On a setup optimization problem for interval orders

Information Processing North-Holland

Letters

9 November

44 (1992) 5 l-55

On a setup optimization for interval orders Ahmad

1992

problem

Sharary

Mathematics Department, College of Science, King Saud University, P.O. Box 2455, Riyadh 11451, Saudi Arabia

Nejib Zaguia Computer Science Department, University of Ottawa, 34 George Glinski, Ottawa, Ontario, Canada KIN 6N5 Communicated by F. Dehne Received 2 April 1992

Abstract Sharary, 51-55.

A. and N. Zaguia,

On a setup optimization

problem

for interval

orders,

Information

Processing

Letters

44 (1992)

We prove that the jump number problem for interval orders is reducible to the maximum clique problem in a certain class of graphs. As applications, we solve the jump number problem for the interval orders with no special cycles, and we give an approximation for the number of jumps of any interval order.

Keywords: Combinatorial

problems;

interval

orders;

linear

Introduction Let P be a finite ordered set. A linear extension L of P is a total ordering x1 x2 . . . x, of P preserving the relations on P, that is, xi
Correspondence to: Dr. N. Zaguia, Computer Science Department, University of Ottawa, 34 George Glinski, Ottawa, Ontario, Canada KIN 6N5. 0020.0190/92/$05.00

0 1992 - Elsevier

Science

Publishers

extension;

jump number;

maximum

clique

denoted by s(P, L), and the jump number of P, denoted by s(P), is the minimum of s(P, L) taken over all linear extensions of P. A linear extension L of P with s(P, L) =s(P) will be called optimal. The jump number problem is to compute s(P) and to find an optimal linear extension of P. We denote by O(P) the set of all optimal linear extensions of P. Actually, the jump number problem is a precedence constrained scheduling problem written in the language of ordered sets. A set of tasks subject to precedence constraints is to be scheduled. Every task scheduled after another task which is not constrained to precede it requires a “setup” or “jump” which leads to a fixed additional cost. The problem is to find a schedule which minimizes the number of jumps.

B.V. All rights reserved

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The jump number problem has been shown to be NP-hard even for bipartite ordered sets, (see [l]>. Nevertheless efficient methods for minimizing jumps have been found for large classes of ordered sets such as N-free orders [7], cycle-free orders [3], orders with bounded width [2], and bipartite orders of dimension two [8]. An ordered set P is an interval order if each element x of P is mapped onto an interval 1, on the real line such that for all x and y in P, x < y in P if and only if Z, completely precedes 1, (i.e. sup 1, G inf I,). An equivalent definition is that an ordered set P is an interval order if and only if P does not contain a subset {a, 6, c, d} such that a < b and c < d are the only comparabilities among these elements. The complexity status of the jump number problem of interval orders was an open question for some time. Recently, Mitas [6] has proved that it is an NP-complete problem. Moreover, there are approximation algorithms for the jump number problem of interval orders (see [5,6]). In this paper too, we are interested in the case of interval orders. Our main result is to show that finding the number of jumps of an interval order P is equivalent to finding the size of a maximumsize clique in a graph whose vertices are the covering pairs of P. As an application of this result, we shall solve the jump number problem

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9 November

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for interval orders which contain no special cycles, by giving a formula for the jump number. [This generalizes an algorithm of Faigle and Shrader [4] which solves the case of interval orders with no 4-crowns as suborders.] Also, we shall obtain a lower bound and an upper bound for the number of jumps of any interval order P, in terms of some parameters of P. Throughout this paper (P, <>, or simply P, denotes a finite ordered set. We write a - b whenever a is not comparable to b. We say that a covers b or b is covered by a, denoted by b- < a, if b < a and if b < c < a then a = c. We call a an upper cover of b. Let (P, <1 1 be an ordered set. Consider the set E(P) = {(x, y>: x- <1 y), the set of all covering pairs of P. Define a relation
if and only if

(a, 6) = (c, d) or a <, d and c 4, b. Actually we shall prove (E(P), =g2) is an ordered set whenever P is an interval order. Now let P be an interval order. Let e = (x, y) and e’ = (2, t) belong to E(P). We call (e, e’) a forbidden pair if y = z and e-C, e’. For every x in P, we shall write Q, = {y E P: y -x}. Notice that e(x, y) and e’ = (z, t> is a forbidden pair if and only if y = z and E(Q,> f $. Moreover, if e(x, y> and e’ = (y, t) then E(Q,) = {e”: e <2 e” <* e’}. For instance, Fig. 1 illustrates an interval order P and the corresponding ordered set E(P), where we have denoted ( y, a) by 1, (y, b) by 2, and so on, and where dotted lines join forbidden pairs. Our first result is to show that there is a relationship between s(P) and the maximum size of a chain in E(P) with no forbidden pairs. Theorem 1. Let P be an interval order. Then s(P) = I P I - k - 1, where k is the maximum size of a chain in E(P) with no forbidden pairs.

P

E(P)

Fig. 1. 52

If we delete from the undirected comparability graph of E(P) all edges connecting the forbidden pairs, we obtain a graph which we denote by G(P). Thus the problem of finding s(P) is equivalent to the maximum clique problem in G(P). Notice that the maximum clique problem is known

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e

d

C a Hh Fig. 2.

to be NP-complete in general. Although it can be shown that E(P) is an N-free order, however, as far as we know, G(P) does not belong to any known class of graphs for which the maximum clique problem is solvable in polynomial time. For an ordered set P, the length of P, denoted by h(P), is the maximum size of a chain in P. As a first application of Theorem 1, we shall obtain the following lower bound and upper bound for s(P). Theorem 2. Let P be an interval order. Then ]PI-k-l). An ordered set P is called C-free if it contains no subset {a, b, c, d, e} where a
Proofs of the results Before theorems, lemmas.

we get to the proof of each of the we shall give a sequence of necessary

Lemma 3. If P is an interval order, then (E(P), <,) is an ordered set. Proof. Clearly, <2 is reflexive and antisymmetric. Let (a, b) + (c, d) and (c, d) <2 (x, y). If b <1 x

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then (a, b) <2 (x, y). If b -x then comparing the covering edges (a, b) and (x, y) in the interval order P we conclude that a <, y; hence (a, b) + (x, y). Now assume that x
=

c

(lC+l)=

IPI-s(L)-1.

l
Therefore,

s(L) = I P I - I K I - 1.

q

Lemma 5. Let P he an interval order, and let K be a chain in E(P) with no forbidden pairs. Then K induces a linear extension L of P such that s(L) < IPI-IKl+m-1. Proof. Let K be the chain e, c2 e2 c2 . . . c2 ek, and let e’i = {x, y) whenever ei = (x, y). Define a relation < on lJ 1Cigkti as follows: x
IP-

UIGIGktjl+t<

IPI-(k-m+t)+t =IPI-k+m.

Therefore,

s(L) < I P I -k + m - t.

q

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Proof of Theorem 1. It is an immediate quence of Lemma 4 and Lemma 5.

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Proof of Theorem 2. By Lemma 4, I P I -k - 1~ s(P). Let K = {e, <2 . . . <* ek} be a chain in E(P) of size k, and let (yi, zi) ,..., (y,, z,> be all forbidden pairs of K. Clearly yi f yj and zi Z z, for 1 < i,j < m, i #j; but it is possible that z, = y, for some i and j. A forbidden pair (yi, zi> is said to be of first kind if there is a forbidden pair ( yj, zj> such that either z, = yj or yi = zj. Let A be a subchain of K which consists exactly of the elements of all {yi, ZJ such that ( yi, z,) is a forbidden pair of the first kind, and let B be the subchain K -A. Let t be the least number of indices for which zi = y, for some i and j, and such that when we delete the elements corresponding to these indices from A we obtain a subchain A’ without forbidden pairs. Notice that t < [u/2] where a = I A]. Let K’ be the chain which consists of the elements of A’ and the elements of B, and let r be the number of forbidden pairs of K’. Then r < [b/21 where b= IBI, and IK’I = IKI-tak-[a/21. Now we apply Lemma 5 to K ‘. Indeed, if L’ E L(P) is a linear extension induced by K’, then noting that k = a + b, we have s(L’)
I PI -(k - [a/21) + [b/2] - 1

< IPI-[k/21-1.

•I

Lemma 6. Let P be a C-free interval order. Let (a, b) and (6, c) be a forbidden pair in E(P). Then there exists either a minimal element (x, y) of E(QJ such that (a, x) is a covering edge or a maximal element (x, y) of ECQ,) such that (Y, c> is a covering edge. Proof. Since (a, b) and (b, c) is a forbidden pair, E(Qb) # @. Let (x, y> be a minimal element of E(QJ. Then x -b and y -b. Now, if a <, x then a- in E(Qh). On the other hand, if x -a then a be a maximal element of E(Q,). For a similar reason as above if z1<, c then 54

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v- <, c, hence if L!- c then u <, c. Clearly a <, c’ and x <, c. Since (u, c) a2 (x, y> then necessarely x <, c. Thus {a, x, b, c, v] will contradict 0 the fact that P is C-free. Lemma 7. Let P be a C-free interval order. Let (a, b) and (b, c) be a forbidden pair in E(P). Let (x, y) be a minimal element of E(Q,) such that (a, x) E E(P) (respectively a maximal element of E(QJ such that (y, c) E E(P)). Then (a, b) and (a, x) (respectively (b, c) and (y, c)) have exactly the same down set and the same upper set in E(P). Proof. Let (a, 6) - (a, x) then a<,~ and u<,x. But u-b (if b<,u then b <, x which contradicts x E Qh, and if u <, b then (a, 6) 4:z (u, v) which contradicts (a, b) c2 (u, c)) and thus u E Qh. Since u is not a minimal element of E(Q,), which contradicts the choice of (x, y>. Therefore, (a, x> cz (u, c). Now assume that (u, v> c2 (a, b). If v -a then either u . Thus a <, v and u 4, x. If x <1 u then (a, b) cz (u, v). Hence, let u -x. If a . Since u -x then u 4, a. So we may assume that u - a. If u <, b then u <1 y, and {a, u, x, y, b] contradicts the fact that P is C-free. Thus u b, b and consequently (a, b) . 0 Lemma 8. Let P be a C-free interval order. Let K={x, c2 ... c2 xk} be a maximal chain in E(P). Assume that (9 (x1, xi) is a forbidden pair for some i < j, (ii) (x,~, x,) is not a forbidden pair for every m
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Then there exists either XI E E(P) such that

is a chain in which (xl, x,,,) is not a forbidden pair for i < m
C2 x,i_, -Cz x; C2 xi+1 -C2 *. . c2 Xk

is a chain in which (x,,,, xi) is not a forbidden pair for m is a forbidden some m . Case 2: (x, y) is a maximal element of E(QJ and (y, c) E E(P). Notice that, by Lemma 7, (b, c) and (y, c) have the same down set and the same upper set in E(P). Let xi = (y, c). Clearly, (x,., x,‘) is not a forbidden pair because b f y. Assume that (xm, x:) is a forbidden pair for some

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m < i. Then x, = (t, y) for some t E P. Since (x, y) E E(Qb) then a
References [l] V. Bouchitte and M. Habib, Some NP-completeness properties about linear extensions, Order 4 (1987) 143-154. [2] C.J. Colbourn and W.R. Pulleyblank, Minimizing setup in ordered sets of fixed width, Order 1 (1985) 225-229. [3] D. Duffus, I. Rival and P. Winkler, Minimizing setups for cycle-free ordered sets, Proc. Amer. Math. Sot. 85 (1982) 509-513. [4] U. Faigle and R. Schrader, A setup heuristic for interval orders, Oper. Res. Lett. 4 (1985) 185-188. [5] S. Felsner, A 3/2-approximation algorithm for the jump number of interval orders, Order 6 (1990) 325-334. [6] J. Mitas, Tackling the jump number of interval orders, Order 8 (1991) 115-132. [7] I. Rival, Optimal linear extensions by interchanging chains, Proc. Amer. Math. Sm. 83 (1983) 387-394. [8] G. Steiner and L.K. Stewart, A linear algorithm to find the jump number of 2-dimensional bipartite partial orders, Order 3 (1987) 359-367.

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