Cyclic Ordering is NP-Complete

The cyclic ordering problem is to recognize whether a collection of cyclically ordered triples of elements of a set T is derived from an arrangement of all the elements of T on a circle. This problem is shown to be NP-complete. A cyclic ordering of a set T = {l,. ., t ) is essentially an arrangement of the elements of T on a circle. A specific definition is as follows (see [4]). Two linear orders, (a, , . . . , a , ) and (b,, . . ., b,), on T are called cyclically equivalent if there exists a number q, 1 S q S t, such that ,u 1 = (v 1 -k q ) (mod t ) implies a, = b,. A cyclic ordering of T is an equivalence class of linear orders on T modulo cyclic equivalence; the equivalence class containing (a , , . . ., a,) will be denoted by a , a 2 . . a,. Cyclic ordering is the following recognition problem. The input is a set A of cyclically ordered triples (abbreviated COT's) out of T. The property to be recognized is: There is a cyclic ordering of T from which all the COT's in A are derived; A is called consistent if it has this property. Evidence for the hardness of cyclic ordering was given in r4]. On the other hand, the linear analogue of this problem is known to be easy. Specifically, the property that a set of ordered pairs out of T is derived from a linear order on T, is recognizable in linear time (see [3, Section 2.2.31). Our goal here is to prove that cyclic ordering is NP-complete'. Our problem is obviously in NP since it requires not more than polynomial time to verify that a set of COT's is derived from a certain cyclic ordering. In the remainder of the paper we shall show that satisfiability with at most 3 literals per clause (abbreviated ST3) is *The first author was partially supported by Bath Sheva Fund. ' The reader is assumed to be familiar with NP-completeness and related topics (see [1,2]); the notation for satisfiability is taken from (21. 180 Z. Galil, N. Megiddo reducible to cyclic ordering. This will imply, by definition, that cyclic ordering is NP-complete. The input of ST3 consists of clauses x, v y, v z , ( v = I , . . ., p) where {xu, y,, z,} C U = { u , , . . ., u,, u l , . . ., u,}. Without loss of generality assume that if x, E {u,, fi,), y, E {u,, i i , } and zV € {uk, i lk} then i < j < k . With each u, (7 = 1,. . ., r ) we associate a COTa&y,, and with ii, we associate the reverse COTa,y$,. Let A = {a , , p , , y,, . . . ,a, , pr, y,}. It is assumed that the set A has exactly 3r distinct elements. With each clause x v y v z ({x , y, z } C U) we associate a set A " of COT's as follows. Suppose that abc, def, ghi, are the COT's associated with x, y, 2, respectively ({a , b, c, d, e, f , g, h, i } C A ) . Let B = { j , k , I , m , n } be such that A fl B = 0 and assume that the B,-s that correspond to the various clauses x,, v y, v z , are I pairwise disjoint. Let A" = {acj, bjk, ckl, dfj, ejl, flm, gik, hkm, imn, n m l } . Lemma 1. Let S C U be such that u, E S if and only if u,e S. Let x v y v z be any clause. Let A be a set of COT's defined as follows. Every element of A' (the set of COT's associated with x v y v z ) belongs to A ; the COT's associated with the elements of { x , y, z} \ S belong to A ; if a p y is a COT associated with an element of { x , y,z} fl S then a y p belongs to A. Then, S f l { x , y , z } # 0 if and only if A is consistent. Proof. (Only if) The following table proves that A is consistent whenever s ~ { X , Y , Z I Z P ) . s ~ ( X , Y , Z ) A Every element of A is derived from { x } A" U {acb, def, ghi} {Y A" U {abc, dfe, ghi) ( 2 1 A" U {abc, def, gih) {x, Y 1 A" U {acb, dfe, ghi} {x, 21 A" U {acb, def, gih) {Y, 2 ) A" U {abc, dfe, gih} {x, Y , 2 1 A" U {acb, dfe, gih} ackmbdefjlnghi abcjkdmflneghi abcdefjklngimh ackmbdfejlnghi ackmbdefjlngih abcjkdmflnegih acbjkdmflnegih (If) Notice that if S f l { x , y, z ) = 0 then A = A " U {abc, def, ghi}. Thus, it is sufficient to show that A ' U {abc, def, ghi} is inconsistent which would be a contradiction. To that end, consider the following chains of implications: Cyclic ordering is NP-complete acl hl k ckl abc bcj cjk jkl, gik h k m 8mn ghi hik ikm kmn, llm k m n jkl--+ klm + lmn. These are interpreted as follows. Let C be any cyclic ordering of {a, b, c, . . ., n ) from which all the elements of A are derived. Thus, if abc is also derived from C then necessarily (since acj is derived from C ) bcj is derived from C, and this implies that cjk is derived from C (since bjk is derived from C ) , etc. It can be observed that if every element in A U {abc, def, ghi) is derived from C, then lmn is derived from C. However, this is absurd since nml E A ' . Thus, A0 U {abc, def, ghi} is inconsistent and the proof is complete. Corollary 2. Let S be as in Lemma 1. For every v (v = 1, . . ., p ) let A, denote the set A that corresponds to the clause x, v y , v 2,. Under these conditions, S n {x,, y,, 2,) 0 for v = 1 , . . ., p if and only if A U . . . U A, is consistent. Proof. The "if" part is immediate from Lemma 1. We shall prove the "only if" part. It follows from the "only if" part of Lemma 1 that each A, is derived from a cyclic ordering C, of the set of elements appearing in the COT'S of A,. We claim that there is a cyclic ordering Ctr of the set A such that the restriction of each C, to elements of A is derived from C,,. Specifically, this cyclic ordering of A is 8162 . . . S3, where ( L 2 , L l , S3r) = (ar, &, 7,) if UT E S and ( L 2 , &, , S3,) = (a,, A,, p,) if u, E S. This follows from our choice of the ordering of variables in each clause, the specific orderings shown in our table, and the fact that u, E S &Sf S. Since the B,-s are pairwise disjoint and none of them intersects A, it follows that Co can be extended to a cyclic ordering C of A U B 1 U . . . U B, such that every COT of A U . . . U A, is derived from C. Theorem 3. Let A 0, denote the set A associated with the clause x, v y, v z , (V = 1,. . . ,p ) . Then the conjunction (xl v y, v 2,) A . . . A (x, v y, A z,) is satisfiable if and only if the set A U . . . U A o, is consistent. . Proof. (Only if) If the conjunction is satisfiable then, by definition, there exists an S C U such that u, E S U,Sf S and S n {x,, y,,, z y } # 0 for v = 1, . . . , p . Corollary 2 implies that A :' U . . . U A is consistent. (If) Suppose that A: U . . . U A is consistent and let C be an appropriate cyclic ordering of A U B , U . . . U B,. Let S C U be the set of all x E U such that the C O T which is associated with x is not derived from C. Obviously, u, E S eJ U,ff S. 182 Z. Galil, N. Megiddo Furthermore, it follows from Lemma 1 that for every v ( v = 1,. . . , p ) S fl {x,, y,, 2 , ) # 8, since not all the COT's associated with x,, y,, z , are derived from C. This proves that the conjunction is satisfiable. 0 We have thus reduced ST3 to cyclic ordering. Note that for ST3 with p clauses the corresponding cyclic ordering has not more than lop COT's.