Geodesic graphs Yunus Akkaya Rommy Marquez Heather Urban

Geodesic graphs Yunus Akkaya Rommy Marquez Heather Urban Marlana Young

Definitions • G = (V, E) ▫ 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 ▫ 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 v ≤x w → x – v – w u ≤x v → 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) □ P 3(H)

K 2 Product proof • Prove, P(x, y) (K □ G) = Px (K) □ Py (G) • Let K be K 2 with vertices a & b • Let G be a connected graph • K 2 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) □ P 3(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.

Work in progress • Generalizing the K 2 proof to fit Kn • Generalizing the proof to fit all graphs

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/~dc 340/EGT 1. 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.