Scheduling N Burgers for a k-Burger Grill : Chromatic Numbers With Restrictions

The chromatic number has a well-known interpretation in the area of scheduling. If the vertices of a finite, simple graph are committees, and adjacency of two committees indicates that they must never be in session simultaneously, then the chromatic number of the graph is the smallest number of hours during which the committees/vertices of the graph may all have properly scheduled meetings of one continuous hour each. Slivnik [3] showed that the fractional chromatic number can be similarly characterized. In that characterization, the meetings are allowed to be broken into a finite number of disjoint intervals. Here we consider chromatic numbers definable similarly, with the additional restriction that there are only k meeting rooms available; that is, at each instant no more than k committees can be in session.

The seminal problem from which this inquiry grew was cooked up over a dozen years ago by the second author: assuming that each hamburger patty must be grilled for one unit of time on each side, what is the minimum number of time units needed to grill 3 burgers on a 2-burger grill (a grill on which only two burgers will fit at one time)?It's not hard to find a way of completing the task in 3 time units, during which each patty side gets exactly one time unit of grilling, and the grill is in full use (two burgers on) at all times-assuming flipping and moving patties require no time at all.It is intuitively clear that 3 time units is the minimum called for, but a satisfactory proof requires more than intuition.
The real question behind this problem is not about grilling burgers, important as that is, but rather: what is an interesting general problem of which these burger-grilling problems are special cases?It is the question that we can now answer.
All graphs here are finite and simple.A scheduling function on a graph G is a function ϕ on V (G) such that for all u, v ∈ V (G), (i) ϕ(v) is a finite union of open intervals on the real line; (ii) if uv ∈ E(G) then ϕ(u) ∩ ϕ(v) = ∅; and (iii) the measure of ϕ(v), i.e., the sum of lengths of its maximal subintervals, is 1.
The idea here has to do with an issue briefly discussed in the Abstract.If the vertices of G are committees such that uv ∈ E(G) implies that committees u and v must never be in session simultaneously, then a scheduling function gives a proper coloring of G with meeting times for these committees, such that each committee gets a total of one unit of meeting time.Some may find it disturbing that these meetings can be broken into intervals separated from each other.Clearly it is impractical to allot a committee 45 minutes on one fine morning and then 13 seconds some time later.We will have some consolation on this issue later in this paper, but we are leaving most of that territory unexplored.(0, 1) (0, 1) (1,2) (1, 2) (2,3) (2, 3) Figure 1: Scheduling 3 burgers for a 2-burger grill Suppose that k is a positive integer.A scheduling function ϕ on G obeys restriction k if and only if, for each real number t, |{v ∈ V (G) | t ∈ ϕ(v)}| ≤ k.Note that because of requirement (ii) in the definition of scheduling function, if ϕ is such a function on G and t ∈ R, then {v ∈ V (G) | t ∈ ϕ(v)} is an independent set of vertices in G. Therefore, if k ≥ α(G) = vertex independence number of G, then restriction k is no restriction at all; it is obeyed by every scheduling function on G.
In actual scheduling of committee meetings, obedience to restriction k means that no more than k meetings may every be in session simultaneously.Clearly such a constraint may arise naturally; there may be only k meeting rooms available.In the problem of 3 burgers on a 2-burger grill, the graph involved is 3K 2 , the disjoint union of 3 edges.The 6 vertices are the 6 sides of the 3 hamburgers, with two vertices adjacent if and only if they are the two sides of the same burger patty.The restriction that our scheduling functions on this graph must obey is k = 2, arising from the size of the grill.
The fractional chromatic number, χ f , has many equivalent definitions, most of which can be found in [2]; but here is one, due to T. Slivnik [3], which does not appear there: χ f (G) = inf[T > 0; there is a scheduling function ϕ on G such that for each v ∈ V (G), ϕ(v) ⊆ (0, T )].(It is not immediately obvious, but the inf is min for every G.) For a graph G and a positive integer k, we define the fractional chromatic number of G with restriction k to be χ f (G; k) = inf[T > 0; there is a scheduling function ϕ on G which obeys restriction k such that for each v ∈ V (G), ϕ(v) ⊆ (0, T )] (We shall see later that the inf in this definition is a min, which, in view of the obvious fact that χ f (G; k) = χ f (G) for all k ≥ α(G), will establish the same for Slivnik's characterization of χ f (G).) Two other obvious observations regarding the restricted fractional chromatic number: χ f (G; 1) = |V (G)| for all graphs G, and for all graphs G and positive integers k.
The 3-burgers-on-a-two-burger-grill problem is to determine χ f (3K 2 ; 2).As the reader has probably already worked out, we can see that χ f (3K 2 ; 2) ≤ 3 by the scheduling function indicated in Figure 1, which obeys restriction 2 and finishes the burgers in 3 time units: But how to give a clean proof of the obvious, that χ f (3K 2 ; 2) ≥ 3? For a graph G and a positive integer k, we define α(G; k) = min(k, α(G)).
Proposition 1.1.Suppose that G is a graph and k is a positive integer.Then Because ϕ is a scheduling function on G obeying restriction k, with ϕ(u) ⊆ (0, T ) for all u ∈ V (G), we have Therefore χ f (3K 2 ; 2) ≥ 6 2 = 3, confirming intuition.We will soon see that χ f (NK , and all positive integers N. The proof will show how to schedule N burgers for a k-burger grill, k ≤ N, so that all the cooking is done in 2N k time units.
First, an improvement of the bound on This a well known lower bound on χ f (G).We define the Hall ratio with restriction k to be For any graph G and positive integer k, χ f (G; k) ≥ ρ(G; k).
Proof.For any subgraph H of G, the restriction to V (H) of any scheduling function on G which obeys restriction k is a scheduling function on H which obeys restriction k.Therefore Taking the max over all (induced) H on the right gives the result.
We shall finish this section by showing that χ f (NK p ; k) = N p k for all positive integers N, p, and k ≤ N = α(NK p ). Further, our proof will show that for all such N, p, k, a scheduling function ϕ obeying restriction k can be found for NK p such that each set ϕ(v) is the union of no more than 2 open intervals; we can make ϕ(v) a single open interval for each vertex v if and only if k divides Np.Proof.Note that c ≤ b and ac = bd imply that a ≥ d.Let the vertices of A be u 1 , . . ., u a , and the vertices of B be 1, . . ., b, thought of as the congruence classes in the integers mod b.Let u i be adjacent to the congruence classes (i−1)c + 1, . . ., ic, mod b.Since c ≤ b, these are distinct, so there are no double edges in the bipartite graph thus defined.Each vertex in A has degree c in this graph.As r varies from 0 to a−1 and t varies from 1 to c, independently, the integer rc + t varies injectively over 1, . . ., ac = bd.Therefore each congruence class mod b is wandered over exactly d times; that is, each vertex in B has degree d. k .Further, an optimal scheduling function ϕ on G which obeys restriction k can be found such that for every v ∈ V (G), ϕ(v) is the union of no more than two intervals; such a ϕ can be found such that ϕ(v) is a single interval (of length 1) for each v ∈ V (G) if and only if k divides Np.
Proof.We construct a bipartite graph H as in the proof of Lemma 1.2 with bipartition A, B, satisfying |A| = N, |B| = Np, with every vertex of A having degree kp and every vertex of B having degree k.It is easy to see that the hypotheses of Lemma 1.2 are satisfied.
By appeal to Corollary 1.3 and the proof of Lemma 1.2, each neighbor set of a vertex v ∈ A is a list of kp consecutive congruence classes mod (b = Np).Divide this list into p sublists of k consecutive congruence classes mod b.All but, possibly, one of these will be a block of consecutive integers in the labeling of B as {1, . . ., b} and any one sublist not so favored will be the union of two sublists, {t, . . ., b} ∪ {1, . . ., k + t − (b + 1)}, for some t Let the N vertices of A correspond to the N cliques K p of which G is the disjoint union.Let the Np vertices of B correspond to the intervals (0, Let the p sublists of each neighbor set of vertices in A be assigned to the p vertices of the clique to which the vertex corresponds, and let each vertex of G be assigned the union of the intervals to which it is adjacent in H, together with any endpoints that need to be included to make "meeting periods" of each vertex into one open interval or the union of two open intervals.For instance, if a vertex of G is assigned vertices ( t k , t+1 k ), . . ., ( t+k−1 k , t+k k ) of B, for some t ∈ {0, . . ., Np − k}, then the meeting time assigned to the vertex will be ( t k , t k + 1).Since degrees in B are all equal to k, restriction k is obeyed by the scheduling function thus defined, and clearly every vertex of G is assigned a meeting time of measure 1.
Since the scheduling function schedules all meetings within (0, N p k ), we have then the scheduling function described assigns a single interval of length 1 to each vertex of G. Now suppose that there is a scheduling function ϕ on G, obeying restriction k, which assigns to each v ∈ V (G) a single subinterval of length 1 of (0 and |V (G)| = Np, we see from the proof of Proposition 1.1, with T = Np/k, that outside of a set of measure 0 (in fact, outside of a finite set), each point of (0, N p k ) must lie in exactly k of these intervals.Therefore, exactly k of these intervals must have left hand endpoint 0, and then exactly k must have left hand endpoint 1/k, etc.So Np/k is an integer.Theorem 1.4 takes care of optimal scheduling of N burgers of any shape, so long as they each have p sides, and each side is to be grilled the same unit of time, with the maximum number of separate grilling intervals (1 or 2) per side minimized, on any size grill.What are not dealt with are (1) minimizing the number of instances of 2 separated grilling intervals, when k ∤ Np, and (2) when k ∤ Np, minimizing the maximum, or perhaps the sum, or perhaps some other measure of an average, of the lengths of intervals between the intervals of a 2-interval grilling.Indeed, the scheduling function described in the proof of Theorem 1.4 gives a worst possible result for (2): if k ∤ N p , the interval between the end of the first session and the begining of the 2 nd session, of any 2-session meeting/grilling, is N p k − 1, the maximum possible.
We propose these optimization problems as topics for minds finer than ours.
a disjoint union ("sum") of cliques of orders p 1 , . . ., p N ?We will deal with this problem in section 3.

Alternative definitions of χ f (G; k)
In this section we will recapitulate some of the theory of the fractional chromatic number for the restricted fractional chromatic number.For those readers who are familiar with the theory of the fractional chromatic number, the main results of this section will be unsurprising, and their proofs will be recognizable as obvious modifications of standard proofs in the theory of the fractional chromatic number.But there may be readers with an interest in this paper who are not well-acquainted with the theory of the fractional chromaic number; and even among those who are knowledgeable in this area, the scheduling characterization of the fractional chromatic number may not be terribly familiar.Therefore we think we need to have this hard slog of a section in this paper, but, in conformance with shrewd suggestions of the referee(s) and the editor(s), we are softening the blow by hereby warning readers that they do not necessarily need to pay much attention to this section, and by consigning the more difficult proofs in this section to an Appendix at the paper's end.
If S is the union of a finite number of pairwise disjoint bounded open intervals in R = (−∞, ∞), let m(S) denote the number of (maximal) intervals constituting S, and let µ(S) denote the measure of S, the sum of the lengths of the m(S) intervals whose union is S. Lemma 2.1.Suppose that ϕ is a function on V (G) satisfying all the requirements to be a scheduling function on G except (iii), in place of which ϕ satisfies Then there is a scheduling function φ on G satisfying, for each shrink some or all of the intervals whose union is ϕ(v) to non-empty subintervals with union φ(v) satisfying µ( φ(v)) = 1.
For a positive integer k and a graph G, let If k ≥ α(G), we simplify: I k (G) = I(G), the collection of all independent sets of vertices in G.If G is fixed in the discussion we may use I k to stand for I k (G).
A fractional proper coloring of G satisfying restriction k is a function f : The proofs of Theorem 2.2 and the Corollaries to follow are in the Appendix.
Corollary 2.3 (of Theorem 2.2 and its proof).The inf in the definition of χ f (G; k) is really a min.Furthermore, χ f (G; k) is a rational number, and a scheduling function ϕ on G obeying restriction k, with ϕ(v) ⊆ (0, χ f (G; k)) for all v ∈ V (G), can be found such that for each v ∈ V (G) ϕ(v) is a finite union of open intervals with rational endpoints.
Note that the range of any such h is contained in [0, 1].It follows, by an argument similar to that for fractional proper colorings satisfying restriction k, that { v∈V (G) h(v) | h is a fractional clique in a G satisfying restriction k} has a largest element, which we will call the fractional clique number with restriction k and denote ω f (G; k).
Remark: there is an infelicity in our terminology which we will just have to live with.A fractional clique in G is what we would call here a fractional clique satisfying restriction α(G).But then if 1 ≤ k < α(G) there will be fractional cliques satisfying restriction k, by our definition, which are not, in fact, fractional cliques.The constant function 1/k is one such.Be warned.
For m, r positive integers, with m > r, let [m] = {1, . . ., m} and let [m] The r-fold chromatic number of G under restriction k is the smallest m such there is an (m, r)-coloring of G which satisfies restriction k.We denote this number by χ (r) (G; k).Theorem 2.5.
The proof of Theorem 2.5 is in the Appendix.

Calculating χ f (G; k)
If G and H are graphs, let G + H denote the disjoint union of G and H, and G ∨ H = Ḡ + H the join of G and H, obtained by taking disjoint copies of G and H, together with all Proof.Applying Corollary 2.3, we can suppose that ϕ and ψ are scheduling functions obeying restriction k on G and H, respectively, with v∈V (G) ϕ(v) ⊆ (0, χ f (G; k)) and u∈V (H) ψ(u) ⊆ To prove the reverse inequality, we invoke Theorem 2.2, as well as its proof.Let be a fractional proper coloring of G satisfying restriction k, such that Therefore, the restriction of f to I k (G) is a fractional proper coloring of G satisfying restriction k; the same statement holds with G replaced by H. Therefore For any positive integer m, Km denotes the complement of K m , the empty graph on m vertices.For r ≥ 2 and positive integers m 1 , . . ., m r , the complete r-partite graph with parts of sizes n 1 , . . ., n r , denoted K n 1 , . . ., n r , is Kn (It is well understood that both ∨ and + can be viewed as commutative and associative binary operations on "unlabeled" graphs, meaning isomorphism classes of graphs.)Corollary 3.2.For any positive integers r ≥ 2, n 1 , . . ., n r , and k, Proof.The result will follow by induction on r if we show that χ f ( Km ; k) = max( m k , 1), for all positive integers m and k.Since Km = mK 1 the result follows from Theorem 1.4 when k ≤ m, and when • V (H), one can do no better than to find scheduling functions ϕ 1 , ϕ 2 obeying restriction k on G and H separately, with Therefore, to obtain an optimal scheduling function obeying restriction k for K n 1 ,...,nr , one can "concatenate" scheduling functions obeying restriction k for each Kn i , and a recipe for obtaining these is given in the proof of Theorem 1.4 (take p = 1, N = n i ; if k > n i , replace k by n i ).The scheduling function ϕ on G = K n 1 ,...,nr obtained in this way will have the property that for each v ∈ V (G), ϕ(v) is the union of no more than two open intervals; if k | n i , i = 1, . . ., r, then arrangements can be made so that ϕ(v) is a single open interval, for each v ∈ V (G).We doubt that k | n i , i = 1, . . ., r, is a necessary condition for each committee to be assigned a single open interval by an optimal scheduling function, but we shall leave the question open for now.
We leave it as an enjoyable exercise to verify that if Every graph G is a spanning subgraph of some K n 1 ,...,nr , r = χ(G), and therefore Corollary 3.2 supplies upper estimates of χ f (G; k) for all k.Of course, χ f (G; 1) = |V (G)| is known, and therefore uninteresting: it is for 2 ≤ k ≤ α(G) that we are after χ f (G; k).
For instance, if P is the Petersen graph, by properly coloring P with 3 colors we see that P is a subgraph of K 3,3,4 and , therefore, that χ f (P, k) ≤ 2 max(1, 3 k ) + max(1, 4 k ).For k = 1, we have, as is always the case, χ f (P ; 1) = |V (P )| = |V (K 3,3,4 )| = 10.But for k = 2, 3, we also have equality by Corollary 2.4: the constant function 1/2 on V (P ) is a fractional clique in P satisfying restriction 2, so (1, 2) Figure 2: A scheduling function on P with all meetings contained in (0, 5 2 ) and each meeting assignment a union of 1 or 2 open intervals and the same argument with the constant function 1/3 shows that χ f (P ; 3) = 10/3.When k = 4 = α(P ) the upper estimate gives χ f (P ) = χ f (P ; 4) ≤ 3, and here we do not have equality: it can be seen in a number of different ways that χ f (P ) = 5/2.(See below for one of those ways.) Note that optimal scheduling functions obeying restrictions 2 and 3 on K 3,3,4 are also optimal scheduling functions obeying the same restriction on P .It follows that in each case there is an optimal scheduling function ϕ on P such that for each v ∈ V (P ), ϕ(v) is the union of no more than two open intervals.The same holds for k = 4 = α(P ), as shown in Figure 2.
Incidentally, the scheduling function shown in Figure 2 shows that χ f (P ) ≤ 5 2 .On the other hand, χ f (P ) ≥ |V (P )| α(P ) = 10 4 = 5 2 .We finish this section with the fulfillment of the promise at the end of section 1.
Theorem 3.4.Suppose that N, k, and p 1 ≤ • • • ≤ p N are positive integers, and Proof.First we shall show that for some t ≥ 1 and integers s 1 , . . .s t , each no greater than some corresponding p i .Since α(H; k) = min(t, k), to maximize |V (H)| α(H;k) for fixed t we may as well take s 1 , . . ., s t to be p N −t+1 , . . ., p N .If t ≤ k then ]. Now we will show that χ f (G; k) = ρ(G; k).We already have this result when k ≤ N and p 1 = • • • = p N , by Theorem 1.4.We also have this result when k ≥ N: in this case, clearly V (G) can be partitioned into p N sets in It is worth noticing that each vertex of G gets scheduled for one continuous unit of time by an optimal scheduling function obeying restriction k, if N ≤ k.
Fortified by the truth of the theorem when N ≤ k, when k ≤ N and p 1 = • • • = p N , and also when k = 1 (χ f (H; 1) = |V (H)| 1 for all graphs H), we take an arbitrary k ≥ 2 and show that χ f (G; k) = ρ(G; k) for all choices of N and 1 ≤ p 1 ≤ • • • ≤ p N by induction on N i=1 p i .We may as well suppose that k < N and p 1 < p N .
First suppose that Then, taking into account that p 1 < p N , p N appears fewer than k times in the sequence p 1 , . . ., p N .Form S ∈ I k (G), |S| = k, by taking one vertex from each of the cliques is possible because it could be that p N −k+1 = 1), with q M = p N − 1 and . By the induction hypothesis, there is a scheduling function on G − S which obeys restriction k, with ϕ(v) ⊆ (0, p N − 1) for every v ∈ V (G − S) = V (G)\S.Extend ϕ to V (G) by scheduling every vertex in S for the interval (p N − 1, p N ), and we have a scheduling function on G which shows that χ f (G; k) ≤ p N = ρ(G; k).Now suppose that p If p N appears no more than k times in the list p 1 , . . ., p N , then an argument similar to that just above works: form S ∈ I k (G) by taking one vertex from each of the k largest K p i comprising G, apply the induction hypothesis to G − S, and wind up with a scheduling function on G which obeys restriction k and schedules all vertices within the interval (0, ).So suppose that p N occurs at least k + 1 times on the list p 1 , . . ., p N ; i.e. p N −k = p N .
Let H = NK 1 be the subgraph of G induced by a subset of V (G) obtained by choosing one vertex from each in the list q 1 , . . .q M as many times as p N does in the list p 1 , . . ., p N , and that number is at least k+1.Therefore, q M < q 1 +•••+q M k .Applying the induction hypothesis to G − V (H), and noting that \V (H).By Theorem 1.4, H = NK 1 can be properly scheduled with obedience to restriction k with all assignments contained in the interval 4 More chromatic numbers and some problems Suppose that k and q are positive integers, and G is a graph.We define the chromatic number of G with restriction k and allowance q to be χ(G; k, q) = inf[T ; there is a scheduling function ϕ on G which obeys restriction k, with ϕ(v) ⊆ (0, T ) for every v ∈ V (G), such that ϕ(v) is the union of no more than q open intervals].
This definition can be regarded as a refinement of a special case of a definition in [1], which is the definition of what we would denote here by χ(G; α(G), 1); that is, in [1] we are scheduling with an allowance 1, but no restriction.
Lemma 4.1 will seem self-evident to many, but there is actually a proof of it in [1].

Problem schema
1.For given G and k ≤ α(G), determine a(G; k), as defined in Lemma 4.2(a).
2. For given G and q, determine b(G; a), as defined in Lemma 4.2(b).
For any graph G, a(G; 1) = 1.It would be of interest to determine a(G; α(G)), the smallest integer q such that G can be properly scheduled within (0, χ f (G)) with each vertex scheduled for no more than q separate meeting sessions.Call this number a(G), for short.As mentioned above, we have a(P ) = 2, and from the discussion preceding, a(K p 1 +• • •+K pn ) = 1, as well.
Regarding Problem scheme 2, it would be of special interest to find b(G; 1) for each G, and also b(G; a(G; α(G)), which is the smallest value of k such that χ f (G; k) = χ f (G).Let us call this number b(G), for short.We would expect that b(G) = α(G), at least for most G, and the results and discussion in section 3 show that this is indeed the case for the Petersen graph and the complete multipartite graphs.But if , and this k may well be less than N = α(G).
By Lemma 2.1 there is a scheduling function g on G such that for each v ∈ V (G), g(v) ⊆ g(v) ⊆ (0, χ * (G; k)); since such a g must obey restriction k, it follows that χ f (G; k) ≤ χ * (G; k).Now suppose that ǫ > 0 and that ϕ is a scheduling function on G obeying restriction k such that ϕ(v) ⊆ (0, χ f (G; k) + ǫ) for all v ∈ V (G).Let I 1 , . . ., I p be open subintervals of (0, χ(G; k) + ǫ) such that each set ϕ(v), v ∈ V (G), is a union of some of the I j , and let J 1 , . . ., J q be pairwise disjoint open intervals such that each J i is contained in some I j , and each I j is, except possibly for finitely many points, a union of some of the J i .Define g on V (G) by g(v) = {r|Jr⊆ϕ(v)} J r .Then for each v ∈ V (G), g(v) ⊆ ϕ(v) ⊆ (0, χ f (G; k) + ǫ) and ϕ(v)\g(v) is finite.Clearly g is a scheduling function on G obeying restriction k, because ϕ is such a function.Note that for each v ∈ V (G) and each r, either J r ⊆ g(v) or g(v) ∩J r = ∅.

Proof of Corollary 2.3
In the first part of the proof of Theorem 2.2 the fact that χ * (G; k) is a minimum is used to produce a scheduling function g on G obeying restriction k such that g(v) ⊆ (0, χ * (G; k)) for each v ∈ V (G).Since χ * (G; k) = χ f (G; k), this shows that the infimum that χ f (G; k) is defined to be is actually achieved-i.e., is a minimum.
The other conclusions of this corollary are straightforward from the theory of linear programming, for χ * (G; k) is the value of the linear program: minimize S∈I k (G) f (S) subject to the constraints f (S) ≥ 0 for all S ∈ I k (G) and Mf ≥ 1, in which 1 is column vector of |V (G)| ones, f is a column vector of the values f (S), S ∈ I k (G), in some order, and M is, with respect to the same ordering on I k , a V (G) − I k (G) incidence matrix, with rows indexed by the vertices of G, columns by the sets S ∈ I k (G), with entries The inequality Mf ≥ 1 means that each entry of Mf is at least 1.

Lemma 1 . 2 .
Suppose that a, b, c, d are positive integers such that c ≤ b and ac = bd.Then there is a simple bipartite graph with bipartition A, B, |A| = a, |B| = b, such that each vertex in A has degree c, and each vertex in B has degree d.

Corollary 1 . 3 .
(of the proof of Lemma 1.2) Let a, b, c, d be as in Lemma 1.2, and let H be the bipartite graph described in the proof of Lemma 1.2, with bipartition A, B. Let the vertices of B = {1, . . ., b} be thought of as ordinary integers.Then for each u ∈ A, N H (u) is either a block {t, . . ., t+c−1}, for some t satisfying 1 ≤ t ≤ b−c+1, of consecutive integers, or the union {t, . . ., b} ∪ {1, . . ., c + t − (b + 1)}, for some t satisfying b − c + 2 ≤ t ≤ b, of two blocks of consecutive integers.Theorem 1.4.Suppose that k ≤ N and p are positive integers, and G = NK p .Then χ f (G; k) = N p

Corollary 3 .
5. (of Theorem 1.4 and the proof of Theorem 3.4)If G = K p 1 + • • • + K p N thenfor every positive integer k there is a scheduling function ϕ on G which obeys restriction k, with all schedules contained in (0, ρ(G; k)), such that for each v ∈ V (G), ϕ(v) is the union of one or two open intervals.
from the proof of Theorem 3.4 we have that a(G; k) = 1 if N ≤ k, and if k < N then a(G; k) ≤ 2. (The proof involves induction on p 1 + • • • + p N , as in the proof of Theorem 3.4.)We do not know exactly for which k < N and p 1 , . . ., p N we have a(G; k) = 1.
3.1 we have that for arbitrary graphs G and H and positive integers k, a(G ∨ H; k) = max[a(G; k), a(H; k)]From this and Theorem 1.4 we have that a (K n 1 ,...,np ; k) ≤ 2, with equality if and only if for some i ∈ {1, . . ., p}, k ≤ n i and k ∤ n i .