SPANNING STAR TREES IN REGULAR GRAPHS Jerrold W. Grossman Department of Mathematical Sciences Oakland University Rochester, MI 48309-4401 e-mail:
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Abstract For a subset W of vertices of an undirected graph G, let S(W ) be the subgraph consisting of W , all edges incident to at least one vertex in W , and all vertices adjacent to at least one vertex in W . If S(W ) is a tree containing all the vertices of G, then we call it a spanning star tree of G. In this case W forms a weakly connected but strongly acyclic dominating set for G. We prove that for every r ≥ 3, there exist r-regular n-vertex graphs that have spanning star trees, and there exist r-regular n-vertex graphs that do not have spanning star trees, for all n sufficiently large (in terms of r). Furthermore, the problem of determining whether a given regular graph has a spanning star tree is NP-complete. AMS Subject Classification (1991): Primary: 05C35 Secondary: 05C05, 05C85 Key words: dominating set, weakly connected, strongly acyclic, regular graph §1. Introduction In this paper we continue the study of weakly connected, strongly acyclic domination in graphs, introduced in [2]. The reader should consult that paper for the original motivation for looking at this variation, namely a problem on Euler tours in planar graphs. Some related results can be found in [1]. All graphs G = (V, E) in this paper are finite and connected and contain no loops or parallel edges. If W is a subset of V , then by S(W ), the subgraph weakly induced by W , we mean the subgraph consisting of W , all edges incident to at least one vertex in W , and all vertices adjacent to at least one vertex in W . In other words, S(W ) is the union of the closed stars at the vertices in W , but is in general not an induced subgraph. We seek a W for which S(W ) satisfies three conditions. First, it must include all of V , so that W is a dominating set for G. Second, we want S(W ) to be connected; thus W will be a “weakly connected” dominating set. (This is less demanding than the condition that W be a “connected dominating set”, i.e., that the subgraph induced by W be connected.) Finally, we require that S(W ) contain no cycles; this makes W a “strongly acyclic” dominating set. (This is more demanding than the condition that W be an “acyclic dominating set”, i.e., that the subgraph induced by W be acyclic.) If these three conditions hold, then we call S(W ) a spanning star tree. This paper concerns the existence of spanning star trees in regular graphs. Let us call a connected graph good if it has a spanning star tree and bad if it does not. In §2 we characterize 1
precisely those values of n for which r-regular good graphs on n vertices exist. In §3 we do the same for bad graphs. It was shown in [2] that the problem of finding spanning star trees in arbitrary graphs, or even determining whether they exist, is computationally difficult. These problems remain hard even if the graph is regular; specifically, we show in §4 that the problem of determining whether a given regular graph is good is NP-complete. §2. Good regular graphs We will show by construction that for n larger than about r2 , there exists an n-vertex, r-regular graph having a spanning star tree. Furthermore, we characterize the smaller values of n for which such graphs fail to exist. Obviously 1-regular connected graphs are uninteresting, and 2-regular connected graphs are just cycles, which are clearly good (take W to be all but two consecutive vertices), so we assume henceforth that r ≥ 3. We need slightly different hypotheses depending on the parity of r. THEOREM 2.1. For r ≥ 3 there exists an r-regular good graph having n vertices for every n ≥ r2 − 2r + 3 if r is even, and for every even n ≥ r2 − 3r + 4 if r is odd. Proof : The proof is constructive; we “grow” a spanning star tree S(W ) by adding one new vertex at a time to W . Once the tree is finished, we put in additional edges to make the graph r-regular. Throughout we assume that n is to be even if r is odd. To start, we put vertex v1 of degree r into W . This puts r + 1 vertices into S(W ), namely v1 and the r vertices to which it is to be adjacent. Next we want to put the star of another vertex, v2 , into S(W ). We can do this in at least two ways to maintain connectivity and acyclicity. First, we can choose v2 to be outside of S(W ) at this stage, with exactly one neighbor currently in S(W ), a vertex of degree 1 in S(W ). In that case (which we call “type A”), this step adds r vertices to S(W ), namely v2 and r − 1 new vertices which will be adjacent to it. Alternatively, we can choose v2 to be a vertex of degree 1 in S(W ). Then this step (“type B”) adds r − 1 new vertices to S(W ), namely the other neighbors of v2 . We proceed in this manner until S(W ) contains n vertices. If we made type A additions s times and type B additions t times, then n = r + 1 + sr + t(r − 1); here s and t are arbitrary nonnegative integers. At this point our graph has s + t + 1 vertices of degree r, has s vertices of degree 2, and has n − 2s − t − 1 vertices of degree 1. Furthermore, none of the vertices of degree less than r are adjacent. We must put in some additional edges to make the result r-regular. If r is odd, then s is necessarily even since n is. Add a matching to the s vertices of degree 2 and a cycle to the n−2s−t−1 vertices of degree 1. Now every vertex of degree less than r has degree 3. It is then a simple matter to put in additional cycles to bring each vertex degree up to r. Similarly, if r is even, then add a matching to the n − 2s − t − 1 vertices of degree 1, bringing the degree of every vertex not in W up to 2, and then add cycles as before. Thus there exists an r-regular good graph on n vertices as long as n can be written as r + 1 + sr + t(r − 1). It is elementary to see that if n satisfies the stated bounds, then s and t can be chosen to achieve this. A little reflection shows that the construction we give in the proof of Theorem 2.1 (successively adding vertices to W to maintain S(W ) as a tree) provides essentially the only way one can build an r-regular good graph. [There is actually a bit more freedom in 2
the choice of the successive vi ’s, but we claim that the additional freedom is not needed. If there is a vertex v of degree k < r present in S(W ), with k not necessarily equal to 1, then (type B) we can choose v to add to W , thereby adding r − k new vertices of degree 1 to S(W ); or (type A) we can add to W a new vertex of degree r and attach it with an edge to v, increasing v’s degree to k + 1 and adding r new vertices in all to S(W ). In order to perform a type B operation on a vertex of degree k > 1, however, we first would have to perform the type A operation k − 1 times, adding r(k − 1) + r − k = (r − 1)k new vertices in all; and we could have added this many vertices by performing the degree 1 type A operation k times. Therefore, to achieve a desired number of vertices, we need never use the type B operation for k > 1, and certainly we never need the type A operation for k > 1. Thus the added freedom does not help at all in terms of the number of new vertices added to our graph, and we can ignore it.] As a result, if n cannot be obtained as r + 1 + sr + t(r − 1), then there is no r-regular good graph on n vertices. It is clear that the numbers that can be obtained in this way are r + 1, 2r through 2r + 1, 3r − 1 through 3r + 1, . . . , and in general kr − (k − 2) through kr + 1. Thus we have the following characterization. THEOREM 2.2. For r ≥ 3 there exists an r-regular good graph on n vertices if and only if n = kr + 1 − j for some k ≥ 1 and some j with 0 ≤ j < k (and n is even if r is odd). The size of the “gaps” between the successive intervals of attainable values of n decreases as k increases, and it is simple counting to see that the bounds in Theorem 2.1 are best possible. Moreover, the total number of unattainable values of n is given by the following corollary. COROLLARY 2.3. Let r ≥ 3. If r is even, then the number of different values of n > r for which there does not exist an r-regular good graph on n vertices is (r−1)(r−2)/2. If r is odd then the number of different even values of n > r for which there does not exist an r-regular good graph on n vertices is (r − 1)(r − 3)/4. §3. Bad regular graphs In most cases, our construction of bad graphs is based on certain small graphs that prevent any graph with one of them as a subgraph from having a spanning star tree. Special constructions are needed for certain small values of n. In light of the introductory comments in §2, we again assume that r ≥ 3. Of course Kr+1 is the only r-regular graph with fewer than r + 2 vertices, and it is good. Thus the following result is best possible. THEOREM 3.1. For every r ≥ 3, there exists an r-regular bad graph having n vertices for every n ≥ r + 2 (with n even for odd r), except for r = 3, n = 6. Proof : We deal first with the case r = 3. From [2] we know that there are only six bad graphs on 6 vertices, and none of them is 3-regular. For n = 10 one can give an ad hoc argument that the pentagonal prism is bad, and in fact a computer search reveals that it is the only bad 3-regular graph with 10 vertices. Now let H be the complete graph on 4 vertices with edge uv omitted. If n is a multiple of 4 greater than 4, then we can form a 3
graph on n vertices by stringing together two or more copies of H into a closed “necklace”, adding edges from u in each copy of H to v in the next copy. It is easy to show that this graph has no spanning star tree. Finally for n ≥ 14 and congruent to 2 modulo 4, we can take this necklace on n − 2 vertices and replace one of its “beads” by the graph consisting of a 4-cycle with vertex u adjacent to two consecutive vertices and vertex v adjacent to the other two. The r = 4 case is a little messy. The approach is similar to the necklace construction used above for r = 3. A bead of size 5 is formed by taking a K5 and removing edge uv to provide the leads to other beads. A bead of size 6 is formed by taking a K6 and removing a matching and then removing one more edge uv to provide the leads. It is easy to check that at most one vertex from any bead can be used in a W that weakly induces a spanning star tree. On the other hand, at least one of the u or v vertices would have to be used in order that S(W ) be connected, and we are soon led to the conclusion that such a necklace cannot contain a spanning star tree. So if n is any sum of three or more 5’s and 6’s, then there is a 4-regular bad graph on n vertices. This is true of all n ≥ 20. For 6 ≤ n ≤ 19, consider the 4-regular graph consisting of the integers 0 through n − 1, with two of them adjacent if they differ by 1 or 2, modulo n. It is not hard to see that as long as n 6= 9, 13, or 17, this graph is bad. (Put vertex 0 into W ; then vertices 1, 2, 3, n − 1, n − 2, and n − 3 are excluded. Without loss of generality, in order to make S(W ) connected, put vertex 4 into W , and so on.) Note that 17 = 5 + 6 + 6. So only the cases n = 9 and n = 13 remain. A general construction given for n = 2r + 1 in the final paragraph of this proof takes care of n = 9. Finally, for n = 13 one can show by brute force that the graph with edges ag, ah, ai, aj, bg, bh, bi, bj, ch, ci, cj, ck, dh, dk, dl, dm, ei, ek, el, em, f j, f k, f l, f m, gl, and gm is bad. Now assume that r ≥ 5 is odd. Let H be the graph on the r + 2 vertices v, u0 , u1 , . . . , ur , with edges between every pair of these vertices except u0 v, ur v, and u2i−1 u2i for i = 1, 2, . . . , (r − 1)/2. Note that the degree of every vertex in H is r, except that vertex v has degree r − 1. Now form a regular graph G on n vertices, where n ≥ 2r + 4 and is even, by taking H together with a graph on n − r − 2 vertices in which every vertex has degree r except for vertex w, which has degree r − 1, and joining w to v. It is easy to check that G cannot have a spanning star tree S(W ), because if v were in W then u0 could not get dominated, and if v were not in W , then there is no way to dominate all the vertices in H and still have S(W ) connected and acyclic. Similarly, for even r ≥ 6, let H consist of the r+2 vertices v, u0 , u1 , . . . , ur , with edges between every pair of them except u0 v, ur−1 v, ur v, and u2i−1 u2i for i = 1, 2, . . . , (r − 2)/2. Note that the degree of every vertex in H is r, except that vertex v has degree r − 2. Now form a regular graph G on n vertices, where n ≥ 2r + 3, by taking H together with a graph on n − r − 2 vertices in which every vertex has degree r except for vertices w1 and w2 , which have degree r − 1, and joining w1 and w2 to v. Again, one can verify that G cannot have a spanning star tree. The argument so far, together with the observation from Theorem 2.2 that all rregular graphs on n vertices are bad if r + 2 ≤ n ≤ 2r − 1, proves the theorem except for n = 2r, n = 2r + 2, and, for even r, n = 2r + 1. To get an r-regular bad graph on 2r + 2 vertices, take Kr+1,r+1 minus a matching. To get an r-regular bad graph on 2r vertices, 4
r ≥ 5, start with Kr,r on blue vertices ui and red vertices vi , for i = 1, 2, . . . , r, add the cycles u1 u2 . . . ur u1 and v1 v2 . . . vr v1 , and remove the cycle u1 v1 u2 v2 . . . ur vr u1 . Finally, for n = 2r + 1 when r is even, take the graph whose vertices are the integers 0 through n − 1, with adjacencies between v and v ± 1, v ± 3, v ± 5, . . . , v ± (r − 1), all arithmetic done modulo n. The verification that these graphs are bad is left to the reader. §4. Algorithmic questions In [2] it was shown that it is NP-hard to find a spanning star tree in an arbitrary graph (or even to determine whether a spanning star tree exists). One might hope that the problem is easier for regular graphs. Here we show that it is not. THEOREM 4.1. The problem of determining whether a given regular graph is good is NP-complete. Proof : Clearly this problem is in the class NP. The proof of completeness is by reduction from the same problem for arbitrary connected graphs. Suppose we wish to determine whether a given graph G has a spanning star tree. We will construct a regular graph G′ with odd vertex degree r, which we can take to be at most one greater than the maximum degree of G. First, attach pendant edges to G to bring the degree of each vertex of G up to r. Second, for each pendant vertex v so created, attach edges vvi for 1 ≤ i ≤ r − 1, edges v1 wj for 1 ≤ j ≤ r − 1, and enough edges joining some of the vi ’s and wj ’s (as we r−1 r−1 did in the proof of Theorem 2.1) to make the subgraph induced by {v} ∪ {vi }i=1 ∪ {wj }j=1 regular of degree r, except that vertex v has degree r − 1. Then the new portions of G′ neither help nor hinder the search for a spanning star tree S(W ) of G, because one can always include each v and each v1 in W . Therefore G is good if and only if G′ is good. Clearly the construction is polynomial in the size of G. We end with two open questions. First, our construction in the proof of Theorem 4.1 resulted in a graph that was not 2-connected. Will adding higher connectivity assumptions renders the problem easier (or, in the context of the results of §3, make it more difficult to find bad graphs)? Second, if we fix r, is the problem of determining whether a given r-regular graph is good still NP-complete?
References [1] Jean Dunbar et al., Weakly connected domination in graphs, to appear. [2] Jerrold W. Grossman, Dominating sets whose closed stars form spanning trees, to appear.
This is a non-final draft, as of March 29, 1995 5
