# Uniquely C 4 -Saturated Graphs

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1 Uniquely C 4 -Saturated Graphs Joshua Cooper, John Lenz, Timothy D. LeSaulnier, Paul S. Wenger, Douglas B. West January 26, 2010 Abstract For a fixed graph H, a graph G is uniquely H-saturated if G does not contain H, but the addition of any edge from G to G completes exactly one copy of H. Using a combination of algebraic methods and counting arguments, we determine all the uniquely C 4 -saturated graphs; there are only ten of them. 1 Introduction For a fixed graph H, a graph G is H-saturated if G does not contain H but joining any nonadjacent vertices produces a graph that does contain H. Let P n, C n, K n denote the path, cycle, and complete graph with n vertices, respectively. The study of H-saturated graphs began when Turán [5] determined the n-vertex K r -saturated graphs with the most edges. In the opposite direction, Erdős, Hajnal, and Moon [1] determined the n-vertex K r - saturated graphs with the fewest edges. A survey of results and problems about the smallest n-vertex H-saturated graphs appears in [4]. A graph G is uniquely H-saturated if G is H-saturated and the addition of any edge joining nonadjacent vertices completes exactly one copy of H. The graphs found in [1] are uniquely K r -saturated. For example, consider H = C 3. Every C 3 -saturated graph has diameter at most 2. All trees with diameter 2 are stars and are uniquely C 3 -saturated. A uniquely C 3 - saturated graph G cannot contain a 3-cycle or a 4-cycle, so such a graph that is not a tree Research of all authors partially supported by NSF grant DMS EMSW21-MCTP: Research Experience for Graduate Students. Mathematics Dept., Univ. of South Carolina, Columbia, SC; Mathematics Dept., Univ. of Illinois, Urbana, IL; addresses Research supported by NSA grant H

2 has girth 5. Every graph with girth 5 and diameter 2 is uniquely C 3 -saturated. The graphs with diameter d and girth 2d + 1 are the Moore graphs. Hoffman and Singleton [2] proved that besides odd cycles there are only finitely many Moore graphs, all having diameter 2. Thus, except for stars, there are finitely many uniquely C 3 -saturated graphs. Ollmann [3] determined the C 4 -saturated n-vertex graphs with the fewest edges, but few of these are uniquely C 4 -saturated. An exception is the triangle K 3 ; whenever n < V (H), vacuously K n is uniquely H-saturated. In this paper we determine all the uniquely C 4 - saturated graphs. Theorem 1. There are precisely ten uniquely C 4 -saturated graphs. In the list, the only example with girth 5 is the 5-cycle. The others are small trees or contain triangles; all have at most nine vertices. The sense in which uniquely C k -saturated graphs can be viewed as generalizing the Moore graphs of diameter 2 is reflected in our proof. The structure and techniques of the paper are very similar to the eigenvalue approach used to prove both the Hoffman-Singleton result on Moore graphs and the Friendship Theorem, which states that a graph in which any two distinct vertices have exactly one common neighbor has a vertex adjacent to all others (see Wilf [7]). Structural arguments are used to show that under certain conditions the graphs in question are regular. Counting of walks then yields a polynomial equation involving the adjacency matrix, after which eigenvalue arguments exclude all but a few graphs. The graphs that result from the Friendship Theorem consist of some number of triangles sharing a single vertex; such graphs are uniquely C 5 -saturated. Thus, unlike for C 4, there are infinitely many uniquely C 5 -saturated graphs. Wenger [6] has shown that except for small complete graphs, the friendship graphs are the only uniquely C 5 -saturated graphs. 2 Structural Properties Our graphs have no loops or multi-edges. A k-cycle is a cycle with k vertices, and we define a k-path to be a k-vertex path. A path with endpoints x and y is an x,y-path. For a vertex v in a graph G, the neighborhood N(v) is {u V (G): uv E(G)}. The kth neighborhood N k (v) is {u V (G): d(u,v) = k}, where the distance d(u,v) is the minimum length of a u, v-path. The diameter of a graph is the maximum distance between vertices. The degree d(v) of a vertex v in a graph G is the number of incident edges. We begin with basic observations about the structure of uniquely C 4 -saturated graphs. 2

3 Lemma 2. The following properties hold for every uniquely C 4 -saturated graph G. (a) G is connected and has diameter at most 3. (b) Any two nonadjacent vertices in G are the endpoints of exactly one 4-path. (c) G contains no 6-cycle and no two triangles sharing a vertex. Proof. If x and y are nonadjacent vertices in G, then the edge xy completes a 4-cycle. Thus G contains an x,y-path of length 3. Since G is uniquely C 4 -saturated, x and y are the endpoints of exactly one 4-path. Opposite vertices on a 6-cycle would be the endpoints of two 4-paths if nonadjacent and would lie on a 4-cycle if adjacent. The same is true for nonadjacent vertices in the union of two triangles sharing one vertex. The union of two triangles sharing two vertices contains a 4-cycle. Lemma 3. If G is uniquely C 4 -saturated and V (G) 3, then G has girth 3 or 5. Proof. If G contains a triangle, then G has girth 3, so we may assume that G is triangle-free. Hence there are vertices x and y with d(x,y) = 2; let z be their unique common neighbor. By Lemma 2, there is a 4-path joining x and y. If it contains z, then G contains a triangle. Otherwise, x and y lie on a 5-cycle. Since G is C 4 -free, it follows that G has girth 5. If G has maximum degree at most 1, then G is K 1 or K 2, and these are uniquely C 4 - saturated. We may assume henceforth maximum degree at least 2. Lemma 3 then allows us to break the study of uniquely C 4 -saturated graphs into two cases: girth 3 and girth 5. 3 Girth 5 Lemma 4. If G is a uniquely C 4 -saturated graph with girth 5, then G is regular. Proof. Let u and v be adjacent vertices, with d(u) d(v). Since G is triangle-free, N(v) is an independent set, and hence the 4-paths joining neighbors of v do not contain v. If d(u) < d(v), then by the pigeonhole principle two of the unique 4-paths from u to the other d(v) 1 neighbors of v begin along the same edge uu incident to u. Each of these two paths continues along an edge to v to form distinct 4-paths from u to v. Since N(v) is independent, u is not adjacent to v, so this contradicts Lemma 2. We conclude that adjacent vertices in G have the same degree. Since G is connected, it follows that G is k-regular. We now show that exactly one uniquely C 4 -saturated graph has girth 5. 3

4 Theorem 5. The only uniquely C 4 -saturated graph with girth 5 is C 5. Proof. Let G be a uniquely C 4 -saturated n-vertex graph with girth 5. By Lemma 4, G is regular; let k be the vertex degree. Let A be the adjacency matrix of G, let J be the n-by-n matrix with every entry 1, and let 1 be the n-vector with each coordinate 1. If x and y are nonadjacent vertices of G, then by Lemma 2 there is one x,y-path of length 3 and no other walk of length 3 joining x and y. If x and y are adjacent, then there are 2k 1 walks of length 3 joining them. If x = y, then no walk of length 3 joins x and y, because G is triangle-free. This yields A 3 = (J A I) + (2k 1)A, or J = A 3 (2k 2)A + I. Because J is a polynomial in A, every eigenvector of A is also an eigenvector of J. Since G is k-regular, 1 is an eigenvector of A with eigenvalue k. Also 1 is an eigenvector of J with eigenvalue n. This yields the following count of the vertices of G: n = k 3 (2k 2)k + 1 = k 3 2k 2 + 2k + 1. We have observed that every eigenvector of A is also an eigenvector of J. Since J has rank 1, we conclude that Jx = 0x when x is an eigenvector of A other than 1. If λ is the corresponding eigenvalue of A, then J = A 3 (2k 2)A + I yields 0 = λ 3 (2k 2)λ + 1. (1) It follows that A has at most three eigenvalues other than k. Let q denote the polynomial in (1). Being a cubic polynomial, it factors as q(λ) = λ 3 (2k 2)λ + 1 = (λ r 1 )(λ r 2 )(λ r 3 ). (2) It follows that r 1 + r 2 + r 3 = 0. (3) Suppose first that two of these roots have a common value, r. From (3), the third is 2r, and we have λ 3 (2k 2)λ + 1 = (λ r) 2 (λ + 2r) = λ 3 3r 2 λ + 2r 3. By equating coefficients, r equals both (1/2) 1/3 (irrational) and (2k 2)/3 (rational). Hence q has three distinct roots. Suppose next that q has a rational root. The Rational Root Theorem implies that 1 and 1 are the only possible rational roots of q. If 1 is a root, then k = 1 and G does not have girth 5. If 1 is a root, then k = 2 and G = C 5. 4

5 Hence we may assume that q has three distinct irrational roots. In this case we will obtain a contradiction. Index the eigenvalues so that the multiplicities a, b, and c of r 1, r 2, and r 3 (respectively) satisfy a b c. Letting p A be the characteristic polynomial of A, Combining (2) and (4) yields p A (λ) = (λ k)(λ r 1 ) a (λ r 2 ) b (λ r 3 ) c. (4) p A (λ) = (λ k)(λ 3 (2k 2)λ + 1) a (λ r 2 ) b a (λ r 3 ) c a. Because A has integer entries, p A (λ) Q[λ]. By applying the division algorithm, p = rs and p,r Q[λ] imply s Q[λ]. Hence (λ r 2 ) b a (λ r 3 ) c a Q[λ]. Since q(λ) is a monic cubic polynomial in Q[λ] with three irrational roots, it is irreducible and is the minimal polynomial of r 1, r 2, and r 3 over Q. Thus q divides (λ r 2 ) b a (λ r 3 ) c a if c > a. In that case, since r 1 is a root of q, it is also a root of (λ r 2 ) b a (λ r 3 ) c a. We conclude that c = a, and all three eigenvalues have the same multiplicity. The trace of A is 0, so k + ar 1 + ar 2 + ar 3 = k + a(r 1 + r 2 + r 3 ) = Tr(A) = 0. (5) Together, (3) and (5) require k = 0. Thus q cannot have three distinct irrational roots when G has girth 5. 4 Girth 3 We now consider uniquely C 4 -saturated graphs with a triangle. The next lemma gives a structural decomposition. For a set S V (G), let d(x,s) = min{d(x,v): v S}, let N(S) = {v V (G): d(v,s) = 1}, and let N k (S) = {v V (G): d(v,s) = k}. Lemma 6. Let S be the vertex set of a triangle in a graph G, with S = {v 1,v 2,v 3 }. For i {1, 2, 3}, let V i = N(v i ) S, and let V i = N 2 (v i ) N(S). Let R = N 3 (S). If G is uniquely C 4 -saturated, then G has the following structure: (a) V i V j = when i j; (b) each vertex in V i has exactly one neighbor in V i ; (c) V i V j = when i j; (d) no edges join V i and V j when i j; (e) N(S) is independent; (f) each V i induces a matching; (g) each vertex in R has exactly one neighbor in each V i. 5

6 v 1 V 1 V 1 v 2 V 2 V 2 R v 3 V 3 V 3 Figure 1: Structure of uniquely C 4 -saturated graph with a triangle. Proof. Since G has diameter 3, we have described all of V (G). Figure 1 makes it easy to see most of the conclusions. The prohibition of 4-cycles and of triangles with common vertices implies (a), (b), and (e). The prohibition of 6-cycles implies (c) and (d). Given these results, (f) is implied by the existence of a unique 4-path joining v i to each vertex of V i. For (g), each vertex in R is joined by a unique 4-path to each vertex in S; it can only reach v i quickly enough by moving first to a vertex of V i, and uniqueness of the 4-path prohibits more than one such neighbor. The main part of the argument is analogous to the regularity, walk-counting, and eigenvalue arguments in Lemma 4 and Theorem 5. Theorem 7. If G is a C 4 -saturated graph with a triangle, then R = in the partition of V (G) given in Lemma 6. Proof. If R, then each set V i and V i in the partition is nonempty. We show first that G is regular, then show that each vertex lies in one triangle, and finally count 4-paths to determine the cube of the adjacency matrix and obtain a contradiction using eigenvalues. Consider V i and V j with i j. A vertex x in V i reaches each vertex of V j by a unique 4-path, passing through R and V j. By Lemma 6(g), each vertex of R has one neighbor in V j, so each edge from x to R starts exactly one 4-path to V j. By Lemma 6, the other neighbors of x are one each in V i and V i, so d(x) = V j + 2. Since the choice of i and j was arbitrary, we conclude that each vertex of N 2 (S) S has degree a + 2, where a = V 1 = V 2 = V 3. For x V i and y V j with j i, the unique 4-path joining x to any neighbor of y in V j must pass through V i and R. By Lemma 6(g), these paths use distinct vertices in R; since 6

7 G has no 6-cycle through y, they also use distinct vertices in V i. Hence d(x) d(y). By symmetry, all vertices of N(S) have the same degree; let this degree be b + 1. Consider r R. By Lemma 6(g), 4-paths from r to V i may visit another vertex in R and then reach V i in exactly one way, or they may go directly to V i, traverse an edge within V i, and continue to V i. The total number of such paths is [d(r) 3]+1, and this must equal V i. Hence d(r) = a + 2. Since V i = a and d(x) = b + 1 for x V i, Lemma 6 yields V i = ab. Consider x V i and j i. Each 4-path from x to V j starts with an edge in V i, ends with an edge in V j, or uses two vertices in R. Since each vertex in N 2 (S) has a neighbors in R, there are a paths of each of the first two types. Since each vertex of R has degree a + 2, with three neighbors in N 2 (S), there are a(a 1) paths of the third type. Since these paths reach distinct vertices of V j, and every vertex of V j is reached, V j = a(a + 1). Hence a(a + 1) = ab, and b = a + 1. Since every vertex of G has degree a + 2 or b + 1, we conclude that G is k-regular, where k = a + 2. We show next that every vertex of G lies in a triangle. If v lies in no triangle, then N(v) is independent, and having unique 4-paths from N 2 (v) to v forces N 2 (v) to induce a 1-regular subgraph. Since N 2 (v) = k(k 1), there are ( k 2) edges induced by N 2 (v). Each 4-path with both endpoints in N(v) has internal vertices in N 2 (v). Since there are ( k 2) such pairs of endpoints and each edge within N 2 (v) extends to exactly one such path, no edge within N 2 (v) lies in a triangle with a vertex of N(v). Thus each neighbor of v also lies in no triangle. We conclude that neighboring vertices both do or both do not lie in triangles. By induction on the distance from S, every vertex lies in a triangle. By Lemma 2, each vertex lies in exactly one triangle. With A being the adjacency matrix of G, the matrix A 3 again counts walks of length 3. Since each vertex is on one triangle, each diagonal entry is 2. Since G is k-regular, entries for adjacent vertices are 2k 1, and by unique C 4 -saturation the remaining entries equal 1. Hence A 3 = J + (2k 2)A + I, and again J is expressible as a polynomial in A: J = A 3 (2k 2)A I. Again 1 is an eigenvector of A with eigenvalue k and of J with eigenvalue n. All other eigenvalues of A satisfy p(λ) = 0, where p(λ) = λ 3 (2k 2)λ 1. Arguing as in the proof of Theorem 5, p(λ) cannot be irreducible over Q. If λ is rational, then λ = ±1, and k {1, 2}. However, R requires k 3. 7

8 Having shown that R =, we now consider instances with N 2 (S). Lemma 8. Let G be a uniquely C 4 -saturated graph with a triangle having vertex set S. If N 2 (S), then G is one of the three graphs in Figure 2. v 1 F 1 v 1 F 2 v 1 F 3 Figure 2: Examples having a vertex at distance 2 from a triangle. Proof. Let S = {v 1,v 2,v 3 }. In the partition defined in Lemma 6, a 4-path joining V i and V j must pass through R. Since R =, we conclude that only one of {V 1,V 2,V 3} is nonempty; by symmetry, let it be V 1. Since G has diameter 3, we have V 2 = V 3 =. By Lemma 6(f), V 1 induces a matching. By Lemma 6(b), every vertex of V 1 thus has degree 2. Consider w V 1 with neighbors u and v in V 1. If u and v are not adjacent, then a 4-path joining them must use w and the neighbor in V 1 of one of them. Thus if w has three pairwise nonadjacent neighbors in V 1, then at least two of them have neighbors in V 1 that are also neighbors of w. This yields two triangles containing w, contradicting Lemma 2. We conclude that w cannot have more than three neighbors in V 1. If w V 1 has three neighbors in V 1, then two of them (say x and y) are adjacent. The only 4-paths that can leave x or y for other vertices of V 1 end at the remaining neighbor of w or its mate in V 1. Hence G = F 2. If w V 1 has two neighbors in V 1, then they are adjacent, and no 4-paths can join them to other vertices of V 1. Hence G = F 3. In the remaining case, every vertex of V 1 has at most one neighbor in V 1. Since any two vertices of V 1 are joined by a 4-path through an edge within V 1, there can only be two vertices in V 1, and G = F 1. One case remains. Lemma 9. If G is a uniquely C 4 -saturated graph having a triangle S adjacent to all vertices, then G consists of S and a matching joining S to the remaining (at most three) vertices. Proof. We have assumed N 2 (S) =. Since 4-paths joining vertices in V i must pass through V i, each V i has size 0 or 1. Since V i V j = (Lemma 6(a)), G is as described. 8

9 We can now prove Theorem 1. Theorem 1. There are exactly ten uniquely C 4 -saturated graphs. Proof. Trivially, K 1, K 2, and K 3 are uniquely C 4 -saturated. With girth 5, there is only C 5, by Theorem 5. With girth 3, Lemma 8 provides three graphs when some vertex has distance 2 from a triangle, and Lemma 9 provides three when there is no such vertex. References [1] P. Erdős, A. Hajnal, and J. W. Moon, A problem in graph theory. Amer. Math. Monthly 71 (1964), [2] A. J. Hoffman and R. R. Singleton, On Moore graphs with diameters 2 and 3. IBM J. Res. Develop. 4 (1960), [3] L. T. Ollmann, K 2,2 -saturated graphs with a minimal number of edges. Proc. Third SE Conf. on Combinatorics, Graph Theory, and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1972), [4] O. Pikhurko, Results and open problems on minimum saturated hypergraphs. Ars Combin. 72 (2004), [5] P. Turán, Eine Extremalaufgabe aus der Graphentheorie. Mat. Fiz. Lapok 48 (1941), [6] P. S. Wenger, Uniquely C k -saturated graphs. In preparation. [7] H. S. Wilf, The friendship theorem. In Combinatorial Mathematics and its Applications (Proc. Conf., Oxford, 1969) (Academic Press, 1971),

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