Not AvailableSupported in part by O.N.R. Contract N00014-67-A-0204-0016 and supported in part by the U.S. Army Research Office (Durham) under contract DAHCO4-70-C-0058

Kleitman, Daniel J.

Magnanti, Thomas L.

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bat;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ F-Sf;Xt~~~~~f;SBS4  ~ ~ M S;~:;*i ffdDSu 0k0fA'f{ 0u4 ri0 0f;t 00oflo- 0: sg: 'sz:k :g;::fC;:f : : .07 z:Sz;gS ~i\D::: df:0;0 t:>r:0 W0;f;0u  f:SE 0 C EL .El- 0 ' ' .: ............ '. :-. '. :'........................... ,r--- : :-:-~- -werk- g -- pe-:-- - - -... ; .. = . - ..... .... .. . . ........ > ..... . ,, , ...  .... . . ,... ...ki . . . e.; CU n, ;; ; f f:: f0900 00 00-ff- D l: :0 0 00: n.00 . ?70 20-,&XT -. ff 000~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-- WE . S . .tV - . . . - -. ,., . - - - -, , . . ., . . WS . .. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~lM :ASSCUE TTS I NSTITUTE.n0XOF TEC :;;nStxi000aHNOLOGY r0 >0 ::- -iiOn the Number of Latent Subsetsof Intersecting CollectionsbyD. J. Kleitman* and T. L. Magnanti**OR 012-72October, 1972Supported in part by the Office of Naval Researchunder Contract N00014-67-A-0204-0016.** Supported in part by the U. S. Army Research Office(Durham) under contract DAHCO4-70-C-0058.ON THE NUMBER OF LATENT SUBSETSOF INTERSECTING COLLECTIONSD. J. KleitmanMassachusetts Institute of Technology, Math. Dept.and**T. L. MagnantiMassachusetts Institute of TechnologySloan School of ManagementAbstractGiven two collections F1 and F2 of sets each member of oneintersecting each member of the other, let the collections oflatent sets F i=1,2 consist of the sets that are contained in members1of F i but that are not themselves members of F i . If lower caseletters indicate the size of the collections we then haveflf2 - flf2 This result is used to prove that a self-intersectingsubfamily F of a simplicial complex G having the property thatany element of F contains s1 or s 2 can be no larger than thelesser of the number of elements of G containing sl and thenumber containing s2. Certain extensions and a related con-jecture of Chvatal are described.* Supported in part by O.N.R. Contract N00014-67-A-0204-0016** Supported in part by the U.S. Army Research Office (Durham)under contract DAHCO4-70-C-0058.-2-IntroductionTwo collections F and F of subsets of a given finiteF 2set S are called intersecting if each set in F1 has a non-empty intersection with every set in F2. Latent subsets F.Jof F are defined to be those subsets of S which are a subsetof a set in F but which are not themselves in F, i.e.,F. = {A C S: A C A1 Fj and A F}.J -- -- LWe let fj and fj denote respectively the number of sets in F.and F. In this note, we prove that if F1 and F2 are intersecting then the following equality is satisfied:fLfLflf2 flf2 . (1)We begin by developing a canonical form for F1 and F2which preserves both their size and intersection property anddoes not increase the number of latent subsets of either col-lection. Then we prove the asserted inequality for the inter-secting collections in canonical form. We conclude by givingseveral extensions of the inequality and an application.Canonical FormIn 1], one of the authors introduced a canonical formfor intersecting collections which he used to obtain boundson the number of sets in certain collections. We show that-3-the same technique may be used for our latent subset problem.Let us order the elements of S as l, s2,..., sn. Wedefine the following set of mappings mj, for 1 < j < n-l,acting on subsets of S:SA + - if sj A, sj+1 AI 5~ 5 j+l j+lmj (A) = A otherwisewhere A + s-sj+ is used in place of A{sj} - {sj+l}. Thej j+l j+lsame convention will be applied for one element sets throughoutthis paper. The mappings m "push" elements of A toward the]lower ordered elements.For any collection F of subsets of S, we define mj(F)acting on F by:mj )am(A) if m (A) ! Fm.(F)(A) ={A if m (A) F.Beginning with two intersecting collections F1 and F2 ofsubsets of S, it is shown in [1] that m(F 1) and m(F 2) areintersecting and that after a finite number of repeated appli-cations of ml, m2,..., mn_1 the resulting collections, calledthe canonical form for F1 and F2, will be invariant underevery m. transformation.3_____II_______ _-4-We now note that m(F), which has the same number ofmembers as F, has no more latent subsets than F. SupposeA is a latent subset of m(F) and not of F. Then, by thenature of the m transformation, that it affects only sets con-taining one and not both of sj and sj+l, we can conclude thatA must contain one and not both of these elements. We shallshow that under the given circumstances the set A' obtained fromA by removing the one of s, sj+l contained in A and insertingthe other, must have been a latent subset of F and is not one ofmj(F). This is all we need prove here.Two cases can be distinguished. If A contains s, itcannot be in F nor can A' be. It is latent in m(F) becausesome B satisfying BA lies in that family, and not in F. Butthen the member of F whose image under mj was B contains A',which is therefore latent in F. And A' can be latent in m (F),only if A is latent in F. For, if C in m(F) contains A',Jthen the set C' obtained from C by interchanging sj with sj+ 1(which may be C itself if both are present in it) must be inF, and will contain A.If A contains sj+l rather than s, then A must be latentin m(F) and not in F through the fact that it is in F and notin m(F) and is contained in some B that lies in m(F). Butthen A must be in m(F) and not in F, and must be containedin B' (defined as C' above) which must be in F. Thus A' islatent in F but not in m (F) which was to be proven.-5-Finally, given two intersecting collections F 1 and F 2 incanonical form it will be convenient to partition the familiesas:F. = G.UHUI. (j=l,2) (2)where(i) sn A for any A e G1 .(ii) if A H1, then sn A and A - sn intersectsevery set in F 2;(iii) if A I1 , then sn A and A - sn is disjointfrom a set in F2and similarly for F.2'For j=l,2 let Hj={A-sn: A H} Ij={A-sn: A I}. Weobserve that the sets in 1 and 2 may be paired in the sensethat if A c I1 then S-A-snC 2 and conversely. Furthermore, ifA E 1(I2 ), then A intersects every set in F 2(F1) except forS-A. To prove both assertions simply note that if B F 2,AAB = 0 and BCS-A-sk then by adding sk to A one obtains amember of F1 (since F1 is in canonical form) that is disjointfrom B.Main ResultWe now prove our main result. If F is any collection ofsets we use notation from the introduction letting F L denoteits latent subsets and f its cardinality. We assume throughoutthis section that S is a given finite set.-6-Theorem 1:Let F 1 and F2 be two intersecting collections of sub-sets of S. ThenLLflf2 > flf2 . (3)~1 2 - 1 2Proof 1:By the results of the previous section, the theorem istrue if it holds for the canonical form for F and F2. Thuswe assume that F1 and F2 are in canonical form and that eachcollection has been partitioned as in (2). If either F1 orF2 is empty the result is obvious; thus we assume that flf2 > 1.We use induction on n the number of elements in S. Fora given value of n we use induction on i1 = i2. If n = 1, theleft and right hand sides of (3) are both equal to one (theempty set is latent in each collection). Suppose that n=k.If i=i 2 =O, then G1 and H1 both intersect each of G2 and H2.Thus by induction on n,LL LLg lg 2 - glg2 , hlh2 > hh 2(4)LLgi~h > giHj i j {1,2}.Note that if T is a latent subset of Hj, then T + sn is a latentL L Lsubset of H so that f > g + .. Thus the result follows in) J~J _)j+h __ _ 1__1___11___1___1__II·_-_-·_ -.1.11--·-·1 Y-PI-XI·--I----_III I···. II--..-7-this case by adding the four inequalities in (4) and usingfj gj + E.Next assume that i=i2=p and suppose without loss ofgenerality that (f + f) < (f2 + f). Let T I1. By ourobservations concerning the canonical form of F1 and F2,S - T c 12 and T intersects every set in F2 other than S - T.Define F1 = FlU{T}, F2 = F 2 - {S - T}. Then F1 and F2are intersecting and T + sn is in H2 not I2' Also f = f +l(since T is latent in F1 and not F1) andf 2 > f2 - 1 (possiblyL L(S-T)cF2 whereas (S-T) ~ F2). But il=p-l, so by induction:lf2 > fl f2thus,(fl - 1) (f + 1) > (f + 1) (f2 - 1)orfl f2flf 2 ) (f +fL) ff 22 -1 2 22+ fl >_1 i .This completes both inductions and proves the theorem.We note that with the hypothesis of Theorem 1 the relatedconjecture that fl + f2 > f + f is not valid. As a counter-example, let F 1 consist of the set {l, s 2} and F2 the sets{s l}' {s2} , {sl' 1 , l' s2}1, 2 31s3}, {s2 , s3}, , s2 , s31}1_ 1_1 I I_ ---·------ ·IIC·^--·LIIIIIPIII*L-8-The following results are easy consequences of Theorem 1.For any collection F of subsets of a given set S we letFL ={ACS: AA 1 £ F and A F} denote the latent supersets ofF.Corollary:(a) Let F1,..., Fr be pairwise intersecting collections ofsubsets of S. Let 0 be a real valued function definedon Rr(r 1)/2 that is non-decreasing in each argument.L~L LL Then 0(f~~f2,4L lf ) r-l r2 fiThen 0(fLf  .. f.f.,..., f f )>0(f ,..fr-lfr) ije{1,...,r}. In particular,r rL L rr. f f > Er fLrj 1 f.f. and i f > f .i=l i=l(b) Let F be a collection of subsets of S with the propertythat no two sets in F are disjoint. Then fL > f.(c) Let F1 and F2 be two collections of subsets of S withthe properties:(i) no set in F1 intersects any set in F2, andL' L' ff(ii) A F1 implies S-A F2. Then f f2 fLf2Proof:(a) and (b) are clear. (c) follows by considering theintersecting collections Pj j=l,2 whose elements are complements(with respect to S) of the sets in Fj.____ _1_1_1___1 __^_1___ _1_·__11 ^_____ I^II  I·PI·-DLIX·X. LII-.-- ---.- II~. 1 | - 1: : _ ___ . _ .__ _ .__ ...... I.1 _.___._.-9-The following result is an application of Theorem 1.If F is interpreted as a collection of committees each chairedby one of two men, it shows that an equal number of sub-committees may be chosen with a common chairman.Theorem 2:Let F be an intersecting collection (i.e. no two setsin F are disjoint) of subsets of S with the property that eachset in F contains at least one of the elements sl, s2 C S. Thenthere is an intersecting collection F' subordinate to F (i.e.,A F' implies that for some A ACA1 e F) that satisfies:(i) f' > f(ii) either s1 is contained in every set in F' ors2 i .Proof:If either {s1} or {s2} is a member of F there isnothing to prove; we therefore assume otherwise; that sj1 ' F, j =1,2.Let G = {A F: s. A, A},and let Gj = {A-sj: A Gj}j=1,2.By the hypothesis on F, G1 and G2 are intersecting, thusL L by Theorem 1, glg2 > glg2 so that g > g2 say. The set F',LF'=GU(JG 3 U{A+sl: A G1} with G 3 = F - G1 - G2 , then satisfiesthe conditions above.The analog of Theorem 2 when each member of F contains oneof three elements sl, s2, s3 is not valid. As a counterexample,-10-let F consist of the sets (s, s4, s5), (, S4, s6),(sj, 5, s6) j = 1,2,3. Note that in this example we canselect F', f' > f, with the property that each of its memberscontains s4.A conjecture to this effect has been proposed by V.Chvatal.Conjecture:Let F be a collection of subsets of a finite set Ssuch that X F, YX=>YcF. Then there exists on s S andan intersecting collection F each member of which contains ssuch that for any intersecting subfamily G of F,fs g.The result above settles this conjecture whenever any maximalcardinality sub family of F contains at least one of two elements ofS; and for any F, for all G which contain at least one of two elementsof S.Note added: A recent result of Chvatal (preprint) settlesthe above conjecture when F is in the canonicalform defined above." . ~-- .I .-11-References1. D. J. Kleitman, "On a Conjecture of Milner on k-Graphswith Non-Disjoint Edges", J. Comb. Th., 5 (1968), 153-156.2. V. Chvatal "Intersecting Subfamilies of HereditaryFamilies" (To be published)

On the Number of Latent Subsets of Intersecting Collections

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On the Number of Latent Subsets of Intersecting Collections

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