8b-Graphs-C - To prove that G1 and G2 are not isomorphic....

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Unformatted text preview: To prove that G1 and G2 are not isomorphic. To show that graph G1= (V1, E1) and G2 = (V2, E2) are not isomorphic. (Necessary Condition) | V1 | = | V2| and | E1| = | E2|. (Necessary Condition) G1 and G2 have the same degree sequences. (Necessary Condition) For every proper subgraph G* of G, there is a proper subgraph of G2 that is isomorphic to G*. (Necessary Condition) The complement graph of G1 is isomorphic to the complement graph of G2. Remark: The complement graph of a simple graph G has the same vertices as G. Two vertices are adjacent in the complement if and only if they are not adjacent in G. (Necessary Condition) G1 has a simple circuit (path) of length k, the G2 also has a simple circuit (path) of length k. (Necessary Condition) G1 has a cycle of length k, the G2 also has a cycle of length k. Example: To prove that G1 and G2 are not isomorphic. u1 u8 u7 u6 u5 u1 u8 u7 u6 u5 u2 v8 v7 v6 v5 u2 u1 u8 u2 u3 u4 u7 u6 u3 u4 v1 G1 u5 v2 G2 u3 u4 Complement graph of G1 v3 v4 Complement graph of G2 These two complement graphs are not isomorphic because the left complement graph has two cycles of length 4. The right complement graph has only one cycle of length 8. 8.5: Euler & Hamilton Paths Definition 1 (P. 578): An Euler circuit in a graph G is a simple circuit containing every edge of G. An Euler path in G is a simple path containing every edge of G. Definition 1 (P. 583): A Hamilton circuit is a circuit that traverses each vertex in G exactly once. A Hamilton path is a path that traverses each vertex in G exactly once. Leonhard Euler 1707-1783 William Rowan Hamilton 1805-1865 Example 1 (P. 578): Which of the following graphs have an Euler circuits? Which of the following graphs have an Euler path? Solution: G1 has an Euler circuit a, e, c, d, e, b, a Neither G2 or G3 has an Euler circuit. Both of G1 and G3 have an Euler path. e.g. G3 has an Euler path a, c, d, e, b, d, a, b. (Sure, G1 has a Euler path. Why?) G2 has no Euler path. Example 2 (P. 579): Which of the following graphs have an Euler circuits? Which of the following graphs have an Euler path? Solution: H2 has an Euler circuit. e.g. a, g, c, b, g, e, d, f, a Both of H1 and H3 have no Euler circuit. H3 has an Euler path c, a, b, c, d, b. (Sure, H2 has a Euler path.) Some Useful Theorems A connected multigraph has an Euler circuit iff each vertex has even degree. (Theorem 1) A connected multigraph has an Euler path (but not an Euler circuit) iff it has exactly 2 vertices of odd degree. (Theorem 2) If (but not only if) G is connected, simple, has n 3 vertices, and v deg(v) n/2, then G has a Hamilton circuit. It is a dense graph. Seven Bridges Problem The city of Knigsberg (now Kaliningrad) in the East of Prussia lies where the New Pregel and Old Pregel Rivers join to form the Pregel river. The city is built on the island A as well as the parts of mainland B, C, D and in the 18 century there were seven bridges built as indicated. The Problem is " Is it possible to walk, starting at any point in the city, and traverse each bridge once and only once, and return to the original point?" deg(C) = 3 deg(A) = 5 deg(B) = 3 deg(D) = 3 Euler Path Theorems Theorem 1 (P. 581): A connected multigraph has an Euler circuit if and only if each vertex has even degree. Proof: ( ) The circuit contributes 2 to degree of each node. ( ) By construction using algorithm on p. 580-581 Theorem 2 (P. 582): A connected multigraph has an Euler path (but not an Euler circuit) iff it has exactly 2 vertices of odd degree. One of these 2 vertices is the start vertex, the other is the end vertex. Constructing Euler Circuit Begin with any arbitrary node. Construct a simple path from it till you get back to start. Repeat for each remaining subgraph, splicing results back into original cycle. (@e ?) Many puzzles ask you to draw a picture in a continuous motion without lifting a pencil so that no part of the picture is retraced. Example : Find a Euler circuit for the following graph. Solution: It has a Euler circuit since all it vertices have even degree. 1. Suppose we start from v1 and form a circuit v1e7v2e6v5e5v2e8v1 2. Eliminate the circuit v1e7v2e6v5e5v2e8v1 from the original graph. Graph G' which doesn't contain the circuit v1e7v2e6v5e5v2e8v1 Graph G 3. Either v2 or v5 could serves as the new start vertex of the circuit. v1e7v2e6v5e5v2e8v1 Suppose we start from v2 (rewrite the circuit): v2e6v5e5v2e8v1e7v2 The new extension might be v2e3v4e4v2e10v3e9v2, making the larger circuit v2e6v5e5v2e8v1e7v2e3v4e4v2e10v3e9v2 4. Eliminate the circuit v2e3v4e4v2e10v3e9v2 from graph G'. Graph G" which doesn't contain the edges in circuit v2e6v5e5v2e8v1e7v2e3v4e4v2e10v3e9v2 Graph G Graph G' which doesn't contain the edges in circuit v1e7v2e6v5e5v2e8v1 5. In G", v4 is the only vertex connected to the circuit v2e6v5e5v2e8v1e7v2e3v4e4v2e10v3e9v2 , so we rotate the circuit and write it as v4e4v2e10v3e9v2e6v5e5v2e8v1e7v2e3v4 6. The circuit now can be extended by adding v4e1v5e2v4. Thus, all edges are included and the construction is completed. 7. The Euler circuit is v4e4v2e10v3e9v2e6v5e5v2e8v1e7v2e3v4e1v5e2v4. Example 3 (P. 581): Find a Euler circuit for the following graph. (e @ ?) Solution: It has a Euler circuit since all it vertices have even degree. 1. Suppose we start from a and form a circuit a,b,d,c,b,e,i,f,e,a. 2. Eliminate the circuit a,b,d,c,b,e,i,f,e,a from the original graph to form the graph G'. 3. Either d, f, or i could serves as the new start vertex of the circuit. Suppose we start from d (rewrite the circuit): d,c,b,e,i,f,e,a,b,d, 4. The circuit now can be extended by adding d,g,h,j,i,h,k,g,d. Thus, all edges are included and the construction is completed. The Euler circuit is d,c,b,e,i,f,e,a,b,d,g,h,j,I,h,k,g,d. Example : Find a Euler circuit for the following graph. Solution: It has a Euler circuit since all it vertices have even degree. 1. Suppose we start at G and find a Euler circuit G, h, E, d, C, e, F, g, E, j, H, k, G 2. Only E could serves as the new start vertex of the circuit. Rewrite the circuit. E, d, C, e, F, g, E, j, H, k, G, h, E The circuit now can be extended by adding E, c, B, a, A, b, D, f, E. It makes a larger circuit as follows. E, d, C, e, F, g, E, j, H, k, G, h, E, c, B, a, A, b, D, f, E. 3. Only H could serves as the new start vertex of the circuit. Rewrite the circuit. H, k, G, h, E, c, B, a, A, b, D, f, E, d, C, e, F, g, E, j, H. The circuit now can be extended by adding H, m, J, l, H. It makes a larger circuit as follows. H, k, G, h, E, c, B, a, A, b, D, f, E, d, C, e, F, g, E, j, H, m, J, l, H. Example 4 (P.582): Which graphs have an Euler path? Solution: G1 has exactly 2 vertices of odd degree, namely b and d. b and d are the endpoints of the Euler path. one of the paths is d,a,b,c,d,b. G2 has exactly 2 vertices of odd degree, namely b and d. b and d are the endpoints of the Euler path one of the paths is b,a,g,f,e,d,c,g,b,c,f,d. G3 has 6 vertices of odd degree. G3 has no Euler path. Hamilton Circuit & Round-the-World Puzzle Can we traverse all the vertices of a dodecahedron, visiting each once? 12- Dodecahedron puzzle Equivalent graph Pegboard version The game was marketed in 1859, accompanied by a printed leaflet of instructions. It also appeared in a solid dodecahedron from under the title A Voyage Round the World, with the vertices representing cities Brussels, Canton, Delhi, . . . , Zanzibar. Hamilton sold the idea of the Icosian game to a wholesale dealer of games and puzzles for 25. The name Hamiltonian cycle can be regarded as a misnomer, since Hamilton was not the first to look for cycles which pass through every vertex of a graph. Knight's Tour Problem: Can a knight visit each square of a chessboard by a sequence of knight's moves, and finish on the same square as it began? associated graph of the chessboard There is no knight's tour on a 44 chessboard. There is no knight's tour on a nn chessboard where n is odd. In 1759, Euler describe a systematic approach to find the Hamiltonian tour in his paper. The solution is particularly interesting, because if we write the order of the moves, as in the right-hand diagram, we get a magic square in which the number in each row or column have 260. Example 5 (P. 584): Which of the simple graphs have a Hamilton circuit or if not, a Hamilton path? Solution: G1 has a Hamilton circuit a,b,c,d,e,a. G2 has no Hamilton circuit. G2 has a Hamilton path a,b,c,d. G3 has neither a Hamilton circuit nor a Hamilton path. Example 6 (P. 585): Show that neither the following graphs has a Hamilton circuit. Solution: G has a vertex of degree 1, namely e. H has no Hamilton circuit. ( All of vertices a,b,e,d have degree of 2. Every edge incident with these vertices must be part of any Hamilton circuit. (Every Hamilton circuit have to contain all four edges incident with c. Impossible!) Example 7 (P. 585): Kn has a Hamilton circuit whenever n 3. Solution: We can start from any vertex. If (but not only if) G is connected, simple, has n 3 vertices, and v deg(v) n/2, then G has a Hamilton circuit. Hamiltonian Path Theorems Theorem 3 (P. 586) Dirac's theorem: If (but not only if) G is connected, simple, has n 3 vertices, and v deg(v) n/2, then G has a Hamilton circuit. Gabriel Andrew Dirac (19251984) was a Sweden mathematician. dense enough Theorem 4 (P. 586) Ore's theorem: If G is connected, simple, has n 3 nodes, and deg(u) + deg(v) n for every pair u, v of non-adjacent nodes, then G has a Hamilton circuit. ystein Ore (1899 - 1968) was a Norwegian mathematician. Hamilton-Circuit is NP-complete Let HAM-CIRCUIT be the problem: Given a simple graph G, does G contain a Hamiltonian circuit? This problem has been proven to be NP-complete! This means, if an algorithm for solving it in polynomial time were found, it could be used to solve all NP problems in polynomial time. 8.6: Shortest-Path Problems Dijkstra's Shortest Path Algorithm destination origin 1st nearest node A O nearest node B 2nd nearest node rd T How to find the shortest path? TDBA-O EBAO Dijkstra's Algorithm Edsger Wybe Dijkstra 1930-2002 Dijkstra's Shortest Path Algorithm Example 1 (P.595): Find the shortest path from a to z. 2nd neares vertex Solution: Find the 1st nearest vertex: (a d) = min {(a b), (a d)} Find the 2st nearest vertex: (a b) = 4, (a d) = 4, (d e) = 3 (a b) = min { (a d) , 2 + (d e) } b is the 2st nearest vertex. z aa e d b 1st nearest 3rd neare vertex Example 1 (P.597): Find the shortest path from a to z. Example: Find the shortest path from a to z. The Traveling Salesman Problem All permutations 120! n n! (10i n n! (10i ) 0! 1! 2! 3! 4! 5! 6! 7! 8! 9! 7 4 ) 284 567 304 625 12 9 18 16 32 36 59 81 81 121 105 169 132 225 228 441 265 529 367 784 389 841 435 961 483 1089 508 1156 697 1681 726 1764 944 2410 =1 =1 =2 =6 = 24 = 120 = 720 = 5,040 = 40, 320 = 962,880 11! 12! 13! 14! 15! 16! 17! 18! 19! 20! = 39,916,800 = 479,001,600 = 6,227,020,800 = 87, 178,291,200 = 1,307,674,368,000 = 20,922,789,888,000 = 355,687,428,096,000 = 6,402,373,705,728,000 = 121,645,100,408,832,000 = 2,432,902,008,176,640,000 Web Pages The Traveling Salesman Problem http://www.tsp.gatech.edu/ http://www.densis.fee.unicamp.br/~moscato/TSPBIB_home.html http://www.research.att.com/~dsj/chtsp/ Books 1. E. L. Lawler, J.K. Lenstra, A.H. G. Rinnooy Kan, and D.B. Shamoys (ed.), "The Traveling Salesman Problem: A Guided Tour of Combinatorial Optimization" John Wiley & Sons, New York. 2. G. Gutin and A.P. Punnen (eds.), The Traveling Salesman Problem and Its Variations, Kluwer, 2002, 848 pp. USD 85/EUR 93/GBP , http://www.wkap.nl/prod/b/1-4020-0664-0 May, 2004 Keld Helsgaun 24,978 cities in Sweden a tour of length 855,597 Traveling Salesman Problem with 24,978 cities in Sweden Tour Length 855618 855612 855610 855602 855597 Date September 4, 2001 September 20, 2001 September 30, 2001 March 16, 2003 March 18, 2003 algorithm Tour Merging LKH LKH Merge Hybrid Genetic LKH Research Team Cook and Seymour Helsgaun Helsgaun Hung Dinh Nguyen Keld Helsgaun 8.7: Planar Graphs Definition 1 (P.604): A graph is called a planar graph if it can be drawn in the plane without any edges crossing. Such a drawing is called a planar representation of the graph. Example: K3,3 is not planar. Example 1 (P.604): Is K4 planar? Example 2 (P.604): Is Q3 planar? Example 3 (P.604): Is K3,3 planar? Euler's Formula Theorem 1 (P.606) Euler's Formula: G = (V, E) is a connected planar simple graph with |V | = v and |E| = e. Let r be the number of regions in a planar representation of G. Then r = e v + 2. r=6 v=7 e = 11 e v + 2 = 11 7 + 2 = 6 = r Euler's Formula Corollary 1 (P.607): G = (V, E) is a connected planar simple graph with |V | = v 3 and |E| = e. Let r be the number of regions in a planar representation of G. Then e 3v 6. Corollary 2 (P.607): If G = (V, E) is a connected planar simple graph, then G has a vertex of degree not exceeding five. Corollary 3 (P.609): If G = (V, E) is a connected planar simple graph with |V | = v 3 and |E| = e and has no circuit of length 3, then e 2v 4. Theorem: Euler's Formula (P. 606) Let G = (V, E) be a connected, planar simple graph with |V| = v, |E| = e. Let r be the number of regions in a planar embedding of G (including the region on the outside). Then r=ev+2 r = e v +2 r= 4 v=4 e=6 Tetrahedron Cube Octahedron Dodecahedron Example 4 (P.607): Planar simple graph G = (V, E) with |V | = 20, deg(v) = 3, v V. Find r the number of regions in a planar representation of G. Solution: Total # of degree of vertices = 3 20 = 60. 2e = 60 e =30. r = e v + 2 = 30 20 + 2 = 12. Example 5 (P.608): Show that K5 is not planar by using Corollary 1. Proof: If K5 is planar, then the following inequality must hold. e 3v 6 v = 5, e = 10, 3v 6 = 9. The inequality e 3v 6 does not hold. Thus, K5 is not planar. Example 6 (P.609): Show that K3,3 is not planar by using Corollary 3. Proof: K3,3 has no circuit of length of 3. If K3,3 is planar, then the following inequality must hold e 2v 4 v = 6, e = 9, 2v 4 = 8. The inequality e 2v 4 does not hold. Thus, K3,3 is not planar. Homeomorphic elementary subdivision elementary subdivision elementary subdivision homeomorphic Kuratowski's Theorem Theorem 2 (P.610): Kuratowski's Theorem A graph is nonplanar if and only if it conatins a subgraph homeomorphic to K3,3 or K5. Example 7 (P.610): Which of the following graphs is planar? Kazimierz Kuratowski 1896-1980 Example 8 (P.610): Is the Petersen graphs Fig-14(a) planar? Solution: Delete b and the edges that have b as an endpoint. Sphere and Plane In applications, the sphere is the most important surface on which graphs are drawn. Theorem: A graph can be drawn without edge-crossings in the plane if and only if it can be drawn without edge-crossings in the sphere. Mathematically, the sphere (and plane) are by far the easiest surfaces for graph drawing problem. Theorem (Jordan Curve Theorem): Every closed curve in the sphere (plane) separates the sphere (plane) into two regions. Theorem (Schnfliess) Each side of the separation of the sphere by a closed curve is topologically equivalent to a disk. Remark: The Schnfliess Theorem does not hold in dimensions greater than two. 8.8: Graph Coloring Definition 1 (P.614): A coloring of a graph is the assignment of a color to each vertex so that no two adjacent vertices are assigned the same color. Definition 2 (P.614): The chromatic number of a graph is the minimum number of colors needed for a coloring of this graph. Four Color Problem 3 1 2 6 4 4 5 3 1 5 2 6 The Four Color Problem originated in 1852, and was finally solved in 1976, when Appel and Haken showed it was true indeed that every map could be colored with four or fewer colors. It might be seem that this should have ended interest in the Four Color Problem; however, such was not the case, due to the unusual nature of the solution (proof). Appel and Haken solved the Four 3 Color Problem by dividing the problem into nearly two thousand cases, according the to the arrangements of countries within a map. To determine the 5 possible w...
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This note was uploaded on 05/22/2008 for the course INFO 3720 taught by Professor Gomes during the Spring '07 term at Cornell University (Engineering School).

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