Sum of Squares of Degrees in a Graph

1 Introduction P Let G(v, e) be the set of all simple graphs with v vertices and e edges and let P2 (G) = d2i denote the sum of the squares of the degrees, d1 , . . . , dv , of the vertices of G. The purpose of this paper is to finish the solution of an old problem: 1. What is the maximum value of P2 (G), for a graph G in G(v, e)? 2. For which graphs G in G(v, e) is the maximum value of P2 (G) attained? Throughout, we say that a graph G is optimal in G(v, e), if P2 (G) is maximum and we denote this maximum value by max(v, e). These problems were first investigated by Katz [Ka] in 1971 and by R. Ahlswede and G.O.H. Katona [AK] in 1978. In his review of the paper by Ahlswede and Katona. P. Erdős [Er] commented that “the solution is more difficult than one would expect.” Ahlswede and Katona were interested in an equivalent form of the problem: they wanted to find the maximum number of pairs of different edges that have a common vertex. In other words, they want to maximize the number of edges in the line graph L(G) as G ranges over G(v, e). That these two formulations of the problem are equivalent follows from an examination of the vertex-edge incidence matrix N for a graph G ∈ G(v, e): trace((N N T )2 ) = P2 (G) + 2e trace((N T N )2 ) = trace(AL (G)2 ) + 4e, where AL (G) is the adjacency matrix of the line graph of G. Thus P2 (G) = trace(AL (G)2 ) + 2e. (trace(AL (G)2 ) is the twice the number of edges in the line graph of G.) Ahlswede and Katona showed that the maximum value max(v, e) is always attained at one or both of two special graphs in G(v, e). They called the first of the two special graphs a quasi-complete graph. The quasi-complete graph in G(v, e) has the largest possible complete subgraph Kk . Let k, j be the unique integers such that µ ¶ k+1 e= − j, where 1 ≤ j ≤ k. 2 The quasi-complete graph in G(v, e), which is denoted by QC(v, e), is obtained removing the j edges, (k − j + 1, k + 1), (k − j + 2, k + 1), . . . , (k, k + 1), from the complete graph on the k + 1 vertices, 1, 2, . . . , k + 1, and adding v − k − 1 isolated vertices. The other special graph in G(v, e) is the quasi-star, which we denote by QS(v, e). This graph has as many dominant vertices as possible. (A dominant vertex is one with maximum degree v − 1.) Perhaps the¡ easiest way to describe QS(v, e) is to say that it is the graph complement of QC(v, e0 ), ¢ v 0 where e = 2 − e. Define the function C(v, e) to be the sum of the squares of the degree sequence of the quasicomplete graph in G(v, e), and define S(v, e) to be the sum of the squares of the degree sequence of the quasi-star graph in G(v, e). The value of C(v, e) can be computed as follows: Let e = ¡k+1¢ 2 − j, with 1 ≤ j ≤ k. The degree sequence of the quasi-complete graph in G(v, e) is d1 = · · · = dk−j = k, dk−j+1 = · · · = dk = k − 1, dk+1 = k − j, dk+2 = · · · = dv = 0. Hence C(v, e) = j(k − 1)2 + (k − j)k 2 + (k − j)2 . (1) Since QS(v, e) is the complement of QC(v, e0 ), it is straightforward to show that S(v, e) = C(v, e0 ) + (v − 1)(4e − v(v − 1)) (2) from which it follows that, for ¡fixed ¢ v, the function S(v, e) − C(v, e) is point-symmetric about the v middle of the interval 0 ≤ e ≤ 2 . In other words, ¡ ¢ S(v, e) − C(v, e) = − S(v, e0 ) − C(v, e0 ) . It also follows from Equation (2) that QC(v, e) is optimal in G(v, e) if and only if QS(v,¡e0¢) is optimal in G(v, e0 ). This allows restrict our attention to values of e in the interval [0, v2 /2] ¡v¢ us ¡to ¢ or equivalently the interval [ 2 /2, v2 ]. On occasion, we will do so but we will always state results for all values of e. As the midpoint of the range of values for e plays a recurring role in what follows, we denote it by µ ¶ 1 v m = m(v) = 2 2 and define k0 = k0 (v) to be the integer such that µ ¶ µ ¶ k0 k0 + 1 ≤m< . 2 2 To state the results of [AK] we need one more notion, that of the distance from µ ¶ k0 b0 = b0 (v) = m − . 2 (3) ¡k0 ¢ 2 to m. Write We are now ready to summarize the results of [AK]: Theorem 2 [AK] max(v, e) is the larger of the two values C(v, e) and S(v, e). Theorem 3 [AK] max(v, e) = S(v, e) if 0 ≤ e < m − v2 and max(v, e) = C(v, e) if m + v2 < e ≤ Lemma 8 [AK] If 2b0 ≥ k0 , or 2v − 2k0 − 1 ≤ 2b0 < k0 , then C(v, e) ≤ S(v, e) for all C(v, e) ≥ S(v, e) for all 0 ≤ e ≤ m and µ ¶ v m≤e≤ . 2 ¡v¢ 2 If 2b0 < k0 and 2k0 + 2b0 < 2v − 1, then there exists an R with b0 ≤ R ≤ min {v/2, k0 − b0 } such that C(v, e) ≤ S(v, e) for all 0≤e≤m−R C(v, e) ≥ S(v, e) for all m−R≤e≤m C(v, e) ≤ S(v, e) for all C(v, e) ≥ S(v, e) for all m≤e≤m+R µ ¶ v m+R≤e≤ . 2 Ahlswede and Katona pose some open questions at the end of [AK]. “Some strange numbertheoretic combinatorial questions arise. What is the relative density of the numbers¡ v¢ for which R = 0 [max(v, e) = S(v, e) for all 0 ≤ e < m and max(v, e) = C(v, e) for all m < e ≤ v2 ]?” This is the point of departure for our paper. Our first main result, Theorem 1, strengthens Ahlswede and Katona’s Theorem 2; not only does the maximum value of P2 (G) occur at either the quasi-star or quasi-complete graph in G(v, e), but all optimal graphs in G(v, e) are related to the quasi-star or quasi-complete graphs via their so-called diagonal sequence. As a result of their relationship to the quasi-star and quasi-complete graphs, all optimal graphs can be and are described in our second main result, Theorem 2. Our third main result, Theorem 6, is a refinement of Lemma 8 in [AK]. Theorem 6 characterizes the values of v and e for which S(v, e) = C(v, e) and gives an explicit expression for the value R in Lemma 8 of [AK]. Finally, the “strange number-theoretic combinatorial” aspects of the problem, mentioned by Ahlswede and Katona, turn out to be Pell’s Equation y 2 − 2x2 = ±1. Corollary 3 answers the density question posed by Ahlswede and Katona. Before stating the new results, we summarize work on the problem that followed [AK]. The authors of these papers were often unaware of the work in [AK]. A generalization of the problem of maximizing the sum of the squares of the degree sequence was investigated by Katz [Ka] in 1971 and R. Aharoni [Ah] in 1980. Katz’s problem was to maximize the sum of the elements in A2 , where A runs over all (0, 1)-square matrices of size n with precisely j ones. He found the maxima and the matrices for which the maxima are attained for the special cases where there are k 2 ones or where there are n2 − k 2 ones in the (0, 1)-matrix. Aharoni [Ah] extended Katz’s results for general j and showed that the maximum is achieved at one of four possible forms for A. If A is a symmetric (0, 1)-matrix, with zeros on the diagonal, then A is the adjacency matrix A(G) for a graph G. Now let G be a graph in G(v, e). Then the adjacency matrix A(G) of G is a v × v (0, 1)-matrix with 2e ones. But A(G) satisfies two additional restrictions: A(G) is P symmetric, and 2 all diagonal entries are zero. However, the sum of all entries in A(G) is precisely di (G)2 . Thus our problem is essentially the same as Aharoni’s in that both ask for the maximum of the sum of the elements in A2 . The graph-theory problem simply restricts the set of (0, 1)-matrices to those with 2e ones that are symmetric and have zeros on the diagonal. Olpp, apparently unaware of the work of Ahlswede and Katona, reproved the basic result that max(v, e) = max(S(v, e), C(v, e)), but his results are stated in the context of two-colorings of a graph. He investigates a question of Goodman [Go1, Go2]: maximize the number of monochromatic triangles in a two-coloring of the complete graph with a fixed number of vertices and a fixed number of red edges. Olpp shows that Goodman’s problem is equivalent to finding the two-coloring that maximizes the sum of squares of the red-degrees of the vertices. Of course, a two-coloring of the complete graph on v vertices gives rise to two graphs on v vertices: the graph G whose edges are colored red, and its complement G0 . So Goodman’s problem is to find the maximum value of P2 (G) for G ∈ G(v, e). Olpp shows that either the quasi-star or the quasi-complete graph is optimal in G(v, e), but he does not discuss which of the two values S(v, e), C(v, e) is larger. He leaves this question unanswered and he does not attempt to identify all optimal graphs in G(v, e). In 1999, Peled, Pedreschi, and Sterbini [PPS] showed that the only possible graphs for which the maximum value is attained are the so-called threshold graphs. The main result in [PPS] is that all optimal graphs are in one of six classes of threshold graphs. They end with the remark, “Further questions suggested by this work are the existence and uniqueness of the [graphs in G(v, e)] in each class, and the precise optimality conditions.” Some of the results we prove below follow from results in [AK]. We reprove them here for two reasons: First, the new proofs introduce techniques that are used to prove the extensions of the results of [AK]. Second, they make the paper self-contained. We will point out at the appropriate places when a result is not new. 2 2.1 Statements of the main results Threshold graphs All optimal graphs come from a class of special graphs called threshold graphs. The quasi-star and quasi-complete graphs are just two among the many threshold graphs in G(v, e). The adjacency matrix of a threshold graph has a special form. The upper-triangular part of the adjacency matrix of a threshold graph is left justified and the number of zeros in each row of the upper-triangular part of the adjacency matrix does not decrease. We will show adjacency matrices using “+” for the main diagonal, an empty circle “◦” for the zero entries, and a black dot, “•” for the entries equal to one. For example, the graph G whose adjacency matrix is shown in Figure 1(a) is a threshold graph in G(8, 13) with degree sequence (6, 5, 5, 3, 3, 3, 1, 0). By looking at the upper-triangular part of the adjacency matrix, we can associate the distinct partition π = (6, 4, 3) of 13 with the graph. In general, the threshold graph Th(π) ∈ G(v, e) corresponding to a distinct partition π = (a0 , a1 , . . . , ap ) of e, all of whose parts are less than v, is the graph with an adjacency matrix whose upper-triangular part is left-justified and contains as ones in row s. Thus the threshold graphs in G(v, e) are in one-to-one correspondence with the set of distinct partitions, Dis(v, e) of e with all parts less than v: X Dis(v, e) = {π = (a0 , a1 , . . . , ap ) : v > a0 > a1 > · · · > ap > 0, as = e} We denote the adjacency matrix of the threshold graph Th(π) corresponding to the distinct partition π by Adj(π). Peled, Pedreschi, and Sterbini [PPS] showed that all optimal graphs in a graph class G(v, e) must be threshold graphs. Lemma 1 [PPS] If G is an optimal graph in G(v, e), then G is a threshold graph. Thus we can limit the search for optimal graphs to the threshold graphs. Actually, a much larger class of functions, including the power functions, dp1 + · · · + dpv for p ≥ 2, on the degrees of a graph are maximized only at threshold graphs. In fact, every Schur convex function of the degrees is maximized only at the threshold graphs. The reason is that the degree sequences of threshold graphs are maximal with respect to the majorization order among all graphical sequences. See [MO] for a discussion of majorization and Schur convex functions and [MR] for a discussion of the degree sequences of threshold graphs. 2.2 The diagonal sequence of a threshold graph To state the first main theorem, we must now digress to describe the diagonal sequence of a threshold graph in the graph class G(v, e). Returning to the example in Figure 1(a) corresponding to the distinct partition π = (6, 4, 3) ∈ Dis(8, 13), we superimpose diagonal lines on the adjacency matrix Adj(π) for the threshold graph Th(π) as shown in Figure 1(b). (a) (b) Figure 1: The adjacency matrix, Adj(π), for the threshold graph in G(8, 13) corresponding to the distinct partition π = (6, 4, 3) ∈ Dis(8, 13) with diagonal sequence δ(π) = (1, 1, 2, 2, 3, 3, 1). The number of black dots in the upper triangular part of the adjacency matrix on each of the diagonal lines is called the diagonal sequence of the partition π (or of the threshold graph Th(π)). The diagonal sequence for π is denoted by δ(π) and for π = (6, 4, 3) shown in Figure 1, δ(π) = (1, 1, 2, 2, 3, 3, 1). The value of P2 (Th(π)) is determined by the diagonal sequence of π. Lemma 2 Let π be a distinct partition in Dis(v, e) with diagonal sequence δ(π) = (δ1 , . . . , δt ). Then P2 (Th(π)) is the dot product P2 (Th(π)) = 2δ(π) · (1, 2, 3, . . . , t) = 2 t X iδi . i=1 For example, if π = (6, 4, 3) as in Figure 1, then P2 (Th(π)) = 2(1, 1, 2, 2, 3, 3, 1) · (1, 2, 3, 4, 5, 6, 7) = 114, which equals the sum of squares of the degree sequence (6, 5, 5, 3, 3, 3, 1) of the graph Th(π). Theorem 2 in [AK] guarantees that one (or both) of the graphs QS(v, e), QC(v, e) must be optimal in G(v, e). But there may be other optimal graphs in G(v, e), as the next example shows. The quasi-complete graph QC(10, 30), which corresponds to the distinct partition (8, 7, 5, 4, 3, 2, 1) is optimal in G(10, 30). The threshold graph G2 , corresponding to the distinct partition (9, 6, 5, 4, 3, 2, 1) is also optimal in G(10, 30), but is neither quasi-star in G(10, 30) nor quasi-complete in G(v, 30) for any v. The adjacency matrices for these two graphs are shown in Figure 2. They have the same diagonal sequence δ = (1, 1, 2, 2, 3, 3, 4, 4, 4, 2, 2, 1, 1) and both are optimal. Figure 2: Adjacency matrices for two optimal graphs in G(10, 30), QC(10, 30) = Th(8, 7, 5, 4, 3, 2, 1) and Th(9, 6, 5, 4, 3, 2, 1), having the same diagonal sequence δ = (1, 1, 2, 2, 3, 3, 4, 4, 4, 2, 2, 1, 1) We know that either the quasi-star or the quasi-complete graph in G(v, e) is optimal and that any threshold graph with the same diagonal sequence as an optimal graph is also optimal. In fact, the converse is also true. Indeed, the relationship between the optimal graphs and the quasi-star and quasi-complete graphs in a graph class G(v, e) is described in our first main theorem. Theorem 1 Let G be an optimal graph in G(v, e). Then G = Th(π) is a threshold graph for some partition π ∈ Dis(v, e) and the diagonal sequence δ(π) is equal to the diagonal sequence of either the quasi-star graph or the quasi-complete graph in G(v, e). Theorem 1 is stronger than Lemma 8 [AK] because it characterizes all optimal graphs in G(v, e). In Section 2.3 we describe all optimal graphs in detail. 2.3 Optimal graphs Every optimal graph in G(v, e) is a threshold graph, Th(π), corresponding to a partition π in Dis(v, e). So we extend the terminology and say that the partition π is optimal in Dis(v, e), if its threshold graph Th(π) is optimal in G(v, e). We say that the partition π ∈ Dis(v, e) is the quasi-star partition, if Th(π) is the quasi-star graph in G(v, e). Similarly, π ∈ Dis(v, e) is the quasi-complete partition, if Th(π) is the quasi-complete graph in G(v, e). We now describe the quasi-star and quasi-complete partitions in Dis(v, e). First, the quasi-complete graphs. Let v be a positive integer and e an integer such that 0 ≤ e ≤ There exists unique integers k and j such that µ ¶ k+1 e= − j and 1 ≤ j ≤ k. 2 ¡v¢ 2 . The partition π(v, e, qc) := (k, k − 1, . . . , j + 1, j − 1, . . . , 1) = (k, k − 1, . . . , b j, . . . , 2, 1) corresponds to the quasi-complete threshold graph QC(v, e) in Dis(v, e). The symbol b j means that j is missing. To describe the quasi-star partition π(v, e, qs) in Dis(v, e), let k 0 , j 0 be the unique integers such that µ ¶ µ 0 ¶ v k +1 e= − + j 0 and 1 ≤ j 0 ≤ k 0 . 2 2 Then the partition π(v, e, qs) = (v − 1, v − 2, . . . , k 0 + 1, j 0 ) corresponds to the quasi-star graph QS(v, e) in Dis(v, e). In general, there may be many partitions with the same diagonal sequence as π(v, e, qc) or π(v, e, qs). For example, if (v, e) = (14, 28), then π(14, 28, qc) = (7, 6, 5, 4, 3, 2, 1) and all of the partitions in Figure 3 have the same diagonal sequence, δ = (1, 1, 2, 2, 3, 3, 4, 3, 3, 2, 2, 1, 1). But none of the threshold graphs corresponding to the partitions in Figure 3 is optimal. Indeed, if the quasicomplete graph is optimal in Dis(v, e), then there are at most three partitions in Dis(v, e) with the same diagonal sequence as the quasi-complete graph. The same is true for the quasi-star partition. If the quasi-star partition is optimal in Dis(v, e), then there are at most three partition in Dis(v, e) having the same diagonal sequence as the quasi-star partition. As a consequence, there are at most six optimal partitions in Dis(v, e) and so at most six optimal graphs in G(v, e). Our second main result, Theorem 2, entails Theorem 1; it describes the optimal partitions in G(v, e) in detail. Figure 3: Four partitions with the same diagonal sequence as π(14, 28, qc) Theorem 2 Let v be a positive integer and e an integer such that 0 ≤ e ≤ the unique integers satisfying µ ¶ k+1 e= − j, with 1 ≤ j ≤ k, 2 and ¡v¢ 0 0 2 . Let k, k , j, j be µ ¶ µ 0 ¶ v k +1 e= − + j 0 , with 1 ≤ j 0 ≤ k 0 . 2 2 Then every optimal partition π in Dis(v, e) is one of the following six partitions: 1.1 π1.1 = (v − 1, v − 2, . . . , k 0 + 1, j 0 ), the quasi-star partition for e, 1.2 π1.2 = (v − 1, v − 2, . . . , 2k 0 \ − j 0 − 1, . . . , k 0 − 1), if k 0 + 1 ≤ 2k 0 − j 0 − 1 ≤ v − 1, 1.3 π1.3 = (v − 1, v − 2, . . . , k 0 + 1, 2, 1), if j 0 = 3 and v ≥ 4, 2.1 π2.1 = (k, k − 1, . . . , b j, . . . , 2, 1), the quasi-complete partition for e, 2.2 π2.2 = (2k − j − 1, k − 2, k − 3, . . . 2, 1), if k + 1 ≤ 2k − j − 1 ≤ v − 1, 2.3 π2.3 = (k, k − 1, . . . , 3), if j = 3 and v ≥ 4. On the other hand, partitions π1.1 and π2.1 always exist and at least one of them is optimal. Furthermore, π1.2 and π1.3 (if they exist) have the same diagonal sequence as π1.1 , and if S(v, e) ≥ C(v, e), then they are all optimal. Similarly, π2.2 and π2.3 (if they exist) have the same diagonal sequence as π2.1 , and if S(v, e) ≤ C(v, e), then they are all optimal. A few words of explanation are in order regarding the notation for the optimal partitions in Theorem 2. If k 0 = v, then j 0 = v, e = 0, and π1.1 = ∅. If k 0 = v − 1, then e = j 0 ≤ v − 1, and π1.1 = (j 0 ); further, if j 0 = 3, then π1.3 = (2, 1). In all other cases k 0 ≤ v − 2 and then π1.1 , π1.2 , and π1.3 are properly defined. If j 0 = k 0 or j 0 = k 0 − 1, then both partitions in 1.1 and 1.2 would be equal to (v − 1, v − 2, . . . , k 0 ) and (v − 1, v − 2, . . . , k 0 + 1, k 0 − 1) respectively. So the condition k 0 + 1 ≤ 2k 0 − j 0 − 1 merely ensures that π1.1 6= π1.2 . A similar remark holds for the partitions in 2.1 and 2.2. By definition the partitions π1.1 and π1.3 are always distinct; the same holds for partitions π2.1 and π2.3 . In general the partitions πi.j described in items 1.1-1.3 and 2.1-2.3 (and their corresponding threshold graphs) are all different. All the exceptions are illustrated in Figure 4 and are as follows: For any v, if e ∈ {0, 1, 2} or e0 ∈ {0, 1, 2} then π1.1 = π2.1 . For any v ≥ 4, if e = 3 or e0 = 3, then π1.3 = π2.1 and π1.1 = π2.3 . If (v, e) = (5, 5) then π1.1 = π2.2 and π1.2 = π2.1 . Finally, if (v, e) = (6, 7) or (7, 12), then ¡v¢ π1.2 = π2.3 . Similarly, if (v, e) = (6, 8) or (7, 9), then π1.3 = π2.2 . For v ≥ 8 and 4 ≤ e ≤ 2 − 4, all the partitions πi.j are pairwise distinct (when they exist). Figure 4: Instances of pairs (v, e) where two partitions πi.j coincide In the next section, we determine the pairs (v, e) having a prescribed number of optimal partitions (and hence graphs) in G(v, e). 2.4 Pairs (v, e) with a prescribed number of optimal partitions. In principle, a given pair (v, e), could have between one and six optimal partitions. It is easy to see that there are infinitely many pairs (v, e) with¡ only one optimal partition (either the quasi-star or ¢ the quasi-complete). For example the pair (v, v2 ) only has the quasi-complete partition. Similarly, there are infinitely many pairs with exactly two optimal partitions and this can be achieved in many different ways. For instance if (v, e) = (v, 2v − 5) and v ≥ 9, then k 0 = v − 2, j 0 = v − 4 > 3, and S(v, e) > C(v, e) (c.f. Corollary 2). Thus only the partitions π1.1 and π1.2 are optimal. The interesting question is the existence of pairs with 3,4,5, or 6 optimal partitions. Often, both partitions π1.2 and π1.3 in Theorem 2 exist for the same pair (v, e); however it turns out that this almost never happens when they are optimal partitions. More precisely, Theorem 3 If π1.2 and π1.3 are optimal partitions then (v, e) = (7, 9) or (9, 18). Similarly, if π2.2 and π2.3 are optimal partitions then (v, e) = (7, 12) or (9, 18). Furthermore, the pair (9, 18) is the only one with six optimal partitions, there are no pairs with five. If there are more than two optimal partitions for a pair (v, e), then S(v, e) = C(v, e), that is, both the quasi-complete and the quasi-star partitions must be optimal. In the next two results, we describe two infinite families of partitions in Dis(v, e), and hence graph classes G(v, e), for which there are exactly three (four) optimal partitions. The fact that they are infinite is proved in Section 9. Theorem 4 Let v > 5 and k be positive integers that satisfy the Pell’s Equation (2v − 3)2 − 2(2k − 1)2 = −1 (4) ¡ ¢ and let e = k2 . Then (using the notation of Theorem 2), j = k, k 0 = k + 1, j 0 = 2k − v + 2, and there are exactly three optimal partitions in Dis(v, e), namely π1.1 = (v − 1, v − 2, . . . , k + 2, 2k − v + 2) π1.2 = (v − 2, v − 3, . . . , k) π2.1 = (k − 1, k − 2, . . . , 2, 1). The partitions π1.3 , π2.2 , and π2.3 do not exist. Theorem 5 Let v > 9 and k be positive integers that satisfy the Pell’s Equation (2v − 1)2 − 2(2k + 1)2 = −49 (5) ¡ ¢ and e = m = 21 v2 . Then (using the notation of Theorem 2), j = j 0 = 3, k = k 0 , and there are exactly four optimal partitions in Dis(v, e), namely π1.1 = (v − 1, v − 2, . . . , k + 1, 3) π1.3 = (v − 1, v − 2, . . . , k + 1, 2, 1) π2.1 = (k − 1, k − 2, . . . , 4, 2, 1) π2.3 = (k − 1, k − 2, . . . , 4, 3). The partitions π1.2 and π2.2 do not exist. 2.5 Quasi-star versus quasi-complete In this section, we compare S(v, e) and C(v, e). The main result of the section, Theorem 6, is a theorem very much like Lemma 8 of [AK], with the addition that our results give conditions for equality of the two functions. If e = 0, 1, 2, 3, then S(v, e) = C(v, e) for all v. Of course, if e = 0, e = 1 and v ≥ 2, or e ≤ 3 and v = 3, there is only one graph in the graph class G(v, e). If e = 2 and v ≥ 4, then there are two graphs in the graph class G(v, 2): the path P and the partial matching M , with degree sequences (2, 1, 1) and (1, 1, 1, 1), respectively. The path is optimal as P2 (P ) = 6 and P2 (M ) = 4. But the path is both the quasi-star and the quasi-complete graph in G(v, 2). If e = 3 and v ≥ 4, then the quasi-star graph has degree sequence (3, 1, 1, 1) and the quasi-complete graph is a triangle with degree sequence (2, 2, 2). ¡Since P2 (G) = 12 for both of these graphs, both are optimal. Similarly, ¢ S(v, e) = C(v, e) for e = v2 − j for j = 0, 1, 2, 3. ¡ ¢ Now, we consider the cases where 4 ≤ e ≤ v4 − 4. Figures 5, 6, 7, and 8 show the values of the difference S(v, e) − C(v, e). When the graph is above the horizontal axis, S(v, e) is strictly larger than C(v, e) and so the quasi-star graph is optimal and the quasi-complete is not optimal. And when the graph is on the horizontal axis, S(v, e) = C(v, e) and both the quasi-star and the quasi-complete graph are optimal. Since the function S(v, e) − C(v, e) is central ¡v¢ symmetric, we shall consider only the values of e from 4 to the midpoint, m, of the interval [0, 2 ]. 180 195 190 164 171 147 153 129 136 120 105 110 Figure 5 shows that S(25, e) > C(25, e) for all values of e: 4 ≤ e < m = 150. So, when v = 25, the quasi-star graph is optimal for 0 ≤ e < m = 150 and the quasi-complete graph is not optimal. For e = m(25) = 150, the quasi-star and the quasi-complete graphs are both optimal. Figure 5: S(25, e) − C(25, e) > 0 for 4 ≤ e < m = 150 69 66 60 55 50 45 36 39 Figure 6 shows that S(15, e) > C(15, e) for 4 ≤ e < 45 and 45 < e ≤ m = 52.5. But S(15, 45) = C(15, 45). So the quasi-star graph is optimal and the quasi-complete graph is not optimal for all 0 ≤ e ≤ 52 except for e = 45. Both the quasi-star and the quasi-complete graphs are optimal in G(15, 45). Figure 6: S(15, e) − C(15, e) > 0 for 4 ≤ e < 45 and for 45 < e ≤ m = 52.5 Figure 7 shows that S(17, e) > C(17, e) for 4 ≤ e < 63, S(17, 64) = C(17, 64), S(17, e) < C(17, e) 91 81 78 70 66 58 55 45 for 65 ≤ e < m = 68, and S(17, 68) = C(17, 68). Figure 7: S(17, e) − C(17, e) > 0 for 4 ≤ e ≤ 63 and 65 ≤ e < m = 68 171 162 148 153 133 136 117 120 105 91 82 100 Finally, Figure 8 shows that S(23, e) > C(23, e) for 4 ≤ e ≤ 119, but S(23, e) = C(23, e) for 120 ≤ e ≤ m = 126.5. Figure 8: S(23, e) − C(23, e) > 0 for 4 ≤ e ≤ 119, S(23, e) = C(23, e) for 120 ≤ e < m = 126.5 These four examples exhibit the types of behavior of the function S(v, e) − C(v, e), for fixed v. The main thing that determines this behavior is the quadratic function ¢ 1¡ 1 − 2(2k0 − 3)2 + (2v − 5)2 . 4 (The integer k0 = k0 (v) depends on v.) For example, if q0 (v) > 0, then S(v, e) − C(v, e) ≥ 0 for all values of e < m. To describe the behavior of S(v, e) − C(v, e) for q0 (v) < 0, we need to define q0 (v) := R0 = R0 (v) = where 8(m − e0 )(k0 − 2) , −1 − 2(2k0 − 4)2 + (2v − 5)2 µ ¶ k0 e0 = e0 (v) = = m − b0 2 Our third main theorem is the following: Theorem 6 Let v be a positive integer 1. If q0 (v) > 0, then S(v, e) ≥ C(v, e) for all S(v, e) ≤ C(v, e) for all 0 ≤ e ≤ m and ¡ ¢ m ≤ e ≤ v2 . S(v, e) = C(v, e) if and only if e, e0 ∈ {0, 1, 2, 3, m}, or e, e0 = e0 and (2v − 3)2 − 2(2k0 − 3)2 = −1, 7. 2. If q0 (v) < 0, then C(v, e) ≤ S(v, e) for all 0 ≤ e ≤ m − R0 C(v, e) ≥ S(v, e) for all m − R0 ≤ e ≤ m C(v, e) ≤ S(v, e) for all C(v, e) ≥ S(v, e) for all m ≤ e ≤ m + R0 ¡ ¢ m + R0 ≤ e ≤ v2 . S(v, e) = C(v, e) if and only if e, e0 ∈ {0, 1, 2, 3, m − R0 , m} 3. If q0 (v) = 0, then S(v, e) ≥ C(v, e) for all S(v, e) ≤ C(v, e) for all 0 ≤ e ≤ m and ¡ ¢ m ≤ e ≤ v2 . S(v, e) = C(v, e) if and only if e, e0 ∈ {0, 1, 2, 3, e0 , ..., m} The conditions in Theorem 6 involving the quantity q0 (v) simplify and refine the conditions in [AK] involving k0 and b0 . The condition 2b0 ≥ k0 in Lemma 8 of [AK] can be removed and the result restated in terms of the sign of the quantity 2k0 + 2b0 − (2v − 1) = −2q0 (v). While [AK] considers only the two cases q0 (v) ≤ 0 and q0 (v) > 0, we analyze the case q0 (v) = 0 separately. It is apparent from Theorem 6 that S(v, e) ≥ C(v, e) for 0 ≤ e ≤ m − αv if α > 0 is large enough. Indeed, Ahlswede and Katona [AK, Theorem 3] show this for α = 1/2, thus establishing an inequality that holds for all values of v regardless √ of the sign of q0 (v). We improve this result and show that the inequality holds when α = 1 − 2/2 ≈ 0.2929. √ Corollary 1 Let α =¡1− ¢ 2/2. Then S(v, e) ≥ C(v, e) for all 0 ≤ e ≤ m−αv and S(v, e) ≤ C(v, e) v for all m + αv ≤ e ≤ 2 . Furthermore, the constant α cannot be replaced by a smaller value. Theorem 3 in [AK] can be improved in another way. The inequalities are actually strict. Corollary 2 S(v, e) > C(v, e) for 4 ≤ e < m−v/2 and S(v, e) < C(v, e) for m+v/2 < e ≤ ¡v¢ 2 −4. 2.6 Asymptotics and density We now turn to the questions asked in [AK]: What is the relative density of the positive integers v for which max(v, e) = S(v, e) for 0 ≤ e¡ <¢ m? Of course, max(v, e) = S(v, e) for 0 ≤ e ≤ m if and only if max(v, e) = C(v, e) for m ≤ e ≤ v2 . Corollary 3 Let t be a positive integer and let n(t) denote the number of integers v in the interval [1, t] such that max(v, e) = S(v, e), for all 0 ≤ e ≤ m. Then √ n(t) = 2 − 2 ≈ 0.5858. t→∞ t lim 2.7 Piecewise linearity of S(v, e) − C(v, e) The diagonal sequence for a threshold ¡ ¢ graph helps explain the behavior of the difference S(v, e) − C(v, e) for fixed v and 0 ≤ e ≤ v2 . From Figures 5, 6, 7, and 8, we see that S(v, e) − C(v, e), regarded as a function ¡of¢ e, is piecewise ¡ ¢ linear ¡ ¢ and the ends of the intervals on which the function is linear occur at e = 2j and e = v2 − 2j for j = 1, 2, . . . , v. We prove this fact in Lemma 10. For now, we present an example. Take v = 15, for example. Figure 6 shows linear behavior on the intervals ¡ ¢[36, 39], [39, 45], [45, 50], [50, 55], [55, 60], [60, 66], and [66, 69]. There are 14 binomial coefficients 2j for 2 ≤ j ≤ 15: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105. ¡ ¢ The complements with respect to 15 2 = 105 are 104, 102, 99, 95, 90, 84, 77, 69, 60, 50, 39, 27, 14, 0. The union of these two sets of integers coincide with the end points for the intervals on which S(15, e) − C(15, e) is linear. In this case, the function is linear on the 27 intervals with end points: 0, 1, 3, 6, 10, 14, 15, 21, 27, 28, 36, 39, 45, 50, 55, 60, 66, 69, 77, 78, 84, 90, 91, 95, 99, 102, 104, 105. These special values of e correspond to special types of quasi-star and quasi-complete graphs. ¡¢ If e = 2j , then the quasi-complete graph QC(v, e) is the sum of a complete graph on j vertices ¡¢ and v − j isolated vertices. For example, if v = 15 and j = 9, and e = 92 = 36, then the uppertriangular ¡v¢ ¡part ¢ of the adjacency matrix for QC(15, 21) is shown on the left in Figure 9. And if j e = 2 − 2 , then the quasi-star graph QS(v, e) has j dominant vertices and none of the other v − j vertices are adjacent to each other. For example, the lower part of the adjacency ¡ ¢ triangular ¡12¢ matrix for the quasi-star graph with v = 15, j = 12, and e = 14 − = 39, is shown on the right 2 2 in Figure 9. quasi-complete partition π=(8,7,6,5,4,3,2,1) π=(9,7,6,5,4,3,2,1) π=(9,8,6,5,4,3,2,1) π=(9,8,7,5,4,3,2,1) π=(14,13,9) π=(14,13,10) π=(14,13,11) π=(14,13,12) e = 36 e = 37 e = 38 e = 39 quasi-star partition Figure 9: Adjacency matrices for quasi-complete and quasi-star graphs with v = 15 and 36 ≤ e ≤ 39 As additional dots are added to the adjacency matrices for the quasi-complete graphs with e = 37, 38, 39, the value of C(15, e) increases by 18, 20, 22. And the value of S(15, e) increases by 28, 30, 32. Thus, the difference increases by a constant amount of 10. Indeed, the diagonal lines are a distance of five apart. Hence the graph of S(15, e) − C(15, e) for 36 ≤ e ≤ 39 is linear with a slope of 10. But for e = 40, the adjacency matrix for the quasi-star graph has an additional dot on the diagonal corresponding to 14, whereas the adjacency matrix for the quasi-complete graph has an additional dot on the diagonal corresponding to 24. So S(15, 40) − C(15, 40) decreases by 10. The decrease of 10 continues until the adjacency matrix for the quasi-complete graph contains a complete column at e = 45. Then the next matrix for e = 46 has an additional dot in the first row and next column and the slope changes again. 3 Proof of Lemma 2 Returning for a moment to the threshold graph Th(π) from Figure 1, which corresponds to the distinct partition π = (6, 4, 3), we see the graph complement shown with the white dots. Counting white dots in the rows from bottom to top and from the left to the diagonal, we have 7,5,2,1. These same numbers appear in columns reading from right to left and then top to the diagonal. So if Th(π) is the threshold graph associated with π, then the set-wise complement of π (π c ) in the set {1, 2, . . . , v − 1} corresponds to the threshold graph Th(π)c —the complement of Th(π). That is, Th(π c ) = Th(π)c . The diagonal sequence allows us to evaluate the sum of squares of the degree sequence of a threshold graph. Each black dot contributes a certain amount to the sum of squares. The amount depends on the location of the black dot in the adjacency matrix. In fact all of the dots on a particular diagonal line contribute the same amount to the sum of squares. For v = 8, the value of a black dot in position (i, j) is given by the entry in  + 1 3  1 + 3   1 3 +   1 3 5   1 3 5   1 3 5   1 3 5 1 3 5 the following matrix:  5 7 9 11 13 5 7 9 11 13   5 7 9 11 13   + 7 9 11 13   7 + 9 11 13   7 9 + 11 13   7 9 11 + 13  7 9 11 13 + This follows from the fact that a sum of consecutive odd integers is a square. So to get the sum of squares P2 (Th(π)) of the degrees of the threshold graph associated with the distinct partition π, sum the values in the numerical matrix above that occur in the positions with black dots. Of course, an adjacency matrix is symmetric. So if we use only the black dots in the upper triangular part, then we must replace the (i, j)-entry in the upper-triangular part of the matrix above with the sum of the (i, j)- and the (j, i)-entry, which gives the following matrix:       E=      + 2 + 4 6 8 10 12 6 8 10 12 14 + 10 12 14 16 + 14 16 18 + 18 20 + 22 + 14 16 18 20 22 24 26 +       .      (6) Thus, P2 (Th(π)) = 2(1, 2, 3, . . .) · δ(π). Lemma 2 is proved. 4 Proofs of Theorems 1 and 2 Theorem 1 is an immediate consequence of Theorem 2 (and Lemmas 1 and 2). And Theorem 2 can be proved using the following central lemma: Lemma 3 Let π = (v − 1, c, c − 1, . . . , b j, . . . , 2, 1) be an optimal partition in Dis(v, e), where e − (v − 1) = 1 + 2 + · · · + c − j ≥ 4 and 1 ≤ j ≤ c < v − 2. Then j = c and 2c ≥ v − 1 so that π = (v − 1, c − 1, c − 2, . . . , 2, 1). We defer the proof of Lemma 3 until Section 5 and proceed now with the proof of Theorem 2. The proof of Theorem 2 is an induction on v. Let π be an optimal partition in Dis(v, e), then π c is optimal in Dis(v, e0 ). One of the partitions, π, π c contains the part v − 1. We may assume without loss of generality that π = (v − 1 : µ), where µ is a partition in Dis(v − 1, e − (v − 1)). The cases where µ is a decreasing partition of 0, 1, 2, and 3 will be considered later. For now we shall assume that e − (v − 1) ≥ 4. Since π is optimal, it follows that µ is optimal and hence by the induction hypothesis, µ is one of the following partitions in Dis(v − 1, e − (v − 1)): 1.1a µ1.1 = (v − 2, . . . , k 0 + 1, j 0 ), the quasi-star partition for e − (v − 1), 1.2a µ1.2 = (v − 2, . . . , 2k 0 \ − j 0 − 1, . . . , k 0 − 1), if k 0 + 1 ≤ 2k 0 − j 0 − 1 ≤ v − 2, 1.3a µ1.3 = (v − 2, . . . , k 0 + 1, 2, 1), if j 0 = 3, 2.1a µ2.1 = (k1 , k1 − 1, . . . , jb1 , . . . , 2, 1), the quasi-complete partition for e − (v − 1), 2.2a µ2.2 = (2k1 − j1 − 1, k1 − 2, k1 − 3, . . . 2, 1), if k1 + 1 ≤ 2k1 − j1 − 1 ≤ v − 2, 2.3a µ2.3 = (k1 , k1 − 1, . . . , 3), if j1 = 3, where e − (v − 1) = 1 + 2 + · · · + k1 − j1 ≥ 4, with 1 ≤ j1 ≤ k1 . In symbols, π = (v − 1, µi.j ), for one of the partitions µi.j above. For each partition, µi.j , we will show that (v − 1, µi.j ) = πs.t for one of the six partitions, πs.t , in the statement of Theorem 2. The first three cases are obvious: (v − 1, µ1.1 ) = π1.1 (v − 1, µ1.2 ) = π1.2 (v − 1, µ1.3 ) = π1.3 . Next suppose that µ = µ2.1 , µ2.2 , or µ2.3 . The partitions µ2.2 and µ2.3 do not exist unless certain conditions on k1 , j1 , and v are met. And whenever those conditions are met, the partition µ2.1 is also optimal. Thus π1 = (v − 1, µ2.1 ) is optimal. Also, since e − (v − 1) ≥ 4, then k1 ≥ 3. There are two cases: k1 = v − 2, k1 ≤ v − 3. If k1 = v − 2, then µ2.2 does not exist and ½ π2.1 , if µ = µ2.1 (v − 1, µ) = π1.1 , if µ = µ2.3 . If k1 ≤ v − 3, then by Lemma 3, π1 = (v − 1, k1 − 1, . . . , 2, 1), with j1 = k1 and 2k1 ≥ v − 1. We will show that k = k1 + 1 and v − 1 = 2k − j − 1. The above inequalities imply that µ ¶ k1 + 1 = 1 + 2 + · · · + k1 ≤ e 2 µ ¶ k1 + 1 = − k1 + (v − 1) 2 µ ¶ µ ¶ k1 + 1 k1 + 2 < + (k1 + 1) = . 2 2 But k is the unique integer satisfying ¡k¢ 2 ≤e< ¡k+1¢ 2 . Thus k = k1 + 1. It follows that µ ¶ k+1 e = (v − 1) + 1 + 2 + · · · + (k − 2) = − j, 2 and so 2k − j = v. We now consider the cases 2.1a, 2.2a, and 2.3a individually. Actually, µ2.2 does not exist since k1 = j1 . If µ = µ2.3 , then µ = (3) since k1 = j1 = 3. This contradicts the assumption that µ is a partition of an integer greater than 3. Therefore µ = µ2.1 = (k1 , k1 − 1, . . . , jb1 , . . . , 2, 1) = (k − 2, k − 3, . . . 2, 1), since k1 = j1 and k = k1 + 1. Now since 2k − j − 1 = v − 1 we have ¡ ¢ ¡ ¢ ½ π2.1 if e = v2 or e = v2 − (v − 2) π = (2k − j − 1, k − 2, k − 3, . . . 2, 1) = . π2.2 otherwise. Finally, if µ is a decreasing partition of 0, 1, 2, or 3, then either π = (v − 1, 2, 1) = π1.3 , or π = (v − 1) = π1.1 , or π = (v − 1, j 0 ) = π1.1 for some 1 ≤ j 0 ≤ 3. Now, we prove that π1.2 and π1.3 (if they exist) have the same diagonal sequence as π1.1 (which always exists). This in turn implies (by using the duality argument mentioned in Section 3) that π2.2 and π2.3 also have the same diagonal sequence as π2.1 (which always exists). We use the following observation. If we index the rows and columns of the adjacency matrix Adj(π) starting at zero instead of one, then two positions (i, j) and (i0 , j 0 ) are in the same diagonal if and only if the sum of their entries are equal, that is, i + j = i0 + j 0 . If π1.2 exists then 2k 0 − j 0 ≤ v. Applying the previous argument to π1.1 and π1.2 , we observe that the top row of the following lists shows the positions where there is a black dot in Adj(π1.1 ) but not in Adj(π1.2 ) and the bottom row shows the positions where there is a black dot in Adj(π1.2 ) but not in Adj(π1.1 ). (v − k 0 − 2, v − 1) (v − 1 − k 0 , v − 2) ... ... (v − k 0 − t, v − 1) (v − 1 − k 0 , v − t) ... ... (v − k 0 − (k 0 − j 0 ), v − 1) (v − 1 − k 0 , v − (k 0 − j 0 )). Each position in the top row is in the same diagonal as the corresponding position in the second row. Thus the number of positions per diagonal is the same in π1.1 as in π1.2 . That is, δ (π1.1 ) = δ (π1.2 ). Similarly, if π1.3 exists then k 0 ≥ j 0 = 3. To show that δ (π1.1 ) = δ (π1.3 ) note that the only position where there is a black dot in Adj(π1.1 ) but not in Adj(π1.3 ) is (v − 1 − k 0 , v − 1 − k 0 + 3), and the only position where there is a black dot in Adj(π1.3 ) but not in Adj(π1.1 ) is (v − k 0 , v − 1 − k 0 + 2). Since these positions are in the same diagonal then δ (π1.1 ) = δ (π1.3 ). Theorem 2 is proved. 5 Proof of Lemma 3 There is a variation of the formula for P2 (Th(π)) in Lemma 2 that is useful in the proof of Lemma 3. We have seen that each black dot in the adjacency matrix for a threshold graph contributes a summand, depending on the location of the black dot in the matrix E in (6). For example, if π = (3, 1), then the part of (1/2)E that corresponds to the black dots in the adjacency matrix Adj(π) for π is     + • • • + 1 2 3    + • ◦ + 3 ,   Adj((3, 1)) =      + ◦ + + + Thus P2 (Th(π)) = 2(1 + 2 + 3 + 3) = 18. Now if we index the rows and columns of the adjacency matrix starting with zero instead of one, then the integer appearing in the matrix (1/2)E at entry (i, j) is just i + j. So we can compute P2 (Th(π)) by adding all of the positions (i, j) corresponding to the positions of black dots in the upper-triangular part of the adjacency matrix of Th(π). What are the positions of the black dots in the adjacency matrix for the threshold graph corresponding to a partition π = (a0 , a1 , . . . , ap )? The positions corresponding to a0 are (0, 1), (0, 2), . . . , (0, a0 ) and the positions corresponding to a1 are (1, 2), (1, 3) . . . , (1, 1 + a1 ). In general, the positions corresponding to at in π are (t, t + 1), (t, t + 2), . . . , (t, t + at ). We use these facts in the proof of Lemma 3. Let µ = (c, c − 1, . . . , b j, . . . , 2, 1) be the quasi-complete partition in Dis(v, e − (v − 1)), where 1 ≤ j ≤ c < v − 2 and 1 + 2 + . . . + c − j ≥ 4. We deal with the cases j = 1, j = c, and 2 ≤ j ≤ c − 1 separately. Specifically, we show that if π = (v − 1 : µ) is optimal, then j = c and π = (v − 1, c − 1, . . . , 2, 1), with 2c ≥ v − 1. Arguments for the cases are given below. (7) 5.1 j = 1 : µ = (c, c − 1, . . . , 3, 2) Since 2 + 3 + . . . + c ≥ 4 then c ≥ 3. We show that π = (v − 1 : µ) is not optimal. In this case, the adjacency matrix for π has the following form: 0 1 2 .. . c−1 c c+1 .. . 0 1 2 ··· + • • ··· + • ··· + .. . c • • • • ··· v − 1 • ··· • ◦ ··· ◦ + • • ◦ ··· + ◦ ◦ ··· + ◦ ··· .. . v−1 5.1.1 ◦ ◦ ◦ .. . ◦ + 2c ≤ v − 1 Let π 0 = (v − 1, 2c − 1, c − 2, c − 3, . . . , 3, 2). The parts of π 0 are distinct and decreasing since 2c ≤ v − 1. Thus π 0 ∈ Dis(v, e). The adjacency matrices Adj(π) and Adj(π 0 ) each have e black dots, many of which appear in the same positions. But there are differences. Using the fact that c−1 ≥ 2, the first row of the following list shows the positions in which a black dot appears in Adj(π) but not in Adj(π 0 ). And the second row shows the positions in which a black dot appears in Adj(π 0 ) but not in Adj(π): (2, c + 1) (3, c + 1) · · · (c − 1, c + 1) (1, c + 2) (1, c + 3) · · · (1, 2c − 1) (c − 1, c) (1, 2c) For each of the positions in the list, except the last ones, the sum of the coordinates for the positions is the same in the first row as it is in the second row. But the coordinates of the last pair in the first row sum to 2c − 1 whereas the coordinates of the last pair in the second row sum to 2c + 1. It follows that P2 (π 0 ) = P2 (π) + 4. Thus, π is not optimal. 5.1.2 2c > v − 1 Let π 0 = (v − 2, c, c − 1, . . . , 3, 2, 1). Since c < v − 2, the partition π 0 is in Dis(v, e). The positions of the black dots in the adjacency matrices Adj(π) and Adj(π 0 ) are the same but with only two exceptions. There is a black dot in position (0, v − 1) in π but not in π 0 , and there is a black dot in position (c, c + 1) in π 0 but not in π. Since c + (c + 1) > 0 + (v − 1), π is not optimal. 5.2 j = c : µ = (c − 1, . . . , 2, 1) Since 1 + 2 + · · · + (c − 1) ≥ 4, then c ≥ 4. We will show that if 2c ≥ v − 1, then π has the same diagonal sequence as the quasi-complete partition. And if 2c < v − 1, then π is not optimal. The adjacency matrix for π is of the following form: 0 1 .. . 0 1 2 ··· c + • • ··· • • ··· + • .. . c ··· v − 1 • ◦ ◦ + • ◦ ··· ◦ + ◦ ··· ◦ + ··· .. . ◦ + v−1 5.2.1 • 2c ≥ v − 1 The quasi-complete partition in G(v, e) is π 0 = (c + 1, c, . . . , b k, . . . , 2, 1), where k = 2c − v + 2. To see this, notice that 1 + 2 + · · · + c + (c + 1) − k = 1 + 2 + · · · + (c − 1) + (v − 1) for k = 2c − v + 2. Since 2c ≥ v − 1 and c < v − 2, then 1 ≤ k < c and π 0 ∈ Dis(v, e). To see that π and π 0 have the same diagonal sequence, we again make a list of the positions in which there is a black dot in Adj(π) but not in Adj(π 0 ) (the top row below), and the positions in which there is a black dot in Adj(π 0 ) but not in Adj(π) (the bottom row below): (0, c + 2) (0, c + 3) · · · (0, c + t + 1) · · · (1, c + 1) (2, c + 1) · · · (t, c + 1) (0, v − 1) · · · (v − c − 2, c + 1). Each position in the top row is in the same diagonal as the corresponding position in the bottom row, that is, 0 + (c + t + 1) = t + (c + 1). Thus the diagonal sequences δ(π) = δ(π 0 ). 5.2.2 2c < v − 1 In this case, let π 0 = (v − 1, 2c − 2, c − 3, . . . , 3, 2). And since 2c − 2 ≤ v − 3, the parts of π 0 are distinct and decreasing. That is, π 0 ∈ Dis(v, e). Using the fact that c − 2 ≥ 2, we again list the positions in which there is a black dot in Adj(π) but not in Adj(π 0 ) (the top row below), and the positions in which the is a black dot in Adj(π 0 ) but not in Adj(π): (2, c) (3, c) · · · (c − 1, c) (c − 2, c − 1) (1, c + 1) (1, c + 2) . . . (1, 2c − 2) (1, 2c − 1). All of the positions but the last in the top row are on the same diagonal as the corresponding position in the bottom row: t + c = 1 + (c − 1 + t). But in the last positions we have (c − 2) + (c − 1) = 2c − 3 and 1 + (2c − 1) = 2c. Thus P2 (π 0 ) = P2 (π) + 6 and so π is not optimal. 5.3 1 < j < c : µ = (c, c − 1, . . . , b j, . . . , 2, 1) 0 1 .. . ··· ··· c−1 c c+1 c+2 0 1 2 + • • + • ··· • • • ··· • • • ◦ ◦ c−j c−j+1 .. . v−1 We will show that π = (v − 1, c, c − 1, . . . , b j, . . . , 2, 1) is not optimal. The adjacency matrix for π has the following form: • • ◦ ··· ◦ .. . c−1 c c+1 .. . • ◦ ◦ ··· ◦ + • ◦ ◦ ··· ◦ + ◦ ◦ ··· ◦ + ◦ ··· ◦ + .. . + v−1 There are two cases. 5.3.1 2c > v − 1 \ Let π 0 = (v − r, c, c − 1, . . . , j + 1 − r, . . . , 2, 1), where r = min(v − 1 − c, j). Then r > 1 because j > 1 and c < v − 2. We show that π 0 ∈ Dis(v, e) and P2 (π 0 ) > P2 (π). In order for π 0 to be in Dis(v, e), the sum of the parts in π 0 must equal the sum of the parts in π: 1 + 2 + . . . + c + (v − r) − (j + 1 − r) = 1 + 2 + . . . + c + (v − 1) − j. And the parts of π 0 must be distinct and decreasing: v − r > c > j + 1 − r > 1. The first inequality holds because v−1−c ≥ r. The last two inequalities hold because c > j > r > 1. Thus π 0 ∈ Dis(v, e). The top row below lists the positions where there is a black dot in Adj(π) but not in Adj(π 0 ); the bottom row lists the positions where there is a black dot in Adj(π 0 ) but not in Adj(π): (0, v − 1) (c − j + r − 1, c + 1) ··· ··· (0, v − t) (c − j + r − t, c + 1) ··· ··· (0, v − r + 1) (c − j + 1, c + 1). Since r > 1 then the lists above are non-empty. Thus, to ensure that P2 (π 0 ) > P2 (π), it is sufficient to show that for each 1 ≤ t ≤ r − 1, position (0, v − t) is in a diagonal to the left of position (c − j + r − t, c + 1). That is, 0 < [(c − j + r + 1 − t) + (c + 1)] − [0 + (v − t)] = 2c + r − v − j, or equivalently, v − 2c + j − 1 ≤ r = min(v − 1 − c, j). The inequality v −2c+j ≤ v −1−c holds because j < c, and v −2c+j ≤ j holds because v −1 < 2c. It follows that π is not an optimal partition. 5.3.2 2c ≤ v − 1 Again we show that π = (v − 1, c, c − 1, · · · , b j, · · · , 2, 1) is not optimal. Let π 0 = (v − 1, 2c − 2, c − 2, · · · , j[ − 1, · · · , 2, 1). The sum of the parts in π equals the sum of the parts in π 0 . And the partition π 0 is decreasing: 1 ≤ j − 1 ≤ c − 2 < 2c − 2 < v − 1. The first three inequalities follow from the assumption that 1 < j < c. And the fourth inequality holds because 2c ≤ v − 1. So π 0 ∈ Dis(v, e). The adjacency matrices Adj(π) and Adj(π 0 ) differ as follows. The top rows of the following two lists contain the positions where there is a black dot in Adj(π) but not in Adj(π 0 ); the bottom row lists the positions where there is a black dot in Adj(π 0 ) but not in Adj(π). List 1 (2, c + 1) (1, c + 2) ··· ··· (t, c + 1) (1, c + t) ··· ··· (c − j, c + 1) (1, 2c − j) List 2 (c − j + 1, c) (1, 2c − j + 1) ··· ··· (c − j + t, c) (1, 2c − j + t) ··· ··· (c − 1, c) (1, 2c − 1). Each position, (t, c + 1) (t = 2, . . . , c − j), in the top row in List 1 is in the same diagonal as the corresponding position, (1, c + t), in the bottom row of List 1. Each position, (c − j + t, c) (t = 1, . . . , j − 1), in the top row of List 2 is in a diagonal to the left of the corresponding position, (1, 2c − j + t) in the bottom row of List 2. Indeed, (c − j + t) + c = 2c − j + t < 2c − j + t + 1 = 1 + (2c − j + t). And since 1 < j, List 2 is not empty. It follows that P2 (π 0 ) > P2 (π) and so π is not a optimal partition. The proof of Lemma 3 is complete. 6 Proof of Theorem 6 and Corollaries 1 and 2 The ¡notation in this section changes a little from that used in Section 1. In Section 1, we write ¢ e = k+1 − j, with 1 ≤ j ≤ k. Here, we let t = k − j so that 2 µ ¶ k e= + t, (8) 2 with 0 ≤ t ≤ k − 1. Then Equation (1) is equivalent to C(v, e) = C(k, t) = (k − t)(k − 1)2 + tk 2 + t2 = k(k − 1)2 + t2 + t(2k − 1). (9) Before proceeding, we should say that the abuse of notation in C(v, e) = C(k, t) should not cause confusion as it will be clear which set of parameters (v, e) vs. (k, t) are being used. Also notice that if we were to expand the range of t to 0 ≤ t ≤ k, that is allow t = k, then the representation of e in Equation (8) is not unique: µ ¶ µ ¶ k k+1 e = +k = + 0. 2 2 But the value of C(v, e) is the same in either case: C(k, k) = C(k + 1, 0) = (k + 1)k 2 . Thus we may take 0 ≤ t ≤ k. We begin the proofs now. At the beginning of Section 2.5, we showed that S(v, e) = C(v, e) for e = 0, 1, 2, 3. Also note that, when m is an integer, Diff(v, m) = 0. We now compare S(v, e) with C(v, e) for 4 ≤ e < m. The first task is to show that S(v, e) > C(v, e) for all but a few values of e that are close to m. We start by finding upper and lower bounds on S(v, e) and C(v, e). Define U (e) = U (k, t) = √ e( 8e + 1 − 1) and µµ ¶ ¶³ ´ p k (2k − 1)2 + 8t − 1 . +t 2 The first lemma shows that U (e) is an upper bound for C(v, e) and leads to an upper bound for S(v, e). Arguments used here to obtain upper and lower bounds are similar to those in [Ni]. Lemma 4 For e ≥ 2 C(v, e) ≤ U (e) and S(v, e) ≤ U (e0 ) + (v − 1)(4e − v(v − 1)). It is clearly enough to prove the first inequality. The second one is trivially obtained from Equation (2) linking the values of S(v, e) and C(v, e). ¡ ¢ ¡ ¢ Proof. We prove the inequality in each interval k2 ≤ e ≤ k+1 2 . So fix k ≥ 2 for now. We make yet another change of variables to get rid off the square root in the above expression of U (k, t). Set t(x) = (x2 − (2k − 1)2 )/8, for 2k − 1 ≤ x ≤ 2k + 1. Then U (k, t(x)) − C(k, t(x)) = ¡ ¢ 1 (x − (2k − 1))((2k + 1) − x) x2 + 4(k − 2)(k + x) − 1 , 64 which is easily seen to be positive for all k ≥ 2 and all 2k − 1 ≤ x ≤ 2k + 1. Now define L(e) = L(k, t) = √ e( 8e + 1 − 1.5) and µµ ¶ ¶³ ´ p k (2k − 1)2 + 8t − 1.5 . +t 2 The next lemma shows that L(e) is a lower bound for C(v, e) and leads to a lower bound for S(v, e). Lemma 5 For e ≥ 3 C(v, e) ≥ L(e) and S(v, e) ≥ L(e0 ) + (v − 1)(4e − v(v − 1)). Proof. As above, set t(x) = (x2 − (2k − 1)2 )/8, 2k − 1 ≤ x ≤ 2k + 1, and x(k, b) = 2k + b, −1 ≤ b ≤ 1. Then C(k, t(x(k, b))) − L(k, t(x(k, b))) = µ ¶ 2(k + 1) 2 4k(22k − 49) + 13 1 1 2 2 b (b + 4k − 4) + (4k − 7) b + + 64 32 4k − 7 64(4k − 7) This expression is easily seen to be positive for k ≥ 3. We are now ready to prove that S(v, e) > C(v, e) for 0 ≤ e ≤ m for all but a few small values and some values close to m. Lemma 6 Assume v ≥ 5. For 4 ≤ e < v we have C(v, e) < S(v, e). √ Proof. ¡As¢ we showed above in Lemma 4, e( 8e + 1 − 1) is an upper bound on C(v, e) for all 1 ≤ e ≤ v2 . Furthermore, it is easy to see that for 1 ≤ e < v we have S(v, e) = e2 + e. In fact, the quasi-star graph is optimal for 1 ≤ e < v. The rest is then straightforward. For 4 ≤ e, we have 0 < (e − 3)(e − 1) = (e + 2)2 − (8e + 1). Taking square roots and rearranging some terms proves the result. Lemma 7 Assume v ≥ 5. For v ≤ e ≤ m − 0.55v we have S(v, e) > C(v, e). Proof. Assume that 0 ≤ e ≤ m. Let e = m − d with 0 ≤ d ≤ m. By Lemmas 4 and 5, we have S(v, e) − C(v, e) L(e0 ) + (v − 1)(4e − v(v − 1)) − U (e) p p (m + d) 8(m + d) + 1 − (m − d) 8(m − d) + 1 ¶ ¶ µµ m 5 d+ . − 4(v − 1) + 2 2 ≥ = We focus on the first two terms. Set h(d) = p p (m + d) 8(d + m) + 1 − (m − d) 8(m − d) + 1. (10) By considering a real variable d, it is easy to see that h0 (d) > 0, h(2) (0) = 0, and h(3) (d) < 0 on the interval in question. Thus h(d) is concave down on 0 ≤ d ≤ m. We are comparing h(d) with the line (4(v − 1) + 5/2)d + m/2 on the interval [0.55v, m − v]. The concavity of h(d) allows to check only the end points. For d = m − v, we need to check ´ 1 ¡ p √ ¢ 1 ³ v (v − 3) 4v 2 − 12v + 1 − 2 8v + 1 > v 4v 2 − 21v + 7 2 4 It is messy and elementary to verify this inequality for v ≥ 9. For d = 0.55v we need to check ¶p µ 2 ¶p µ 2 v v + 0.3v − 0.8v 2v 2 + 2.4v + 1 − 2v 2 − 6.4v + 1 > v(2.325v − 0.95). 4 4 This inequality holds for v ≥ 29. This time the calculations are a bit messier, yet still elementary. For 4 < v ≤ 28, we verify the result directly by calculation. In Section 1, we introduced the value e0 = ¡k0 ¢ 2 . We now define e1 = f1 = f2 = µ ¶ k0 − 1 2 µ ¶ µ ¶ v k0 + 1 − 2 2 µ ¶ µ ¶ v k0 + 2 − . 2 2 The next lemma shows that those binomial coefficients and their complements are all we need to consider. Lemma 8 e1 , f2 < m − 0.55v. As a consequence S(v, e) > C(v, e) for all 4 ≤ e ≤ max{e1 , f2 }. We need a small result on the relationship between k0 and v first. The upper bound will be used later in this section. √ Lemma 9 2 2 Proof. Since ¡ ¢ v − 21 − ¡k0 ¢ 2 ≤m≤ 1 2 < k0 < ¡k0 +1¢ 2 √ 2 2 v + 21 . − 21 , we have 2k0 (k0 − 1) ≤ v 2 − v ≤ 2k0 (k0 + 1) − 2. Thus 2(k0 − 1/2)2 ≤ (v − 1/2)2 + 1/4 ≤ 2(k0 + 1/2)2 − 2. That is, √ sµ √ sµ ¶ ¶ 2 2 1 2 9 1 1 2 1 1 + − ≤ k0 ≤ + + . v− v− 2 2 4 2 2 2 4 2 The result follows using (v − 1/2)2 < (v − 1/2)2 + 9/4 and (v − 1/2)2 + 1/4 < v 2 . Proof of Lemma 8. Note that e1 = e0 − (k0 − 1) ≤ m − (k0 − 1) and f2 = f1 − (k0 + 1) < m − (k0 + 1) < m − (k0 − 1). Hence, it is enough to show that 0.55v < (k0 − 1). This follows from the previous lemma for v ≥ 12. For 5 ≤ v ≤ 11, we verify the statement by direct calculation. Next, we show that the difference function Diff(v, e) = S(v, e) − C(v, e) ¡ ¢ is piecewise linear on the intervals induced by the binomial coefficients k2 , 2 ≤ k ≤ v, and their ¡ ¢ ¡ ¢ complements v2 − k2 , 2 ≤ k ≤ v. In Section 2.7, we show a specific example. ¡ ¢ ¡ ¢ Lemma 10 As a function of e, the function Diff(v, e) is linear on the interval max{ k2 , v2 − ¡l+1¢ ¡k+1¢ ¡v¢ ¡ l ¢ 2 } ≤ e ≤ min{ 2 , 2 − 2 }. The line has slope − Proof. If e = ¢ 1¡ 1 − (2k − 3)2 − (2l − 3)2 + (2v − 5)2 . 4 ¡k+1¢ 2 − j with 1 ≤ j ≤ k, then it is easy to see from Equation (1) that µ ¶ k C(v, e + 1) − C(v, e) = 2e − 2 + 2k = 2e − k(k − 3). 2 Using Equations (2) and (9), we find that, if e0 = S(v, e + 1) − S(v, e) = ¡l¢ + c, 1 ≤ c ≤ l, then µ ¶ µ ¶ v l 2e + 4(v − 1) − 2 − 2l + 2 + 2. 2 2 2 (11) We now have (S(v, e + 1) − C(v, e + 1)) − (S(v, e) − C(v, e)) = k(k − 3) + l(l − 3) − (v − 1)(v − 4) + 2 = − ¢ 1¡ 1 − (2k − 3)2 − (2l − 3)2 + (2v − 5)2 . 4 The conclusion follows. Since we already know that Diff(v, e) > 0 for 4 ≤ e ≤ max{e1 , f2 }, and Diff(v, e) = 0 for e = 0, 1, 2, 3, or m, we can now focus on the interval I1 = (max{e1 , f2 }, m). The only binomial coefficients or complements of binomial coefficients that can fall into this interval are e0 and f1 . There are two possible arrangements we need to consider 1. e1 , f2 < e0 ≤ f1 < m and 2. f1 < e0 ≤ m. The next result deals with the first arrangement. Lemma 11 If e0 ≤ f1 < m, then q0 (v) > 0. Furthermore, S(v, e) ≥ C(v, e) for 0 ≤ e ≤ m with equality if and only if e = 0, 1, 2, 3, or m; or e = e0 and (2v − 3)2 − 2(2k0 − 1)2 = −1, 7. Proof. e0 ≤ f1 implies e0 ≤ m − k0 /2. By Lemma 9, we conclude that for v > 12, 4q0 (v) = 1 − 2(2k0 − 3)2 + (2v − 5)2 = 16(m − e0 ) − 16(v − k0 ) + 8 ≥ = 24k0 − 16v + 8 √ 24( 2/2(v − 1/2) − 1/2) − 16v + 8 √ √ (12 2 − 16)v − (6 2 + 4) > 0. ≥ For smaller values, we verify that q0 (v) > 0 by direct calculation. ¡ ¢ If e = f1 in Equation (9), and since e0 ≤ f1 < m, then k = k0 and t = f1 − k20 . Using Equation (2), Diff(v, f1 ) = (m − f1 )q0 (v) >¡0. Similarly, since f2 < e0 ≤ f1 , then for e = e00 in Equation (9), ¢ we have k = k0 + 1 and t = e00 − k02+1 . Again, using Equation (2), Diff(v, e0 ) = = (v 2 − 3v − 2k02 + 2k0 + 2)(v 2 − 3v − 2k02 + 2k0 )/4 2 2 2 (12) 2 ((2v − 3) − 2(2k0 − 1) + 1)((2v − 3) − 2(2k0 − 1) − 7)/64. Notice that Diff(v, e0 ) ≥ 0 since both factors in (12) are even and differ by 2. Equality occurs if and only if (2v − 3)2 − 2(2k0 − 1)2 = −1 or 7. Finally, observe that Diff(v, e1 ) > 0 and Diff(v, f2 ) > 0 by Lemmas 7 and 8, and e1 and f2 are both less than f1 . Hence Diff(v, e) ≥ 0 for e ∈ [max{e1 , f2 }, m] follows from the piecewise linearity of Diff(v, e). The rest follows from Lemma 7. Now we deal with the case f1 < e0 . There are three cases depending on the sign of q0 (v). All these cases require the following fact. If f1 < e0 , then for e0 ≤ e ≤ m in Equation ¡k0 ¢ ¡k0 ¢ (9), k = k0 0 0 and t = e − 2 . Since f1 < e ≤ m, for e in Equation (9), k = k0 and t = e − 2 . Thus, using Equation (2), Diff(v, e) = (m − e)q0 (v) (13) whenever f1 < e0 ≤ e ≤ m. This automatically gives the sign of Diff(v, e) near m. By the piecewise linearity of Diff(v, e) given by Lemma 10, the only thing remaining is to investigate the sign of Diff(v, f1 ). Lemma 12 Assume f1 < e0 and q0 (v) > 0. Then S(v, e) ≥ C(v, e) for 0 ≤ e ≤ m, with equality if and only if e = 0, 1, 2, 3, m. Proof. First, note that e1 ≤ f1 < e0 < m, since e1 > f1 occurs only if m = e0 and thus q0 (v) = 2 − 4(v − k0 ) < 0. For e0 ≤ e < m, by Equation (13), Diff(v, ¡ e) ¢= (m − e)q0 (v) > 0. Furthermore, if e = f1 in Equation (9), then k = k0 − 1 and t = f1 − k02−1 . Thus, by Equation (2), Diff(v, f1 ) = (−4k04 + 16k03 + 4v 2 k02 − 12vk02 − 8v 2 k0 + 4k0 − v 4 + 6v 3 + v 2 − 6v)/4, and Diff(v, f1 ) − Diff(v, e0 ) = (2k02 − v 2 + v)(−2 − 2k02 + 8k0 + v 2 − 5v)/2. The first factor is positive for v ≥ 15. This follows √ f1 < e0 . The second factor is positive √ because 2 from the fact that v < 2k0 + ( 2 + 1)/2 by Lemma 9, and −2 − 2k0 + 2k0 + v 2 − v ≥ 0 because e1 ≤ f1 : For v ≥ 15, −2 − 2k02 + 8k0 + v 2 − 5v = (−2 − 2k02 + 2k0 + v 2 − v) + 2(3k0 − 2v) ≥ 2(3k0 − 2v) > 0, Since Diff(v, e0 ) > 0, then Diff(v, f1 ) > 0 for v ≥ 15. The only case left to verify satisfying the conditions of this lemma is v = 14. In this case, f1 = 36 and Diff(14, 36) = 30 > 0. The previous two lemmas provide a proof of part 1 of Theorem 6. Lemma 13 Assume f1 < e0 and q0 (v) = 0. Then S(v, e) ≥ C(v, e) for 0 ≤ e ≤ m with equality if and only if e = 0, 1, 2, 3, e0 , e0 + 1, . . . , m. Proof. For e0 ≤ e ≤ m, by Equation (13), Diff(v, e) = (m − e)q0 (v) = 0. As in the previous lemma, for v ≥ 15 Diff(v, f1 ) − Diff(v, e0 ) = (2k02 − v 2 + v)(−2 − 2k02 + 8k0 + v 2 − 5v)/2 > 0 and thus Diff(v, f1 ) > 0. The only value of v < 15 satisfying the conditions of this lemma is v = 6 with f1 = 5, and Diff(6, 5) = 4 > 0. The previous lemma provides a proof for part 3 of Theorem 6. Lemma 14 Assume f1 < e0 ≤ m and q0 (v) < 0. Then S(v, e) ≥ C(v, e) for 0 ≤ e ≤ m − R0 and S(v, e) ≤ C(v, e) for m − R0 ≤ e ≤ m with equality if and only if e = 0, 1, 2, 3, m − R0 , m. Proof. For e0 ≤ e < m, by Equation (13), Diff(v, e) = (m − e)q0 (v) < 0. This time it is possible that f1 < e1 . In this case, by Lemmas 7 and 8, we know that Diff(v, f1 ), Diff(v, e1 ) > 0. Also, m = e0 and R0 = 0, implying Diff(v, e0 ) = 0 and Diff(v, e) > 0 for all e1 ≤ e < e0 = m − R0 = m. If e1 ≤ f1 , by Lemma 10, Diff(v, e) is linear as a function of e on the interval [f1 , e0 ]. Let −q1 (v) be the slope of this line. Since e1 < f1 < e0 ≤ m, then k = k0 and l = k0 in Lemma 10. Thus q1 (v) = (−1 − 2(2k0 − 4)2 + (2v − 5)2 )/4 = q0 (v) + 2k0 − 4 and Diff(v, f1 ) = (m − e0 )q0 (v) + (e0 − f1 )q1 (v). The line through the two points (e0 , Diff(v, e0 )) and (f1 , Diff(v, f1 )) crosses the x-axis at m − R0 . We now show that f1 < m − R0 < e0 , which in turn proves that Diff(v, f1 ) > 0. We have m − R0 = = q0 (v) q1 (v) 2k0 − 4 . m − (m − e0 ) q1 (v) e0 + (m − e0 ) (14) (15) Since e0 ≤ m and v > 4, then 1 k0 ≤ + 2 sµ ¶ sµ ¶ v 1 v−2 + <2+ , 4 2 2 (16) which is equivalent to q1 (v) > 0. Thus m − R0 < e0 by Equation (14). To prove f1 < m − R0 , according to Equation (15), we need to show µ ¶ k0 + 1 2k0 − 4 < (m − e0 ) − m. q1 (v) 2 After multiplying by q1 (v), the last inequality becomes ¶ µµ ¶ ¶ µ µ ¶ ¡ ¢ k0 + 1 k0 + 1 k0 (2k0 − 4) < − m (v − 2)(v − 3) − 2(k0 − 2)2 , m− + 2 2 2 which is equivalent to k0 (2k0 − 4) 2 < µµ ¶ ¶ k0 + 1 − m ((v − 2)(v − 3) − 2(k0 − 2)(k0 − 3)) . 2 ¡ ¢ Since f1 < e0 we know that k0 /2 < k02+1 − m. Also, Inequality (16) is equivalent to 2k0 − 4 < (v − 2)(v − 3) − 2(k0 − 2)(k0 − 3). Multiplying these two inequalities yields the result. The previous lemma provides a proof of part 2 of Theorem 6. The expression for m − R0 is sometimes an integer. Those v < 1000 for which m − R0 is an integer are 14, 17, 21, 120, 224, 309, 376, 393, 428, 461, 529, 648, 697, and 801. In the remaining part of this section, we prove Corollaries 1 and 2. Lemma 15 Assume that v > 4 and q0 (v) < 0. Then R0 ≤ αv where α = 1 − √ 2/2. Proof. We show that R0 ≤ αv for v > 4. Recall that R0 = (m − e0 )(2k0 − 4) . q1 (v, k0 ) Thus we need to show αvq1 (v, k0 ) − (m − e0 )(2k0 − 4) > 0. ¡ ¢ Define the function h(x) = αvq1 (v, x) − (m − x2 )(2x − 4). The interval for x is limited by the condition that q0 (v) < 0 which implies that √ √ 5 2 3 2 v− + < k0 . i1 := 2 4 2 √ Furthermore, since e0 ≤ m, we know that i2 := ( 2/2)v+1/2 > k0 . We show that h(x) is increasing on I := [i1 , i2 ]. Note that, since v > 4, √ h00 (x) = −6 − (4 − 2 2)v + 6x > 0 for x ∈ I. Hence h(x) is concave up on I. Furthermore µ ¶ ³ √ ´ 2 15 √ 11 √ 73 0 >0 2 v− 2+ h (i1 ) = 3 − 2 2 v + −10 + 2 4 8 for v ≥ 11, and hence h(x) > = h(i1 ) ³ √ ´ √ ´´ 1 ³³ −72 + 58 2 v + 23 6 − 5 2 > 0 32 for v ≥ 11. The only values of v greater than 4 and smaller than 11 for which q0 (v) < 0 are v = 7, 10. The result is easily verified in those two cases. How good is the bound R0 ≤ αv? Suppose there is a parameter β such that R0 ≤ βv with β < α. Assume q0 (v) = −2. There are pinfinitely many values of v for which this is true (see Section 9). In all of those cases k0 (v) = 1/2 (9 + (2v − 5)2 )/2 + 3/2. We have the following √ √ 2β − 2 + 1 ≥ 0 (βvq1 (v) − (m − e0 )(2k0 − 4))/v 2 → ´ as v → ∞. Thus β ≥ α and hence α the greatest number for which the bound on R0 holds. Since S(v, e) ≥ C(v, e) for all 1 ≤ e ≤ m − R0 , we proved Corollary 1. To prove Corollary 2, we need to investigate the other non-trivial case of equality in Theorem 6. It occurs when e = e0 and (2v − 3)2 − 2(2k0 − 1)2 = −1, 7. Notice that this implies m − e0 = = ¢ 1 ¡ (2v − 1)2 − 2(2k0 − 1)2 + 1 16 v−1 v or . 2 2 There are infinitely many values of v such that (2v − 3)2 − 2(2k0 − 1)2 = −1, and infinitely many values of v such that (2v − 3)2 − 2(2k0 − 1)2 = 7 (see Section 9). Thus the best we can say is that S(v, e) > C(v, e) for all 4 ≤ e < m − v/2, and Corollary 2 is proved. 7 Proof of Corollary 3 Recall that for each v, k0 (v) = k0 is the unique positive integer such that µ ¶ µ µ ¶ ¶ k0 + 1 k0 1 v < ≤ . 2 2 2 2 It follows that −1 ≤ (2v − 1)2 − 2(2k0 − 1)2 , and (2v − 1)2 − 2(2k0 + 1)2 ≤ −17. (17) Let us restrict our attention to the parts of the hyperbolas Hlow : (2v − 1)2 − 2(2k − 1)2 = −1, Hhigh : (2v − 1)2 − 2(2k + 1)2 = −17 that occupy the first quadrant as shown in Figure 10. Then each lattice point, (v, k0 ) is in the closed region bounded by Hlow below and Hhigh above. Furthermore, the sign of the quadratic form (2v − 5)2 − 2(2k − 3)2 + 1 determines whether the quasi-star graph is optimal in G(v, e) for all 0 ≤ e ≤ m. By Theorem 6, if (2v − 5)2 − 2(2k − 3)2 + 1 ≥ 0, then S(v, e) ≥ C(v, e) (and the quasi-star graph is optimal) for 0 ≤ e ≤ m. Thus, if the lattice point (v, k) is between Hhigh and the hyperbola H : (2v − 5)2 − 2(2k − 3)2 = −1, then the quasi-star graph is optimal in G(v, e) for all 0 ≤ e ≤ m. But if the lattice point (v, k0 ) is between H and Hlow , then there exists a value of e in the interval 4 ≤ e ≤ m such that the quasi-complete graph is optimal and the quasi-star graph is not optimal. Of course, if the lattice point (v, k0 ) is on H, then¡ the ¢ quasi-star graph is optimal for all 0 ≤ e ≤ m but the quasi-complete graph is also optimal for k20 ≤ e ≤ m. Apparently, the density limit n(v) v→∞ v lim from Corollary 3 depends on the density of lattice points (v, k) in the region between Hhigh and H. v 25 20 15 10 5 k 5 10 15 20 Figure 10: Hyperbolas (2v−1)2 −2(2k−1)2 = −1, (2v−1)2 −2(2k+1)2 = −17, (2v−5)2 −2(2k−3)2 = −1 √ We can give a heuristic argument to suggest that the limit is 2 − 2. The asymptotes for the three hyperbolas are ¶ µ √ 3 5 = 2 k− A:v− 2 2 ¶ µ √ 1 1 = 2 k− Alow : v − 2 2 ¶ µ √ 1 1 = , 2 k+ Ahigh : v − 2 2 and intersect the k-axis at k = klow = khigh = √ 6−5 2 4 √ 2− 2 4 √ −2 − 2 . 4 The horizontal distance between Ahigh and Alow is klow − khigh = 1 and the horizontal distance between Ahigh and A is k − khigh = 2 − √ 2. To make the plausibility argument rigorous, we need a theorem of Weyl [We, Satz 13, page 334], [Ko, page 92]: For any real number r, let hri denote the fractional part of r. That is, hri is the unique number in the half-open interval [0, 1) such that r − hri is an integer. Now let β be an irrational real number. Then the sequence hnβi, n = 1, 2, 3, . . ., is uniformly distributed in the interval [0, 1). In our problem, the point (v, k0 ) is between the hyperbolas Hlow and Hhigh and, with few exceptions, (v, k0 ) is also between the asymptotes Alow and Ahigh . To be precise, suppose that (v, k0 ) satisfies Inequalities (17). We need an easy fact from number theory here. Namely that y 2 − 2x2 ≡ −1 (mod 8) for all odd integers x, y. Thus 2(2k0 − 1)2 < (2v − 1)2 < 2(2k0 + 1)2 , unless (2v − 1)2 − 2(2k0 − 1)2 = −1. (These are the exceptions.) But for all other points (v, k0 ) we have ¶ ¶ µ µ √ 1 1 1 √ < v − < 2 k0 + . 2 k0 − 2 2 2 Thus and so √ µ ¶ 1 1 2 v− + − k0 < 1 0< 2 2 2 *√ µ + √ µ ¶ ¶ 1 1 1 1 2 2 v− + − k0 = v− + . 2 2 2 2 2 2 Next, consider the condition q0 (v, k0 ) ≥ 0, which is equivalent to (2v − 5)2 − 2(2k0 − 3)2 ≥ −1. Unless (2v − 5)2 − 2(2k0 − 3)2 = −1, q0 (v, k0 ) ≥ 0 is equivalent to + *√ µ ¶ √ 1 1 2 v− + > 2 − 1. 2 2 2 To summarize, if (v, k0 ) does not satisfy either of these Pell’s Equations (2v − 1)2 − 2(2k0 − 1)2 = −1, then q0 (v, k0 ) ≥ 0 if and only if √ 2−1< (2v − 5)2 − 2(2k0 − 3)2 = −1, + *√ µ ¶ 1 1 2 v− + < 1. 2 2 2 From Weyl’s Theorem, we know that the fractional part in the above inequality is uniformly distributed in the interval [0, 1). Since the density of the values of v for which √ (v, k0 ) is a√solution to one of the Pell’s Equations above is zero, then limv→∞ n(v)/v = 1 − ( 2 − 1) = 2 − 2. The proof of Corollary 3 is complete. 8 Proofs of Theorems 3, 4, and 5 We first prove Theorem 3. If π1.2 and π1.3 are optimal partitions, then according to Theorem 2, j 0 = 3, k 0 ≥ j 0 + 2 = 5, and so v ≥ 2k 0 − j 0 ≥ 7. In addition ¡ ¢ the quasi-star partition ¡v¢ is optimal, that¡is,¢ S(v, e) ≥ C(v, e). Thus by Corollary 2, either e ≥ v2 − 3 or e ≤ m + v/2 = 2 /2 + v/2. If ¡ ¢ v v 1 v 0 0 0 e ≥ 2 − 3 and since j = 3, then k ≤ 3 contradicting k ≥ 5. Thus e ≤ 2 2 + 2 . Since 2k 0 − 3 ≤ v ¡ ¢ ¡ 0 ¢ and e = v2 − k 2+1 + 3, then µ ¶ µ 0 µ ¶ ¶ k +1 (v + 3) /2 + 1 1 v v v ≤ 3+ + ≤ + . 2 2 2 2 2 2 Therefore 7 ≤ v ≤ 13. In this range of v, the only pairs (v, e) that satisfy all the required inequalities are (v, e) = (7, 9) or (9, 18). Using the relation between a graph and its complement described below Equation (2), we conclude that if π2.2 and π2.3 are optimal partitions, then (v, e) = (7, 12) or (9, 18). As a consequence, we see that the pair (9, 18) is the only candidate to have six different optimal partitions. This is in fact is the case. The six graphs and partitions are depicted in Figure 11. Also, the pairs (7, 9) and (7, 12) are the only candidates to have five different optimal partitions. For the pair (7, 9), the partitions π1.1 , π1.2 , π1.3 , π2.1 and π2.2 all exist and are optimal. However, π1.3 = π2.2 . Thus the pair (7, 9) only has four distinct optimal partitions. Similarly, for the pair (7, 12) the partitions π1.1 , π1.2 , π2.1 , π2.2 and π2.3 all exist and are optimal, but π1.2 = π2.3 . So there are no pairs with five optimal partitions, and thus all other pairs have at most four optimal partitions. Moreover, S(v, e) = C(v, e) is a necessary condition to have more than two optimal partitions, since any pair other than (7, 9) or (7, 12) must satisfy that both π1.1 and π2.1 are optimal. The proof of Theorem 3 is complete. Figure 11: (v, e) = (9, 18) is the only pair with six different optimal graphs. For all graphs, P2 (Th(πi.j )) = max(v, e) = C(v, e) = S(v, e) = 192 ¡ ¢ ¡ ¢ In Theorem 4, e = k2 = k+1 − k and thus j = k. Note that, if v > 5 and k satisfy Equation (4), 2 ¡ ¢ ¡ ¢ then k +2 < v < 2k −1, and so k ≥ 4. Thus e = v2 − k+2 2 +(2k +2−v) with 4 ≤ 2k +2−v ≤ k +1, that is, k 0 = k + 1 and j 0 = 2k + 2 − v. Hence, π1.1 = (v − 1, v − 2, . . . , k + 2, 2k + 2 − v) and π2.1 = (k − 1, ..., 1) (which always exist) are different because 2k + 2 − v ≥ 4 > 1. The partition π1.2 = (v − 2, ..., k) exists because k ≤ v − 3, and it is different to π2.1 because k ≥ 4 > 1 (π1.2 6= π1.1 by definition). Finally, the partitions π1.3 , π2.2 , and π2.3 do not exist because j 0 = 2k + 2 − v ≥ 4, k + 1 > k − 1 = 2k − j − 1, and j = k ≥ 4, respectively. Theorem 4 is proved. ¡ ¢ ¡ ¢ − 3. Moreover, since v > 9, then k > Now, if v and k satisfy Equation (5), then 21 v2 = k+1 2¡ ¡ ¢ ¢ ¡ ¢ ¡ ¢ − 3 = v2 − k+1 (v + 3)/2. Hence, in Theorem 5, e = m = 21 v2 = k+1 + 3, with k ≥ 3 2 2 0 0 because v > 1. That is, k = k and j = j = 3. Thus π1.1 = (v − 1, v − 2, ..., k + 1, 3), π1.3 = (v − 1, v − 2, ..., k + 1, 2, 1), π2.1 = (k − 1, k − 2, ..., 4, 3), and π2.3 = (k − 1, k − 2, ..., 4, 2, 1) all exist and are different because k = v does not yield a solution to (5). Also π1.2 and π2.2 do not exist because 2k − j − 1 = 2k 0 − j 0 − 1 = 2k − 4 > v − 1. Theorem 5 is proved. 9 Pell’s Equation Pell’s Equation V 2 − 2J 2 = P, (18) where P ≡ −1 (mod 8), appears several times in this paper. For example, a condition for the equality of S(v, e) and C(v, e) in Theorem 6 involves the Pell’s Equation (2v−5)2 −2(2k0 −3)2 = −1. And in Theorem 5, we have (2v − 1)2 − 2(2k + 1)2 = −49. There are infinitely many solutions to each of these equations. In each instance, V and J in Equation (18) are positive odd integers and P ≡ −1 (mod 8) . The following lemma describes the solutions to the fundamental Pell’s Equation. Lemma 16 [HW] All positive integral solutions of V 2 − 2J 2 = −1 are given by (19) √ √ √ V + J 2 = (1 + 2)(3 + 2 2)n , where n is a nonnegative integer. It follows from the lemma that if (V, J) is a solution to Equation (19), then both V and J are odd. We list the first several solutions to Equation (19): V J 1 1 7 5 41 29 239 169 1393 985 Now let us consider the equation (2v − 3)2 − 2(2k − 1)2 = −1 from Theorem 4. Since all of the positive solutions (V, J) consist of odd integers, the pair (v, k) defined by v= V +3 , 2 k= J +1 2 are integers and satisfy Equation (4). Thus there is an infinite ¡ ¢family of values for v > 5 such that there are exactly 3 optimal partitions in Dis(v, e) where e = k2 . The following is a list of the first three values of v, k, e in this family: 22 15 105 v k e 121 85 3570 698 493 121278 Next, consider Equation (5) from Theorem 5 and the corresponding Pell’s Equation: V 2 − 2J 2 = −49. √ A simple argument using the norm function, N (V + J 2) = V 2 − 2J 2 shows that all positive integral solutions are given by √ √ √ √ √ √ √ V + J 2 = (1 + 5 2)(3 + 2 2)n , (7 + 7 2)(3 + 2 2)n , or (17 + 13 2)(3 + 2 2)n , where n is a nonnegative integer. The first several solutions are 1 5 V J 7 7 17 13 23 17 49 35 103 73 137 . 97 k= J −1 2 Thus the pairs (v, k), defined by v= V +1 , 2 satisfy Equation¡(5). ¢ The first three members, (v, k, e) of this infinite family of partitions Dis(v, e) with v > 9, e = v2 /2, and exactly 4 optimal partitions are: v k e 12 8 33 25 17 150 52 36 663 69 48 1173 The Pell’s Equation 4q0 (v) = (2v − 5)2 − 2(2k0 − 3)2 + 1 = 0 (20) appears in Theorem 6. Here again there are infinitely many solutions to the equation (2v − 5)2 − 2(2k − 3)2 = −1 starting with: v k 2 1 2 2 3 1 3 2 6 4 23 16 122 . 86 The proof of Corollary 1 requires infinitely many solutions to the equation q0 (v) = −2, which is equivalent to the Pell’s Equation (2v − 5)2 − 2(2k − 3)2 = −9. (21) All its positive integral solutions are given by √ √ √ V +5 J +3 , , k= V + J 2 = (3 + 3 2)(3 + 2 2)n , 2 2 where n is a nonnegative integer. The first several solutions are v= 3 2 v k 12 8 63 44 360 254 2091 1478 The proof of Corollary 2 requires infinitely many solutions to the Pell’s Equation (2v − 3)2 − 2(2k − 1)2 = 7, (22) and infinitely many solutions to the Pell’s Equation (2v − 3)2 − 2(2k − 1)2 = −1. All positive integral solutions to (22) are given by √ √ √ V +3 J +1 , , v= k= V + J 2 = (3 + 2)(3 + 2 2)n , 2 2 where n is a nonnegative integer. The first several solutions are v k 3 1 4 2 8 5 15 10 39 27 (23) √ √ (5 + 3 2)(3 + 2 2)n , 80 56 We have shown that Equation (23) has infinitely many solutions, as it is the same equation that appears in Theorem 4. However, in Corollary 2, k must be k0 , the unique integer that satisfies Inequality (3). This condition is also necessary for Equations (20), (21), and (22). In other words, we must show that for v large enough, every solution (v, k) to one of the Equations (20), (21), or (22), satisfies Inequality (3). 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