GRAPH THEORY LECTURE 3 STRUCTURE AND REPRESENTATION PART B
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1 GRAPH THEORY LECTURE 3 STRUCTURE AND REPRESENTATION PART B Abstract. We continue 2.3 on subgraphs. 2.4 introduces some basic graph operations. 2.5 describes some tests for graph isomorphism. Outline 2.3 Subgraphs, continued 2.4 Some Graph Operations 2.5 Tests for Non-Isomorphism 2.6 Matrix Representations (read text) 2.7 More Graph Operations (read text) 1
2 2 GRAPH THEORY LECTURE 3 STRUCTURE AND REPRESENTATION PART B 3. Subgraphs, continued Def 3.1. The center of a graph G, denoted Z(G), is the subgraph induced on the set of central ertices of G (see 1.4). Example Each ertex of the graph in Fig 3.8 is labeled by its eccentricity Figure 3.8: A graph whose center is a 7-cycle. Since the minimum eccentricity is 3, the ertices of eccentricity 3 lie in the center. Remark 3.1. In 2.4, we show that any graph can be the center of a graph.
3 GRAPH THEORY LECTURE 3 STRUCTURE AND REPRESENTATION PART B 3 Local Subgraphs Def 3.2. The (open) local subgraph (or (open) neighborhood subgraph) of a ertex is the subgraph L() induced on the neighbors of. Example Figure 3.9 shows a graph and the local subgraphs of three of its ertices. G u w u u u w w w L (u ) L( ) L(w ) Figure 3.9: A graph G and three of its local subgraphs.
4 4 GRAPH THEORY LECTURE 3 STRUCTURE AND REPRESENTATION PART B Theorem 3.3. Let f : G H be a graph isom and u V G. Then f maps the local subgraph L(u) of G isomorphically to the local subgraph L(f(u)) of H. Proof. This follows, since f maps V L(u) bijectiely to V L(f(u)), and since f is edge-multiplicity presering. Example In Fig 3.10, is G = H? G H Figure 3.10: Two 4-regular 8-ertex graphs. Consider Each local subgraph for G of Fig 3.10 is isom to 4K 1. Each local subgraph for H is isom to P 4. Thus, G = H. Alternatiely, we obsere that α(g) = 4, but α(h) = 2, and also that ω(g) = 2, but ω(h) = 3.
5 GRAPH THEORY LECTURE 3 STRUCTURE AND REPRESENTATION PART B 5 Components Def 3.3. A component of G is a maximal conn subgraph of G. Example The 7-ertex graph in Figure 3.11 has four components. Figure 3.11: A graph with four components. Reiew from 1.4 A ertex is said to be reachable from ertex u if there is a walk from u to. Def 3.4. In a graph G, the component of a ertex, denoted C(), is the subgraph induced by the subset of all ertices reachable from. Notation The # of components of a graph G is denoted c(g). Computational Note: Partitioning a small graph into components is triial. But larger graphs that are specified by some computer representation require a computer algorithm. These component-finding algorithms are deeloped in Chapter 4 as straightforward applications of our graph-traersal procedures.
6 6 GRAPH THEORY LECTURE 3 STRUCTURE AND REPRESENTATION PART B Deleting Vertices or Edges 4. Some Graph Operations Def 4.1. The ertex-deletion subgraph G is the subgraph induced by the ertex-set V G {}. That is, V G = V G {} and E G = {e E G : / endpts(e)} c c a b u d h f g a b u G w G-w Figure 4.1: The result of deleting ertex w from graph G.
7 GRAPH THEORY LECTURE 3 STRUCTURE AND REPRESENTATION PART B 7 Def 4.2. The edge-deletion subgraph G e is the subgraph induced by the edge-set E G {e}. That is, V G e = V G and E G e = E G {e} a b u d c h f g a b u d c h g G w G-f w Figure 4.2: The result of deleting edge f from graph G.
8 8 GRAPH THEORY LECTURE 3 STRUCTURE AND REPRESENTATION PART B Network Vulnerability Def 4.3. A ertex-cut in a graph G is a ertex-set U such that G U has more components than G. Def 4.4. A cut-ertex (or cutpoint) is a ertex-cut consisting of a single ertex. u w x y z Figure 4.3: A graph with four cut-ertices.
9 GRAPH THEORY LECTURE 3 STRUCTURE AND REPRESENTATION PART B 9 Def 4.5. An edge-cut in a graph G is a set of edges D such that G D has more components than G. Def 4.6. A cut-edge (or bridge) is an edge-cut consisting of a single edge. b a r t s c Figure 4.4: A graph with three cut-edges.
10 10 GRAPH THEORY LECTURE 3 STRUCTURE AND REPRESENTATION PART B The Graph-Reconstruction Problem Def 4.7. Let G be a graph with V G = { 1, 2,..., n }. Then the ertexdeletion subgraph list of G is the list of the subgraphs G 1,..., G n Def 4.8. The reconstruction deck of G is its ertex-deletion subgraph list, with no labels on the ertices. We regard each indiidual ertex-del subgraph as being a card in the deck. u G y x w G-u G- G-w G-x G-y Figure 4.6: A graph and its ertex-deletion subgraph list. Def 4.9. The graph-reconstruction problem is to decide whether two non-isomorphic simple graphs with three or more ertices can hae the same reconstruction deck.
11 GRAPH THEORY LECTURE 3 STRUCTURE AND REPRESENTATION PART B 11 Remark 4.1. We obsere that the 2-ertex graphs K 2 and 2K 1 (two nonadjacent ertices) hae the same deck. K 2 2K 1 Remark 4.2. The graph-recon problem would be easy to sole if the ertexdel subgraphs included the ertex and edge names. Example 4.5. For graph G of Fig 4.6, the graph-recon problem asks whether G is the only graph (up to isomorphism) that has the following deck. Figure 4.7: Is G the only graph with this deck?
12 12 GRAPH THEORY LECTURE 3 STRUCTURE AND REPRESENTATION PART B Reconstruction Conjecture: Let G and H be two graphs with V G = { 1, 2,..., n } and V H = {w 1, w 2,..., w n } with n 3, and with the same reconstruction deck, i.e., such that Then G = H. G i = H wi for each i = 1..., n The following results (next page) indicate that some information about the original graph can be deried from its ertex-deletion subgraph list. This information is a big help in soling small reconstruction problems.
13 GRAPH THEORY LECTURE 3 STRUCTURE AND REPRESENTATION PART B 13 Theorem 4.4. The # of ertices and the # of edges of a graph G can be calculated from its ertex-deletion subgraph list. Proof. If the deck has n cards, then clearly V G = n. Moreoer, each edge appears only in the n 2 subgraphs that do not contain either of its endpoints. It follows that the sum E G counts each edge n 2 times. Thus, 1 E G = E G n 2 Example 4.5, continued: For the deck in Figure 4.6, reproduced immediately below, the alue of the sum is 1( ) = 3 3 = 9 which is indeed the number of edges in the graph G = K 5 e of Figure 4.6, from which the deck was generated.
14 14 GRAPH THEORY LECTURE 3 STRUCTURE AND REPRESENTATION PART B Corollary 4.5. The degree sequence of a graph G can be calculated from its reconstruction deck. Proof. First calculate E G. For each card, the degree of the missing ertex is the difference between E and the number of edges on that card. Example 4.5, continued: For the deck in Figure 4.7, the degrees are 4 = = = = = 9 6
15 GRAPH THEORY LECTURE 3 STRUCTURE AND REPRESENTATION PART B 15 Corollary 4.6. Any regular graph can be reconstructed from its reconstruction deck. Proof. By Cor 4.5, we could determine from the deck that the graph is regular (all the cards would hae the same number of edges) and its degree d of regularity. On any card in the deck, there would be d ertices of degree d 1. To reconstruct the graph, simply join those ertices to the missing ertex. Remark 4.3. B. McKay [Mc77] and A. Nijenhuis [Ni77] hae shown, with the aid of computers, that a counterexample to the reconstruction conjecture would hae to hae at least 10 ertices.
16 16 GRAPH THEORY LECTURE 3 STRUCTURE AND REPRESENTATION PART B Adding Edges or Vertices Def Adding an edge between two ertices u and w of a graph G means creating a supergraph, denoted G {e}, with ertex-set V G and edgeset E G {e}, where e is a new edge with endpoints u and w. a b u c d h f g + e a b u c d h f g w e w Figure 4.8: Adding an edge e with endpoints u and w. Def Adding a ertex to a graph G, where is a new ertex not already in V G, means creating a supergraph, denoted G {}, with ertex-set V G {} and edge-set E G.
17 GRAPH THEORY LECTURE 3 STRUCTURE AND REPRESENTATION PART B 17 Graph Union The iteratie application of ertex addition results in the secondary graph operation called graph union. Def The (graph) union of two graphs G = (V, E) and G = (V, E ) is the graph G G, whose ertex-set and edge-set are the disjoint unions, respectiely, of the ertex-sets and edge-sets of G and G. Example 4.9. Figure 4.9: The graph union K 3 K 3 of two copies of K 3. Def The n-fold self-union ng is the iterated disjoint union G G of n copies of the graph G. (e.g. Fig 4.9 is 2K 3.)
18 18 GRAPH THEORY LECTURE 3 STRUCTURE AND REPRESENTATION PART B Joining a Vertex to a Graph Def If a new ertex is joined to each ertex of a graph G, then the resulting graph is called the join of to G or the suspension of G from, and is denoted G +. Def The n-wheel W n is the join K 1 + C n of a single ertex and an n-cycle. (The n-cycle forms the rim of the wheel, and the additional ertex is its hub.) If n is een, then W n is called an een wheel; if odd, then W n is called an odd wheel. + = Figure 4.10: The 5-wheel W 5 = K 1 + C 5. The next proposition hints at the general usefulness of the join construction.
19 GRAPH THEORY LECTURE 3 STRUCTURE AND REPRESENTATION PART B 19 Proposition 4.7. Let H be a graph of diameter d. graph G of radius d of which H is the center. Then there is a Proof. First construct a supergraph Ĥ by joining two new ertices u and to eery ertex of H, but not to each other; that is, Ĥ = ((H + u) + ) u Then form the supergraph G by attaching new ertices u 1 and 1 to ertices u and, respectiely, with edges uu 1 and 1, as in Figure Thus, G = (Ĥ 2K 1) + uu where the two copies of K 1 are the two triial graphs consisting of ertices u 1 and 1. It is easy to erify that, in graph G, each ertex of H has eccentricity 2, ertices u and hae eccentricity 3, and u 1 and 1 hae eccentricity 4. Thus, H is the center of G. u u 1 G H 1 Figure 4.11: Making graph H the center of a graph G.
20 20 GRAPH THEORY LECTURE 3 STRUCTURE AND REPRESENTATION PART B Edge-Complementation Def Let G be a simple graph. Its edge-complement (or complement) G has the same ertex-set, but two ertices are adjacent in G iff they are not adjacent in G. G G Figure 4.12: A graph and its complement. Remark 4.4. The edge-complement of the edge-complement is the original graph, i.e., G = G. The following theorem formulates precisely the sense in which the independence number of a graph and the clique number are complementary. Theorem 4.8. Let G be a simple graph. Then ω(g) = α(g) and α(g) = ω(g) Remark 4.5. Other graph operations are defined in 2.7.
21 GRAPH THEORY LECTURE 3 STRUCTURE AND REPRESENTATION PART B Tests for Non-Isomorphism Def 5.1. A graph inariant (or digraph inariant) is a property of graphs (digraphs) that is presered by isomorphisms. We established in 2.1 that the number of ertices, the number of edges, and the degree sequence are the same for any two isomorphism graphs, so they are graph inariants. A Local Inariant The following theorem proides additional necessary criteria for isomorphism. Theorem 5.1. Let f : G H be a graph isomorphism, and let V G. Then the multiset of degrees of the neighbors of equals the multiset of degrees of the neighbors of f(). Proof. Immediate consequence of Cor w G H Figure 5.1: Non-isom graphs with the same degree seq.
22 22 GRAPH THEORY LECTURE 3 STRUCTURE AND REPRESENTATION PART B Distance Inariants Def 5.2. Let W = 0, e 1, 1,..., e n, n be a walk in the domain G of a graph isom f : G H. Then the image of walk W is the walk f(w ) = f( 0 ), f(e 1 ), f( 1 ),..., f(e n ), f( n ) in the codomain graph H. Theorem 5.2. The isomorphic image of a graph walk W is a walk of the same length. Proof. This follows directly from the edge-mult-presering property of an isomorphism.
23 GRAPH THEORY LECTURE 3 STRUCTURE AND REPRESENTATION PART B 23 Corollary 5.3. The isomorphic image of a trail, path, or cycle is a trail, path, or cycle, respectiely, of the same length. Proof. This is an immediate consequence of Thm 5.2 and the bijectiity of the isomorphism. Corollary 5.4. For each integer l, two isomorphic graphs must hae the same # of trails (paths) (cycles) of length l f(010) additional classroom example f(000) f(110) f(100) f(111) f(011) 010 f(101) f(001) Corollary 5.5. The diameter, the radius, and the girth are graph inariants. Proof. This is an immediate consequence of Cor 5.3 and the bijectiity of the isomorphism.
24 24 GRAPH THEORY LECTURE 3 STRUCTURE AND REPRESENTATION PART B Example 5.4. Figure 5.2 shows the circular ladder CL 4 and the Möbius ladder ML 4, which are 8-ertex graphs. Since both graphs are 3-regular, it is pointless to apply isomorphism tests based on ertex degree CL 4 ML 4 3 Figure 5.2: Non-isom 3-regular graphs. In CL 4, we obsere two kinds of symmetry: rotational symmetry; the isom that swaps the inner and outer cycles. For ML 4, we obsere that j j + 1 mod 8 is an automorphism. Thus both graphs are ertex-transitie. The max distance from ertex 0 in CL 4 is 3, to ertex 6. The max distance from ertex 0 in ML 4 is 2. Thus, they are not isomorphic. We may also obsere that α(cl 4 ) = 4, but α(ml 4 ) = 3.
25 GRAPH THEORY LECTURE 3 STRUCTURE AND REPRESENTATION PART B 25 Subgraph Presence Theorem 5.6. The number of distinct subgraphs of a gien isom type within a graph is a graph inariant. Proof. Let f : G 1 G 2 be a graph isomorphism and H a subgraph of G 1. Then f(h) is a subgraph of G 2, of the same isomorphism type as H. The bijectiity of f establishes the inariant. Example 5.5. Whereas A and C hae no K 3 subgraphs, B has two, D has four, and E has one. Thus, Thm 5.6 implies that the only possible isomorphic pair is A and C. Howeer, graph C has a 5-cycle, but graph A (bipartite) does not. A B C D E Figure 5.3: Fie mutually non-isom, 8-ertex, 3-reg graphs. Alternatiely, we see that A and C are the only pair with the same multiset of local subgraphs. We could distinguish this pair by obsering that α(a) = 4 and α(c) = 3.
26 26 GRAPH THEORY LECTURE 3 STRUCTURE AND REPRESENTATION PART B Edge-Complementation Theorem 5.7. Let G and H both be simple graphs. They are isomorphic iff their edge-complements are isomorphic. Proof. By definition, a graph isomorphism necessarily preseres non-adjacency as well as adjacency. Example 5.6. The edge-complement of the left graph consists of two disjoint 4-cycles, and the edge-complement of the right graph is an 8-cycle. 2C 4 C 8 Figure 5.4: Two relatiely dense, non-isomorphic 5-regular graphs and their edgecomplements.
27 GRAPH THEORY LECTURE 3 STRUCTURE AND REPRESENTATION PART B Supplementary Exercises Exercise 10 Show that the two graphs in Fig 6.1 are not isom. Figure 6.1: Exercise 11 Decide which pairs are isomorphic. A B C Figure 6.2: Exercise 19 Reconstruct the graph whose deck is in Fig 6.3. Figure 6.3:
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