Beyond Degree Choosability

Let $G$ be a connected graph with maximum degree $\Delta$. Brooks' theorem states that $G$ has a $\Delta$-coloring unless $G$ is a complete graph or an odd cycle. A graph $G$ is \emph{degree-choosable} if $G$ can be properly colored from its lists whenever each vertex $v$ gets a list of $d(v)$ colors. In the context of list coloring, Brooks' theorem can be strengthened to the following. Every connected graph $G$ is degree-choosable unless each block of $G$ is a complete graph or an odd cycle; such a graph $G$ is a \emph{Gallai tree}. This degree-choosability result was further strengthened to Alon--Tarsi orientations; these are orientations of $G$ in which the number of spanning Eulerian subgraphs with an even number of edges differs from the number with an odd number of edges. A graph $G$ is \emph{degree-AT} if $G$ has an Alon--Tarsi orientation in which each vertex has indegree at least 1. Alon and Tarsi showed that if $G$ is degree-AT, then $G$ is also degree-choosable. Hladky, Kral, and Schauz showed that a connected graph is degree-AT if and only if it is not a Gallai tree. In this paper, we consider pairs $(G,x)$ where $G$ is a connected graph and $x$ is some specified vertex in $V(G)$. We characterize pairs such that $G$ has no Alon--Tarsi orientation in which each vertex has indegree at least 1 and $x$ has indegree at least 2. When $G$ is 2-connected, the characterization is simple to state.


Introduction
Brooks' theorem is one of the fundamental results in graph coloring. For every connected graph G, it says that G has a ∆-coloring unless G is a complete graph K ∆+1 or an odd cycle. When we seek to prove coloring results by induction, we often want to color a subgraph H where different vertices have different lists of allowable colors (those not already used on their neighbors in the coloring of G − H). This gives rise to list coloring. Vizing [15] and, independently, Erdős, Rubin, and Taylor [5] extended Brooks' theorem to list coloring. They proved an analogue of Brooks' theorem when each vertex v has ∆ allowable colors (possibly different colors for different vertices). Erdős, Rubin, and Taylor [5] and Borodin [3] strengthened this Brooks' analogue to the following result, where a Gallai tree Gallai tree is a connected graph in which each block is a complete graph or an odd cycle.
Theorem A. If G is connected and not a Gallai tree, then for any list assignment L with |L(v)| = d(v) for all v ∈ V (G), graph G has a proper coloring ϕ with ϕ(v) ∈ L(v) for all v.
The graphs in Theorem A are degree-choosable degreechoosable . It is easy to check that every Gallai tree is not degree-choosable. So the set of all connected graphs that are not degree-choosable are precisely the Gallai trees. Hladký, Krá ' l, and Schauz [7] extended this characterization to the setting of Alon-Tarsi orientations.
For any digraph D, a spanning Eulerian subgraph is one in which each vertex has indegree equal to outdegree. The parity of a spanning Eulerian subgraph is the parity of its number of edges. For an orientation of a graph G, let EE (resp. EO) denote the number of even (resp. odd) spanning Eulerian subgraphs. An orientation is Alon-Tarsi if G has a proper coloring ϕ from any list assignment L such that |L(v)| = f (v) for all v ∈ V (G). Alon and Tarsi [1] used algebraic methods to prove the following theorem for choosability. Later, Schauz [13] strengthened the result to paintability, which we discuss briefly in Section 5.
Theorem B. For a graph G and f : V (G) → N, if G is f -AT, then G is also f -choosable.
In this paper we characterize those graphs G with a specified vertex x that are not f -AT, where f (x) = d(x) − 1 and f (v) = d(v) for all other v ∈ V (G). All such graphs are formed from a few 2-connected building blocks, by repeatedly applying a small number of operations. Most of the work in the proof is spent on the case when G is 2-connected. This result is easy to state, so we include it a bit later in the introduction, as our Main Lemma. Near the end of Section 4, with a little more work we extend our Main Lemma, by removing the hypothesis of 2-connectedness, to characterize all pairs (G, h x ) that are not AT. This result is Theorem 4.5.
This line of research began with Gallai, who studied the minimum number of edges in an n-vertex k-critical graph G. Since G has minimum degree at least k − 1, clearly |E(G)| ≥ k−1 2 n. Gallai [6] improved this bound by classifying all connected subgraphs that can be induced by vertices of degree k − 1 in a k-critical graph. By Theorem A, all such graphs are Gallai trees. Here, we consider graphs G that are critical with respect to Alon-Tarsi orientation. Specifically, G is not (k − 1)-AT, but every proper subgraph is; such graphs are k-AT-critical. The characterization of degree-AT graphs shows that, much like k-critical graphs, in a k-AT-critical graph G, every connected subgraph induced by vertices of degree k − 1 must be a Gallai tree. Our main result characterizes the subgraphs that can be induced by vertices of degree k − 1, together with a single vertex of degree k. Thus, it is natural to expect that this result will lead to improved lower bounds on the number of edges in n-vertex k-AT-critical graphs.
Similar to that for degree-AT, our characterization remains unchanged in the contexts of list-coloring and paintability, as we show in Section 5. We see a sharp contrast when we consider graphs G with two specified vertices x 1 and x 2 that are not f -AT, where f (x i ) = d(x i )−1 for each i ∈ {1, 2} and f (v) = d(v) for all other v ∈ V (G). For Alon-Tarsi orientations, we have more than 50 exceptional graphs on seven vertices or fewer. Furthermore, the characterizations for list-coloring, paintability, and Alon-Tarsi orientations all differ.
We consider graphs with vertices labeled by natural numbers; that is, pairs (G, h) where G is a graph and h : V (G) → N. We focus on the case when h(x) = 1 for some x and h(v) = 0 for all other v; we denote this labeling as h x . hx We say that (G, h) is AT When H is an induced subgraph of G, we simplify notation by referring to the pair (H, h) when we really mean H, h↾ V (H) .
Given a pair (G, h) and a specified edge e ∈ E(G), when we stretch e stretch e , we form (G ′ , h ′ ) from (G, h) by subdividing e twice and setting h ′ (v i ) = 0 for each of the two new vertices, v 1 and v 2 (and h ′ (v) = h(v) for all other vertices v). In Section 2, we prove a Stretching Lemma, which shows that if (G, h) is not AT and e ∈ E(G), then stretching e often yields another pair (G ′ , h ′ ) that is also not AT. Thus, stretching plays a key role in our main result.
It is easy to check that the three pairs (G, h) shown in Figure 1 are not AT (and we do this below, in Proposition 1.1). Let D D be the collection of all pairs formed from the graphs in Figure 1 by stretching each bold edge 0 or more times. The Stretching Lemma implies that each pair in D is not AT. Our Main Lemma is that these are the only pairs (G, h x ), where G is 2-connected and neither complete nor an odd cycle, such that (G, h x ) is not AT, for some vertex x ∈ V (G).
The characterization of degree-choosable graphs has been applied to prove a variety of graph coloring results [2,4,10,11,14]. Likewise, we think our main results in this paper may be helpful in proving other results for Alon-Tarsi orientations, such as giving better lower bounds on the number of edges in k-AT-critical graphs. To conclude this section, we show that each pair in D is not AT.
Proof. For each pair (G, h x ) ∈ D, we construct a list assignment L such that |L(x)| = d(x)−1 and |L(v)| = d(v) for all other v ∈ V (G), but G has no proper coloring from L. Now (G, h x ) is not AT, by the contrapositive of Theorem B.
Let (G, h x ) be some stretching of the leftmost pair in Figure 1. Assign the list {1, 2, 3} to each of the vertices on the unbolded triangle and assign the list {1, 2} to each other vertex. If G has some coloring from these lists, then vertex x, labeled 1 in the figure, must get color 1 or 2; by symmetry, assume it is 1. Along each path from x to the triangle, colors must alternate 2, 1, . . .. Each of the paths from x to the triangle has odd length; thus, color 1 is forbidden from appearing on the triangle. So G has no coloring from L. Now let (G, h x ) be some stretching of the center pair in Figure 1. The proof is identical to the first case, except that each path has even length, so if x gets color 1, then color 2 is forbidden on the triangle.
Finally, consider the rightmost pair in Figure 1. Here d(x) = 4 and d(v) = 3 for all other v ∈ V (G). Thus, it suffices to show that G is not 3-colorable. Assume that G has a 3-coloring and, by symmetry, assume that x is colored 1. Now colors 2 and 3 must each appear on two neighbors of x. Thus, the two remaining vertices must be colored 1. Since they are adjacent, this is a contradiction, which proves that G is not 3-colorable.

Subgraphs, subdivisions, and cuts
When Hladký, Krá ' l, and Schauz characterized degree-AT graphs, their proof relied heavily on the observation that a connected graph G is degree-AT if and only if G has some induced subgraph H such that H is degree-AT. Below, we reprove this easy lemma, and also extend it to our setting of pairs (G, h x ).
Subgraph Lemma. Let G be a connected graph and let H be an induced subgraph of G. If H is degree-AT, then also G is degree-AT. Similarly, if x ∈ V (H) and Proof. Suppose that H is degree-AT, and let D ′ be an orientation of H showing this. Extend D ′ to an orientation D of G by orienting all edges away from H, breaking ties arbitrarily, but consistently. Now every directed cycle in D is also a directed cycle in D ′ (and vice versa), so G is degree-AT. The proof of the second statement is identical. The proof of the third statement is similar, but now if some edge xy has endpoints equidistant from H, then xy should be oriented into x.
Recall that, given a pair (G, h) and a specified edge e ∈ E(G), when we stretch e, we form (G ′ , h ′ ) from (G, h) by subdividing e twice and setting h ′ (v i ) = 0 for each of the two new vertices, v 1 and v 2 (and h ′ (v) = h(v) for all other vertices v). By repeatedly stretching edges, starting from the three pairs in Figure 1, we form all pairs (G, h x ), where G is 2-connected and (G, h x ) is not AT. The following lemma will be useful for proving this.
Proof. Suppose e = u 1 u 2 and call the new vertices v 1 and v 2 so that G ′ contains the induced path u 1 v 1 v 2 u 2 . For (1), let D be an orientation of G showing that (G, h) is AT. By symmetry we may assume u 1 u 2 ∈ E(D). Form an orientation D ′ of G ′ from D by replacing u 1 u 2 with the directed path u 1 v 1 v 2 u 2 . We have a natural parity preserving bijection between the spanning Eulerian subgraphs of D and D ′ , so we conclude that ( As above, we have a parity preserving bijection between the spanning Eulerian subgraphs of D and D ′ , so we conclude that (G, h) is AT. So suppose instead that G ′ contains neither of the directed paths u 1 v 1 v 2 u 2 and u 2 v 2 v 1 u 1 . Now no spanning Eulerian subgraph of D ′ contains a cycle passing through v 1 and v 2 . So, the spanning Eulerian subgraph counts of D ′ are the same as those of Given a pair (G, h) that is not AT, the Stretching Lemma suggests a way to construct a larger graph G ′ such that (G ′ , h ′ ) is not AT. In some cases, we can also use the Stretching Lemma to construct a smaller graph G such that ( G, h) is not AT. Specifically, we have the following.
Proof. The first statement is immediate from the Stretching Lemma. Now we prove the second. Suppose (G, h) satisfies the hypotheses. Applying part (2) of the Stretching Lemma shows that either By hypothesis, the former is false. Thus, the latter is true.
With standard vertex coloring, we can easily reduce to the case where G is 2-connected. If G is a connected graph with two blocks, B 1 and B 2 , meeting at a cutvertex x, then we can color each of B 1 and B 2 independently, and afterward we can permute colorings to match at x. For Alon-Tarsi orientations, the situation is not quite as simple. However, the following lemma plays a similar role for us.
Proof. We begin with the first statement. For each i ∈ {1, 2}, choose an orientation D i of Hence G is f -AT. Now we prove the second statement. Suppose that G is f -AT and choose an orientation Since the indegree of x in D is the sum of the indegree of x in D 1 and the indegree of x in D 2 , the lemma follows.

Degree-AT graphs and an Extension Lemma
Recall that our Main Lemma relies on a characterization of degree-AT graphs. As we mentioned in the introduction, a description of degree-choosable graphs was first given by Borodin [3] and Erdős, Rubin, and Taylor [5]. Hladký, Krá ' l, and Schauz [7] later extended the proof from [5] to Alon-Tarsi orientations. This proof relies on Rubin's Block lemma, which states that every 2-connected graph G contains an induced even cycle with at most one chord, unless G is complete or an odd cycle. For variety, and completeness, we include a new proof; it extends ideas of Kostochka, Stiebitz, and Wirth [9] from list-coloring to Alon-Tarsi orientations. For this proof we need the following very special case of a key lemma in [8]. When vertices x and y are adjacent, we write x ↔ y; otherwise x ↔ y.
Proof. Order the vertices of H with z 1 first and z 2 second so that every vertex, other than z 1 , has at least one neighbor preceding it. Orient each edge of H from its earlier endpoint toward its later endpoint. Orient xz 1 into z 1 and orient all other edges incident to x into x. Let D be the resulting orientation.
. So, we just need to check that EE(D) = EO(D).
Since xz 1 is the only edge of D leaving x, and D − x is acyclic, every spanning Eulerian subgraph of D that has edges must have edge xz 1 . Consider an Eulerian subgraph A of D containing xz 1 . Since z 1 has indegree 1 in A, it must also have outdegree 1 in A. We show that A has a mate A ′ of opposite parity. If z 2 ∈ A then z 1 z 2 w ∈ A, for some w, so we form A ′ from A by removing z 1 z 2 w and adding z 1 w. If instead z 1 z 2 / ∈ A, then z 2 ∈ A and z 1 w ∈ A for some w ∈ N H [z 1 ] − z 2 , so we form A ′ from A by removing z 1 w and adding z 1 z 2 w. Hence exactly half of the Eulerian subgraphs of D that contain edges are even. Since the edgeless spanning subgraph of D is an even Eulerian subgraph, we conclude that EE(D) = EO(D) + 1. Hence G is f -AT.
We use the previous lemma to give a new proof of the characterization of degree-AT graphs.

Lemma 3.2. A connected graph G is degree-AT if and only if it is not a Gallai tree.
Proof. We begin with the "only if" direction. Neither odd cycles nor complete graphs are degree-choosable. Thus, by Theorem B, they are not degree-AT. By induction on the number of blocks, Lemma 2.2 implies that no Gallai tree is degree-AT. Now, the "if" direction. Suppose there exists a connected graph that is not a Gallai tree, but is also not degree-AT. Let G be such a graph with as few vertices as possible. Since G is not degree-AT, no induced subgraph H of G is degree-AT by the Subgraph Lemma. Hence, for any v ∈ V (G) that is not a cutvertex, G − v must be a Gallai tree by minimality of |G|.
If G has more than one block, then for endblocks B 1 and B 2 , choose noncutvertices w ∈ B 1 and x ∈ B 2 . By the minimality of |G|, both G − w and G − x are Gallai trees. Since every block of G appears either as a block of G − w or as a block of G − x, every block of G is either complete or an odd cycle. Hence, G is a Gallai tree, a contradiction. So instead G has only one block, that is, G is 2-connected. Further, G − v is a Gallai tree for all v ∈ V (G).
Let v be a vertex of minimum degree in G. Since G is 2-connected, d G (v) ≥ 2 and v is adjacent to a noncutvertex in every endblock of G − v. If G − v has a complete block B with noncutvertices x 1 , x 2 where v ↔ x 1 and v ↔ x 2 , then we can apply Lemma 3.1 to conclude that G is degree-AT, a contradiction. So, v must be adjacent to every noncutvertex in every complete endblock of G − v.
Suppose d G (v) ≥ 3. Now no endblock of G − v can be an odd cycle of length at least 5 (G would have vertices of degree 3 and also d G (v) ≥ 4, contradicting the minimality of d G (v)). Let B be a smallest complete endblock of G − v. Now for a noncutvertex x ∈ V (B), we have d G (x) = |B| and hence d G (v) ≤ |B|. If G − v has at least two endblocks, then 2(|B| − 1) ≤ |B|, so d G (v) ≤ |B| = 2, a contradiction. Hence, G − v = B and v is joined to B, so G is complete, which is a contradiction.
Thus, we have d G (v) = 2. Suppose G − v has at least two endblocks. Now it has exactly two and v is adjacent to one noncutvertex in each. Neither of the endblocks can be odd cycles of length at least five, since then we can get a smaller counterexample by the Stretching Lemma. Since v is adjacent to every noncutvertex in every complete endblock of G −v, both endblocks must be K 2 . But now either G = C 4 (which is degree-AT, by orienting the cycle consistently) or we can get a smaller counterexample by the Stretching Lemma. So, G − v must be 2-connected. Since G − v is a Gallai tree, it is either complete or an odd cycle. If G−v is not complete, then we can get a smaller counterexample by the Stretching Lemma. So, G −v is complete and v is adjacent to every noncutvertex of G −v; that is, G is complete, a contradiction.

When h is 1 for one vertex
In this section, we prove our Main Lemma. For a graph G and x ∈ V (G) recall that h x : V (G) → N is defined as h x (x) = 1 and h x (v) = 0 for all v ∈ V (G − x). We classify the connected graphs G such that (G, h x ) is AT for some x ∈ V (G). We begin with the case when G is 2-connected, which takes most of the work. At the end of the section, we extend our characterization to all connected graphs.
We will show that for most 2-connected graphs G and vertices x ∈ V (G), the pair (G, h x ) is AT. Specifically, this is true for all pairs except those in D, defined in the introduction. In view of the Subgraph Lemma, for a 2-connected graph G and x ∈ V (G), to show that (G, h x ) is AT it suffices to find some induced subgraph H such that (H, h x ) is AT. The subgraphs H that we consider all have d H (x) = 0 or d H (x) ≥ 3. This motivates the next lemma, which allows us to reduce to the case d G (x) ≥ 3.  consists of two vertices joined by three internally disjoint paths, P 1 , P 2 , and P 3 . When we write h x for a θ-graph, we always assume that d(x) = 3. We will see shortly that if H is a θ-graph with d H (x) = 3, then (H, h x ) is AT. Thus, the Subgraph Lemma implies that if (G, h x ) is not AT, then G has no induced θ-graph H with d H (x) = 3. A T -graph T -graph is formed from vertices x, z 1 , z 2 , z 3 , by making the z i pairwise adjacent, and joining each vertex z i to x by a path P i (where the P i are disjoint). Equivalently, a T -graph is formed from K 4 by subdividing each of the edges incident to x zero or more times.
Similar to the proof characterizing degree-AT graphs in [7], our approach in proving our Main Lemma is to find an induced subgraph H such that (H, h x ) is AT, and apply the Subgraph Lemma. Thus, we need the following lemma about pairs (H, h x ) that are AT. Proof. In each case, we give an AT orientation D of G such that d − Case (i). Orient the edges of each path P i consistently, with P 1 and P 2 into x and P 3 out of x; this orientation satisfies the degree requirements. Further, it has exactly three spanning Eulerian subgraphs, including the empty subgraph. Thus, EE + EO is odd, so EE = EO.
Case (ii). Let P 1 and P 2 be two paths with opposite parities. As before, orient the edges of each path consistently, with P 1 and P 2 into x and P 3 out of x. Orient the three additional edges as − − → z 1 z 2 , − − → z 2 z 3 , and − − → z 3 z 1 . The resulting digraph D has four spanning Eulerian subgraphs, 3 of one parity and 1 of the other. Note that the empty subgraph and the Further, the parities are the same for the two subgraphs consisting of the directed cycles xP 3 z 3 z 1 P 1 and xP 3 z 3 z 1 z 2 P 2 . So, EE = EO.
Case (iii). The simplest instance of this case is when G = K 5 − e. Now (G, h x ) is AT by Lemma 3.1. In fact, that proof gives the stronger statement that there exists an orientation D satisfying the degree requirements such that EE(D) = EO(D)+1. In particular, EE+EO is odd. To handle larger instances of this case, we repeatedly subdivide edges incident to x and orient each of the resulting paths consistently, and in the direction of the corresponding edge in D. The resulting orientation satisfies the degree requirements. Further, the sum EE + EO remains unchanged, and thus odd. Hence, still EE = EO.  Proof. We can assume that G is not AT; otherwise, we are done by the Subgraph Lemma. By symmetry, assume P is a subpath of P 3 . First, we get an orientation of G with indegree at least 1 for all vertices and d − (x) = 2. Orient P 1 from z 1 to x, P 2 from z 2 to x, P 3 from x to z 3 , and the triangle as − − → z 1 z 2 , − − → z 2 z 3 , and − − → z 3 z 1 . To get an orientation of G ′ , orient the new path P ′ consistently, and opposite of P . Now the only directed cycle containing edges of P ′ is P ′ P . Since the Eulerian subgraph counts are equal for G, they differ by 1 for G ′ . Now we can prove Case (2) of our Main Lemma. For reference, we restate it.
Proof. When (G, h x ) ∈ D the lemma holds by Proposition 1.1. Now let G be 2-connected, choose x ∈ V (G) with d(x) ≥ 3, and suppose that (G, h x ) / ∈ D. Since G − x is connected, let H ′ be a smallest connected subgraph of G − x containing three neighbors of x; call these neighbors w 1 , w 2 , and w 3 . Consider a spanning tree T of H ′ . Since H ′ is minimum, each leaf of T is among {w 1 , w 2 , w 3 }. If T is a path, then H ′ is also a path. Otherwise, T is a subdivision of K 1,3 . Let s be the vertex with d T (s) = 3. If E(G) − E(T ) has any edge with both ends outside of N(s), then we can delete some vertex in N(s) and remain connected, contradicting the minimality of H ′ . Similarly, if N(s) contains at least two edges, then H ′ − s still connects, w, y, and z. Now let H be the subgraph of G induced by V (H ′ ) ∪ {x}. Note that H is either a θ-graph (if H ′ is a tree) or a T -graph (if H ′ has one extra edge in N(s)).
If H is a θ-graph, then (G, h x ) is AT, by Lemma 4.2.i and the Subgraph Lemma. So assume H is a T -graph. Let z 1 , z 2 , z 3 be the vertices of degree 3 (other than x), and let P 1 , P 2 , and P 3 denote the paths from x to z 1 , z 2 , and z 3 ; when we write V (P i ), we exclude x and z i , so possibly V (P i ) is empty for one or more i ∈ {1, 2, 3}. If any two of P 1 , P 2 , and P 3 have lengths with opposite parities, then we are done by Lemma 4.2.ii; so assume not. Now (H, h x ) ∈ D, so we can assume that V (G − H) = ∅. Choose u ∈ V (G − H), and let H u be a minimal 2-connected induced subgraph of G that contains V (H) ∪ {u}. By the Subgraph Lemma and Lemma 3.2, G − x is a Gallai tree. Thus, so is H u − x; in particular, the block B u of H u − x containing u is complete or an odd cycle. Therefore, we either have

.2.i and the Subgraph Lemma). So (i) happens for every vertex in
contain an induced copy of Figure 3(a); hence, (G, h x ) is AT by Lemma 4.2.iii and the Subgraph Lemma.
Assume instead that (ii) happens for every vertex in V (G−H), including u. By symmetry, So we conclude that u ↔ x, which requires j = ℓ − 1, by the minimality of H. Hence, H u is formed from a T -graph by adding a vertex u that is adjacent to x and also to the vertices of a K 2 endblock D u of H − x. Suppose there are distinct vertices u 1 , u 2 ∈ V (G − H) adjacent to vertices of the same K 2 endblock. Now G contains an induced copy of Figure 3(a), so (G, h x ) is AT by Lemma 4.2.iii and the Subgraph Lemma. Thus, each K 2 endblock has at most one such u.
Let t be the number of K 2 endblocks in H − x. By construction, t ≤ 3; this implies that |V (G − H)| ≤ t ≤ 3. If t = 0, then G = H = K 4 , which contradicts that G is not complete. If t = 1, then G = H u , for the unique u ∈ V (G − H); this is the Moser spindle, shown in Figure 1(c). So, assume that t ∈ {2, 3}. By symmetry, assume that for each i ∈ {1, 2} there exists u i such that V (B u i ) ⊆ P i ∪ {z i }. Now the subgraph induced by {u 2 } ∪ V (H − P 1 ) is reducible by Lemma 4.3. So, again we are done by the Subgraph Lemma.
Taken together, Lemmas 4.4 and 4.1, with Lemma 3.2, prove our Main Lemma. However, this characterizaton requires that G be 2-connected. Now we extend our result to the more general case, when G need only be connected. We use the following two definitions. Let G be a graph, x a vertex of G, and B a block of G. An x-lobe of G is a maximal subgraph A such that A − B is connected, and A includes a single vertex of B.  if G has a proper coloring from its lists L whenever L is such that |L(x)| = d(x) − 1 and |L(v)| = d(v) for all other v; otherwise (G, h x ) is not choosable. A pair being paintable paintable pair is defined analogously. We characterize all pairs (G, h x ), where G is connected and (G, h x ) is not choosable (resp. not paintable). In fact, we will see that these characterizations, for both choosability and paintability, are identical to that for pairs that are not AT.
For completeness, we include the following definition of f -paintable. Schauz [12] gave a more intuitive (yet equivalent) definition, in terms of a two player game. We say that G is Since all pairs (G, h x ) that are AT are also both paintable and choosable, it suffices to show that every pair (G, h x ) that is not AT is also not choosable (here we use that if a pair is paintable, then it is also choosable).
Theorem 5.1. For every connected graph G, the pair (G, h x ) is not choosable if and only if (G, h x ) is not AT. Thus, the same characterization holds for pairs that are not paintable.
Proof. As noted above, every pair that is AT is also choosable and paintable. Thus, it suffices to show that each pair (G, h x ) in Theorem 4.5 is not choosable.
To show that Gallai trees are not degree-choosable, assign to each block B a list of colors L B such that |L B | = d B (x) for each x ∈ V (B); further, for all distinct blocks B 1 and B 2 , we require that L B 1 and L B 2 are disjoint. For each v ∈ V (G), let L(v) = ∪ B i ∋v L B i . To show that G is not colorable from these lists, we use induction on the number of blocks. Let B be an endblock and x a cutvertex in B. Let G ′ = G \ (V (G) − x). Since B is complete or an odd cycle, B has no coloring from L B . Thus any coloring ϕ of G from L does not use L B on x. Hence, ϕ gives a coloring ϕ ′ of G ′ from its lists L ′ , where L ′ (x) = L(x) \ L B and L ′ (v) = L(v) for all v ∈ V (G) \ V (B). This coloring ϕ ′ of G ′ contradicts the induction hypothesis. Thus, G has no coloring from L.
Here we use a similar approach. Consider a pair (G, h x ) that satisfies one of Cases (1)-(5) in Theorem 4.5. We show that (G, h x ) is not choosable. Case (1) is immediate by the previous paragraph. Case (2) is immediate, since |L(x)| = 0. For Case (3), give lists to the Gallai tree of G − x as above; now let L(x) = {c} for some new color c, and add c to the list of each neighbor of x. Again G cannot be colored from L. For Case (4), assign lists to V (B) as in Proposition 1.1 and to the other blocks as above. Again, G has no coloring from these lists. Finally, consider Case (5). Assign lists for all blocks outside of A as above, and assign lists for A as above in Case (2), (3), or (4).
To conclude this section, we consider labelings h x,y , where h x,y (x) = h x,y (y) = 1 and h x,y (v) = 0 for all other v ∈ V (G). We show that the set of pairs (G, h x,y ) that are not AT differs from the set of that are not paintable. Further, both sets differ from the set of pairs that are not choosable. It suffices to give a pair (G 1 , h x,y ) that is choosable but not paintable and a second pair (G 2 , h x,y ) that is paintable but not AT.  Figure 4 is choosable, but not paintable. The pair (G 2 , h x,y ) on the right in Figure 4 is paintable, but not AT.
Proof. Let (G 1 , h x,y ) denote the pair on the left, where x and y are the vertices labeled 1. Let (G 2 , h x,y ) denote the pair on the right, where x and y are the vertices labeled 1. We first show that (G 1 , h x,y ) is choosable. Let L denote the list assignment. If there exists c ∈ L(x) ∩ L(y), then use c to color x and y, and color the remaining vertices greedily. So suppose there does not exist such a color c. Let z be a vertex in both triangles and note that there exist c ∈ (L(x) ∪ L(y)) \ L(z). By symmetry, assume that c ∈ L(x). Color x with c, and color G 1 − x greedily, starting with the vertex of degree 2 and ending with z.
We now show that (G 1 , h x,y ) is not paintable. Let S be the vertices of one triangle. By definition, there must be I ⊆ S such that G 1 − I is f ′ -paintable, where f ′ (v) := f (v) for v ∈ V (G 1 ) − S and f ′ (v) := f (v) − 1 for v ∈ S − I. I must have one vertex, w. There are two choices for w; either w is in two triangles or not. If w is not in two triangles, then G 1 − w is a triangle with a pendant edge, where the vertices on the triangle all have list size 2, so G 1 − w is not paintable. If w is one of the vertices in two triangles, then G 1 − w is a 4-cycle with list sizes alternating 1, 2, 1, 2. Again G 1 − I is not paintable (nor choosable).
To see that (G 2 , h x,y ) is not AT, note that any good orientation would need indegrees summing to at least 7, but G 2 has only 6 edges. Now we show that (G 2 , h x,y ) is paintable. Note that G 2 is isomorphic to K 2,3 , the complete biparite graph. Call the parts X and Y , with |X| = 2 and |Y | = 3. If S includes at least two vertices of X or at least two vertices of Y , take I to be an independent set of size at least 2. It is easy to check that G − I is paintable, since it induces either an independent set or a path, where each endvertex has more colors than neighbors. So assume that S contains at most one vertex from each of X and Y . If S contains a vertex of X, then color it. The resulting graph is paintable, since it is a claw, K 1,3 , with at most one leaf having a single color and all other vertices having two colors. Finally, suppose S contains only a single vertex of Y . Let I = S. The resulting graph is C 4 , which is degree-paintable (since it is degree-AT).
A graph is unstretched unstretched if it has no induced path u 1 v 1 v 2 u 2 where d(v 1 ) = d(v 2 ) = 2 (as in Corollary 2.1). We finish with the following question.
Question. Are there only finitely many unstretched, 2-connected graphs G such that (G, h x,y ) is not choosable (resp. paintable, AT)? More generally, let h x 1 ,...,x k be a labeling that assigns 1 to vertices x 1 , . . . , x k and 0 to all others. Are there only finitely many unstretched, 2connected graphs G such that (G, h x 1 ,...,x k ) is not choosable (resp. paintable, AT)?