Atoll decompositions of graphs

AtoI I Decompositions of Graphs

Fred Buckley' ST: JOHNS UNIVERSITY

STATEN ISLAND, NY 10301

ABSTRACT

An island decomposition of a graph G consists of a set of vertex-disjoint paths which cover the vertex set of G. If the endpoints of the paths are mutually nonadjacent, then w e have an atoll decomposition. We characterize graphs requiring two paths in an island decomposition yet having no atoll decomposition. Results are given relating atoll decompositions to cutpoints and Hamilton ian blocks.

1. INTRODUCTION

An island decomposition (ID) of a graph G is a set of vertex-disjoint paths (islands) which cover the vertices of G. Every graph has an ID, since we can let each vertex be an island We use i( G) to denote the minimum number of islands in any I D of G. The problem of determining i( G) for various classes of graphs has been examined by Boesch et al. [ 11 and Slater [ 111. Skupien [ 101 examined sufficient conditions for G to have the property that i( G) I n. The problem of "bridging" the islands by adding as few edges as possible so that the resulting graph contains a Hamiltonian cycle was studied by Goodman and Hedetniemi [ 31 and Goodman et al. 14-61.

An atoll decomposition (AD) is an ID for which the endpoints of distinct islands are mutually nonadjacent and the endpoints of a single island are adjacent only when that island is K,. In this case, we call the islands atolls. Let A be the set of all graphs which have an AD. If G E A , let a ( @ be the minimum number of atolls in an A D of G. Clearly, i( G) I a( G) since each atoll decomposition is an island decomposition.

A graph for which a(G) = 1 is a graph having a Hamiltonian path which does not extend to a Hamiltonian cycle. (We stress that G may have a Hamiltonian cycle; however, at least one of its Hamiltonian paths does not extend to a Hamiltonian cycle.) This is what sparked our interest in this area. We will be considering minimum path decompositions in which none of the paths extend to cycles. Before going any further, we give an example. In part

'Current address: Baruch College (CUNY) New York, NY 10010.

Journal of Graph Theoty, Vol. 6 ( 1 982) 31 7-324 o 1982 by John Wiley & Sons, Inc. CCC 0364-9024/82/030317-08$01.80

318 JOURNAL OF GRAPH THEORY

(i) ( i i)

FIGURE 1

(i) of Figure 1, we display an AD for a graph Part (ii) of Figure 1 depicts, for that same graph, an ID which is not an AD.

The concept of ADS was developed in 191, where Samowitz conjectured that two certain conditions [( 1) and (2) below) characterize graphs G for which i(G) = 2 and G has no AD. We show that co.njecture to be false. However, with the addition of a third condition, we get the desired characterization. Our characterization relates ADS to cutpoints and Hamiltonian blocks.

We now give some notation which we will use to describe certain paths. If P is a path and v1 and v2 are vertices within P, then v,-P-v, is the subpath of P between v I and v2. If v3 is adjacent to v2 and not in v,-P-v,, then v,-P-v,,v, is the extension of the first path by the edge ~ 2 ~ 3 . Occasionally we will replace the path P by a cycle C. In those cases, the direction taken around C will be forced by the vertex v3 used to extend the path v,-C-v,.

2. GRAPHS NOT HAVING AN ATOLL DECOMPOSITION

In this section, we characterize graphs which do not have an atoll decomposition but which have a minimum island decomposition into either one or two islands. A graph is randomly Hamiltonian if for every vertex v any path beginning at v can be extended to a Hamiltonian cycle.

Lemma 1 (Chartrand and Kronk [2]). A graph G is randomly Hamiltonian if and only if G is one of the graphs C, or Kp (p L 3 ) or Kn,n ( n 2 2).

If a( G) = 1 then i( C ) is also 1, since i( C ) I a( G). Therefore, if a( G) = 1, then G must have at least one Hamiltonian path which does not extend to a Hamiltonian cycle; that is, G is not randomly Hamiltonian.

Theorem 1 (Samowitz 191). G is randomly Hamiltonian if and only if G g A and i(C) = 1.

ATOLL DECOMPOSITIONS OF GRAPHS 31 9

Theorem 2. G 4 A and i( C ) = 2 if and only if (1) G has two components and one contains a Hamiltonian path, while the

other is randomly Hamiltonian, (2) G is connected and contains a cutpoint v such that G - v has three

components, each of which is randomly Hamiltonian, or ( 3 ) G is connected and contains three mutually adjacent cutpoints u, t', and

w, each of whose removal produces two components, and a n y two of the cutpoints, say u and v, have the property that G - (u, 4 has three components, each of which is a (randomly Hamiltonian) clique.

In [9], Samowitz conjectured that conditions (1) and (2) were sufficient. The corona K3 0 K3 is a counterexample to that conjecture. We defer the proof of Theorem 2 until we have completed two more needed lemmas. A minimum island decomposition (MID) of G is an ID of G containing i( C ) islands. A paper by Ore [ 81 dealt with vertex-disjoint path coverings of G, in which the collection of paths in a cover contains a maximum number of edges. These are clearly equivalent to MIDs. Ore derived a theorem involving degree conditions for the existence of a Hamiltonian path.

Lemma 2. If G is connected, G g A and i( G') = 2 , then each island in a MID has at least three vertices.

ProoJ: Suppose (P, P') is a MID of G and P' = K , orK2. Clearly, at least one of P and P must have at least three vertices, therefore P does. Let v , and v, be endpoints of P and let v' f V(P). Since i( C ) = 2 and G 4 A , we have vIv' and v,v' 4 E( G'), whereas vl vz E E( C). Thus vl-P-vz, v, is a cycle. Since G is connected, a vertex of P must be adjacent to a vertex of this cycle. But this implies that i( C ) = 1, a contradiction.

The graph G consisting of three copies of K3 with an additional vertex adjacent to one vertex from each of the K,s is the extremal graph for Lemma 2.

H

Lemma 3. If G is connected, G 4 A and i(C) = 2 , then G contains a cutpoint

ProoJ: Suppose G has only one block. Then each pair of edges in G is contained in a common cycle. Let (P, P ') be a MID for G . By Lemma 2, P and P' each have at least three vertices. Let t',, v2, and v; and v; be the endpoints o f P and P ', respectively. Since G g A , these vertices cannot form an independent set. Also, since i (G) = 2, the endpoints of P cannot be adjacent to the endpoints of P. Thus the endpoints of one of the paths, say P , must be adjacent. Therefore, v,-P-v2, v , is a cycle. Call this C. Since G is connected, G 4 A and i (G) = 2 , we now have Y ; V ; 4 E(G) and no vertex on C can be adjacent to any of the first three vertices on either end of P'. Consider an edge within P and an edge within P'. Since G has only one


block, these edges are contained in a common cycle. Therefore there are two distinct vertices a , and a2 on P which are adjacent to distinct vertices a', and a;, respectively on P '. This implies that P ' must have at least eight vertices. We consider three possibilities for a; and a;: (4) there is exactly one vertex b between a', and a; on P ', ( 5 ) there are at least two vertices between a', and a; on P ' , and ( 6 ) vertices a', and a; are successive vertices on P '. If (4) holds, let d , and d2 be neighbors, respectively, of a , and a2 on C.

Consider the I D (d,-C-al, a',-P I+',; b-P '-v;). Since G 4 A and i(G) = 2, we have bv; E E(G). But then the island &around C-a2, a;-P '+;, b-P '-v', is a spanning path, contradicting i(G) = 2. Thus (4) cannot hold.

If ( 5 ) holds, let d , and d2 be neighbors, respectively, of a, and a, on C, and let b', and b; be the neighbors, respectively, of a', and a;, between a; and a; on P '. Considering the same I D as in case (4), with b replaced by b', , we find that b',v; E E(G) . By the symmetry present, we have b;v', E E(G). Letf' be the neighbor (which one can show must exist) of b', other than a', on P I . The island d2-around C-a2, a;-P ' -v;, b;-P ' - v ; , b;-P '-f' is a spanning path, contradicting i(G) = 2. Thus ( 5 ) cannot hold.

If ( 6 ) holds, we consider the island v;-P '-a',, al-C-a2, a;-P '-v;. There must be at least three vertices of C not covered by this path. Let c, and c2 be the neighbors of a, and a2 not covered by the path and let e , be the other neighbor of c1 on C. We have clc2 EE(G) . Let d , and d2 be the other neighbors of a, and a2 on C, and let e2 be the neighbor of d2 distinct from a2 on C. (The figure for this part is Fig. 2 with no vertices between a; and a; on P '.) By the symmetry present, dld2 E E(G). Consider the I D (e2-C-d,, d2 , a2, a;-P'-v;; el-C-c2, c,, a , , a',-P'-v',). Since v; and v; are not adjacent to any vertices of C , v', v; 4 E( G) and G 4 A , one has el e2 E E( G). But then v;-P'- a;, a2, d2, d,-C-e2, el-C-c2, c,, a , , a',-P'-v', is a spanning path, which implies i(G) = 1, a contradiction. Thus ( 6 ) cannot hold, and G contains a cutpoint. W

Proof (of Theorem 2). Sufficiency is not difficult to establish; therefore, we only prove necessity. That is, we show that if G E A and i( G) = 2, then ( 1 ), (2), or (3) holds.

Suppose that G $? A and i( G') = 2. Clearly G has at most two components. If G has two components, then it follows easily from Theorem 2 that (1) holds. Suppose then that G has only one component. By Lemma 3, G has a cutpoint. Since i( G) = 2, one finds that for any v, G - v has at most three components. Thus for some vertex v, G - u has two or three components. Let (P, P) be an island decomposition for G whose islands have endpoints v,, v2 and v',, v; , respectively. The endpoints of one of these paths must be adjacent. Thus, suppose v;v; E E(G) and let the cycle v;-P '-v;, v', be called C. Since G is connected, some vertex v of P is adjacent to a vertex z of C. This implies that v,v2 4 E(G) . Lemma 2 implies that there are at least three

ATOLL DECOMPOSITIONS OF GRAPHS 321

FIGURE 2

vertices on each side of v within P. Let u and w be neighbors of v and s and t be neighbors of u and w, so that u,-P-s, u, v, w, f-P-v2 equals P. Since G 4 A and i(G) = 2, we find vlu and v2w E E(G).

Let m , and m2 be the neighbors of z on C . Suppose uw E E( G). Let C , be the cycle vI-P-u,uI and C2 be the cycle v2-P-w,u2. Consider the decomposition (s-around C,-u, w-around C2-t; v , z-around C ) . Since G g A and i(G) = 2, we have vm, and vm2 E E(G). Then considering the path v,-P- u, w-P-v, and the various paths spanning V(C) U (v), we find that V(C) U (v ) induces a clique. From this it follows that v is the only vertex on P that is adjacent to a vertex of C.

If there were an edge besides uw, uv, and uw from a vertex on v,-P-v to a vertex on v2-P-v, one would find that every four successive vertices on C, form a cycle. From this one could easily find either a spanning path or a pair of paths to show that G E A , both contradictions. Thus uv, uw, and vw are the only such edges.

As a final step in the case where uw EE(G) , we consider the island decomposition (u,-P-u, w, v, z-around C; f-P-u,). One has tv, E E( G) and C, must have at least four vertices. Then considering the path v,-P-v, z-around C-m, and the various paths spanning C,, one finds that V(C,) induces a clique. In a similar manner V(C,) induces a clique. Thus, if uw E E( G), then u, u, and w are three mutually adjacent cutpoints and ( 3 ) holds.

Suppose that uw $? E(G'). We show that there can be no edge from a vertex in C, to a vertex in C - z. Suppose that there were such an edge with k and K as the endpoints. By Lemma 2, there must be at least three vertices between K and z on C. Let n, and n, be the neighbors, other than z, of m, and m,, respectively. Let j , and j , be the neighbors of K on C so that K, j,-C-n,,


FIGURE 3

m, is a path on C (see Fig. 3). Let k* be a neighbor of k on C1. Consider the island decomposition (k*-around C,-k, k', j2-C-m2;j,-C-m,, z, Y-P-v,). Since v2 is not adjacent to any vertex of C , i(G) = 2, and G 4 A , we have k*m2 E E(G). But then k-around C,-k*, m2-around C-z, v-P-v, is a spanning path, a

contradiction. Thus there is no edge from C1 to C - z . By the symmetry present, one has that the only possible adjacencies between distinct cycles C , , C2, and C are the edges uz and zw. If one of these is present, then by the arguments given for the previous case, where uw E E( c), we find that with appropriate relabeling (3) holds.

If uw, uz, and wz g E ( @ , then one considers whether C,, C,, or C has chords forming odd cycles. If so, then one can show that the vertices of the cycle involved must induce a clique. One can also show that if one of the cycles has a chord and all of its chords form even cycles, then the vertices of that cycle must induce the bipartite graph K-,,. It follows from Lemma 1 that the subgraphs induced by the vertices of each cycle are randomly Hamiltonian. Therefore, (2) holds. W

Theorem 3. If G E A and i( c) = 2, then a( G) = 2.

The proof of Theorem 3 uses techniques similar to those of Theorem 2, therefore we omit the proof.

3. CONCLUDING REMARKS

One can extend Theorem 3 for small values of i(c). However, existing proofs consist of lengthy case by case assaults. The difficulty in obtaining the

ATOLL DECOMPOSITIONS OF GRAPHS 323

general result is that the natural approach would be induction. Unfortunately, it is possible to have i(G) = n and G E A , yet upon removing an island from a MID, one may find that the remaining graph is not in A. However, the evidence appears to support the validity of the following:

Conjecture. If G E A, then i( G) = a( c).

In this article, we have been concerned with minimum atoll decompositions, At the other extreme is the concept of maximum atoll decompositions. Slater [ 1 1 1 considered maximum island decompositions in which endpoints of distinct islands are required to be nonadjacent. If we further require that no islands can be extended to cycles (that is, the islands are atolls), we have a maximum atoll decomposition. Investigation of maximum atoll decompositions may lead to interesting relationships with other types of existing graph decomposition problems.

References

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[2] G. Chartrand and H.V. Kronk, Randomly traceable graphs. SIAMJ. Appl. Math. 16 ( 1968) 696-700.

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[4] S.E. Goodman, S.T. Hedetniemi, and P.J. Slater, Advances on the hamiltonian completion problem. J. Assoc. Compuf. Mach. 22 ( 1 975)

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Atoll decompositions of graphs