from v1 and traverse to v2, and v3 and so on to vn, and finally from vn and v1. This is a Hamiltonian circuit Theorem 2.4: In a complete graph with n vertices there are (𝒏 − 𝟏)/𝟐 edge –disjoint Hamiltonian circuits, if n is an odd number ≥ 𝟑 Proof: Note that a complete graph has 𝑛 (𝑛−1) 2 edges and a Hamiltonian cycle in 𝐾𝑛 contains only n edges. Therefore, the maximum number of edge disjoint Hamiltonian cycle is 𝑛 −1 2 Now, assume that 𝑛 ≥ 3 and is odd. Construct a sub graph 𝐺 of 𝐾𝑛 as explained below: The vertex 𝑣1 is placed at the centre of a circle and the remaining 𝑛 − 1 vertices are placed on the circle, at equal didtances along the circle such that the angle made at the 360 centre by two points is 𝑛 −1 degrees. The vertices with odd suffixes are placed along the upper half of the circle and the vertices with even suffixes are placed along the lower half circle. Then draw edges 𝑣𝑖 𝑣𝑖+1 , where 1 ≤ 𝑖 ≤ 𝑛 , with the meaning that 𝑣𝑛+1 = 𝑣1 are drawn as shown in the figure. 30 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Clearly the reduced graph 𝐺1 is a cycle covering all vertices of 𝐾𝑛 . That is if we rotate 360 the vertices along the curve for 𝑛 −1 degrees, we get another Hamiltonian sub graph 𝐺2 of 𝐾𝑛 , which has no common edges with 𝐺1 In a similar way, rotate the polygonal pattern clockwise by 360 𝑛 −1 degrees. After 𝑛 − 1 𝑡 rotation, all vertices will be exactly as in the figure 1. Therefore 𝑛 − 1 rotation are valid. But, it can be noted that the cycle 𝐺𝑖 obtained after the 𝑖th rotation and the cycle 𝐺𝑛 −1 +𝑖 are isomorphic graphs, because all the vertices in the upper half 2 cycle in 𝐺𝑖 will be in the lower half cycle in 𝐺𝑛 −1+𝑖 and vice versa, in the same order. That is, we have now that there are 𝑛 −1 2 2 distinct such non-isomorphic edge-disjoint cycles in 𝐾𝑛 . Hence, the number of edge-disjoint Hamiltonian cycles is 31 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor 𝑛 −1 2 Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Seating Arrangement Problem Problem: Nine members of a new club meet each day for lunch at a round table. They decide to sit such that every member has different neighbours at ach lunch. How many days can this arrangement last? Seating arrangement of nine members in a round table can be represented by means of a graph. Each vertex represents a member and an edge joining two vertices represents the relationship of sitting next to each other We can solve the seating arrangement at round table problem using the Theorem 2.4, as follows Representing a member x by a vertex and the possibility of his sitting next to another member y by an edge between x and y, we construct a graph G Since every member is allowed to sit next to any other member, G is a complete graph of nine vertices- nine being the number of people to be aeated around the table. Every seating arrangement around the table is clearly a Hamiltonian circuit. The first day of their meeting they can sit in any order, and it will be a Hamiltonian circuit H1. The second day, if they are to sit such that every member must have didderent neighbours, we have to find another Hamiltonian circuit H 2 in G, with an enitirely different set of edges from those in H1 ; that is H1 and H2 are edge disjoint Hamiltonian circuits. From Theorem 2.4, the number of edge disjoint Hamiltonian circuit in G is four ; therefore only four such arrangements exixts among people Dirac’s Theorem for Hamiltonicity Theorem: Every graph G with n ≥ 3 vertices and minimum degree d(G) ≥ n/2 has a Hamiltonian cycle. Proof: Suppose that 𝐺 = (𝑉, 𝐸) satisfies the hypothesis of the theorem. Then G is connected, since otherwise the degree of any vertex in a smallest component C of G would be 𝑛 𝑛 atmost 𝐶 − 1 < 2 , contradicting the hypothesis 𝛿 𝐺 ≥ 2 Let 𝑃 = 𝑥0 𝑥1 … 𝑥𝑘 be the longest path in G, as seen in the figure given below: 32 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Note that the length of P is k. Since P cannot be extended to a longer path, all the neighbours of 𝑥0 lie on P. Assume the contradictory. Let y be an adjacent vertex of 𝑥0 which is not in P Then the path 𝑃′ = 𝑦𝑥0 𝑥1 … 𝑥𝑘 is a path of length 𝑘 + 1, contradicting the hypothesis that P is the longest path in G Similarly, we note that and all the neighbours of 𝑥𝑘 will also lie on P, unless we reach at a contradiction as mentioned above Hence, at least vertices 𝑥1 𝑥2 … 𝑥𝑘 are adjacent to 𝑥0 . Another way of saying the second part of the last sentence is: atleast least 𝑛 2 of the vertices 𝑥0 𝑥1 … 𝑥𝑘−1 are adjacent to 𝑥𝑘 , and at least 𝑛 2 𝑛 2 of the of the vertices 𝑥𝑖 ∈ 𝑥0 , … , 𝑥𝑘−1 are such that 𝑥0 𝑥𝑖+1 ∈ 𝐸. Combining both statements and using the pigeon-hole principle,we see that there is some 𝑥𝑖 with 0 ≤ 𝐼 ≤ 𝐾 − 1, 𝑥𝑖 𝑥𝑘 ∈ 𝐸 and 𝑥0 𝑥𝑖+1 ∈ 𝐸 Consider the cycle 𝐶 = 𝑥0 𝑥𝑖+1 𝑥𝑖+2 … 𝑥𝑘−1 𝑥𝑘 𝑥𝑖 𝑥𝑖−1 … 𝑥1 𝑥0 as given in the following graph We claim that the above cycle C is a Hamiltonian cycle of G. Otherwise, since G is connected, there would be some vertex 𝑥𝑗 of C adjacent to a vertex y not in C, so that 𝑒 = 𝑥𝑖 𝑦 ∈ 𝐸. But then we could attach e to a path ending in 𝑥𝑗 containing k edges of C, constructing a path in G longer than P( see the graph given below), which is contradiction to the hypothesis that P is the longest path in G. Therefore C must cover all vertices of G and hence it is a Hamiltonian cycle in G. 33 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Travelling Salesman problem Problem: A salesman is required to visit a number of cities during a trip. Given the distance between the cities, in what order should he travel so as to visit every city precisely once and return home, with the minimum mileage travelled? We can represent the cities by vertices and the roads between them by edges in a graph. In this graph, with every edge 𝑒𝑖 there is associated a real number 𝑤(𝑒𝑖 ). Such a graph is called a weighted graph; 𝑤 𝑒𝑖 being the weight of edge 𝑒𝑖 In this problem, if each of the cities has a road to every other city, we have a completed weighted graph . This graph has numerous Hamiltonian circuits and we are to pick the one that has the smallest sum of distances Theoretically , the problem of travelling salesman can always be solved by enumerating all 𝑛 − 1 !/2 Hamiltonian circuits, calculating the distance traveled in each and then picking the shortest one. Directed Graph A directed graph or digraph G consists of a set of vertices 𝑉 = 𝑣1 , 𝑣2 , … , a set of edges 𝐸 = 𝑒1 , 𝑒2 , … and a mapping 𝜑 that maps every edge onto some ordered pair of vertices (𝑣𝑖 , 𝑣𝑗 ) In digraph, vertices are represented by points and edges by a line segment between vi and vj with an arrow directed from vi to vj The vertex vi, which edge ek is incident out of is called the initial vertex of ek The vertex vj, which edge ek is incident into is called the terminal vertex of ek A digraph is also referred to as oriented graph. The number of edges incident out of a vertex vi is called the out-degree (outvalence or outward demidegree) of vi and is written 𝑑 +(𝑣𝑖 ) The number of edges incident into vertex vi is called the in-degree (in valence or inward demidegree) of vi and is written 𝑑 − (𝑣𝑖 )\ 34 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics In any digraph, sum of all in degree is equal to the sum of all out-degree, each sum being equal to the number of edges in G 𝑛 𝑛 𝑑 + 𝑣𝑖 = 𝑖=1 𝑑 − 𝑣𝑖 = 𝑒 𝑖=1 An isolated vertex is a vertex in which the in-degree and the out-degree are both equal to zero A vertex in a digraph is called pendant if its is of degree one 𝑑 + 𝑣𝑖 + 𝑑 − 𝑣𝑖 = 1 Types of Digraphs Simple Digraphs: A digraph that has no self loop or parallel edges is called a simple digraph Asymmetric Digraphs: Digraphs that have atmost one directed edge between a pair of vertices, but contains self loops are called asymmetric or antisymmetric Symmetric Digraphs: Digraphs in which for every edge (a,b), there is also an edge (b,a) 35 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics A digraph that is both simple and symmetric is called a simple symmetric digraph. Similarly, a digraph that is both simple and asymmetric is called a simple asymmetric digraph Complete Digraphs: Two types: Complete symmetric and complete asymmetric A complete symmetric digraph is simple digraph in which there is exactly one edge directed from every vertex to every other vertex A complete asymmetric digraph of n vertices contains 𝑛(𝑛 − 1) edges A complete asymmetric digraph is an asymmetric digraph in which there is exactly one edge between every pair of vertices A complete asymmetric digraph of n vertices contains 𝑛(𝑛 − 1)/2 edges It is also called a tournament or a complete tournament Balanced Digraph: A digraph is said to be balanced if for every vertex vi the in degree equals the out degree. 𝑑 + 𝑣𝑖 = 𝑑 − 𝑣𝑖 . Balanced Digraph is also referred to as pseudosymmetric digraph or iso graph Balanced Digraph is said to be regular if every vertex has same in degree and out degree as every other vertex Digraphs and Binary Relations Reflexive relation: A relation R on a set X that satisfies 𝑥𝑖 𝑅𝑥𝑖 for every 𝑥𝑖 ∈ 𝑋 is called a reflexive relation. The digraph of a reflexive relation will have a self loop at every vertex. Such a digraph representing a reflexive binary relation on its vertex set 36 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics may be called a reflexive digraph. A digraph in which no vertex has a self loop is called an irreflexive digraph Symmetric Relation: For some relation R it may happen that for all 𝑥𝑖 and 𝑥𝑗 , if 𝑥𝑖 𝑅𝑥𝑗 holds then 𝑥𝑗 𝑅𝑥𝑖 also holds. Such a relation is called a symmetric relation. The digraph of a symmetric relation is a symmetric digraph because for every directed edge from vertex 𝑥𝑖 to 𝑥𝑗 there is a directed edge from 𝑥𝑗 to 𝑥𝑖 Transitive Relation: A relation R is said to be transitive if for any three elements𝑥𝑦 , 𝑥𝑗 and 𝑥𝑘 in the set , 𝑥𝑖 𝑅𝑥𝑗 and 𝑥𝑗 𝑅𝑥𝑘 always imply 𝑥𝑖 𝑅𝑥𝑘 . A digraph representing a transitive relation is called a transitive directed graph. Equivalence relation: A binary relation is called an equivalence relation if it is reflexive, symmetric and transitive. The graph representing equivalence relation is called equivalence graph 37 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Points to Remember Euler graph: If some closed walk in a graph contains all the edges of the graph, then the walk is called an Euler Line and a graph that contain Euler line is called Euler graph Unicursal line or open Euler line: An open walk that includes all edges of a graph without retracing any edge is a an open Euler line or Unicursal line. A graph that has a unicursal line is called a unicursal graph Arbitrarily Traceable Graphs: A vertex v in an Euler graph have a property that an Euler line is always obtained when one follows any walk from vertex v. Such a graph is called an arbitrarily traceable graph from vertex v. Hamiltonian Circuits: A circuit in a connected graph G is said to be Hamiltonian if it includes every vertex of G Hamiltonian Path: If we remove any one edge from a Hamiltonian circuit , we are left with a path. This path is called a Hamiltonian Path The length of a Hamiltonian path in a connected graph of n vertices is n-1 Complete Graph: A simple graph in which there exists an edge between every pair of vertices is called complete graph. Dirac’s Theorem for Hamiltonicity: Every graph G with n ≥ 3 vertices and minimum degree d(G) ≥ n/2 has a Hamiltonian cycle. Directed Graphs: A directed graph or digraph G consists of a set of vertices 𝑉 = 𝑣1 , 𝑣2 , … , a set of edges 𝐸 = 𝑒1 , 𝑒2 , … and a mapping 𝜑 that maps every edge onto some ordered pair of vertices (𝑣𝑖 , 𝑣𝑗 ) The number of edges incident out of a vertex vi is called the out-degree (outvalence or outward demidegree) of vi and is written 𝑑 +(𝑣𝑖 ) The number of edges incident into vertex vi is called the in-degree (in valence or inward demidegree) of vi and is written 𝑑 − (𝑣𝑖 ) Types of Digraphs: Simple graphs, Asymmetric digraphs, Symmetric digraphs, Complete digraphs, Balance digraphs 38 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics PROBLEMS 1. You are given a 10-piece domino set whos titles have the following set of dots: (1,2); (1,3); (1,4); (1,5); (2,3); (2,4); (2,5); (3,4); (3,5); (4,5). Discuss the possibility of arranging the tiles in a connected series such that one number on a title always touches the same number on its neighbour (Hint: Use a five-vertex complete graph and see if it is an Euler graph) 2. Is it possible to move a knight on a chess board such that it completes every permissible move exactly once? A move between two squares is counted as one regardless of the direction in which it is made (Hint: Is the graph is unicursal) 3. A round robin tournament(when every player plays against every other) among n players (n being an even number) can be represented by a complete graph of n vertices. Discuss how you would schedule the tournaments to finish in the shortest possible time. 4. Draw a graph that has a Hamiltonian path but does not have a Hamiltonian circuit. 5. Show that neither of the graphs in the figure has a Hamiltonian path 6. Show that the graph of a rhombicdodecahedron (with eight vertices of degree three and six vertices of degree four) has no Hamiltonian path 39 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics 7. Draw a graph in which an Euler line is also a Hamiltonian circuit. What can you say about such graphs in general? 8. Is it possible , starting from any of the 64 squares of the chess board, o move a knight such that it occupies every square exactly once and return to the initial position? if so, give one such tour. (Hint: Look for a Hamiltonian circuit in the graph) 9. Prove that a graph G with n vertices always has a Hamiltonian path if the sum of the degrees of every pair of vertices vi,vj in G satisfies the condition 𝑑 𝑣𝑖 + 𝑑 𝑣𝑗 ≥ 𝑛 − 1 (Hint: First show that G is connected. then use induction on path length in G) 10. Using the result of the above problem, show that in a dancing ring of n children it is always possible to arrange the children so that everyone has a friend at each side if every child enjoys friendship with atleast half the children PREVIOUS YEAR UNIVERSITY QUESTIONS 1. With suitable example explain arbitrarily traceable graphs 2. State Dirac’s theorem for hamiltonicity and why it is not a necessary condition for a simple graph to have a Hamiltonian circuit. 3. Differentiate between symmetric and asymmetric digraphs with examples and draw a complete symmetric digraph of four vertices. 4. Define Euler line, Euler graph and Hamiltonian Circuit 5. Mention the difference between an Euler graph and Unicursal graph 6. Prove that in a complete graph with n vertices there are (𝑛 − 1)/2 edge –disjoint Hamiltonian circuits, if n is an odd number ≥ 3 OR 7. Prove that “ In a complete graph with n vertices, there are (𝑛 − 1)/2 edge disjoint Hamiltonian circuits, if n is an odd number≥ 3𝑛 8. Prove that a given connected graph G is an Euler graph if and only if all vertices of G are of even degree 9. Define Hamiltonian graph.Find an example of a non-Hamiltonian graph with a Hamiltonian path. 1. Define the following with example i) Isomorphic digraph ii) Complete symmetric digraph 2. Explain the travelling salesman problem OR Explain the travelling salesman problem using the concept of Hamiltonian circuit 3. Define Euler graph. 4. Define connected graph and components of a graph 5. Prove that a given connected graph G is an Euler graph if and only if all vertices of G are of even degree 6. Explain the travelling salesman problem 40 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics OR Explain the travelling salesman problem using the concept of Hamiltonian circuit 7. Define Euler graph. 8. What is the necessary and sufficient condition for a graph to be Euler? And also prove it. 9. Prove that the number of odd degree vertices in a graph is always even. 10. Discuss the Konigsberg Bridge problem.Is there any solution to the problem? Justify your answer. 11. Define connected graph and components of a graph 12. For a Eulerian graph G, prove the following properties. i) The degree of each vertex of G is even. ii) G is an edge-disjoint union of cycles. 13. State Travelling-Salesman Problem and how TSP solution is related with Hamiltonian Circuits? WORK SHEET 1. Draw the graphs for the following: a. Graph that is Euler and Hamiltonian b. Graph that is Euler but not Hamiltonian c. Graph that is Hamiltonian but not Euler 2. Consider the graph G given below: Define Euler graph. Is G an Euler? If yes, write an Euler line from G. 3. Define Hamiltonian circuits and paths with examples. Find out the number of edgedisjoint Hamiltonian circuits possible in a complete graph with five vertices. 4. For the following graph, find the shortest path between fromv1 to v4. Also find a Euler circuit. 5. 19 students in a nursery school play a game each day, where they hold hands to form a circle. For how many days can they do this, with no students holding hands with the same playmates more than once? Substantiate your answer with graph theoretic concepts. 6. Show that in any group of two or more people, there are always two with exactly the same number of friends inside the group. 41 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics MODULE III Trees A tree is a connected graph without any circuits. A tree is a simple graph(without anu self loppor parallel edges) Applications The genealogy of a family can be represented by means of a tree(Family Tree) The sorting of mail according to zip code and the sorting of punched cards are done according to a tree(Decision tree or Sorting tree). Sorting or Decision Trees are frequently used in Computer Applications and Switching Theory Properties of Tree A graph with n vertices is called a tree if G is connected and is circuitless G is connected and has n-1 edges G is circuitless and has n-1 edges There is exactly one path between every pair of vertices in G G is a minimally connected graph 42 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Theorem 3.1: There is one and only one path between every pair of vertices in a tree, T. Proof: Since T is a connected graph, there must exist atleast one path between every pair of vertices in T. Now suppose that between two vertices a and b of T there are two distinct paths. The union of these paths will contain a circuit and T cannot be a tree Theorem 3.2: If in a graph G there is one and only one path between every pair of vertices, G is a tree Proof: Existance of a path between every pair of vertices assures that G is connected. A circuit in a graph implies that there is atleast one pair of vertices a,b such that there are two distinct paths between a and b. Since G has one and only one path between every pair of vertices, G can have no circuit. Therfore G is a tree. Theorem 3.3: A tree with n vertices have n-1 edges Proof: Theorem 3.4: A graph is a tree if and only if it is minimally connected Proof: In a tree, all vertices are connected together with minimum number of edges. A connected graph is said to be minimally connected if removal of any one edge from it disconnects the graph. 43 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics A minimally connected graph cannot have a circuit; otherwise we could remove one of the edges in the circuit and still leave the graph connected. Thus a minimally connected graph is a tree Conversely, if a connected graph G is not minimally connected, there must exist an edge 𝑒𝑖 in G such that 𝐺 − 𝑒𝑖 is connected. Therefore, 𝑒𝑖 is in some circuit, which implies that G is not a tree. Hence the theorem Theorem 3.5: A graph with n vertices, n-1 edges, and no circuit is connected Proof: To interconnect n distinct points , the minimum number of line segment needed is 𝑛 − 1. The resulting structure, according to Theorem 3.3 is a tree We know that a connected graph with n vertices and without any circuit has 𝑛 − 1 edges. We also know that a graph with n vertices which has no circuit and has 𝑛 − 1.edges is always connected(i.e., it is a tree). Hence the theorem Pendant Vertices in a Tree Pendant vertex is defined as a vertex of degree one Concept of pendant vertex can be used in computer programmes as data tree Theorem 3.6: In any tree (with two or more vertices), there are atleast two pendant vertices Proof: 44 Pendant vertices are vertex of degree one. For a tree of n vertices we have n-1 edges and hence 2(𝑛 − 1) degrees to be divided among n vertices Since no vertex can be of zero degree, we must have atleast two vertices of degree one in a tree. Hence the theorem. Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Distance and Centres in a Tree Distance, d(vi , vj) between any two vertices vi and vj is the length of the shortest path (i.e., number of edgesin the shortest path) between them. Metric: (Note:Funtion f(x,y) is used to compute the distance between two vertices x and y) A function that satisifies the following three conditions is called a metric Nonnegativity: 𝑓 𝑥, 𝑦 > 0 𝑎𝑛𝑑 𝑓 𝑥, 𝑦 = 0 𝑖𝑓 𝑎𝑛𝑑 𝑜𝑛𝑙𝑦 𝑖𝑓 𝑥 = 𝑦 Symmetry: 𝑓 𝑥, 𝑦 = 𝑓 𝑦, 𝑥 Triangle inequality: 𝑓 𝑥, 𝑦 ≤ 𝑓 𝑥, 𝑧 + 𝑓 𝑧, 𝑦 𝑓𝑜𝑟 𝑎𝑛𝑦 𝑧 The distance between the vertices of a connected graph is a metric Proof: Distance d(vi , vj) between two vertices of a connected graph satisfies nonnegativity and symmetry Since d(vi , vj) is the length of the shortest path between vertices vi and vj, this path cannot be longer than another path between vi and vj, which goes through a specified vertex vk. Hence 𝒅 𝒗𝒊 , 𝒗𝒋 ≤ 𝒅 𝒗𝒊 , 𝒗𝒌 + 𝒅 𝒗𝒌 , 𝒗𝒋 Therfore the distance between the vertices of a connected graph is a metric Eccentricity of a vertex 𝑬(𝒗): The eccentricity of a vertex 𝑬(𝒗) in a graph G is the distance from 𝒗 to the vertex farthest from 𝒗 in G 𝑬 𝒗 = 𝒎𝒂𝒙𝒗𝒊,,𝒗 𝒅(𝒗, 𝒗𝒊 ) Centre of G: A vertex with minimum eccentricity in graph G is called a centre of G 45 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Theorem 3.7: Every tree has either one or two centres Proof: The maximum distance,𝑚𝑎𝑥 𝑑 𝑣𝑖 , 𝑣 from a given vertex 𝑣 to any other vertex 𝑣𝑖 occurs only when 𝑣𝑖 is a pendant vertex. With this observation, let us startwith a tree T having more than one two vertices. Tree T must have two or more pendant vertices. Delete all the pendant vertices from T. The resulting graph 𝑇 ′ is still a tree.. Removal of all pendant vertices from Tuniformly reduced the eccentricity of the remaining vertices (i.e., vertices of 𝑇 ′ ) by one. Therefore all vertices that T had as centers will still remain centres in 𝑇 ′ . From 𝑇 ′ we can again remove all pendant vertices and get another tree 𝑇 ′ ′ We continue this process until there is left either a vertex (which is the centre of T) or an edge (whose end vertices are the two centres of T). Thus the theorem. If a tree has two centres , the two centres must be adjacent Radius and Diameter of a Tree: The eccentricity of a centre in a tree is defined as the radius of the tree. The diameter of a tree T is defined as the length of the longest path in T Rooted and Binary Trees Rooted Trees: A tree in which one vertex (called the root) is distinguished from all the other is called a rooted tree. In a diagram of a rooted tree, the root is generally marked distinctly. All rooted trees with four vertices are shown in figure where the roots are enclosed in a small triangle 46 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Binary Trees: A binary tree is defined as a tree in which there is exactly one vertex (root) is of degree two, and each of the remaining vertices is of degree one or three. Every binary tree is a rooted tree since the vertex of degree two is distinct from all other vertices, this vertex serves as a root. Most straightforward application of binary tree is in search procedures Properties of Binary Tree The number of vertices n in a binary tree is always odd (Proof: This is because there is exactly one vertex of even degree, and the remaining n-1 vertices are of odd degree. Since the number of vertices of odd degree is even in a graph, n-1 is even. Hence n is odd) Number of pendant vertices in a binary tree T is 𝑝 = 𝑛 +1 2 Proof: Let p be the number of pendant vertices in a binary tree T. Number of vertices of degree three = 𝑛 − 𝑝 − 1. Number of edges in T = 1 𝑝+3 𝑛−𝑝−1 +2 𝑛+1 = 2 2 Internal Vertex A non pendant vertex in a tree is called an internal vertex of a tree 47 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics The number of internal vertices in a binary tree is one less than the number of pendant vertices Level of a vertex in a tree In a binary tree a vertex vi is said to be at level li if vi is at a distance of li from the root Root is at level 0 A 13 vertex , four level binary tree is shown in figure. The number of vertices at levels 1, 2, 3 and 4 are 2, 2, 4 and 4 respectively Maximum number of vertices in a k-level binary tree is 20 + 21 + 22 + ⋯ + 2𝑘 ≥ 𝑛 Height of the tree The maximum level, lmax, of any vertex in a binary tree is called the height of the tree Minimum possible height of an n vertex binary tree is 𝑚𝑖𝑛 𝑙𝑚𝑎𝑥 = log 2 𝑛 + 1 − 1 Maximum possible height of an n vertex binary tree is 𝑚𝑎𝑥 𝑙𝑚𝑎𝑥 = 48 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor 𝑛 −1 2 Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Construct a binary tree of given n such that the farthest vertex is as far as possible from the root, we have exactly two vertices at each level except at level 0 Question: Construct a binary tree for 7 nodes such that the farthest vertex is as far as possible from the root(binary tree with max height) and find the height of the tree Question: Construct a binary tree with 15 nodes such that the tree must have a minimum height. Also find the height of the tree Path length (External path length) of a tree It can be defined as the sum of the path lengths from root to all pendant vertices (or sum of the levels of all pendant vertices) Path length related to the execution time of an algorithm The path length of the binary tree in the following figure is 1 + 3 + 3 + 4 + 4 + 4 + 4 = 23 49 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics . Qn: Find the path length of trees for the following figure: 𝒑𝒂𝒕𝒉 𝒍𝒆𝒏𝒈𝒕𝒉 = 𝟐 + 𝟐 + 𝟑 + 𝟑 + 𝟑 + 𝟑 = 𝟏𝟔 𝒑𝒂𝒕𝒉 𝒍𝒆𝒏𝒈𝒕𝒉 = 𝟏 + 𝟐 + 𝟑 + 𝟒 = 𝟓 + 𝟓 = 𝟐𝟎 On counting trees Arthur Cayley used a connected graph to represent and count the structural isomers of the saturated hydrocarbons𝐶𝑘 𝐻2𝑘 +2 . Corresponding to their chemical valencies, a carbon atom was represented by a vertex of degree four and a hydrogen atom by a vertex of degree one The total number of vertices in such a graph is 𝑛 = 3𝑘 + 2 50 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics And total number of edges is 1 𝑒 = (𝑠𝑢𝑚 𝑜𝑓 𝑑𝑒𝑔𝑟𝑒𝑒𝑠) 2 1 = 4𝑘 + 2𝑘 + 2 2 = 3𝑘 + 1 Since the graph is connected and the number of edges is one less than number of vertices, it is a tree. Thus the problem of counting structural isomers of a given hydrocarbon becomesthe problem of counting trees Labeled Graph A graph in which each vertex is assigned a unique label or name (no two vertices have the same label) is called labeled graph 51 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Prufer Code We can represent tree using prufer code or prufer sequence Algorithm for generating Prufer sequence 1. Take any tree, 𝑇 ∈ 𝑇𝑛 whose vertices are labelled from 1 to n 2. Take the vertex with the smallest label whose degree is equal to 1, delete it from the tree and write down the value of its only neighbour 3. Repeat the process with the new, smaller tree. Continue until only one edge remains Algorithm for constructing tree using prufer sequence(Decoding) 1. Find the smallest number from 1 to n that is not in the sequence P and attach the vertex with that number to the vertex with the first number in P.(n=2+number of elements in P) 2. Remove the first number of P from the sequence. Repeat this process considering only the numbers whose vertex have not yet attained their correct degree 3. Do this until there are no numbers left in P. Remember to attach last number in P to vertex n 52 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Cayley’s Theorem: The number of labeled trees with n vertices 𝒏 ≥ 𝟐 𝒊𝒔 𝒏𝒏−𝟐 Proof: Algorithm for generating Prufer sequence 1. Take any tree, 𝑇 ∈ 𝑇𝑛 whose vertices are labelled from 1 to n 2. Take the vertex with the smallest label whose degree is equal to 1, delete it from the tree and write down the value of its only neighbour 3. Repeat the process with the new, smaller tree. Continue until only one edge remains Spanning Trees A tree T is said to be a spanning tree of a connected graph G if Tis a sub graph of G and T contains all vertices of G Spanning tree is also referred to as a skeleton or scaffolding of G Since spanning tree are the largest trees(with maximum number of edges) among all trees in G, it is also called maximal tree subgraph or maximal tree of G A disconnected graph with k components has a spanning forest consists of k spanning tree.(Forest: Collection of trees) Theorem: Every connected graph has atleast one spanning tree Proof: If G has no circuit, it is its own spanning tree If G has a circuit , delete an edge from the circuit. This will still leave the graph connected. 53 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics If there are more circuits, repeat the operation till an edge from the last circuit is deleted- leaving a connected, circuit free graph that contains all the vertices of G. Thus we have Every connected graph has atleast one spanning tree. Branch and Chord of a spanning tree An edge in a spanning tree T is called a branch of T An edge of G that is not in a given spanning tree T is called a chord of T Theorem: With respect to any of the spanning trees, a connected graph of n vertices and e edges has n-1 branches and e-n+1 chords Rank and Nullity of a graph The rank and nullity of a graph G with n verticres and e edges is defined as 𝑟𝑎𝑛𝑘, 𝑟 =𝑛−𝑘 𝑛𝑢𝑙𝑙𝑖𝑡𝑦, 𝜇 =𝑒−𝑛+𝑘 For a connected graph or a spanning tree, k=1. Therefore 𝐫𝐚𝐧𝐤 𝐨𝐟 𝐆 = 𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐛𝐫𝐚𝐧𝐜𝐡𝐞𝐬 𝐢𝐧 𝐚𝐧𝐲 𝐬𝐩𝐚𝐧𝐧𝐢𝐧𝐠 𝐭𝐫𝐞𝐞 𝐨𝐟 𝐆 = 𝑛 − 1 𝐧𝐮𝐥𝐥𝐢𝐭𝐲 𝐨𝐟 𝐆 = 𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐜𝐡𝐨𝐫𝐝𝐬 = 𝑒 − 𝑛 + 1 𝐫𝐚𝐧𝐤 + 𝐧𝐮𝐥𝐥𝐢𝐭𝐲 = 𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐞𝐝𝐠𝐞𝐬 𝐢𝐧 𝐆 Nullity of a graph is also referred to as its cyclomatic number or first Betti number Fundamental Circuits A circuit formed by adding a chord to a spanning tree is called a fundamnetal circuit Theorem: A connected graph G is a tree if and only if adding an edge between any two vertices in G creates exactly one circuit Proof: 54 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics If we add an edge between any two vertices of a tree G, a circuit is created. This is because there already exists one path between any two vertices of a tree; adding an edge between them creates an additional path , and hence a circuit. let us now consider a spanning tree T in a connected graph G. Adding any one chord to T will create exactly one circuit.(Fundamental circuit) Therefore A connected graph G is a tree if and only if adding an edge between any two vertices in G creates exactly one circuit. Finding all spanning tree of a graph We can generate one spanning tree from another, through addition of chords and deletion of an appropriate branch, is called a cyclic interchange or elementary tree transformation Distance between two spanning trees Distance between two spanning trees Ti and Tj of a graph G is defined as the number of edges of G present in one tree but not in other. Ring sum of two spanning trees Ti and Tj of G is the sub graph of G containing all edges of G that are either in Ti or Tj but not in both 𝟏 𝒅 𝑻𝒊 , 𝑻𝒋 = 𝑵(𝑻𝒊 ⨁𝑻𝒋 ) 𝟐 𝒘𝒉𝒆𝒓𝒆 𝑵 𝒈 𝒅𝒆𝒏𝒐𝒕𝒆𝒔 𝒕𝒉𝒆 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒆𝒅𝒈𝒆𝒔 𝒊𝒏 𝒈𝒓𝒂𝒑𝒉 𝒈 𝒅 𝑻𝒊 , 𝑻𝒋 is the minimumnumber of cyclic interchanges involved in going from 𝑻𝒊 to 𝑻𝒋 55 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Theorem:The distance between the spanning tree of a graph is metric. That is it satisfies 𝒅 𝑻𝒊 , 𝑻𝒋 ≥ 𝟎 𝒂𝒏𝒅 𝒅 𝑻𝒊 , 𝑻𝒋 = 𝟎 𝒊𝒇 𝒂𝒏𝒅 𝒐𝒏𝒍𝒚 𝒊𝒇 𝑻𝒊 = 𝑻𝒋 𝒅 𝑻𝒊 , 𝑻𝒋 = 𝒅 𝑻𝒋 , 𝑻𝒊 𝒅 𝑻𝒊 , 𝑻𝒋 ≤ 𝒅 𝑻𝒊 , 𝑻𝒌 + 𝒅 𝑻𝒌 , 𝑻𝒋 Points to Remember A tree is a connected graph without any circuits. There is one and only one path between every pair of vertices in a tree, T. A tree with n vertices have n-1 edges A graph is a tree if and only if it is minimally connected A graph with n vertices, n-1 edges, and no circuit is connected Pendant vertex is defined as a vertex of degree one In any tree (with two or more vertices), there are atleast two pendant vertices Distance, d(vi , vj) between any two vertices vi and vj is the length of the shortest path (i.e., number of edgesin the shortest path) between them. Metric: A function that satisifies the following three conditions is called a metric Nonnegativity: 𝑓 𝑥, 𝑦 > 0 𝑎𝑛𝑑 𝑓 𝑥, 𝑦 = 0 𝑖𝑓 𝑎𝑛𝑑 𝑜𝑛𝑙𝑦 𝑖𝑓 𝑥 = 𝑦 Symmetry: 𝑓 𝑥, 𝑦 = 𝑓 𝑦, 𝑥 Triangle inequality: 𝑓 𝑥, 𝑦 ≤ 𝑓 𝑥, 𝑧 + 𝑓 𝑧, 𝑦 𝑓𝑜𝑟 𝑎𝑛𝑦 𝑧 The distance between the vertices of a connected graph is a metric The eccentricity of a vertex 𝑬(𝒗) in a graph G is the distance from 𝒗 to the vertex farthest from 𝒗 in G Centre of G: A vertex with minimum eccentricity in graph G is called a centre of G Every tree has either one or two centres If a tree has two centres , the two centres must be adjacent The eccentricity of a centre in a tree is defined as the radius of the tree. The diameter of a tree T is defined as the length of the longest path in T A tree in which one vertex (called the root) is distinguished from all the other is called a rooted tree. A binary tree is defined as a tree in which there is exactly one vertex (root) is of degree two, and each of the remaining vertices is of degree one or three. The number of vertices n in a binary tree is always odd 𝑛 +1 Number of pendant vertices in a binary tree T is 𝑝 = 2 A non pendant vertex in a tree is called an internal vertex of a tree In a binary tree a vertex vi is said to be at level li if vi is at a distance of li from the root Root is at level 0 The maximum level, lmax, of any vertex in a binary tree is called the height of the tree Path length can be defined as the sum of the path lengths from root to all pendant vertices (or sum of the levels of all pendant vertices) A tree T is said to be a spanning tree of a connected graph G if Tis a sub graph of G and T contains all vertices of G 56 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics PROBLEMS 1. Cite three different situations that can be represented by trees. 2. Show a tree in which its diameter is not equal to twice the radius. Under what condition does this inequality hold? Elaborate. 3. Suppose you are given eight coins and are told that seven of them are of equal weight, and one coin is either heavier or lighter than the rest. You are provided with an equal arm balance, which you may use only three times, for comparing coins. Sketch a strategy in the form of decision trees for identifying the nonconforming coin, as well as for finding out whether it is heavier or lighter than the rest. 4. Show that a path is its own spanning tree 5. Prove that a pendant edge in a connected graph G is conatine din every spanning tree of G 6. What is the nullity of a complete graph of n vertices? PREVIOUS YEAR UNIVERSITY QUESTIONS 1. List down any two properties of trees and also prove the theorem: A graph is a tree if and only if it is a minimally connected 2. Define spanning tree and distance between two spanning trees. 3. Consider the tree T, given below Label the vertices of T appropriately and find the center and diameter of T 4. What is eccentricity of a node? How is it used in finding the center of a graph? 5. Let G = (V,E) be a connected graph, and let T = (V, S) be a spanning tree of G. Let e = (a, b) be an edge of G not in T. Prove that, for any edge f on the path from a to b in T, (V, (S∪{e}) −{f}) is another spanning tree for G 6. Define spanning trees. Consider the graph G given below and obtain any three spanning trees from G. Calculate the number of distinct spanning trees possible from a complete graph with n vertices. 7. Define rank and nullity of a graph G 8. Define a Metric and prove that distance is a metric 9. Explain Center and radius of a tree 57 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics 10. Prove that a tree with 𝑛 vertices has 𝑛 − 1 edges 11. Show that the distance between vertices of a connected graph is a metric. 12. Discuss the center of a tree with suitable example. 13. Write notes on the fundamental circuit. 14. Prove that in a tree T(V,E),|V|=|E|+1. 15. Define spanning tree with example. 16. Explain the difference between a general tree and a binary tree 17. Show a tree in which its diameter is not equal to twice the radius. Under what conditions does this inequality hold? Elaborate. 18. Prove that every connected graph has at least one spanning tree 19. Prove that every tree has either one or two centers 20. Prove that an arborescence is a tree in which every vertex other than the root has an in-degree of exactly one 21. Show that the number of labeled trees with n vertices n n-2 22. Define spanning tree. Show that the distance between the spanning tree of a graph is a metric 1. 2. 3. 4. 5. 6. 7. WORK SHEET Write notes on the fundamental circuit. Prove that in a tree T(V,E),|V|=|E|+1. Define spanning tree with example. Show that the distance between vertices of a connected graph is a metric. Discuss the center of a tree with suitable example. List down any two properties of trees and also prove the theorem: A graph is a tree if and only if it is a minimally connected. Consider the tree T, given below Label the vertices of T appropriately and find the center and diameter of T. 8. Define spanning trees. Consider the graph G given below and obtain any three spanning trees from G. Calculate the number of distinct spanning trees possible from a complete graph with n vertices. 58 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics 9. Let G = (V,E) be a connected graph, and let T = (V, S) be a spanning tree of G. Let e = (a, b) be an edge of G not in T. Prove that, for any edge f on the path from a to b in T, (V, (S∪{e}) −{f}) is another spanning tree for G 10. Find all spanning trees in the following graph: 11. Draw two different binary tree with 5 vertices having maximum number of leaves 12. A tree has five vertices of degree 2, three vertices of degree 3 and four vertices of degree 4. How many vertices of degree one does it have? 13. Define binary trees. Draw binary trees with 11 vertices: a. having maximum height b. with minimum height 14. How many labeled trees can be constructed using 4 vertices? Show all the labeled trees with four vertices 15. Sketch all (unlabeled) binary trees with six pendant vertices. Find the path length of each. [Hint: Distribute the 11 vertices (because n=6+5) among different levels. Observe that at level 0 has exactly one vertex, level 1 has exactl two vertices; level 2 can have either two or four vertices; and so on. There are six such trees, and two of them are shown in the following figure.] 59 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics 16. Draw all trees of n labeled vertices for n=1, 2, 3, 4 and 5 17. Draw all trees of n unlabeled vertices for n=1, 2, 3, 4 and 5 18. Draw all unlabeled rooted trees of n vertices for n=1, 2, 3, 4 and 5 19. It can be shown that there are only six different trees of six vertices. Two such trees are given in the figure. Draw the other four. 20. Sketch all spanning trees of the graphs in the figure 60 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics MODULE IV Cut Set and Cut Vertices In a coonected graph G, a cut-set a set of edges whose removal from G leaves G disconnected, provided removal of no proper subset of these edges disconnects G Cut-set is also referred to as minimal cut-set, proper cut-set, simple cut-set or cocycle From the figure, 𝑎, 𝑐, 𝑑, 𝑓 , 𝑎, 𝑏, 𝑔 , 𝑎, 𝑏, 𝑒, 𝑓 , 𝑑, , 𝑓 , 𝑘 are set of cut-sets. But , 𝑎, 𝑐, , 𝑑 is not a cut-set because one of its proper subsets , 𝑎, 𝑐, is a cut-set A cut-set always “cuts” a graph into two. Therefore , a cut-set can also be defined as a minimal set of edges in a connected graph whose removal reduces the rank of the graph by one. From the figure , the rank of the graph in figure (b) is four one less than that of the graph in figure (a) If we partition all the vertices of a connected graph G into two mutually exclusive subsets , a cut-set is a minimal number of edges whose removal from G destroys all paths between these two sets of vertices From the above figure, cut set , 𝑎, 𝑐, 𝑑, 𝑓 connects vertex set , 𝑣1 , 𝑣2 , 𝑣6 with 𝑣3 , 𝑣4 , 𝑣5 Since removal of any edge from a tree breaks the tree into two parts, every edge of a tree is a cut-set Properties of Cut-Set Theorem: Every cut-set in a connected graph G must contain at least one branch of every spanning tree of G Proof: In a given connected graph G, let Q be a minimal set of edges containing atleast one branch of every spanning tree of G. Consider G-Q , the sub graph that remains after removing the edges in Q from G. Since the sub graph G-Q contains no spanning tree of G, G-Q is disconnected. 61 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Also, since Q is a minimal set of edges with this property, any edge e from Q returned to G-Q will create at least one spanning tree. Thus the sub graph G-Q+e will be a connected graph. Therefore, Q is a minimal set of edges whose removal from G disconnects G. This, by definition , is a cut set. Hence the theorem. Theorem: In a connected graph G, any minimal set of edges containing at least one branch of every spanning tree of G is a cut set Theorem: Every circuit has even number of edges in common with any cut set Proof: Consider a cut set S in a graph G. Let the removal of S partition the vertices of G into two (mutually exclusive or disjoint)subsets V1 and V2. Consider a circuit Γ in G. If all the vertices in Γ are entirely within vertex set V1(or V2), the number of edges common to S and Γ is zero; that is 𝑁 𝑆 ∩ Γ = 0, an even number If, on the other hand, some vertices in Γ are in V1 and some in V2, we traverse back and forth between the sets V1 and V2 as we traverse the circuit. Because of the closed nature of a circuit, the number of edges we traverse between V 1 and V2 must be even. And since every edge in S has one end in V1 and the other in V2, and no other edge in G has this property (of separating sets V1 and V2), the number of edges common to S and Γ is even. Fundamental Cut Sets A cut set S containing exactly one branch of a tree T is called a Fundamental cut-set with respect to T Consider a spanning tree T of a connected graph G. Take any branch b in T. Since 𝑏 is a cut set in T, 𝑏 partitions all vertices of T into two disjoint sets- one at each end of b Consider the same partition of vertices in G, and the cut set S in G that corresponds to this partition. Cut set S will contain only one branch b of T, and the rest of the edges in S are chords with respect to T. 62 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Such a cut set is called Fundamental cut set or basic cut set. In the above figure, a spanning tree T and all five of the fundamental cut-sets with respect to T are shown(broken lines “cutting” through each cut set). The ring sum of any two cut sets in a graph is either a third cut set or an edge disjoint union of cut sets Fundamental Circuits and Cut Sets Consider a spanning tree T in a given connected graph G.. Let ci be a chord with respect to T, and let the fundamental circuit made by ci be called Γ, consisting of k branches 𝑏1, 𝑏2, … , 𝑏𝑘 , in addition to the chord ci; that is 𝚪 = 𝒄𝒊 , 𝒃𝟏, 𝒃𝟐, … , 𝒃𝒌, is a fundamental circuit with respect to T. Every branch of any spanning tree has a fundamental cut set associated with it. Let S i be the fundamental cut-set associated with bi, consisting of q chords in addition to the branch b1; that is 𝐒𝐢 = 𝒃𝟏 , 𝒄𝟏, 𝒄𝟐, … , 𝒄𝒒, is a fundamental cut-set with respect to T With respect to a given spanning tree T, a chord ci that determines a fundamental circuit 𝚪 occurs in every fundamental cut-set associated with the branches in 𝚪 and in no other. With respect to a given spanning tree T, a branch bi that determines a fundamental cut-set S is contained in every fundamental circuit associated with the chords in S and in no others. Edge Connectivity The number of edges in the smallest cut set (i.e., cut set with fewest number of edges) is defined as the edge connectivity of G Edge connectivity of a connected graph can be defined as the minimum number of edges whose removal (i.e., deletion ) reduces the rank of the graph by one. The edge connectivity of a tree is one 63 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Vertex Connectivity Vertex connectivity of a connected graph G is defined as the minimum number of vertices whose removal from G leaves the remaining graph disconnected. Vertex connectivity of a tree is one ` 64 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Separable Graph: Aconnected graph is said to be separable if its vertex connectivity is one All other connected graphs are called nonseparable In a separable graph a vertex whose removal disconnects the graph is called a cutvertex, a cut node or an articulation point The maximum vertex connectivity one can achieve with a graph G of n vetices and e edges (𝑒 ≥ 𝑛 − 1) is the integral part of the number 2𝑒 𝑛; that is 2𝑒 𝑛 Every cut-set in a non separable graph with more than two vertices contains atleast two edges A vertex v in a connected graph G is a cut-vertex if and only if there exist two vertices x and y in G such that every path between x and y passes through v Theorem: The edge connectivity of a graph G cannot exceed the degree of the vertex with the smallest degree in G Proof: Let vertex vi be the vertex with the smallest degree in G. Let 𝑑(𝑣𝑖 ) be the degree of vi. Vertex vi can be separated from G by removing the 𝑑(𝑣𝑖 ) edges incident on vertex vi. Hence the theorem. Theorem: The vertex connectivity of any graph G can never exceed the edge connectivity of G Proof: Let 𝛼 denote the edge connectivity of G. Therefore, there exists a cut-set S in G with 𝛼 edges. Let S partition the vertices of G into subsets V1 and V2. By removing atmost 𝛼 vertices from V1 or V2 on which the edges in S are incident, we can effect the removal of S(together with all other edges incident on these vertices) from G. Hence the theorem. 65 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics 1-Isomorphism A separable graph consists of two or more nonseparable subgraphs. Each of the largest nonseparable subgraphs is called a block. Two graphs G1 and G2 are said to be 1-isomorphic if they become isomorphic to each other under the following operations o “Split” a cut-vertex into two vertices to produce two disjoint sub graphs If G1 and G2 are two 1-isomorphic graphs, the rank of G1 equals the rank of G2 and the nullity of G1 equals the nullity of G2 2-Isomorphism 2-Connected Graph: Graph with vertex connectivity of two In a 2-connected graph G let vertices x and y be a pair of vertices whose removal from G will leave the remaining graph disconnected. Suppose that we perform the following operation 2 on G: Operation 2: “Split” the vertex x into x1 and x2 and the vertex y into y1 and y2 such that G split into g1 and g2 . Let vertices x1 and y1 go with g1 and x2 and y2 with g2. Now rejoin the graphs g1 and g2 by merging x1 with y2 and x2 with y1. Operation 1: “Split” a cut-vertex into two vertices to produce two disjoint sub graphs Two graphs are said to be 2-ismorphic if they become isomorphic after undergoing operation 1 or operation 2 or both operation any number of times. 66 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Circuit Correspondance: Two graphs G1 and G2 are said to have circuit correspondence if they meet the following condition: There is one-to-one correspondence between the edges of G1 and G2 and a one to one correspondence between the circuits of G1 and G2, such that a circuit in G1 formed by certain edges of G1 has a corresponding circuit in G2 formed by the corresponding edges of G2, and vice cersa. Isomorphic graphs have circuit correspondence 2- isomorphic graph also have circuit correspondance Planar Graph A graph G is said to be planar if there exists some geometric representation of G which can be drawn on a plane such that no two of its edges intersect. The geometric represnttaion of planar graph is also called embedding. An embedding of a planar graph G on a plane is called a plane representation of G A graph that cannot be drawn on a plane without a crossover between edges is called non planar Kuratowski’s Graphs Two specific non planar graphs These are called Kuratowski’s graphs, after the Polish mathematician Kasimir Kuratowski, who discovered their unique property 67 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Kuratowski’s Graphs are A complete graph with five vertices Regular connected graph with six vertices and nine edges Two common geometric representation of Kuratowski’s second graph is shown in the above figure Properties of Kuratowski’s Graphs 1) 2) 3) 4) Both are regular graphs Both are non planar Removal of one edge or a vertex makes each a planar graph Kuratowski’s first graph is the non planar graph with the smallest number of vertices and Kuratowski’s second graph is the non planar graph with the smallest number of edges. Thus both are the simplest non planar graphs Different Representation of Planar Graphs Straight line representation Plane representation Embedding on a sphere Straight Line Representation Any simple planar graph can be embedded in a plane such that every edge is drawn as a straight line segment This representation is only used for simple graph because self loop and parallel edges cannot be represented by straight line 68 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Plane Representation A plane representation of a graph divides the plane into regions (also called windows, faces, or meshes). A region is characterized by the set of edges (or the set of vertices) forming its boundary. Infinite Region: The portion of the plane lying outside a graph embedded in a plane. Sucha region is called the infinite, unbounded, outer, or exterior region for that particular plane representation. Embedding on a Sphere To eliminate the distinction between finite and infinite region , a planar graph is often embedded in the surface of a sphere. It is accomplished by stereographic projection of a sphere on the plane Stereographic Projection Put the sphere on the plane and call the point of contact SP(south pole). At point SP, draw a straight line perpendicular to the plane and let the point where line intersects the surface of the sphere be called NP( north pole) Now corresponding to any point p on the plane, there exists a unique point p’ on the sphere and vice versa, where p’ is the point at which the straight line from point p to point NP intersects the surface of the sphere. 69 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Thus there is a one-to-one correspondence between the points of the sphere and the finite points on the plane, and points at infinity in the plane correspond to the point NP on the sphere. Theorem: A graph can be embedded in the surface of a sphere if and only if it can be embedded in a plane Proof: A planar graph embedded in the surface of a sphere divides the surface into different regions. Each region on the sphere is finite, the infinite region on the plane having been mapped onto the region containing the point NP. Now it is clear that by suitably rotating the sphere we can make any specified region map onto the infinite region on the plane. Euler’s Theorem Euler’s formula gives the number of regions in any planar graph Theorem: A connected planar graph with n vertices and e edges has e-n+2 regions Proof: We can disregard a self loop or pararllel edges because it simply adds one region to the graph and simultaneously increases the value of e by one We can also disregard all edges that do not form boundaries of any region. Adding of any such edge increases( or decreases) e by one and increases (or decreases) n by one, keeping the quanitiy e-n unaltered Since any simple planar graph can have a plane representation such that each edge is a straight line, any planar graph can be drawn such that each region is a polygon(a polygonal net) Let the polygonal net representing the given graph consist of f regions or faces, and let 𝑘𝑝 be the number of p-sided regions. Since each edge is on the boundary of exactly two regions, 3. 𝑘3 + 4. 𝑘4 + 5. 𝑘5 + … + 𝑟. 𝑘𝑟 = 2. 𝑒 𝐸𝑞𝑛 1 where 𝑘𝑟 is the number of polygons with maximum edges. 70 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Also, 𝑘3 + 𝑘4 + 𝑘5 + ⋯ + 𝑘𝑟 = 𝑓 𝐸𝑞𝑛(2) Sum of angles subtended at each vertex in the polygonal net = 2𝜋𝑛 𝐸𝑞𝑛(3) Sum of all interior angles of a p-sided polygon = 𝜋 𝑝 − 2 Sum of exterior angles of a p-sided polygon = 𝜋 𝑝 + 2 𝐸𝑞𝑛 3 𝑏𝑒𝑐𝑜𝑚𝑒𝑠 Sum of angles subtended at each vertex in the polygonal net = Sum of all interior angles+ Sum of exterior angles of a p-sided polygon = 𝜋 3 − 2 . 𝑘3 + 𝜋 4 − 2 . 𝑘4 + 𝜋 5 − 2 . 𝑘5 + … + 𝜋 𝑟 − 2 . 𝑘𝑟 + 4𝜋 = 𝜋 2𝑒 − 2𝑓 + 4𝜋 𝐸𝑞𝑛(4) 𝐹𝑟𝑜𝑚 𝐸𝑞𝑛 3 𝑎𝑛𝑑 𝐸𝑞𝑛 4 𝜋 2𝑒 − 2𝑓 + 4𝜋 = 2𝜋𝑛 2𝜋 𝑒 − 𝑓 + 2 = 2𝜋𝑛 𝑒−𝑓+2 =𝑛 Therefore number of regions is 𝑓 =𝑒−𝑛+2 In any simple graph, connected planar graph with f regions, n vertices, and e edges 𝒆 > 2 , the following inequality holds: 𝟑 𝒆≥ 𝒇, 𝟐 𝒆 ≤ 𝟑𝒏 − 𝟔 Proof: Since each region is bounded by at least three edges and each edge belongs to exactly two regions, 2𝑒 ≥ 3𝑓 , 𝑒≥ 3 𝑓 2 Substituting for f from Euler’s formula in in the above inequality 𝑒≥ 3 𝑒−𝑛+2 2 𝑒 ≤ 3𝑛 − 6 The inequality 𝒆 ≤ 𝟑𝒏 − 𝟔 is often useful in finding out of a graph is non planar Question: Check whether the graph 𝑲𝟓 , a complete graph of five vertices is planar Ans: 𝒏 = 𝟓, 𝒆 = 𝟏𝟎, 𝟑𝒏 − 𝟔 = 𝟗 < 𝑒 71 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Thus the graph violates the inequality, and hence it is not planar Prove that Kuratowski’s first graph is non planar Ans: Kuratowski’s first graph is a complete graph with five vertices 𝒏 = 𝟓, 𝒆 = 𝟏𝟎, 𝟑𝒏 − 𝟔 = 𝟗 < 𝑒 Thus the graph violates the inequality, and hence it is not planar Prove that Kuratowski’s second graph is non planar Ans: Kuratowski’s second graph is a regular connected graph with 6 vertices and 9 edges. To prove the non planarity of Kuratowski’s second graph , we can make use of the additional fact that no region in this graph can be bounded with fewer than four edges. Hence, if this graph were planar, we would have 𝟐𝒆 ≥ 𝟒𝒇, And substituting for f from Euler’s formula 𝟐𝒆 ≥ 𝟒 𝒆 − 𝒏 + 𝟐 𝟐×𝟗 ≥ 𝟒 𝟗−𝟔+𝟐 𝟏𝟖 ≥ 𝟐𝟎 Thus the graph violates the inequality, and hence it is not planar Unique Embedding If all possible embeddings on a sphere no two are distinct, the graph is said to have a unique embedding on a sphere. The following figure shows unique embedding Detection of Planarity We can check the planarity of a graph through Elementary Reduction. It consists of the following steps 72 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Step 1: Since a disconnected graph is planar if and only if each of its components is planar, we need consider only one component at a time. Also a separable graph is planar if and only if each of its block is planar. Therefore, for the given arbitrary graph G, determine the set 𝐺 = 𝐺1 , 𝐺2 , … , 𝐺𝑘 𝑤𝑒𝑟𝑒 𝑒𝑎𝑐𝐺𝑖 𝑖𝑠 𝑎 𝑛𝑜𝑛 𝑠𝑒𝑝𝑎𝑟𝑎𝑏𝑙𝑒 𝑏𝑙𝑜𝑐𝑘 𝑜𝑓 𝐺 Thus we have to check each 𝐺𝑖 for planarity Step 2: Elimination of a vertex of degree two by merging two edges in series doesnot affect planarity. Therefore, eliminate all edges in series . [In a graph two edges are said to be in series if they have exactly one vertex in common and if this vertex is of degree two) Step 3: Since addition or removal of self loops doesnot affect planarity, remove all self loops Step 4: Since parallel edges also do not affect planarity, eliminate edges in paralle by removing all but one edge between every pair of vertices Repeated application of step2, 3 and 4 will reduce the graph into a new graph which has any one of the following property: 1. Graph has a single edge or 2. A complete graph of four vertices 3. A nonseparable, simple graph with 𝒏 ≥ 𝟓 and 𝒆 ≥ 𝟕 For the first two cases, the graph is planar. For the third case we need to use Euler’s theorem and planarity inequality for planarity detection 73 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Homomorphic Graphs Two graphs are said to be homomorphic if one graph can be obtained from the other by the creation of edges in series( i.e., by insertion of vertices of degree two) or by the merger of edges in series. Geometric Dual Consider the plane representation of a graph in Fig (a), with six regions or faces 𝐹1 , 𝐹2 , 𝐹3 , 𝐹4 , 𝐹5 and 𝐹6 . Let us place six points 𝑝1 , 𝑝2 , … , 𝑝6 one in each of the regions. Next let us join these six points according to the following procedure If two regions 𝐹𝑖 and 𝐹𝑗 are adjacent(i.e., have a common edge), draw a line joining points 𝑝𝑖 and 𝑝𝑗 that intersects the common edge between 𝐹𝑖 and 𝐹𝑗 exactly once. 74 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics If there is more than one edge common between 𝐹𝑖 and 𝐹𝑗 , draw one line between points 𝑝𝑖 and 𝑝𝑗 for each of the common edges. For an edge e lying entirely in one region, say 𝐹𝑘 , draw a self-loop at points 𝑝𝑘 intersecting e exactly once. By this procedure we obtain a new graph 𝐺 ∗ (broken lines in the above figure) consisting of six vertices 𝑝1 , 𝑝2 , … , 𝑝6 and edges joining these vertices. Such graph 𝐺 ∗ is called a dual or geometric dual of G Relationship between G and its dual 𝑮∗ An edge forming a self loop in G yields a pendant edge in 𝑮∗ A pendant edge in G yields a self loop in 𝑮∗ Edges that are in series in G produce parallel edges in 𝑮∗ Paralle edges in G produce edges in series in 𝑮∗ Number of edges constituting the boundary of a region𝐹𝑖 in G is equal to the degree of the corresponding 𝑝𝑖 in 𝑮∗ and vice versa 6. Graph 𝑮∗ is also embedded in a plane, so it is planar 7. G and 𝑮∗ are called self dual graphs 8. If 𝑛, 𝑒, 𝑓, 𝑟, 𝜇 denotes the number of vertices, edges, regions, rank and nullity of a connected planar graph G and if 𝑛∗ , 𝑒 ∗ , 𝑓 ∗ . 𝑟 ∗ , 𝜇 ∗ are the corresponding numbers in dual graph 𝑮∗ , then 𝑛∗ = 𝑓 𝑒∗ = 𝑒 𝑓∗ = 𝑛 𝑟∗ = 𝜇 𝜇∗ = 𝑟 1. 2. 3. 4. 5. Combinatorial Dual Two planar graphs 𝐺 and 𝑮∗ are said to be combinatorial duals of each other if there is one to one correspondence between the edges in G and 𝑮∗ such that a set of edges in 𝐺 forms a circuit if and only if the corresponding set in 𝑮∗ forms a cut set Two planar graphs 𝐺 and 𝑮∗ are said to be combinatorial duals of each other if there is one to one correspondence between the edges in 𝐺 and 𝑮∗ such that if g is a ny sub graph of 𝐺1 and h is the corresponding subgraph in 𝑮∗ then 𝒓𝒂𝒏𝒌 𝒐𝒇 𝑮∗ − 𝒉 = 𝒓𝒂𝒏𝒌 𝒐𝒇𝑮∗ − 𝒏𝒖𝒍𝒍𝒊𝒕𝒚 𝒐𝒇 𝒈 75 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Theorem: There is a one to one correspondence between the edges in G and G* such that a set of edges in G forms a circuit if and only if the corresponding set in G*forms a cut set Proof: Let us consider planar representation of a planar graph. Let us also draw dual G* of G. Then consider an arbitrary circuit Γ in G. Clearly, Γ will form some closed simple curve in the plane representation of G- dividing the plane into two areas Thus the vertices of G * are partitioned into two non empty, mutually exclusive subsets- one inside Γ and the other outside That is set of edges Γ* in G* corresponding to the set Γ in G is a cut set in G * Likewise it is apparent that corresponding to a cut-set S* in G*, there is a unique circuit consisting of the corresponding edge-set S in G such that S is a circuit. This proves the theorem. Theorem: A graph has a dual if and only if it is planar Proof: Use proof by contradiction. Inorder to prove the theorem, we have only to prove that a nonplanar graph does not have a dual. Let G be a nonplanar graph. Then according to Kuratowski’s theorem, G contains K5 or K3,3 or a graph homomorphic to either of these. Thus if we can show that neither K5 nor K3,3 has a dual, we have proved the theorem. This we shall prove by contradiction as follows: Suppose that K3,3 has adual D. Observe that the cut-sets in K3,3 corresponding to circuits in D and vice versa. Since K3,3 has no cut-set consisting of two edges, D has no circuit consisting of two edges. That is , D contains no pair of parallel edges,Since every circuit in K3,3 is of length four or six, D has no cutsets with less than four edges. Therefore, the degree of every vertex in Dis at least four.As D has no parallel edges and the degree of every vertex is atleast four, D must have atleast five verticeseach of degree four or more. That is D must have atleast (5 × 4)/10 = 10 edges. This is a contradiction, because K3,3 has 9 edges and so must its dual. Thus K 3,3 cannot have a dual Suppose that the graph K5 has a dual H. Note that K5 has 10 edges no pair of parallel edges no cut-set with two nedges cut set with only four or six edges Consequently, graph H must have 10 edges no vertex with degree less than three 76 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics no pair of parallel edges circuits of length four and six only Now graph H contains a hexagon(a circuit of length six) and no more than three edges can be added to a hexagon without creating a circuit of length three or a pair of parallel edges Since both of these are forbidden in H and H has 10 edges, there must be atleast seven vertices in H. the degree of each of these vertices is atleast three.. This leads to H having atleast 11 edges. a contradiction Dual of Sub Graph Let G be a planar graph and G* be its dual. Let a be an edge in G, and the corresponding edge in G* be a*. Suppose that we delete an edge a from G and then try to find the dual of G –a. If edge a was on the boundary of two regions, removal of a would merge these two regions into one. Thus the dual of 𝐺 − 𝑎 * can be obtained from G * by deleting the corresponding edge a* and then fusing the two end vertices of a* in G*- a* If the edge a is not the boundary, a* forms a self loop.In that case G*- a* is the same as 𝐺 − 𝑎 * Thus if a graph G has a dual G*, the dual of any subgraph of G can be obtained by successive application of this procedure Self Dual Graphs If a planar graph G is isomorphic to its own dual , it is called a self dual graphs Thickness and Crossing of Planar graphs Thickness: Least number of planar subgraphs in G is called thickness of G Thickness of planar graph is one Thickness of Kuratowski’s graph is two Crossings Least number of crossing or intersectionsnecesssary inorder to draw the graph in the plane Crossing number of planar graph is zero Crossing number of Kuratowski’s graph is one Points to Remember Cut Set: In a coonected graph G, a cut-set a set of edges whose removal from G leaves G disconnected, provided removal of no proper subset of these edges 77 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics disconnects G Every edge of a tree is a cut-set Every cut-set in a connected graph G must contain at least one branch of every spanning tree of G Every circuit has even number of edges in common with any cut set A cut set S containing exactly one branch of a tree T is called a Fundamental cut-set with respect to T The number of edges in the smallest cut set (i.e., cut set with fewest number of edges) is defined as the edge connectivity of G The edge connectivity of a tree is one Vertex connectivity of a connected graph G is defined as the minimum number of vertices whose removal from G leaves the remaining graph disconnected. Vertex connectivity of a tree is one Aconnected graph is said to be separable if its vertex connectivity is one. All other connected graphs are called nonseparable In a separable graph a vertex whose removal disconnects the graph is called a cutvertex, a cut node or an articulation point Two types of Isomorphism: 1 isomorphism, 2-Isomorphism 2-Connected Graph: Graph with vertex connectivity of two n-Connected Graph: Graph with vertex connectivity of n A graph G is said to be planar if there exists some geometric representation of G which can be drawn on a plane such that no two of its edges intersect. The geometric represnttaion of plananr graph is also called embedding. Kuratowski’s Graphs A complete graph with five vertices Regular connected graph with six vertices and nine edges Different Representation of Planar Graphs Straight line representation Plane representation Embedding on a sphere Euler’s Theorem A connected planar graph with n vertices and e edges has e-n+2 regions In any simple graph, connected planar graph with f regions, n vertices, and e edges 𝒆 > 2 , the following inequality holds: 𝟑 𝒆≥ 𝒇, 𝟐 𝒆 ≤ 𝟑𝒏 − 𝟔 If all possible embeddings on a sphere no two are distinct, the graph is said to have a unique embedding on a sphere. The following figure shows unique embedding Homomorphic Graphs Two graphs are said to be homomorphic if one graph can be obtained from the other by the creation of edges in series( i.e., by insertion of vertices of degree two) or by the merger of edges in series. Combinatorial Dual Two planar graphs 𝐺 and 𝑮∗ are said to be combinatorial duals of each other if there is one to one correspondence between the edges in G and 𝑮∗ such that a set of edges in 𝐺 forms a circuit if and only if the corresponding set in 𝑮∗ forms a cut set Two planar graphs 𝐺 and 𝑮∗ are said to be combinatorial duals of each other if there 78 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics is one to one correspondence between the edges in 𝐺 and 𝑮∗ such that if g is a ny sub graph of 𝐺1 and h is the corresponding subgraph in 𝑮∗ then 𝒓𝒂𝒏𝒌 𝒐𝒇 𝑮∗ − 𝒉 = 𝒓𝒂𝒏𝒌 𝒐𝒇𝑮∗ − 𝒏𝒖𝒍𝒍𝒊𝒕𝒚 𝒐𝒇 𝒈 Self Dual Graphs If a planar graph G is isomorphic to its own dual , it is called a self dual graphs Thickness: Least number of planar subgraphs in G is called thickness of G Thickness of planar graph is one Thickness of Kuratowski’s graph is two Crossings Least number of crossing or intersectionsnecesssary inorder to draw the graph in the plane Crossing number of planar graph is zero Crossing number of Kuratowski’s graph is one PREVIOUS YEAR UNIVERSITY QUESTIONS 1. Explain self dual with an example 2. Prove the statement: Every cut-set in a connected graph G must also contain at least one branch of every spanning tree of G 3. List down the properties stating the relationship between the edges of graph G and its dual G* 4. Define cut set. Find any four cut sets from the graph G given below and also find the edge connectivity of G. 5. Define vertex connectivity and draw a graph with an articulation point. 6. State Euler’s Theorem (formula). 7. Draw two Kuratowski’s graphs and also prove that Kuratowsk’s first graph is non planar using appropriate inequality. 8. Draw the geometric dual (G*) of the graph G given below and also check whether G and G* are self dual or not, substantiate your answer clearly? 79 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics 9. Prove that the ring sum of any two cut-sets in a graph is either a third cut-set or an edge-disjoint union of cut-sets. 10. Prove that a connected planar graph with n vertices and e edges has e-n+2 regions. 11. Consider the following graph G and any one of its spanning trees, T.List all fundamental circuits and fundamental cut-sets with respect to T. 12. State and prove Euler’s theorem involving number of regions, edges and vertices in a planar graph 13. A connected planar graph has 10 vertices each of degree 3. Into how many regions does the representation of planar graph split the plane 14. Define the terms thickness and crossings with an example 15. Explain Planar Graph and show that K5 is not planar 16. Show that a graph can be embedded in the surface of a sphere if and only if it can be embedded in a plane 17. Let G be a planar graph with𝑣 ≥ 3 vertices and 𝑒 edges. Then prove that 𝑒 ≤ 3𝑣 − 6 18. What are the relationship between a planar graph G and its dual G *?Find the geometric dual of the following graph 9. Prove that a complete graph of five vertices is non planar 10. Explain the procedure to obtain the geometric dual of a given graph 11. Prove that every planar graph contain at least one vertex of degree ≤ 5 12. Prove that a graph has a dual if and only if it is planar 13. State and prove Euler Theorem 14. What do you mean by geometric dual of a graph? Illustrate with examples 15. A necessary and sufficient condition for two planar graphs G1 and G2 to be duals of each other is as follows: There is one-to-one correspondence between the edges in G1 forms a circuit if and only if the corresponding set in G2 forms a cut –set. Prove it geometrically. (Consider a graph with atleast 6 vertices and 6 faces) 16. Differentiate Geometric dual with Combinatorial dual 17. Prove that the complete graph K5 is non planar 80 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics 18. If a connected planar graph G has n vertices, e edges and r regions, then prove that 𝑛−𝑒+𝑟 = 2 19. Describe the methods used to determine the planarity of a graph 20. What is the necessary condition for a graph to have geometric dual? List the characteristics of graphs that are geometric duals of each other. Give two examples of graphs having geometric duals. 21. Using Euler’s formula show that Kwatowski’s two graphs are not planar 22. Prove that a graph has a dual if and only if it is planar WORKSHEET 1. Explain self dual with an example 2. Prove the statement: Every cut-set in a connected graph G must also contain at least one branch of every spanning tree of G 3. List down the properties stating the relationship between the edges of graph G and its dual G* 4. Define cut set. Find any four cut sets from the graph G given below and also find the edge connectivity of G. 5. Define vertex connectivity and draw a graph with an articulation point. 6. State Euler’s Theorem (formula). 7. Draw two Kuratowski’s graphs and also prove that Kuratowsk’s first graph is non planar using appropriate inequality. 8. Draw the geometric dual (G*) of the graph G given below and also check whether G and G* are self dual or not, substantiate your answer clearly? 9. Prove that the ring sum of any two cut-sets in a graph is either a third cut-set or an edge-disjoint union of cut-sets. 10. Prove that a connected planar graph with n vertices and e edges has e-n+2 regions. 11. Consider the following graph G and any one of its spanning trees, T.List all fundamental circuits and fundamental cut-sets with respect to T. 81 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE 82 HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics MODULE V MATRIX REPRESENTATION OF GRAPHS Incidence Graph Let G be a graph with n vertices, e edges, and no self loops. Define an n by e matrix 𝐴 = [𝑎𝑖𝑗 ],whose n rows correspond to the n vertices and the e columns correspond to the e edges, as follows: 𝑎𝑖𝑗 = 1, 𝑖𝑓 𝑗𝑡 𝑒𝑑𝑔𝑒 𝑒𝑗 𝑖𝑠 𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑡 𝑜𝑛 𝑡𝑒 𝑖 𝑡 𝑣𝑒𝑟𝑡𝑒𝑥 𝑣𝑖, and 𝑎𝑖𝑗 = 0, 𝑜𝑡𝑒𝑟𝑤𝑖𝑠𝑒 Such a matrix A is called the vertex-edge incidence matrix, or simply incidence matrix. Matrix A for a graph G is sometimes also written as A(G) The matrix contains only two elements, 0 and 1. Such a matrix is called a binary matrix or a (0,1) -matrix Properties of Incidence Matrix 1) Since every edge is incident on exactly two vertices, each column of A has exactlt two 1’s. 2) The number of 1’s in each row equals the degree of the corresponding vertex. 3) A row with all 0’s represents an isolated vertex. 83 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics 4) Parallel edges in a graph produce identical columns in its incidence matrix 5) If a graph G is disconnected and consists of two components 𝑔1 and 𝑔2 , the incidence matrix A(G) of graph G can be written in a block diagonal form as 𝐴(𝑔1 ) 0 𝐴 𝐺 = 0 𝐴(𝑔2 ) where 𝐴 𝑔1 and 𝐴(𝑔2 ) are the incidence matrices of components 𝑔1 and 𝑔2 . 6) Permutation of any two rows or columns in an incidence matrix simply corresponds to relabeling the vertices and edges of the same graph. Theorem: Two graphs G1 and G2 are isomorphic if and only if their incidence matrices A(G1) and A(G2) differ only by permutations of rows and columns Rank of the Incidence Matrix Let G be a graph and let A(G) be its incidence matrix. Now each row in A(G) is a vector over GF(2) in the vector space of graph G. Let the row vectors be denoted by A1, A2, . . ., An. Then, 𝐴1 𝐴2 . 𝐴 𝐺 = . . 𝐴𝑛 Since there are exactly two ones in every column of A, the sum of all these vectors is 0 (this being a modulo 2 sum of the corresponding entries). Thus vectors A1, A2, . . ., An are linearly dependent. Therefore, rank A < n. Hence, rank A ≤ n−1. Theorem: If A(G) is an incidence matrix of a connected graph G with n vertices, the rank of A(G) is n-1 Pro of: Let G be a connected graph with n vertices and let the number of edges in G be m. Let A(G) be the incidence matrix and let A1, A2, . . ., An be the row vector of A(G).Then, 𝐴1 𝐴2 . 𝐴 𝐺 = . . 𝐴𝑛 Clearly, rank A(G) ≤ n−1. (1) Consider the sum of any m of these row vectors, m ≤ n−1. Since G is connected, A(G) cannot be partitioned in the form 𝐴(𝐺1 ) 0 𝐴 𝐺 = 0 𝐴(𝐺2 ) 84 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics such that 𝐴(𝐺1 ) has m rows and 𝐴(𝐺2 ) has n−m rows. Thus there exists no m×m submatrix of A(G) for m≤ n−1, such that the modulo 2 sum of these m rows is equal to zero. As there are only two elements 0 and 1 in this field, the additions of all vectors taken m at a time for m = 1, 2, . . ., n−1 gives all possible linear combinations of n−1 row vectors. Thus no linear combinations of m row vectors of A, for m ≤ n−1, is zero. Therefore, rank A(G) ≤ n−1. (2) Combining (1) and (2), it follows that rank A(G) = n−1. Theorem: Rank of A(G) is n-k, if G is disconnected graph with n vertices and k components Reduced Incidence Matrix Let G be a connected graph with n vertices and m edges. Then the order of the incidence matrix A(G) is n×m. Now, if we remove any one row from A(G), the remaining (n−1) by m submatrix is of rank (n−1). Thus the remaining (n−1) row vectors are linearly independent. This shows that only (n−1) rows of an incidence matrix are required to specify the corresponding graph completely, because (n−1) rows contain the same information as the entire matrix. Such an (n−1) × m matrix of A is called a reduced incidence matrix and is denoted by Af. The vertex corresponding to the deleted row in Af is called the reference vertex. Theorem: The reduced incidence matrix of a tree is non-singular. Proof: A tree with n vertices has n−1 edges and also a tree is connected. Therefore, the reduced incidence matrix is a square matrix of order n−1, with rank n−1. Thus the result follows. Now a graph G with n vertices and n −1 edges which is not a tree is obviously disconnected. Therefore the rank of the incidence matrix of G is less than (n−1). Hence the (n−1)×(n−1) reduced incidence matrix of a graph is non-singular if and only if the graph is a tree. Circuit Matrix Let the graph G have m edges and let q be the number of different circuits in G. The circuit matrix 𝐵 = [𝑏𝑖𝑗 ]𝑞×𝑚 of G is a (0, 1) − matrix of order q × m, with bij = 1, if the ith cycle includes jth edge bij = 0, otherwise. The circuit matrix B of a graph G is denoted by B(G). 85 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics The graph G1 has four different cycles Z1 = {e1, e2}, Z2 = {e3, e5, e7}, Z3 = {e4, e6, e7} and Z4 = {e3, e4, e6, e5}. Properties of Circuit Matrix 1) A column of all zeros corresponds to a non circuit edge, that is, an edge which does not belong to any circuit. 2) Each row of B(G) is a circuit vector. 3) A circuit matrix has the property of representing a self-loop and the corresponding row has a single one. 4) The number of ones in a row is equal to the number of edges in the corresponding circuit. 5) If the graph G is separable (or disconnected) and consists of two blocks (or components) g1 and g2, then the cycle matrix B(G) can be written in a block-diagonal form as 𝐵(𝑔1) 0 𝐵 𝐺 = 0 𝐵(𝑔2) where B(g1) and B(g2)are the cycle matrices of g1 and g2 6) Permutation of any two rows or columns in a circuit matrix corresponds to relabeling the circuits and the edges. 7) Two graphs G1 and G2 have the same circuit matrix if and only if G1 and G2 are 2isomorphic Theorem: If G is a graph without self-loops, with incidence matrix A and circuit matrix B whose columns are arranged using the same order of edges, then every row of B is orthogonal to every row of A, that is ABT = BAT = 0 (mod2), where AT and BT are the transposes of A and B respectively. Proof: 86 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Consider a vertex v and circuit ℸ in a graph G. Either v is in ℸ or it is not. If v is not inℸ, there is no edge in the circuit ℸ that is incident on v. On the other hand, if v is inℸ, the number of edges in the circuit ℸ that are incident on v is exactly two. Consider the ith row in A and the jth row in B. Since the edges are arranged in the same order, the non zero entries in the corresponding position occurs only if a particular edge is incident on the ith vertex and also in the jth circuit. If the ith vertex is not in the jth circuit, there is no such non zero entry, and the dot product of the two rows is zero. If the ith vertex is in the jth circuit, there will be exactly two 1’s in the sum of the product of two arbitrary rows. Since the dot product of two arbitrary rows one from A and other from B is zero. Hence the theorem. Fundamental Circuit Matrix A sub matrix of a circuit matrix in which all rows correspond to a set of fundamental circuit is called a fundamental circuit matrix B f If n is the number of vertices, m the number of edges in a connected graph G, then Bf is an (m−n+1)×m matrix because the number of fundamental circuitss is m−n+1, each fundamental circuit being produced by one chord. A matrix Bf thus arranged has the form 𝐵𝑓 = 𝐼𝜇 : 𝐵𝑡 where Iμ is an identity matrix of order μ = m−n+1 and Bt is the remaining μ ×(n−1) submatrix, corresponding to the branches of the spanning tree. 𝑍1 = 𝑒1 , 𝑒2 , 𝑒4 , 𝑒7 𝑍2 = 𝑒3 , 𝑒4 , 𝑒7 𝑍3 = 𝑒5 , 𝑒6 , 𝑒7 Theorem: If B is a circuit matrix of a connected graph G with n vertices and m edges, then rank B = m−n+1. Proof: Let A be the incidence matrix of the connected graph G. Then ABT ≡ 0(mod2). Using Sylvester’s theorem, we have rank A+ rank BT ≤ m 87 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics so that rank A+ rank B ≤ m. Therefore, rank B ≤ m− rank A. As rank A = n−1, we get rank B ≤ m−(n−1)= m−n+1. But, rank B ≥ m−n+1. Combining, we get rank B = m−n+1. Theorem: If B is a circuit matrix of a disconnected graph G with n vertices, m edges and k components, then rank B = m−n+k. Proof: Let B be the circuit matrix of the disconnected graph G with n vertices, m edges and k components. Let the k components be G1, G2, ...,Gk with n1, n2, ..., nk vertices and m1, m2, . . .,mk edges respectively Then 𝑛1 + 𝑛2 + ⋯ + 𝑛𝑘 = 𝑛 and 𝑚1 + 𝑚2 + ⋯ + 𝑚𝑘 = 𝑚 Let 𝐵1 , 𝐵2 ,…, 𝐵𝑘 be the circuit matrices of G1, G2, ...,Gk 𝐵1 (𝐺1 ) 0 0 … 0 0 𝐵1 (𝐺1 ) 0 … 0 0 0 𝐵1 (𝐺1 ) … 0 Then 𝐵 𝐺 = . . . . . . . . . . . . 0 0 0 … 𝐵1 (𝐺1 ) We know rank𝐵𝑖 = 𝑚𝑖 − 𝑛𝑖 + 1, 𝑓𝑜𝑟 1 ≤ 𝑖 ≤ 𝑘 Therefore, 𝑟𝑎𝑛𝑘 𝐵 = 𝑟𝑎𝑛𝑘 𝐵1 + ⋯ + 𝑟𝑎𝑛𝑘 𝐵𝑘 = 𝑚1 − 𝑛1 + 1 + ⋯ + (𝑚𝑘 − 𝑛𝑘 + 1) = 𝑚1 + ⋯ + 𝑚𝑘 + 𝑛1 + ⋯ + 𝑛𝑘 + 𝑘 𝒓𝒂𝒏𝒌 𝑩 = 𝒎 − 𝒏 + 𝒌 Cut Set Matrix Let G be a graph with m edges and q cutsets. The cut-set matrix C = [cij]q×m of G is a (0,1)-matrix with 𝟏, 𝒊𝒇 𝒊𝒕𝒉 𝒄𝒖𝒕𝒔𝒆𝒕 𝒄𝒐𝒏𝒕𝒂𝒊𝒏𝒔 𝒋𝒕𝒉 𝒆𝒅𝒈𝒆 𝒄𝒊𝒋 = 𝟎, 𝑶𝒕𝒉𝒆𝒓𝒘𝒊𝒔𝒆 Example Consider the graphs shown in Figure. In the graph G1, E = {e1, e2, e3, e4, e5, e6, e7, e8}. The cut-sets are c1 = {e8}, c2 ={e1, e2}, c3={e3, e5}, c4 = {e5, e6, e7}, c5 = {e3, e6, e7}, c6 = {e4, e6}, c7 = {e3, e4, e7} and c8 = {e4, e5, e7}. 88 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Thus the cut-set matrices are given by Properties of Cut Set Matrix 1) The permutation of rows or columns in a cut-set matrix corresponds simply to renaming of the cut-sets and edges respectively. 2) Each row in C(G) is a cut-set vector. 3) A column with all zeros corresponds to an edge forming a self-loop. 4) Parallel edges form identical columns in the cut-set matrix. 5) In a non-separable graph, since every set of edges incident on a vertex is a cutset,therefore every row of incidence matrix A(G) is included as a row in the cut-set matrix C(G). That is, for a non-separable graph G, C(G) contains A(G). For a separable graph,the incidence matrix of each block is contained in the cut-set matrix. 6) rank C(G)≥rank A(G). Therefore, for a connected graph with n vertices, rank C(G)≥ n−1. Theorem: If G is a connected graph, then the rank of a cut-set matrix C(G) is equal to the rank of incidence matrix A(G), which equals the rank of graph G. Proof: Let A(G), B(G) and C(G) be the incidence, circuit and cut-set matrix of the connected graph G. Then we have rank C(G) ≥ n−1. 89 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Since the number of edges common to a cut-set and a cycle is always even, every row in C is orthogonal to every row in B, provided the edges in both B and C are arranged in the same order.Thus, BCT =CBT ≡ 0 (mod 2). Now, applying Sylvester’s theorem to the above equation, we have rank B+ rank C ≤ m. For a connected graph, we have rank B = m−n+1. Therefore, rank C ≤m- rank B = m−(m-n−1)= n-1 So, rank C = n−1. Path Matrix Let G be a graph with m edges, and u and v be any two vertices in G. The path matrix for vertices u and v denoted by P(u, v) = [pij]q×m, where q is the number of different paths between u and v, is defined as 𝒑𝒊𝒋 = 𝟏, 𝟎, 𝒊𝒇 𝒋𝒕𝒉 𝒆𝒅𝒈𝒆 𝒍𝒊𝒆𝒔 𝒊𝒏 𝒕𝒉𝒆 𝒊𝒕𝒉 𝒑𝒂𝒕𝒉 𝑶𝒕𝒉𝒆𝒓𝒘𝒊𝒔𝒆 Clearly, a path matrix is defined for a particular pair of vertices, the rows in P(u, v) correspond to different paths between u and v, and the columns correspond to different edges in G. Example The different paths between the vertices v3 and v4 are p1 = {e8, e5}, p2 = {e8, e7, e3} and p3 = {e8, e6, e4, e3} The path matrix for v3, v4 is given by 90 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Properties of Path matrix. 1) A column of all zeros corresponds to an edge that does not lie in any path between u and v. 2) A column of all ones corresponds to an edge that lies in every path between u and v. 3) There is no row with all zeros. 4) The ring sum of any two rows in P(u, v) corresponds to a cycle or an edge-disjoint union of cycles. Theorem: If the columns of the incidence matrix A and the path matrix P(u, v) of a connected graph are arranged in the same order, then under the product (mod 2). APT (u, v) = M, where M is a matrix having ones in two rows u and v, and the zeros in the remaining n−2 rows Adjacency Matrix (Connection Matrix) 1) Let V =(V, E) be a graph with V ={v1, v2, . . . , vn}, E ={e1, e2, . . . , em} and without parallel edges. The adjacency matrix of G is an n×n symmetric binary matrix X = [xij] defined over the ring of integers such that 1, 𝑖𝑓𝑡𝑒𝑟𝑒 𝑖𝑠 𝑎𝑛 𝑒𝑑𝑔𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑖 𝑡 𝑎𝑛𝑑 𝑗𝑡 𝑣𝑒𝑟𝑡𝑒𝑥 𝑥𝑖𝑗 = 0, 𝑂𝑡𝑒𝑟𝑤𝑖𝑠𝑒 2) Example: Consider the graph given in the figure The adjacency matrix of G is given by 91 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Properties of Adjacency Matrix 1) The entries along the principal diagonal of X are all zeros if and only if the graph has no self-loops. However, a self-loop at the ith vertex corresponds to xii = 1 2) If the graph has no self-loops, the degree of a vertex equals the number of ones in the corresponding row or column of X. 3) Permutation of rows and the corresponding columns imply reordering the vertices. We note that the rows and columns are arranged in the same order. Therefore, when two rows are interchanged in X, the corresponding columns are also interchanged. Thus two graphs G1 and G2 without parallel edges are isomorphic if and only if their adjacency matrices X(G1) and X(G2) are related by 𝑋 𝐺2 = 𝑅−1 𝑋 𝐺1 𝑅 where R is a permutation matrix. 4) A graph G is disconnected having components G1 and G2 if and only if the adjacency matrix X(G) is partitioned as 𝑋 𝐺1 0 𝑋 𝐺 = 0 𝑋 𝐺2 where 𝑋 𝐺1 and 𝑋 𝐺2 are respectively the adjacency matrices of the components G1 and G2. Obviously, the above partitioning implies that there are no edges between vertices in G1 and vertices in G2. 5) If any square, symmetric and binary matrix Q of order n is given, then there exists a graph G with n vertices and without parallel edges whose adjacency matrix is Q. Edge sequence An edge sequence is a sequence of edges in which each edge, except the first and the last, has one vertex in common with the edge preceding it and one vertex in common with the edge following it. A walk and a path are the examples of an edge sequence. An edge can appear more than once in an edge sequence. In the graph of following Figure , v1e1v2e2v3e3v4e4v2e2v3e5 v5, or e1e2e3e4e2e5 is an edge sequence. Powers of Adjacency Matrix (X) Let us multiply by itself the adjacency matrix of the simple graph. The result is another symmetric matrix 𝑋 2 92 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE 𝑋2 HandOut and WorkSheet of CS309 Graph Theory And Combinatorics = number of ones in the product of ith row and jth column (or jth row) of X 𝑖𝑗 = number of positions in which both ith and jth rows of X have ones = number of vertices that are adjacent to both ith and jth vertices = number of different paths of length two between ith and jth vertices Also, 𝑋 2 𝑖𝑖 = number of ones in the ith row (or column) of X = degree of the corresponding vertex. Theorem: Let X be the adjacency matrix of a simple graph G. Then ij th entry in Xr is the number of different edge sequence of r edges between vertices vi and vj Proof: We prove the result by using induction on k. The result is trivial for k = 0 and 1. Since X2 = X.X, X2 is a symmetric matrix, as product of symmetric matrices is also symmetric. For k = 2, i ≠ j, we have 𝑋 2 𝑖𝑗 = number of ones in the product of ith row and jth column (or jth row) of X = number of positions in which both ith and jth rows of X have ones = number of vertices that are adjacent to both ith and jth vertices = number of different paths of length two between ith and jth vertices Also, 𝑋 2 𝑖𝑖 = number of ones in the ith row (or column) of X = degree of the corresponding vertex. This shows that 𝑋 2 𝑖𝑗 is the number of different paths and therefore different edge sequences of length 2 between the vertices vi and vj. Thus the result is true for k = 2. Assume the result to be true for k, so that 𝑋 𝑘 𝑖𝑗 = number of different edge sequences of length k between vi and vj . 93 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics We have, 𝑋 𝑘 +1 𝑖𝑗 j = 𝑋𝑘 𝑖𝑗 = 𝑛 𝑟=1 𝑋𝑘 𝑖𝑟 𝑋 𝑟𝑗 = 𝑛 𝑟=1 𝑋𝑘 𝑖𝑟 𝑥𝑟𝑗 Now, every vi −vj edge sequence of length k+1 consists of a vi −vr edge sequence of length k, followed by an edge vtvj. Since there are 𝑋 𝑘 𝑖𝑟 such edge sequences of length k and xrj such edges for each vertex vr, the total number of all vi −vj edge sequences of length k+1 is 𝑛𝑟=1 𝑋 𝑘 𝑖𝑟 𝑥𝑟𝑗 . This proves the result for k+1 also. In a connected graph, the distance between two vertices vi and vj (for i≠ 𝒋) is k, if and only if k is the smallest integer for which the i,j th entry in xk is nonzero. If X is the adjacency matrix of a graph G with n vertices, and 𝒀 = 𝑿 + 𝑿𝟐 + 𝑿𝟑 + ⋯ . +𝑿𝒏 , 𝒕𝒉𝒆 𝒓𝒊𝒏𝒈 𝒐𝒇 𝒊𝒏𝒕𝒆𝒈𝒆𝒓𝒔 ∀ 𝒏 ≥ 𝟐 , then G is disconnected if and only if there exists at least one entry in matrix Y that is zero Proof: We have, 𝑦𝑖𝑗 = [𝑌]𝑖𝑗 = 𝑋 𝑖𝑗 + 𝑋 2 𝑖𝑗 + ⋯ + 𝑋 𝑛 −1 𝑖𝑗 Since 𝑋 𝑘 𝑖𝑗 denotes the number of distinct edge-sequences of length k from vi to vj , yij = number of different vi −vj edge sequence of length 1 + number of different vi −vj edge sequences of length 2+. . .+ number of different vi −vj edge sequences of length n−1. Therefore, yij = number of different vi −vj edge sequence of length less than n. Now let G be connected. Then for every pair of distinct i, j there is a path from v i to vj. Since G has n vertices, this path passes through atmost n vertices and so has length less than n. Thus, 𝑦𝑖𝑗 ≠ 0 for each i, j with 𝑖 ≠ 𝑗. Conversely, for each distinct pair i, j we have 𝑦𝑖𝑗 ≠ 0. Then from above, there is at least one edge sequence of length less than n from vi to vj. This implies that vi is connected to vj. Since the distinct pair i, j is chosen arbitrarily, G is connected. Relationship between A(G) and X(G) If a graph G has neither self-loops nor parallel edges( Simple graph), both A(G) and X(G) contains the entire information. ie., X(G)=A(G). Adjacency Matrix and Isomprphism 94 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics Points to Remember Matrix representation of graphs: Adjacency Matrix Incidence Matrix Circuit Matrix Fundamental Circuit Matrix Path Matrix Cut Set Matrix Incidence Matrix Two graphs G1 and G2 are isomorphic if and only if their incidence matrices A(G1) and A(G2) differ only by permutations of rows and columns If A(G) is an incidence matrix of a connected graph G with n vertices, the rank of A(G) is n-1 Rank of A(G) is n-k, if G is disconnected graph with n vertices and k components The reduced incidence matrix of a tree is non-singular. Circuit Matrix If G is a graph without self-loops, with incidence matrix A and circuit matrix B whose columns are arranged using the same order of edges, then every row of B is orthogonal to every row of A, that is ABT = BAT = 0 (mod2), where AT and BT are the transposes of A and B respectively. If B is a circuit matrix of a connected graph G with n vertices and m edges, then rank B = m−n+1. If B is a circuit matrix of a disconnected graph G with n vertices, m edges and k components, then rank B = m−n+k. Cut set Matrix If G is a connected graph, then the rank of a cut-set matrix C(G) is equal to the rank of incidence matrix A(G), which equals the rank of graph G. Path Matrix If the columns of the incidence matrix A and the path matrix P(u, v) of a connected graph are arranged in the same order, then under the product (mod 2). APT (u, v) = M, where M is a matrix having ones in two rows u and v, and the zeros in the remaining n−2 rows Powers of Adjacency matrix Let X be the adjacency matrix of a simple graph G. Then ijth entry in Xr is the number of different edge sequence of r edges between vertices vi and vj In a connected graph, the distance between two vertices vi and vj (for i≠ 𝑗) is k, if and only if k is the smallest integer for which the i,j th entry in xk is nonzero. If X is the adjacency matrix of a graph G with n vertices, and 𝑌 = 𝑋 + 𝑋 2 + 𝑋 3 + ⋯ . +𝑋 𝑛 , 𝑡𝑒 𝑟𝑖𝑛𝑔 𝑜𝑓 𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑠 ∀ 𝑛 ≥ 2 , then G is disconnected if and only if there exists at least one entry in matrix Y that is zero For a simple graph, A(G)=X(G) 95 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics PROBLEMS 1. Write the incidence matrices for the labelled simple graphs shown below. Put the incidence matrix of the graph in the block diagonal form. 2. Show that for a simple disconnected graph of k components, n vertices, and e edges the ranks of matrices A,B, and C are n-k, e-n+k, and n-k respectively 3. Label the edges of the graph in the figure and write down its circuit matrix B. 4. Write down the path matrix P(v1,v6) for the graph in the figure PREVIOUS YEAR UNIVERSITY QUESTIONS 1. Write the properties of Incidence matrices. 2. List down any four properties of adjacency matrix 3. Construct an adjancy matrix(X) for the following graph and also mention how the concept of edge sequencs is described with X3(no need to find X3 from X) 4. Find the incidence matrix of the following graph 5. Define the adjacency matrix and incidence matrix representation of a graph with an example 6. Prove the theorem: If A(G) is an incidence matrix of a connected graph G with n vertices, the rank of A(G) is n-1 7. Describe with examples the usage of incidence matrix to find two graphs (g1 and g2) are isomorphic. 8. Define cut-set matrix with an example and list down any four properties of cut set matrix 96 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics 9. If B is a circuit matrix of a connected graph G with e edges and n vertices, then show that the number of linearly independent rows in B = e-n+1 10. Define the circuit-matrix B(G) of a connected graph G with n vertices and e edges. Prove that the rank of B(G) is e-n+1. 11. Define the adjacency matrix X(G) of a graph. Let X(G) be adjacency matrix of a simple graph G, then prove that ijth entry in Xr is the number of different edge sequences of r edges between vertices vi and vj. 12. Draw the adjacency graph for the following adjacency matrix. 13. Write the fundamental circuit matrix with respect to the spanning tree shown in heavy lines for the following graph. 14. Define the incidence matrix of a graph G.Prove that the rank of an incidence matrix of a connected graph with n vertices is n-1. 15. Draw the graph represented by the following incidence matrix. WORK SHEET 1. Incidence matrix and Adjacency matrix of a graph will always have same dimensions? a) True b) False Ans: 2. The column sum in an incidence matrix for a simple graph is __________ a) depends on number of edges b) always greater than 2 c) equal to 2 d) equal to the number of edges Ans: 3. A matrix whose rows are the rows of the unit matrix, but not necessarily in their natural order, is called: a) adjacency matrix b) adjacency matrix of pseudo graph c) adjacency matrix of simple graph d) permutation matrix Ans: 97 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics 4. An ____________in a graph is a finite alternating sequence of vertices and edges, beginning and ending with vertices, such that each edge is incident on the vertices preceding and following it. Ans: 5. Two graphs are isomorphic, if and only if their vertices can be labelled in such a way that the corresponding adjacency matrices are ___________. Ans: 6. Find the adjacency and incidence matrix of the following graph: 7. Represent the following graphs using adjacent matrices: 8. Represent the incidence matrix for the following graphs: 9. Draw the graphs with the given adjacency matrix: 98 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics 10. Draw the graphs represented by the following incidence matrices: 11. Write the properties of Incidence matrices. 12. List down any four properties of adjacency matrix 13. Construct an adjancy matrix(X) for the following graph and also mention how the concept of edge sequencs is described with X3(no need to find X3 from X) 14. Find the incidence matrix of the following graph 15. Define the adjacency matrix and incidence matrix representation of a graph with an example 16. Prove the theorem: If A(G) is an incidence matrix of a connected graph G with n vertices, the rank of A(G) is n-1 17. Describe with examples the usage of incidence matrix to find two graphs (g1 and g2) are isomorphic. 18. Define cut-set matrix with an example and list down any four properties of cut set matrix 19. If B is a circuit matrix of a connected graph G with e edges and n vertices, then show that the number of linearly independent rows in B = e-n+1 20. Define the circuit-matrix B(G) of a connected graph G with n vertices and e edges. Prove that the rank of B(G) is e-n+1. 99 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor Dept of CSE HandOut and WorkSheet of CS309 Graph Theory And Combinatorics 21. Define the adjacency matrix X(G) of a graph. Let X(G) be adjacency matrix of a simple graph G, then prove that ijth entry in Xr is the number of different edge sequences of r edges between vertices vi and vj. 22. Draw the adjacency graph for the following adjacency matrix. 23. Write the fundamental circuit matrix with respect to the spanning tree shown in heavy lines for the following graph. 24. Define the incidence matrix of a graph G.Prove that the rank of an incidence matrix of a connected graph with n vertices is n-1. 25. Draw the graph represented by the following incidence matrix. s 100 Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor