Sharon G. Boswell. Defence Science & Technology Org., Land Operations Div ... James A. MacDougall. Mathematics Department, University of Newcastle,. NSW ...
Minimally Path-Saturated Graphs Sharon G. Boswell Defence Science & Technology Org., Land Operations Div. - Canberra, PO Box 1500, Salisbury, SA, Australia 5108 Roger B. Eggleton Mathematics Department, Illinois State University, Normal, IL 61790-4520, USA James A. MacDougall Mathematics Department, University of Newcastle, NSW, Australia 2308 Abstract: A finite or infinite graph G is minimally P,-saturated if every edge added to G creates a new m- ath, while every edge deleted from G leaves a gra h which no longer has gat property. For rn = 3 and 4 we determine all su esa~hs.For m 2 5 we determine all rninimallv P,-saturated cvcles. paths"aid unions of paths. The minimal P,-sat;ra&on thresholds fo;
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weight classes of caterpillars and related graphs are also determined. A convenient notation for finite trees is introduced to aid the exposition. 1.Introduction Paths in graphs have been much studied because of their role in determining fundamental structural properties (connectivity, diameter, shortest routes, and so on). In this paper we examine the presence of paths of fixed length, and determine graphs of several classes which are critical with respect to the presence of such paths, in a sense that we shall now make precise. All graphs considered are simple (no loops or multiple edges), but not necessarily finite; for standard notation and terminology we will follow [I]and [6].
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Given graphs G and HI we say that G is H-free if it contains no copies of HI that is, no subgraph of G is isomorphic to H. Lf G is H-free then it is marimally H-pee if adding any new edge e to G creates a copy of HI that is, for every new edge e there is a subgraph isomorphic to H in G+e: clearly any such subgraph must contain e. More generally, whether or not G is H-free we say that G is H-hamrated if adding any new edge e to G creates a copy of H. In particular, complete graphs trivially satisfy this condition, so are H-saturated. Suppose G is H-saturated. Then it is minimally H-saturated if no spanning proper subgraph of G is H-saturated; this is equivalent to the condition that for every edge e of G the graph G-e is not H-saturated. Note that all maximally H-free graphs are minimally H-saturated, but a minimally H-saturated graph may very well contain copies of H. Hence, depending upon the choice of HI typically the class of minimally H-saturated graphs is strictly larger than the class of maximally H-free graphs. For example, if H = P g is CONGRESSUS NUMERANTIUM 138 (1999), pp.97-117
the path of order 5 and G = Cg is the cycle of order 5, then G is minimally H-saturated but it is certainly not H-free (Fig. 1).
P5 in C5+ac
No P5 contains ce inC5-ab +ce
Figure 1: A 5-cycle Cg is minimally Pg-saturated. Maximally triangle-free or minimally triangle-saturated graphs correspond to the case H = K3. Such graphs have been much studied: for example, a general study is given in [3], and constructions for families of minimally triangle-saturated graphs are given in [2,4]. Again, some recent work [5] is related to the more general case H = K,.
In this paper we study H-saturated and minimally H-saturated graphs in the case where H is a path Pm of fixed order m 2 3. Note that we shall use P, to denote the path of order m and length m-1, whereas various other authors use this notation to denote the path of length m and order m+l. 2. Paths of order 3 To begin we. determine which graphs are P3-saturated, and which among these are minimally P3-saturated. Lemma 2.1: Evely connectedgraph is P3-saturated
Proof. Trivially every complete graph is P3-saturated, so it suffices to consider any incomplete connected graph G. Let T be any spanning tree of G. Adding any new edge e creates a cycle Cr of order r 2 3 in T+e. The edge e belongs to a path P3 in the cycle Cy SO T is P3-saturated. Hence G is P3-saturated.
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Theorem 2.2: The P3-saturated graphs are precisely those graphs with at most one isolated vertex. Proof. The connected case follows from Lemma 2.1. Let G be any graph with at least two components. Each component of G is P3-saturated, by the lemma. If G has two isolated vertices an edge added between them
does not create a P3 so G is not P3-saturated. However, if the union of any two components of G has at least three vertices, any new edge added between two components is adjacent to an edge in one of those components, so creates a new path Pg in G. Hence G is P3-saturated. Now consider minimally Pg-saturated graphs. For simplicity, let us define a star to be any tree of diameter at most 2. There is a unique star of any order; in particular, the stars of orders 1 , 2 and 3 are Pi = K1, P2 = K2 and P3.
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Theorem 2.3: The minimally P3-saturated graphs are precisely the unions of stars in which exactly one has order 1 or all have order 2. Proof. Let G be a minimally P3-saturated graph. Theorem 2.2 implies
that G has at most one isolated vertex but deletion of any edge produces a spanning subgraph with at least two isolated vertices. Clearly G contains no cycle, since deleting an edge from a cycle does not create any new isolated vertices, so every component of G is a tree. Likewise G contains no path P4, since deletion of the middle edge of a P4 would not create any new isolated vertices. Hence every tree in G has diameter at most 2, so every component of G is a star. Suppose G has a component S of order greater than 2. Deleting any edge e from S would produce an isolated vertex and a star of order at least 2. But G-e is not P3-saturated, so must have more than one isolated vertex, hence G must have an isolated vertex. If G has no isolated vertex (a star of order I), all components must be stars of order 2; otherwise one component of G is a star of order 1,and any others must be stars of order at least 2. Clearly, every such union of stars is minimally P3-saturated. Corollary 2.4: The maximally P3-free graphs are precisely those of the form A or PI u A, where A is any union of paths P2. 3. Paths of order 4: the saturated graphs
In this section we determine which graphs are Pq-saturated. Those which are minimally Pqsaturated are determined in Section 5. Lemma 3.1: P3 is the only connected graph which is not Pq-saturated. ,
Proof. Adding an edge to P3 produces the cycle C3, which is Pq-free, so P3 is not Pq-saturated. We claim that every other connected graph G is Pq-saturated. Complete graphs are trivially Pq-saturated, so suppose G is incomplete and has order greater than 3. Let T be a spanning tree of G. Adding any new edge e creates a cycle Cr of order r 2 3 in T+e. If r 2 4 the edge e belongs to a path Pq in the cycle. If r = 3, some vertex v is adjacent
to the cycle in'T+e, because T is connected and has order greater than 3, so v and the three vertices of the cycle lie on a path Pq containing the edge e in T+e. Hence T is Pq-saturated, so G is Pq-saturated.
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Theorem 3.2: The Pq-saturated graphs are precisely those graphs in which P3 is not a component, and if an isolated vertex is present then no other component is a star. Proof. The single component case follows from Lemma 3.1, which also shows that a disconnected graph cannot be Pq-saturated if any of its components is a Pg. Let G be any disconnected graph in which no component is a Pg. By the lemma, each component of G is Pq-saturated. If G has an isolated vertex v and any other component is a star S then G is not Pq-saturated, because an edge added between v and a central vertex of S (which is unique unless S has order 2) does not create a Pq. On the other hand, if G has two components of order at least two, any new edge added between them is adjacent to an edge in each, so creates a new path Pq. Finally, if G has an isolated vertex v and a component A which is not a star, an edge added between v and any vertex w of A creates a new Pq, since some vertex x of A is at distance at least 2 from w. Hence G is Pq-saturated unless it has an isolated vertex and another component which is a star. The theorem now follows.
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To discuss minimally Pq-saturated graphs, it is convenient to have a systematic notation for certain trees which arise in the analysis, so we shall first develop such a notation in the next section. 4. Constructible trees Here we describe briefly how finite trees can be constructed, and derive from this a systematic notation for all finite trees. We assume that the vertices of any finite path are labelled by positive integers beginning with 1, so that the labels of the vertices incident with any edge of the path are consecutive integers.
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A stem is any finite path. Every stem is a constructible tree, and its root is the vertex labeled 1. If A is a constructible tree with stem P, and B is a constructible tree with stem P,, then the constructible tree Tformed by attaching B by s to A at v is the tree, denoted by A(v:s, B), formed from A u B by adding an edge between the vertex labelled v in the stem of A and the vertex labelled s in the stem of B. The stem and root of T are the stem and root of A.
Theorem 4.1: Eve~yfinitetree is constructible, with any maximal path as stem. Proof. Let T be any finite tree of order n, and let P be any maximal path in T. If T has no vertex of degree greater than 2 then T = P, so T is a stem,
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hence a constructible tree. If T has a vertex of degree greater than 2 then n 1 4 , and we may assume inductively that the theorem holds for all smaller trees. Since T is connected and not a path, P has a vertex v of degree greater than 2 in T. Let w be a neighbour of v not on P. Then T-vw is a pair of trees A, B each of order less than n, with v in A and w in B. Hence A is a constructible tree with P as stem, and B is a constructible tree with any maximal path through w as stem. Thus T = A(r:s, B), where r and s are the labels of v and w in their stems, so T is constructible, with P as stem.
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Corollary 4.2: The constructible trees are precisely the finite trees. Proof. Clearly every constructible tree is a finite tree, so the result follows immediately from Theorem 4.1. 0
If A, B, B' are constructible trees and T := A(r:s, B), the notation defined above implies that the constructible tree T' := T(r1:s',B') can be denoted by T' = A(r:s, B)(r1:s',B'). We define T' := A(r:s, B; rf:s', B') to condense the concatenated parenthetical expressions. This condensed notation correctly reflects the fact that the stem and root of T f are the stem and root of A. Note also that T' is isomorphic to the constructible tree T" := A(rf:s', B'; r:s, B), so with slight abuse of notation we can write A(r':sl, B'; r:s, B) = A(r:s, B; r':sf, B'). Iterating condensed notation leads to representations of the form A&), where A is a constructible tree and L is a list, each entry of which is a triple of the form r:s, B, where r and s are positive integers and B is a constructible tree. If L has consecutive terms r:s, B; r:st, Bf; ... corresponding to several attachments at the same vertex r, then L can be abbreviated by writing r:{s, B; s f , B'; ...} for these terms. For the empty list a, the condensed notation is A := A(a). Theorem 4.3: Every finite tree T is a constructible tree of the form P(L), where P is any maximal path in T and L is a finite list, possibly empty, each entry of which is a triple of the fomz r:s, B, where r and s are positive integers and B is a constructible tree. Proof. Every finite path P is a constructible tree P = P ( o ) . Let T be any finite tree which is not a path, so T has order n 14, and assume inductively that the theorem holds for all smaller trees. If P is any maximal path in T I then T is a constructible tree of the form T = A(r:s, B), where A and B are constructible trees and A has P as stem, by Theorem 4.1. But A has order less than n, so A = P(L) for some list L of the type described in the theorem, so with condensed notation T = P(L)(r:s, B ) = P ( L f ) , where L' is the concatenated list L; r:s, B. Thus T has a representation of the required f o m . C] In view of Theorem 4.3, a particular way of constructing finite trees from paths is of special interest. The construction corresponds to
specdying constructible trees by multilevel path lists, defined as follows. The empty list o is a multilevel path list for any finite path P,. If r, s and t are any positive integers with r Im and s It, and L is any multilevel path list for Pt, then r:s, Pt(L) is a multilevel path list for P,. If L and L' are any multilevel path lists for Pm, then the concatenated list L; L' is also a multilevel path list for Pm. Evidently a multilevel path list for a finite path P is just a list of finite paths together with positive integers specifying vertices of attachment. At the "top" (outermost) level, the attachments are to the path P itself. At "lower" (more nested) levels, the attachments are to appropriate paths within the list. The following result now emerges from iteration of Theorem 4.3. Corollary 4.4: Every finite tree T has a representation of the fomz P(L), where P is any maximal path in T and L is a multilevel path list for P. Let T be any finite tree, and let P be any maximal path in T. The proofs of Theorems '4.1 and 4.3 yield an easy recursive algorithm for representing T as a constructible tree P(L) with an appropriate multilevel path list L. If T is a path then T = P and L = o. Otherwise, let A, B, ... be the trees comprising the forest T-V(P), where V(P) is the vertex set of P. Then T = P(r:s, A; r':sl, B; ...),' where r, r', ... are appropriate labels for attachment vertices on P, and s, st, ... are appropriate labels for attachment vertices on maximal paths in A, B, ... respectively. Iterate this procedure for those trees A, B, ... which are not paths. Let Pt be the maximal path chosen as stem of A, and let L(A) denote the corresponding multilevel path list for Pt, so A = Pt(L(A)). Then T = P(L), where the consecutive terms of the desired multilevel path list L are of the form r:s, Pt(L(A)). Thus a multilevel path list representation is easy to produce for any given finite tree, and from such a representation the corresponding tree is easily connstructed. However, note that a given tree will usually have several multilevel path list representations. For example (Fig. 2): P3(2:3, P5(2:1, Pi; 3:2, Pg; 4:1, Pi)) = P5(2:1, Pi; 3:(2, Pg; 2, Pg}; 4:1, Pi). Terms at the "lowest" (most nested nonempty) levels of a multilevel path list are of the form r:s, Pt and it is natural to further abbreviate them, but readability must be maintained. We shall use the convention that s, Pt can be replaced by s[t], and s[tIn represents n-fold iteration of s[t] for any integer n 2 0. Further, l[t] can be replaced by [t]. Thus we . obtain (Fig. 2):
Figure 2: A tree constructed from five paths by attachment. It is convenient to extend the notation to include infinite trees. Here it suffices to let exponents take all possible cardinal values so, for example, stars of the form P2(2:[l]n) include all infinite stars as well as all finite stars of order at least 2, and trees of the form Pq(2:[lIn; 3:[3.]), with diameter 3 and stem vertex 3 of degree 2, include a l l those in which stem vertex 2 has any infinite degree as well as all those in which it has any finite degree at least 2. With this systematic notation available, let us now determine which graphs are minimally Pq-saturated. 5. Minimally P4-saturated graphs
We begin with several lemmas which help us close in on an explicit description of these graphs. There are 23 different trees of order 8, and we shall now show that several of them have a key role to play in characterizing minimally Pq-saturated graphs. Lemma 5.1: A tree T contains a forest of two components neither of which is a star if and only if T contains at least one of the trees Pg, P7(3:[1]) or Pg(3:[1]; 4:[1]).
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Proof, Each of the specified trees (included in Fig. 3) contains a forest of two components neither of which is a star, so every tree which contains any of the specified trees has the same property. Conversely, suppose the tree T has an edge ab such that neither component of T-ab is a star. Let A and B be the components containing the vertices a and b respectively. If a is a leaf of A and v is its neighbour, A must have a vertex which is not adjacent to v since A is not a star; hence a is an end
vertex of a P4 in A. If a is not a leaf of A it has at least two neighbours, and A must have a vertex not adjacent to a since A is not a star; hence a is an inner vertex of a Pq in A. Similarly for b in B. Let P be a Pq through a in A, and P' be a Pq through b in B. Then T contains ( P u P')+ab, which in every case is one of the three specified trees. [I
Figure 3: Ten special trees of order 8.
Lemma 5.2: A tree T contains a forest of two components, neither of which is a star of order 3 or less, fi and only if T contains at least one of the ten trees:
Proof. Each of the specified trees (Fig. 3 ) contains a forest of the desired type, so every tree which contains any of the specified trees has the same property. Conversely, suppose the tree T has an edge ab such that neither component of T-ab is a star of order 3 or less. Let A and B be the components containing the vertices a and b respectively. The case in which neither component is a star is covered by Lemma 5.1, so we may now suppose A is a star, of order at least 4. Then a is either a leaf or the centre of a star S of order 4. If B is not a star, then b is either an end vertex or an inner vertex of a path P of order 4 in B. Then T contains the subgraph ( S u P)+ab, which in every case is one of the four trees P ~ ( ~ : [ I~I )~, ([1]2), 2 : P ~ ( z : [ I ]4:[1]) ; or p5(2:[1]2;3:[1.1). Otherwise B is a star and b is either a leaf or the centre of a star S' of order 4. Then T contains the subgraph (S u S')+ab, in every case one of the three trees Pg(2:[1];5:[1]),p5(2:[112;4:[1])or p4(2:[112; 3:[112). [7
Lemma 5.3: A tree T of order at least 5 contains a spanning forest PI u A, such that no component of A is a star, if and only if T itself is not a star. Proof. We distinguish five cases, based primarily on the diameter of T. Case 1: T is a star. Lf e is any edge of T then T-e = PI u S where S is a star. It follows that T has no spanning forest of the specified type. Case 2: T has diameter 3. There is a maximal path Pq in T, and any other vertex of T must be adjacent to an inner vertex of this path. Hence T contains a tree Pq(2:[1]) and the leaves of any such subgraph are leaves of T, But P4(2:[l]) contains a forest PI u Pq, where PI is a leaf of T, so T contains a spanning forest PI u T', where T' is a nonstar tree. Case 3: T has finite diameter, at least 4. There is a path Pg in T with an end vertex which is a leaf of T. This Pg contains P I u Pq, where Pi is a leaf of T, so T has a spanning forest PI u T', where T' is a nonstar tree. Case 4: T has infinite diameter and at least one leaf. As in Case 3, T contains a Pg with an end vertex which is a leaf of T, so as before T has a spanning forest P I u T', where T' is a nonstar tree. Case 5: T has infinite diameter and no leaves. Let u be any vertex of T and let A := T-u. Let T' be any component of A, and let v be its vertex of attachment to u. Since T has no leaves, v cannot be the only vertex of T'. Lf w is.any other vertex of T', its degree in T' is the same as its degree in T, so w is not a leaf of TI. Hence T' is a tree of order greater than one, with at most one leaf, so T' has infinite diameter. Therefore PI u A is a spanning forest of T, where PI = u and A is a forest in which no component is a star. This completes demonstration of the lemma. C]
It follows from Lemma 5.3 and Theorem 3.2 that the only trees which are minimally Pq-saturated are stars of order different from 3. This fact is subsumed in the theorem we shall now prove, characterizing the minimally Pq-saturated graphs. In the proof a role is played by trees having a leaf adjacent to a vertex of degree 2; for brevity, we call such a leaf exposed. Theorem 5.4: The minimally Pq-saturated graphs are the graphs of theform
A or A u Pl u A u A' or A u P2(2:[1Ir)u Z u B u B', . for any cardinal r + 1; any union A of complete graphs K3; any union E of stars each of order 2 or at least 4; any union A of trees each of diameter 3 or 4; any union A' of trees each of diameter 5 or 6 and not containing Pg(3:[1];4:[1])or P7(3:[l]); any union B of trees each of diameter 3 and of the form Pq(Z:['l]; 3:[lIn)for any cardinul n 2 0; and any union B' of trees each of diameter 4 and of the form Pg or Pg(2:[l];3:{[llm;2[3In);4 : [ l ]for ) cardinals m, n 2 0.
Proof. Let G be any minimally Pq-saturated graph, and let G A be the subgraph of G remaining after deletion of every component isomorphic to K3. Suppose G A has a component H which is not a tree. If H has a
spanning tree S which is a star and e is any edge of H not in S, then S+e has order at least 4 and is a connected spanning subgraph of H containing a unique cycle C3. Let e' be any edge of the cycle different from e. Then S+e-e' is a spanning tree which is not a star. But any connected graph always has a spanning tree, so in fact H has a spanning tree H' which is not a star. But GA satisfies the conditions in Theorem 3.2, so replacing H by H' gives a proper spanning subgraph which satisfies those conditions, so is Pq-saturated. This contradicts the fact that GA is minimally Pqsaturated, so it follows that every component of GA is a tree. Further restrictions on GA depend upon whether any component is a star. We consider three cases. Case 1: One component of GA is an isolated vertex. Let G Y:= GA - Pi1 and let T be any component of GY. Because GA is minimally Pq-saturated and has an isolated vertex, Theorem 3.2 implies that T is a tree of diameter at least 3 containing none of the three trees specified in Lemma 5.1. Since T does not contain Pa, its diameter is at most 6. Thus T is a tree of diameter 3 or 4, or a tree of diameter 5 or 6 not containing Pg(3:[1]; 4:[1]) or P7(3:[l]).
Let H be any graph of the form A u P1 u A u A', where A, A and A' are as specified in the theorem. Let e be any edge of H. By Theorem 3.2, if e is in A then H-e has a component P3, so is not Pqsaturated; if e is in A u A' then H-e is not Pq-saturated, since Lemma 5.1 implies that a component of H-e-PI is a star. Thus H is minimally Pq-saturated. Hence, the graphs of the form A u P1 u A u A' are precisely the minimally Pq-saturated graphs with an isolated vertex. Case 2: One component of GA is a star P2(2:[lIr) for some cardinal r ;t 1. Then GA has no isolated vertex, by Theorem 3.2. Any component T of GA which is not a star is a tree of diameter at least 3. Since GA is minimally Pq-saturated and has no isolated vertex, Theorem 3.2 implies that T contains none of the ten trees specified in Lemma 5.2. In particular, T cannot contain Pg so its diameter is less than 7.
If the order of T is different from 5 it cannot have an exposed leaf. For otherwise it contains a spanning subgraph T-e = P2 u T' where T' is a tree of order different from 3, whence the contradiction that G A - ~ is Pq-saturated, by Theorem 3.2. Thus if it has an exposed leaf, T must be Pq(2:[1]) or Pg.
Subcase 2a: T has diameter 6. Then T contains a maximal path P7 but not P7(2:[L]),by Lemma 5.2, so each end vertex of the P7 is an exposed leaf in T. But T cannot have an exposed leaf, so no component T has diameter 6. Subcase 2b: T has diameter 5. Then T contains a maximal path Pg but not Pg(2:[1]; 5:[1]), by Lemma 5.2, so at least one end vertex of the Pg is an exposed leaf in T, a contradiction. Hence no component T has diameter 5. Subcase 2c: T has diameter 4, so contains a maximal path Pg. If T contains p5(2:[l12) its order is at least 7 and vertex 5 of the Pg is an exposed leaf in T, because T cannot contain ~g(2:[:1]~; 4:[1]), by Lemma 5.2. But T has no exposed leaf, so does not contain p5(2:[112). If it contains Pg(2:[1]) then to avoid having an exposed leaf it must contain Pg(2:[1]; 4[1]), so T = P5(2:[1]; 3:{[lIm;2[3In];4[1]) for cardinals m, n 2 0. Otherwise T = Pg. Subcase 2d: T has diameter 3, so contains a maximal path Pq. If T contains p4(2:[1l2)its order is at least 6 and vertex 4 of the Pq cannot be an exposed leaf in T. But T cannot contain p4(2:[l12;3:[112), by Lemma 5.2, so T = P4(2:[1]; 3:[lIn) for some cardinal n 2 2. Otherwise T = Pq(2:[l]) or Pq(2:[1]; 3:[1]), but not Pq, which has order 4 and exposed leaves. Let H be any graph of the form A u P2(2:[3.Ir) u C u B u B' for some cardinal r # 1, with A, C, B and B' as specified in the theorem. Let e be any edge of H. Lemma 5.2 and the above analysis imply that H-e either has a component P3, or else has two components Pi and P2(2:[1Ir). Theorem 3.2 then implies H-e is not Pq-saturated, so H is minimally Pqsaturated. Hence the graphs of the form A u P2(2:[1.Ir) u C u B u B', for any cardinal r # 1, are precisely the minimally Pq-saturated graphs with one component which is a star of order at least 2. Case 3: No component of GAis a star. In this case we claim that GAis empty, so G is simply a union of any number of components K3. For suppose GAis nonempty. Let T be any component of GA. Since T is not a star, its order is at least 4. If T = Pq, deleting its middle edge produces 2P2 so the resulting proper spanning subgraph of GAis Pq-saturated, by Theorem 3.2, contradicting the choice of GA. Hence T has order greater than 4. By Lemma 5.3, deleting a suitable set of edges from T produces a spanning forest P i u A, where no component of A is a star, so the resulting proper spanning subgraph of GAis Pq-saturated, by Theorem 3.2, again a contradiction. Therefore GA is empty. This completes the proof of the theorem.
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Corollary 5.5: The maximally Pq-free graphs are precisely the graphs of the form A u Pi or A u C, where A is any union of complete graphs K3 and C is any union of stars each of order 2 or at least 4.
6. Minimally path-saturated cycles
For the remainder of this paper we shall study Pm-saturated and minimally Pm-saturated graphs when m 2 5. In view of Section 5 we shall not seek to explicitly determine all minimally P,-saturated graphs for any case with m 15. Instead we shall confine our analyses to particular classes of graphs. Let us begin by determining which cycles C, are Pm-saturated, and which of these are minimally Pm-saturated. For n 1 4 , the cycle Cn has pairs of nonadjacent vertices and any added edge e is adjacent to four edges of C, in Cn+e. Iff is any edge adjacent to e, the graph C,+e-f is a cycle with a pendant path. Let f be
Figure 4: Adding an edge to an n-cycle creates a new n-path. the unique edge of the cycle which is adjacent to e and the attached path (Fig. 4). Deleting f leaves the spanning n-path Pn= Cn+e-f-f. (In fact, there are two disjoint choices for the set If, f )so adding any edge to C, creates two new subgraphs Pn.)Hence Theorem 6.1: I f m 2 5 and n 14, the cycle C, is P,,,-saturated
just ifn 1m.
Hamiltonian graphs are graphs with a spanning cycle, so Corollary 6.2: I f m 2 5, any hamiltonhn graph of order n 2 m is PKsaturated. Next we identify the minimally Pm-saturated cycles. Deleting any edge from a cycle leaves a path, and adding an edge to a path in general creates a cycle with two pendant paths attached at adjacent vertices (if the added edge was incident with an end vertex, the corresponding pendant path is null). Fix integers r 10, s 11, t 1 0 and let A be the graph formed by adding an edge e to a path of order r+s+t+2, so that e is incident with the vertices r+l from one end of the path and t+l from the other (Fig. 5). If the two vertices of A incident with e are deleted, the components produced are paths of orders r, s and f, so if r
and t are positive the three maximal paths in A which contain e are those on the vertices of e and two of these three paths. Thus, the maximal paths in A which contain e have orders r+s+2, r+t+2 and s+t+2. Now assign r := L(m-3)/2J, s := L(m-2)/21, t := L(m-3)/2_1 for any fixed m 2 5. Then the maximum order of any path containing e in A is
At least two of the three maximal paths containing e have this order.
Figure 5: The graph A. Putting q = s+2 and r = t, the extremal graph just described gives us Lemma 6.3: Let q := L(m+2)/2Jand r := L(m-3)/2J, with m 2 5, and let the graph A be a q-cycle with two adjacent vertices a and b, on each of which a pendant r-path is attached. The maximum order of any path in A which contains the edge ab is m-1. Theorem 6.4: For m 2 5 and n 2 4, the cycle C, is minimally Pm-saturated if and only if L(3m-5)/212 n 2 in. Proof. For C, to be Pfiaturated we need n 2 m, by Theorem 6.2. Deletion of any edge of C, leaves P,, so minimality requires that there be some edge e which can be added to P, without forming a subgraph P, which contains e. By Lemma 6.3, the edge e = ab has this property when
n = q+2r = 2r+s+2 = 2 L(m3)/2J + L(m-z)/zJ
+ 2 = L(3rn-~)/zJ.
Still with n 2 m , now choose any n < L(3m-5)/2J. Modify the graph A b truncating one or both pendant paths, removing a total of L(3m-5)/2 - n vertices: clearly the new graph itill has no subgraph Pm which contains the edge e = ab. Hence C, is minimally Pm-saturated when m I n -< L(3m-5)/21.
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Next if n 2 L(3m-5)/2J + 1 = L(3m-3)/2JYthen Cn is P,-saturated, by Theorem 6.1. Consider the graph A in Lemma 6.3. Adding a total of n - L(3m-5) / 21 vertices to a pendant path of A lengthens a maximum path containing e = ab, so e belongs to a subgraph P,. Any maximum path containing e is lengthened if one end of e is fixed at a and the other is chosen closer to a or further from a, or if e is translated along P,, or if both modifications are combined. Thus every choice of e in P,+e belongs to a P,, so C, is not minimally P,-saturated when n 2 L(3m-3)/21. In contrast, no cycle is minimally Pg-saturated, and C3 is the only cycle which is minimally Pq-saturated, by Theorems 2.3 and 5.4. (In fact, Cg is trivially minimally Pm-saturated for every m 2 4.) 7. Minimally path-saturated paths Now we consider which paths P, are P,-saturated, of these are minimally P,-saturated, for m 2 5.
and which
In the proof of Theorem 6.4, the discussion about adding vertices to the critical graph A in Lemma 6.3 implies: Theorem 7.1: Form 2 5 and n 2 3, the path P, is P,-saturated n 2 L(3m-3)/21.
ifand only if
A graph is path-hamiltortian if it has a spanning path, so Corollary 7.2: For m 2 5, path-hamiltoniangraphs of order n are Pmsaturated provided n 2 L(3m-3) /21. Next let us determine which P,-saturated paths are minimally P,-saturated. Deleting an edge from a path produces two paths, and adding an edge incident with both paths creates a tree with at most four
Figure 6: The graph H.
i
in
leaves. Fix integers q 2 r 2 0 and s 2 t 2 0, and let H be the tree formed by adding an edge e between a path of order q+r+l and a path of order s+t+l, so that e is incident with vertex q+l of the first path and with vertex s+l of the second path (Fig. 6). Deleting the two vertices of H incident with e leaves paths of orders q, r, s and t, so if all these are positive the four maximal paths containing e in H have orders q+s+2, q+t+2, r+s+2 and r+t+2. In any case the maximum order of any path containing e in H is equal to q+s+2. Now assign q := m-k-3, r := m-k-3, s := k, t := k, with m 2 5 and 0 I k I m-3. Then q+s+2 = m-1, so we have Lemma 7.3: Let q := m-k-3 and s := k, with m 1 3 and 0 I k I m-3, and let Hk be the tree of order 2m-4 comprising an edge e incident with the midvertices of two paths, of orders 2q+l and 2s+l. The marimum order of any path containing e in Hk is m-1. The extremal graphs Hk play a role for minimally Pm-saturated paths which is analogous to that played for cycles by the extremal graph A in Lemma 6.3. Theorem 7.4: For m 2 5 and n 2 3, the path P, is minimally P,-saturated only if L(3rn-3)/211 n If(m), where A m ) = 2m-4 if5 5 m 5 7, andflm) = 2m-5 ifm 2 8.
if and
Proof. The path Pzm-5 is Pm-saturated when 2m-5 1L(3m-3)/2], hence when m Z 6, by Theorem 7.1. For any edge e' the graph P2,-5-e' comprises two paths, one of odd order and one of even order. Append an extra vertex v to the end of the even path to form P q + l u P2s+l, where q and s are integers satisfying q 2 s 2 0 and q+s = m-3. Now adding an edge e incident with the midvertices of both paths produces the tree Hs. Thus for any edge e' of Pzrn-5 we can choose an edge e so that P2,-5-ef+e = Hs-v, where v is a leaf of Hs. By Lemma 7.3, no m-path in Hs contains the edge e, so adding e to P2,-5-e' does not create an m-path. Hence Pzrn-5 is minimally Pm-saturated if m 2 6. Now consider the path Pem-4, which is Pm-saturated if m 1 5, by Theorem 7.1. When P2m-q-e' comprises two paths of odd order, analysis like the above shows there is an edge e such that Pzm-4-e1+e = Hs for some s, so adding e does not create an m-path, by Lemma 7.3. The alternative case is when Pzm-4-e' comprises two paths of even order, say P2m-4-e' = Pzr u Pzs where the integers r and s satisfy r 2 s 2 0 and r+s = m-2. If s 2 2 we can add any edge e to the smaller path P2, without creating a Pm because P2,+e has order 2s < m. If s = 1 then P2m-4-e' = P2rn-6 u P2. No edge can be added to P2, and adding an edge e between P2rn-6 and P2 produces a maximal path of order at least m, since each of the two central vertices of Pzm-6 is the end vertex of a subgraph Pm-2.
Theorem 7.1 shows P2m-6 is Pm-saturated i f 2m-6 1L(3m-3)/21, hence when m 18 adding an edge e to P2m-6 always creates a new m-path. If 5 I m I 7, the larger component of P2m-6 v P2 is not Pm-saturated so a suitable choice of e does not create an m-path in P2rn-6t-e. In summary, we have shown that P z m 4 is minimally Pm-saturated precisely when the spanning subgraph P2m-6 v P2 is not Pm-saturated, which occurs just when5ImI7. To complete the proof we need to check that no path of order more than 2 m 4 is minimally Pm-saturated. When m 1 5 aTld n 12m-3, the path P,-i is Pm-saturated, by Theorem 7.1, so evidently Pn-1 v PI is Pm-saturated, and hence the spanning supergraph P, is not minimally Pm-saturated.
0
The complete graphs P1 and P2 are minimally Pm-saturated (trivially) for every m 1 3 . They are the only minimally Pm-saturated paths when m = 3 or 4, by Theorems 2.3 and 5.4. 8. Minimally path-saturated unions of paths
Every spanning proper subgraph of a path is a union of paths, so if a particular path is Pm-saturated but not minimally so, any spanning subgraph which is minimally Pm-saturated must be a union of paths. We now determine which unions of paths are Pm-saturated, and which are minimally so. Note that a disconnected graph is Pm-saturated if and only if each incomplete component is Pm-saturated, and every pair of components is P,-saturated. Routine computation for appropriate graphs H (Fig. 6 ) shows Lemma 8.1: For m 1 5, a new path of order at least m is created whenever any edge is added between (a) two paths of order at least L(3m-3)/2] ; (b) Pi and a path of order at least M ; or (c) P2 and a path of order at least 2m-6. Theorem 8.2: For m 2 5, any union of two or more paths is Pmsaturated ifand only if it falls into one of the following cases: (a) all components huve order at least L(3m-3)/2J; (b) one component is Pi and all others huve order at least 2m4; or (c) one component is P2 and all others have order at least g(m), where g(m) = L(3m-3)/21 for 5 < m S 7 and g(m) = 2m-6 far m 1 8. Proof. For m 2 5, let G be a Pm-saturated union of two or more paths. Since any incomplete component must be Pm-saturated, such components have order at least L(3rn-3)/21, by Theorem 7.1. Now consider possible complete components of G.
-
Case 1:Components PI. Since P2m-5 u PI is not Pm-saturated, by Lemma 7.3, it follows that P, u PI is not Pm-saturated if 1 I n 12m-5 so if G has a component PI then every other component has order at least 2m-4. But 2m-4 2 L(3m-3)/21. Therefore every other component is incomplete and Pm-saturated. Case 2: Components P2. Since P2m-7 u P3 is not Pm-saturated, by Lemma 7.3, it follows that P, u P2 is not Pm-saturated if 1 1n I2m-7, so if G has a component P2 then every other component has order at least 2m-6. If 5 I m I 7 then every other component is incomplete and Pm-saturated, so has order at least L(3m-3)/21, which exceeds 2m-6. If m 2 8 then every other component has order 2m-6 2 L(3m-3)/21 so is incomplete and Pm-saturated. Conversely, with Lemma 8.1 it is routine to venfy that if G falls in any of the cases specified in the theorem, it is Pm-saturated. Corollary 8.3: Form 2 5, a union of two or more path-hamiltonian graphs is P,saturated ifand only ifit falls into one of the following cases: (a) all components have order at least L(3m-3)/21; ( b ) one component is PI and all others have order at least 2m-4; or ( c ) one component is P2 and all others have order at least g(m), where g(m) = L(3m-3)/21 for 5 I m I 7 and g(m) = 2m-6 for m 2 8. Note that Corollary 8.3 admits as components any path-hamiltonian graphs, not just paths. Theorem 8.2 also allows us to determine which unions of paths are minimally Pm-saturated. Theorem 8.4: For m 2 5, a union of two or more paths is minimally P,saturated ifand only if it falls into one of the following cases: (a) each component has order at least L(3m-3)/21 and at mostf(m), where flm) = 2m-4 for 5 I m I 7 andflm) = 2m-5 for m 2 8; one component is PI and all others have order at least 2m-4 and at most 4m-9; (c) one component is P2 and all others have order at least g(m) and at most 2g(m)-1, where g(m) = L(3m-3)/21 for 5 5 m 5 7 and g(m) = 2m-6 for m 2 8.
6)
Proof. Deletion of an edge from a disjoint union of paths always produces a disjoint union of paths. Hence if G is a P,-saturated disjoint union of paths, it is not minimally Pm-saturated precisely when some edge e can be selected so that G-e falls in one of the cases specified in Theorem 8.2. It is now routine to venfy the details stated in the present theorem. The corresponding results for m = 3 or 4 are contained in Theorems 2.3 and 5.4, respectively. The unions of paths which are minimally P3saturated are all those of the form A or Pl u A, where A is a union of
paths P2. The unions of paths which are minimally Pq-saturated are all those of the form Pi u A or P2 u B, where A is a union of paths each of order 4,5,6 or 7, and B is a union of paths each of order 2 or 5. 9. Caterpillars and other appendaged paths For any integer n 2 3, an n-cateppillar is an n-path Pn with one or more P i attachments to its inner vertices, so the stem Pn is a maximal path and any other vertex is a leaf. The weight of an n-caterpillar is the maximum number of Pi attachments at any stem vertex. We shall now investigate which caterpillars are Pm-saturated. Let P* be a Pm-saturated n-caterpillar of weight r. If P* has no PI attachment at some inner vertex v of its stem and we attach Pi = w at v, the resulting graph P*+w is still an n-caterpillar of weight r. Add an edge e to P*+w . If e is not incident with w, it creates an m-path in P*+e, so creates an m-path in P*+w+e. If e is incident with w, say e = uw, there are two cases. If u is not adjacent to v, then uv can be added to P* and creates an m-path P, so adding e =uw to creates P-uv+uwv, an (m+l)-path formed by subdividing the edge uv in P with the vertex w. On the other hand, if u is adjacent to v, then adding e = uw to P*+w creates Pn-uv+uwv, where Pn is the stem of P*. This is an (n+l)-path formed by subdividing the edge uv in Pn with the vertex w, so P*+w is Pm-saturated if n 2 m-1. Similarly, simply subdividing any stem edge uv of P* produces an (n+l)-caterpillar P*-uv+uwv, of the same weight, which is also Pm-saturated. Hence Lemma 9.1: Let P* be a P,,+aturated n-caterpillar of weight r. Then (a) attaching Pi to any stem vertex v of degree 2 produces an n-caterpillar P*(v:Pi) of weight r which is also Pm-saturated ifn 2 m-1; (b) simply subdividing a stem edge of P*producesan (n+l)-caterpillar of weight r which is also Prsaturated.
In general, the set of all n-caterpillars with weight r is quite varied, so we shall in fact determine whether each such set contains any member which is not Pm-saturated. To achieve this it is helpful to extend our terminology to sets of graphs. If C is any set of graphs, then C is Pm-saturated if every graph G E C is Pm-saturated; equivalently, C is not Prsaturated if at least one G E E is not P,-saturated. Theorem 9.2: With m 1 5 , the set of n-caterpillars of weight r is Prsaturated if and only if r = 1 and n 2 L(3m-3)/21, or r 2 2 and n 2 f(m), where flm) = L(3m-3)/21 if5 Im I 7, and Am) = 2m-6 ifm 2 8. Proof. No path of order less than L(3m3)/2J can be Pm-saturated, by Theorem 7.1. For any weight r 2 1 the caterpillar P,( L(m-1)/21 : [1Ir) is
not P,-saturated if n = L(3m-5)/2J1 because adding an edge between the stem vertices L(m-l)/zJ and m-1 does not create an m-path. But for any n 2 L(3m-3)/2J and any k satisfying 1 < k < n, adding any edge to the weight 1 caterpillar Pn(k: [I]) does create a P,, so every n-caterpillar of weight 1 with n 2 L(3m-3)/2J is P,-saturated. Let f(m) := m a { L(3m-3)/2J1 2rn-6). For k satisfying 1 < k
