On the Alon-Tarsi Number and Chromatic-choosability of Cartesian Products of Graphs

We study the list chromatic number of Cartesian products of graphs through the Alon-Tarsi number as defined by Jensen and Toft (1995) in their seminal book on graph coloring problems. The Alon-Tarsi number of $G$, $AT(G)$, is the smallest $k$ for which there is an orientation, $D$, of $G$ with max indegree $k\!-\!1$ such that the number of even and odd circulations contained in $D$ are different. It is known that $\chi(G) \leq \chi_\ell(G) \leq \chi_p(G) \leq AT(G)$, where $\chi(G)$ is the chromatic number, $\chi_\ell(G)$ is the list chromatic number, and $\chi_p(G)$ is the paint number of $G$. In this paper we find families of graphs $G$ and $H$ such that $\chi(G \square H) = AT(G \square H)$, reducing this sequence of inequalities to equality. We show that the Alon-Tarsi number of the Cartesian product of an odd cycle and a path is always equal to 3. This result is then extended to show that if $G$ is an odd cycle or a complete graph and $H$ is a graph on at least two vertices containing the Hamilton path $w_1, w_2, \ldots, w_n$ such that for each $i$, $w_i$ has a most $k$ neighbors among $w_1, w_2, \ldots, w_{i-1}$, then $AT(G \square H) \leq \Delta(G)+k$ where $\Delta(G)$ is the maximum degree of $G$. We discuss other extensions for $G \square H$, where $G$ is such that $V(G)$ can be partitioned into odd cycles and complete graphs, and $H$ is a graph containing a Hamiltonian path. We apply these bounds to get chromatic-choosable Cartesian products, in fact we show that these families of graphs have $\chi(G) = AT(G)$, improving previously known bounds.


Introduction
In this paper all graphs are finite, and all graphs are either simple graphs or simple directed graphs. Generally speaking we follow West [22] for terminology and notation. List coloring is a well known variation on the classic vertex coloring problem, and it was introduced independently by Vizing [21] and Erdős, Rubin, and Taylor [6] in the 1970's. In the classic vertex coloring problem we wish to color the vertices of a graph G with as few colors as possible so that adjacent vertices receive different colors, a so-called proper coloring. The chromatic number of a graph, denoted χ(G), is the smallest k such that G has a proper coloring that There are few results in the literature regarding the list chromatic number of the Cartesian product of graphs. In 2006, Borowiecki, Jendrol, Král, and Miškuf [4] showed the following.
Here col(G), the coloring number of a graph G, is the smallest integer d for which there exists an ordering, v 1 , v 2 , . . . , v n , of the elements in V (G) such that each vertex v i has at most d − 1 neighbors among v 1 , v 2 , . . . , v i−1 . The coloring number is a classic greedy upper bound on the list chromatic number, and it immediately implies that ∆(G) + 1 is an upper bound on the list chromatic number where ∆(G) is the maximum degree of G. Vizing [21] extended this by proving the list coloring version of Brooks' Theorem.
Theorem 2 ( [21]). Suppose that G is a connected graph with maximum degree ∆(G). If G is neither a complete graph nor an odd cycle, then χ ℓ (G) ≤ ∆(G).
Borowiecki et al. [4] construct examples where the upper bound in their theorem is tight. Specifically, they show that if k ∈ N and G is a copy of the complete bipartite graph K k,(2k) k(k+k k ) , then χ ℓ (G G) = χ ℓ (G) + col(G) − 1. On the other hand, there are examples where the upper bounds from Theorems 1 and 2 are not tight. For example, suppose that G is a copy of C 2k+1 P n where n ≥ 3. Since χ ℓ (C 2k+1 ) = col(C 2k+1 ) = 3 and χ ℓ (P n ) = col(P n ) = 2, Theorem 1 tells us that χ ℓ (G) ≤ 4. Similarly, Theorem 2 tells us χ ℓ (G) ≤ 4, yet we will show below that χ ℓ (G) = 3. Another, more dramatic, example where Theorems 1 and 2 do not produce tight bounds is when we are working with the Cartesian product of two complete graphs. Suppose m ≥ n ≥ 2, and note that K m K n is the line graph for the complete bipartite graph K m,n . So, by Galvin's celebrated result ( [7]): m = χ(K m K n ) = χ ℓ (K m K n ). However, Theorem 1 only yields an upper bound of χ ℓ (K m K n ) ≤ m + n − 1 since χ ℓ (K m ) = col(K m ) = m. Similarly, Theorem 2 only tells us Alon [1] showed that for any graph G, col(G) ≤ 2 O(χ ℓ (G)) . Combining this result with Theorem 1 implies that we have an upper bound on χ ℓ (G H) in terms of only the list chromatic numbers of the factors. Borowiecki et al. [4] conjecture that a much stronger bound holds: there is a constant A such that χ ℓ (G H) ≤ A(χ ℓ (G) + χ ℓ (H)). While we will not address this conjecture in this paper, we will present results that are improvements on Theorem 1 when the factors in the Cartesian product satisfy certain properties. Our main aim in this note is to illustrate how to utilize the classic Alon-Tarsi Theorem in these situations to get better bounds on the AT-number and consequently the list chromatic number.

Alon-Tarsi Number
Suppose graph D is a simple digraph. We say that E is a circulation contained in D if E is a spanning subgraph of D and for each We say a circulation is even (resp. odd) if it has an even (resp. odd) number of edges. Classic results in graph theory tell us that a circulation is a digraph which consists of Eulerian components. This means a circulation can be decomposed into directed cycles. Alon and Tarsi [2] use algebraic methods, subsequently called the Combinatorial Nullstellensatz, to obtain a remarkable relationship between a special orientation of a graph and a certain graph polynomial. The result is the celebrated Alon-Tarsi Theorem.
Theorem 3 (Alon-Tarsi Theorem). Let D be an orientation of the simple graph G. Suppose that L is a list assignment for G such that |L(v)| ≥ d + D (v) + 1. If the number of even and odd circulations contained in D differ, then there is a proper L-coloring for G. In addition, if the maximum indegree of D is m and the number of even and odd circulations contained in D differ, then χ ℓ (G) ≤ m + 1.
Recently, further implications of the Alon-Tarsi Theorem have appeared in the literature. Before mentioning one of these implications we need some terminology. For a graph G suppose that for each v ∈ V (G), k tokens are available at v. Two players called the marker and remover then play the following game on the graph G. For each round, the marker marks a non-empty subset, M , of vertices on the graph which uses one token for each marked vertex. The remover then selects a subset of vertices, I ⊆ M , to remove such that I is an independent set of vertices in G. The marker wins by marking a vertex that has no tokens, and the remover wins by removing all the vertices from the graph. A graph is said to be k-paintable if the remover has a winning strategy when k tokens are available at each vertex. The paint number or online choice number of G, χ p (G), is the smallest k such that G is k-paintable. We have that χ ℓ (G) ≤ χ p (G), and there exist graphs where χ ℓ (G) < χ p (G) (see [5] and [18]).
Schauz [19] showed that if D is an orientation of G with maximum indegree m such that D satisfies the hypotheses of the Alon-Tarsi Theorem, then χ p (G) ≤ m + 1. So, whenever we use the Alon-Tarsi theorem to bound the list chromatic number of a graph, we actually get the same bound on the paint number of the graph which is a stronger result.
The Alon-Tarsi number of G, AT (G), is the smallest k for which there is an orientation, D, of G with max indegree k − 1 such that the number of even and odd circulations contained in D are different (i.e. the smallest k for which the hypotheses of the Alon-Tarsi Theorem are satisfied). Jensen and Toft [12] first defined and suggested this graph invariant for study. Hefetz [11] studied the AT-number and showed, among other results, that AT (H) ≤ AT (G) whenever H is a subgraph of G. Recently, Zhu [23] showed that the AT-number of all planar graphs is at most 5, improving the classic list coloring bound on planar graphs.
In summary we know that: In general all these inequalities can be strict, our aim is to find classes of graphs where they all are equal -a stronger form of chromatic-choosability. We study this phenomenon for Cartesian products of graphs.

Applying the Alon-Tarsi Theorem
In this section we will start by proving that the Cartesian product of an odd cycle and path is chromatic-choosable. Then, we prove several generalizations of this argument. Suppose that G is the Cartesian product of an arbitrary odd cycle on 2k + 1 vertices and an arbitrary path on n vertices. Throughout the proof of Theorem 4 we assume that we form G as follows. Suppose we place the vertices of C 2k+1 around a circle so that two vertices are adjacent if and only if they appear consecutively along the circle, and we name the vertices in counterclockwise fashion around the circle as: v 1 , v 2 , . . . , v 2k+1 (we also call this ordering of the vertices cyclic order ). Similarly we name the vertices of P n as: w 1 , w 2 , . . . , w n so that two vertices are adjacent if and only if they appear consecutively in this list (we say that this ordering of the vertices is in order ).
Proof. The result is obvious when n = 1. So, we assume that n ≥ 2. Now, we orient the edges of G = C 2k+1 P n as follows. For each of the n copies of C 2k+1 in G we orient the edges of each copy in counterclockwise fashion. Also we orient the edges of the form Now, let G * be the graph obtained from G by adding an extra edge in G with endpoints (v 1 , w 1 ) and (v 2 , w n ). Since G is a subgraph of G * , we have that AT (G) ≤ AT (G * ). We will now show that AT (G * ) ≤ 3. To do this we form digraph D * by orienting the edges of G * so that all the edges of G * that are in G are given the same orientation as in D and the edge {(v 1 , w 1 ), (v 2 , w n )} is oriented so that (v 1 , w 1 ) is the head and (v 2 , w n ) is the tail. We call this oriented edge e * . Note that for each ( To show the number of even and odd circulations contained in D * differ, we will prove that the number of circulations contained in D * is odd. We will refer to the n oriented copies of C 2k+1 in D * as the base cycles. We name the base cycles: B 1 , . . . , B n so that B i is the oriented copy of C 2k+1 such that all the vertices in B i have second coordinate w i . We note that the only directed cycles contained in D * are the base cycles and cycles that contain the edge e * . Let C be the set of all circulations in D * . We let H contains all the edges of at least one of the base cycles or H does not include any edge from at least one of the base cycles . Suppose H ∈ A. We may assume H contains all the edges from the base cycles: B a 1 , . . . , B am and does not include any edges from the base cycles B b 1 , . . . , B bt (Note: at most one of these two lists of base cycles may be empty). Now, we form another circulation H ′ ∈ A from H as follows. We delete all the edges of the base cycles: B a 1 , . . . , B am and we add all the edges of the base cycles: B b 1 , . . . , B bt . We immediately note that H = H ′ and this mapping gives us a way to pair up distinct elements of A. Thus, |A| is an even number. Now, let B = C − A. By the definition of C and A we have that B contains all the circulations of D * that contain at least one edge from each base cycle, but do not include all the edges of any base cycle. Suppose that H ∈ B. We know that H may be decomposed into directed cycles. No directed cycle in the decomposition of H can be a base cycle. Since the only directed cycles in D * are the base cycles and cycles containing e * , we may conclude that H is a single cycle containing e * that contains at least one edge from each base cycle. This means that we can view each element of B as a directed cycle starting with vertex (v 1 , w 1 ) and ending with the edge e * . Now, for q, r ∈ Z, 0 ≤ q ≤ n − 2, and 0 ≤ r ≤ 2k, let e (2k+1)q+r be the directed edge with tail (v r+1 , w q+1 ) and head (v r+1 , w q+2 ). For 0 ≤ i ≤ n − 2, we let: Intuitively, E i consists of the directed edges from the copies of P n that connect vertices in i=0 is a subsequence of the finite sequence of edges: . We call (e a i ) n−2 i=0 a level subsequence of edges if e a i ∈ E i for each i, a 1 = 0, and a i ≡ a i+1 mod 2k + 1 for each i ≤ n − 3. Now, let Q be the set of all level subsequences of edges, and let R be the set of all level subsequences of edges that do not have the edge e (2k+1)(n−2)+1 as the last edge in the sequence.
We will now construct a bijection between B and R. Given a sequence of edges in R, (e a i ) n−2 i=0 , there is a unique cycle in D * starting at (v 1 , w 1 ) and ending with the edge e * that includes each edge in (e a i ) n−2 i=0 . To form this cycle, simply follow a portion of the base cycle B 1 to get from (v 1 , w 1 ) to e a 0 . The fact that a 1 = 0 guarantees that we must traverse at least one edge of B 1 . Then, for each 0 ≤ i ≤ n − 3, follow a portion of the base cycle B i+2 to get from edge e a i to e a i+1 . The fact that a i ≡ a i+1 mod 2k + 1 for each i ≤ n − 3 guarantees that we must traverse at least one edge of B i+2 . Finally, follow a portion of the base cycle B n to get from e a n−2 to e * . The fact that R contains all level subsequences that do not end with the edge incident to e * guarantees we must traverse at least one edge of B n . We immediately notice from this construction that the cycle we form contains at least one edge from each base cycle, but does not include all the edges of any base cycle. Thus, the cycle we form is in B. So, we have a function from R to B. To see that this function has an inverse suppose that H ∈ B. We know that we can view H as a directed cycle starting with (v 1 , w 1 ) and ending with the edge e * . By the way in which D * is oriented we know that H must alternate between edge(s) from B i and an edge from E i−1 for 1 ≤ i ≤ n − 1. Then, cycle H has edge(s) from B n followed by e * . Let (u k ) n−2 k=0 be the ordered sequence of edges in H that are in ∪ n−2 i=0 E i . We immediately have that u k ∈ E k for 0 ≤ k ≤ n − 2 by the way the paths are oriented. Since H contains at least one edge from each base cycle, u 0 = e 0 , and if u k = e a ′ and u k+1 = e b ′ , then a ′ ≡ b ′ mod 2k + 1 for 0 ≤ k ≤ n − 3. So, (u k ) n−2 k=0 is a level subsequence of edges by definition. Finally, since H contains at least one edge from B n , we know that u n−2 = e (2k+1)(n−2)+1 . So, (u k ) n−2 k=0 ∈ R. We now have that our function maps (u k ) n−2 k=0 to H and the inverse maps H to (u k ) n−2 k=0 . Thus, we have a bijection between B and R and we conclude that |B| = |R|. Now, we will show that |R| is an odd number. We note that |Q| = (2k) n−1 since forming an element of Q leaves one with 2k distinct choices at each step. We let d r,n be the number of elements in Q that have a final edge with an index congruent to r mod 2k + 1 (Note: n is the number of vertices in our path and 0 ≤ r ≤ 2k). The following recursive relationships are immediate: d r,n = i∈{0,...,2k},i =r d i,n−1 for n ≥ 3 where d 0,2 = 0 and d i,2 = 1 for each i ∈ {1, . . . 2k}. An easy inductive argument shows that d 0,n is even and d i,n is odd for each i ∈ {1, . . . 2k} whenever n ≥ 2. Since we may conclude that |R| is an odd number which immediately implies |B| is an odd number.
Since |C| = |A| + |B|, we have that |C| is odd. Thus, D * contains an odd number of circulations. This means that the number of even circulations in D * does not equal the number of odd circulations in D * . So AT (G * ) ≤ 3 and hence AT (G) ≤ 3. We also have that 3 ≤ χ(G). So, AT (G) = 3, and we are done.
Remark. It is easy to notice that in our proof of Theorem 4 we have proven something slightly stronger with regard to list coloring. Specifically we proved that if we have a list assignment, L, for G that assigns two colors to all but one vertex in the first base cycle and three colors to all other vertices of G, then there is a proper L-coloring for G.
Having proven Theorem 4, it is easy to now classify the list chromatic number of the Cartesian product of an arbitrary cycle and path. Erdős et al. [6] classified all graphs with list chromatic number equal to 2. Let Θ(l 1 , . . . , l k ) with branch vertices u and v be the graph that is the union of k pairwise internally disjoint u, v-paths of lengths l 1 , . . . , l k .
Proof. Statement (i) follows from Theorem 4. For statement (ii) notice that when n ≥ 2, C 2k+2 P n contains more than one cycle and no cut edges. Moreover, all the vertices in C 2k+2 P n have degree at least 3 which means that C 2k+2 P n is not Θ(2, 2, 2t) for any t ∈ N. So, Theorem 5 implies that 3 ≤ χ ℓ (C 2k+2 P n ). By Theorem 5, we also have that even cycles have list chromatic number equal to 2. So, Theorem 1 implies χ ℓ (C 2k+2 P n ) ≤ 2 + 2 − 1 = 3.
We now give a natural generalization of the argument of Theorem 4.
Theorem 7. Suppose that G is a complete graph or an odd cycle with |V (G)| ≥ 3. Suppose H is a graph on at least two vertices that contains a Hamilton path, w 1 , w 2 , . . . , w m , such that w i has at most k neighbors among w 1 , . . . , w i−1 . Then, AT (G H) ≤ ∆(G) + k.
Before we prove this theorem a couple of remarks are worth making. We require m ≥ 2, since we know χ ℓ (G H) = ∆(G) + 1 when m = 1. Now, suppose G and H satisfy the hypotheses of the Theorem. We have that χ ℓ (G) = col(G) = ∆(G) + 1 and col(H) − 1 ≤ k. Proof. Suppose G has n vertices. We name the vertices of G (cyclically if G is a cycle): v 1 , v 2 , . . . , v n . G contains an odd cycle, C, as an induced subgraph. If G is an odd cycle let C = G, and if G is a complete graph let C be the subgraph of G induced by the vertices v n−2 , v n−1 , and v n . For the remainder of the proof assume that C has 2k + 1 vertices where k ∈ N (Note: we know k = 1 when G is a complete graph). We also let P be the Hamilton path, w 1 , w 2 , . . . , w m , contained in H. Now, consider the graph G H. We form digraph D from this graph by orienting its edges as follows. We begin by orienting each of the m copies of C in a counterclockwise fashion. Then, for each edge in a copy of G and not in a copy of C with endpoints (v r , w u ) and (v s , w u ) with s > r, we orient the edge so that (v r , w u ) is the tail and (v s , w u ) is the head. Finally, for each edge in a copy of H with endpoints (v u , w r ) and (v u , w s ) with s > r, we orient the edge so that (v u , w r ) is the tail and (v u , w s ) is the head.
From D we form the digraph D * by adding a directed edge with tail (v n−2k+1 , w m ) and head (v n−2k , w 1 ). Note that D * is a simple digraph since m ≥ 2. We will refer to the edge we added as e * . We immediately note that by the conditions placed on G and H, we have that . Similar to the proof of Theorem 4, we will now prove that the number of circulations in D * is odd. We will refer to the m oriented copies of C in D * as the base cycles. We name the base cycles: B 1 , . . . , B m so that B i is the oriented copy of C such that all the vertices in B i have second coordinate w i . We note that the only cycles in D * are the base cycles and the cycles that contain e * . Let S be the subgraph of D * that is made up of the m oriented copies of C in D * , the oriented copies of P that have first coordinate v n−2k , v n−2k+1 , . . . , and v n , and the edge e * . Notice that S is C P plus edge e * oriented as it is in the proof of Theorem 4. Let C be the set of all circulations in D * . We define A as in the proof of Theorem 4. By the same argument as in the proof of Theorem 4, we see that |A| is even. Now, let B = C − A. As in the proof of Theorem 4, we have that any K ∈ B is a single cycle containing e * that contains at least one edge from each base cycle. Moreover, any K ∈ B must be completely contained in S (since any vertex of the form (v ℓ , w u ) with l < n − 2k is in no cycles of D * and any cycle in D * that contains an oriented edge of a copy of H that is not in a copy of P must leave out at least one base cycle).
Thus, by the same argument as in the proof of Theorem 4, we have that |B| is odd. Thus, |C| is odd, and we have the desired result by the Alon-Tarsi Theorem.
Remark. Note that our proof also works when G is any graph that contains an induced odd cycle C such that every maximum degree vertex in G is either in C or adjacent to C. However, this is not particularly useful for bounding χ ℓ (G H), since when G is not an odd cycle or complete graph Theorems 2 and 1 yield: Now, we present another extension of Theorem 4. We first need some definitions and notation. Suppose that G is an arbitrary graph, and T is some subset of V (G). We write G[T ] for the subgraph of G induced by the vertices in T . Now, suppose that G 1 and G 2 are two arbitrary vertex disjoint graphs. The join of the graphs G 1 and G 2 , denoted G 1 ∨ G 2 , is the graph consisting of G 1 , G 2 , and additional edges added so that each vertex in G 1 is adjacent to each vertex in G 2 . When a graph, G, consists of G 1 , G 2 , and some set of additional edges (possibly empty) that have one endpoint in V (G 1 ) and the other endpoint in V (G 2 ) we say that G is a partial join of G 1 and G 2 .
Theorem 8. Suppose that G is a graph with the property that there exists a partition, {S 1 , . . . , S m }, of V (G) such that for each i G[S i ] is an odd cycle. For each i ≥ 2 suppose each vertex in S i has at most ρ neighbors in ∪ i−1 j=1 S j (we let ρ = 0 in the case that m = 1). Then, AT (G P k ) ≤ 3 + ρ, for any k ∈ N.
Note that we obtain Theorem 4 when m = 1. We also notice that when m = 2, G is a partial join of two odd cycles. When m = 3, G is a partial join of the odd cycle G[S 3 ] and G[S 1 ∪ S 2 ] (where G[S 1 ∪ S 2 ] is itself the partial join of two odd cycles). And so on.
Proof. First, for each i, we index the vertices of G[S i ] in cyclic order as: v i,1 , v i,2 , . . . , v i,m i where we know that m i is odd for each i. If e ∈ E(G) is an edge with one endpoint in S i and one endpoint in S j with i = j we refer to e as a connecting edge.
We now turn our attention to the case where k = 1. In this case, we must show that AT (G) ≤ 3 + ρ. We form an oriented version, D, of the graph G as follows. For each odd cycle, G[S i ], orient the edges of the cycle so that v i,l is the tail and v i,l+1 is the head for 1 ≤ l ≤ m i − 1. Then, orient the final edge of the cycle so that v i,1 is the tail and v i,m i is the head. Finally, for each connecting edge, e ∈ E(G), with one endpoint in S i and one endpoint in S j with i < j orient e so that its tail is in S i and its head is in S j . One may now note that for each v ∈ S i , d − D (v) ≤ 2 + ρ. We now claim that D is acyclic. First, note that no oriented connecting edge can be in a cycle in D since there is no way in D to get from a vertex in S j to a vertex in S i when i < j. This means that if there is a cycle, C, in D there must exist an i such that the vertices of C are a subset of the vertices of S i . However, D[S i ] is acyclic by construction. So, no such C can exist. This means that D has one even circulation (the circulation with no edges) and zero odd circulations. Then AT (G) ≤ 3 + ρ as desired.
We now turn our attention to the case where k ≥ 2. Let H = P k . We note that G H is made of the disjoint union of graphs: plus the connecting edges in each of the k copies of G. Let G * * be the graph formed from G H by adding an edge in G[S i ] H for each i as we did in the proof of Theorem 4 to form G * . We call the newly formed subgraph of G * * consisting of G[S i ] H plus an additional edge, M i for each i. We form an oriented version, D, of G * * as follows. For each subgraph M i of G * * , we orient this subgraph just as we oriented D * in the proof of Theorem 4. Finally, we orient all the connecting edges in all the copies of G just as we did in the case of k = 1.
One may note that for each v ∈ V (D), we have that d − D (v) ≤ 2 + ρ. Similar to the case where k = 1, we note that no connecting edge from any copy of G is in a cycle contained in D. This means that if C i is the number of circulations in the oriented version of M i , the number of circulations in D is equal to Finally, we can easily combine the idea of the proofs of Theorems 7 and 8 to obtain the following.
Theorem 9. Suppose that G is a graph with the property that there exists a partition, {S 1 , . . . , S m }, of V (G) such that for each i |S i | ≥ 3 and G[S i ] is an odd cycle or a complete graph. For each i ≥ 2 suppose each vertex in S i has at most ρ i neighbors in ∪ i−1 j=1 S j , and let ρ 1 = 0. For each i ≥ 1 we let: Suppose H is a graph on at least two vertices that contains a Hamilton path, w 1 , w 2 , . . . , w n , such that w i has at most k neighbors among w 1 , . . . , w i−1 . Then, AT (G) ≤ α and AT (G H) ≤ α + k − 1.
Both Theorems 8 and 9 are sharp and give improvements over existing bounds as shown in examples in the next section.

Some Examples
In this section we present some examples where our results from Section 2 improve upon known bounds for the list chromatic number. We let the kth power of graph G, denoted G k , be the graph with vertex set V (G) where two vertices are adjacent if their distance in G is at most k. It is easy to see that P r n , 1 ≤ r ≤ n − 1, satisfies χ(P r n ) = χ ℓ (P r n ) = col(P r n ) = r + 1. We already know Theorem 7 is sharp by r = 1 below.
Proof. Let G = C 2k+1 and H = P r n . Name the vertices of H: w 1 , w 2 , . . . , w n so that w i is the i th vertex of the underlying path on n vertices contained in H. Clearly, H contains the Hamilton path: w 1 , w 2 , . . . , w n , and for each i, w i has at most r neighbors among w 1 , w 2 , . . . , w i−1 . Thus, by Theorem 7, we have that AT (G H) ≤ 2 + r.
We note that Theorem 1 only yields that χ ℓ (G H) ≤ r + 3. Also, Theorem 2 yields χ ℓ (G H) ≤ 2 + ∆(P r n ), and r + 1 ≤ ∆(P r n ) when n ≥ 3 and r ≤ n − 2. So, the above example improves upon known bounds on the list chromatic number when n ≥ 3 and r ≤ n − 2. We suspect that C 2k+1 P r n is often chromatic-choosable, but improving upon our upper bound with the Alon-Tarsi Theorem seems difficult. For example, it would be impossible to find an orientation of C 2k+1 P 2 n with max indegree of 2 for large values of n. Before we move on to an example for Theorems 8 and 9, let us note a couple of other facts implied by Theorem 7: AT (K n P m ) = n, for n ≥ 3; and n − 2 ≤ AT (C 2k+1 H n ) ≤ n − 1 where H n = K n − E(P 4 ) for n ≥ 5.
First, consider the case where m ≥ 3. Let {S 1 , S 2 } be the partition of V (G) where S 1 = V (G 1 ) and S 2 = V (G 2 ). Note G[S 1 ] is a complete graph, and G[S 2 ] is an odd cycle. Also, each vertex in S 2 has exactly m neighbors in S 1 . So, by Theorem 9 we have that AT (G P n ) ≤ m + 3.
Finally, suppose m = 1, 2. Let G (m) be a partial join of G 3 = C 3 and G 2 so that m vertices in V (G 3 ) are adjacent to all the vertices in V (G 2 ), and the other 3 − m vertices in V (G 3 ) are not adjacent to any of the vertices in V (G 2 ). Now, let {S 3 , S 4 } be the partition of V (G (m) ) where S 3 = V (G 3 ) and S 4 = V (G 2 ). Note G (m) [S 3 ] and G (m) [S 4 ] are odd cycles, and each vertex in S 4 has exactly m neighbors in S 3 . So, by Theorem 8 we have that AT (G (m) P n ) ≤ m + 3. The result follows since G P n is a subgraph of G (m) P n .
There is still much to be discovered about the list chromatic number of the Cartesian product of graphs. Aside from the important conjectures proposed in [4], this paper also provides us with some interesting questions. Specifically, an ambitious question would be: Can we determine when G H will be chromatic-choosable based upon some property of the factors? We further explore this question in a following paper [14]. Simpler questions, motivated by Corollaries 6 and 10, include: For what graphs G, is G P n chromatic-choosable? When is C 2k+1 P r n chromatic-choosable? For the first of these questions one may conjecture based upon the results of this paper that if G is chromatic-choosable and χ(G) ≥ 3, then G P n chromatic-choosable. However, this conjecture is false since one can construct a 3assignment to show that 3 < χ ℓ (C 2 6 P 2 ).