Yunus Akkaya Rommy Marquez Heather Urban Marlana Young Geodesic graphs Yunus Akkaya Rommy Marquez Heather Urban Marlana Young

Definitions G = (V,E) Bipartite graph V = the set of all vertices in G EXAMPLE: V={A,B,C,D} E= the set of all edges in G EXAMPLE: E={(A,B), (A,C), (B,C), (B,D), (C,D)} Bipartite graph a connected graph whose vertices can be divided into two sets, where the only edges are from one set to the other. Characterized as not having odd cycles

Definitions Geodesic Distance Diameter Antipodal vertices The shortest path between any two vertices. Distance d(x,y) The length of a geodesic Diameter d(G) Length of the longest geodesic Antipodal vertices d(x,y) = d(G) Are the endpoints of the diameter

Helpful Information G is a connected graph If x, u and v є G, then u ≤x v → u is on a x-v geodesic poset = a set of vertices that are partially ordered

Poset Theorem V = the set of all the vertices in G Theorem: V + the relation ≤x = a poset, (V, ≤x ) Prove ≤x is a partial order ≤x is reflexive ≤x is antisymmetric ≤x is transitive

≤x is reflexive G is a connected graph u must be in x-u geodesic Since u ≤x u is true, ≤x is reflexive

≤x is antisymmetric Prove u ≤x v and v ≤x u → u = v u ≤x v → x – u – v d(x,u) ≤ d(x,v) v ≤x u → x – v – u d(x,v) ≤ d(x,u) d(x,v) = d(x,u) u and v are the same distance away from x & they are in each others geodesic u = v ≤x is antisymmetric

≤x is transitive Prove u ≤x v and v ≤x w → u ≤x w u ≤x v → x – u – v v ≤x w → x – v – w x – u – v – w Therefore, u ≤x w ≤x is transitive

Geodesic graph (V, ≤x ) is a poset Geodesic graph Px (G) All vertices that are found in graph G are also found in Px (G) Px(G) is defined to be the Hasse diagram of (V, ≤x ) x and y є V edge between x and y if x < y or y < x & there is no vertex between them

Basic geodesic graph

Basic geodesic graph

Double Geodesic From a connected graph G, find Px (G) Since Px (G) contains all the shortest paths between x and all other vertices: Px (Px(G)) = Px (G)

Properties of a Geodesic Graph For any graph G: Px(G) is bipartite for all x є G If the graph G is bipartite or an even cycle: Px(G) = G for all x є G If the graph G is an odd cycle: Px(G), for all x є G, will be missing an edge e є E(G)

The Product of Two Graphs G □ H = Cartesian product of two graphs G and H Vertices are in the form (x,y) for all x є G and for all y є H. (x,y) is adjacent to (u,v) in G □ H (where x & u є G and y & v є H), iff: x is adjacent to u in G and y = v y is adjacent to v in H and x = u

Example:

Open Question We looked to prove that for any graphs H and G: Px (G) □ Py (H) = P(x,y) (G □ H) As was suggested by our advisor Dr. Louis Friedler

Bipartite product proof Prove, Px (G) □ Py (H) = P(x,y) (G □ H) Let G & H be bipartite graphs By the properties of a bipartite geodesic graph: Px(G) = G & Py(H) = H where x є G & y є H The Cartesian product G □ H is bipartite P(x,y)(G □ H) must also be bipartite Hence, Px (G) □ Py (H) = P(x,y) (G □ H)

Example: PC(G) □ P3(H)

P(C,3)(G □ H)

K2 Product proof Prove, P(x,y) (K □ G) = Px (K) □ Py (G) Let K be K2 with vertices a & b Let G be a connected graph K2 is bipartite, so we know by the properties of bipartite graphs that Px (K) = K for all x є K Two cases for G: Case 1: If G is bipartite we know by the Bipartite Product Proof that P(x,y) (K □ G) = Px (K) □ Py (G) must be true for all x є K and all y є G.

Case 2: Px (K) □ Py (G) If G has an odd cycle we know from the properties of geodesic graphs that an edge e є E(G) will not be in Py (G) . K □ G have vertices of the form (a,c) and (b,c) for all c є G. Hence, Px (K) □ Py (G) will only be missing the edges ((a,c),(a,d)) and ((b,c),(b,d)), where c and d are the endpoints of the edge e.

Example: PA(K) □ P3(G)

Case 2 continued… P(x,y) (K □ G) K □ G will only have an odd cycle if it is within G As before, we know that an edge e є E(G) will not be in Py (G) . Therefore, P(x,y) (K □ G) is missing the edges ((a,c),(a,d)) and ((b,c),(b,d)), where c and d are the endpoints of edge e.

Example: P(A,3)(K □ G)

Conclusion Hence, P(x,y) (K □ G) = Px (K) □ Py (G) is true for all G.

Final thoughts We believe Px (G) □ Py (H) = P(x,y) (G □ H) is true for all G and H. Closely related questions involve the geodetic number of Cartesian graph products. The geodetic number is the length of the diameter. Yunus

Work in progress Generalizing the K2 proof to fit Kn Generalizing the proof to fit all graphs Yunus

Questions?

References [1] Bresar, Bostjan; Klavzar, Sandi; Horvat, A.T., On the geodetic number and related metric sets in Cartesian product graphs, Mathematics Subject Classification (2000). [2] Conlon, David, Extremal graph theory, <http://www.dpmms.cam.ac.uk/~dc340/EGT1.pdf>, 2011 Oct. 11. [3] Gross, Jonathan L.; Yellen, Jay, Handbook of Graph Theory, CRC Press, Boca Raton, 2004. [4] Santhakumaran, A. P.; Titus, P., Geodesic Graphs, ARS Combinatoria (2011), 75-82. [5] V. G. Vizing, The Cartesian product of graphs, Vyc. Sis. 9 (1963), 30–43. [6] West, Douglas B., Introduction to Graph Theory, (Second Edition), Prentice Hall, Upper Saddle River NJ, 2001.