Spanning Trees with Many Leaves in Graphs without Diamonds and Blossoms

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1 Spanning Trees ith Many Leaes in Graphs ithot Diamonds and Blossoms Pal Bonsma Florian Zickfeld Technische Uniersität Berlin, Fachbereich Mathematik Str. des 7. Jni 36, 0623 Berlin, Germany Abstract It is knon that graphs on n ertices ith minimm degree at least 3 hae spanning trees ith at least n/4 + 2 leaes and that this can be improed to (n + 4)/3 for cbic graphs ithot the diamond K 4 e as a sbgraph. We generalize the second reslt by proing that eery graph ith minimm degree at least 3, ithot diamonds and certain sbgraphs called blossoms, has a spanning tree ith at least (n+4)/3 leaes, and generalize this frther by alloing ertices of loer degree. We sho that it is necessary to exclde blossoms in order to obtain a bond of the form n/3 + c. We se the ne bond to obtain a simple FPT algorithm, hich decides in O(m) + O (6.75 k ) time hether a graph of size m has a spanning tree ith at least k leaes. This improes the best knon time complexity for Max Leaf Spanning Tree. Introdction In this paper e stdy spanning trees ith many leaes. We proe a ne extremal reslt, and apply it to obtain a fast FPT algorithm for the related decision problem MaxLeaf. We first introdce the extremal problem and explain or contribtion. Throghot this paper G is assmed to be a simple and connected graph on n 2 ertices. Other graphs may be mlti-graphs, disconnected, or a K. The minimm ertex degree of G is denoted by δ(g). Vertices of degree are called leaes. Linial and Strteant [2] and Kleitman and West [] shoed that eery graph G ith δ(g) 3 has a spanning tree ith at least n/4+2 leaes, and that this bond is best possible. The paper [] also improes on this bond for graphs of higher minimm degree. The examples shoing that n/4 + 2 is best possible for graphs of minimm degree 3 all consist of cbic diamonds connected in a cyclic manner. A diamond is the graph K 4 mins one edge, and an indced diamond sbgraph of a graph G is a cbic diamond if its for ertices all hae degree 3 in G, see Figre (a). Since these examples are ery restricted it is natral to ask if better bonds can be obtained hen diamonds are forbidden as sbgraphs. This qestion as ansered by Griggs, Kleitman and Shastri [0] for cbic graphs, hich are graphs here eery ertex has degree 3. Spported by the Gradate School Methods for Discrete Strctres in Berlin, DFG grant GRK 408 Spported by the Stdienstiftng des detschen Volkes

2 (a) (c) Figre : A cbic diamond (a), a 2-necklace, and a 2-blossom (c). They sho that a cbic graph G ithot diamonds alays admits a spanning tree ith at least n/3 + 4/3 leaes. For minimm degree 3 the folloing bond is proed in [2]. A graph G ith δ(g) 3, ithot cbic diamonds, contains a spanning tree ith at least 2n/7+2/7 leaes. Both bonds are best possible for their respectie classes. In [2] it is conjectred that the folloing statement holds, hich old improe the bond 2n/7+2/7 ith only a minor extra restriction, and old also generalize the reslt for cbic graphs from [0]: Eery graph G ith δ(g) 3 and ithot 2-necklaces contains a spanning tree ith at least n/3 + 4/3 leaes. Informally speaking, a 2-necklace is a concatenation of k diamonds ith only to otgoing edges, see Figre. In Section 2 e disproe this conjectre by constrcting graphs ith δ(g) = 3 ithot 2- necklaces, hich do not admit spanning trees ith more than 4n/3+2 leaes. On the positie side, e proe that the statement is tre after only exclding one more ery specific strctre, called a 2-blossom, see Figre (c). Precise definitions of 2-necklaces and 2-blossoms are gien in Section 2. So e proe that graphs G ith δ(g) 3 ithot 2-necklaces or 2-blossoms hae a spanning tree ith at least n/3+4/3 leaes. In fact e generalize this statement een frther by remoing any restriction on the minimm degree. The reslting statement is gien in Theorem, hich is or main reslt. Let V 3 (G) denote the set of ertices in G ith degree at least 3 and n 3 (G) its cardinality. V 3 (G) Let l(t) be the nmber of leaes of a graph T. n 3 (G) Theorem. Let G be a simple, connected graph on at least to ertices hich contains neither 2-necklaces nor 2-blossoms. Then, G has a spanning tree T ith { 4/3 if δ(g) 3 l(t) n 3 (G)/3 + 2 if δ(g) 2. Section 2 shos that Theorem is also best possible for non-cbic graphs. Withot proof e remark that also other reslts can be generalized in a similar ay, e.g. it is not hard to extend the proof in [] to proe that all graphs G hae a spanning tree ith at least (n n 2 (G))/4 + c leaes, here n 2 (G) denotes the nmber of ertices of degree 2 in G. Or proof of Theorem is constrctie and can be trned into a polynomial time algorithm for the constrction of a spanning tree. The main technical contribtion of this paper is that e proe this generalization of the statement in [0], and improement of the statement in [2], ithot a proof as lengthy as the proofs in these to papers. This is made possible by extending the techniqes and proofs from [0]. In Section 3 e arge that the long case stdy in [0] actally proes a strong ne lemma, hich e se as an important step in the proof of Theorem. We share the opinion expressed in [0] that a shorter proof of the bond for cbic graphs might not exist. Therefore sing that reslt in order to proe the more general statement seems appropriate. We no explain the conseqences that Theorem has for FPT algorithms (short for fixed parameter tractable) for the folloing decision problem. l(t) 2

3 Max-Leaes Spanning Tree (MaxLeaf): INSTANCE: A graph G and integer k. QUESTION: Does G hae a spanning tree T ith l(t) k? It is knon that MaxLeaf is N P-complete, see [9]. When choosing k as a parameter, an algorithm for this problem is called an FPT algorithm if its complexity is bonded by FPT f(k)g(n), here g(n) is a polynomial. See [8] and [5] for introdctions to FPT algorithms. algorithm f(k) is called the parameter fnction of the algorithm. Usally, g(n) ill trn ot to be parameter a lo degree polynomial, ths to assess the speed of the algorithm it is mainly important fnction to consider the groth rate of f(k). Thogh since MaxLeaf is N P-complete, f(k) ill most likely alays be exponential. Bodlaender [] constrcted the first FPT algorithm for MaxLeaf ith a parameter fnction of roghly (7k 4 )!. Since then, considerable effort has been pt in finding faster FPT algorithms for this problem, see e.g. [4, 7, 3, 6, 2]. The papers [3, 6, 2] also establish a strong connection beteen extremal graph-theoretic reslts and fast FPT algorithms. In [3], the bond of n/4+2 from [] mentioned aboe is sed to find an FPT algorithm ith parameter fnction O ( ( ) 4k k ) O (9.49 k ). Here the O notation ignores O polynomial factors. With the same techniqes the bond of 2n/7+2/7 mentioned aboe is is trned into the so far fastest algorithm, ith a parameter fnction in O ( ( 3.5k) k ) O (8.2 k ) ([2]). Similarly Theorem yields a ne FPT algorithm for MaxLeaf, presented in Section 4. Theorem 2. There exists an FPT algorithm for MaxLeaf ith time complexity O(m) + O (6.75 k ), here m denotes the size of the inpt graph and k the desired nmber of leaes. This algorithm is the fastest FPT algorithm for MaxLeaf at the moment, both optimizing the dependency on the inpt size and the parameter fnction. It simplifies the ideas introdced by Bonsma, Breggemann and Woeginger [3] and is also significantly simpler than the other recent fast FPT algorithms. Hardly any preprocessing of the inpt graph is needed, since Theorem is already formlated for a ery broad graph class. We end in Section 5 ith a discssion of possible extensions and frther conseqences of Theorem. 2 Obstrctions for Spanning Trees ith Many Leaes 2. Diamond Necklaces, Blossoms and Floers As mentioned in the introdction 2-necklaces hae been identified as an obstrction for the existence of spanning trees ith n/3 + c leaes in graphs ith minimm degree 3, see [] and [2]. In this section e sho that they are not the only sch obstrction. We start by precisely defining 2-necklaces and 2-blossoms. The degree of a ertex in a graph G is denoted by d G () and by d() if ambigities can be exclded. A ertex of a sbgraph H of G ith d H () < d G () is called a terminal of H. terminal Definition (2-Necklace). The graph K 4 mins one edge is called a diamond and denoted diamond by N. The degree 3 ertices are the inner ertices of the diamond. inner For k 2 the diamond necklace N k is obtained from the graph N k and a ertex disjoint ertices N by identifying a degree 2 ertex of N ith a degree 2 ertex of N k. Ths, N k has again diamond to degree 2 ertices, hich are denoted by c and c 2. necklace An N k sbgraph of G is a 2-necklace if it only has c and c 2 as terminals, hich both hae 2-necklace degree 3 in G. See Figres (a) and. If G contains an N this ay, this N sbgraph is also called a cbic diamond of G. 3 cbic diamond

4 Diamond necklaces ill also be called necklaces for short. In the corse of stdying the leafy tree problem e fond that the sbgraphs defined next are also an obstacle for the existence of spanning trees ith n/3 + c leaes. Definition 2 (2-Blossom). The graph B on seen ertices shon in Figre 2 (a) is the blossom graph. A blossom sbgraph B of G is a 2-blossom if c and c 2 are its only terminals, and they both hae degree 3 in G, see Figre 2. c 2 (a) a a 2 c b a 3 a 4 (c) Figre 2: A blossom graph (a), a 2-blossom, and a floer (c). If G contains a 2-blossom B, only the ertex b has degree 4 in G, and the remaining ertices of B hae degree 3 in G. The to otgoing edges of a 2-necklace respectiely a 2-blossom may in fact be the same edge, in that case G is jst a 2-necklace respectiely 2- blossom pls one additional edge. The next proposition shos ho many leaes can be gained ithin a blossom. (a) c c 2 c c 2 Figre 3: Spanning trees restricted to a blossom Proposition. Let G be a graph ith a blossom sbgraph B that has c and c 2 as its only terminals. A spanning tree T of G exists ith maximm nmber of leaes, sch that E(T) E(B) has one of the forms in Figre 3. Proof. Consider a spanning tree T of G ith maximm nmber of leaes. We may distingish the folloing to cases for E(T) E(B): either this edge set indces a tree, or it indces a forest ith to components, one containing c and the other containing c 2. In the first case, at most three non-terminal ertices of B can be leaes of T, since a path from c to c 2 contains at least to internal ertices. In addition, if one of c and c 2 is a leaf of T, then T can be seen to hae at most to non-terminal ertices of B among its leaes. Since c and c 2 together form a ertex ct of G, one of them is not a leaf in T. It follos that replacing E(T) E(B) by the edge set in Figre 3 (a) does not decrease the nmber of leaes. Since this edge set forms again a spanning tree of B, the reslting graph is a spanning tree of G. No sppose E(T) E(B) forms to components. At most for non-terminal ertices of B can be leaes of T. If one of c and c 2 is a leaf of T, then T can hae at most three non-terminal ertices of B among its leaes. One of c and c 2 is not a leaf in T, and ths it follos again that replacing E(T) E(B) by the edge set in Figre 3 does not decrease the nmber of leaes, hile maintaining a spanning tree of G. We no present a family of graphs ith minimm degree 3 hich do not contain diamond necklaces bt do not hae spanning trees ith n/3 + c leaes. 4

5 Definition 3 (Floer, Floerbed). The floer graph is the graph on thirteen ertices shon floer graph in Figre 2 (c). The floerbed R i of length i consists of i floers, connected in a cyclic manner, see Figre 4. floerbed Formally, R i is constrcted by starting ith i disjoint floers, and adding i edges in sch a ay that the graph is connected and has minimm degree 3. Figre 4 shos the floerbed R 5. The solid edges sho a spanning tree ith 4n/3 + 2 leaes, hich e ill sho to be optimal. Figre 4: The floerbed R 5. The solid edges sho a tree ith maximm nmber of leaes. Proposition 2. The floerbed R i has no spanning tree ith more than 4n/3 + 2 leaes. Proof. Let F be a floer in R i containing a blossom B, here c and c 2 are the terminals of B. The neighbor of c j not in B is called f j (j =, 2). We ill arge that no spanning tree T of R i has more than for leaes among V (B) {f, f 2 }. Proposition shos that ithot loss of generality e may assme that E(T) E(B) has one of the to forms in Figre 3. If it has the first form, then one of f and f 2 may be a leaf of T, bt not both since together they form a ertex ct of R i. If E(T) E(B) has the second form, then f and f 2 are both ct ertices of T, so neither can be a leaf. No e consider the other ertices of R i that may be leaes in T. Let C be the cycle in R i that joins the i floers, that is, the facial cycle of length 2i in Figre 4. Sppose V (C) is a leaf of T. In G, all ertices of C except one are ct ertices, so T may hae at most one other ertex of C as a leaf. It follos that at most to ertices from C can be leaes in a spanning tree T of R i. The remaining ertices of R i that e hae not considered yet (to for eery floer) are ct ertices of R i and therefore not leaes of T. Smmarizing, any spanning tree has at most to leaes in C, and at most for additional leaes for eery floer. The statement follos. 2.2 Tightness of the Bond The bond n/3+4/3 for cbic graphs is shon to be tight in [0]. Infinitely many examples are gien ith no more than n/3 + 2 leaes. On the other hand it is shon that there exists only one graph that ensres that the additie term 4/3 can not be increased: the 3-dimensional cbe Q 3, hich has eight ertices and only admits for leaes. Becase the bond is best possible for cbic graphs, or bond is best possible as ell. Bt also graphs ith arbitrarily many ertices of higher and loer degree can be constrcted hich do not admit more than n 3 /3 + 2 leaes. Figre 5 (a) shos sch an example ith many degree 2 and degree 4 ertices (hich is closely related to one of the examples from [0]). 5

6 The reason that the additie term cannot be increased to 2 is again only one example: Figre 5 shos a graph on n = 7 ertices that only admits 4 = n 3 /3 + 5/3 leaes. This graph ill be called G 7 in the remainder. This graph is in fact a blossom pls to edges; deleting any edge beteen to degree 4 ertices yields a 2-blossom. An additie constant of 5/3 is possible for non-cbic graphs, bt e ill not proe this statement in this ersion of the paper. (a) Figre 5: Non-cbic extremal graphs. G 7 3 Proof of the Main Theorem This section is deoted to the proof of the main theorem. We first sketch the proof and gie an oerie of the different ingredients that ill be sed. First e introdce a nmber of redction rles in Section 3.. These redction rles are applied to the graph G ntil an irredcible graph G is obtained. These rles hae the property that if the main theorem holds for eery component of G, it also holds for G. In the next sections, e therefore only hae to consider irredcible graphs. In Section 3.2 e arge that the proofs from [0] for cbic graphs in fact sho that if a non-spanning forest F of G is gien, that contains all ertices of G of degree at least 4, then one of the trees of F can be extended to a larger tree hile maintaining the proper leaf ratio. Finally, in Section 3.3 e sho ho to obtain this starting forest F that coers all high degree ertices, hile haing enogh leaes. We se these tools in Section 3.4 to proe Theorem. 3. Redcible Strctres In this section e introdce a nmber of redction rles. The proof of the main theorem relies on locally extending a forest ntil it becomes spanning hile garanteeing a certain nmber of leaes for eery intermediate forest. The redctions help to delay the treatment of some sbstrctres hich cannot be readily handled dring the extension process and they also simplify the case stdy in the main proof. Ignoring rles that disconnect the graph, the main idea behind the redction rles is as follos. A graph G is redced to a graph G ith n 3 (G) n 3 (G ) = k, sch that eery spanning tree of G can be trned into a spanning tree of G ith at least k/3 additional leaes. This preseres the desired leaf ratio. Lemma 3 states this idea more precisely. We no gie the necessary definitions. A ertex ith degree at most to ill be called a goober. We adopt this notion from [0], althogh there it is defined differently. In [0] goober goobers are those ertices of degree at most to reslting from a redction rle. We obsere that nohere in the proofs the extra strctral information hich this definition may proide is actally sed. Hence goobers may simply be defined as e do here. The important gain is that no e do not hae to reqire graphs to hae minimm degree 3 in or statements. 6

7 One important conention is that goobers are alays defined ith respect to the hole graph, that is hen e consider a sbgraph H of G, a ertex of H is a goober if d G () 2. In or figres, hite ertices indicate goobers. A high degree ertex is a ertex of degree at least 4. We first repeat the seen redction rles defined in [0], and then introdce fie ne rles hich are designed to handle strctres containing higher degree ertices. While the first seen rles are defined in [0] for graphs ith maximm degree 3 e define them for arbitrary graphs, bt the ertices on hich they act mst hae the same degrees as in the original definition. The seen redction rles from [0] consist of graph operations on certain strctres, and conditions on hen they may be applied. Figre 6 shos the operations. The black ertices all hae degree 3, and goobers are shon as hite ertices. Dashed edges are present in the reslting graph if and only if they exist in the original graph. The nmbers aboe the arros indicate the decrease in n 3, and the nmbers belo the arros indicate the nmber of leaes that can be gained in a spanning tree hen reersing the redction. () 6 2 (2) 0 0 high degree ertex (3) 3 (4) 6 2 (5) (6) (7) 2 4 Figre 6: The seen lo degree redction rles The folloing restrictions are imposed on the application of these rles (see Section 3 of [0]): Redctions (), (3), (4) and (5) may not be applied if the to otgoing edges from the left side, or the to otgoing edges from the right side, share a non-goober end ertex. (An otgoing edge from the left and an otgoing edge from the right may share a non-goober end ertex.) Redction (7) may not be applied if any pair of otgoing edges shares an end ertex. In other ords, a rle may not be applied if it old introdce mlti-edges incident ith non-goobers, or if it old introdce a diamond. These seen redction rles ill be called the lo-degree redction rles. We define an inariant that exhibits the properties hich shold be maintained hile doing graph redctions. Definition 4 (Inariant). A graph H is said to satisfy the inariant if: H is connected, or eery component of H contains a goober, and lo-degree redction rles satisfy the inariant 7

8 eery component of H is either simple or it is a K 2 + e, and H contains neither 2-necklaces nor 2-blossoms. The redction rles are applied in the indction step in the proof of or main theorem; this inariant states the important properties that shold be presered in the redction process. Lemma. Let G be obtained from G by the application of a lo-degree redction rle. If G satisfies the inariant then so does G. Proof. Note that the redction rles ()-(6) only introdce goobers as ne ertices and the only ne edges are incident to these goobers. Frthermore all other ertex degrees remain nchanged. Hence these redctions cannot introdce 2-necklaces or 2-blossoms. Rle (7) cannot introdce a 2-blossom since a 2-blossom cannot share a ertex ith a triangle indced by three ertices of degree three. This is not tre for 2-necklaces, bt if rle (7) introdces a 2-necklace, to of the otgoing edges share an end ertex, contradicting the condition for applying rle (7). For all of the rles that may disconnect the graph, it is clear that both ne components ill contain a goober. So the only ay in hich one of the redctions might iolate the inariant is by introdcing mltiple edges. Bt sing the imposed restrictions it can be seen that mltiple edges can only be introdced beteen to goobers, giing a K 2 + e. We no introdce fie ne redction rles, hich e call the high-degree redction rles. high-degree Each rle again consists of a graph operation and conditions on the applicability. Figre 7 redction shos the graph operations for the fie rles. rles The encircled ertices are the terminals, hich may hae frther incidences, nlike the other ertices. None of the ertices in the figres may coincide, bt there are no restrictions on otgoing edges sharing end ertices. The nmbers aboe the arros indicate the decrease in n 3, and the nmbers belo the arros indicate the nmber of leaes that can be gained in a spanning tree hen reersing the redction. Since (R4) mst disconnect a component, this notion is not releant for (R4); this rle ill be treated separately belo. (R) (R3) 3 (R2) (R5) 0 0 (R4) t t Figre 7: The high-degree redction rles. The folloing restrictions are imposed on the applicability of these operations to a graph G. First, none of the redction rles may be applied if it introdces a ne 2-necklaceor 2-blossom. In addition, the folloing rle-specific restrictions are imposed. Let cc(h) denote the nmber cc(h) of connected components of a graph H. (R) d G () 4. 8

9 (R2) d G () 4 and d G () 4. (R3) cc(g ) = cc(g), the edge is not in G, and in addition d G () 3, or d G () 3, or both. (R4) cc(g ) > cc(g), that is G is not connected. (R5) d G () 4, d G () 4, and may not be a bridge. A bridge is an edge hose deletion increases the nmber of components. In the remainder, e bridge ill call a redction rle admissible if it can be applied ithot iolating one of the imposed admissible conditions. In particlar the condition that no 2-necklaces or 2-blossoms are introdced ill be important. Since the high-degree redctions rles are defined sch that no 2-necklaces or 2-blossoms can be introdced, it is easy to see that the folloing lemma holds: Lemma 2. Let G be obtained from G by the application of a high-degree redction rle. If G satisfies the inariant then so does G. Obsere that in particlar, (R5) seems conterprodctie hen the goal is to find spanning trees ith many leaes, bt it is sefl to keep the case analysis in the proof of Lemma 6 simple. Definition 5 (Redcible). A graph G is redcible if one of the lo-degree or high-degree redcible redction rles can be applied, and irredcible otherise. irredcible Griggs et al [0] call a graph irredcible if none of the lo-degree redction rles can be applied. Clearly, a graph that is irredcible according to or definition is also irredcible according to their definition, so e may apply their lemmas for irredcible graphs also sing the aboe definition of irredcibility. Note that irredcible graphs satisfying the inariant are simple becase of redction rle (2). Components ith only one ertex ill be called triial components in the seqel. We no sho that e can reerse all of these redction rles hile maintaining spanning trees for eery component, haing the proper nmber of leaes. For the lo-degree redction rles, this lemma as implicitly proed in [0]. So for the detailed tree reconstrctions e refer to [0], bt e do repeat the main idea behind the proof here. triial components Lemma 3 (Reconstrction Lemma). Let G be the reslt of applying a redction rle to a connected graph G and α 0. If G has k non-triial components C,...,C k, hich all hae a spanning tree ith at least n 3 (C i )/3 + α leaes, then G has a spanning tree T ith l(t) n 3 (G)/3 + αk 2(k ). Note that the redction rles create at most to components, that is k 2. We se this lemma ith α = 4/3 if G is connected, and ith α = 2 otherise. Proof. Sppose the applied rle as a lo-degree redction rle. Note that cc(g ) is either or 2. If G is connected, then its spanning tree can be trned into a spanning tree of G ith (n 3 (G) n 3 (G ))/3 more leaes. To proe this, it is shon in Section 3 of [0] for eery rle ho to adapt the tree of G for G (tree reconstrctions). This already proes the statement if cc(g ) =. If cc(g ) = 2, then applying the same tree reconstrctions yields a spanning forest of G consisting of to trees, ith again (n 3 (G) n 3 (G ))/3 more leaes in 9

10 total. If both components of G are non-triial (k = 2), then the to reslting trees of G can be connected to one spanning tree T by adding one edge, losing at most to leaes. In that case e hae: l(t) n 3 (G )/3 + 2α + (n 3 (G) n 3 (G ))/3 2 = n 3 (G)/3 + αk 2(k ). If exactly one of the to components is triial (k = ) then the applied rle mst be Rle (2) or (3). In this case, it can be checked that after the tree reconstrction, one edge can be added ithot decreasing the nmber of leaes; one leaf is lost bt an isolated ertex becomes a leaf. Then e hae: l(t) n 3 (G )/3 + α + (n 3 (G) n 3 (G ))/3 = n 3 (G)/3 + αk 2(k ). If both components of G are triial (k = 0), then the applied rle as (2), and G = K 2, for hich the statement holds: 2(k ) = 2. This proes the lemma hen a lo-degree redction rle is applied. (R) (R2) (R3) (R4) t t Figre 8: Spanning tree constrctions hen reersing the ne redction rles No e consider the high-degree redction rles. Note that rles (R), (R2), (R3) and (R5) do not increase the nmber of components, so k =. So for (R5) e do not hae to change the spanning tree of G. For (R), (R2) and (R3), Figre 8 shos ho to gain at least one additional leaf in eery case, hich sffices since each of these rles decreases n 3 by at most three. Here it is essential that (R3) is admissible only if it creates at most one goober. Dashed edges in the figre are present on the right if and only if they are present on the left. Symmetric cases are omitted in the figre. Note that none of the terminals of the operations can lose leaf stats, except in the second reconstrction for (R3). This is compensated by gaining to ne leaes here. So in eery case enogh leaes are gained to maintain the ratio. Recall that (R4) is only admissible if it disconnects G into to components, hich ill be non-triial, so k = 2. Figre 8 shos ho to constrct a spanning tree for G from the to spanning trees for the components, ithot decreasing the total nmber of leaes. Hence the nmber of leaes of the reslting tree is at least n 3 (G )/3 + 2α = n 3 (G)/3 5/3 + 2α > n 3 (G)/3 + 2α 2 = n 3 (G)/3 + αk 2(k ) This proes the lemma for all redction rles. The folloing property of irredcible graphs sbstantially simplifies sbseqent proofs. Here G 7 denotes the graph from Figre 5. 0

11 Lemma 4 (Edge Deletion). Let G be an irredcible graph not eqal to G 7 ith adjacent ertices and. If d() = d() = 4, then is a bridge, or one of, becomes an inner ertex of a cbic diamond pon deletion of the edge. Proof. Sppose for the sake of contradiction a non-bridge edge exists, beteen ertices of degree 4, sch that none of, becomes an inner ertex of a diamond pon deletion of. Since G is irredcible, no redction rle is admissible. Clearly, this mst mean that a 2-necklace or 2-blossom is introdced hen is deleted, that is hen (R5) is applied to. In either case, e ill derie a contradiction to the irredcibility of G. Claim The graph G does not contain a 2-necklace N. Sppose for the sake of contradiction that G does contain a 2-necklace N. Consider N as a sbgraph of G (so is conted toards the degrees of and ). We first treat the case that N consists of at least to diamonds. If one of the diamonds in N contains three ertices of degree 3, e can se rle (R), see Figre 9 (a). So no e may assme that one diamond on the end of the necklace contains as one of the three ertices not shared ith the next diamond, and the diamond on the other end of the necklace contains this ay. If is a ertex ith degree 2 in N, then (R2) can be applied, see Figre 9. This does not introdce a 2-necklace since the degree 4 ertex is part of N on the other end. Becase is part of a diamond, this can also not introdce a 2-blossom. In the remaining case, both and are internal ertices of their respectie diamonds. No it is admissible to apply (R5) to a different edge incident ith, see Figre 9 (c), here the dashed edge is the deleted one. This does not introdce a 2-necklace or 2-blossom: becomes part of a triangle that is indced by degree 3 ertices, for hich all otgoing edges hae different end ertices. Sch a triangle cannot be part of a 2-blossom or 2-necklace. The other end ertex of the deleted edge is still part of a diamond after deletion, and ths is not part of a 2-blossom. It is not part of a 2-necklace since is in this part of the necklace. This concldes the case here N consists of at least to diamonds. (a) (R) (R2) Figre 9: Redctions hen a long 2-necklace is created. (c) No sppose N consists of a single diamond. If is an inner ertex of this diamond, then cannot be part of the same diamond since e are dealing ith simple graphs. This is then the case e exclded by assmption, see Figre 0 (a). So ithot loss of generality is one of the ertices that hae degree 2 in the diamond. No rle (R) or (R2) is admissible, depending on hether is also in the diamond, see Figres 0 and (c). This does not introdce a 2-blossom or 2-necklace, since in the case in Figre 0, a triangle containing a goober is introdced, and in the case in Figre 0 (c), has degree 4 and a goober at distance 2. Note that also no parallel edges are introdced: in the case in Figre 0 (c) the edges leaing the diamond are distinct, that is deleting does not gie a K 4, since in that case old hae been a bridge. This shos that it is admissible to apply either (R) or (R2), hich contradicts the irredcibility of G.

12 (a) (R2) (c) (R) Figre 0: Redctions hen a cbic diamond is created. Claim 2 The graph G does not contain a 2-blossom B. For B e se the ertex labels from Figre (a). The degree 4 ertex of B is labeled b, its terminals are called c-ertices, and the remaining for ertices are called its a-ertices. No consider B as a sbgraph of G (so is conted toards the ertex degrees). Since d G () = 4 = d G () neither of them is eqal to b, since b has degree 4 een after the deletion of. (a) a a 2 c b c 2 b a 3 a 4 Figre : The blossom B after deleting. If is an a-ertex, say ithot loss of generality = a, then it is admissible to delete the edge connecting to b instead, see Figre. We arge that this does not introdce a 2-blossom or 2-necklace. Figre 2 shos the possible reslts of deleting b in more detail, depending on the position of. First sppose a 4. After deleting b, b becomes part of (a) (c) (d) (e) Figre 2: Possible reslts of deleting b. a triangle that does not share a ertex ith another triangle, since e assmed a 4. It follos that b is neither part of a 2-necklace, nor of a 2-blossom. The ertex may be part of a triangle (hen = c 2 or hen is not in B bt adjacent to c ), bt sch a triangle is not part of a diamond, hence is not part of a 2-necklace. Finally e arge that is not part of a 2-blossom: since b is not part of a 2-blossom, its neighbor a 2 is not part of a 2-blossom B nless it is a terminal of B. In that case it is not part of a triangle, bt its neighbor c 2 is, hich is impossible. Hence a 2 is not part of a 2-blossom. Then if is part of a 2-blossom B, it mst be a terminal of B, and ths not part of a triangle, bt its neighbor c mst be part of a triangle. This is again not possible. This concldes the proof that if a 4, deleting b is an admissible application of (R5). No e need to consider the case case that = a 4 and = a, see Figre 2 (e). Deleting b does not introdce a 2-necklace, bt there is exactly one ay in hich it may introdce a 2-blossom, hich has as its central degree 4 ertex. Figre 3 shos this case, the bold edges indicate the ne blossom. Bt no it can be seen that the original graph, hich incldes b, 2

13 b Figre 3: Deleting b yields a blossom. is exactly G 7, a contradiction ith or assmption. We conclde that if is an a-ertex and G G 7, in eery case the edge b can be deleted by an admissible application of (R5). It remains to consider the case that is a c-ertex. Then, (R3) cold be sed, see Figre 4 (a) and. The bold edges indicate the strctre redced by (R3). If in case (a) a 2-necklace is introdced, old be an inner ertex of one of its diamonds, bt that is not possible since d() = 4. In case no 2-necklace can be introdced, since is part of at most one triangle. In neither case a 2-blossom is introdced. (a) (R3) (R3) Figre 4: More redctions if a 2-blossom is created. We hae ths deried a contradiction to the irredcibility of the graph for all cases here deleting old not be an admissible application of (R5), hich proes the lemma. 3.2 Using the Reslt for Cbic Graphs to Proe Tree Extendibility Using or obseration that the reslts in [0] hold hen goobers are simply defined as ertices ith degree at most 2, e may restate Theorem 3 from [0] as follos. Theorem 3. Eery irredcible graph G of maximm degree exactly 3 and ithot cbic diamonds has a spanning tree ith at least n 3 (G)/3 + α leaes, here α = 4/3 if G is cbic and α = 2 otherise. We gie a short oerie of the proof of this statement, as it appears in [0]. For a sbgraph T of G, in addition to l(t) the folloing ales are considered. By n G (T) e n G (T) denote the nmber of non-goober ertices of G that are in V (T). By l d (T) e denote the l d (T) nmber of dead leaes of T, that is leaes of T, hich hae no neighbor in V (G) \ V (T). dead leaes The ale of T is defined as 2.5l(T)+0.5l d (T) n G (T). First it is shon in [0] that a tree ale T ith ale at least 4 can alays be fond, and een one ith ale at least 5.5 if H contains at least one goober. Next, it is shon that eery non-spanning tree T can be extended, that is a tree spergraph T of T can be fond ith ale at least the ale of T. This part of the proof consists of a rather inoled case stdy. The extensions can be repeated ntil a spanning tree is fond, in hich case all leaes are dead. Reriting the ale expression, and ronding p the start ale then yields Theorem 3. We obsere that nohere in the case stdy that proes extendibility any information abot the crrent tree T is sed; loosely speaking, only information abot the part of H otside of T is sed. In particlar, the fact that T is connected is neer sed in the proof, and neither are pper bonds on degrees of ertices already inclded in T. 3

14 We ill no define the leaf potential of a sbgraph hich generalizes the aboe definition of the ale of a tree, and e ill formalize the notions extendible and otside. Using these ne notions a sefl lemma can be formlated, hich e conclde is proed, bt not stated in [0]. Definition 6 (Leaf-Potential). The leaf-potential of a sbgraph F G is P G (F)= 2.5l(F) + leafpotential 0.5l d (F) n G (F) 6cc(F). P G (F) If ambigities are exclded in the context e simply rite P(F). Definition 7 (Extendible). Let F be a sbgraph of a graph G. Then F is called extendible extendible if there exists an F ith F F G and P G (F ) P G (F). Aboe e already informally mentioned the sbgraph of G otside a sbgraph F G. Considering the proof in [0], e see that this graph may formally be defined as an edge indced graph as follos. Definition 8 (Graph Otside F). Let F be a non-spanning sbgraph of G. The sbgraph of G otside of F is F C = G[{ E(G) : V (F)}]. The bondary of F is V (F) V (F C ) Note that no edges beteen to ertices that are both in V (F) are inclded in F C. If G is clear from the context e call F C the graph otside F. Expressed sing these definitions, the case stdy in [0] yields the folloing lemma. Lemma 5 (Extension Lemma). Let G be a connected irredcible graph, and let F G sch that F C has maximm degree 3 and contains no cbic diamonds. Then F is extendible. sbgraph of G otside of F F C bondary graph otside F 3.3 Groing Trees arond High Degree Vertices The prpose of this section is to proe Lemma 6, hich gros trees arond high degree ertices and yields a graph satisfying the assmptions of Lemma 5. Lemma 6 is the core of or proof of Theorem. We denote the set of ertices of G hich are not in F by V (F) = V (G) \ V (F). The neighborhood N() of a ertex is the set of all ertices adjacent to, and the closed neighborhood of is N[] = N() {}. Expanding a ertex V (G), hich is an operation Expanding on a sbgraph F of G, yields a ne sbgraph ith ertex set V (F) N[], and edge set E(F) { : N[]\V (F)}. So all nely added neighbors of become leaes, and may lose leaf stats. The nmber of components increases by one if and only if V (F). Expanding a list of ertices means expanding the ertices in the gien order. We adopt the short-hand notation (x, y, z):= 2.5y + 0.5z x from [0] to express the (x, y, z) change in P G hen extending a graph F to a ne graph F. Let l denote l(f ) l(f), and define l d and n G analogosly. So hen F and F hae the same nmber of components, the extension is alid if and only if ( n G, l, l d ) 0, and if a ne component is introdced e need ( n G, l, l d ) 6. For the sake of simpler notation, instead of riting e.g. ( n G, l, l d ) (4, 3, ) = 4, e ill simply rite (4, 3, ) = 4. Hence the three parameter ales need not be exactly n G, l and l d bt reflect the orst case scenario. That is, the change in the leaf potential that e proe is to be read as a loer bond for the actal change. 4

15 Lemma 6 (Start Lemma). Let G be an irredcible graph not eqal to G 7 and F a (possibly empty) sbgraph of G, sch that F C contains at least one ertex of degree at least 4, and contains neither 2-necklaces or 2-blossoms. Then, F is extendible. Proof. First sppose F is not the empty graph. If there is a ertex on the bondary of F hich is not a leaf, then F can be obtained by expanding. There is no leaf lost since as not a leaf, and the nely added ertices are leaes. So the agmentation ineqality is satisfied: (k, k,0) 0. Hence, e may assme in the remainder that only leaes of F hae neighbors in V (F), or in other ords, all ertices on the bondary of F are leaes of F. (A) (0, 0, 0) = 0 (A4) (A2) (i +, i, 0) 0.5 if i (A5) (A3) (, 0, i) 0 if i 2 (i + 2, i, j) 0 if i, j (A6) (i + 3, i, 0) 0 if i 2 (i + 2, i, 0) if i 2 (A7) (i + 3, i, ) 0.5 if i 2 Figre 5: Simple agmentations of an existing sbgraph The next step is the attempt to agment F sing the operations (A)-(A7), see Figre 5. Conentions for this figre are that encircled ertices belong to V (F), solid edges sho the expansion and ertex degrees shon are to be nderstood as loer bonds. Dead leaes are marked ith a cross. All of the expansions in the figre extend F ithot creating a ne connected component, and satisfy ( n G, l, l d ) 0. Ths the reslting graph F is an extension as claimed in the lemma. Together these agmentation rles yield the folloing claim. Claim 0 The sbgraph F is extendible, if a ertex in V (F) has a goober neighbor in V (F) or at least to neighbors in V (F), or if there is a high-degree ertex V (F) at distance at most to from F. If a goober from V (F) is adjacent to F then (A) can be applied. If a ertex in V (F) has at least to neighbors in V (F), (A2) can be applied. So from no on e ill assme eery ertex in V (F) has at most one neighbor in V (F), and this neighbor is not a goober. If a high-degree ertex in V (F) is adjacent to a ertex in V (F), (A3), (A4) or (A5) can be applied. The creation of the dead leaes in (A3) and (A4) follos from the fact that (A2) cannot be applied anymore. If a high-degree ertex in V (F) has distance to from a ertex in V (F), (A6) or (A7) can be applied. The rest of the proof ill handle the more complicated case hen F is the empty graph, or the only high-degree ertices in V (F) are at a larger distance from V (F). We then introdce a ne component for F. This is more complicated becase adding a frther component comes at a certain cost, more precisely e need that the ne component satisfies ( n G, l, l d ) 6. 5

16 The rest of the proof is diided into three more claims. The first one handles the easiest cases, and the second one handles all cases except those here eery degree 4 ertex is the common ertex of to edge-disjoint triangles. This final case is then taken care of in the third claim. Throghot the proof e assme, sometimes implicitly, that none of the sitations that hae been handled earlier can occr. Claim Let V (F), d() 4, and N(). In the folloing for sitations F is extendible: d() 5, or d() = 4 and is a goober, or d() = d() = 4, or d() = 4 and N[] N[]. First note that no ertex in N[] or N[] is part of F by Claim 0. If d() 5, expanding yields (k +, k, 0) 6.5, since k 5. For d() = 4 and a goober, expanding gies (4, 4, 0) = 6. No sppose d() = d() = 4. If is a bridge, expanding and yields (8, 6, 0) = 7. Otherise, Lemma 4 shos that either or, say, becomes the inner ertex of a cbic diamond pon deletion of the edge. (Note that e assmed G G 7, so Lemma 4 may be applied.) Ths, either has to neighbors not in N[] and expanding, yields (7, 5, ) = 6 (see Figre 6 (a)), or shares to neighbors ith in hich case expanding yields (5, 4, 3) = 6.5 (see Figre 6 ). So no e may assme that all neighbors of hae degree 3. If N[] N[] then either the niqe ertex N[]\N[] has to neighbors not in N[], in hich case expanding, gies (7, 5, ) = 6, see Figre 6 (c), or there is another ertex x N[] ith N[x] N[], and is expanded to obtain (5, 4, 2) = 6, see Figre 6 (d). (a) (c) (d) x Figre 6: Figres for Claim. Smmarizing, e may no assme that V (F) contains no ertices of degree at least 5, and if it contains a ertex of degree 4, all neighbors of hae degree 3 and hae either one or to neighbors not in N[]. Claim 2 If V (F) contains a ertex ith d() = 4 and a ertex N() hich has to neighbors a, b N[], then F is extendible. We denote the other three neighbors of by x, y, z. If one of a, b, x, y, z has all of its neighbors in {a, b} N[], e hae (7, 5, ) = 6 by expanding,, see Figre 7 (a). If a or b is a goober e obtain (6, 5, 0) 6.5, see Figre 7. If a or b is adjacent to a ertex c V (F), then expanding, ill make c a dead leaf and yields (7, 5, ) 6, see Figre 7 (c). If one of a, b, x, y, z has at least to neighbors not in N[] {a, b}, e obtain (9, 6, 0) = 6 by expanding, and this ertex, see Figre 7 (d). Hence e may assme that a, b, x, y, z each hae exactly one neighbor otside N[] {a, b}. This neighbor is not part of F. Since they all hae degree at least 3, these fie ertices mst indce three edges. This implies that one of a, b has degree 4 since e already kno that 6

17 (a) (c) (d) y x a b a b c a b a b Figre 7: Figres for Claim 2. x, y, z hae degree 3. We may assme ithot loss of generality that d(a) = 4 and d = 3. We distingish to cases depending on hether a is adjacent to b or not. The statement a is adjacent to b is denoted by a b. We denote the neighbor of x otside of N[] {a, b} by a b x, and similarly a, b, y, z are defined. Case. a is adjacent to x and y hile b is adjacent to z. Consider expanding, x, z. All ertices in {a, b,, y, x, z } are adjacent to at least one of, x, z, ths e hae (9, 6, ) = 6.5 nless x = z, see Figre 8 (a). By an analogos argment ith y in the place of x e may no assme that x = z = y. Then, expanding, x yields (7, 5, ) = 6, since y becomes a dead leaf, see Figre 8. (a) x x z z z y z x y b a b a Figre 8: Figres for Claim 2, Case Case 2. a is adjacent to b and x hile y is adjacent to z. If x a, expanding a, x, yields (9, 6, ) = 6.5, see Figre 9 (a), so e may assme that x = a =: c, and this creates a sitation symmetric in b and c. By Claim e hae that b c. No first sppose b c. Then expanding b, c, x, yields (0, 6, 3) = 6.5, proided b has a neighbor d other than y, z, see Figre 9. Note that d V (F) is not possible since agmentation (A6) cold hae been applied instead. (a) z b y x a x a z y b a x c b d (c) z y b a x y c (R3) z a b b b Figre 9: Figres for Claim 2, Case 2 If N(b ) = {b, c, y}, then (R3) is admissible, see Figre 9 (c). Since b becomes a goober this cannot introdce a 2-necklace. Hence it mst be that b y, z and the graph has (9, 5, 5) = 6, see Figre 20 (a). This concldes the cases ith b c. The case b c can be exclded becase then (R3) old be admissible, see Figre 20. Note that this cannot create a 2-necklace inoling y, z since then (R2) old hae been admissible. 7

18 (a) y z b a x c z y b a x c b (R3) z y a x c b Figre 20: Additional figres for Claim 2, Case 2. This concldes the proof of Claim 2. Smmarizing Claims 0,, and 2, e may no assme that all neighbors of a degree 4 ertex V (F) hae degree 3, and hae exactly one neighbor not in N[]. In other ords, is the common ertex of to edge-disjoint triangles, see Figre 2. p q p s q r Figre 2: The bo tie sbgraph s r Claim 3 If the graph otside F contains a ertex ith d() = 4 sch that all its neighbors hae degree 3 and one neighbor otside N[], then F is extendible. We denote the neighbors of by p, q, r, s and assme that p q and r s. The neighbor of p otside N[] is denoted by p and similarly q, r, s are defined, see Figre 2. We split the proof of the claim into three cases. Case. p = q If p has degree 3, e can apply (R), see Figre 22 (a). So no ithot loss of generality p has degree 4. Then by Claim 2, p is also part of to edge-disjoint triangles. So if p = r then also p = s. In that case e can expand p, to obtain (6, 4, 4) = 6, see Figre 22. So no p r, p s. Consider applying (R2) to the diamond consisting of p, p, q,, see Figre 22 (c). If this introdces a 2-necklace, (R) cold hae been applied to the diamond on the other end of this necklace. It cannot introdce a 2-blossom since the triangles of a 2-blossom contain a degree 4 ertex. (a) p (R) p (c) p (R2) p Case 2. p = r Figre 22: Figres for Claim 3, Case Note that d(p) = 3 by Claim 2. Also q s since the graph does not contain 2-blossoms and the case that d(q = s) = 4 is again exclded by Claim 2. Ths, (R3) is admissible, see Figre 23. 8

19 p q p q s s Figre 23: The figre for Claim 3, Case 2 Case 3. p, q, r, s pairise different. In this case either (R4) or (R3) is admissible: if is a ct ertex, (R4) may be sed (it increases the nmber of components). Otherise, e may assme ithot loss of generality that p and s are in same connected component of G, and (R3) can be applied ithot disconnecting the graph. See Figres 24 (a) and. (a) p s (R4) p s p s (R3) p s q r q r q r q r Figre 24: Figres for Claim 3, Case 3 This concldes all possible cases: heneer the sbgraph of G otside of F contains a high degree ertex, e hae shon that G is either redcible, or F is extendible. 3.4 The Proof of the Main Reslt This section is deoted to combining the tools deeloped in the last three sbsection in order to proe Theorem, hich e repeat here for conenience. Theorem. Let G be a simple, connected graph on at least to ertices hich contains neither 2-necklaces nor 2-blossoms. Then, G has a spanning tree T ith { 4/3 if δ(g) 3 l(t) n 3 (G)/3 + 2 if δ(g) 2. Proof. We proe the statement by indction. For or indction hypothesis e actally proe that the aboe statement holds for eery connected graph hich satisfies the inariant. Then the statement follos for simple graphs. First sppose G is irredcible. If G has maximm degree exactly 3, Theorem follos immediately from Theorem 3. If G has maximm degree at most 2, G has a spanning tree ith at least to leaes (note that e assmed that G is not a K ), hich sffices. If G = G 7, then a spanning tree ith 4 = n 3 (G)/3+5/3 leaes can be obtained. So e may no assme that G contains at least one high degree ertex, and is not eqal to G 7. We start ith an empty sbgraph F of G, hich has P G (F) = 0. The Start Lemma (Lemma 6) shos that, as long as there is at least one high degree ertex not in F, e can extend F hile maintaining P G (F) 0. When all high degree ertices are inclded in F, the Extension Lemma (Lemma 5) can be applied iteratiely, ntil a spanning sbgraph F is obtained ith P G (F ) 0. Withot loss of generality, e may assme that F is a forest; cycles can be broken ithot decreasing the nmber of leaes. Since all leaes of a spanning sbgraph are dead e dedce 0 P G (F ) = 3l(F ) n 3 (G) 6cc(F ) = l(f ) n 3 (G)/3 + 2cc(F ). 9

20 We can no add cc(f ) edges to F to obtain a spanning tree, losing at most 2(cc(F ) ) leaes, so the reslting tree has at least n 3 (G)/3 + 2 leaes. It remains to consider the case that G is redcible (the indction step). Some redction rle is admissible, and the redced graph G again satisfies the inariant, by Lemmas and 2. First sppose G is connected. None of the redction rles remoe goobers, so if δ(g) 2, then δ(g ) 2, and by indction G has a spanning tree ith at least n 3 (G )/3 + 2 leaes. Lemma 3 then shos that G has a spanning tree ith at least n 3 (G)/3+2 leaes. Similarly, if δ(g) 3 then it follos that G has a spanning tree ith at least n 3 (G)/3 + 4/3 leaes. No sppose the redction rle yields a disconnected graph G. Then, by the definition of the redction rles, eery reslting component has a goober. So by indction, eery non-triial component C of G has a spanning tree ith at least n 3 (C)/3 + 2 leaes. Ths Lemma 3 implies that G has a spanning tree ith at least n 3 (G)/3 + 2 leaes. 4 A fast FPT Algorithm for MaxLeaf In this section e present a fast and relatiely simple FPT algorithm for MaxLeaf, hich ses Theorem as an essential ingredient. The other to ingredients are a short preprocessing step, consisting of to redction rles, and an enmeratie procedre, hich is similar to the one introdced in [3], and also applied in [2]. We start by presenting the to redction rles that constitte the preprocessing phase. The redction rles for the FPT algorithm are different from the rles sed in Section 3. It is important that they yield an eqialent instance of the decision problem MaxLeaf, bt there are no conditions on the ratio beteen the decrease in ertices and in possible leaes. Recall that in 2-necklaces and 2-blossoms both terminals hae degree 3 in G. The rles e introdce no also redce diamonds and blossoms hose to terminals hae arbitrary degree. Hoeer the to terminals of the sbgraph mst still be the to ertices that hae degree 2 in the diamond necklace or blossom itself. Sch a sbgraph of G ill be called a 2-terminal diamond respectiely a 2-terminal blossom. Rle (F) in Figre 25, hich resembles rle (R2), 2-terminal redces 2-terminal diamonds. Since the 2-necklace N k consists of k 2-terminal diamonds it is redced as ell by rle (F). Rle (F2) in Figre 25 redces 2-terminal blossoms. The next lemma proes the correctness of these rles. diamond 2-terminal blossom k k (F) (F2) Figre 25: To redction rles for an instance (G, k) of MaxLeaf. Lemma 7. Let G be the reslt of applying redction (F) or (F2) to G. Then (G, k ) is a YES-instance for MaxLeaf if and only if (G, k) is a YES-instance for MaxLeaf. Proof. First consider the case here the applied rle as (F), redcing a 2-terminal diamond D of G ith terminals and. Consider a spanning tree T of G ith at least k leaes. Obsere that in T, e can alays replace the set of edges E(T) E(D) by one of the to sets shon on the right in Figre 26 (a), or a symmetric set, ithot decreasing the nmber of leaes. Then by replacing it by the corresponding strctre on the left, e obtain a spanning tree T of G ith at least k leaes. Note here that the terminals and remain leaes in 20