TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 205, 1975 ASYMPTOTIC ENUMERATION OF PARTIAL ORDERS ON A FINITE SET BY D. J. KLEITMANi1) AND B. L. ROTHSCHILD(2) ABSTRACT. By considering special cases, the number Pn of partially ordered sets on a set of n elements is shown to be (1 + 0(\¡n))Qn, where Qn is the number of partially ordered sets in one of the special classes. The number Qn can be estimated, and we ultimately obtain '»-('+0©)C5lC)(7)<''-^-^)- 1. Introduction. It is known [7] that the logarithm (base 2, as all logarithms in this paper will be) of the number Pn of partial orders (or equivalently of T0 topologies) on a set of tj elements is T22/4 + o(tj2). In this paper we show that Pn is asymptotically equal to Qn, the number of partial orders in a certain special class which is characterized in a simple way. It will follow that log Pn = «2/4 + 3n/2 + 0(log tj). An explicit asymptotic formula for Pn will be given, but it is a bit messy. In [5] the number Gn of "graded" partially ordered sets is enumerated. Since the partial orders counted by Qn turn out to be graded, the number arrived at in [5] is also asymptotically equal to Pn. The computation of Gn [5], then, is asymptotically applicable to Pn. The methods used here are similar to those used in [7] for obtaining the asymptotic estimate for log Pn, but here they are somewhat more delicate and more complicated. We use induction to show that Pn < Qn(l + 0(1/«)), while Qn < Pn by definition. The proof is accomplished by obtaining all partial orders on tj + 1 elements from those on tj or fewer elements in certain specified ways. In each of these ways, except those corresponding to Qn + x, we obtain only an asymptotically negligible number of partial orders. The partial orders corresponding to Qn can be described as follows. They consist of three sets Lx, L2, L3 with \LX\, \L3\ = tj/4 + o(n), \L2\ = ti/2 + o(tj). Each element in L¡ "covers" only elements in L¡_x (see below for definitions). And finally, each element behaves in the "average" way. That is, each element in Received by the editors November 2, 1973 and, in revised form, January 21, 1974. AMS (MOS) subject classifications (1970). Primary 05A15. (x) Work supported in part by ONR Grant 0014-67-A-0204-0063. (2) Work supported in part by NSF Grant GP-33580X. Copyright © 1975, American Mathematical Society 205 206 D. J. KLEITMAN AND B. L. ROTHSCHILD L¡ is covered by (asymptotically) half of those in Li+l and covers half of those inL¡_x. 2. Terminology. We represent a partial order P on a set of tj elements by its unique Hasse diagram, also denoted by P. This is a directed graph with the elements of P as vertices and a single directed edge from a to ft if and only if a covers b in P (a covers ft if a > ft and a > c>b implies c = ft). Distinct partial orders have distinct diagrams. Thus we let Pn denote both the number of partial orders and the number of diagrams on n elements, as directed graphs diagrams are characterized by the exclusion of two types of configurations. Namely, a directed graph is a diagram of a partial order if and only if it contains no set of (directed) edges Ex, E2, • • • , Ek, E0 such that the terminal vertex of E¡ is the initial vertex of El+X, 1 m(l + m~3/8)/2. Let Bm + i = \B(V)\. D(V): Diagrams with a (v, ß>set with \C(Q)\ < t?j(1 - ttj~3/8)/2. Let Dm + l = \D(V)l E(V): Diagrams with at least 30 nonempty levels. Let Em + X = |Zr(I0l. F(V): Diagrams with a (v, Q)-set with \C(Q)\ > m(l - m~3,8)/2, such that the smallest set R of vertices incident with every edge connecting two vertices in V- (M U C(Q)) satisfies \R\ > 2m3/4. Let Fm + X = \F(V)\. G(V): Diagrams not in E(V) with a (v, Q)-set satisfying ttj(1 - ttj~3/8)/2 < \C(Q)\ < ttj(1 + T72-3/8)/2; with a set R of at most 2ttj3/4 vertices the deletion of which leaves no edge connecting two vertices of V— ({v} U C(Q) U R); and such that the smallest set S of vertices incident with all edges of C(Q) satisfies \S\>2m1'8. LetGm + 1 = \G(V)\- H(V): Diagrams P satisfying the following properties: There is a set TQ of at most 3T7i7yl8 vertices such that P - T0 is bipartite; the parts U and W of P - T0 have at least ttj/2 - 3ttj7/8 and at most t?j/2 + 3tt27/8 vertices; P- T0 has at most 29 levels, Lx, L2, • • • , Lk, k < 29; there is some level L¡ and a set {x, y} in P - T0 such that {x, y} is a good set in P - T0 and L¡ n C(x) n C(y) = 0; 208 D. J. KLEITMAN AND B. L. ROTHSCHILD either \L¡ CtU\> m15'16 and {x, y} C W, or \L¡ DW\> m15/16 and {x, y} E U. UtHm + x = \H(V)\. I(V): Diagrams P not in A(V) with a set T of at most 102ttj1s/16 vertices satisfying the following properties: P- T = Lx\l L2V L3 where Lx, L2, L3 are the levels of P- T (L3 possibly empty); both \LX U L3\ and \L2\ are between ttj/2 - 102T?jls/16 and ttj/2 + 102ttj1s/16; every two vertices in W = L2 have a common adjacent vertex in U= Lx U L3, and vice versa; there is a t E T adjacent both to a vertex x in Lx U L3 and a vertex y EL2. Let Im + X = \I(V)\. J(V): Diagrams P with a set T of at most 102ttj15/16 vertices satisfying the following properties: P - T is a three level bipartite diagram with parts U and W; ttj/2 - 102m15/16 < \U\ v all of the following in- equalities hold: (1) log (An + JPn) „)„_1)„) 2«1/2 of them) adjacent to u, u either covers or is covered by all vertices in a set ß of [tj1'2] vertices. Thus we have a (v, Q)-set in P. By B and D, \C(Q)\ must be between tj(1 - n-3/8)/2 and «(1 + tj_3/8)/2. By F and G there are sets R and S with IK| < 2tj3/4 and \S\ < 2tj7/8 such that P - (R U S) has no edges between two vertices of V - ({v} U C(Q) U R U S) = W', or between two vertices of (C(Q) U {v}) - (R U S) = U'. Thus P-(RUS) is bipartite with parts U' and W' and at most 29 levels. \R U S\ < 3tj7/8. Suppose there is a pah xx, yx of vertices such that [xx, yx} is a bad set in P - (R U S). Then consider P - (R U S U {xx, yx}). At least (tj - 3tj7/8)1/2 of the vertices of P - (R U S U {xv yx}) are in lower levels than in P - (R U S). Now suppose {x2, y2} is bad in P - (R U S U {xx,yx}). Consider P - (fiUSU {xx, yx, x2, y2}). At least (ti - 3tj7/8)1/2 vertices in P - (R U S U {xx, yx, x2, y2}) are in lower levels than in P - (R U S U {xx, yx}). We con- tinue this process one step at a time, choosingxx, yv x2, y2, • • • , xm, ym with {f/> y¡} a bad set in P - (R U 5 U [xx, yx, • " , x*v y¡-X})- We proceed until we can choose no more bad pairs {x, y}. For/ >,•_,} I < 3tj7/8. Since each vertex can decrease its level at most 28 times during this process, and since tj3/4(tj - 3n1ls)xl2 > 28(tt + 1), the total number ttj of steps before the process stops is at most tj3/4. Let T0 = R U S U {xx, yx, • • • ,xm, ym}. Then P - T0 is bipartite with parts U' - T0 and W' - T0, where |y0| < 3tj7/8, and \W' - T0\ and \U' - T0\ are between tj/2 - 3tt7/8 and tt/2 + 3tj7/8. Let the levels of P - T0 be Lx, L2, " ' , Lk, k < 29. Let ft be the largest value of/ such that either \L, n U'\ > tj15/16 or \L¡ n W'\ >nxs'16, say \Lh n U'\ > tj1s/16. Then we claim ft < 3. For suppose h> 4. Let u>x>z>.y be in levels ft, ft - 1, ft - 2, ft - 3, respectively, with u E Lh C\ U'. Then rj6IC' and z G ZJ', since P - T0 is bipartite. Now by /Z, C(x) n C( v) DLH^0; let «' G 0(x) n C(>») n /,,-. Then «' >x > z > j, and u' also coversy, since it is ENUMERATION OF PARTIAL ORDERS 211 adjacent and in a higher level. But then u, x, z,y form an excluded configura- tion, a contradiction. Hence ft < 3. For Z = 1, 2, 3, we claim that if \L¡ n U'\ > nxs/X6 (respectively \L¡ n W\ > Tils/16), then L, n W' = 0 (respectively L¡ n U' = 0). For let x e /,,. n W' (respectively x E L¡ n Z7'). By H, as above, x must be adjacent to some vertex in L¡ H U' (respectively L¡ n W1), a contradiction since no two vertices in L¡ can be adjacent. Thus if \L¡\ > 2tj15/16, L¡ must be entirely in U' or entirely in W'. Since \U' - T0\ and |W' - T0\ are both at least u/2 - 3tj7/8, and since \LÁ < 2t21S/16 for/ > 3, we must have at least one of Lx, L2, L3 entirely in U' with at least tj15'16 vertices, and one entirely in W' with at least tj15'16 vertices. In particular, L2 is entirely in W' or entirely in U'. For if not, Lx would be entirely in U' or entirely in W'. But then L2 would be entirely in W' or entirely in U', since every vertex in L2 is adjacent to some vertex of Lx. Similarly, since L2 E W' or L2 Ç U', we must have L3 E U' or L3 Ç W'. Now either \LX\ < 2nX5/X6 or Lx EU' oiLxE W'. Let T= T0 U \Jj>3L¡ U ¿x if \LX\< 2nxs/l6, and T=T0U \Jj>3L¡ ii \Lx\>2nX5IX(> (and thus Lx Çf/'orZ,, ç W')- iri < 60TJ15/16, and /> - T is bipartite with parts U = U'- T and W = W - T, with \U\ and |W| between n/2 - 60«ls/16 and n/2 + ÓOtj15'16. P-T has two or three levels, Lx, L2, L3 (where L3 =0 in the two level case). Finally U = Lx U ¿3, W = Z,2, or Í/ = L2, IV = Zj U ¿3. We assume, without loss of generality, ¿,U¿3= U, L2 = W. By construction of U and W, and by //, every two vertices of U axe adjacent to a common vertex of W, and vice versa. Also |¿1|>2tj15/16. By /, no vertex in T is adjacent both to vertices of W and U. Hence T is divided into two subsets, Ttj and Tw, where C(TW) C\W = 0. 0(7^) n {/ = 0. By /, no two vertices of Tu are adjacent, and no two vertices of Tw are adjacent. Thus P itself is bipartite with parts U U Tv and W U 7V Suppose Z G T, x, y G /,,-, i=l,2 or 3, and t is adjacent to x and to y. Then either t covers both x and y or f is covered by both. For if t covers x and V covers t, by construction of Z7 and W we can let u be a vertex in L¡_x or Li+X to which both x and y are adjacent. Then t, x, y, v form an excluded configura- tion, a contradiction. Suppose t E T is adjacent to vertices in x and y in different levels of P - T. Then these levels must be Lx and L3 (by /). x and .y are adjacent to a common vertex v in L2. Let x be the vertex in Lx, y E L3. Then y covers v, v covers x, and we must have that y covers t and r covers x, or else x, ,y, u, í form an ex- cluded configuration. So if t is adjacent to vertices in Lx and L3, it covers those in Lx and is covered by those in Ly All vertices of T, therefore, are in one of the following subsets: 212 D. J. KLEITMAN AND B. L. ROTHSCHILD T¡+ = {t \t covers only vertices in L¡ and is covered by none in V - T}, TJ = {t 11 is covered only by vertices in L¡ and covers none in V - T} for i = 1, 2, 3, T2 = {t\t covers vertices in Lx and is covered by vertices in L3, and is adjacent to no others in V-T}. We now claim that P = (77) V (Lx U T2) V (L2 U Tx+ U T3 U T°) V (L3 U T+) V (7-+). To show this we must show that no vertex of T3 is adjacent to any in T2, no vertex of Tx~ is adjacent to any in T2 , and all other adjacencies are in the "right direction." (We already know that there are no edges between two vertices of rf U Tx+ U 77 U T3+ U T% or between two vertices of T2 U T2+.) If x S r^, y S 77, are adjacent, there can be no edge between C(x)■- T and C(y) - T or we would have an excluded configuration. Thus by K, x and .y cannot be adjacent. Similarly, x ETX and y ET2 cannot be adjacent. Now suppose x E T¡~, y E TJ+ x, 1 = 1 or 2. If x coversy, C(x) - T and C(y) - T must have no edge between them, or an excluded configuration would result. Hence, by K, x cannot cover y. Similarly, if x E T¡, y E T¡~+x, Z = 1 or 2, then y cannot cover x. Finally, if x E T2 and y E T2 (oty E T2, x ET2, resp.) and if x covers y (respectively x is covered by y), then no vertex of C(x) - T is adjacent to any vertex of C(y) - T, or an excluded configuration would result. Thus, by K, x cannot cover y (respectively y cannot cover x). This completes the elimination of all possible connections which would contradict the claim. Thus we have shown that P = (77) V (77 U Lx) V (L2 U Tx+ U 77 U r2°) V (L2 U 7+) V T2/2-102T215/16, we must have |Z,3| n/2-n31'32; or if J+ # 0, \L3\ > n/2-n31'32. By M, neither of these possibilities occurs. Hence Tx = T3 = 0. This gives us P = Sx V 52 V 53, where SX=LXU T2, S2=I2ur2°U Ti+UT3, 53 = L3ur2+. Finally, then, by AT and 0, P must be in Q(V). In particular, since every vertex of S2 covers some (at least tj/8 - tj7'8) vertices of ENUMERATION OF PARTIAL ORDERS 213 Sj, and every vertex of S3 covers some (at least tt/4 - t27/8) vertices of S2, the S¡ are in fact levels in P. Part I of the proof of the theorem is now complete. 5. Proof of theorem. II. We shall use the results of the last section, together with the lemma, to show that.P„ <(1 + 0(l/72))ß„. First we show thatP„ <(1 + 0(llri))Xn. Let v be the number guaranteed by the lemma, and let N = max(2i>, 109). Let C0 be a number large enough so that Pn < (1 + (C0)¡n)Xn for all n „ < (1 + C/n)Xn for alln. The proof of this claim is by induction on tj. For tj < N it is true by choice of C We assume that it is true for all tj < ttj, for some m>N, and show that Bm +1 ^ 0 + C/(m + l))Xm + x as well. Since, by the last section, we have Pm + x < Am + i +Bm + i +^m + i +•" + #«+i + Xm+l> we need onlV ^ow that (Am + i + "' +Nm + OI(Xm + 0 < cKm + 0- To do this we sha11 employ the inequalities of the lemma to show that each of the terms Am + X¡(Xm + j), Bm + xliXm + x), • • • , Nm + xl(Xm + x) is at most 1/13 • C/(ttj + 1) (there are 13 terms here). These arguments are all similar, and we illustrate a few typical ones. (D^B+l.ds+i ^L^S- <2-/4(l + C/TT2)2-'"/2 + sl0^<-4l • ^, Xm + X *m XmXm + \ 772+1 1J Dm + l _ Dm+1 m-jm1/2] m-[mxl2] Xm Xm + 1 Pm-\mW\Xm-\m*tl\ Xm-[mlf2\+l Xfn + 1 (3) <2m3l2l2-'Am9l8il +_Ç._\ 2-(m/2-lm1/2]/2)([m1/2l + l) i/2-%m9/8f/] m — \wi * [m1'2] . 2S([m1/2] +1)log TTJ <^ C 13 TT7 + r (13) Nm + 1 Xm + X <2_(,ogm)2/6> bythelemmaj N^ aiSO) by the definitions of 0(V), X(V) and Q(V), we have Xn+X = On + x + Qn+l. These facts lead to 214 D. J. KLEITMAN AND B. L. ROTHSCHILD < f 1 + 1 )ßn + i for « sufficiently large. This establishes .Pn = (1 + 0(l¡n))Qn and completes the proof of the theorem. 6. Proof of lemma. We let V be a set of tt + 1 vertices. We recall our convention, that all statements in inequalities asserted below are meant to be valid only for n sufficiently large. More specifically, for each statement below there is a number N' such that the statement is valid for tj > N'. We can then let v be the maximum of all of these N'. The numbered paragraphs below cor- respond to the numbered inequalities of the lemma. For the sake of brevity, we include only a few of the cases in detail. The rest of them are represented only by the final inequalities of the arguments, from which one can obtain a hint as to the order in which things are constructed. We choose one simple case, and the most complicated ones to do in detail. (1) This inequality is Lemma 2 of [7]. It is proved like those below. (2) This and (3) below correspond to Lemmas 3 and 4 of [7]. We obtain all diagrams in B(V) from diagrams on tj vertices by choosing v E V, choosing a diagram on V - {v}, and then adjoining v to the diagram so as to satisfy the conditions for B(V). We will obtain upper bounds for the num- ber of possible choices by counting some possibilities which cannot satisfy the conditions for B(V) as well as all those that do. This is the general method used below, where instead of just choosing a single v, we may need to choose a sub- set S E V, a diagram on V - S, and then to adjoin S to the diagram. To obtain diagrams in B(V), then, we choose vE V (n + 1 ways to choose v) and a diagram on V - {v} (at most Pn ways to do this). Now v will be taken as the vertex of a (v, ß)-set. To connect v, we first choose a level for v to be in (at most « + 1 possibil- ities). Then we choose ß (at most „C[„i/2j ways, where many of the sets of [tj1'2] vertices in V - {it} included in this number will not be valid candidates for ß, depending on which diagram was chosen for V - {v}). The directions of the connections between v and ß are now determined, because in order to be in ENUMERATION OF PARTIAL ORDERS 215 B(V), the levels of vertices in ß must be unaffected by the addition of it. Hence v covers all vertices of ß if its level is higher than all of them, and is covered by all vertices if its level is lower. (One of these two possibilities must occur, or we would have an invalid choice for Q.) Since ß will satisfy conditions for B(V), we have \C(Q)\ > n(l + n~3/s)/2. Since there are no triangles, v can be connected to at most «(1 - ti_3/8)/2 remain- ing vertices. Since v must be a good vertex, at most tj1'2 of these vertices can have then levels affected by the addition of v. The vertices which can have their levels changed can be chosen in at most '"¿n'(Kl-»-'">«l)<«W2?''' /=o ^ l 1 /2 ways, v can then be connected to these vertices in at most 3" ways (the 3 is for the choices: v covers x, x covers it, and x and v not connected). Finally, the other vertices to which v can be connected can be chosen in at most 2"*1-" ^2 ways. The directions of these connections are determined by the levels relative to the level of v. All connections of it are now completed. Multiplying all these numbers of possible choices gives the following: log (^A < l0g(TJ + 1) + l0g(7J + 1) + log ("/2] + log (72(72/2)" 1/2) + nl'2l0g 3 + 72(1 -T2"3/8)/2 < 2 log (77 + 1) + T21/2 log TJ + log TJ + T21/2 log Tl/2 + TJ1/2 log 3+ T2/2- 72S/8/2 <«/2-TI5/8/4. This completes (2). We used one of two basic relations here, which will be used repeatedly below. .eg (;,,,) <„° iog. The other is Stirling's formula or the normal approximation, which gives (recall, for tj sufficiently large, here depending on the ß): log ( " J < - T2(oi log a + (1 - a) log(l - a)) for 0 < ß < a < lA, and ß fixed (independent of ti). 216 D. J. KLEITMAN AND B. L. ROTHSCHILD loS (r „ ' ^M,1 < " - M™2 for 0 < ot < 0 < 1, and 0 fixed. \[tj(1 -a)/2]J 2"2/4 (also from paragraph (14) below). Thus log|/(HI/P„<-ri33/17/500. We obtain diagrams in l"(V) as follows: We choose t (n + 1 ways) and a diagram on V- {t} so that there is a set T' with T = T' U {t} satisfying con- ditions for I(V) (at most Pn ways, and at most 1„1S/16]J Jl027 ways to choose T'). Then, since every two vertices in U (respectively, W) are adjacent to a common vertex of W (respectively, Z7), t can be connected to each level in only one direction or an excluded configuration results (at most 23 choices for directions for t). í can be connected to r'at most 3lo2"ls'16 ways. To satisfy the conditions on I(V) there must be vertices x E U and y E W adjacent to t (at most ti2 ways to choose x and y, and 4 ways to connect them to t). By A, t must be adjacent to tj/64 other vertices at least, and of those to at least T216/17 in U, or n16'11 in W, say W. There are at most 1ti/2 + 102tj15/16]\ [„16/17! ) ways to choose a set S of [ti16'17] vertices in W to be included in C(t) n W. But by the conditions for l"(V), \C(S) C\ U\> n/2 - ti/300. x must thus be in U - (C(S) n U) and C(x) n W must be at least nl65, by A. The remaining vertices of C(t) must be chosen from ((W-S)- C(x)) U(U- C(S)), of which there are at most (n/2 + lOV5'16 - [ir16'17] -tj/65 +TJ/2 + lOV5'16 -n/2 + ri/300) -1"/21-["/41(2["/41 _ tfn/2] ways. We only get diagrams counted by Xn here, and we get no diagrams twice. Thus log X„ > log (^ + log (" " j^1) + [T2/2] (72 - [T2/2] - [T2/4] ) + [n/2] [72/4] + [n/2] log(l - 1/2Í"/4!) + (n - [n/2] - [tj/4]) log(l - 1/2Í"/2») >l0g(l/T2 • 2") + l0g(l/T2 • 2"/2) + (TJ - [ti/2])[t2/2] - 1 > T22/4 + 3T2/2 - 3 log 72. We find an upper bound for Xn equally easily. Let V be a set of tt elements. Diagrams in X(V') are obtained as follows. We choose S2 (at most 2" ways); Sx from V - S2 (at most 2"/2+log " ways); and connections from Sx U S3 to S2 (at most 2" I4 ways). This gives log(X„) < ti2/4 + 3n/2 + log n. Together with the lower bound for Xn+X we get \og(Xn/Xn + x) < - n/2 + 5 log 72. (12) log(Mn + x)<(n + l)2/4 + (tj + 1) + 2 log« + 4«31/32log«. From the lower bound on!n + 1 we get log(M„ + X/Xn + x) <-n/4. ENUMERATION OF PARTIAL ORDERS 219 (13) We obtain diagrams in N(V) by considering two cases. In the first case, we suppose that not both of the inequalities (72 + l)/2 - log tj < \S2\ < (n + l)/2 + log « hold. This gives a number N'n+X of choices with ^og(N'n+1)<2(n + 1)+ 1 -\S2\ + \S2\(n + 1 -\S2\) O3) ^(T2 + l)2 , 3(72 4-1) 1 <■ 4 ■ 2 In the second case, we assume that fOogTî)2 T2 + 1 . ^ ip i ^72 + 1 , , —2-log n < |521 <—2— +1°g"> and that either ls«l>ZHrL + »1/2log7j or |sy for 1 = 1 or 3. This gives a number N¡¡+ x of choices with l0g«+1) = 2 + (72 + 1) + log TJ + (T2 + l)2/4 [(« + l)/2 + log 72] (14) + log , \[(" + l)/44-«1/2logn] < (72 + l)2/4 + 3(tj + l)/2 - y4 (log TJ)2. We get n+X/ \ Xn + X 108 O <* ""fe" <- 5 <*■>*• log -fr1" < l0g(» + 1) + 25t21/2 log T2 + log TJ "n (15) 8\U(« + l)/4 + 727/8] + 2(«+1)/4+ni/2log„ + 1A(" + 1)/4+«1/2/1°g"A\ log " V [(» + D/8 + «7/8] )) * 2 IO" ' 7. Proof of corollary. For a set K of tj vertices, let 7(10 De the class of diagrams P such that P = LXV L2V L3, where Lx, L2, L3 are the levels of P. 220 D. J. KLEITMAN AND B. L. ROTHSCHILD Let Yn = \Y(V)\. We obtain all diagrams in Y(V) by first choosing Lx, then L2 and then connecting L2 to Lx and L3 to L2. There are exactly (2IL2' - 1) ways to connect each vertex in L3 to L2 (there must be at least one connection since L3 is a level), and exactly (2 * - 1) ways to connect each vertex of L2 to Lv This gives Yn - t (") Z (" y '") (2'' - D'C - I)""'"-''. Since Y(V) > Q(V), we have Y„