Threshold Weights and Measures

Chapter 15 Threshold Weights and Measures 15.1

Introduction

In this chapter we consider three ways, in addition to the threshold dimension, to gauge by how much a given graph differs from being a threshold graph. The threshold weight of a graph was introduced by Wang and Williams [WW91] and also considered by Mahadev and Peled [MP91]. We assign a non-negative weight to each vertex and each edge of the graph, and define the score of a set of vertices as the sum of the weights of its vertices and of the edges induced by these vertices. We wish to choose the weights so that the score of each stable set is smaller than the score of each non-stable set, and the sum of the edge weights is minimal. Under a suitable normalization, the minimum value is the threshold weight of the graph, and by this definition it is zero if and only if the graph is a threshold graph. We discuss the threshold weight in Section 15.2. The threshold measure of a graph was introduced by Peled and Simeone [PS89a]. Here one assigns non-negative weights just to the vertices, as in the definition of a threshold graph, and considers that the score of a set of vertices is the sum of the weights of its vertices. In general we cannot make the score of each stable set smaller than the score of each non-stable set. Instead, we normalize the weights by requiring that the smallest score of a non-stable set be 1, and minimize the largest score of a stable set. The minimum value, which is smaller than 1 if and only if the graph is threshold, is the threshold measure of the graph. We discuss the threshold measure in Section 15.3. The threshold gap of a graph was 375

Threshold Weights and Measures

376

introduced by Hammer, Ibaraki and Simeone [HIS78, HIS81]. It is actually a property of the degree sequence of the graph, namely half of the smallest distance in the L 1 n o r m between the degree sequence and any threshold sequence of the same length. Thus it vanishes if and only if the given degree sequence is a threshold sequence. Arikati and Peled lAP94] introduced the majorization gap of the degree sequence as the minimum number of reverse unit transformations required to transform it into a threshold sequence. It turns out that the threshold gap is the same as the majorization gap. We discuss the threshold and majorization gaps in Section 15.4.

15.2

Threshold Weights

In this section, based on the work of Wang and Williams [WW91] and Mahadev and Peled [MP91], we define the threshold weight of a graph and discuss its properties. As we mentioned in the Introduction, to define the threshold weight of a graph G = (V, E), we assign a weight w~ >_ 0 to each v e r t e x i E V and a w e i g h t we >_ 0 to each edge e E E. The score o f a s e t S C_ V of vertices is computed as ~ i e v wi + ~eeE(S) we, where E(S) is the set of edges with both endpoints in S. We require that the score of each stable set be smaller than the score of each non-stable set. This requirement has the normalized form in which the score of each stable set is at most t - 1 and the score of each non-stable set is at least t for some threshold parameter t. By the non-negativity of the weights, it is sufficient to require this only for maximal stable sets and minimal non-stable sets (two adjacent vertices). We wish to minimize the sum of the edge weights, and its minimum value is defined as the threshold weight of G. We are thus led to the following formal definition.

Definition 15.2.1 The threshold weight W(G) of a graph G = (V, E) is the optimum value of the following linear programming problem. rain

~

w~

eEE

s.t.

~ wi ~ t - 1

for each maximal stable set S C_ V

iES

wi + wj + we >_ t for each edge e = ij E E

w>O.

(15.1)

15.2

Threshold Weights

377

An assignment (w; t) is called a feasible ( o p t i m a l ) t h r e s h o l d a s s i g n m e n t when it is feasible (optimal) for the linear programming problem (15.1). In general (15.1) involves an exponential number of constraints, since there are as many maximal stable sets. The graph G satisfies W(G) - 0 if and only if G is a threshold graph, since W(G) - 0 means that all the edge weights are zero. The assignment of zero vertex weights and unit edge weights is always feasible with t - 1, and this feasible threshold assignment has an objective function value equal to IEI, the number of edges. Therefore 0 < w ( a ) _< ]El for every graph G, where one extreme, 0, characterizes the threshold graphs. We call a graph G - (V, E) at the other extreme, i.e., one satisfying W ( G ) - IEI, ~ h~avy graph. In this sense, the heavy graphs are the most non-threshold graphs (of course, if E - O, then G is both a threshold graph and a heavy graph, but this is the only such case). A feasible threshold assignment remains feasible when restricted to an induced subgraph, and hence W ( H ) <_ W(G) for every induced subgraph H of G. Moreover, if C 1 , . . . , Ck are the connected components of G, then for the same reason ~ =k1 W(C~) <_ W(G). Recall that two edges are said to be in conflict if their endpoints induce an alternating 4-cycle. L e m m a 15.2.2 Let e and f be edges in conflict in G. Then every feasible threshold assignment (w; t) for G satisfies w~ + wf >_ 2. Consequently, if G is a non-threshold graph, then W(G) > 2. P r o o f . Let e - p q and f wp + wT _< t - 1;

rs, where pr and qs are nonedges. Then

wq+ws
wp + wq + w~ > t;

w~ + w~ + wf >_ t

and therefore

w~ + w f > 2 t - wp - w q - w ~ - w ~ > _ 2 .

D e f i n i t i o n 15.2.3 Let (w; t) be a feasible threshold assignment of a graph G. The edges e satisfying w~ < 1 are called the light edges of (w; t). Part (a) of Figure 15.1 illustrates a feasible threshold assignment whose light edges do not form a threshold graph, although the graph induced by their endpoints is threshold. Part (b) illustrates an optimal threshold assignment

Threshold Weights and Measures

378

(its optimality follows from Lemma 15.2.2) whose light edges form a threshold graph, but the graph induced by their endpoints is not threshold. However, the following lemma shows that the light edges of an optimal threshold assignment form a threshold subgraph (not necessarily an induced one). Figure 15.1" (a) a feasible threshold assignment whose light edges do not form a threshold graph. (b) an optimal threshold assignment the endpoints of whose light edges do not induce a threshold graph. t-4 t-5

1

1

0

3

0

A

w

./

x_

w

2

0

2

4 A

0

2

0

A

w

3

2

(b) L e m m a 15.2.4 Let ( w ; t ) be an optimal threshold assignment of a graph G - (V,E), and let L be the set of light edges of ( w ; t ) . Then (V,L) is a threshold graph. P r o o f . If (V, L) is not a threshold graph, it has an alternating 4-cycle with edges pq and rs and nonedges pr and qs. The nonedges pr and qs may be nonedges of G, or edges of G with weight at least 1. If pr is a nonedge of G, we have wp + w~ _< t - 1 by the constraints of (15.1), and if pr is an edge with wp~ _> 1, we have the same conclusion, otherwise wp~ could be decreased below 1, contradicting the optimality of the threshold assignment. Similarly wq + w~ _< t - 1. On the other hand, wp + wq + Wpq ~ t together with wpq < 1 imply that w p + w q > t - 1, and similarlyw~ + w ~ > t - 1. These four inequalities are inconsistent. ..

15.2

T h r e s h o l d Weights

379

The following result follows directly from the constraints (15.1). It is very useful in proving that certain graphs possessing symmetry properties are heavy. L e m m a 15.2.5 Let ab be an edge and cd a nonedge of a graph G. Every feasible threshold assignment (w; t) for G satisfying wc + Wd > Wa + Wb must satisfy Wab >_ 1. In particular, if e -- xy is an edge, zy is a none@e, and Wz > wx, then we > 1. The following corollary is an example of applying Lemma 15.2.5. C o r o l l a r y 15.2.6 All cycles of length 4 or more and their complements are heavy. Proof. Let G be the k-cycle Ck. Consider any optimal threshold assignment (w;t) of G. Each of the k cyclic permutations of the vertex weights of w around the cycle accompanied by a corresponding permutation of the edge weights is again an optimal threshold assignment. The average (w*; t) of these k optimal threshold assignments is yet another optimal threshold assignment (this follows from the convexity of the set of optimal threshold assignments). But (w*;t) enjoys the property that its vertex weights are constant. Let e be an arbitrary edge of G and let f be a nonedge adjacent to e, which exists since k > 4. By applying Lemma 15.2.5 to (w*; t), we see that w~ > 1. Thus G is heavy. The proof for the complements of cycles is similar. 9 The heavy graphs cannot be characterized by forbidden configurations, since they are not closed under taking induced subgraphs. In fact, every graph with at least one edge has an induced threshold subgraph with one edge, which is not heavy. The following theorem gives a convenient characterization of heavy graphs in terms of the optimum value of a linear programming problem having fewer variables and constraints than (15.1). T h e o r e m 15.2.7 A graph G [El

>

max

(V, E) is heavy if and only if

E dizi

iEv

s.t.

for each maximal stable set iES

Scv

z>_O,

(15.2)

T h r e s h o l d W e i g h t s and M e a s u r e s

380

where di is the degree of vertex i. Proof. The linear programming dual of (15.1) is max

~-~ ys S

s.t. S

e

Vi C V S ~i

(15.3)

e ~i

y~ _< 1

Ve C E

y>_O. The constraints of (15.3) imply that E s ys <_ IEI, equality holding if and only if y~ = 1 for all e C E. By the definition of heavy graphs and the duality theorem of linear programming, we thus have that G is heavy if and only if the following system of linear inequalities is solvable. ys -

IEI

S

~--~Ys >_ di

VicV

S ~i

y>_O. In other words, again using linear programming duality, if and only if IE]

_> min E Ys

=

max

s

s.t.

~ ys >_di Vi

E dizi iEv

s.t.

~ z i <_ 1 VS

S~i

iES

y_O

z_>O.

A direct proof of Theorem 15.2.7 not using LP duality can be found in [MP91]. We apply Theorem 15.2.7 to show that several types of graphs are heavy. L e m m a 15.2.8 Ps (the path on 5 vertices) is heavy.

15.2

Threshold Weights

381

P r o o f . Denote the vertices as 1 , . . . , 5 along the path. Let z be feasible for (15.2). Then Zl + z3 + zs _< 1, z2 + z4 _< 1, and z3 _< 1. After multiplying these inequalities by 1, 2, and 1, respectively, and adding, we obtain Zl + 2 ( z 2 + z3 + z4) + zs _< 4, which is the condition for a heavy graph in Theorem 15.2.7.

T h e o r e m 1 5 . 2 . 9 If G - ( V , E) is a regular graph and x(G) < G is heavy.

IVI/2,

th~

P r o o f . Let r be the degree of every vertex, and let S 1 , . . . , S X be a partition of V into X - x(G) stable sets. If z is feasible for (15.2), then x k=l iESk

iEV

which is the condition of Theorem 15.2.7 for G to be heavy. T h e o r e m 1 5 . 2 . 1 0 Let G = (V,E) have a proper vertex-coloring in which each color has at least two vertices and all the vertices of the same color have the same degree. Then G is heavy. In particular, the complete k-partite graph K ~ .....~nk is heavy if mj >_ 2 for each j, the perfect matching inK2 is heavy if m >_ 2, and so on.

P r o o f . Let '--q'l,-.., Sk be a partition of V into stable sets, and let Dj be the common degree of the vertices of Sj for each j - 1 , . . . , k. If z is feasible for (15.2), then k

iEV

j=l

k

Dy

E Dj

iCSj

j=l

d /2 -IEI, iEV

which is the condition of Theorem 15.2.7 for G to be heavy. 9 Recall that N ( x ) denotes the open neighborhood of a vertex x, namely the set of all neighbors of x. The next theorem has a construction that preserves heaviness. T h e o r e m 15.2.11 Let G be a heavy graph, let x be a vertex of G, and let H be obtained from G by adding a new vertex y such that N(y) C N ( x ) in H (y is not adjacent to x). Then H is heavy. In particular, if v is a non-isolated vertex of a heavy graph and we add a new vertex of degree 1 adjacent to v, the graph remains heavy.

Thre s hold Weights and Measures

382

P r o o f . Let di denote the degree of vertex i C V ( H ) in H. The degree of vertex i E V(G) in G is then given by

d'i

if i ~ N(y) _ f di, d i - 1, if i c N(y).

In particular, d2 - d~ >_ dy. Let z be a feasible solution of (15.2) for H. Then the following constitutes a feasible solution of (15.2) for G:

, {

zi -

zi, if/=fix z~ + zu, i f i - x .

By applying the constraints of (15.2) to z on H and Theorem 15.2.7 to z / on G, we obtain

E

di Zi --

iEV(H)

iq~N(y)U{x,y}

<<=

E

iq~N(y)U{x,y}

E

iq~N(y)W{x,y}

~

i~x,y

dizi nt- E dizi + iEN(y) ieN(y)

d',z~ + ~

d~z~ + dyzy ieN(y)

d'~z, + IN(v)I + d'~"

iEn(y)

d'~z~+ IN(y)I + d'~z"

_< IE(o)I + IN(y)i- IE(H)I. This is the condition for H to be heavy by theorem 15.2.7. T h e o r e m 15.2.12 Every forest having at least two non-singleton connected components is heavy. P r o o f . The forest can be obtained from a matching inK2 with m _> 2, which is heavy by Theorem 15.2.10, by repeatedly adding vertices of degree 1, which preserves heaviness by Theorem 15.2.11. 9 Recall that a star is a tree with at most one vertex of degree larger than 1. It is a threshold graph, and so its threshold weight is zero. A bistar is as tree with exactly two vertices of degree larger than 1; in other words, it is obtained from two disjoint stars by joining their centers. In the following, we show that the triangle-free graphs are heavy, with the exception of stars and bistars plus isolated vertices.

15.2

Threshold Weights

Theorem

15.2.13 I f G -

383

(V, E) is a bistar, then W ( G ) - IE[ - 1.

P r o o f . Let a and b be the vertices of degrees rn + 1 and n + 1, respectively (i.e., a is adjacent to b and to rn additional vertices of degree 1, and similarly for b). To show that W ( G ) = IEI- 1 = rn + n, we exhibit feasible solutions of the dual linear programming problems (15.1) and (15.3) that achieve the same objective function value. The solution of (15.1) has t = 2, w~ = 1 for each edge e with the exception ofw~b = 0, and wi = 0 with the exception of w~ = Wb = 1. The edge constraints are clearly satisfied, and so are the maximal stable set constraints, as is easy to verify for all the maximal stable sets, namely S~ = N ( a ) , Sb = N(b) and So = V - {a, b}. The objective function value is rn + n. The solution of (15.3) has y~ = 1 for each edge e with the exception of W~b = 0, and Yso = n, YSb = rn, YSo = 0. Again it is easy to verify the constraints and the fact that the objective function value is rn + n. .. Isolated vertices are immaterial to the question whether a graph is heavy, so in the next theorem we assume that there are none. T h e o r e m 15.2.14 Let G be a triangle-free graph without isolated vertices, which is not a star or a bistar. Then G is heavy. P r o o f . Let (w; t) be an optimal threshold assignment for G - (V, E), and let L be the set of light edges for it. If L - O, then clearly G is heavy, so assume that L 7~ 0 . By Lemma 15.2.4, the edges of L form a triangle-free threshold graph T, which is necessarily a star. Assume first that L consists of a single edge e. Since G is triangle-free, has no isolated vertices and is not a star or a bistar, it has an edge f in conflict with e, and by Lemma 15.2.2 w~ + w/ _> 2. Every other edge has weight of at least 1, and so the sum of the edge weights is at least ]E I and G is heavy. Now assume that L has more than one edge, and thus T is a star with more than one edge. Let a be the center of T. Since G is triangle-free, one of the cases below must hold. C a s e 1" a is on some cycle C of length 4. Let a, b, c, d be the vertices along C (b and d may or may not be in T), let V l , . . . , vk be the vertices of T other than b, d that are adjacent to b, and let U l , . . . , ut be the other vertices of T. Then a, c, b, d, V l , . . . , vk induce a complete bipartite graph K2,k+2, which is heavy by Theorem 15.2.10. By adding to it U l , . . . , ul we obtain an induced subgraph H of G in which U l , . . . , ul have degree 1, and so H is heavy by Theorem 15.2.11. Since (w; t) remains a feasible threshold assignment when

384

Threshold Weights and Measures

restricted to the induced subgraph H, the sum of the edge weights of the edges of H is at least IE(H)]. Since g contains all the light edges, every other edge has weight at least 1, and therefore the sum of all the edge weights of G is at least IE(G)I, so G is heavy. C a s e 2" a is on some cycle C of length 5, but not on any cycle of length 4. Then C is a chordless cycle and so is heavy by Corollary 15.2.6. The vertices of T not already on C are not adjacent to any vertex of C except for a, since G is triangle-free and a is not on any cycle of length 4. As in Case 1, these vertices and the vertices of C induce a heavy subgraph H of G containing all the light edges, and we proceed as before. C a s e 3" a is not on any cycle of G of length smaller than 6, and there are paths a, p, q and a, r, s having only the vertex a in common. Then q, p, a, r, s induce a chordless path of length 5, which is heavy by Lemma 15.2.8, and the vertices of T not already on it are not adjacent to any vertex of it except for a. As before, these vertices and the vertices of the path induce a heavy subgraph H of G containing all the light edges, so G is heavy. C a s e 4" a is not on any cycle of G of length smaller than 6, and there are no paths as in Case 3. If G has an edge e in a different connected component than the one containing T, then e and any light edge induce a matching 2K2, which is heavy by Theorem 15.2.10. Hence by Theorem 15.2.11 T and e form a heavy induced subgraph H of G containing all the light edges, and we are done as before. Therefore we may assume that G is connected. Since G is not a star or a bistar and we are in Case 4, there is a chordless path of length 4 starting from a. The vertices of this path, together with any vertex of T not already on it, induce a chordless path of length 5, and the argument continues as in Case 3. .. a clique C c_ V of G - (V,E) is called a big clique when IE(C)I > ] E ( V - C)I, where E ( X ) denotes the set of edges with both ends in X C_ V. The following lemma characterizes graphs possessing a big clique. L e m m a 15.2.15 A graph G - (V,E) has a big clique if and only if the following inequalities have an integral solution.

dizi > [El

where di is the degree of i E V

iEV

for each maximal stable set S C V iES

z>O.

(15.4)

15.2

Threshold Weights

385

P r o o f . Let z be integral and satisfy (15.4). Then each zi is 0 or 1, and the set C = {i C V : zi = 1} is a clique. Therefore

[El < ~ & z ~ - ~ & - 21E(C)I + IBI, iEV

iEC

where B is the set of all edges joining C to V - C. Therefore IE(C)I > ] E ( V - C)I , so C is a big clique. Conversely, if C is a big clique of G, its characteristic vector z satisfies (15.4) by a similar argument, tt C o r o l l a r y 1 5 . 2 . 1 6 A graph with a big clique is not heavy. P r o o f . This is immediate from Theorem 15.2.7 and Lemma 15.2.15. 9 It was conjectured that the converse of Corollary 15.2.16 holds. Indeed it holds for triangle-free graphs by Theorem 15.2.14, since stars and bistars have a big clique. However, it fails even f o r / Q - f r e e and Ks-free graphs, as the following examples show. Let Gk,~ denote the graph obtained from the k-cycle Ck by adding r dominating vertices. Then G9,2 is not heavy, and yet has no big clique. To see this, denote the successive vertices of C9 by u o , . . . , us and the two dominating vertices by vo, Vl. Then every maximal clique of G9,2 = (V, E) has the form C = {v0, Vl, Uj, Uj+ 1} (addition mod 9), and ]E(C)[ = 6 = [E(V - C)], so C is not a big clique. To show that G9,2 is not heavy, set zi = 1 for i = v0, Vl and zi = 1/4 for i = u 0 , . . . , Us. It is easy to verify that z is a feasible solution of (15.2) with objective function value larger than IE]. By a similar argument Gk,T is not heavy and has no big clique for each r -2,3mod4

if k - 3 + (~+2). The above counterexample is Ks-free, but not K4-free. There are also K4-free counterexamples. Consider the 5-wheel G5,1 and attach a path of length 2 to one of its vertices. Call the resulting graph an umbrella if the path is attached to the center, and a kite if it is attached to another vertex. It is easy to see that both the umbrella and the kite have no big cliques. Both of them are also not heavy by Theorem 15.2.7, as can be seen from the (optimum) solution of (15.2): zi = 1 if i is the center of the wheel, zi = 1/2 if i is any other vertex of the wheel, and zi = 0 if i is one of the other two vertices. The conjecture does hold for perfect graphs, as shown by the next theorem. Theorem

1 5 . 2 . 1 7 Every non-heavy perfect graph has a big clique.

Thre s hold Weights and Measures

386

P r o o f . Let G = (V, E) be a non-heavy perfect graph. By Theorem 15.2.7, since G is not heavy, inequalities (15.4) have a solution. This means that the optimum solutions of the linear programming problem (15.2) have a value larger than ]El, and hence this linear programming problem has a basic feasible solution z with value larger than IEI. Since G is perfect, every basic feasible solution is integer by Chvgtal's theorem [Chv75], so z must be integer. Hence G has a big clique by Lemma 15.2.15. m To close this section, we mention the following open p r o b l e m s . 1. We do not know if the problem of recognizing a heavy graph belongs to NP, nor do we know if it belongs to co-NP. 2. The problem of recognizing a graph with a big clique is in NP, but we do not know if it is in P. 3. Within the class of non-heavy graphs, which graphs, in addition to perfect graphs, have an integer optimal solution to the linear program (15.2)?

15.3

Threshold Measures

This section is based on the work of Peled and Simeone [PS89a]. By the definition of threshold graphs, a graph is threshold if and only if we can assign a non-negative weight to each vertex so that the total weight of each stable set of vertices is smaller than the total weight of each non-stable set of vertices. By the non-negativity of the weights it is enough to require this separation just for maximal stable sets and minimal non-stable sets (pairs of adjacent vertices). In a suitable normalization of the weights, this means that a graph G = (V, E) is a threshold graph if and only if there exist vertex weights wv >_ 0, v C V and a threshold r < 1 such that ~ v e S Wv <_ r for each maximal stable set S and w~ + Wv _> 1 for each edge u v C E . In other words, the minimum of r is smaller than 1 when subject to the maximal stable set and the minimal non-stable set constraints and the non-negativity of w. We therefore regard the minimum of r subject to the above constraints as a certain measure of the non-thresholdness of G. This motivates the following definition.

15.3

387

Threshold Measures

1 5 . 3 . 1 The t h r e s h o l d m e a s u r e v ( G ) of a graph G - (V, E) is the optimal value of the following linear programming problem.

Definition

min

s.t.

r

~ wv <_ ~-

for each maximal stable set S c_ V

yES

(15.5)

wu + w, >_ l for each edge uv C E w>O.

To emphasize the dependence of the optimum w on G, we sometimes denote it by w(G). Thus the optimum value of 15.5 is smaller than 1 if and only if G is a threshold graph. In that case, the minimum can be found by the very efficient and elegant algorithm of Section 1.3. The optimum value is equal to 1 if and only if G is a pseudothreshold graph. We have seen in Section 14.2 an algorithm to recognize these graphs and a characterization of their structure, from which it follows that when the optimum of (15.5) is 1, there is always an optimum solution for which the components of w have values of 0, 1/2 and 1. Figure 15.2 illustrates optimal solutions of (15.5) for a threshold graph and a non-threshold graph. Like (15.1), the linear programming problem (15.5) has exponentially many constraints, since in general there are as many maximal stable sets. This indicates that finding T(G) may be hard. We show below that this is indeed the case. We begin our general discussion of the threshold measure by determining the threshold measure of a disjoint union of two or more graphs. 15.3.2 w(G1 [..J G2) --~ (w(G1),w(G2)). Consequently T(G1 U G2) -- 7"(G1) -k- T(G2). This generalizes immediately to the union of any number of graphs.

Theorem

P r o o f . For k = 1,2, it is convenient to denote by Ak and Bk the edge-vertex incidence matrix and the maximal-stable-set-vertex incidence matrix of Gk. In other words, Ak has a column for each vertex and a row for each edge of Gk, and the corresponding entry is 1 or 0 according as the vertex is or is not an endpoint of the edge. Similarly Bk has a column for each vertex and a row for each maximal stable set, and the corresponding entry is 1 or 0 according

Threshold Weights and Measures

388

Figure 15.2" Threshold measures of (a) a threshold graph and (b) a non-threshold graph. Tz

8 9

5 4

T-1 4

1

1

9

l

9

12 / '

~ ~1

(b) as the vertex belongs or does not belong to the maximal stable set. We also let e k and fk denote all-1 column vectors of appropriate dimensions. Then the linear programming problem (15.5) for Gk can be written as follows, where tt k and u k denote row vectors of dual variables corresponding to the vector constraints of (15.6)" min #k

s.t.

Vk

A k w k >_ ek - - B k w k + rk.f k > 0 Wk>_O.

(15.6)

The linear programming dual of (15.6) is the following" T(Gk) lVk

-

max

13,k e k

s.t.

ttkAk -- v k B k <_ 0

vkf

- 1

tt k >_ O,

(15.7)

~'k >_ O.

To formulate the corresponding linear programming problems for G1 U G2, we use the fact that its maximal stable sets are precisely the unions of maximal

15.3

Threshold Measures

389

stable sets of G1 and G2. If i and j index the maximal stable sets of G1 and G;, respectively, then the maximal independent set constraint of (15.5) for G1 tO G2 reads ( B l W l ) i -~- (B2w2) j ~ T. We can therefore write (15.5) for G1 [-J G2 as follows, where jttl, jM2, Pl, P2 and uij are the dual variables to the corresponding constraints of (15.8), and where ~1 and rl2 are auxiliary column vectors introduced for convenience: r(GIUG2)

--

rain T s.t.

/-t2 Pl P2 Pij

A l W l ~ el A2w2 k e2 - B l W l + T~I -- 0 - B 2 w 2 + r12 - 0

(15.8)

--(?]1)i- (?]2)j -t'- T ~ 0 W 1 ~ 0,

W2 ~ 0.

The linear programming dual of (15.8) is as follows" 7-(G1 [,.J G2) Wl w2 (?]1)i

-

max s.t.

I-I,l e l -~ ~2e2 ~t 1A1 - PlB1 < 0 t t 2 A 2 - p2B2 < 0 ( P l ) i - E j 17ij = 0 (m)j

T

-

aj

= 0

~ i j vii = 1 tt 1 > 0 , tt 2 > 0 ,

aj >_ O.

The theorem asserts that if wl, r~ and w~, r~ are optima of (15.6) for k - 1,2, respectively, then (w~, w~), r~ + r~ defines an optimum of (15.8) when the auxiliaries rl~, rl~ are determined from the equations of (15.8). To show this, note that the constraints of (15.8) are satisfied, and then consider optima t t ; , v ; of (15.7). They satisfy the following complementary slackness conditions: tt*k(Akw*k -- ek) = 0 (15.10) v*k(--Bkw*k + r[~f k ) = O.

(15.11)

We define variables for (15.9) as follows:

It is easy to see that these variables define a feasible solution of (15.9). They also satisfy all the required complementary slackness conditions: (15.10),

390

Threshold Weights and Measures

(15.11), and the condition .;

+

-(,;),-

- 0

By theorem 15.3.2 we may assume without loss of generality that G is connected. We now consider how the threshold measure of a join of two graphs is related to their threshold measures. Unlike the case of the union, we only give estimates, for a reason that will be shown soon. T h e o r e m 1 5 . 3 . 3 Let G1 and G2 be disjoint graphs with largest stable set sizes OL1 and a2, respectively. Then C~lt~2 ~1 -t'- C~2

~ T(G1 (~ G 2 ) ~

~1 max(o~ 1 , c~2 ) 9

P r o o f . The upper bound for T(G10 G2) follows from the fact that assigning to every vertex a weight of g1 and taking T -- 71 max(a1, a2) defines a feasible solution of (15.5) for G1 | G2. To prove the lower bound, let Sk be a stable set of size ak in Gk for k = 1, 2. For every feasible solution w, T of (15.5) for G1 @G2, Sk possesses a vertex vk satisfying Wvk <_T/ak. The non-stable set {Vl, V2} satisfies wvl + WV2 ~ 1, which gives the desired bound. ,, C o r o l l a r y 15.3.4 If ~ 1 - - OL2 phic), then T(Ga | G 2 ) - a/2.

--

OL

(in particular, if G1 and G2 are isomor-

An immediate consequence of Corollary 15.3.4 is that it is hard to compute the threshold measure in general. Theorem

15.3.5 It is NP-hard to compute the threshold measure of a graph.

P r o o f . This follows from a(G) = 2T(G | G). " Unlike T(G~U G2), T(G1 @ G 2 ) i s not a function of T(G1) and r(G2). Indeed, one can easily find graphs G, G' satisfying T(G) = T(G') but a(G) 5r a(G'), as illustrated in Figure 15.3. Now if there were a function f satisfying r(G1 | G2) = f(r(G1), T(G2)), we would have the absurdity

f(T(G), T(G)) -- T(G 9G) - a(G)2 r a(G')2

=

a') -

15.3

391

Threshold Measures

Figure 15.3" Graphs with the same threshold measure but different largest stable set sizes. Optimal solutions of (15.5) are shown. r-1

c~-2

r-1

0

0

,iw

,w

1

0

0

,~,

1

1

c~-4 0

0

w

1

In the following we show that, in contrast to the situation described by Theorem 15.3.5 for general G, if G is bipartite, the linear programming problem (15.5), although still having exponentially-many essential constraints, is equivalent to a small, structured linear program with a symmetric coemcient matrix having 2's on the main diagonal and 0's and l's elsewhere. This small linear program can then be solved in polynomial time by the well-known methods of Khachian [Kha79] and Karmarkar [Kar84]. T h e o r e m 15.3.6 If G is a bipartite graph with an edge-vertex incidence matrix A, then

T(G)

-- min s.t.

eTpt A A r t t >_ e tt >_ O,

(15.12)

where e denotes an all-1 column vector of an appropriate dimension. Moreover, if It is an optimum of (15.12), then w - ATtt is an optimum of (15.5). Note that A A T is the edge adjacency matrix of G (or equivalently the adjacency matrix of the line-graph of G) with 2's along the main diagonal. P r o o f . By theorem 15.3.2 we may assume without loss of generality that G is connected, and in particular has no isolated vertices. Denote by B the maximal stable set vs. vertex incidence matrix of G and by f an all-1 column vector of an appropriate dimension. Then the definition (15.5) of r(G) can

392

Threshold

Weights

and Measures

be written as follows" T(G)

-

min

T

s.t.

Aw>e

(15.13)

Bw<_rf

w>O. m

The constraint B w <_ ~-f on w and ~- can be expressed as follows" r

>

max

wTb

s.t.

Ab < e

(15.14)

b binary, because A b <_ e and b binary imply that b is the characteristic vector of a stable set of vertices of G, so b is dominated by a row of B, and conversely. Since A has no zero columns, the constraint "b binary" of (15.14) can be relaxed to "b _> 0, integral". Since A is totally unimodular (because G is bipartite), the integrality constraint on b can be dropped. Thus by the duality theorem of linear programming, (15.14) is equivalent to the following: r

>

max

wTb

s.t.

A b <_ e

=

min

eTp,

s.t.

AT# > w

b_>O

(15.15)

#>_0.

Since A T >_ O, the inequality in (15.15) is equivalent to the existence of tt >_ 0 satisfying AT# _> w and r - eTpt. Therefore (15.13) is equivalent to the following: T(G)

--

min

7"

s.t.

Aw>e

(15.16)

ATIt >_ w eT# - r

#>o,

w>O.

The constraints involving w in (15.16), namely A w >_ e, ATtt >_ w and w >_ 0, can be replaced by A A T # >_ e, thus eliminating w. Indeed, this condition is clearly necessary, and conversely, if it holds and tt >_ 0, then w defined by w - A T # will satisfy its constraints in (15.16). This proves (15.12) and the moreover statement of the theorem. ..

15.3

Threshold Measures

393

There are several variations of Theorem 15.3.6. By the symmetry of A A T, the linear programming dual of (15.12) is given by "r(G)

-

max s.t.

err A A T v <_ e v>0.

(15.17)

Further, if G has edges, then w(G) > 0, which permits us to transform the variables/~ and r - e T # of (15.12) according to 1~-#

t T

1 T

and rewrite (15.12) as follows" =

max

t

=

s.t.

AAT)~ >_ te er,~ -- 1 ,~>0

max

min

( A A r,~)i

s.t.

eT~- 1 ,~>0.

(15.18)

Similarly, we can transform the variables v and r - e r u of (15.17) according to p=-- v S - - - -1 T

T

and rewrite (15.17) as follows" 1

=

min

s

s.t.

A A T p <_ se eTp--1 p>O

=

min o

max

"(AATp)j "----

s.t.

e Tp -1 p>_O.

3

(15.19)

The problem (15.19) is the linear programming dual of (15.18). The formulations (15.18) and (15.19) exhibit ~ 1 as the value of a two-person zero-sum game with payoff matrix A A T. The pure strategies of the players are the edges of G, and the payoff to the first player is the number of common endpoints of the two chosen edges. The optimal mixed strategies are the optimal ,~ and p.

:394

T h r e s h o l d W e i g h t s and M e a s u r e s

A vertex cover (or transversal) is a set of vertices meeting every edge. Using (15.17) and essentially reversing the steps in the proof of Theorem 15.3.6, we can obtain the following theorem, which together with Definition 15.3.1 gives a combinatorial rain-max relation for bipartite graphs:

T h e o r e m 15.3.7 If G -

( V , E ) is a bipartite graph with an edge-vertex incidence matrix A, then r ( G ) is the optimal value of the following linear programming problem:

max s.t.

r

~

zv >_ r

for each minimal vertex cover C c_ V

vec

(15.20)

zu+z~
for each edge uv 6 E

z>0.

Moreover, if v is an optimum of (15.17), then z of (15.20).

A T v is an o p t i m u m

P r o o f . By arguments similar to those of Theorem 15.3.6, we obtain the following from (15.17)" r(G) - m a x { r " A z < e, A T v < z, e T v -- r , z > O , v > 0}.

The existence of v > 0 satisfying A T v < z and e T v -- r is equivalent to T <_ m a x { e r r

A9T v <_ z,~' > O}

=

min{zT6

A9 6 >_ e, 6 >_ 0}

=

min{zT6

A9 6 > e , 6 binary}.

This means that with the vertex weights z, every vertex cover has a total weight of r or more, and establishes (15.20). The moreover part follows from the argument leading to the first equation of this proof. [] Figure 15.4 illustrates the optimal solutions of (15.12) and (15.17) for a tree. The optimal w and z are obtained from the illustrated tt and v, respectively, by summing them over the edges incident with each vertex. If a vector tt _> 0 satisfies A A T t t - e, then tt is an optimum of the linear programming problem (15.12) as well as of its dual (15.17). Stated differently, e v e r y / and j in (15.18) and (15.19) realizes the minimum or maximum there. This motivates the following definition.

15.3

395

Threshold Measures

3 Figure 15.4" Illustration of Theorem 15.3.6 for a tree. r - eTlu- eTv- 3"

oI

o

dk

A

0

1

lii1

3 A

1

5

0

A

1

0

3 A

0

A

1

7

D e f i n i t i o n 15.3.8 A bipartite graph with an edge-vertex incidence matrix A is called e q u i t a b l e if there exists a vector tt >_ 0 satisfying A A T t t - e. Clearly a graph is equitable if and only if all its connected components are equitable. Below we characterize the connected equitable bipartite graphs in terms of the maximum weight of a stable set of vertices under certain vertex-weights. The problem of finding a stable set with maximum weight in a bipartite graph has well-known network flow algorithms. We also give a characterization of equitable bipartite graphs in terms of a certain generalization of Hall's condition for the existence of a perfect matching in a bipartite graph. For trees the matrix A A T is non-singular, and we give very simple combinatorial algorithm for solving A A r t t = e for trees, leading to a characterization of equitable trees. We use the convention that the color classes of a bipartite graph are called red and blue. T h e o r e m 15.3.9 Let G be a connected bipartite graph having r >_ 1 red and b >_ 1 blue vertices. Assign a weight of ir to each red vertex and a weight of-g1

396

Threshold

Weights

and Measures

to each blue vertex. Then G is equitable if and only if the largest total weight of a stable set is 1.

P r o o f . Let i and j index the red and blue vertices of G - (V, E), respectively. We can write the linear programming problem (15.12) as follows, using the vertex variables w = ATtt, which by Theorem 15.3.6 constitute an o p t i m u m of (15.5): r(G)

-

min

~#~j i,j

s.t.

E

#ij -- Wi

J E

i

#ij -- Wj

(15.21)

Wi nt- Wj ~ 1

ijcE

#~j >_ 0

ijEE

#~j - 0

ij~E.

If G is equitable, then (15.21) has an o p t i m u m satisfying wi + wj = 1 for all ij C E. Then by the connectivity of G all the wi are equal and all the wj b for all i are equal. Since ~ i wi - ~ i j pij - ~ j wj, it follows that wi - -;-4-g r wj = ~+b for all j, and r(G) - r+b" ~b Since w is a feasible solution of (15.5) every stable set S must satisfy ~ e s Wv < ;~b" This means that with the vertex weights given in the theorem, the total weight of every stable set is 1 or less. Since the stable set of all red vertices has a total weight of 1, the "only if" part of the theorem follows. Conversely, if G satisfies the condition of the theorem, then

1

~

max

(

l~xi+

1 )

7" i

-b j

s.t.

xi nt- xj ~ 1

~xj

ij E E

(15.22)

x binary. Since the constraint matrix of (15.22) is totally unimodular and there are no isolated vertices, the condition "x binary" can be relaxed to x >_ 0. Then

15.3

Threshold

397

Measures

by linear programming duality there exist

)~ij, ij

C E, satisfying

Aij

b

ijEE

1

~j >_ j:ijEE

for all i

r

1

(~5.23)

for all j

i:ijEE

A~j > 0

ij C E.

But every solution of (15.23) must satisfy all its inequalities except for the non-negativity constraints with equality, as can be seen by summing the /-inequalities or the j-inequalities. Therefore the vector A*=

rb A>O r+b -

satisfies AATA * - e, which proves the "if" part. 9 Next we characterize equitable bipartite graphs by a generalization of Hall's condition for the existence of a perfect matching in a bipartite graph. T h e o r e m 15.3.10 Let G be a connected bipartite graph having r >_ 1 red and b >_ 1 blue vertices. Then G is equitable if and only if every set X of red vertices satisfies 1

IN(X)l >_ -~ F ~h~

(15.24)

N ( X ) d ~ o t ~ th~ ~ t of ~ighbo~ of th~ w~tic~ of X .

P r o o f . Denote by R and B the sets of red and blue vertices, and assign weights of !~ and g1 to their vertices, respectively. Condition (15.24) states that the total weight of N ( X ) is at least the total weight of X, for each XcR. To prove the "if" part, let S be any stable set of vertices, and take X - S N R, so that S and N ( X ) are disjoint. Then B includes the set ( S N B ) U N ( X ) , whose total weight is at least the total weight of ( S N B ) U X = S by (15.24). Hence B is a stable set of maximum weight, and so G is equitable by Theorem 15.3.9. To prove the "only if" part, assume that G is equitable, and therefore the weight of every stable set is at most 1 by Theorem 15.3.9. For every set

Threshold Weights and Measures

398

-IXl+w(b-IN(X)l

1 X C R, the set X U ( B - N ( X ) ) i s stable, and hence 1 _> r giving (15.24). 9 From Theorem 15.3.10 and Hall's condition we immediately obtain the following.

Corollary 15.3.11 Let G be a bipartite graph with an equal number of red and black vertices. Then G is equitable if and only if G has a perfect matching. To characterize equitable trees, we need some notation. We continue the convention that i, i0 index red vertices and j, jo index blue vertices. For a tree T and any edge e = ij of T, deleting e from T produces two subtrees. We denote by Tij the subtree containing i and by Tji the subtree containing

j.

P r o p o s i t i o n 15.3.12 Let T = (V, E) be a tree and let x be an assignment of vertex-weights, possibly negative, satisfying

E xi - E xj. i

(15.25)

j

Then there exist unique edge-weights Aij, ij E E satisfying ~ij

-

xi

for all i

-

xj

f o r aU j.

(15.26)

j:ijEE i:ijEE

An explicit formula for the ~iojo = E ieTio3o

xi-

,~ij i8 given by E

xj-

JET/0~ 0

E

xj-

jET30i 0

y~ xi.

(15.27)

iET~0,0

P r o o f . The existence and uniqueness of the Aij follow from an algorithm that solves (15.26) "from the leaves up". That is, choose a leaf (a vertex of degree 1), say i ~, and its unique neighbor j'. Set ~i,j, = xi, as dictated by (15.26), delete i' and the edge i'j' and replace x~ by x j , - x~,. The result is a smaller tree that still satisfies (15.25), to which we can apply induction. The basis of the induction is the one-edge tree, where (15.25) is utilized. To prove the explicit formula (15.27), we show below that the Aij defined by (15.27) satisfy

E

j:iojEE

Xo.

(15.28)

15.3

399

Threshold Measures

Figure 15.5: Illustration of the proof of Proposition 15.3.12. i0

Tjxio

~/j2io

Tjkio

A similar proof holds for the other equation of (15.26). Let j l , . . . ,jk be all the blue neighbors of i0, as in Figure 15.5. Then

k

k(

E ~,oj- E ~,o~- E r=l

j:ioj6E

r=l

j

i

E x~-

j6Tj,.i o

xo)

Zx )

iETj,., o

proving (15.28). " We are now ready for a simple combinatorial characterization of equitable trees. T h e o r e m 1 5 . 3 . 1 3 Let T be a tree having r > 1 red and b >_ 1 blue vertices. For each edge e - iojo of T, define _ ~(~) r

r

b(~)

b '

(15.29)

Tiojo containing io, respectively. Then T is equitable if and only if ~ >_ 0 for each edge e. In that case )~ is an optimum of the threshold measure problem (15.18).

Threshold Weights and Measures

400

P r o o f . Formula (15.29) is obtained from (15.27) by setting x~ xj __ ~. 1 Therefore by Proposition 15.3.12, )~ satisfies 1

A~j -

-

j:ijEE

all i

r

Aij

1 ~

-

1?- and

(15.30) all j,

i:ijEE

where T - (V, E). Thus if )~ _> 0, then )~ satisfies (15.23), and the argument of the "if" part of Theorem 15.3.9 shows that T is equitable. Conversely, if T is equitable, then as in the "only if" part of Theorem 15.3.9, the equations (15.30) admit a solution )~' _> 0. But by Proposition 15.3.12 these 1 equations have a unique solution given by (15.27), where xi - 1 and xj = -g, namely the ,k defined by (15.29). Therefore ,V - ,k and ,k _> 0 [] We conclude this section with two examples of equitable trees. T h e o r e m 15.3.14 Let T be a tree in which all red vertices have degree 2 (equivalently, T is obtained by subdividing each edge of a tree with a new vertex). Then T is equitable. Proof. To see vertex from j

T satisfies b - r + l . A l s o , each edge e - iojo of T satisfies r(e) - b(e). this, use the bijection mapping each red vertex i of Tiojo to the blue j of Tiojo such that i is the immediate successor of j along the path to i0. Therefore formula (15.29) gives A(e)-r(e)

(1 1) r

r + l

>0"

A complete k-ary tree is a rooted tree in which every internal vertex has exactly k children, and all the leaves have the same distance from the root. Theorem

15.3.15 Every complete k-ary tree is equitable.

P r o o f . Let T - (V, E) be a complete k-dry tree, and assume without loss of generality that its root is blue. Let h be the height of T (the distance from the root to each leaf) 9 The total number of vertices of T is IV] - k h]+~ l l l 9 The numbers r and b of red and blue vertices of T are functions of k and h. Specifically, if h is even, then b - P(k, h) "- 1 + k 2 +

]Ch -

Q ( k , h) . -

kh + 2 - 1

k 4 --~-""" + ~h __

IVI - P ( k ,

h) -

]~2

__1

k 2-

1 ~

1

15.4

401

Threshold and Majorization Gaps

and if h is odd, then k h+l -- 1 k:-i

b - R(k,h) "- g ( k , h - 1 ) -

k h+l - 1

- s(k, h) . - IVI - R(k, h) - k k~---=-V To compute the A~ of (15.29) for a given edge e, it is convenient to always use that subtree of T - e that contains the root, using either (15.29) or its analog with red and blue switched. Let 1 be the height of this subtree. If h is even, then

_

P(k,l) Q(k,1) V~(k~h) - Q(k, h)'

R(k,~)

S(k,1)

-d-(-i l h ) - P ( k, h ) '

for k even for k odd,

and if h is odd, then

~ -

P(k, 1) Q(k,l) 5 ( ; , ~ - R(k,h)' R(k,~) S(k,t)

for k even = 0,

for k odd.

In all cases, one obtains Ae _> 0.

15.4

Threshold

and

Majorization

Gaps

In this section we introduce two measures of non-thresholdness of a degree sequence: the threshold gap, based on the work of Hammer, Ibaraki and Simeone [HIS78, HISS1], and the majorizaion gap, based on the work of Arikati and Peled lAP94]. It turns out that these two measures coincide, and we investigate their properties and the relations between them. Recall from Section 3.1 that a vector d - ( d l , . . . , d~) with integer components is called a proper sequence if n - 1 >_ dn >_ " " >_ dl > 0. Recall also the definition of the corrected Ferrers diagram C(d) representing a proper sequence d (Definition 3.1.6), and of the corrected conjugate sequence d' and the corrected Durfee number m of d. Also recall (Theorem 3.2.2) that a proper sequence d is threshold if and only if C(d) is symmetric (i.e., d' - d), in which case C(d) is the adjacency matrix of the unique labeled realization of d. We define a distance between proper sequences as follows.

402

T h r e s h o l d Weights and M e a s u r e s

D e f i n i t i o n 15.4.1 The distance between proper sequences d and e of the same length n is defined as n

l i d - ell -

i=1

(15.31)

I & - e l.

In other words, the distance is just 1/2 of the Ll-norm of the difference d - e . If ~'~'~in._=ldi and ~i~1 ei are even, as is the case when d and e are graphic sequences, then the distance lid- ~11 is an integer. The reason is that

9

di~ei

di>_ei

i

di
Since 2in___l di and ~i~a ei are even, so are ~d,>~, di--~-~d,~, ei, and therefore ~i~1 I d i - ei] is an even integer. We denote by 7~T,~ the set of proper threshold sequences of length n. The threshold gap of d is defined as follows. D e f i n i t i o n 15.4.2 If d is a proper sequence of length n, its t h r e s h o l d g a p is eE D"I'n

lid-

Thus the threshold gap of d is a measure of the non-thresholdness of d. In Subsection 15.4.1 we determine the threshold gap of a proper sequence and all the threshold sequences that realize it. Recall the definitions of the majorization ~ and strict majorization ~relations between sequences of length n (Definition 3.1.1). Recall also the definition of unit transformation (Definition 3.1.2). If a is obtained from b by a unit transformation, we say that b is obtained from a by a reverse unit transformation. By Condition 9 of Theorem 3.1.7, every degree sequence d is majorized by some threshold sequence, and by Condition 8 of Theorem 3.2.2 the majorization is strict if and only if d is not a threshold sequence. From this and from the Muirhead Lemma (Theorem 3.1.3) it follows that if d is any degree sequence, then some threshold sequence can be obtained from d by a finite number of successive reverse unit transformations. This motivates the following definition of the majorization gap of d as a measure of the non-thresholdness of d.

15.4

Threshold and Majorization Gaps

403

D e f i n i t i o n 15.4.3 The m a j o r i z a t i o n g a p R ( d ) of a graphical sequence d is the smallest number of reverse unit transformations required to transform d into a threshold sequence. Thus R(d) = 0 if and only if d is already a threshold sequence. In Subsection 15.4.2 we give a formula for R(d) and show that it is equal to the threshold gap of d. We also determine its m a x i m u m for a fixed number of edges or vertices, as well as exhibiting all the sequences that realize this maximum, which can be thought of as the most non-threshold in this sense. We also discuss the related questions of the difference gap and some others.

15.4.1

The Threshold Gap

Before we discuss the threshold gap, we introduce two binary operations on the proper sequences of length n. D e f i n i t i o n 15.4.4 If d and e are proper sequences of length n, their j o i n and m e e t are defined as the proper sequences d V e and d A e given by (dV e)i

-

max(di, ei)

i - 1,...,n

(dA e)i

-

min(di, ei)

i - 1,...,n.

It turns out that DT-~ forms a lattice under the operations of join and meet, having the largest element ( n - 1, n - 1 , . . . , n - 1) and the smallest element (0, 0 , . . . , 0). This follows from the following proposition. Proposition

15.4.5 DT~ is closed under joins and meets.

P r o o f . For integers k _> 1 and h >_ 0, we denote by S(k, h) the set of the smallest h positive integers excluding k, that is to say

S(k,h)-

{1,...,h}, {1, ,k-l,k+l,...,h+l},

ifhk.

If d is a proper threshold sequence whose unique labeled realization on the vertices 1 , . . . , n is G, then the neighborhood of each vertex i in G is Na(i) = S(i, di). Conversely, if a graph G on the vertices 1 , . . . , n has neighborhoods satisfying this condition, then G is a threshold graph with degree sequence d, so d is a threshold sequence.

404

Threshold

Weights and Measures

We prove the closure of 2)7",~ under joins; the results for meets can be proved similarly or deduced from the closure under complements. Let d, e, C 2)Tn, and let G and H be the threshold graphs on the vertices 1 , . . . , n in which vertex i has degree d~ and e~, respectively. Thus N o ( i ) = S(i, d~) and NH(i) = S(i, ei) for each i. Put f = dV e and F = G tO H. Then for each i we have

NF(i) -- S(i, d~) U S(i, e~) - S(i, max(d~, e~)) - S(i, f~). Hence f is a threshold sequence, i.e., f C ~)']"n. 9 We now define an easily-computed quantity t(d) associated with a proper sequence d, and show that it is equal to the threshold gap of d. D e f i n i t i o n 1 5 . 4 . 6 If d is a proper sequence of length n and corrected Durfee

number rn, we put 1

m

t(d) - -~ ~

7-51

!

d~[ -

Id~ -

1

n

E

!

Id - d l.

( 5.a2)

~=m+l

To justify the equality between the two expressions in the definition of t(d), consider the corrected Ferrers diagram C - C(d), and divide it into four submatrices M, A, B, 0 as illustrated in Figure 15.6. Figure 15.6: The corrected Ferrers diagram of a proper sequence. The submatrix M is full of l's except on its main diagonal; the submatrix O is full of O's. 1

m

m+l

n

M

B

A

O

m

m+l

Since each row and column sum of M is m - 1, d~ - di is equal to the difference between the i-th column sum of A and the i-th row sum of B. Since

15.4

Threshold and Majorization Gaps

405

the l's of A are at the top of the columns and the l's of B are at the left n ~" 1, if Cij r Cji of the rows, we have Id~- di I - ~j=~+~ u~j, where u~y - ~ O, otherwise Consequently m

m

i=1

n

i=1 j = m + l

A similar argument involving the rows of A and the columns of B shows that n

E Id:-d

i=m+l

m

n

l-E E

Uij~

i=1 j = m + l

and therefore the equality in Definition 15.32 has been justified, and we also have an interpretation of t(d) as a measure of the deviation of C(d) from symmetry. Furthermore, we have ~i~=~ di - m ( m - 1) + a + b, where a and b are the numbers of l's in the submatrices A and B, respectively. Therefore if ~in=.l di is even, as is the case when d is graphic, a and b have the same parity, and it follows that t(d) is an integer. As a first step in proving that t(d) is the majorization gap of d, we associate with each proper sequence d of length n with corrected Durfee number m the sequences d and d defined by

d i - { d'i' f o r i - 1 , . . . , r n

di,

for i - m

+ 1,..., n

cli- { di, f o r / - 1 , . . . , r n d}, for i - m + l , . . . , n .

Referring to Figure 15.6, we see that C(d) is obtained from C(d) by replacing the submatrix B with A T, whereas C(d) is obtained by replacing A with B r. Therefore d and d are proper sequences and their Ferrers diagrams are symmetric, i.e., d, d C 7PTn. To find the corrected Durfee numbers r'n - re(d) and rh - re(d) of d and d, we notice that for C - C(d) we always have Cm+l,~n = 0 (for if Cm+l,,~ = 1 one would also have Cm,m+l = 1 by dm >_ din+l, and m could be increased by 1). In contrast, x - C~n,m+l can be either 0 or 1. From the definition of d it is clear that d - C(d) satisfies C',~+~,,~ = 0 and therefore rh - m. On the other hand, d' - C(d) s a t i s f i e s C m + l , m = x and Cm+2,m+l = 0, and therefore rh is m or m + 1 according as x is 0 or 1. We have therefore verified the following formula: ~-m

'

~_{

m, re+l,

if d i n - m - 1 if d i n > m - 1 .

(15.33)

Thre s hold Weights and Measures

406

It follows directly from (15.31) and (15.32) that for each proper sequence d, the proper threshold sequences d and d satisfy

lid- rill - l i d - rill -

t(d).

(15.34)

T h e o r e m 15.4.7 E v e r y proper sequence d of length n satisfies t(d) -

min l i d - cll.

cEDTn

P r o o f . Let g: C 77T~ be optimal, i.e., lid- ~11- mincev~rn lid- ell. Let d be the corrected Ferrers diagram of ~: and rh its corrected Durfee number. First we show that rn _< rh _< rh, where rh is the corrected Durfee number of d, given by (15.33). Assume that, if possible, rh < m. Put h = ~:,~+1 < rh, and define the sequence c by { ~i+1, fori-h+l,...,rh ci ~, for i - rh + 1 g:i, otherwise. Since C is symmetric and h < rh, c is a proper threshold sequence, and its symmetric corrected Ferrers diagram is obtained from ~' by enlarging its corrected Durfee,square from size rh to size rh + 1. We have C~+l - h < c~+1 - r h _< r n - 1 _< dm _< drh+l and therefore Ic~+l - d~+l I < 1~=~+1- d~+a I - ( ~ - h).

(15.35)

In addition, we have in any case that (i-

h+ 1,...,~).

(15.36)

By adding the inequalities (15.35) and (15.36), we obtain IIc-dll < II~-dll, contradicting the optimality of ~:. This proves that m < ~n, and a similar argument shows that rh < rh. We have shown that m < rh < rh, and therefore by (15.33) rh is either m or rh. We show below that l[~:- dll - l i d dll if r~ - m, and it can be similarly shown that I1~- dl[ - l i d - dll if ~ - ~ . Consequently, at least one of J and d is an optimal threshold sequence, and the required conclusion

that l i d - ~11-

t(d) will follow from (15.34).

As stated above, we consider the case ~ - m and have to show that I 1 ~ - d l l - lid-dll. Note that ~ ~nd d ~re proper threshold sequences with the

15.4

T h r e s h o l d a n d M a j o r i z a t i o n Gaps

407

same corrected Durfee n u m b e r m. If g: - d, we are done. Otherwise there must be some i - m + 1 , . . . , n such that g:i r a~i, because rows m + 1 , . . . , n of the s y m m e t r i c corrected Ferrers diagram determine it completely. Let k be the largest such i satisfying g:i > di, if any. Put h - g:k and define the sequence c' by

, ~ c.i-1, ci - ~ ci,

fori-k,h otherwise.

The sequence c' is proper, because by the maximality of k we have g:k > dk _> dk+l >_ ck+l, and by the s y m m e t r y of its corrected Ferrers diagram it is a threshold sequence. Clearly its corrected Durfee n u m b e r is again rn. Since c~ - g : k - 1 >_ dk - dk, we have I c ~ - d k [ - Ig:k--dkl-- 1. Hence I I c ' - d l l _< I]g;- dll because in any case I c ~ - dh] _< Ig:h- dhl + 1. It follows that c' is another optimal threshold sequence and we may replace g: with c'. R e p e a t e d application of this procedure eventually allows us to assume that ci <_ di for i - m + 1 , . . . , n. If we still have g: =/= d, there must be some i - rn + 1 , . . . , n such that ~i < di. Let k be the smallest such i. Put h - g:k and define the sequence c" by ,,

]' g : i + l ,

ci - ~ ci,

fori-k,h+l otherwise.

The sequence c" is again proper, because by the minimality of k we have g:k < dk _< dk-1 _< g:k-1, and by the s y m m e t r y of its corrected Ferrers diagram it is a threshold sequence. Its corrected Durfee n u m b e r is again rn. Since c~ - g:k + 1 _< dk - dk, we have [[c"-d][ _< IIg:- dll as before, so c" is optimal and we may replace g: with c". R e p e a t e d application of this procedure allows us to assume finally that ci - di for i - m + 1 , . . . , n. This implies that g: - d and completes the proof. " For a given proper sequence d of length n, we denote by 2)Tn(d) the set of proper threshold sequences of length n at m i n i m u m distance from d, that is to say, by T h e o r e m 15.4.7, lPT~(d) = {c C 2P7"~ : [ I c - all = t(d)}. We determine this set below. For this purpose we put

d+-dvd,

d--dAd.

Since d, d E 2)Tn, we have d +, d- E 2)Tn by Proposition 15.4.5. It is easy to see that d + - max(d~, d'~) and d~- - min(d~, d'~). To justify (15.32), we have argued that t(d) is equal to one half of the n u m b e r of positions where the submatrix A in Figure 15.6 differs from B T.

408

T h r e s h o l d Weights and M e a s u r e s

It is easy to see that both lid + - d l l quantity, which means that

and lid- - d l l

are equal to the same

d +, d- C DT'n(d).

(15.37)

We need the following l e m m a to determine D'Y~(d). Lemma

1 5 . 4 . 8 Let d be a proper sequence and h, k integers satisfying d-~ <

h <_ d +. Then the following implications hold:

d; < & G+ > d k

G+ > & ---->, d ; < & .

P r o o f . We prove only the first implication for k - rn + 1 , . . . , n; the other cases are similar. The hypothesis d~- < dk implies t h a t d~ < dk and hence d~- - d~ and d + - dk. Therefore the condition d~- < h _< d + of the l e m m a reads d~ < h <_ dk. Since k _> rn + 1, the condition dk _> h is equivalent to d~ _> k - 1 and the condition d~ < h is equivalent to dh < k - 1. Therefore dh < d~, which means dh < d +. 9 We are now ready to characterize ~D'Yn(d). Theorem

1 5 . 4 . 9 Let d be a proper sequence of length n. Then -

{c

Z

7. . d - <

< d + }.

In other words, I)T~(d) is the sublattice of 7?T~ with the largest element d + and the smallest element d-. P r o o f . First we prove that if c C Z)Tn and d- _< c _< d +, then c C Z)'Yn(d). Observe that since d + and d- have the same corrected Durfee n u m b e r rn and c lies between them, c too has the same corrected Durfee n u m b e r m. We m a y assume t h a t c -r d - , for otherwise the result follows from (15.37). Let k be the largest subscript i satisfying d~- < ci. This k must be larger t h a n m, for otherwise ci - d~ for each i > m, which would imply that the threshold sequences c and d- coincide. We assume that dk -- d +, i.e., that d;- < dk; the case dk - d~- is similar. P u t h - ck. Since rn < k, we have h < rn and therefore Ch _> k - 1. Since d~- < h _< d + and d~- < dk, L e m m a 15.4.8 gives us t h a t d + > dh - d~. If D - - C ( d - ) is the corrected Ferrers diagram of d - , then from h - ck > d~- we have D-~h -- O, hence by the s y m m e t r y of D -

15.4

T h r e s h o l d a n d M a j o r i z a t i o n Gaps

we have D~k - 0, and hence d~ < k given by

1

~

409

1 _< Ch. Consider now the sequence c 1 fori-h,k otherwise.

ci--1,

ci - ~ Ci

This sequence is proper because of the s y m m e t r y of C(c) and the fact that if ck - ck+l, then Ck+l > d~- >_ d~-+l , contradicting the maximality of k. By the s y m m e t r y of C(c) and the definition of h it is also clear that C(c 1) is symmetric, so c 1 C DT-n. By Ch > d~ and ck > d~- we also have d- <_ c I _< d + and IIc I - d-II < I I c - d-II. Furthermore, from ck _< d + - dk we have Iclk--dkl -- Ick--dkl+ 1, and from Ch > d-~ - dh we have Iclh--dhl -- Ich--dhl--1; consequently IIc1 - a l l - I I c - all. By repeating this argument, we obtain a finite sequence c - cO,..., c s - d- of proper threshold sequences all at the same distance from d. Therefore c e DT-n(d) by (15.37). Conversely, given that c C D T n ( d ) , we show that c _< d+; the argument for d- _< c is similar. Assume otherwise, and let k be any subscript satisfying ck > d +. Let G and H be the threshold graphs on 1 , . . . ,n satisfying dega(i ) - ci and degH(i ) -- d + for all i. Let h be the largest subscript such that kh is an edge of G. Since degH(k ) < dega(k), kh is not an edge of H. Therefore degH(h ) < dega(h), that is to say d + < Ch. Consider the sequence c' given by , ( ci-1, fori-k,h ci - ~ ci, otherwise. By our familiar argument, c' C DT-n, and by ck > d + > dk and Ch > d + > dh, we have ] [ c ' - dll < I I c - dll , contradicting c e "D'-f'n(d). 9

15.4.2

The Majorization Gap

In order to find an explicit expression for the majorization gap R ( d ) , we use the following notations. For a sequence d - ( d l , . . . , dn), let S k ( d ) denote the k-th partial sum of d, i.e 9,Sk(d) - 2~=1d~. k Note that Sn(d) - Sn(d'). For a proper sequence d - ( d l , . . . , dn), put 6i(d) - (d~i - di) +,

i - 1 , . . . , n,

where x + - max(x, 0), and n -

i=1

Threshold Weights and Measures

410

Our first main result is a formula for the majorization gap, to be proved below"

T h e o r e m 1 5 . 4 . 1 0 For every proper degree sequence d, R(d)

6(d)

-

E x a m p l e . Let d = (2,2,2,2,2). Then d ' = ( 4 , 4 , 2 , 0 , 0 ) and 5(d) = 4. The theorem asserts that R(d) = 2. A reverse unit transformation from 5 to 1 transforms d into f = (3, 2,2,2, 1), and a reverse unit transformation from 4 to 1 transforms f into e = ( 4 , 2 , 2 , 1 , 1 ) . Since e = e', e is a threshold sequence. Before we prove Theorem 15.4.10, we discuss some preliminary results. For any two sequences a = ( a l , . . . , an) and b = ( b l , . . . , bn), we define

i-1,...,n

5i(a, b) - (ai - bi) +, and n

6(a,

-

6 (a, i=1

If a and b are integer sequences, a ~ b, a l _> . " _> a~ and bl > " ' " > bn, we define U ( a , b) to be the m i n i m u m number of unit transformations required to transform a into b. Observe that under these conditions the following are equivalent: (i) a = b; (ii) 5(a, b ) = 0; (iii) U(a, b ) = O. Lemma

15.4.11 Let a -

(al,...,an)

quences such that a ~ b, aa >_ .'. U(a, b) - 5(a, b).

and b - ( b l , . . . , b n ) >_ an and bl >_ ""

be integer se-

> bn.

Then

P r o o f . It is easy to see that if a r b and c is obtained from a by a unit transformation, then 5(a, b)+ 1 _> 5(c, b) _> 5(a, b ) - 1. Thus U(a, b) >_ 5(a, b). Also, as given on page 135 of [MO79], if a r b, we can always perform a unit transformation on a to obtain a c such that c ~ b and 5(c, b) = 5(a, b ) - 1. Thus U(a, b)<_ 5(a, b). ..

15.4

T h r e s h o l d and Majorization Gaps

411

It is easy to check that for integer x, (x+l)+_x+_

{ 1, i f x _ > 0 0, otherwise, 2, if x_>0 1, if x - - 1 0, otherwise,

x +-

(x + 2) +

x+_(x_l)+_

x+-(x-2)

{ 1, i f x > _ l 0, otherwise,

+-

[ 2, if x>_2 / 1, if x - 1 0, otherwise. (15.38)

We use these facts to prove the following lemma.

L e m m a 15.4.12 Let d - (dl,...,dn) be a proper sequence and a s s u m e that e is obtained f r o m d by a reverse unit t r a n s f o r m a t i o n . Then

6(d)- 2 < 6(~) < ~(d)+ 2. Proof. Represent d by its corrected Ferrers diagram and assume that e d - up + U~. Let or-

{ d~+l, d~+2,

ifd~<~r-I ifd~_>Tr-1,

7-

{ do, dp+l,

ifdo<

p

ifdp>p.

Thus the transfer is from row p, column T of C ( d ) to row 7r, column or, and consequently 7r 5r cr and p 5r r. See Figure 15.7. We have four cases. Case 1" 7r 7~ r and p -/ a. Since e - d - U p + u~,, for 1 <_ i <_ n, we have e i - - d i except for e~= d~+l, e o d R - 1 , and e i - - d i except for D

~-d~+l,~=d~ . l

l

6~(~)- (~"

l

l

i. T h ~ n. 6 ~ ( ~ l - ( ~. l

!

d~

~ ) + - ( .d "

e~) + - ( d ~ - d ~ + 1) +. Hence 5~(e) - 5~(d) -

-1, ifd'-d~> 0, otherwise,

5o(e) - 5o(d) -

1, i f d ' - d ~ >_0 0, otherwise.

5~

1, i f d ' p - d p > O 0, otherwise,

Similarly, - 5~

-

-1, ~ ( e ) - 5~(d) -

ifd'~-d~>l O, otherwise.

I

1

l)+,~nd

412

Threshold

Weights and Measures

Figure 15.7: Illustrating a transfer from p to 7r. Solid lines represent l's and possibly . ' s .

T

~T

I

1 ....

I

C a s e 2: 7r 7~ T and p - or. Here, ~ r ( e ) - ~ ( d ) and 8 ; ( e ) - ~ ( d ) Case 1. Also 6 p ( e ) - ( e ' p - ep) + - ( d ' p - d p + 2) +. Hence

8 p ( e ) - 6p(d) -

2, if d'p - dp > 0 1, if d'p - - d p - - - - 1 O, otherwise.

C a s e 3" 7 r - T and p5r a. H e r e S p ( e ) - S p ( d ) )+ - ( d rI - d ~ - 2 ) Case 1. S i n c e 6 r ( e ) - ( G - e/r --2,

6r(e)-6r(d)

-

are as in

-1, O,

and 5o(e)-5~,(d) + we obtain

are as in

if d'~ - dr > 2 if d" - d~ - 1 otherwise.

C a s e 4" 7r - ~- and p - ~r. We now h a v e S r ( e ) - S r ( d ) as in Case 3, and 5 p ( e ) - 6p(d) as in Case 2. To complete the proof, note that 5i(e) - 5i(d) except for i E {~r, p, ~, ~-}. For future reference, note the following from the proof of L e m m a 15.4.12. Necessary and sufficient conditions for 6(e) - 5 ( d ) - 2 in Cases 1, 2, 3 are: 1. if 7r -~ T, p -~ or, then (a) d ' -

d~> 1,

15.4

Threshold and Majorization Gaps

(b)

r

413

0,

(c) f ~ - d ~ < O , (d) s

d~>_ 1.

2. if 7r 7~ r, p - (r, then (a) d ' -

(b)

d~>_ 1,

l

(c) d ' -

d~_> 1.

3. if 7c - r, p -r a, then

d'-

2,

(b) d ' p - d p < O ,

d'-

O.

L e m m a 1 5 . 4 . 1 3 Let d be a proper sequence. For 1 ( i , j (_ n, if di < j - 1, ! then dj < i. P r o o f . The proof is simple. L e m m a 1 5 . 4 . 1 4 Let d C D~ and let p be the largest index i such that di < d~. Then p < n. Further, 1. A s s u m e d'p (_ p -

!

1 and let q - d;. Then dq - p -

!

1 and dq ( p -

2.

2. A s s u m e dpt >_ p and let q - dpt + l. Then dq - p and dql <_ p - 1 . P r o o f . Since Sn_l(d) < Sn_l(d') and Sn(d) - Sn(d'), we have d= >_ d~, so p < n by definition of p. The fact that p < n justifies our mentioning of column p + 1 (of C) below. In both cases q is the position of the last 1 in I ! column p, so Cqp - 1. The assumption on p implies dp+ 1 _< dp+l _< dp < dp. ' 1 < dp, ' implying that Cq,p_t_ 1 r 1. We prove the two parts as Hence dp+ follows" !

1. See Figure 15.8. If d; _< p - 1 , then p > q and Cq,p+l r "k. It follows that Cq,p_t. 1 -- 0, and therefore (since Cqq - * and Cqp - 1) dq - p - 1. Also dp < d; - q gives Cpq - 0, and again C(q, q) - * implies d; _< p - 2.

T h r e s h o l d Weights and M e a s u r e s

414

Figure 15.8: Illustrating Part 1 of Lemma 15.4.14. q

P

~ * ~ 1 0 0

*

!

2. In this case, apply L e m m a 15.4.13 with i - p and j - 1 - dp. T h e n ! j - dip + 1 - q and dq < p, as required. In our case p < q. See Figure 15.9. We have seen t h a t Cq,;+l r 1, b u t Cqp - 1, so dq _> p. If q > p + 1, t h e n Cq,p+l = 0 and dq - p, as required. If q - p + 1, t h e n Cqp - Cp+l,p = 1 and Cqq - , . On the other h a n d , Cq,q+l m u s t be 0 if ! it exists, for otherwise Cp;;+l would be 1, c o n t r a d i c t i n g dq < p. T h u s again dq - p.

Figure 15.9: Illustrating Part 2 of Lemma 15.4.14.

Lemma 1,...,p-

p

q

,

0

1

*

Let d C Dn and let p be as in L e m m a 15.~.1~. FOr r 1, dr >_ p implies dr < dlr.

15.4.15

-

-

15.4

Threshold and Majorization Gaps

415

Figure 15.10" Illustrating the proof of Lemma 15.4.15. r

p

t

~ * ~ I 0

0

*

P r o o f . A s s u m e to the c o n t r a r y t h a t d~ > d'~. Let t - d~ + 1 > p (since d~ > r, t is the position of the last 1 in row r. See Figure 15.10). By c o n s t r u c t i o n C~t - 1, and by a s s u m p t i o n Ct~ - O, so d't ~_ r and dt < r - 1. B u t t h e n dlt > dt c o n t r a d i c t i n g the definition of p. 9 Lemma

15.4.16

Let d be a proper sequence. If there exists a k such that

d'k < d'k+l, then 1. d'k - k - l , d ' k +

~-k;

2. k is unique. P r o o f . 1" If d~ < k - 1 or d~ > k, then d~+ 1 < d~ since the l's of C(d) are left-justified. Hence d~ - k - 1. T h e n again by this reason and the a s s u m p t i o n t h a t d~ < d~+l, we obtain d~+ 1 - k. 2" T h e s a m e p r o p e r t y of C(d) implies t h a t the conditions d~ - i - 1 and d~+ 1 - i are possible for at most one i. ,, 1 5 . 4 . 1 7 (1) Let d C Dn have corrected Durfee number m, and let C(e) be obtained from C(d) by deleting the last 1 in row i, where 1 < i < rn and (a) if i < m, then di > di+l; (b) if i - r n , then dm >_ m. Then e C D , . Further, if d} > d,, then 5(e) > 5(d). (2) Let d G Dn, and let C(e) be obtained from C(d) by adding a 1 to the end of column i, where 1 < i < rn and (a) if i > 1, then d~ < d~_l;

Lemma

416

Threshold Weights and Measures

(b) if i = m, then dm >_ m. Then e e D , . Further, if d~ >_ d~, then 5(e) > 5(d). P r o o f . 1: The a s s u m p t i o n s on i g u a r a n t e e t h a t e is a proper sequence. The c o l u m n of the 1 t h a t is removed is j = 1 + di > m. This implies t h a t e E Dn. Further, if d~ > di, then Cji - 1, and thus dj ~_ i. But the a s s u m p t i o n s on i i m p l y d~ - i. Therefore by equation (15.38),

5(e) - 5(d) - (d'i - di + 1) + - (d'i - di) + + (d~ - dj - 1) + - (d} - dj) + - 1 + O. 2: This is similar to Part 1.

..

Let d c Dn with d~ > 0 and m - 2. Then S(d) <_ n with equality if and only if dl - d2.

2,

Lemma15.4.18

P r o o f . Note t h a t the condition dn > 0 implies n _> 2. Using d2 < dl _< n - 1, d~ - 1, d2 _> 1, and di - 1, d~ - 2, for i - 3 , . . . , d2 + 1, we obtain n

~(d)

-

~ (d'~ - d~)+ i=1

=

(n-

_< ( n -

1-dl)+

-~-(d2- 1)(2-

1)

1 -d2) + + d2- 1

=

n-l-d2+d2-1

"~

rt ~2.

We need three more l e m m a s to prove T h e o r e m 15.4.10. Lemma

1 5 . 4 . 1 9 Let d C D~. A s s u m e that dn > O, d~ < d'a, and let p be the

largest i such that di < d~. Let q be such that dq > dqt and either (a) q < n or (b) q - n , dn > 2. Then Sk(d) < X k ( d ' ) - 1,

for k -

p,...,q-

1.

Since d C Dn, Sk(d) <_ Sk(d'). A s s u m e t h a t , if possible, Sk(d) >_ Sk(d') - 1 for some k satisfying p _< k _< q - 1. W h e n q < n we have

Proof.

X~(d) > S~(d')d~ > d~,

( a s s u m p t i o n on p) ( a s s u m p t i o n on q) ( a s s u m p t i o n on p)

I dq >_ dq-+-i

(d~ < dl ~ d'n - 0 , d. > 0)

d. > d ' + l .

d~ > d~,

i-k+l,...,q-1 i-q+l,...,n-1

15.4

417

Threshold and Majorization Gaps

Adding all these inequalities, we obtain a contradiction, S~(d) > S ~ ( d ' ) + 1. W h e n q - n, the inequality for dq drops, but by assumption the inequality for dn can be strengthened by 1, and the same contradiction is obtained.

1 5 . 4 . 2 0 Let d C Dn, a s s u m e that dl < d'1, dn > O, and let p be as in Lemma 15.~. 19. Then

Lemma

Sk(d) <_ S k ( d ' ) - 1,

k - p,...,n-

1.

P r o o f . We have Sk(d) < Sk(d') since d E Dn. If Sk(d) - Sk(d'), then there exists an i, k < i < n such t h a t d~ < d~, since d~ > d'n and S n ( d ) - Sn(d'). But then i > p, contradicting the definition of p. tt Lemma

15.4.21

Under the assumptions of Lemma 15.~.20:

1. if d'p < p - 1, then Sk(d) < S k ( d ' ) - 1 for k - 1 , . . . , dpI - 1 ; !

2. if dp >_ p, then Sk(d) <_ Sk(d') - 1 for k - 1 , . . . , p -

1.

P r o o f . 1" Let q - d p I < p - 1 . By L e m m a 15.4.14, dq - p - 1 and dq/ < p - 2 , so dq > d'q. I f d i < d} for i - 2 , . . . , q 1, then the assumption dl < d~ implies Sk(d) < S k ( d ' ) - 1 for k - 1 , . . . , q - 1, as required. So assume t h a t there exists an i, 2 < i < q - 1, such that di > d}, and let r be the smallest such i. Since r < q, d~ k dq - p - 1 . But the fact that & > d'~ and L e m m a 15.4.15 imply d~ < p - 1. Therefore d~ - p - 1. We assert t h a t di > d} for i - r , . . . , q . Indeed, for such i, p - 1 d~ > di k dq - p - 1, so di - p - 1. This implies, by the construction of C, t h a t for i - r , . . . , q 1, d'~ > q - 1 > i, and therefore d} > d'~+1 by I I l L e m m a 15.4.16. Hence d~ >_ d~+ 1 >_ ..- >~ dq_ 1 ~_ dq. Thus for i - r , . . . , q, d~ <_ d~ < d~ - p - 1 - d i , proving the assertion. Now for k - 1 , . . . , r 1, the facts dl < d~ and di <_ d~ for i - 2 , . . . , k imply t h a t Sk(d) <_ Sk(d') - 1. Also, for k - r , . . . , q - 1, if Sk(d) - Sk(d'), then Sk+l(d) > Sk+l(d') as dk+l > d~+ 1 by the assertion. But this contradicts d C D~. Hence again Sk(d) <_ S k ( d ' ) - 1. ! 2" Let q - d p + 1 _> p + 1. From L e m m a 15.4.14, dq - p. S i n c e p < q,

Threshold Weights

418

and

Measures

dp _> dq - p. Thus d l > . . . > dp. Using L e m m a 15.4.15, we obtain di _< d~ for i - 1 , . . . , p - 1. This and the assumption d l < d~ complete the proof. 9 P r o o f of T h e o r e m 15.4.10. From L e m m a 15.4.12 we have R(d) >_ [ ~ 1 , as a reverse unit transformation can decrease 5(d) by at most 2. We show that if 5(d) > 0, we can always construct a proper sequence e such that 1. e is obtained from d by a reverse unit transformation, 2. e is a degree sequence, and 3.

= e(d)-

2.

It then follows that 5(d)is even and R(d) - 6(d) 2 " To show that e is a degree sequence we use the Berge Condition of Theorem 3.1.7, i.e., & ( e ) is even and Sk(e)_< Sk(e')for k = 1 , . . . , n . To show that 5(e) = 5 ( d ) - 2 we use the observation made after the proof of L e m m a 15.4.12. Without loss of generality, we may assume that dn > 0, for if dn = 0, then we work with c - ( d l , . . . ,d~-l). We may also assume that d 1 < d~, for if d 1 - d i ( - 7 " t - 1), then we work with c - ( d 2 - 1 , . . . , d ~ - 1). We distinguish two cases. C a s e 1" There exists an i > 1 such that di < d~. Let p be the largest such i. The basic idea is to transfer the 1 at the end of column p to the end of row 1. Let s = dl + 2 be the destination column for the moving 1. We have two subcases now. ! ! C a s e 1.1" dp _< p - 1 . Then q - dp is the source row for the moving 1. Also dq = p - 1 by L e m m a 15.4.14 and hence p is the source column for the moving 1. Define e = d - uq + Ul. Then for i = 1 , . . . , n , (a) ei = di except for e l - - d l q- 1 and eq - - d q - 1, and (b) e Ii - - d Ii except for epI - d Ip - 1 and Cs,

1.

We first prove that e is a degree sequence. Since 5's(e') = S'n(e), it sumces to prove S'k(e) _< S'k(e'), for k = 1 , . . . , s - 1. This is done as follows: Sk(e)=Sk(d)+l_< Sk(d') = S k ( e ' ) for k = 1 , . . . , q - 1 (by L e m m a 15.4.21), Sk(e) = Sk(d) < Sk(d') = Sk(e') for k = q , . . . , p - 1,

15.4

Threshold and Majorization Gaps

Sk(e)=Sk(d)<_Sk(d')-i

419

=Sk(e')

fork=p,...,s-1

(by L e m m a 15.4.20). To show that 5(e) = 5 ( d ) - 2, note that (p, r, 7r, a) = (q, p, 1, s), where p, T, 7r, a are as defined in the proof of L e m m a 15.4.12. By assumption p -r 1. ! Also 0 < dn _< dp < dp i m p l i e s p _ < s - l , and q < p. H e n c e q < s - l , giving q =fi s. Now the necessary and sufficient conditions for 5(e) = 5 ( d ) - 2 are verified as follows" (a) d ~ - d 1 ~ 1 by assumption; (b) d ' q - dq < 0 as dq - p - 1 and d'q _< p - 2 by L e m m a 15.4.14; (c) d'~ - d~ < 0 as d~ _> d~ _> 1 and d'~ - 0 ; and (d) d ' p - dp >_ 1 by definition of p. ! C a s e 1.2" dp > p. From L e m m a s 15.4.20 and 15.4.21 we have Sk(d) < Sk(d'),

k-

1,..., n-

1.

(15.39)

Let q - dp! + 1 be the source row for the moving 1, and define e as in Case 1.1. We first show that Sk(e) _< Sk(e') for k = 1 , . . . , s - 1. It follows from L e m m a 15.4.14 that dq - p and dq' <_ p - 1 . Note t h a t s - 1 - d l + l _< d ~ <_ n - 1. We have: Sk(e) = Sk(d) + 1 <_ Sk(d') = Sk(e') for k = 1 , . . . , p - 1 (by (15.39)), -

_<

+

-

(since ep - dp _< dpI _ _ 1 _ _ epl ) , Sk(e) = S k ( d ) + 1 <_ S k ( d ' ) - 1 = Sk(e') for k = p + 1 , . . . , q 1 (by L e m m a 15.4.19), Sk(e)=Sk(d)<_Sk(d')-i =Sk(e') fork=q,...,s-1 (by (15.39)). To show that 5(e) = 5 ( d ) - 2, observe that again (p, T, 7r, a) = (q,p, 1,s). ! If q r s, then (a), (c), and (d) are as in Case 1.1, and (b) d q - dq < 0 by L e m m a 1 5 . 4 . 1 4 . If q - s, t h e n d ~ - d l >_ 1 a n d d ' p - d , _ > 1 as before, and I / ! dq - dq ~ - 2 since dq - p ~ 2 and dq - d s - O. C a s e 2: i - 1 is the only index with the property di < d}. Then dn - 1, for dn >_ 2 implies d~ - n - 1 - d~ > dl >_ d2, contradicting the assumption of the case. We assert that d~ >_ dl -t- 2. Indeed, we show that d~ - dl ~- 1 implies that S~(d) is odd, contradicting the assumption that d is a degree sequence. We have d~ - d 1 + 1 , d~ <_ di, i-2,...,n-1, d" - d ~ - l - 0 .

420

Threshold Weights and Measures

By adding we obtain ~n(d') < Sn(d) = Sn(d'). Thus all the inequalities above are in fact equalities, i.e., d~ - di for i - 2 , . . . , n - 1. Therefore the sequence c = ( d l , . . . , dn-1,0) has a symmetric corrected Ferrers diagram. This implies that S~(c) is even, which in turn implies that S n ( d ) is odd. This proves the assertion. Let s = d l q- 2 and define e = d - un + U l. Then for i = 1 , . . . , n, ei = di except for el - dl -t- 1 and en - d n - 1 - 0, and e Ii - d Ii except for e II - d II - 1 and e ts - 1. To show that e is a degree sequence, set q = s (_< n - 1), p = 1 and apply L e m m a 15.4.19. Then for k = 1 , . . . , s - 1, Sk(e) = S k ( d ) + 1 <_ S k ( d ' ) - 1 = Sk(e'). Clearly for k = s , . . . , n, Sk(e) <_ Sk(e'). It remains to verify the necessary and sufficient conditions for 5(e) = 5 ( d ) - 2. Note that here (p, T, 7r, ~r) = (n, 1, 1, s). Observe that s ~ n by the assertion. Now (a) d~ - d 1 ~ 2 b y the assertion; (b) d~ - d n = - 1 < 0; and

(~) d'~- d ~ - - d ~ < - d ~ < 0.

We now study the connection between the majorization gap R ( d ) and the threshold gap t(d). Specifically we show that these gaps are equal and every threshold sequence achieving the majorization gap also achieves the threshold gap. Theorem

15.4.22 For every p r o p e r degree sequence d,

t(d)

-

R(d).

P r o o f . Let A - {i " 1 < i < m , d ~ - d i > 0}, A' - {i " 1 < i < m , d } - d ~ < 0}, and B - {i 9m + l <_ i <_ n , d ' ~ - d i >_ 0}. Define a+ - E~eAd'~--d~, a _ -- ~ i e A , d ~ - di, /3+ - ~ i e B d } di. It follows from the equality (15.4.6) that a _ - - / 3 + . Therefore t(d)

=

1

~

=

, 1 1 [di - d~l - ~(a+ - a _ ) - ~(~+ + 3+)

~

-~ ~

AK

~1 ~ iEAuB

(d'~ - d~) - ~1 ~~( d , -,

d~)+ _ 15(d) - R(d),

i=1

where the last equality follows by Theorem 15.4.10. 9 The following result states that all threshold sequences achieving the majorization gap R ( d ) also achieve the threshold gap t(d). Recall that if a ~ b, U ( a , b) denotes the m i n i m u m number of unit transformations required to transform a into b.

15.4

Threshold and Majorization Gaps

421

T h e o r e m 15.4.23 Let d be a proper degree sequence and e a proper threshold sequence such that e ~ d and U(e, d ) - R(d). Then lie- dll- t(d) P r o o f . It suffices to prove lie- dll - R(d) by virtue of Theorem 15.4.22. Since e > d, we have S n ( e ) - Sn(d), and so

( e i - di) ei >di

~

( d i - ei).

(15.40)

ei < di

We then have, where n is the length of d and e, n

lie- d[I -- ~ ~ le~- &l i=1

=

1 ~ ~

1 (~, - <) + ~ ~

ei >di

-

~

(< - ~)

ei
(ei-di)

(by (15.40))

ei>di

=

n

E(~,-

<)+

i=1

=

~(~, d)

=

U(e,d)

=

R(d).

(by Lemma 15.4.11)

We remark that the assumptions that e is threshold, l i e - dll - t(d), and S~(d) - Sn(e) do not imply that e > d. For example consider d (6,6,4,4,3,3,1,1) and e - ( 7 , 6 , 4 , 3 , 3 , 2 , 2 , 1 ) . We now consider the problem of characterizing the degree sequences with m a x i m u m majorization. We show that for a fixed number of edges and a variable number of vertices, R(d) is maximized precisely when d is the degree sequence of a matching plus isolated vertices. For a fixed number n of vertices and a variable number of edges, we exhibit all the degree sequences that maximize R(d), and they turn out to be almost ~-regular. As is customary, we denote the constant sequence (a, a , . . . , a) of length n by a n, the sequence (a, b , . . . , b) by ab n-l, etc. T h e o r e m 15.4.24 Let d be a proper degree sequence with a fixed sum 2q > O. Then R(d) <_ q - 1.

Furthermore, equality holds if and only if d is of the form d - 12qO~.

Threshold Weights and Measures

422

P r o o f . We have

~(d)

-

~ ( d ' , - d~)+ i>l

=

( d '1 -

dl)+

-t-

E ( d t i - di) + i>2

_< d'l -d~ + ~; d'~

(~n~ dl _> d~ ~na d~, d~ _> 0).

i>2

If dl > 2, then 5(d) < Ei>l d ~ - 2 - 2q - 2. If dl - 1, then d - 12q0r, and so 5(d) - 2 q - 2. Conversely, let d satisfy ~ > l ( d } - d ~ ) + - 2 q - 2 . Assume that, if possible, dl>

1. T h e n

(d i - d l ) + 2 (d~-d~)+

dl-dl _< dl _< d~,

--

i>_2.

Adding these inequalities, we obtain 2q - 2q, hence all the above inequalities hold as equalities. Therefore for i >_ 2, if d} > 0, then di - O. But this fails for i - 2, as d~ >_ dl - 2. [] Let us now consider the majorization gap of the degree sequences of a fixed length n. We make use of the following functions:

f(n)

-

L~J

f(n)-l,

[~],

g(n) -

f(n),

ifn-3mod4 otherwise.

We reproduce the definition of Dn for convenience. For positive integers n, Dn - { d -

( d l , . . . , dn)" d is proper, Sk(d) < Sk(d') for k - 1 , . . . , n}.

proper degree sequence of length 5(d) <_ g(n). Further, for n >_ 5, equality holds if and only if

Theorem

15.4.25

Let d be a

for n ~ 3 mod 4

d-

F~I

~

o~

d-l~J

n.

Then

~

n-3 2 '

for n - 3 mod 4 or

To prove Theorem 15.4.25, we find it convenient to prove the following theorem:

15.4

Threshold and Majorization Gaps

Theorem

15.4.26

423

For d C D~, 5(d) <_ f(n). Further, for n > 5, equality

holds if and only if d -

[~-~1 n or d -

We need several results to prove these theorems. about the functions f ( n ) and g(n).

1. f ( n -

1) < f ( n ) for n > 1, f ( n -

2. f ( n ) > n 3. g ( n - 1 )

First, some simple facts

1) < f ( n ) for n > 3;

2 for n _> 2, with strict inequality for n > 5;

< g(n) for n > 4, g(n) > n + l f o r n > 7 .

Next, three lemmas about corrected Ferrers diagrams. L e m m a 1 5 . 4 . 2 7 Let 0 r d C D n . Put s - dl + 1, p - d~s, q - d~p + 1, r - ds. Thus s > _ r e > p > 1 and q > m > r > 1. Assume that the following hold (See Figure 15.11):

1. s > m + 1, and if equality holds, then p < m; 2. q < s ; 3. r < p - r . Then: 1. there exists a sequence e C

Dn

such that

e I

--

dl-1 and 5(e) > 5(d)+2;

2. there exists a sequence f c D~ such that (a) (i) f l

-

dl -

1 or (ii) f l

-

dl

and f'~ - 1;

(b) 5 ( f ) > 5(d) + 1; (c) S n ( f ) and Sn(d) have the same parity. P r o o f . We begin by proving s t a t e m e n t 1. First note that p _< m _< q and r < p. Secondly, we m a y assume that the first r columns of C(d) are full, i.e., d~i - n - 1, i - 1,...,r, (15.41) for otherwise we can fill column 1, then 2, and so on up to r (by adding l's at the end of these columns) without going out of D~, and increase 5(d) at each step, by L e m m a 15.4.17.

424

Threshold Weights and Measures

Figure 15.11: Illustrating L e m m a 15.4.27. r

p

m

q

s

*

0

m

0

Counting in two ways the number of l's in rows 1 , . . . , s and columns r + 1 , . . . , p of C(d), we obtain p

~

d:-(p-r)(q-1)+ i=r+l

(15.42)

(di-r).

i=q+l

Since d e Dn, ~p(d) ~ ~p(d'). This implies, by (15.41) and (15.42), p(s - 1 )

p


~

d'i - r ( n - 1 )

+ (p- r)(q-1)

i=r+l

•

(di - r).

i=q+l

Therefore p(s - q) <_ r(n - 1 - q) + r + EiLq+l (di ~--q+l ( d ~ - r), and it follows that (p-

+

r)(s - q) < r(n - l - s) + p +

~

F) < r ( r t - -

(di - r).

1 - q) + p +

(15.43)

i=q+l !

!

' for if d~ < d~+1 for some By L e m m a 15.4.16, dr+ 1 _> dr+ 2 >_ ..9 > dp, i < p _< q, a contradiction. Also r < i < p, then q - 1 _< d~ < d~+1 -

15.4

425

Threshold and Majorization Gaps

di-s-1

I a <_s-2. andd~+

fori-r+l,...,p,

d~ > d'~,

Hence

i - r + 1,...,p.

(15.44)

I I l l Again by Lemma 15.4.16, dlq+l ~ dq+ 2 ~ . . . ~ ds, f o r if d i < di+ 1 f o r some q < i < s, then d~+1 - i > q >_ m, a contradiction. Therefore for i - q + 1 , . . . , s , di <_ dq+l _< p - 1 and d~ >_ d'~ - p . Hence

d'~ > d,,

i-

q + 1,...,s.

(15.45)

Using (15.41), (15.44) and (15.45), we have q

5(d) - r(n - s) + ~

(d'i - di) + + ~

i=p+l

(d'i - di).

(15.46)

i=q+l

Define a sequence e such that C(e) is obtained from C(d) by making the first p columns full and deleting the s-th column, i.e., , ei-

f n-l, / 0, d~

fori-l,...,p fori-s otherwise

It is easy to check (using s > m + 1) that e C Dn by Lemma 15.4.17. Further,

~(~) -

s--1

~(~,,-

~)+

i=1 q

= p(n-s+l)+

s-1

y~ (d'i - d i ) + +

y~ (d'i - p )

i=p+l

=

i=q+l q

~(~ - ~) + ( p - ~)(~ - ~) + p + ~

(d'~ - d~) + +

i=p+l

(d'~ - p)

(~s d' - p)

i=q+l q

=

r(n-s)+(p-r)(n-s)+p+

Z

(d',- d,) + +

i----p+l

(d'i - di) i--q+l

=

~

( p - di)

i-q+l

~(d) + ( p - ~)(~ - ~) + p -

~ i=q+l

( p - d~)

(by (15.46))

Threshold Weights and Measures

426

=

5(d)+(p-r)(n-s)+p-

)_2 ( P - r ) + i=q+l

=

~(d) + (p - ~)(~ - ~) + p -

~_~ ( d i - r ) i=q+l

(~ - q ) ( p - ~) + ~

(d~-

~)

i=q+l

> 6(d)+ ( p - r)(n - s ) - r(n - 1 - s) = ~(d) + ( p - ~)(~ - ~ ) - r(~ - ~ ) + >

~ ( d ) + ~.

( ~ ~ < p-

(by (15.43))

~)

So 5(e) >_ 5(d)+ 2. This proves Statement 1. We now prove Statement 2. If S'~(e) has the same parity as S~(d) (before the achievement of (15.41)), then f = e has the required properties. If the parities differ, take f = e + u l . In this case C ( f ) can be obtained from C(d) by making the first p columns full and deleting all the l's except the first 1 in the s-th column. So f C Dn by Lemma 15.4.17. The other required properties of e follow from S~(I) = S~(e)+ 1, 5~(f) = 5 1 ( e ) - 1, and 5~(f) = 5~(e) for i = 2,...,n. ,, L e m m a 15.4.28 relaxes one of the assumptions of L e m m a 15.4.27. Lemma

15.4.28

Under the conditions of Lemma 15.4.27 except r <_ p - r"

1. there exists a sequence e G Dn such that

e I

--

d l - 1 and 5(e) >_ 5(d) + 1;

2. there exists a sequence f C D~ such that (a) (i) f l -- dl - 1 or (ii) f a - dl and f~ - 1; (b) in case (i) above, 5(f) > 5(d) + 1; in case (ii) above, 5(f) >_ 5(d); (c) S~(f) and con(d) have the same parity. P r o o f . If r _< p - r, then the results follow from Lemm~ 15.4.27, so assume r > p - r. As in the proof of Lemma 15.4.27, we may assume that (15.41) holds. We also deduce Equation (15.46) as before. Define a sequence e such that C(e) is obtained from C(d) by deleting the s-th column. Then c E D~ by L e m m a 15.4.17 and

~(~)

s--1

-

E(~'~-

~)+

i=1 q

=

r(n+l-s)+

~ i=p+l

s-1

(d'i - d i ) ++ ~ i=q+l

(d'i - d i )

15.4

427

Threshold and Majorization Gaps

=

~(~ - ~) + ~ +

Z

(d~ - d~) + +

i=p+l

=

5(d)+r-p+r

>

~(d).

( d ' , - d,) - d: + d~ i=q+l

(by (15.46) and the definition o f p a n d r )

( ~ ~ > p - ~)

This proves Statement 1. To prove Statement 2, define f from e as in the proof of L e m m a 15.4.27. 9 L e m m a 15.4.29 is a variation of L e m m a 15.4.27. L e m m a 1 5 . 4 . 2 9 Let 0 ~ d C Dn. Put s - da + 1, p - dis, r - ds. A s s u m e that s - m + 1 and p - m (see Figure 15.12). Figure 15.12" Illustrating Lemma 15.4.29.

r

m8

.

m 8

0

-k

0

O.

Then 1. if m > 3, then there exists a sequence and 5(e) > 5(d) + 1;

e C Dn such that el - d l - 1

2. if m > 4, then there exists a sequence f c Dn such that (a) (i) f l = d l -

1 or (ii) f l - - d ,

and f~ = 1;

(b) 5 ( f ) > 5(d)-4- 1; (c) S ~ ( f ) and Sn(d) have the same parity. P r o o f . We begin by proving Statement 1. First, observe that m < n - 1, because m = n - 1 and p = m would imply Sin(d) > Sm(d'), contradicting d E Dn. Secondly, assume without loss of generality that (15.41) holds as in

428

Threshold Weights and Measures

t h e p r o o f of L e m m a 15.4.27. Define a s e q u e n c e e such t h a t C ( e ) is o b t a i n e d f r o m C(d) by m a k i n g t h e first m - 1 c o l u m n s full, if necessary, a n d d e l e t i n g t h e c o l u m n s - rn + 1, i.e.,

ei! m

Then e E

Dn

rtml 07

d~,

~ for i - 1 , . . . , r n for i - m + 1 otherwise.

1

by L e m m a 15.4.17. Using t h e facts d} - n -

di

m, d~ - m - 1 for i - r + 1 , . . . , definition of m a n d r), we o b t a i n

5(d) - r ( n -

1 - m)+ m-

m,

d i 'n +

r < (m-

1 -

m,

1 for i - 1 , . . . , r, 1 (by _< m -

dm+l - r

1)(n - 1 - m ) +

m - r.

Also, 5(e) - ( r n -

1)(n - 1 - ( r n

- 1)) - (rn - 1)(n - 1 - r n ) + m - 1.

It is now clear t h a t if r > 1, t h e n 5(e) > 5(d). If r = 1, t h e s a m e c o n c l u s i o n holds, since m - 1 > 1 a n d n - 1 - m > 0. T h i s proves S t a t e m e n t 1. W e now prove S t a t e m e n t 2. If Sn(d) before t h e a c h i e v e m e n t of (15.41) a n d S~(e) h a v e t h e s a m e parity, t h e n f = e has t h e r e q u i r e d p r o p e r t i e s . O t h e r w i s e t a k e f = e + Um+l. O n c e again, C ( f ) can be o b t a i n e d f r o m C(d) by m a k i n g t h e first m - 1 c o l u m n s full a n d d e l e t i n g all t h e l ' s e x c e p t t h e first 1 in c o l u m n m + 1. T h e r e f o r e f C Dn by L e m m a 15.4.17. F u r t h e r , (~(f) = ( ~ ( e ) - 1, as (~l(f) = ( ~ l ( e ) - 1, a n d 5~(f) = 5~(e) for i = 2 , . . . , r n . Hence 5 ( f ) = (rn - 1)(n - 1 - rn) + m - 2. N o w it is clear t h a t if r > 2, t h e n 5 ( f ) > 5(d). If r - 2, t h e s a m e c o n c l u s i o n holds since m - 1 > r ( b e c a u s e m >_ 4) a n d n - 1 - m > 0. Finally, if r - 1, t h e n t h e conclusion holds again, since

5(f)

-(m-2)(n-l-rn)+n-l-m+m-2 >_

(m-2)(n-l-rn)+m-1

>

n-l-rn+m-1

(asn-2>rn)

(asm>_4andn-l-m>O)

=

W e are now r e a d y to prove T h e o r e m s 15.4.25 a n d 15.4.26.

15.4

Threshold and Majorization Gaps

429

P r o o f o f T h e o r e m 1 5 . 4 . 2 6 . T h e s t a t e m e n t is true for n - 1, so a s s u m e n > 2. If d n - 0, t h e n by induction on n, 8(d) < f ( n 1) < f ( n ) , and f ( n - 1) < f ( n ) for n > 3. We m a y therefore assume t h a t d~ > 0, and c o n s e q u e n t l y m > 2. If m 2, then 5(d) < n - 2 by L e m m a 15.4.18, and hence 8(d) _< f(n). E q u a l i t y holds if and only if d~ - d2 and n - 2 - f ( n ) , which implies n < 4. Hence we m a y assume m _> 3. P u t s - dl + 1, p - d~s, q - dp~ + 1, r - ds, and observe as before t h a t p _< m < q, and m < s Proposition. If s > m + 1, then there exists a sequence e C Dn such that el

-

-

dl

-

1 and 5(e) > 5(d). Consequently we may assume that s - m.

To prove the proposition, first assume t h a t q > s. Define a sequence e such t h a t C(e) is o b t a i n e d from C(d) by m a k i n g the first p columns full, and deleting the s-th column. T h e n e C D~ and 5(e) > 5(d) by L e m m a 15.4.17. Now a s s u m e q < s. T h e n r d~ < dq+l < p - l , and h e n c e r < p. I f s > m + l or p < m, t h e n the required sequence e exists by L e m m a 15.4.28. If s - m + 1 and p - m, t h e n the required sequence e exists by L e m m a 15.4.29. This proves the proposition, and we m a y assume t h a t s - m. F u r t h e r , we m a y a s s u m e t h a t the first m - 1 columns of d are full, for otherwise we m a k e t h e m full w i t h o u t leaving D~, t h e r e b y increasing 8(d) by L e m m a 15.4.17. Then d - ( m - 1) n ,

5(d)- (m-

1)(n-

1 -(m-

1)),

and 5(d) reaches a m a x i m u m when n--1

m-

2 ' n ~ 7-1,7,

1 -

Therefore 5(d)_<

__tl )2, ~(~-1),

nodd neven.

(15.47)

n odd neven.

This m e a n s t h a t 5(d) <_ f(n). Further, for n _> 5, equality holds if and only if d - ( m - 1) n, where m - 1 is given by E q u a t i o n (15.47), i.e., if and only if d-[~-~l

Proof of Theorem

n

or

d - L - ~ - ! j n.

1 5 . 4 . 2 5 . We use the n o t a t i o n E~ - {d E D ~ - S~(d) even},

Threshold Weights and Measures

430

i.e., En(d) is the set of all proper degree sequences of length n. Since E~ C_ Dn, we have 5(d) _< f( n ) by Theorem 15.4.26. For n ~ 3 mod 4, f ( n ) - g(n) and hence 5(d) _< g(n). The cases of equality for d C Dn are when d - [ _ ~ ] n or d - [~-~]~. Since these d belong to E~ when n ~ 3 m o d 4 , all the conclusions of Theorem 15.4.25 are established in this case. However, for n = n-l~2 is odd, whereas (~(d)is even by Theorem 15.4.10, since 3 rood 4, f ( n ) - (--5-~ d is a degree sequence. Therefore (~(d) < f ( n ) - 1 - g(n) for n - 3 mod 4. It remains to track down the cases of equality for n - 3 mod 4. Thus from now on, we assume that n - 3 rood 4, n _> 7, and ~i(d) - g(n). If d n - - 0 , then ~(d) <_ g ( n - 1) < g(n), contradicting our assumption, so we may assume that dn > 0, and therefore m > 2. Also m - 2 implies, by L e m m a 15.4.18, that ~i(d) _< n - 2 < g(n), again contradicting our assumption, so we assume that m > 3 . P u t s - d l + l . Proposition. If s > m + 2, then there exists a sequence f E En such that

(~(f) > ~(d), contradicting ~(d) - g(n). Consequently we may assume that s-m ors-m+ l. We prove the proposition by showing that, when s > m + 2, 1. if d's - 1, then there exists an f E En such that 5(f) > 5(d); 2. if d'~ >_ 2, then there exists an f E E~ such that 6(f) >_ 5(d), and if equality holds, then fl - d~ (so that f and d have the same s) and f;-1. C a s e 1" d's - 1. The basic idea is to work with the sequence ( d l 1, d 2 , . . . , d~) and introduce a 1 at the end of the first row if the parity becomes odd. Put t - s - l , p - d ~ t , q d tp + l . T h e n p _ < m_< q. Assume that q >_ t. Define a sequence f such that C ( f ) is obtained from C(d) by deleting the last 1 in column s - 1 and the 1 in column s. Then f C E~ and 5(f) > 5(d) by L e m m a 15.4.17. Therefore we may assume q < t. Put r = dr. See Figure 15.13. C a s e 1.1: t > m + 1 or p < m. Define a sequence c such that C(c) is obtained from C(d) by deleting the last 1 in the first row. Note that c E D~ by L e m m a 15.4.17 and, further, c satisfies all the hypotheses of L e m m a 15.4.28. By the latter, there exists a sequence g E Dn such that 5(g) >_ 5(c) + 1. The required sequence f is defined by f

_ J" g, (g~ + 1 , 9 2 , . . . , g ~ ) ,

\

Sn(g) even S~(g) odd.

15.4

Threshold and Majorization Gaps

431

Figure 15.13" Illustrating Case 1 of the Proposition in the proof of Theorem 15.4.25. r

m

p

m

ts

.jrm

0

To see this, recall that in the proof of L e m m a 15.4.28, C(g) is obtained from C(c) by making the first p columns full and deleting the t-th column (if r _< p - r ) , or by making the first r columns full and deleting the t-th column (if r > p - r). Note that when S~(g) is odd, C(f) could be obtained from C(c) by making the appropriate columns full and then deleting all but the first 1 in column t. Therefore f C Dn in this case by L e m m a 15.4.17. Clearly f E Dn also holds when S~(g) is even. Further, Sn(f) is even in both cases. Thus f C E~. From the construction of c and f it is clear that 5(c) - (5(d)+ 1 and (5(f) _> ( 5 ( g ) - 1. Therefore 5(f) >_ 5 ( g ) - 1 _> (5(c) - 5 ( d ) + 1. 1.2" t - m + 1 and p - m. This is similar to Case 1.1, except that here we use L e m m a 15.4.29 instead of L e m m a 15.4.28. Case

C a s e 2" 2 _< d', _< rn. Put p - d'~ and q - d'p + 1. If q _> s, then we may remove the last two l's in the s-th column without leaving En and thereby increase (5(d) by L e m m a 15.4.17. We therefore assume that q < s. Then the required f exists by L e m m a 15.4.28.

This completes the proof of the proposition and hence we may assume that s - m or s - m + 1.

Threshold Weights

432

and

Measures

It is convenient here to define a new function. Let a _> 6 be a fixed integer such t h a t a - 2 m o d 4. For 0 _< k _< a, define k(a

h~(k) -

k(a-

-- k),

k)-

k

l,

even

k odd,

i.e.,

h~(k)- 2 L"~-"~ 2 J. Note t h a t the m a x i m u m of ha(k) occurs at k -

~2

m a x i m u m is ~2-4 4 " Proposition. if and only if

1, 7, 7 + 1~, ~

If s - m or s - m + 1, then 5(d) _< h n - l ( m -

d

-

(m-l)

d

-

(m -

~

and the

1), with equality

formodd

1) n - 1 ( m -

2)

or

d - m(m

-

1) n - 1

f o r m even.

To prove the proposition, consider first the case t h a t s - m. If columns 1 , . . . , m - 1 are m a d e full in this order, then d stays in Dn and 5(d) increases at each step by L e m m a 15.4.17. For odd m, the resulting d - ( m - 1) n also belongs to E~, and so it is the only sequence of E~ satisfying s - m t h a t m a x i m i z e s 5(d). T h e m a x i m u m in this case equals ( m - 1 ) ( n - 1 - ( m 1)) hn-1 (m1 ) . For even m, the above d does not belong to E~, and so the only sequence of E~ satisfying s - m t h a t maximizes 5(d) is the previous sequence in the filling-up process, n a m e l y d - ( m - 1 ) ~ - l ( m - 2). T h e m a x i m u m in this case equals ( m - 1)(n - 1 - ( m 1 ) ) - 1 - h~-i (m - 1). Now consider the case t h a t s - m + 1. We assert t h a t if d'~ _> 2, t h e n there exists f C E~ such t h a t 5( f ) > 5(d), and consequently we m a y a s s u m e t h a t d'~- 1. To prove the assertion we distinguish two cases. ! C a s e 1" 2 _< d~ < m. P u t p - d~, r - d~, q - dp + 1. We m a y a s s u m e t h a t q < s, for otherwise the last two l's in the s-th column of d m a y be r e m o v e d w i t h o u t leaving E~, t h e r e b y increasing 5(d) by L e m m a 15.4.17. T h u s q - m. If r _< p - r , then the required f exists by L e m m a 15.4.27, so we a s s u m e t h a t r > p - r. Using dli > rn and di - rn for i -- 1 , . . . , r, d~ - m - 1 and di - rn ' for i - r + 1 , . . . , p, d~ - di - r n - 1 for i - p + 1 , . . . , m , din+ 1 - - p, d m + l - r and d} - 0 for i - rn + 2 , . . . , n , we obtain 5(d) - ~ ( d ' i - di) + p i=1

r.

(15.48)

15.4

Threshold

and Majorization

Gaps

433

Let f be a sequence such that C(f) is obtained from C(d) by (a) deleting the s-th column and (b) adding a 1 at the end of the (r + 1)-st column if p is odd. Then f C En by L e m m a 15.4.17 and 7-

5(f)

>

~-~(d'~- ( d ~ - 1)) i=1 r

=

r

i=1

>

5(d).

(by (15.48) a n d r > p - r )

C a s e 2" d~, - m. If m > 4, then the required f exists by L e m m a 15.4.29. Now assume the special case m - 3. Then d - 32d'2-21n-d'2-1 with d~ > 2 (see Figure 15.14). Figure 15.14: Illustrating a special case in the proof of Theorem 15.4.25. m

* 1

1 1

1 1 * 1 1 0

1

1

m s

:

d;-t-1

*

:

1 1 1

n

8

1 1

0

1

If d~ - 2, then Sn(d) - n + 6 is odd, since n - 3 mod 4, contradicting d C En. Therefore d~ >_ 3. Hence 5(d) - ( n - 4 ) + + ( d ; - 3 ) + + ( 2 3) + + ( 3 - 2 ) + - n+d~-6 _< 2 n - 7 , and writing n - 4 k + 3 , we have 5(d) <_ 8 k - 1 < 4k 2 + 4 k - ( 2 k + 1) 2 - 1 - g(n), contradicting the assumption This completes the proof of the assertion, and consequently we may assume that d ~, - 1. If m is odd, then make the first m - 1 columns of C(d) full and delete the last 1 in the first row without leaving En, thereby increasing 5(d) by L e m m a 15.4.17. Therefore m may be assumed to be even. Then it is easy to

Threshold Weights and Measures

434

check as before that d = m ( m - 1)n-1 is the only sequence in En satisfying s =rn + 1 and d~ = 1 that maximizes 5(d), and 5 ( d ) = hn-1( r n - 1). This proves the proposition, and so 5(d) _< h ~ _ l ( r n - 1). The function h ~ _ l ( m - 1) reaches its m a x i m u m when rn-

1 - n-1 2

1'

or

rn-

1 --

n -21

'

or

m--l--

_~ +1.

The m a x i m u m i s ( _ ~ ) 2 _ 1 - g(n). By the previous proposition it is achieved only by the following sequences d: when r n -

1 - n-x 2

when r n -

1 - -~

1 (rn odd)" (rn even)"

d-

(

or d when r n -

)nl ~

2 ---fn + l (~__A1 ) ~-1

1 - - ~ + 1 (rn Odd)"

See [AP94] for related results about the difference gap and the fact that every degree sequence d such that R(d) - 1 has a realization that is both bithreshold and cobithreshold.