Abstract. We show that a proper coloring of the diagram of an interval order I may require 1 +

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1 Colorings of Diagrams of Interval Orders and -Sequences of Sets STEFAN FELSNER 1 and WILLIAM T. TROTTER 1 Fachbereich Mathemati, TU-Berlin, Strae des 17. Juni 135, 1000 Berlin 1, Germany, partially supported by the DFG, felsner@math.tu-berlin.de Bell Communications Research, 445 South Street L-367, Morristown, NJ 0796, U.S.A., and Department of Mathematics, Arizona State University, Tempe AZ 8587, U.S.A. wtt@bellcore.com Abstract. We show that a proper coloring of the diagram of an interval order I may require 1 + dlog height(i) e colors and that + dlog height(i) e colors always suce. For the proof of the upper bound we use the following fact: A sequence C 1 ; : : : ; C h of sets (of colors) with the property () C j 6 C i?1 [ C i for all 1 < i < j h. can be used to color the diagram of an interval order with the colors of the C i. We construct -sequences of length n? + n?1 using n colors. The length of -sequences is bounded by n?1 + n?1 and sequences of this length have some nice properties. Finally we use -sequences for the construction of long cycles between two consecutive levels of the Boolean lattice. The best construction nown until now could guarantee cycles of length (N c ) where N is the number of vertices and c 0:85. We exhibit cycles of length 1 N. 4 Mathematics Subject Classications (1991). 06A07, 05C35. Key Words. Interval order, diagram, chromatic number, Hamiltonian path, Boolean lattice. 1. Introduction and Overview For a nonnegative integer, let I be the interval order dened by the open intervals with endpoints in f1; : : :; g. It has height? 1 and is isomorphic to the canonical interval order of this height (see Furedi, Hajnal, Rodl and Trotter [1] for canonical interval orders). Two vertices v and w in I are a cover, denoted by v w, exactly if the right endpoint of the interval of v equals the left endpoint of the interval of w. The diagram D I of I is thus recognized as the shift graph G( ; ) (see [1] for shift graphs). In general we denote by D I the diagram of an interval order I, and we denote the chromatic number of the diagram by (D I ). We include the (well-nown) proof of the next lemma since we will need similar methods in later arguments.

2 S. FELSNER AND W. T. TROTTER LEMMA 1.1. (D I ) = dlog height(i ) e = Proof. Suppose we have a proper coloring of D I with colors f1; : : :; cg. With each point i associate the set C i of colors used for the intervals having their right endpoint at i. Note that C 1 = ;. For 1 i < j, we have C j 6 C i ; otherwise the interval (i; j) would have the same color as some interval (l; i). This proves that all of the subsets C i of f1; : : :; cg are distinct; therefore c and c. A coloring of D I using colors can be obtained by the following construction. Tae a linear extension of the Boolean lattice B and let C i be the i th set in this list. Assign to the interval (i; j) any color from C j n C i. A coloring obtained in this way is easily seen to be proper. We derive a remar for later use and a theorem from this construction. REMARK 1.. In a coloring of D I is incident with an interval of each color. which uses exactly colors, every point i f1; :::; g Proof. The crucial fact here is that every subset of f1; : : :; g is the C i for some i. Now choose any i f1; : : :; g and a color c f1; : : :; g, we have to show that an interval of color c is incident with i. If c C i, then this is immediate from the denition of C i. Otherwise, i.e., if c 6 C i, then there is a j c > i such that C jc = C i [ fcg and the interval (i; j c ) is colored c. With the next lemma we improve the lower bound: There are interval orders I with (D I ) 1 + log (height(i)). Compared with Lemma 1.1, this is a minor improvement, but we feel it worth stating, since later we will prove an upper bound of +log (height(i)) on the chromatic number of the diagram of I. LEMMA 1.3. For each there is an interval order I such that (D I ) 1 + log height(i ) = Proof. Tae I as the order obtained from I (see Lemma 1.1) by removing the intervals of odd length, i.e., the interval order dened by the open intervals (i; j) with i; j n 1; : : :; o and j? i 0 (mod ). The height of I is?1? 1 which is the height of I?1 ; however, as we are now going to prove, a proper coloring of I requires at least colors. Note that two intervals (i1; j1) and (i; j) with j1 i induce an edge in the diagram of I if either j 1 = i or j1 = i? 1. In I we nd an isomorphic copy of I?1 consisting of the intervals (i; j) with both i and j odd. Call this the odd I?1. The even I?1 is dened by the interval (i; j) with i and j even. Let C i be the set of colors used for intervals with right end-point i? 1, and

3 COLORINGS OF DIAGRAMS 3 let D i be the set of colors used for intervals with right end-point i. From Lemma 1.1, we now that if both the odd and the even copy only need? 1 colors, then the C i and the D i have to form linear extensions of the Boolean lattice B?1. Now dene C i as the set of colors used for intervals with left-endpoint i? 1. From Remar 1., we now that C i is exactly the complement of C i. With the corresponding denition, D i and D i are seen to be complementary sets as well. Note that a proper coloring requires C i \ D i = ;. We therefore have C i D i. A similar argument gives D i C i+1. Altogether we nd that the C i have to be a linear extension of B?1 with C i C i+1 for all i. This is impossible. The contradiction shows that at least colors are required. Now we turn to the upper bound which we view as the more interesting aspect of the problem. THEOREM 1.4. If I is an interval order, then (D I ) + log height(i) Proof. In this rst part of the proof, we convert the problem into a purely combinatorial one. The next section will then deal with the derived problem. Let I = (V; <) be an interval order of height h, given together with an interval representation. For v V, let (l v ; r v ] (left open, right closed) be the corresponding interval. With respect to this representation, we distinguish the `leftmost' h-chain in I. This chain consists of the elements x 1 ; : : :; x h where x i has the leftmost right-endpoint r v among all elements of height i. It is easily checed that x1; : : :; x h is indeed a chain. Now let r i = r xi be the right endpoint of x i 's interval and dene a partition of the real axis into blocs. The i th bloc is B(i) = [r i ; r i+1 ): This denition is made for i = 0; : : :; h with the convention that B(0) extends to minus innity and B(h) to plus innity. In some sense these blocs capture a relevant part of the structure of I. This is exemplied by two properties. The elements v with r v B(i) are an antichain for each i. This gives a minimal antichain partition of I. If r v B(j), then l v B(i) for some i less than j. Suppose we are given a sequence C1; : : :; C h of sets (of colors) with the following property () C j 6 C i?1 [ C i for all 1 < i < j h. A sequence with this property will henceforth be called an -sequence. The -sequence C1; : : :; C h may be used to color the diagram D I with the colors occurring in the C i.

4 4 S. FELSNER AND W. T. TROTTER The rule is: to an element v V with l v B(i) and r v B(j) assign any color from C j n (C i?1 [ C i ). This set of colors is nonempty by the property of the sequence C i, since i < j. We claim that a coloring obtained this way is proper. Assume to the contrary that there is a covering pair w v such that w and v obtain the same color. Let r w B() and l v B(i). Since w v, we now that i. Due to our coloring rule, we now that the color of w is an element of C and the color of v is not contained in C i?1 [ C i ; hence < i? 1. This, however, contradicts our assumption that w v, since l xi B(i? 1) and l v r xi = r i gives w < x i < v. We have thus reduced the original problem to the determination of the minimal number of colors which admits a -sequence of length h. We will demonstrate in the next j section, Lemma.1 and Lemma.3, how to construct a -sequence of length n? + n+1 using n colors. This will complete the proof of the theorem. j In the third section we give an upper bound of n?1 + n+1 for the maximal length of a -sequence. From the proof, we derive some further properties -sequences of this length necessarily satisfy. Finally we apply the construction of long -sequences to the problem of nding long cycles between two consecutive levels of the Boolean lattice. A famous instance of this problem is the question whether there is a Hamiltonian cycle between the middle two levels of the Boolean lattice (see e.g. Kierstaed and Trotter [] or Savage [3]. The best constructions nown until now could guarantee cycles of length (N c ) where N is the number of vertices and c 0:85. We exhibit cycles of length 4 1 N.. A Construction of Long -Sequences Let t(n; ) denote the maximal length of a sequence C i of sets satisfying: (1) C i f1; : : :; ng, () jc i j = and () if i < j then C j 6 C i?1 [ C i. LEMMA.1. t(n; ) n? 1 Proof. The sequences actually constructed will have the additional property (4) jc i?1 [ C i j = + 1 for all i. The proof is by induction. For all n and = 1 or = n the claim is obviously true. Now suppose that two -sequences as specied have been constructed on f1; : : :; n?1g: rst a sequence of -sets A = A1; : : :; A s of length?? n? s = + 1, and second a sequence n? of (? 1)-sets B = B 1 ; : : :; B t of length t = + 1?1 + 1.

5 COLORINGS OF DIAGRAMS 5 Property (4) guarantees that there is a permutation of the colors such that A s = B1 [ B. Now let Ai ; if 1 i s C i = Bi?s+1 [ fng; if s +? 1 i s + t? 1. n?1 The length of the new sequence is s + t? 1 = + 1. Properties (1) and () are obviously true for the sequence C i and property (4) is true for both the A and the B sequence. These observations and the choice of give property (4) for the C sequence. It remains to verify property. If i < j < s + 1, this property is inherited from the A sequence. If s + 1 < i < j, it is inherited from the B sequence. In case i < s + 1 j, we have n C j and n 6 C i?1 [ C i. The remaining case is s + 1 = i < j. Here the choice of and the sacrice of B1 show that C s [ C s+1 = A s [ B [ fng = B 1 [ B [ fng. Again the property can be concluded from this property for the B sequence. For = and = n? 1, we can prove that the inequality of Lemma.1 is tight, but in general the value of t(n; ) is open. PROBLEM.. Determine the true value of t(n; ). Let T (n) denote the maximal length of a sequence C i of sets satisfying: (1) C i f1; : : :; ng and () if i < j then C j 6 C i?1 [ C i. LEMMA.3. T (n) X n odd n? 1 ( + 1 ) = n? n Proof. Let L(n; ) be the (n; ){sequence constructed in the preceding lemma. We claim that L = L 1(n; 1) L 3(n; 3) L 5(n; 5) : : : with appropriate permutations j is a -sequence of subsets of f1; : : :; ng. The 's can be found recursively. 1 = id and if? has been determined, then is chosen as a permutation such that the last set of the sequence L? (n; {) is a subset of the rst set of L (n; ). Let Ci be the i th set in the sequence L. We now chec property. If the three sets C i?1, C i and C j are in the same subsequence L (n; ), then the property is inherited from this subsequence. If C i L (n; ) and Cj L 0 (n; 0 ) with 0?, then jc i?1 [ C i j < Cj is a consequence of property (4) for the subsequence L (n; ), and gives the claim in this case. There remains the situation where C i?1 is the last set of its subsequence. The choice of the gives C i?1 C i and the property reduces to C j 6 C i, which is obvious. The length of L is the sum over the length of the L j (n; ) used in L. This is the + 1 with odd, which is n? + n+1 sum over? n?1.

6 6 S. FELSNER AND W. T. TROTTER 3. The Structure of Very Long -Sequences. THEOREMj 3.1. Let C = C1; : : :; C t be a -sequence of subsets of f1; : : :; ng. t n?1 + n+1 Proof. We start with some denitions. For 1 i t? 1, let Then S i = f S : C i+1 S C i [ C i+1 g (1) and s i = js i j. Observe that with r i = jc i n C i+1 j we have the equation s i = r i? 1: () We now prove two important properties of the sets S i S i \ S j = ; if i 6= j. Assume to the contrary that S S i \ S j and let i < j. From the denition of the S i, we obtain C j+1 S C i [ C i+1 which contradicts the property of the sequence C. C \ S i = ; for all i. Assume C j S i. If j i, then C i+1 C j gives a contradiction. If j = i + 1, note that C i+1 6 S i from the denition. If j > i + 1, the contradiction comes from C j C i [ C i+1. Therefore, C and the S i are pairwise disjoint subsets of B n. This gives the inequality X t?1 n t + s i (3) i=1 We now partition the indices f1; : : :; t? 1g into three classes I1 = fi : jc i j = jc i+1 jg; note that i I1 implies s i 1. I = fi : jc i j < jc i+1 jg; trivially s i 0 for i I. I 3 = fi : jc i j > jc i+1 jg; note that if i I 3, then the corresponding s i is relatively large, i.e., s i jc ij?jc i+1 j+1? 1. This estimate is a consequence of Equation and the fact that C i+1 has to contain an element not contained in C i. First we investigate the case I 3 = ;. This condition guarantees that the sizes of the sets in C is a nondecreasing sequence. Since B n has n + 1 levels, the size of the sets in C can increase at most n times, i.e., jij n and ji1j t? 1? n. It follows that: n t + X ii 1 s i + X ii s i t + ji1j t + (t? 1? n) j This gives t n + (n + 1); hence t n?1 + n+1 in this case. The case I3 6= ; is somewhat more complicated. Let the number of descending steps be d and I 3 = fi 1 ; : : :; i d g. Let m ij denote the number of levels the sequence is decreasing

7 COLORINGS OF DIAGRAMS 7 when going from C ij to C ij +1, i.e., m ij = jc ij j? jc ij +1j and s ij m i j +1? 1. Again we can estimate the size of I, namely ji j n + P d j=1 m i j. It follows that: n t + X ii 1 s i + X ii s i + X ii 3 s i dx t + ji 1 j + ( m i j +1? 1) j=1 t + ((t? 1)? jij? ji3j) + m i j +1? d j=1 dx t + (t? 1? n? m ij? d) + m i j +1? d j=1 j=1 dx dx Comparing this with the calculations made for the case I3 = ;, we nd that t j n?1 + n+1 m i j > m ij? ; hence the above inequality can never hold. would require? P d j=1 m i j? d + P d j=1 m i j For each j, we have REMARK. Let T (n) = +j n?1 n+1 be the upper bound from the theorem. We have seen that a -sequence C of length T (n) can only exist if I3 = ;. Moreover, the following conditions follow from the argument given for Theorem 3.1. (1) There are exactly n increasing steps, i.e., jij = n. () If i I 1, then s i = 1, i.e., any two consecutive sets of equal size have to be a shift: C i+1 = (C i n fxg) [ fyg with x C i and y 6 C i. If i I then s i = 0, i.e., if jc i j < jc i+1 j, t en there is a containment C i C i+1. (3) Every element of B n is either an element of C or appears as the unique element of some S i, i.e., as C i [ C i+1. From this observations, we obtain an alternate interpretation for a sequence C of length T (n) in B n. In the diagram of B n, i.e., the n-hypercube, consider the edges (C i ; C i+1 ) for i I and for i I1 the edges (C i ; T i ) and (T i ; C i+1 ) where T i is the unique member of S i, i.e., T i = C i [ C i+1. This set of edges is a Hamiltonian path in the hypercube and respects a strong condition of being level accurate. After having reached the th level for the rst time the path will never come bac to level? (see Figure 1 for an example, the bullets are the elements of a very long -sequence).

8 8 S. FELSNER AND W. T. TROTTER Figure 1 A level accurate path in B4. PROBLEM 3.. Do sequences of length T (n) exist for all n? We are hopeful that such sequences exist. Our optimism stems in part from computational results. The number of sequences starting with ;; f1g; fg; : : :fng is 1 for n 4, 10 for n = 5, 13 for n = 6 and there are thousands of solutions for n = 7. The next case n = 8 could not be handled by our program, but Marus Fulme (personal communication) wrote a program which also resolved this case armatively. 4. Long Cycles between Consecutive Levels in B n Let B(n; ) denote the bipartite graph consisting of all elements from levels and + 1 of the Boolean lattice B n. A well nown problem on this class of graphs is the following: Is B( + 1; ) Hamiltonian for all? Until now it was nown that this is the case for 9. Since the problem seems to be very hard, some authors have attempted to construct long? cycles. The best results (see Savage [3]) lead to cycles of length (N c ) +1 where N = is the number of vertices of B( + 1; ) and c 0:85. THEOREM 4.1. In B(n; ), there is a cycle of length 4 max n? 3? 1 + 1; n? 3 n?? + 1 Proof. Note that the graphs? B(n; ) and B(n; n??1) are isomorphic, it thus suces to n?3 exhibit a cycle of length 4?1 + 4 in B(n; ). To this end, tae a -sequence C 1; : : :; C? t of (? 1)-sets on f1; : : :; n? g. From Lemma.1, we now that n?3 t?1 + 1 can be achieved. Now consider the following set of edges in B(n; )

9 COLORINGS OF DIAGRAMS 9 C i [ fng ; C i [ C i+1 [ fng C i [ C i+1 [ fng ; C i+1 [ fng C t [ fng ; C t [ fn? 1; ng for 1 i < t, and C i [ fn? 1g ; C i [ C i+1 [ fn? 1g C i [ C i+1 [ fn? 1g ; C i+1 [ fn? 1g C 1 [ fn? 1g ; C 1 [ fn? 1; ng for 1 i < t, and C t [ fn? 1; ng ; C t [ fn? 1g for 1 i < t, for 1 i < t, C 1 [ fn? 1; ng ; C 1 [ fng The proof that this set of edges in fact determines a cycle of length 4t in B(n; ) is straightforward. With a simple calculation on binomial coecients, we obtain a nal theorem THEOREM 4.. There are cycles in B( + 1; ) of length at least 1 4 N.,. References [1] Z. Furedi, P. Hajnal, V. Rodl and W. T. Trotter, Interval Orders and Shift Graphs, Proceeding of the Hajnal/Sos Conference on Combinatorics, Budapest, 1991, to appear. [] H. A. Kierstead and W. T. Trotter, Explicit Matchings in the Middle Two Levels of the Boolean Lattice, Order 5(1988), [3] C. Savage, Long Cycles in the Middle Two Levels of the Boolean Lattice, preprint, 1990.