P3-Connected Graphs of Minimum Size

P3-Connected Graphs of Minimum Size David C. Fisher, Kathryn Fraughnaugh, and Larry Langley Department of Mathematics University of Colorado at Denver Denver, CO 80217-3364, U.S.A. A graph is C4 -free if it does not contain any 4-cycles. A C4 -free graph is C4 -saturated if adding any edge creates a 4-cycle. Ollmann showed that an n-node C4 -saturated graph has at least 32 (n ? 2) edges. A graph is P3 -connected if there is a path of length 3 between each pair of nonadjacent nodes. We show that an n-node P3 -connected graph has at least 23 (n ? 2) edges. Since all C4 -saturated graphs are P3connected, this generalizes Ollmann'sresults. We also nd all n-node 3 P3 -connected graphs with 2 (n ? 2) edges.

Let Pk be a path on k edges. Two nodes are Pk -connected if they are connected by a Pk . A graph is Pk -connected if all pairs of nonadjacent nodes are Pk -connected. The only P1-connected graphs are complete graphs. A graph is P2-connected if and only if its diameter is at most 2. In general, a Pk -connected graph has diameter at most k. However, a diameter k graph need not be Pk -connected. For example, Pk for k  3 has diameter k and yet is not Pk -connected. Let Ck be a cycle with k edges. A graph without Ck as a (not necessarily induced) subgraph is called Ck -free. A Ck -saturated graph is a maximal Ck free graph in the sense that adding any edge creates a Ck . If adding an edge between two nodes creates a Ck , then a Pk?1 must connect the nodes. So a Ck -saturated graph is Pk?1-connected. However, a Pk?1-connected graph need not be Ck -saturated (e.g., Figure 3 (d) and (e)). Assume throughout that n is the number of nodes and e is the number of edges in a graph. Ollmann [1] answered the question: What is the minimum number of edges in a C4 -saturated graph? He showed e  23 (n ? 2) for all graph. He also described all C4-saturated graphs with   C4-saturated e = 32 (n ? 2) showing that, given integrality, the bound is exact for n > 4. Later, Tuza [2] gave a much simpler proof of the bound and description. Theorem 1 answers the question: What is the minimum number of edges in a P3 -connected graph? The answer is again e  32 (n ? 2). However since all C4 -saturated graphs are P3 -connected, this is a generalization of Ollmann's result. Theorem 2 describes the class of P3-connected graphs  with e = 23 (n ? 2) . In addition to the three found by Ollman (Figure 3 (a), (b), and (c)), two new families are given (Figure 3 (d) and (e)). Graphs in these two families have an odd number of nodes, and are P3-connected but not C4-saturated. 1

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Proof. Since it is connected, G has a spanning tree T with n ? 1 edges. An edge of G ? T can form a C3 with at most two edges of T. So G ? T has at least 12 (n ? 1) edges. 2 3 Theorem 1. Let G be a P3-connected graph. Then e  2 (n ? 2). Proof. This is trivially true for n = 1. So assume n  2. Let  be the minimum degree of the nodes of G. Since G is connected,  > 0. If   3, then e  23 n giving the result. So assume  is 1 or 2. Case 1:  = 1. Let v be a degree 1 node with neighbor w. Let T be

the breadth- rst tree of G rooted at v. Let X and Y be the nodes that are distance 2 and 3, respectively, from v. Since all nodes are distance at most three from v, all nodes other than w and v are in X or Y . Thus jX j + jY j = n ? 2. Because x 2 X is not adjacent to v, there is a path v; w; q; x. Since q is distance 2 from v, we have q 2 X and hence the nodes in X have degree at least 1 within X. Let Y1 be the degree 1 nodes in Y and k = jY1j. Let Y2 = Y ? Y1 . Let X1 be the nodes of X adjacent to Y1 and let X2 = X ? X1 . Then for all y1 ; y2 2 Y1 , there is a path y1 ; x1; x2; y2 where x1; x2 2 X1 . It is easy to see that each node in X1 is adjacent to exactly one node in Y1 , and every pair of nodes in X1 are adjacent. Thus jX1 j = k and X1 is a clique (see Figure 1). Since each node in X2 and Y2 is incident to at least one edge not in T, the number of edges not in T is     k + n ? 2 ? 2k  n ? 4 : j X j X 1j 2j + jY2j = e(G ? T)  2 + 2 2 2 2 2

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Figure 2. The tree formed in Case 2 of Theorem 1. Thus e  21 (n ? 4) + n ? 1 = 23 (n ? 2). Case 2:  = 2 and a degree 2 node is not in a C3 . Let v have degree 2 with nonadjacent neighbors w1 and w2 . Let T be the breadth- rst tree of G rooted at v. Let X and Y be the nodes which are distance 2 and 3, respectively, from v (see Figure 2). Since a P3 connects v to x 2 X, there is an edge incident to x which is not in T. Further since  = 2, each y 2 Y is also incident to an edge not in T. So at least 21 (jX j + jY j) = 12 (n ? 3) edges are not in T. Therefore e  n ? 1 + 12 (n ? 3) = 21 (3n ? 5). Case 3:  = 2 and each degree 2 node is in a C3 . Let S be the subgraph of all C3's with one or more degree 2 nodes. Let S1 ; S2 ; : : :; Sm be the components of S. Lemma 1 shows that e(Si )  32 (jSi j? 1). Since the degree 2 nodes of two components are P3-connected, there is? an
edge between every pair of components of S. So there are at least m2 edges between components of S. Then the number of edges in the induced subgraph induced by the nodes of S is     m m X X e(hS i)  e(Si ) + m2  23 (jSi j ? 1) + m2 i=1 i=1 2 = 23 jS j + m2 ? 2m  32 jS j ? 2: Since all degree 2 nodes are in S, nodes not in S have degree 3 or more. 2 Thus e  32 jS j ? 2 + 23 (n ? jS j) = 32 n ? 2. 3  Theorem 2. Let G be a P3 -connected graph with e = 2 (n ? 2) . Then G is in one of the families of graphs in Figure 3.

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Figure 3 { Families of 3 -connected graphs with 32 ( ? 2) edges. The dots indicate an arbitrary number (including zero) of P

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Case 1a: k = 1. Let X1 = fx1g and Y1 = fy1g. Since every 1node in X and Y2 has degree at least one in G ? T, we have that e(G ? T)  2 (jX j + jY2 j) = 1 (n ? 3). So e(G ? T) = 1 (n ? 3) and hence n is odd and the edges of 2 2 G ? T form a matching on X [ Y2 . Since each node of X is adjacent to a node of X, the edges of G ? T form a matching on X and hence also form a matching on Y2. Let x2 2 X2 so that x1 x2 is an edge. Since x1 is not adjacent to X2 ? fx2 g, if y 2 Y is not adjacent to x1 or x2 , then y is not connected to y1 by a P3. So all nodes of Y2 are adjacent to either x1 or x2. 4

Let y2 ; y3 2 Y2 be such that y2 y3 is an edge and suppose y2 x1 and y3 x2 are edges. Then y2 is not adjacent to x2 otherwise y2 would be incident with two edges not in T. Similarly y3 is not adjacent to x1 . Under these circumstances, y2 is not P3-connected to y1 which is a contradiction. So the ends of any edge in Y are either both adjacent to x1 or both adjacent to x2. Thus G is in Family (c) where w; x1; x2 is the central C3, and vw and x1y1 are pendant edges. Case 1b: k = 2. Let X1 = fx1; x2g and Y1 = fy1; y2g where x1y1 and x2y2 are edges. Note that x1x2 must be an edge as well. Assume n is even. Here e(G ? T) = 21 (n ? 4). For the inequalities to be equalities, each of the n ? 4 nodes in X and Y2 has degree 1 in G ? T, and there are no edges between X1 and X2 . Then an argument similar to Case 1a shows that G is in Family (a) where w; x1; x2 is the central C3 and vw, x1y1 and x2y2 are pendant edges. Now assume n is odd. Here e(G ? T) = 21 (n ? 3). Since jX j + jY2j = n ? 4 and every node in X [ Y2 has degree at least 1 in G ? T, the subgraph of G ? T restricted to X [ Y2 consists of a P2 and isolated edges. Since each node of X is adjacent to another node in X in G ? T and x1x2 is an edge in G ? T, there are ve possibilities for the P2. If the P2 is x1; x2; x3 where x3 2 X2 , then by arguments similar to Case 1a, we have that G is in Family (e) where w; x1; x2 is the central C3, vw, x1y1 and x2 y2 are pendant edges, and x3 is adjacent to both w and x2 . If the P2 is x3; x4; x5 where x3; x4; x5 2 X2 , then again by arguments similar to Case 1a, G is in Family (d) where w; x1; x2 is the central C3 , vw, x1y1 and x2y2 are pendant edges, and x3; x4; x5 is the P2 adjacent to w. If the P2 is x1 ; x2; y3 where y3 2 Y2 , then x1y3 must be an edge so that y3 is P3-connected to y2 . So by arguments similar to Case 1a, G is in Family (e) where w; x1; x2 is the central C3, vw, x1y1 and x2y2 are pendant edges, and y3 is adjacent to both x1 and x2. If the P2 is x3; x4; y3 where y3 2 Y2, then y3 cannot be P3-connected to both y1 and y2 . So this possibility is impossible. Finally, if the P2 is y3 ; y4 ; y5 where y3 ; y4 ; y5 2 Y2 , then by arguments similar to Case 1a, G is in Family (d) where w; x1; x2 is the central C3, vw, x1y1 and x2y2 are pendant edges, and y3 ; y4 ; y5 form the P2 which is adjacent to either x1 or x2 . Case 2:  = 2 and a degree 2 node is not in a C3 . Since e  21 (3n ? 5), we have that n is odd. Let v, w1, w2 , X, Y , and T be as de ned. All nodes save v, w1 and w2 have degree 1 in G ? T. Since v is P3-connected to each x 2 X, every x 2 X has a neighbor in X. Let Xi be the neighbors of wi in X. Since both w1 and w2 have degree at least 2, we have Xi 6= ;. Since the nodes of X have degree 1 in X, the edges in X form a matching and X1 \ X2 = ;. Since each node in Y is incident to an edge in G ? T, the edges in Y also form a matching, and each node in Y is adjacent to exactly 5

one node in X. Suppose there exists a pair of nodes x1 2 X1 and x2 2 X2 where x1 has no neighbor in X2 and x2 has no neighbor in X1 . Since x1x2 62 E, there is a path x1; a; b; x2. Our choice of x1 and x2 guarantees that a 62 X2 and b 62 X1 . Since the edges within X form a matching, we also cannot have both a 2 X1 and b 2 X2 . So at least one of a and b is not in X. Without loss of generality, assume a 62 X. If a = w1 , then b 2 X1 , a contradiction. Thus a 2 Y . Since no node in Y is adjacent to two nodes in X, we have b 2 Y . But since a and b have degree 2, then b is not P3-connected to x1, a contradiction. Thus there is not a pair of nodes of this type. Suppose jX1 j > 1 and jX2 j > 1. Then at least two edges are between X1 and X2 , for otherwise a pair of the type precluded above would occur. Let x1x3 and x2 x4 be these edges with x1; x2 2 X1 and x3; x4 2 X2 . Since xi can be adjacent to only one node in X, we have x1 nonadjacent to x2 and x3 nonadjacent to x4 . Arguments similar to the ones above show that there are paths x1; y1 ; y2; x2 and x3; y3 ; y4; x4 where yi 2 Y . Since each node of Y has degree 2, the yi are distinct, the neighbors of y2 are y1 and x2 , the neighbors of y3 are y4 and x3, and there are no edges between fy1 ; x2g and fy4 ; x3g. Thus y2 is not P3-connected to y3 , a contradiction. So either jX1 j = 1 or jX2j = 1. Without loss of generality, assume X1 = fx1 g and X2 = fx2; x3; : : :; xkg and where x1x2 is an edge. Let Y1 be the nodes of Y adjacent to x1. Recall that each node of Y is adjacent to exactly one node in X and the edges of Y form a matching. Assume jX2 j = 1. Let Y2 = Y ? Y1 . Since w1 is P3-connected to nodes of Y1 , the edges in Y1 form a matching. Similarly, the edges in Y2 form a matching. So G is in Family (b) with v; w1; x1; x2; w2 forming the central C5 with only x1 and x2 having pendant C3's. Otherwise jX2 j > 1. Let y1 2 Y1 . Let y2 be the neighbor of y1 in Y . If y2 x1 is an edge, then y2 is not P3-connected to x3. If y2x2 is an edge, then y2 is not P3-connected to w2. If y2 xi is an edge for i > 2, then y2 is not P3-connected to x1 because x2 is not adjacent to xi. So y2 is not adjacent to X and hence y1 cannot exist. Therefore, x1 has degree 2 and is not adjacent to Y . Then replacing v by w1 in the above argument we get that G is in Family (b). 2

References 1. L.T. Ollmann, K2;2-saturated graphs with a minimalnumber of edges, Proceedings of the Third Southeast Conference on Combinatorics, Graph Theory, and Computing (1972) 367-392. 2. Z. Tuza, C4-saturated graphs of minimum size, Acta Universitatis Carolinae{Mathematica et Physica (1989) 161-167. 6