Toughness and spanning trees in $K_4$-minor-free graphs

TL;DR

This paper proves that connected $K_4$-minor-free graphs with certain toughness conditions contain spanning trees with bounded degrees, extending known results for general graphs by leveraging the structure of these specific graph classes.

Contribution

It establishes new sufficient conditions for the existence of spanning $f$-trees in $K_4$-minor-free graphs, strengthening previous results for broader classes of graphs.

Findings

Connected $K_4$-minor-free graphs have spanning $f$-trees under specific component conditions.

Toughness conditions imply the existence of spanning $k$-trees in these graphs.

The results do not extend to all planar graphs, highlighting the importance of the $K_4$-minor-free restriction.

Abstract

For an integer $k$, a $k$-tree is a tree with maximum degree at most $k$. More generally, if $f$ is an integer-valued function on vertices, an $f$-tree is a tree in which each vertex $v$ has degree at most $f(v)$. Let $c(G)$ denote the number of components of a graph $G$. We show that if $G$ is a connected $K_4$-minor-free graph and $$ c(G-S) \;\le\; \sum_{v \in S} (f(v)-1) \quad\hbox{for all $S \subseteq V(G)$ with $S \ne \emptyset$} $$ then $G$ has a spanning $f$-tree. Consequently, if $G$ is a $\frac{1}{k-1}$-tough $K_4$-minor-free graph, then $G$ has a spanning $k$-tree. These results are stronger than results for general graphs due to Win (for $k$-trees) and Ellingham, Nam and Voss (for $f$-trees). The $K_4$-minor-free graphs form a subclass of planar graphs, and are identical to graphs of treewidth at most $2$, and also to graphs whose blocks are series-parallel. We provide examples to show that the inequality above cannot be relaxed by adding $1$ to the right-hand side, and also to show that our result does not hold for general planar graphs. Our proof uses a technique where we incorporate toughness-related information into weights associated with vertices and cutsets.

Equations

$$c(G-S)\le \sum _{v\in S}(f(v)-1)for\; allS\subseteq V(G)withS\ne \varnothing$$

$$c(G-S)$$

$$c(G-S)\le \sum _{v\in S}(f(v)-1)+1for\; allS\subseteq V(G).$$

$$c(G-S)\le \sum _{v\in S}(f(v)-1)for\; allS\subseteq V(G)withS\ne \varnothing .$$

$$c(G-S)\le \sum _{v\in S}(f(v)-1)for\; everyS\subseteq XwithS\ne \varnothing .$$

$$c(G-U)\ge \sum _{uv\in E(J)}(c(G,uv)-2)+\sum _{w\in U}(c(G-w)-1)+|U|.$$

$$c(G-U)\ge \sum _{uv\in E(J)}(c(G,uv)-2)+|U|.$$

$$|{\mathcal{C}}_x|\ge \sum _{uv\in E(J_i^x)}(c(G_i,uv)-2)+|U_i^x|-1.$$

$$c(G_i-U_i)$$

$$c(G-U)$$

$$c(G-U)=c(B-U)+\sum _{w\in U}(c(G-w)-1).$$

$$c(G,uv)=c(B,uv)for\; alluv\in E(J).$$

$$c(B-U)\ge \sum _{uv\in E(J)}(c(B,uv)-2)+|U|.$$

$$c(G-U)$$

$$c(G-S)\le \sum _{v\in S}(f(v)-1)for\; allS\subseteq V(G)withS\ne \varnothing$$

$$\omega (F,u)+\omega (F,v)$$

$$\sum _{F\in \mathcal{F}(u)}\omega (F,u)$$

$$\begin{array}[]{ll}\gamma(sF)=c(G,F)-2&\hbox{if $sF\in A_{1}$,}\\ \gamma(Fu)=\infty&\hbox{if $Fu\in A_{2}$, \quad and}\\ \gamma(ut)=f(u)-c(G-u)-1&\hbox{if $ut\in A_{3}$.}\end{array}$$

$$\omega (F,u)+\omega (F,v)=\phi (Fu)+\phi (Fv)=\phi (sF)=\gamma (sF)=c(G,F)-2$$

$$\sum _{F\in \mathcal{F}(u)}\omega (F,u)=\sum _{Fuentersu}\phi (Fu)=\phi (ut)\le \gamma (ut)=f(u)-c(G-u)-1$$

$$[S,T]=[\{s\}\cup {\mathcal{F}}_1\cup W_1,{\mathcal{F}}_2\cup W_2\cup \{t\}]=[\{s\},{\mathcal{F}}_2]\cup [{\mathcal{F}}_1,W_2]\cup [W_1,t]$$

$$\gamma (S,T)=\sum _{F\in {\mathcal{F}}_2}\gamma (sF)+\sum _{w\in W_1}\gamma (wt)=\sum _{\{u,v\}\in {\mathcal{F}}_2}(c(G,uv)-2)+\sum _{w\in W_1}(f(w)-c(G-w)-1).$$

$$\sum _{w\in W_1}(f(w)-c(G-w)-1)<\sum _{\{u,v\}\in {\mathcal{F}}_1}(c(G,uv)-2),$$

$$\sum _{\{u,v\}\in {\mathcal{F}}_1}(c(G,uv)-2)+\sum _{w\in W_1}(c(G-w)-1)>\sum _{w\in W_1}(f(w)-2).$$

$$\sum _{uv\in E(J)}(c(G,uv)-2)+\sum _{w\in U}(c(G-w)-1)>\sum _{w\in U}(f(w)-2).$$

$$c(G-U)$$

$$\omega (F,u)+\omega (F,v)$$

$$\sum _{F\in \mathcal{F}(u)}\omega (F,u)+c(G-u)-1$$

$$r_i{s}_i\in E(T)ifc(G,r_i{s}_i)=\mathrm{0,}and{r}_i{s}_i\notin E(T)ifc(G,r_i{s}_i)=1.$$

Full text

[TABLE]

then $G$ has a spanning $f$-tree. Consequently, if $G$ is a $\frac{1}{k-1}$-tough $K_{4}$-minor-free graph, then $G$ has a spanning $k$-tree. These results are stronger than results for general graphs due to Win (for $k$-trees) and Ellingham, Nam and Voss (for $f$-trees). The $K_{4}$-minor-free graphs form a subclass of planar graphs, and are identical to graphs of treewidth at most $2$, and also to graphs whose blocks are series-parallel. We provide examples to show that the inequality above cannot be relaxed by adding $1$ to the right-hand side, and also to show that our result does not hold for general planar graphs. Our proof uses a technique where we incorporate toughness-related information into weights associated with vertices and cutsets.

Keywords: toughness, spanning tree, $K_{4}$-minor-free, series-parallel, treewidth.

1 Introduction

All graphs considered are simple and finite. Let $G$ be a graph. We denote by $d_{G}(v)$ the degree of vertex $v$ in $G$. For $S\subseteq V(G)$ the subgraph induced on $V(G)-S$ is denoted by $G-S$; we abbreviate $G-\{v\}$ to $G-v$. The number of components of $G$ is denoted by $c(G)$. The graph is said to be $t$-tough for a real number $t\geq 0$ if $|S|\geq t\cdot c(G-S)$ for each $S\subseteq V(G)$ with $c(G-S)\geq 2$. The toughness $\tau(G)$ is the largest real number $t$ for which $G$ is $t$-tough, or $\infty$ if $G$ is complete. Positive toughness implies that $G$ is connected. If $G$ has a hamiltonian cycle it is well known that $G$ is $1$-tough.

In 1973, Chvátal [3] conjectured that for some constant $t_{0}$, every $t_{0}$-tough graph is hamiltonian. Thomassen (see [2, p. 132]) showed that there are nonhamiltonian graphs with toughness greater than $3/2$. Enomoto, Jackson, Katerinis and Saito [8] showed that every 2-tough graph has a $2$-factor ($2$-regular spanning subgraph), but also constructed $(2-\varepsilon)$-tough graphs with no $2$-factor, and hence no hamiltonian cycle, for every $\varepsilon>0$. Bauer, Broersma and Veldman [1] constructed $(\frac{9}{4}-\varepsilon)$-tough nonhamiltonian graphs for every $\varepsilon>0$. Thus, any such $t_{0}$ is at least $\frac{9}{4}$.

There have been a number of papers on toughness conditions that guarantee the existence of more general spanning structures in a graph. A $k$-tree is a tree with maximum degree at most $k$, and a $k$-walk is a closed walk with each vertex repeated at most $k$ times. Note that a spanning 2-tree is a hamiltonian path and a spanning 1-walk is a hamiltonian cycle. Jackson and Wormald [10] showed that on a given vertex set a $k$-walk can be obtained from a $k$-tree; conversely, a $(k+1)$-tree can be obtained from a $k$-walk. More generally, if $f:V(G)\to\mathbb{Z}$ then an $f$-tree is a tree with $d_{T}(v)\leq f(v)$ for all $v\in V(T)$, and an $f$-walk is a closed walk that uses every vertex $v$ at most $f(v)$ times.

The first toughness result for spanning trees of bounded degree was by Win.

Theorem 1.1** (Win [13]).**

Suppose $G$ is a connected graph, $k\geq 2$, and

[TABLE]

Then $G$ has a spanning $k$-tree.

Win’s result implies that for $k\geq 3$, $\frac{1}{k-2}$-tough graphs have a $k$-tree (and hence a $k$-walk). Ellingham, Nam and Voss showed that the bound on the degrees need not be constant.

Theorem 1.2** (Ellingham, Nam and Voss [7]).**

Suppose $G$ is a connected graph, $f:V(G)\to\mathbb{Z}$ with $f(v)\geq 2$ for all $v\in V(G)$, and

[TABLE]

Then $G$ has a spanning $f$-tree.

While these conditions are sufficient, and sharp in the sense that the right-hand side of the inequality cannot be increased by $1$, they are not necessary. For a necessary condition, a graph with a spanning $k$-walk (and hence a graph with a spanning $k$-tree) must be $\frac{1}{k}$-tough. However, a stronger necessary condition, which also applies for non-constant degree bounds, can be obtained by counting components in a tree $T$: for any $S\subseteq V(T)$, $c(T-S)\leq\sum_{v\in S}(d_{T}(v)-1)+1$. Applying this to a spanning $f$-tree of a graph $G$ gives the following.

Observation 1.3**.**

Suppose $G$ is a graph with a spanning $f$-tree. Then

[TABLE]

When $f(v)=k$ for all $v$, this condition is slightly weaker than being $\frac{1}{k-1}$-tough.

Recently some stronger versions of Theorem 1.2 have been posted by Hasanvand [9, Subsection 4.2], although they do not seem to change the condition (1.1) in a substantial way. Hasanvand does, however, appear to have made a significant advance for $f$-walks. For $k\geq 3$, as noted above, $\frac{1}{k-2}$-tough graphs have a spanning $k$-tree and hence a spanning $k$-walk. Jackson and Wormald [10] conjectured that in fact $\frac{1}{k-1}$-tough graphs have a spanning $k$-walk. Hasanvand used a clever argument to combine a stronger form of Theorem 1.2 with a result of Kano, Katona and Szabó [11] on the existence of spanning subgraphs with parity conditions, to show the following.

Theorem 1.4** (Hasanvand, [9, Theorem 5.5]).**

Suppose $G$ is a graph, $M$ is a matching in $G$, $f:V(G)\to\mathbb{Z}$ with $f(v)\geq 1$ for all $v\in V(G)$, and

[TABLE]

Then $G$ has a spanning $f$-walk that uses the edges of $M$.

With $f(v)=k$ for all $v$, Theorem 1.4 verifies Jackson and Wormald’s conjecture.

It is unknown whether Theorem 1.2 can be improved for graphs in general by weakening condition (1.1). It is therefore of interest to see whether this can be done for special classes of graphs. In this paper we focus on $K_{4}$-minor-free graphs.

A graph $H$ is a minor of a graph $G$ if a graph isomorphic to $H$ can be obtained from $G$ by edge contractions, edge deletions and vertex deletions; if not, $G$ is $H$-minor-free. Since both $K_{3,3}$ and $K_{5}$ contain a $K_{4}$ minor, $K_{4}$-minor-free graphs are $K_{3,3}$-minor-free and $K_{5}$-minor-free, i.e., planar. The class of $K_{4}$-minor-free graphs includes all series-parallel graphs, constructed by series and parallel compositions starting from copies of $K_{2}$. Duffin [5] gave three characterizations for series-parallel graphs; in particular, he showed that a graph with no cutvertex is $K_{4}$-minor-free if and only if it is series-parallel. Wald and Colbourn [12] showed that $K_{4}$-minor-free graphs are also identical to graphs of treewidth at most $2$.

Let $G$ be a $K_{4}$-minor-free graph with toughness greater than $\frac{2}{3}$. The toughness implies that $G$ has no cutvertex, and in conjunction with $K_{4}$-minor-freeness, also implies that $G$ is $K_{2,3}$-minor-free (see Lemma 3.2). Thus, $G$ is either $K_{2}$ or a $2$-connected outerplanar graph. Thus, a $K_{4}$-minor-free graph with at least three vertices and toughness greater than $\frac{2}{3}$ is hamiltonian. Dvořák, Král’ and Teska [6] showed that every $K_{4}$-minor-free graph with toughness greater than $\frac{4}{7}$ has a spanning $2$-walk, and they constructed a $\frac{4}{7}$-tough $K_{4}$-minor-free graph with no spanning 2-walk.

Our main result is a sufficient condition for a connected $K_{4}$-minor-free graph to have an $f$-tree. We show that in fact a very slight strengthening of the necessary condition (1.2) suffices, subtracting $1$ from the right-hand side. For vertices $x,v$ in a graph, define $\delta_{x}(v)$ to be $1$ if $x=v$, and [math] otherwise.

Theorem 1.5**.**

Let $G$ be a connected $K_{4}$-minor-free graph, and $f:V(G)\to\mathbb{Z}$. Suppose that $z\in V(G)$ and

[TABLE]

Then $G$ has a spanning $(f-\delta_{z})$-tree, i.e., a spanning $f$-tree $T$ such that $d_{T}(z)\leq f(z)-1$.

The condition in Theorem 1.5 is something we refer to frequently; we call it the Component Condition and refer to it as (CC). Condition (CC) with $S=\{v\}$ implies that $f(v)\geq 1$ if $V(G)=\{v\}$, and that $f(v)\geq 2$ for all $v\in V(G)$ if $|V(G)|\geq 2$. Therefore, we do not need to explicitly specify that $f$ is nonnegative. If we take $f(v)=k\geq 2$ for all the vertices, then (CC) just means that $G$ is $\frac{1}{k-1}$-tough.

Corollary 1.6**.**

Let $G$ be a $K_{4}$-minor-free graph, let $k\geq 2$ be an integer, and let $z\in V(G)$. If $G$ is $\frac{1}{k-1}$-tough, then $G$ has a spanning $k$-tree $T$ such that $d_{T}(z)\leq k-1$.

Examples show that we cannot weaken (CC) in Theorem 1.5 by adding $1$ to the right-hand side, which gives (1.2). Let $f(x)\geq 2,f(y)\geq 2,f(z)\geq 2$ be three integers, and let $G$ be the $K_{4}$-minor-free graph obtained from a triangle $(xyz)$ by adding $f(v)-1$ pendant edges at each $v\in\{x,y,z\}$. Set $f(v)=2$ for every $v\in V(G)-\{x,y,z\}$. Then (1.2) is easily verified, but every spanning tree $T$ of $G$ has a vertex $v\in\{x,y,z\}$ with $d_{T}(v)=f(v)+1$.

Also, note that there is no upper bound on the values we can assign to $f$. If we take a specified set of vertices $X$ and let $f(v)=|V(G)|+1$ for all $v\in V(G)-X$, (CC) is automatically satisfied for sets $S$ containing a vertex of $V(G)-X$ (we use this trick in some of our proofs). We can therefore bound the degrees of just the specified vertices.

Corollary 1.7**.**

Let $G$ be a connected $K_{4}$-minor-free graph, $X\subseteq V(G)$, and $f:X\to\mathbb{Z}$. Suppose that $z\in X$ and

[TABLE]

Then $G$ has a spanning tree $T$ such that $d_{T}(v)\leq f(v)$ for all $v\in X$, and $d_{T}(z)\leq f(z)-1$.

Toughness is an awkward parameter to deal with in arguments. It is hard to control the toughness of a subgraph, or of a graph obtained by some kind of reduction from an original graph. This makes it difficult to prove results based on toughness conditions using induction. However, in this paper we provide a way of doing induction using a toughness-related condition, by first, in Section 2, transforming the toughness information into weights associated with vertices and cutsets. This approach seems to be new, and of interest apart from our results for $K_{4}$-minor-free graphs.

Another idea in this paper is dealing with situations where we delete independent sets of vertices separately from situations where we delete sets that may have adjacencies. Loosely, we expect to get more components when we delete an independent set. The concept of ‘fundamental’ graphs in Section 2 is used to implement this distinction. In Section 3 we prove a result for fundamental graphs first, then extend it to more general graphs.

2 Structure of nontrivial 2-cuts

In this section we show how to convert a toughness-related condition into weights associated with vertices and certain cutsets. The results in this section apply to all graphs, not just to $K_{4}$-minor-free graphs.

If $G$ is a graph and $H\subseteq G$ ($H$ may just be a set of vertices) then a bridge of $H$ or $H$-bridge in $G$ is a subgraph of $G$ that is either an edge not in $H$ joining two vertices of $H$ (a trivial bridge), or a component of $G-V(H)$ together with all edges joining it to $V(H)$ (a nontrivial bridge). The set of attachments of an $H$-bridge $B$ is $V(B)\cap V(H)$. For $S\subseteq V(G)$, when no confusion will result, we also refer to the set of attachments of a component of $G-S$, meaning that of the corresponding $S$-bridge. A $k$-cut in $G$ is a set $S$ of $k$ vertices for which $G-S$ is disconnected. If $F=\{u,v\}$ is a set of two vertices in $G$, let $c(G,F)$ or $c(G,uv)$ denote the number of nontrivial $F$-bridges that contain both vertices of $F$. The notation $c(G,uv)$ does not imply that $uv\in E(G)$. A nontrivial $2$-cut or N2C in $G$ is a set $F$ of two vertices with $c(G,F)\geq 3$.

A block is a connected graph with no cutvertex, and a block of $G$ is a maximal subgraph of $G$ that is itself a block. Let $\mathcal{B}$ be the set of blocks and $\mathcal{C}$ the set of cutvertices of $G$. The block-cutvertex tree of a connected graph $G$ has vertex set $\mathcal{B}\cup\mathcal{C}$, and $c\in\mathcal{C}$ is adjacent to $B\in\mathcal{B}$ if and only if the block $B$ contains the cutvertex $c$. A block of $G$ is a leaf block if it is a leaf of the block-cutvertex tree. If we choose a particular root edge $e_{0}$ of $G$, then the block $B_{0}$ containing $e_{0}$ is the root block which we treat as the root vertex of the block-cutvertex tree. Every block other than $B_{0}$ has a unique root vertex, namely its parent in the rooted block-cutvertex tree.

In what follows we try to develop information to help us construct a spanning tree with restricted degrees, by allocating the bridges of each nontrivial $2$-cut $\{u,v\}$ between $u$ and $v$. We begin with some basic properties of nontrivial $2$-cuts.

Lemma 2.1**.**

Let $\{u,v\}$ be an N2C in a graph $G$.

(a)* The graph $G$ has three internally disjoint $uv$-paths.*

(b)* If $S\subseteq V(G)-\{u,v\}$ and $|S|\leq 2$ then $u$ and $v$ lie in the same component of $G-S$.*

(c)* If $\{x,y\}$ is an N2C of $G$ distinct from $\{u,v\}$, then $u$ and $v$ lie in a unique $\{x,y\}$-bridge.*

Proof.

Since $c(G,uv)\geq 3$, there are three different $\{u,v\}$-bridges that attach at both $u$ and $v$ and we can take a path through each, proving (a). Then (b) follows immediately. For (c) we may assume that $u\notin\{x,y\}$. Then the unique $\{x,y\}$-bridge containing $u$ also contains $v$, either because $v\in\{x,y\}$, or by (b) otherwise. ∎

Let $G$ be a connected graph, $\mathcal{F}$ the set of N2Cs of $G$, and $W=\bigcup_{F\in\mathcal{F}}F$. The graph with vertex set $W$ and edge set $\mathcal{F}$ is called the N2C graph of $G$. If $W$ is an independent set in $G$, we call $G$ fundamental.

The following observations come from comparing the components of $G-S$ and of $G^{\prime}-S$ for some subgraph (block or union of $\{x,y\}$-bridges) $G^{\prime}$ of $G$, and using Lemma 2.1.

Observation 2.2**.**

Suppose $G$ is a connected graph with $|V(G)|\geq 2$. Let $u,v\in V(G)$. If $u$ and $v$ are not in a common block of $G$, then $c(G,uv)=1$; if they belong to a common block $B$ then $c(B,uv)=c(G,uv)$.

Thus, the N2Cs of a block $B$ are the N2Cs $\{u,v\}$ of $G$ with $\{u,v\}\subseteq V(B)$, and if $G$ is fundamental then $B$ is also fundamental. Conversely, the set of N2Cs of $G$ is the disjoint union of the sets of N2Cs of its blocks, and if each block is fundamental then $G$ is fundamental.

Observation 2.3**.**

Suppose $\{x,y\}$ is an N2C of a $2$-connected graph $G$. Let $G^{\prime}$ be the union of two or more nontrivial $\{x,y\}$-bridges of $G$ (possibly $G^{\prime}=G$). Then $G^{\prime}$ is $2$-connected. If $u$ and $v$ are not in a common $\{x,y\}$-bridge in $G^{\prime}$, then $c(G,uv)=1$; if they belong to a common $\{x,y\}$-bridge in $G^{\prime}$ and $\{u,v\}\neq\{x,y\}$ then $c(G^{\prime},uv)=c(G,uv)$.

Thus, the N2Cs of $G^{\prime}$ are the N2Cs $\{u,v\}$ of $G$ with $\{u,v\}\subseteq V(D)$ for some $\{x,y\}$-bridge $D$ of $G^{\prime}$, except that $\{x,y\}$ is not necessarily an N2C of $G^{\prime}$. If $G$ is fundamental then $G^{\prime}$ is also fundamental.

Observation 2.3 is not necessarily true if we take $G^{\prime}$ to be a single nontrivial $\{x,y\}$-bridge; then $G^{\prime}$ may not be $2$-connected, and we have $c(G^{\prime},uv)>c(G,uv)$ if $x$ and $y$ are in different components of $G^{\prime}-\{u,v\}$.

Lemma 2.4**.**

Suppose $J$ is a connected subgraph of the N2C graph $H$ of a graph $G$.

(a)* If $S\subseteq V(G)$, $|S|\leq 2$ and $S\cap V(J)=\emptyset$, then $V(J)$ lies in a single component of $G-S$.*

(b)* If $J^{\prime}$ is a connected subgraph of $H$ vertex-disjoint from $J$, then $V(J)$ lies in a single component of $G-V(J^{\prime})$.*

(c)* If $u,v\in V(J)$ are separated in $G$ by a cutvertex $x$ of $G$, then $x$ is also a cutvertex of $J$ separating $u$ and $v$ in $J$. *

Proof.

(a) By Lemma 2.1(b), for each $uv\in E(J)$, $u$ and $v$ are in the same component of $G-S$. Applying this repeatedly, all of $V(J)$ is in a single component of $G-S$.

(b) Let $uv\in E(J)$. By (a), all of $V(J^{\prime})$ lies in the same component of $G-\{u,v\}$, say $C$. Since $\{u,v\}$ is an N2C of $G$, there is a path from $u$ to $v$ in $G-V(C)\subseteq G-V(J^{\prime})$, so $u$ and $v$ are in the same component of $G-V(J^{\prime})$. Applying this repeatedly, all of $V(J)$ is in a single component of $G-V(J^{\prime})$.

(c) If there is a $uv$-path in $J$ that does not contain $x$, then by (a) this path lies in a single component of $G-x$, contradicting the choice of $u$ and $v$. Thus, every $uv$-path in $J$ contains $x$, so $x$ is a cutvertex separating $u$ and $v$ in $J$. ∎

Our overall strategy now is to use our toughness-related condition to assign weights associated with (N2C, vertex) ordered pairs. We show that we can either (a) assign weights satisfying certain conditions, or else (b) find a set that violates our toughness-related condition. The following proposition, giving a lower bound on $c(G-U)$ for certain subsets $U$, will be used to demonstrate (b). The statement of this result and many of our computations involve terms of the form $c(G-u)-1$ and $c(G,uv)-2$. The reader may wonder why we do not simplify these; the answer is that we often think of the graph as having a ‘main’ part consisting of one of the bridges of a cutvertex, or two of the bridges of a $2$-cut, and these terms count the number of ‘extra’ bridges outside the ‘main’ part.

Proposition 2.5**.**

Let $J$ be a connected subgraph of the N2C graph of a connected fundamental graph $G$, and $U=V(J)$. Then

[TABLE]

Proof.

The proof is by induction on $|E(J)|$. It is easy to check that the conclusion is true if $|E(J)|=0$, when $|U|=|V(J)|=1$. So we assume that $|E(J)|\geq 1$. We consider three cases.

Case 1: Suppose $G$ has no cutvertex. Since $|E(J)|\geq 1$, $G$ has at least five vertices, so $G$ is $2$-connected. For the $2$-connected case (2.1) simplifies to

[TABLE]

Let $xy\in E(J)$ and let the $\{x,y\}$-bridges in $G$ be $D_{1},D_{2},\dots,D_{k}$, where $k=c(G,uv)\geq 3$. For each $i$ let $U_{i}=U\cap V(D_{i})\supseteq\{x,y\}$ and $J_{i}=J[U_{i}]-xy$, so that $V(J_{i})=U_{i}$.

Let $G_{i}=D_{i}\cup xa_{i}y$, where $a_{i}$ is a new vertex. Each $G_{i}$ is a minor of $G$, and hence $K_{4}$-minor-free. We may apply Observation 2.3 to both $G_{i}$ and $G$, considered as subgraphs of $G\cup xa_{i}y$. We see that $G_{i}$ is $2$-connected. Suppose that $\{u,v\}\subseteq V(G_{i})$ and $\{u,v\}\neq\{x,y\}$. If $a_{i}\in\{u,v\}$, then $c(G_{i},uv)=1$. If $a_{i}\notin\{u,v\}$ then $c(G_{i},uv)=c(G\cup xa_{i}y,uv)=c(G,uv)$. Thus, $\{u,v\}\neq\{x,y\}$ is an N2C of $G_{i}$ if and only if it is an N2C of $G$ with $\{u,v\}\subseteq V(G_{i})$. It follows that $G_{i}$ is fundamental. Moreover, if $H_{i}$ is the N2C graph of $G_{i}$, we have $E(J_{i})\subseteq E(H_{i})$ and $V(J_{i})\subseteq V(H_{i})\cup\{x,y\}$ (each of $x$ and $y$ is in $J_{i}$, but could possibly be in no N2C of $G_{i}$, in which case it is an isolated vertex of $J_{i}$ and not a vertex of $H_{i}$). Since $xy\notin E(J_{i})$, $|E(J_{i})|<|E(J)|$ for all $i$. (In future we will provide less detail when applying Observation 2.3, but the reasoning will be similar to the reasoning here.)

Claim**.**

We have $\displaystyle c(G_{i}-U_{i})\geq\sum_{uv\in E(J_{i})}(c(G_{i},uv)-2)+|U_{i}|$ for each $i=1,2,\dots k$.

Proof of claim. First suppose that $J_{i}$ is connected. Then $x$ and $y$ are incident with edges of $J_{i}$, so they are contained in N2Cs of $G_{i}$, and so $J_{i}\subseteq H_{i}$. Thus, the claim follows because we can apply (2.2) to $G_{i}-U_{i}$ and $J_{i}$ by induction.

Now suppose that $J_{i}$ is disconnected. Since $J$ is connected, $J_{i}$ must have exactly two components $J_{i}^{x}$ and $J_{i}^{y}$, containing $x$ and $y$ respectively. Let $U_{i}^{x}=V(J_{i}^{x})$ and $U_{i}^{y}=V(J_{i}^{y})$. Consider the components of $G_{i}-U_{i}$ and divide them into three groups: $\mathcal{C}_{x}$ are those all of whose attachments belong to $U_{i}^{x}$, $\mathcal{C}_{y}$ are those all of whose attachments belong to $U_{i}^{y}$, and $\mathcal{C}_{xy}$ are those that have attachments in both $U_{i}^{x}$ and $U_{i}^{y}$.

We claim that $|\mathcal{C}_{xy}|\geq 2$. Obviously $a_{i}\in\mathcal{C}_{xy}$. Take an $xy$-path $v_{0}v_{1}\dots v_{t}$ in $D_{i}$. There is a subpath $P=v_{i}v_{i+1}\dots v_{j}$ such that $v_{i}\in U_{i}^{x}$, $v_{j}\in U_{i}^{y}$ and the internal vertices (if any) of $P$ belong to neither $U_{i}^{x}$ nor $U_{i}^{y}$. Since $G$ is fundamental, $P$ does have internal vertices, which belong to a second component in $\mathcal{C}_{xy}$.

Now we claim that

[TABLE]

If $|U_{i}^{x}|=1$ this just says that $|\mathcal{C}_{x}|\geq 0$, which is trivially true. So we may suppose that $|U_{i}^{x}|\geq 2$, which means that $x$ is in an N2C of $G_{i}$ and $J_{i}^{x}\subseteq H_{i}$. Then $U_{i}^{y}$ is in a single component of $G_{i}-U_{i}^{x}$, either trivially if $|U_{i}^{y}|=1$, or by Lemma 2.4(b) if $|U_{i}^{y}|\geq 2$ so that $J_{i}^{y}$ is a subgraph of $H_{i}$. The component of $G_{i}-U_{i}^{x}$ containing $U_{i}^{y}$ must also contain all components in $\mathcal{C}_{y}\cup\mathcal{C}_{xy}$. On the other hand, each $C\in\mathcal{C}_{x}$ is a separate component of $G_{i}-U_{i}^{x}$. Therefore, $c(G_{i}-U_{i}^{x})=|\mathcal{C}_{x}|+1$. Using this and applying (2.2) to $G_{i}-U_{i}^{x}$ and $J_{i}^{x}$ by induction gives (2.3).

We have an inequality similar to (2.3) for $\mathcal{C}_{y}$. Therefore,

[TABLE]

which proves the claim. $\blacksquare$

Now we prove (2.2) and hence (2.1). For each $i$, $c(D_{i}-U_{i})=c(G_{i}-U_{i})-1$ since we lose the component $a_{i}$. Thus,

[TABLE]

Case 2: Suppose $G$ has a cutvertex but all vertices of $J$ lie in a single block $B$ of $G$. If $S\subseteq U$ then each component of $G-S$ is either (a) a component $C$ of $B-S$ together with all $B$-bridges in $G$ that attach at a vertex of $C$, or (b) a component of $G-w$ not containing $B-w$, for some $w\in S$. Thus,

[TABLE]

By Observation 2.2, $B$ is fundamental, $J$ is a subgraph of the N2C graph of $B$, and

[TABLE]

Applying Case 1 to $B$, we know from (2.2) that

[TABLE]

Combining (2.4), (2.5) and (2.6) gives (2.1).

Case 3: Suppose the vertices of $J$ do not all lie in a single block of $G$. Then there is a cutvertex $x$ of $G$ and vertices $u,v$ of $J$ in distinct components of $G-x$. Let $G_{1}$ be the $x$-bridge in $G$ containing $u$, and $G_{2}$ the union of all other $x$-bridges in $G$, which contains $v$.

By Lemma 2.4(c), $x$ is also a cutvertex of $J$ separating $u$ and $v$, and it follows from Lemma 2.4(a) that the vertex set of each $x$-bridge in $J$ lies in some $x$-bridge in $G$. So for $i=1,2$ let $J_{i}$ be the union of the $x$-bridges in $J$ whose vertex sets lie in $G_{i}$. We have $u\in V(J_{1})$, $v\in V(J_{2})$, $V(J_{1})\cap V(J_{2})=\{x\}$, and $J=J_{1}\cup J_{2}$. Since $|V(J_{1})|,|V(J_{2})|\geq 2$, we have $|E(J_{1})|,|E(J_{2})|>0$, and so $|E(J_{1})|,|E(J_{2})|<|E(J)|$.

Let $U_{i}=V(J_{i})$ for $i=1,2$. Thinking of $G$, $G_{1}$ and $G_{2}$ as the unions of their blocks, it follows from Observation 2.2 that the N2Cs of $G_{i}$ are the N2Cs $\{u,v\}$ of $G$ with $\{u,v\}\subseteq V(G_{i})$, $J_{i}$ is a subgraph of the N2C graph of $G_{i}$, $c(G_{i},uv)=c(G,uv)$ for $uv\in E(J_{i})$, and $G_{i}$ is fundamental. Applying (2.1) to $G_{i}-U_{i}$ and $J_{i}$ by induction for each $i$,

[TABLE]

proving (2.1). ∎

Theorem 2.6**.**

Let $\mathcal{F}$ be the set of all N2Cs in a connected fundamental graph $G$, and let $W=\bigcup\mathcal{F}$ be the set of vertices used by $\mathcal{F}$. For each $v\in V(G)$, let $\mathcal{F}(v)=\{F\in\mathcal{F}\;|\;v\in F\}$. If $f:V(G)\to\mathbb{Z}$ satisfies

[TABLE]

then there is a nonnegative integer function $\omega$ on ordered pairs $(F,u)$ with $F\in\mathcal{F}$ and $u\in F$ such that

[TABLE]

Proof.

If $v\in V(G)-W$, then $\mathcal{F}(v)=\emptyset$ so $\sum_{F\in\mathcal{F}(v)}\omega(F,v)=0$. So (2.8) is equivalent to $c(G-v)\leq f(v)-1$ which follows from (CC) by taking $S=\{v\}$. Moreover, (2.7) does not involve any vertices not in $W$. Thus, in constructing $\omega$, the only vertices we need to be concerned with are those in $W$.

We associate with $G$ a network $N$. Its vertex set is $V(N)=\{s,t\}\cup\mathcal{F}\cup W$ where $s,t$ are new vertices. Its arc set $A(N)$ consists of three subsets: $A_{1}=\{sF\;|\;F\in\mathcal{F}\}$, $A_{2}=\{Fu\;|\;F\in\mathcal{F},u\in W,u\in F\}$, and $A_{3}=\{ut\;|\;u\in W\}$. Each arc $a$ has a capacity $\gamma(a)$ defined as follows:

[TABLE]

We claim that a maximum $st$-flow $\varphi$ of value $\Phi=\sum_{sF\in A_{1}}\gamma(sF)$ in $N$ gives a desired way of distributing the weights on N2Cs to the vertices, by taking $\omega(F,u)=\varphi(Fu)$ for all $Fu\in A_{2}$. All arcs in $A_{1}$ must be saturated by such a flow, so flow conservation at a vertex $F\in\mathcal{F}$, where $F=\{u,v\}$, gives

[TABLE]

which verifies (2.7), and flow conservation at a vertex $u\in W$ gives

[TABLE]

which verifies (2.8).

So assume that $N$ does not have a maximum flow of value $\Phi$; we will show that this gives a contradiction. By the Max-Flow Min-Cut Theorem, $N$ has an $st$-cut

[TABLE]

such that $\gamma(S,T)<\Phi$. Here $[Q,R]$ denotes all arcs from $Q$ to $R$, $\mathcal{F}_{1}\cup\mathcal{F}_{2}=\mathcal{F}$, $\mathcal{F}_{1}\cap\mathcal{F}_{2}=\emptyset$, $W_{1}\cup W_{2}=W$, and $W_{1}\cap W_{2}=\emptyset$.

If $\mathcal{F}_{2}=\mathcal{F}$ then $\gamma(S,T)\geq\Phi$ which is a contradiction, so $\mathcal{F}_{2}\neq\mathcal{F}$ and $\mathcal{F}_{1}\neq\emptyset$. Since arcs in $A_{2}$ have infinite capacity and $\gamma(S,T)<\infty$, we must have $[\mathcal{F}_{1},W_{2}]=\emptyset$. Therefore,

[TABLE]

Since $\gamma(S,T)<\Phi=\sum_{sF\in A_{1}}\gamma(sF)=\sum_{\{u,v\}\in\mathcal{F}}(c(G,uv)-2)$, we obtain

[TABLE]

or

[TABLE]

Since $[\mathcal{F}_{1},W_{2}]=\emptyset$, $\bigcup\mathcal{F}_{1}\subseteq W_{1}$. So we may consider the graph $H_{1}$ with vertex set $W_{1}$ and edge set $\mathcal{F}_{1}$. By the Pigeonhole Principle, there is a component $J$ of $H_{1}$ with vertex set $U\subseteq W_{1}$ such that

[TABLE]

Combining this with Proposition 2.5, we get

[TABLE]

giving the required contradiction to (CC). ∎

3 Proof of Theorem 1.5

In this section we use Theorem 2.6 to prove Theorem 1.5. We start with some preliminary results.

Lemma 3.1**.**

Suppose $H$ is a subgraph of a $K_{4}$-minor-free graph $G$. Then every N2C of $H$ is also an N2C of $G$.

Proof.

Suppose $\{u,v\}$ is an N2C of $H$ that is not an N2C of $G$. Then $c(G,uv)<c(H,uv)$ so there must be a path $P$ in $G-\{u,v\}$ between two components $C_{1},C_{2}$ of $H-\{u,v\}$ that attach at both $u$ and $v$, where we may assume that the internal vertices of $P$ (if any) are not vertices of $H$. Let $C_{3}$ be a third component of $H-\{u,v\}$ attaching at both $u$ and $v$. The subgraph $H[\{u,v\}\cup V(C_{1}\cup C_{2}\cup C_{3})]\cup P$ of $G$ contains a $K_{4}$ minor, which is a contradiction. ∎

Lemma 3.2**.**

If $G$ is $K_{4}$-minor-free and $G$ has no N2C, then $G$ is outerplanar.

Proof.

A graph is outerplanar if and only if it is $K_{4}$-minor-free and $K_{2,3}$-minor-free. So assume that $G$ has a $K_{2,3}$ minor. Since $K_{2,3}$ has maximum degree $3$, the existence of a $K_{2,3}$ minor implies that $G$ contains a subdivision $N$ of $K_{2,3}$, consisting of two vertices $s_{1}$, $s_{2}$ of degree $3$ and three internally disjoint $s_{1}s_{2}$-paths of length at least $2$. But now $\{s_{1},s_{2}\}$ is an N2C of $N$ and hence, by Lemma 3.1, an N2C of $G$, which is a contradiction. ∎

Lemma 3.3**.**

For every non-isolated vertex $x$ of a graph $G$, and for every block $B$ of $G$ that contains $x$, there is $xy\in E(B)$ such that $c(B,xy)=c(G,xy)\leq 1$.

Proof.

For each $uv\in E(G)$, let $\mathcal{A}(uv)$ be the set of $\{u,v\}$-bridges of $G$ that attach at both $u$ and $v$. Suppose that $c(G,xy)\geq 2$ for every $xy\in E(B)$. Let $D_{0}\in\mathcal{A}(xy_{0})$, $xy_{0}\in E(B)$, attain the minimum $\min\{|V(D)|\;|\;D\in\mathcal{A}(xy),\;xy\in E(B)\}$. Let $xy_{1}$ be an edge of $D_{0}$ incident with $x$. There is a path from $y_{1}$ to $y_{0}$ in $D_{0}-x$, so there is a cycle containing $xy_{0}$ and $xy_{1}$, showing that $xy_{1}\in E(B)$ also. Hence $c(G,xy_{1})\geq 2$, and we can choose $D_{1}\in\mathcal{A}(xy_{1})$ such that $y_{0}\notin V(D_{1})$. Now for each $z\in V(D_{1})-\{x,y_{1}\}$ there is a path in $D_{1}-x$ from $z$ to $y_{1}$; this path avoids $y_{0}$ so it is also a path in $G-\{x,y_{0}\}$ from $z$ to $y_{1}\in V(D_{0})$, showing that $z\in V(D_{0})$ also. Thus, $V(D_{1})\subseteq V(D_{0})-\{y_{0}\}$, contradicting the minimality of $|V(D_{0})|$. ∎

3.1 Trees in fundamental graphs

We will first show the existence of an $f$-tree in a fundamental $K_{4}$-minor-free graph, which will serve as a base case when we find $f$-trees in general $K_{4}$-minor-free graphs by induction on the number of vertices.

The following fairly technical result is used to translate our weight function from Theorem 2.6 into spanning trees. Recall the definition of root edge, root block and root vertex from Section 2. Note that the special edges $r_{0}s_{0},r_{1}s_{1},\dots,r_{k}s_{k}$ below always exist, by Lemma 3.3.

Theorem 3.4**.**

Let $\mathcal{F}$ be the set of all N2Cs in a connected fundamental $K_{4}$-minor-free graph $G$ with $|V(G)|\geq 2$. For each $v\in V(G)$, let $\mathcal{F}(v)=\{F\in\mathcal{F}\;|\;v\in F\}$. Suppose there are $f:V(G)\to\mathbb{Z}$ and a nonnegative integer function $\omega$ on ordered pairs $(F,u)$ with $F\in\mathcal{F}$ and $u\in F$ such that

[TABLE]

Choose a root edge $r_{0}s_{0}\in E(G)$ such that $c(G,r_{0}s_{0})\leq 1$. Let the blocks of $G$ be $B_{0},B_{1},B_{2},\allowbreak\dots,B_{k}$, where $r_{0}s_{0}\in E(B_{0})$. For each $i=1,2,\dots,k$, let $r_{i}$ be the root vertex of $B_{i}$, and choose $r_{i}s_{i}\in E(B_{i})$ with $c(G,r_{i}s_{i})\leq 1$.

Then $G$ has a spanning tree $T$ such that $d_{T}(v)\leq f(v)$ for all $v\in V(G)$ and $d_{T}(v)\leq f(v)-1$ for all $v\in\{r_{0}\}\cup\{s_{0},s_{1},\dots,s_{k}\}$. Furthermore, for $0\leq i\leq k$,

[TABLE]

Proof.

The proof is by induction on $|V(G)|$. For the basis, if $|V(G)|=2$ then $G=K_{2}$, $\mathcal{F}$ is empty, conditions (3.1) and (3.2) are trivially satisfied, $r_{0}s_{0}$ is the single edge, and we can take $T=G$. So we may assume that $|V(G)|\geq 3$. There are two cases.

Case 1: $G$ has no N2C.

By Observation 2.2 and Lemma 3.2 each block of $G$ is outerplanar and $c(G,r_{i}s_{i})=c(B_{i},r_{i}s_{i})\leq 1$ for $i=0,1,\dots,k$. If $c(G,r_{i}s_{i})=0$ then $r_{i}s_{i}$ is a cutedge of $G$, so $B_{i}=r_{i}s_{i}$, and we take $T_{i}=B_{i}$. Otherwise, $c(G,r_{i}s_{i})=c(B_{i},r_{i}s_{i})=1$, so $B_{i}$ contains a cycle using $r_{i}s_{i}$. Thus, $B_{i}$ is a $2$-connected outerplanar graph, which we may embed in the plane with a hamiltonian cycle $C_{i}$ as its outer face. Since $c(B_{i},r_{i}s_{i})=1$, $r_{i}s_{i}\in E(C_{i})$, so we take $T_{i}=C_{i}-r_{i}s_{i}$. In either case, $T_{i}$ is a hamiltonian path and a spanning tree in $B_{i}$, and hence $T=\bigcup_{i=0}^{k}T_{i}$ is a spanning tree of $G$.

Let us examine degrees in $T$. Since $G$ has no N2C, $\mathcal{F}=\emptyset$, and inequality (3.2) just says that $c(G-v)-1\leq f(v)-2$ for every $v\in V(G)$. Suppose first that $u\in V(G)-\{r_{0},s_{0},r_{1},s_{1},\dots,r_{k},s_{k}\}$. The cutvertices of $G$ are $r_{1},r_{2},\dots,r_{k}$, so $u$ is not a cutvertex of $G$. Hence $u$ lies in a single block, so by construction of $T$, $d_{T}(u)=2\leq f(u)$, as required. Suppose next that $u\in\{r_{1},\dots,r_{k}\}-(\{r_{0}\}\cup\{s_{0},s_{1},\dots,s_{k}\})$. By construction of $T$, $u$ has two incident edges of $T$ from its parent block, and one incident edge from each child block, so $d_{T}(u)=2+(c(G-u)-1)\leq f(u)$, as required. Finally, suppose that $u\in\{r_{0}\}\cup\{s_{0},s_{1},\dots,s_{k}\}$. Then $u$ has one incident edge of $T$ from each block to which it belongs, so $d_{T}(u)=c(G-u)\leq f(u)-1$, as required.

Furthermore, by construction of $T$, for $0\leq i\leq k$ we have $r_{i}s_{i}\in E(T)$ if $c(G,r_{i}s_{i})=0$ and $r_{i}s_{i}\not\in E(T)$ if $c(G,r_{i}s_{i})=1$.

Case 2: $G$ contains an N2C.

Let $\{x,y\}$ be an N2C of $G$, with $\{x,y\}$ contained in a block $B_{m}$. Since $c(G,xy)\geq 3$, we can choose a nontrivial $\{x,y\}$-bridge $D$ that attaches at both $x$ and $y$ so that $r_{0}s_{0},r_{m}s_{m}\notin E(D)$. Let $G_{1}=G-(V(D)-\{x,y\})$, so $G_{1}$ is the union of all $\{x,y\}$-bridges other than $D$, including all bridges that attach only at $x$ or only at $y$. We may assume that the blocks of $G_{1}$ are $B_{0},B_{1},\dots,B_{m-1}$ and the new block $B_{m}^{\prime}$ consisting of all $\{x,y\}$-bridges of $B_{m}$ other than $D\cap B_{m}$. Let $G_{2}=D\cup xay$ where $a$ is a new vertex; $G_{2}$ is a minor of $G$, so it is $K_{4}$-minor-free. Neither $x$ nor $y$ is a cutvertex of $G_{2}$, and $\{x,y\}$ is not an N2C of $G_{2}$ since $c(G_{2},xy)=2$. The blocks of $G_{2}$ are all the blocks of $G$ included in $D$, which are $B_{m+1},B_{m+2},\dots,B_{k}$, and the new block $B_{m}^{\prime\prime}=(D\cap B_{m})\cup xay$. No edge of $B_{m+1},B_{m+2},\dots,B_{k}$ is incident with $x$ or $y$.

We can apply Observations 2.2 and 2.3, breaking $G_{1}$, $G_{2}$ and $G$ up into blocks and into $\{x,y\}$-bridges in $B_{m}^{\prime}$ and $B_{m}^{\prime\prime}$, which are both subgraphs of $B_{m}\cup xay$. It follows that $G_{1}$ and $G_{2}$ are both fundamental. Also, if $\mathcal{F}_{j}$, $j=1$ or $2$, is the set of N2Cs of $G_{j}$, then $\mathcal{F}_{1}$ and $\mathcal{F}_{2}$ are disjoint and $\mathcal{F}=\mathcal{F}_{1}\cup\mathcal{F}_{2}\cup\{\{x,y\}\}$. Possibly $\{x,y\}\in\mathcal{F}_{1}$, but $\{x,y\}\notin\mathcal{F}_{2}$. If $\{u,v\}\in\mathcal{F}_{j}$ and $\{u,v\}\neq\{x,y\}$ then $c(G,uv)=c(G_{j},uv)$.

As $\{x,y\}$ is an N2C, $\omega(\{x,y\},x)\geq 1$ or $\omega(\{x,y\},y)\geq 1$; without loss of generality, assume that $\omega(\{x,y\},x)\geq 1$.

For $F\in\mathcal{F}_{1}\cup\{\{x,y\}\}$ and $u\in F$, recall that $\delta_{x}(x)=1$ and $\delta_{x}(y)=0$, and define

[TABLE]

Then $\omega_{1}(F,u)\geq 0$ whenever it is defined. For $F\in\mathcal{F}_{2}$ and $u\in F$ define $\omega_{2}(F,u)=\omega(F,u)$.

For $u\in V(G_{1})$ define

[TABLE]

For $u\in V(G_{2})$ define

[TABLE]

Claim**.**

For $j=1$ and $2$, (3.1) and (3.2) hold for $f_{j}$ and $\omega_{j}$ in $G_{j}$.

Proof of claim. If $F\in\mathcal{F}_{j}$ and $F\neq\{x,y\}$, then (3.1) for $F$ in $G_{j}$ is just a rewriting of (3.1) for $F$ in $G$. If $u\in V(G_{j})$ and $u\notin\{x,y,a\}$ then (3.2) for $u$ in $G_{j}$ is just a rewriting of (3.2) for $u$ in $G$.

If $F=\{x,y\}\in\mathcal{F}_{1}$ then (3.1) holds for $\{x,y\}$ in $G_{1}$ because

[TABLE]

Suppose $u\in\{x,y\}$. If $\{x,y\}\in\mathcal{F}_{1}$ then $\mathcal{F}(u)=\mathcal{F}_{1}(u)\cup\mathcal{F}_{2}(u)$ and (3.2) for $u$ in $G$ can be rewritten as

[TABLE]

which gives (3.2) for $u$ in $G_{1}$ after cancelling $\delta_{x}(u)+\sum_{F\in\mathcal{F}_{2}(u)}\omega(F,u)$ on both sides. If $\{x,y\}\notin\mathcal{F}_{1}$ then $\{x,y\}$ is included in $\mathcal{F}(u)$ but not in $\mathcal{F}_{1}(u)$, so this inequality is modified by adding a nonnegative term $\omega_{1}(\{x,y\},u)$ on the left, which can be deleted, so (3.2) still holds for $u$ in $G_{1}$. Also, (3.2) holds (with equality) for $u$ in $G_{2}$ by definition of $f_{2}(u)$ and because $\omega_{2}$ is just a restriction of $\omega$ and $c(G_{2},u)=1$.

For $u=a$, $\mathcal{F}_{2}(a)=\emptyset$, $c(G_{2}-a)=1$, and $f_{2}(a)=2$, so (3.2) holds for $a$ in $G_{2}$. $\blacksquare$

In $G_{1}$ we can still choose $r_{0}s_{0}$ as the root edge, and $r_{1}s_{1},r_{2}s_{2},\dots r_{m}s_{m}$ as the other special edges (where $r_{m}s_{m}$ is now in $B_{m}^{\prime}$ instead of $B_{m}$). By the claim we can apply induction to find a spanning tree $T_{1}$. In $G_{2}$ we choose $ax\in E(B_{m}^{\prime\prime})$ as the root edge, and we can still choose $r_{m+1}s_{m+1},\dots,r_{k}s_{k}$ as the other special edges. By the claim we can apply induction to find a spanning tree $T_{2}$. Since $c(G_{2},ax)=1$, we have $ax\notin E(T_{2})$ by (3.3), so $ay\in E(T_{2})$. Thus, $T_{2}^{\prime}=T_{2}-ay$ is a spanning tree of $D=G_{2}-a$.

Now $T_{1}$ and $T_{2}^{\prime}$ both contain $xy$-paths. So $T_{1}\cup T_{2}^{\prime}$ is a spanning subgraph of $G$ containing a single cycle $C$ through $x$ and $y$. Let $yz$ be the edge of $C\cap G_{2}$ incident with $y$. Then $T=(T_{1}\cup T_{2}^{\prime})-yz$ is a spanning tree of $G$.

For all $v\in V(G)$, define $\sigma(v)$ to be $1$ if $v\in\{r_{0}\}\cup\{s_{0},s_{1},\dots,s_{k}\}$, and [math] otherwise. If $u\in V(G)-\{x,y\}$, then $u\in V(G_{j})-\{x,y\}$ for $j=1$ or $2$, and $d_{T}(u)\leq d_{T_{j}}(u)\leq f_{j}(u)-\sigma(u)=f(u)-\sigma(u)$ as required.

We also want to show that $d_{T}(u)\leq f(u)-\sigma(u)$ when $u\in\{x,y\}$. We know that no edge of $B_{m+1},B_{m+2},\dots,B_{k}$ is incident with $u$, so if $\sigma(u)=1$, it is because $u$ is incident with an edge $r_{i}s_{i}$ for $0\leq i\leq m$; this edge belongs to $G_{1}$. Thus, $d_{T_{1}}(u)\leq f_{1}(u)-\sigma(u)$ by construction of $T_{1}$. Since $ax$ is the root edge of $G_{2}$, $d_{T_{2}^{\prime}}(x)=d_{T_{2}}(x)\leq f_{2}(x)-1$, so

[TABLE]

Since we delete two edges $ay$ and $yz$ from $T_{1}\cup T_{2}$ when forming $T$,

[TABLE]

Finally we verify (3.3). For each $i$, $0\leq i\leq k$, $r_{i}s_{i}\in E(G_{j})$ for a unique $j\in\{1,2\}$. If $c(G,r_{i}s_{i})=0$ then $r_{i}s_{i}$ is a cutedge of $G$ and so must belong to the spanning tree $T$. Otherwise, $c(G,r_{i}s_{i})=c(G_{j},r_{i}s_{i})=1$ and so $r_{i}s_{i}\notin E(T_{j})$, and hence $r_{i}s_{i}\notin E(T)$. ∎

Combining Theorems 2.6 and 3.4 we obtain our result on spanning trees in fundamental $K_{4}$-minor-free graphs. The hypotheses (3.1) and (3.2) of Theorem 3.4 are just slightly rewritten versions of the conclusions (2.7) and (2.8) of Theorem 2.6. Note that Theorem 3.4 does not handle the case where $|V(G)|=1$, but that is trivial.

Theorem 3.5**.**

Let $G$ be a fundamental connected $K_{4}$-minor-free graph. Suppose that $f:V(G)\to\mathbb{Z}$ satisfies

[TABLE]

Choose a root edge $r_{0}s_{0}\in E(G)$ such that $c(G,r_{0}s_{0})\leq 1$. Let the blocks of $G$ be $B_{0},B_{1},B_{2},\allowbreak\dots,B_{k}$, where $r_{0}s_{0}\in E(B_{0})$. For each $i=1,2,\dots,k$, let $r_{i}$ be the root vertex of $B_{i}$, and choose $r_{i}s_{i}\in E(B_{i})$ with $c(G,r_{i}s_{i})\leq 1$.

Then $G$ has a spanning tree $T$ such that $d_{T}(v)\leq f(v)$ for all $v\in V(G)$ and $d_{T}(v)\leq f(v)-1$ for all $v\in\{r_{0}\}\cup\{s_{0},s_{1},\dots,s_{k}\}$. Furthermore, for $0\leq i\leq k$, $r_{i}s_{i}\in E(T)$ if $c(G,r_{i}s_{i})=0$, and $r_{i}s_{i}\notin E(T)$ if $c(G,r_{i}s_{i})=1$.

This theorem is quite technical, and only applies to fundamental $K_{4}$-minor-free graphs, but it gives strong results in that situation, particularly for graphs with many cutedges. For example, neither the general result Theorem 1.2 nor our main result Theorem 1.5 can prove that when $G$ is a tree, $G$ has a spanning $f$-tree where $f=d_{G}$ is just the degree function in $G$. However, we can get this result from Theorem 3.5 with $f(v)=d_{G}(v)+1$ for all $v$, $r_{0}s_{0}$ an arbitrary edge, and $\{s_{1},s_{2},\dots,s_{k}\}=V(G)-\{r_{0},s_{0}\}$.

3.2 Trees

in general graphs

In this subsection, we find spanning $f$-trees in general $K_{4}$-minor-free graphs. We will use the Component Condition (CC), as stated in Theorem 1.5, often.

Lemma 3.6**.**

Let $G$ be a 2-connected graph and $f:V(G)\to\mathbb{Z}$. If $G$ and $f$ satisfy (CC), then for every 2-cut $\{u,v\}$ of $G$ with $uv\in E(G)$, $G-uv$ and $f$ also satisfy (CC).

Proof.

Let $G^{\prime}=G-uv$, and let $S\subseteq V(G^{\prime})$ with $S\neq\emptyset$. If $\{u,v\}\cap S\neq\emptyset$ then $G^{\prime}-S=G-S$, and if $u$ and $v$ are in the same component of $G^{\prime}-S$ then $c(G^{\prime}-S)=c((G^{\prime}-S)\cup uv)=c(G-S)$; in either case (CC) holds for $G^{\prime}$, $f$ and $S$. So we may assume that $u,v\notin S$ and $u$ and $v$ are in distinct components of $G^{\prime}-S$.

Let $D_{1},D_{2},\dots,D_{t}$ be the $\{u,v\}$-bridges in $G^{\prime}$, and let $S_{i}=S\cap V(D_{i})$ for each $i$. Then $S=\bigcup_{i=1}^{t}S_{i}$. As $\{u,v\}$ is a $2$-cut, $t\geq 2$. Now $S_{i}$ must separate $u$ and $v$ in $D_{i}$, so the set of components of $D_{i}-S_{i}$ can be written as $\mathcal{A}_{i}\cup\{C_{i}^{u},C_{i}^{v}\}$ where $u\in V(C_{i}^{u})$, $v\in V(C_{i}^{v})$, and $\mathcal{A}_{i}$ contains all other components. The set of components of $G-S_{i}$ is then $\mathcal{A}_{i}\cup\{C_{i}^{u}\cup C_{i}^{v}\cup\bigcup_{j\neq i}D_{j}\}$, so $c(G-S_{i})=|\mathcal{A}_{i}|+1$. The set of components of $G^{\prime}-S$ is $\bigcup_{i=1}^{t}\mathcal{A}_{i}\cup\{\bigcup_{i=1}^{t}C_{i}^{u},\bigcup_{i=1}^{t}C_{i}^{v}\}$. Since $t\geq 2$ and (CC) holds for $G$ and $f$, we have

[TABLE]

and so (CC) holds for $G^{\prime}$, $f$ and $S$. ∎

If every $2$-cut $\{u,v\}$ of a graph $G$ satisfies $uv\notin E(G)$, we say that $G$ is $2$-cut-reduced. If $G$ is $2$-connected and $\{u,v\}$ is a $2$-cut of $G$ with $uv\in E(G)$, then $G-uv$ is still $2$-connected. Therefore, we can apply Lemma 3.6 repeatedly. We may create new $2$-cuts when we do, but we can only create a bounded number, so we obtain the following.

Corollary 3.7**.**

Let $G$ be a 2-connected graph and $f:V(G)\to\mathbb{Z}$. Assume that $G$ and $f$ satisfy (CC). Then $G$ has a $2$-cut-reduced spanning subgraph $G^{\prime}$ such that $G^{\prime}$ and $f$ also satisfy (CC).

We write a $uv$-path $P$ as $P[u,v]$ to emphasize its endvertices. Suppose the block-cutvertex tree of a graph $H$ is a path. Then we write $H=v_{0}B_{1}v_{1}B_{2}v_{2}\dots v_{t-1}B_{t}v_{t}$, where each $B_{i}$ is a block of $H$, $v_{0}\in V(B_{1})-\{v_{1}\}$, $v_{t}\in V(B_{t})-\{v_{t-1}\}$, and each $v_{i}$, $i=1,\dots,t-1$, is a cutvertex of $H$ with $v_{i}\in V(B_{i})\cap V(B_{i+1})$. (If $H$ has only one vertex we take $t=0$ so that $H=v_{0}$.) We say $H$ is a chain of blocks from $v_{0}$ to $v_{t}$. If $t\geq 2$ we say the chain of blocks is nontrivial.

Lemma 3.8**.**

Let $G$ be a $2$-connected graph and $xy\in E(G)$.

(a)* Then $G-xy$ is a chain of blocks from $x$ to $y$.*

(b)* Moreover, if $G$ is a $2$-cut-reduced $K_{4}$-minor-free graph, then $G-xy$ is a nontrivial chain of blocks.*

Proof.

(a) Since $G$ is $2$-connected, if $G-xy$ has a cutvertex then every leaf block of $G-xy$ contains $x$ or $y$ as a non-cutvertex vertex. Hence, $G-xy$ is either a single block or has only two leaf blocks, and is a chain of blocks from $x$ to $y$.

(b) If $G-xy$ is 2-connected, then there are internally disjoint $xy$-paths $P_{1}$ and $P_{2}$ in $G-xy$. Since $G$ is $2$-cut-reduced, $\{x,y\}$ is not a 2-cut of $G$, so there is a path $P$ in $G-\{x,y\}$ connecting $P_{1}$ to $P_{2}$, where we may assume that $|V(P)\cap V(P_{i})|=1$, $i=1,2$. Then $P_{1}\cup P_{2}\cup P\cup xy$ contains a $K_{4}$ minor, which is a contradiction. Hence $G-xy$ is not $2$-connected, so it is a nontrivial chain of blocks. ∎

If $H$ has at least two blocks, $B$ is a leaf block of $H$, and the cutvertex of $H$ in $B$ is $v$, then $H\ominus B$ denotes $H-(V(B)-\{v\})$.

Lemma 3.9**.**

Let $G$ be a $2$-cut-reduced $2$-connected $K_{4}$-minor-free graph. If $G$ is not fundamental and $u\in V(G)$, then there exist $xy\in E(G)$ and a leaf block $B$ of $G-xy$ such that each of $x$ and $y$ is contained in an N2C of $G$, $B$ is fundamental, and $u\in V((G-xy)\ominus B)$.

Proof.

Since $G$ is not fundamental there is at least one edge $xy\in E(G)$ such that both $x$ and $y$ are contained in N2Cs of $G$. By Lemma 3.8, $G-xy$ is a nontrivial chain of blocks from $x$ to $y$. For such an $xy$ there is at least one leaf block $B$ of $G-xy$ so that $u\in V((G-xy)\ominus B)$. Choose such an $xy$ and such a $B$ so that $|V((G-xy)\ominus B)|$ is as large as possible. We may assume that $G-xy=v_{0}B_{1}v_{1}\dots v_{t-1}B_{t}v_{t}$ where $x=v_{0}$, $y=v_{t}$ and $B=B_{1}$.

We claim that $B_{1}$ is fundamental. Suppose not. Then there is $x_{1}y_{1}\in E(B_{1})$ with both $x_{1}$ and $y_{1}$ in N2Cs of $B_{1}$. By Lemma 3.1, both $x_{1}$ and $y_{1}$ are also in N2Cs of $G$. Since $B_{1}$ has N2Cs it has at least five vertices and is $2$-connected. By Lemma 3.8(a), $B_{1}-x_{1}y_{1}$ is a chain of blocks $w_{0}D_{1}w_{1}\dots w_{s-1}D_{s}w_{s}$ (possibly $s=1$, but $s\neq 0$). Choose $h$ and $k$ such that $h\leq k$, $x$ and $v_{1}$ are vertices of $w_{h}D_{h+1}w_{h+1}\dots w_{k-1}D_{k}w_{k}$, and $k-h$ is as small as possible. Since $x\neq v_{1}$ we have $h<k$. We may assume that $x_{1}=w_{0}$, $y_{1}=w_{s}$, $x\in D_{h+1}-\{w_{h+1}\}$ and $v_{1}\in D_{k}-\{w_{k-1}\}$.

By Lemma 3.8, $G-x_{1}y_{1}$ is a nontrivial chain of blocks, which must be $w_{0}D_{1}\dots\allowbreak w_{h-1}D_{h}w_{h}D^{*}w_{k}D_{k+1}\dots\allowbreak D_{s}w_{s}$ where $D^{*}$ is a block $(xD_{h+1}w_{h+1}D_{h+2}\dots\allowbreak w_{k-1}D_{k}v_{1}B_{2}v_{2}\dots\allowbreak v_{t-1}B_{t}y)\allowbreak\cup\allowbreak xy$. Since $G-x_{1}y_{1}$ is nontrivial, either $h>0$ or $k<s$ (hence $s\geq 2$). If $h>0$ then $D_{1}$ is a leaf block of $G-x_{1}y_{1}$, $u\in V((G-xy)\ominus B_{1})=V(B_{2}\cup\dots\cup B_{t})\subset V(D^{*})\subseteq V((G-x_{1}y_{1})\ominus D_{1})$ and $|V((G-x_{1}y_{1})\ominus D_{1})|\geq|V(D^{*})|>|V((G-xy)\ominus B_{1})|$. Thus, $x_{1}y_{1}$ and $D_{1}$ contradict the choice of $xy$ and $B=B_{1}$. We obtain a similar contradiction from $x_{1}y_{1}$ and $D_{s}$ if $k<s$.

Hence, $B=B_{1}$ is fundamental, as claimed, which proves the lemma. ∎

We are now ready to prove Theorem 1.5, which we restate for convenience.

Theorem 1.5.

Let $G$ be a connected $K_{4}$-minor-free graph, and $f:V(G)\to\mathbb{Z}$. Suppose that $z\in V(G)$ and

[TABLE]

Then $G$ has a spanning $(f-\delta_{z})$-tree, i.e., a spanning $f$-tree $T$ such that $d_{T}(z)\leq f(z)-1$.

Proof.

The proof is by induction on $|V(G)|$. The conclusion is true by Theorem 3.5 if $|V(G)|\leq 4$, since $G$ has no N2Cs and is fundamental. So we assume that $|V(G)|\geq 5$. We consider two cases.

Case 1: $G$ has a cutvertex $x$.

Let $G_{1}$ be one $x$-bridge, and let $G_{2}$ be the union of the remaining $x$-bridges. If $z\in V(G_{1})-\{x\}$ let $z_{1}=z$ and $z_{2}=x$, if $z\in V(G_{2})-\{x\}$ let $z_{1}=x$ and $z_{2}=z$, and if $z=x$ let $z_{1}=z_{2}=x$. For every $v\in V(G_{i})-\{x\}$, $i=1,2$, let $f_{i}(v)=f(v)$. Then each $G_{i}$ and $f_{i}$, $i=1$ or $2$, satisfy (CC) for all $S$ that do not contain $x$.

Let $f_{2}(x)$ be the minimum integer such that $G_{2}$ and $f_{2}$ satisfy (CC) for all $S$ containing $x$, and let $f_{1}(x)=f(x)-f_{2}(x)+1$. We claim that $G_{1}$ and $f_{1}$ also satisfy (CC) for sets $S$ containing $x$. If not, there is $U\subseteq V(G_{1})$ with $x\in U$ so that

[TABLE]

By minimality of $f_{2}(x)$ there is $W\subseteq V(G_{2})$ with $x\in W$ so that

[TABLE]

Thus, if $S=U\cup W$ we have

[TABLE]

contradicting (CC) for $G$ and $f$.

Thus, for each $i=1,2$, $G_{i}$ and $f_{i}$ satisfy (CC), so by induction there is a spanning $f_{i}$-tree $T_{i}$ of $G_{i}$ with $d_{T_{i}}(z_{i})\leq f_{i}(z_{i})-1$. Let $T=T_{1}\cup T_{2}$, which is a spanning tree of $G$. If $v\notin\{z,x\}$ then $v\in V(G_{i})-\{x\}$ for some $i$, and $d_{T}(v)=d_{T_{i}}(v)\leq f_{i}(v)=f(v)$. If $z\in V(G_{i})-\{x\}$ then $d_{T}(z)=d_{T_{i}}(z)\leq f_{i}(z)-1=f(z)-1$ since $z=z_{i}$, and $d_{T}(x)=d_{T_{1}}(x)+d_{T_{2}}(x)\leq f_{1}(x)+f_{2}(x)-1=f(x)$ since $x=z_{3-i}$. If $z=x$ then $d_{T}(x)\leq(f_{1}(x)-1)+(f_{2}(x)-1)=f(x)-1$ since $x=z_{1}=z_{2}$. Thus, $T$ is a spanning $f$-tree with $d_{T}(z)\leq f(z)-1$.

Case 2: $G$ has no cutvertex; since $|V(G)|\geq 5$, $G$ is $2$-connected.

We may assume that $G$ is not fundamental, by Theorem 3.5, and that $G$ is $2$-cut-reduced, by Corollary 3.7. By Lemmas 3.8 and 3.9, $G$ contains an edge $xy$ with both $x$ and $y$ in N2Cs of $G$ such that some leaf block $B$ of $G-xy$, which is a nontrivial chain of blocks from $x$ to $y$, is fundamental and $z\in V((G-xy)\ominus B)$. Without loss of generality $x\in V(B)$ and so $G-xy=xB_{1}v_{1}B_{2}v_{2}\dots B_{t}y$ where $t\geq 2$ and $B=B_{1}$. Write $L_{2}=(G-xy)\ominus B_{1}=v_{1}B_{2}v_{2}\dots v_{t-1}B_{t}v_{t}$, so that $z\in V(L_{2})$. Since $x$ and $y$ are in N2Cs, $d_{G}(x),d_{G}(y)\geq 3$ and hence $|V(B_{1})|,|V(L_{2})|\geq 3$.

Let $G_{1}=B_{1}\cup xa_{1}v_{1}$ where $a_{1}$ is a new vertex, and let $G_{2}=L_{2}\cup v_{1}y$. Both $G_{1}$ and $G_{2}$ are minors of $G$ and hence $K_{4}$-minor-free. For $i=1,2$ define $f_{i}(v)=f(v)$ for $v\in V(G_{i})-\{v_{1},a_{1}\}$. Then for $S\subseteq V(G_{2})$ with $v_{1}\notin S$ we have $c(G-S)=c(G_{2}-S)$ and $f_{2}(v)=f(v)$ for all $v\in S$, so (CC) holds for $G_{2}$, $f_{2}$ and $S$. Choose $f_{2}(v_{1})$ to be the minimum integer such that (CC) holds for $G_{2}$ and $f_{2}$, then by minimality there is $W\subseteq V(G_{2})$ with $v_{1}\in W$ so that

[TABLE]

Let $f_{1}(v_{1})=f(v_{1})-f_{2}(v_{1})+2$ and $f_{1}(a_{1})=|V(G_{1})|+1$ (note that our theorems do not require that $f(v)\leq d_{G}(v)$). We claim that (CC) holds for $G_{1}$ and $f_{1}$. This is true for any $S\subseteq V(G_{1})$ with $a_{1}\in S_{1}$, by choice of $f_{1}(a_{1})$. If we have nonempty $S\subseteq V(G_{1})$ with $a_{1},v_{1}\notin S$,

[TABLE]

and so (CC) holds in this situation as well. So if (CC) does not hold for $G_{1}$ and $f_{1}$ there is $U\subseteq V(G_{1})$ with $a_{1}\notin U$, $v_{1}\in U$ and

[TABLE]

Let $S=U\cup W$ then $a_{1}\notin S$, $v_{1}\in S$. Note that $c(G-xy-S)=c(B_{1}-U)+c(L_{2}-W)$. If $x\notin U$ then $c(G-S)\geq c(G-xy-S)-1$ and $c(G_{1}-U)=c(B_{1}-U)$, so

[TABLE]

and if $x\in U$ then $c(G-S)=c(G-xy-S)$ and $c(G_{1}-U)=c(B_{1}-U)+1$, so

[TABLE]

Thus, in either case,

[TABLE]

contradicting (CC) for $G$ and $f$. Thus, $G_{1}$ and $f_{1}$ satisfy (CC).

Now $a_{1}$ does not belong to any N2C of $G_{1}$ since it has degree $2$, so $G_{1}$ is fundamental. Choosing root edge $r_{0}s_{0}=a_{1}x$ and applying Theorem 3.5, there is a spanning $f_{1}$-tree $T_{1}$ of $G_{1}$ with $d_{T_{1}}(x)\leq f_{1}(x)-1$ and $a_{1}x\notin E(T_{1})$. Thus, $a_{1}v_{1}\in E(T_{1})$ and $T_{1}^{\prime}=T_{1}-a_{1}$ is a spanning $f_{1}$-tree of $G_{1}-a_{1}=B_{1}$ with $d_{T_{1}^{\prime}}(v)\leq f_{1}(v)-1$ for $v\in\{x,v_{1}\}$.

Since $|V(B_{1})|\geq 3$, $|V(G_{2})|<|V(G)|$, so by induction there is a spanning $(f_{2}-\delta_{z})$-tree $T_{2}$ of $G_{2}$ (i.e., an $f_{2}$-tree with $d_{T_{2}}(z)\leq f_{2}(z)-1$). Suppose that some such tree $T_{2}$ has $v_{1}y\in E(T_{2})$. Then $T=T_{1}^{\prime}\cup(T_{2}-v_{1}y)\cup xy$ has $d_{T}(v)\leq f_{1}(v)=f(v)$ for $v\in V(B_{1})-\{x,v_{1}\}$, $d_{T}(v)\leq f_{2}(v)-\delta_{z}(v)=f(v)-\delta_{z}(v)$ for $v\in V(L_{2})-\{v_{1},y\}$, and

[TABLE]

Thus, $T$ is our desired spanning tree. We may henceforth assume that every $(f_{2}-\delta_{z})$-tree $T_{2}$ of $G_{2}$ has $v_{1}y\notin E(T_{2})$, and so is a spanning tree of $L_{2}$.

Now define $g_{1}$ on $V(G_{1})$ by $g_{1}(x)=f_{1}(x)+1=f(x)+1$, $g_{1}(v_{1})=f_{1}(v_{1})-1=f(v_{1})-f_{2}(v_{1})+1$, and $g_{1}(v)=f_{1}(v)$ for $v\in V(G_{1})-\{x,v_{1}\}$. Suppose that $G_{1}$ and $g_{1}$ satisfy (CC). By similar reasoning to that for $T_{1}^{\prime}$ above, there is a spanning $g_{1}$-tree $R_{1}^{\prime}$ of $B_{1}$ with $d_{R_{1}^{\prime}}(v)\leq g_{1}(v)-1$ for $v\in\{x,v_{1}\}$. Taking $T=R_{1}^{\prime}\cup T_{2}$ (remembering that $T_{2}$ is now a spanning tree of $L_{2}$) we get a spanning tree of $G$ with $d_{T}(v)\leq g_{1}(v)=f(v)$ for $v\in V(B_{1})-\{x,v_{1}\}$, $d_{T}(v)\leq f_{2}(v)-\delta_{z}(v)=f(v)-\delta_{z}(v)$ for $v\in V(L_{2})-\{v_{1},y\}$, and

[TABLE]

Thus, $T$ is our desired spanning tree. We may henceforth assume that $G_{1}$ and $g_{1}$ do not satisfy (CC).

Let $G_{2}^{+}=L_{2}\cup v_{1}a_{2}y$, where $a_{2}$ is a new vertex; $G_{2}^{+}$ is a minor of $G$ and hence is $K_{4}$-minor-free. Extend $f_{2}$ to $G_{2}^{+}$ by defining $f_{2}(a_{2})=|V(G_{2}^{+})|+1$. Suppose that $G_{2}^{+}$ and $f_{2}$ satisfy (CC). Since $|V(B_{1})|\geq 3$, we have $|V(G_{2}^{+})|<|V(G)|$, and so by induction there is a spanning $(f_{2}-\delta_{z})$-tree $Q_{2}$ of $G_{2}^{+}$. If $Q_{2}$ uses the path $v_{1}a_{2}y$ then $G_{2}$ has a spanning $(f_{2}-\delta_{z})$-tree using $v_{1}y$, which we have assumed does not exist. So $d_{Q_{2}}(a_{2})=1$ and deleting whichever of $v_{1}a_{2}$, $a_{2}y$ is an edge of $Q_{2}$ gives a spanning $(f_{2}-\delta_{z}z)$-tree $Q_{2}^{\prime}$ of $L_{2}$ for which there is $w\in\{v_{1},y\}$ such that $d_{Q_{2}^{\prime}}(w)\leq f_{2}(w)-\delta_{z}(w)-1$. Let $w^{\prime}$ be the vertex of $\{v_{1},y\}$ other than $w$. Now $T_{1}^{\prime}\cup Q_{2}^{\prime}\cup xy$ is a connected spanning subgraph containing a cycle through $x$, $v_{1}$ and $y$. Let $w^{\prime}t$ be an edge of this cycle incident with $w^{\prime}$ and let $T=(T_{1}^{\prime}\cup Q_{2}^{\prime}\cup xy)-w^{\prime}t$, which is a spanning tree in $G$. We have $d_{T}(v)\leq f_{1}(v)=f(v)$ for $v\in V(B_{1})-\{x,v_{1}\}$, $d_{T}(v)\leq f_{2}(v)-\delta_{z}(v)=f(v)-\delta_{z}(v)$ for $v\in V(L_{2})-\{v_{1},y\}$, and

[TABLE]

where we add $1$ to the degrees of $x$ and $y$ due to $xy$, but subtract $1$ from the degrees of $v_{1}$ and $y$ because $\{v_{1},y\}=\{w,w^{\prime}\}$. Thus, $T$ is our desired spanning tree. We may henceforth assume that $G_{2}^{+}$ and $f_{2}$ do not satisfy (CC).

Now $G_{1}$ and $g_{1}$ do not satisfy (CC), although $G_{1}$ and $f_{1}$ do. Suppose (CC) fails for $U\subseteq V(G_{1})$. By definition of $g_{1}$, we see that $G_{1}$, $g_{1}$ and a specific set $S\subseteq V(G_{1})$ do satisfy (CC) provided $v_{1}\notin S$, $a_{1}\in S$, or $\{x,v_{1}\}\subseteq S$ (since $g_{1}(x)+g_{1}(v_{1})=f_{1}(x)+f_{1}(v_{1})$). Thus, $v_{1}\in U$, $a_{1}\notin U$ (so $U\subseteq V(B_{1})$), $x\notin U$, and

[TABLE]

Also $G_{2}^{+}$ and $f_{2}$ do not satisfy (CC), although $G_{2}$ and $f_{2}$ do. Suppose (CC) fails for $W\subseteq V(G_{2})$. We see that $G_{2}^{+}$, $f_{2}$ and a specific set $S\subseteq V(G_{2})$ do satisfy (CC) provided $a_{2}\in S$, or $a_{2}\notin S$ and $c(G_{2}^{+}-S)=c(G_{2}-S)$. The only situation where $a_{2}\notin S$ and $c(G_{2}^{+}-S)\neq c(G_{2}-S)$ is when $\{v_{1},y\}\subseteq S$. Thus, $a_{2}\notin W$ (so $W\subseteq V(L_{2})$), $\{v_{1},y\}\subseteq W$, and

[TABLE]

Letting $S=U\cup W\subseteq V(G)$, we have $x\notin S$ but $\{v_{1},y\}\subseteq S$, and therefore

[TABLE]

which contradicts (CC) for $G$ and $f$. Thus, this final situation never happens, and we can always find the required spanning tree $T$. ∎

4 Concluding remarks

In this section, we show the existence of planar graphs $G$ such that $c(G-S)\leq\sum_{v\in S}(f(v)-1)$ for every $S\subseteq V(G)$ and $S\neq\emptyset$, but where $G$ has no spanning tree $T$ such that $d_{T}(v)\leq f(v)$ for all $v\in V(G)$, where there is an integer $f(v)\geq 2$ for each vertex.

Let $G$ be a $1$-tough planar graph with an integer $f(v)\geq 2$ for each vertex, and let $G^{\prime}$ be obtained by attaching $f(v)-2$ pendant edges to each $v\in V(G)$. Let $f(v)=2$ for $v\in V(G^{\prime})-V(G)$. Since $c(G-S)\leq|S|$ for every $S\subseteq V(G)$ with $S\neq\emptyset$, it is not too hard to show that $c(G^{\prime}-S^{\prime})\leq\sum_{v\in S^{\prime}}(f(v)-1)$ for every $S^{\prime}\subseteq V(G^{\prime})$ with $S^{\prime}\neq\emptyset$. If $G$ has no spanning 2-tree then $G^{\prime}$ has no spanning tree $T$ such that $d_{T}(v)\leq f(v)$ for all $v\in V(G^{\prime})$. Thus, it suffices to show the existence of a 1-tough planar graph $G$ with no spanning 2-tree, i.e., no hamiltonian path.

A vertex in a graph is called a simplicial vertex if the neighbors of this vertex form a clique in the graph. The graph $T$ given in Figure 1, constructed by Dillencourt [4] is 1-tough, non-hamiltonian, and maximal planar. Proposition 5 of that paper states that any path in $T$ connecting any two of the three vertices $A,B,C$ must omit at least one simplicial vertex. In particular, $T$ has no hamiltonian path with both ends in $A,B,C$.

Dillencourt constructed a sequence of 1-tough nonhamiltonian maximal planar graphs as follows. Let $G_{1}=T$. For $n\geq 2$, let $G_{n}$ be obtained from $G_{n-1}$ by deleting every simplicial vertex and replacing it with a copy of $T$. More precisely, for each simplicial vertex $u$, let $x,y,z$ be its neighbors. Delete $u$, insert a copy $T_{u}$ (with vertices $A_{u}$, $B_{u}$, etc.) of $T$ inside the triangle $(xyz)$, and add the edges $A_{u}x,A_{u}y,B_{u}y,B_{u}z,C_{u}z,C_{u}x$.

Suppose that $G_{n}$, $n\geq 2$, has a hamiltonian path $P$. Since $G_{n-1}$ has at least three simplicial vertices, for some such vertex $u$ the copy $T_{u}$ of $T$ in $G_{n}$ contains neither end of $P$. The structure of $G_{n}$ guarantees that $P\cap T_{u}$ must have one of two forms: (1) a single hamiltonian path in $T_{u}$ with ends being two of $A_{u},B_{u},C_{u}$, or (2) a union of two paths $P_{1}$ and $P_{2}$, where $P_{1}$ is a one-vertex path using one of $A_{u},B_{u},C_{u}$, and $P_{2}$ is a path between the other two of $A_{u},B_{u},C_{u}$ using all other vertices of $T_{u}$. In case (2) we can join the vertex of $P_{1}$ to either end of $P_{2}$ to obtain a hamiltonian path in $T_{u}$ with both ends in $A_{u},B_{u},C_{u}$. So either situation means $T$ has a hamiltonian path with ends being two of $A,B,C$, which we know does not happen. Therefore, $G_{n}$ is a $1$-tough maximal planar graph with no hamiltonian path.

We conclude with a general question. Suppose $G$ is a tree, i.e., a connected graph of treewidth at most $1$. Then the sufficient condition (1.1), the necessary condition (1.2), and the even weaker condition

[TABLE]

are all equivalent and all imply that $d_{G}(v)\leq f(v)$ for all $v\in V(G)$, i.e., that $G$ has (in fact, is) a spanning $f$-tree. (When $f(v)=k$ for all $v$, (4.1) just says that $G$ is $\frac{1}{k}$-tough.) If $G$ has treewidth at most $2$, we have a sufficient condition (CC) that is only slightly stronger than the necessary condition (1.2). It therefore seems natural to ask the following.

Question 4.1**.**

Is there a natural strengthening of Theorem 1.2 involving treewidth? More specifically, suppose we have a connected graph $G$ of treewidth $h$, and $f:V(G)\to\mathbb{Z}$. Is there a condition involving $h$ and becoming stronger as $h$ increases, similar to (1.2) or (CC) but weaker than (1.1), that implies that $G$ has a spanning $f$-tree?

Acknowledgement

The authors thank an anonymous referee who pointed out a problem with an earlier version of Proposition 2.5.